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Primary Mathematics  Chapter 5 Measures

Primary Mathematics Workbook 4A Common Core Solutions Chapter 5 Measures Exercise 5.6 Page 163  Exercise 5.6  Problem 1

It is given that

⇒ \(\frac{5}{8}\)day = _____ h.

The unit conversion 1 day = 24h is to be used.

⇒ \(\frac{5}{8} d a y=\frac{5}{8} \times 24 h\)

Multiplying \(\frac{5}{8} \) and 24

⇒ \(\frac{5}{8} \times 24=\frac{5 \times 24}{8}\)

Canceling out common factors

= 5 × 3

= 15

⇒ \(\frac{5}{8}={15}\)

So we have got \(\frac{5}{8}\) day = 15h

The given measure has been converted into an equivalent measure as \(\frac{5}{8}\) day = 15h

Chapter 5 Exercise 5.6 Measures Workbook 4A Answers Page 163  Exercise 5.6  Problem 2

It is given that \(\frac{7}{10}\) m =_____cm.

The unit conversion 1m = 100cm is to be used

Multiplying \(\frac{7}{10}\) and 100 convert to centimeters.

⇒ \(\frac{7}{10}\)

Thus

⇒ \(\frac{7}{10} \times 100\)

⇒ \(\frac{7}{10} \times 100=\frac{7 \times 100}{10}\)

Canceling out a common factor 7 × 10

So we have got 70cm.

The given measure has been converted into an equivalent measure as \(\frac{7}{10}\) = 70cm

Page 163  Exercise 5.6  Problem 3

It is given that \(\frac{9}{20}\) min =____sec.

To find – The equivalent measure.

The unit conversion 1 min = 60 sec is to be used.

We have to multiply  \(\frac{9}{20}\)  and 60 in order to convert it to seconds.

Thus  \(\frac{9}{20}\) × 60 = \(\frac{9×60}{20}\)

Canceling out common factors, 9 × 3

So, we have got  27sec.

The given measure has been converted into an equivalent measure as    \(\frac{9}{20}\) min = 27 sec.

Workbook 4A Chapter 5 Exercise 5.6 Measures Solutions Page 163  Exercise 5.6  Problem 4

It is given that \(\frac{3}{4}\)gal=_____________qt

The unit conversion 1gal = 4qt is to be used

We have multiply \(\frac{3}{4}\) and 4 to convert to quarts.

Thus

⇒ \(\frac{3}{4} \times 4\)

⇒ \( \frac{3}{4} \times \frac{4}{1}\)

⇒ \(\frac{3 \times 4}{4}\)

Canceling common factor 3,

So we have to get 3qt

The given measure has been converted into an equivalent measure as

⇒ \(\frac{3}{4} \mathrm{gal}=3 q t\)

Page 164  Exercise 5.6  Problem 5

It is given that  2​\(\frac{3}{5}\)  m = 2m____ cm

Also, \(\frac{3}{5}\) m = \(\frac{3}{5}\) ×100cm

To write in compound units.

Here, the whole number is 2, so the bigger unit is 2m

The improper fraction is  \(\frac{3}{5}\)  and unit is  \(\frac{3}{5}\)m.

The unit conversion 1m = 100cm is to be used.

Multiplying \(\frac{3}{5}\) and 100 , \(\frac{3}{5}\) ×100  = \(\frac{3×100}{5} \).

Canceling out common factors, 3 × 20.

So, we have got 60cm.

Therefore, final result is 2\(\frac{3}{5}\) m = 2m60cm.

The given measure has been written as a compound unit as below  2\(\frac{3}{5}\) m = 2m60cm.

Page 164  Exercise 5.6  Problem 6

It is given that 4​\(\frac{7}{10}\) L = 4L______ ml.

To write in compound units.

Here, the whole number is 4, so the bigger unit is 4L.

The improper fraction is \(\frac{7}{10}\) and units is \(\frac{7}{10}\)L.

The unit conversion 1L = 1000ml is to be used.

Multiplying \(\frac{7}{10}\) and 1000,\(\frac{7}{10}\) × 1000 = \(\frac{7×1000}{10}\).

Canceling out common factors, 7×100.

So, we have got 700ml

Therefore, the final result is 4​\(\frac{7}{10}\) L = 4L​​ 700ml.

The given measure has been written as a compound unit as below   4​\(\frac{7}{10}\) L = 4L​​ 700ml.

Measures Exercise 5.6 Primary Mathematics Workbook 4A Explanation Page 164  Exercise 5.6  Problem 7

It is given that 3\(\frac{1}{4}\)  h = ____ h ____ min.

To write in compound units.

Here, the whole number is 3, so the bigger unit is 3h.

The improper fraction is \(\frac{1}{4}\) and units is \(\frac{1}{4}\) h.

The unit conversion 1h = 60min is to be used.

Multiplying \(\frac{1}{4}\) and 60 \(\frac{1}{4}\) ×60=\(\frac{1×60}{4}\).

Canceling out common factors, we get 15min.

Therefore, the final result is 3​\(\frac{1}{4}\)h = 3h​​ 15min.

The given measure has been written as a compound unit as below  3​\(\frac{1}{4}\)h = 3h​​ 15min.

Page 164  Exercise 5.6  Problem 8

It is given that 2​​\(\frac{1}{2}\)days =____ days ____ h.

To write in compound units.

Here, the whole number is 2, so the bigger unit is 2 days.

The improper fraction is  ​​\(\frac{1}{2}\) and units is  ​​\(\frac{1}{2}\)day.

The unit conversion 1 day = 24 h is to be used.

Multiplying ​​\(\frac{1}{2}\) and 24, ​​\(\frac{1}{2}\) × 24 = \(\frac{1×24}{2}\)

Canceling out common factors, we have got 12h.

Therefore, the final result is 2​​​\(\frac{1}{2}\) days = 2 days12 h.

The given measure has been written as a compound unit as below    2​​​\(\frac{1}{2}\)days = 2 days12 h.

Common Core Workbook 4A Chapter 5 Measures Exercise 5.6 Help Page 165  Exercise  5.7  Problem 1

It is given that  2​\(\frac{1}{10}\) kg =____ g   2kg= ____ g

\(\frac{1}{10}\)kg = \(\frac{1}{10}\) × 1000g =

​To write the equivalent measures.

Here, the whole number is 2kg

Using conversion 1kg = 1000g, we get it to be equivalent to

⇒ 2kg × 1000 = 2000g.

The improper fraction is  \(\frac{1}{10}\)kg

Using conversion, we get it to be equivalent to

​⇒ \(\frac{1}{10}\)kg=\(\frac{1}{10}\) × 1000g

\(\frac{1}{10}\)kg = 100g
​
Combining these by adding

2000g + 100g = 2100g

So, we have got  2​\(\frac{1}{10}\)

kg = 2100g.

The equivalent measure for the given measure has been written as below 2​\(\frac{1}{10}\) kg = 2100g.

Measures Exercise 5.6 Workbook 4A Detailed Solutions Common Core Page 166  Exercise 5.7  Problem 2

It is given that Brain jogs 3​​\(\frac{1}{8}\) km.

Using conversion 1km = 1000m

3m = 3 × 1000m

3m  = 3000m

The improper fraction is \(\frac{1}{8}\) km

using conversion, we get it to be equivalent to

⇒ \(\frac{1}{8} \mathrm{~km}=\frac{1}{8} \times 1000 \mathrm{~cm}\)

= 125

Combining these by adding

3000m + 125m = 3125m

So, we have to get that the brain jogs 3125m

Brain jogs for \(3 \frac{1}{8} \mathrm{~km}\) and this can expressed in meters as 3125m

Primary Mathematics Workbook 4A Chapter 5 Measures Exercise 5.6 Problem-Solving Page 166  Exercise 5.7  Problem 3

It is given that Peter practices the piano for   1​​\(\frac{3}{4}\)h.

Pablo practices for 125min.

We have to find who practices for a longer time and how much longer.

To compare them, we have to convert 1​​\(\frac{3}{4}\)h.into minutes using the conversion.

After conversion, the person having the greater value in minutes would practice for longer.

To get how much longer, we subtract them.

We have Peter’s time for practice as 1​​​\(\frac{3}{4}\)h.

Let us convert it to minutes first.

Here, the whole number is 1h.

Using conversion 1h = 60min, we get it to be equivalent to 60min.

The improper fraction is  ​​\(\frac{3}{4}\)h.

Using conversion, we get it to be equivalent to
​
⇒  ​​\(\frac{3}{4}\)h × 60min

= 3 × 15min

= 45 min
​
Combining these by adding

60min + 45min = 105min

So, we have got that Peter practices the piano for 105min.

Now, Peter practices piano for 105min.

Pablo practices piano for 125min.

Therefore, it is clear that since 125>105, Pablo practices longer than Peter.

Finding the difference

⇒ 125 − 105 = 20min

So, Pablo practices 20 min longer than Peter.

When it is given that Peter practices the piano for  1​\(\frac{3}{4}\)h and Pablo practices for 125min, we can say that Pablo practices for a longer time and for 20min  longer than Peter.

Primary Mathematics Workbook 4A Chapter 5 Measures Exercise 5.6 Problem Solving Page 166 Exercise 5.7 Problem 4

It is given that 1​\(\frac{1}{2}\) L ____  1050ml.

To compare and fill in the blank.

First, we will convert the quantities on both sides of the blank into a single unit, i.e. milliliters.

Then we will compare them and use the appropriate sign to fill the blank.

Let us convert 1​\(\frac{1}{2}\)Linto milliliters.

Here, the whole number is 1L.

Using conversion 1L = 1000ml, we get it to be equivalent to1000ml

The improper fraction is  \(\frac{1}{2}\)L.

Using conversion, we get it to be equivalent to\(\frac{1}{2}\) × 1000 = 500ml

Combining these by adding

1000ml + 500ml = 1500ml

So, we have got 1500ml.

Now, we have 1500ml _____ 1050ml

Since 1500ml is greater than 1050ml, we can say that

1500ml​​>​​1050ml

Finally, we can fill in the blank as

1​\(\frac{1}{2}\)L​​ > 1050ml

For the given quantities, the comparison has been done and we get 1​​\(\frac{1}{2}\)L​​>​​1050ml.

Primary Mathematics Workbook 4A Chapter 5 Measures Exercise 5.6 Problem Solving Page 166 Exercise 5.7 Problem 5

It is given that 1\(\frac{2}{3}\)h _____ 100min.

To compare and fill the blank.

First, we will convert the quantities on both sides of the blank into a single unit, i.e. minutes.

Then we will compare them and use the appropriate sign to fill the blank.

Let us convert   1\(\frac{2}{3}\)h   into minutes.

Here, the whole number is 1h.

Using conversion 1h = 60min, we get it to be equivalent to60min

The improper fraction is  \(\frac{2}{3}\) h.

Using conversion, we get it to be equivalent to \(\frac{2}{3}\) × 60 = 40min

Combining these by adding

⇒ 60min + 40min = 100min

So, we have got 100min

Now, we have

100min____ 100min

Since both are the same, we can say that

100min = 100min

Finally, we can fill in the blank as

1\(\frac{2}{3}\) h = ​​100 min

For the given quantities, the comparison has been done and we get 1\(\frac{2}{3}\)h = ​​100min.

Page 166 Exercise 5.7 Problem 6

It is given that  2\(\frac{1}{4}\)km_____2500m

To compare and fill in the blank.

First, we will convert the quantities on both sides of the blank into a single unit, i.e. meters.

Then we will compare them and use the appropriate sign to fill in the blank.

Let us convert   2\(\frac{1}{4}\)  km into metres.

Here, the whole number is 2km.

Using conversion 1km = 1000m, we get it to be equivalent to

⇒ 2km × 1000 = 2000km.

The improper fraction is \(\frac{1}{4}\) km.

Using conversion, we get it to be equivalent to

⇒ \(\frac{1}{4}\)km × 1000 = 250m.

Combining these by adding

⇒ 2000m + 250m = 2250m

So, we have got 2250m.

Now, we have

2250m _____ 2500m.

Since 2500m is greater than 2250m, we can say that

⇒ 2250m < 2500m.

Finally, we can fill in the blanks as

2\(\frac{1}{4}\) km < ​​2500m.

For the given quantities, the comparison has been done and we get   2\(\frac{1}{4}\) km <​​ 2500m.

Page 166 Exercise 5.7 Problem 7

It is given that 1\(\frac{1}{20}\) m_____120 cm.

To compare and fill in the blank.

First, we will convert the quantities on both sides of the blank into a single unit, i.e. centimeters.

Then we will compare them and use the appropriate sign to fill the blank.

Let us convert 1\(\frac{1}{20}\) min   to centimetres.

Here, the whole number is 1m.

Using conversion 1m = 100cm, we get it to be equivalent to

⇒ 1m × 100 = 100cm.

The improper fraction is \(\frac{1}{20}\) m.

Using conversion, we get it to be equivalent to

⇒ \(\frac{1}{20}\) m × 100  = 5cm.

Combining these by adding

⇒ 100cm + 5cm = 105 cm

So, we have got 105 cm.

Now, we have 105 cm_____120 cm.

Since 105cmis less than 120cm, we can say that

⇒ 105 cm < 120 cm

Finally, we can fill in the blanks as

1\(\frac{1}{20}\) m < ​​120 cm.

For the given quantities, the comparison has been done and we get 1\(\frac{1}{20}\) m <​​ 120cm.

Page 166 Exercise 5.7 Problem 8

It is given that 1 \(\frac{3}{4}\) ft _____ 20 in.

To compare and fill in the blank.

First, we will convert the quantities on both sides of the blank into a single unit, i.e. inches.

Then we will compare them and use the appropriate sign to fill in the blank

Let us convert 1\(\frac{3}{4}\) ft into inches.

Here, the whole number is 1 ft.

Using conversion 1 ft = 12 in, we get it to be equivalent to

⇒ 1 ft × 12 = 12 in.

The improper fraction is \(\frac{3}{4}\) ft.

Using conversion, we get it to be equivalent to

⇒ \(\frac{3}{4}\) ft × 12 = 9 in.

Combining these by adding

⇒ 12 in + 9 in = 21 in.

So, we have got 21 in.

Now, we have 21 in_____20 in

Since 21 in is greater than 20 in, we can say that

⇒ 21 in > 20 in

Finally, we can fill the blank as 1 \(\frac{3}{4}\) ft >​​ 20 in.

For the given quantities, the comparison has been done and we get 1 \(\frac{3}{4}\) ft >​​ 20 in.

Page 167 Exercise 5.8 Problem 1

It is given that 8months

To express it as a fraction of 1yr

First, we will convert the quantities in terms of months.

Using conversion, we know that 1 year = 12 months.

Hence, the fraction can be written as  \(\frac{8}{12}\)

Expressing it in simplest form by canceling common factors  \(\frac{2}{3}\)

 Therefore, 8 months has been expressed as a fraction of 1yr as   \(\frac{2}{3}\)

Page 167  Exercise 5.8  Problem 2

It is given that 95cm.

To express it as a fraction of 1m.

First, we will convert the quantities in terms of centimeters.

Using conversion, we know that 1m = 100cm.

Hence, the fraction can be written as.\(\frac{95}{100}\)

Expressing it in simplest form by canceling common factors   \(\frac{19}{20}\).

Therefore, 95cm has been expressed as a fraction of 1m as  \(\frac{19}{20}\)

Page 167  Exercise 5.8  Problem 4

It is given that 45min.

To express it as a fraction of 1h.

First, we will convert the quantities in terms of minutes.

Using conversion, we know that 1h = 60 min.

Hence, the fraction can be written as \(\frac{45}{60}\).

Expressing it in simplest form by canceling common factors \(\frac{3}{4}\)

Therefore, 45min has been expressed as a fraction of 1h as \(\frac{3}{4}\)

Page 167  Exercise 5.8  Problem 5

It is given that 15¢.

To express it as a fraction of 1$.

First, we will convert the quantities in terms of cents.

Using conversion, we know that 1$ = 100¢.

Hence, the fraction can be written as \(\frac{15}{100}\)

Expressing it in simplest form by canceling common factors  \(\frac{3}{20}\)

Therefore, 15¢ has been expressed as a fraction of 1$ as \(\frac{3}{20}\).

Page 167  Exercise 5.8  Problem 6

It is given that 650g.

To express it as a fraction of 1kg.

First, we will convert the quantities in terms of grams.

Using conversion, we know that 1kg = 1000g.

Hence, the fraction can be written as  \(\frac{650}{1000}\)

Expressing it in simplest form by canceling common factors   \(\frac{13}{20}\).

Therefore, 650g  has been expressed as a fraction of  1kg as  \(\frac{13}{20}\)

Page 168  Exercise 5.8  Problem 7

It is given that 40min.

To express it as a fraction of 2h.

First, we will convert the quantities in terms of minutes.

Using conversion, we know that 1h = 60 min.

So, we get that

⇒ ​2h = 2 × 60min = 120min
​
Hence, the fraction can be written as  \(\frac{40}{120}\)

Expressing it in simplest form by canceling common factors \(\frac{1}{3}\)

Therefore, 40min has been expressed as a fraction of 2h as \(\frac{1}{3}\).

Page 168  Exercise 5.8  Problem 8

It is given that 8 in.

To express it as a fraction of 3ft.

First, we will convert the quantities in terms of inches.

Using conversion, we know that 1 ft = 12 in.

So, we get that ​3 ft = 3 ×12 in = 36 in
​
Hence, the fraction can be written as \(\frac{8}{36}\)

Expressing it in simplest form by canceling common factors   \(\frac{2}{9}\)

Therefore, 8in has been expressed as a fraction of 3 ft as  \(\frac{2}{9}\).

Page 168  Exercise 5.8  Problem 9

It is given that $3.

To find what fraction of $3 is 90¢.

If we consider x as the fraction, then writing this mathematically, we have $3 × x = 90¢

First, we will convert the quantities in terms of cents.

Using conversion, we know that 1$ = 100¢.

Hence, the fraction can be written as

⇒ 300¢ × x = 90¢

Expressing it in simplest form by canceling common factors, we get

​x = \(\frac{90}{9}\)

x = \(\frac{3}{10}\)

Therefore, 90¢is \(\frac{3}{10}\) fraction of $3.

​Page 168 Exercise 5.8  Problem 10

It is given that Mrs. King bought 2kg of flour and used 750g  for baking bread.

We have to compute the fraction of flour she used.

First, we convert 2kg to grams using conversion 1kg = 1000g

⇒ 2kg = 2 × 1000g

2kg = 2000g
​
Now, the total amount of flour is 2000g.

The amount of flour used is 750g.

So, the fraction of flour used will be  \(\frac{750}{2000}\)

The simplest form would be obtained after canceling common terms as  \(\frac{3}{8}\)

Therefore, the fraction of flour used by Mrs. King is \(\frac{3}{8}\) when she bought 2kg of flour and used 750g for baking bread.

​Page 168 Exercise 5.8  Problem 11

It is given that Mrs. King bought 2kg of flour and used 750g for baking bread.

We have to compute the fraction of flour left with her.

From Exercise 5.8  Problem 10,  we have the total amount of flour as 2000g.

The amount of flour used is 750g.

So, the amount of flour left would be = 2000−750g

= 1250g

The fraction of flour left would be  \(\frac{1250}{2000}\)

Expressing in simplest form by canceling common factor   \(\frac{5}{8}\)

Therefore, the fraction of flour left with Mrs. King is \(\frac{5}{8}\) when she bought 2kg flour and used 750g for baking bread.

Primary Mathematics  Chapter 5 Measures

Primary Mathematics Workbook 4A Common Core Solutions Chapter 5 Measures Exercise 5.1 Page 155  Exercise 5.1 Problem 1

A table with pounds and ounces is given as below

After that, we find the value from the table and then fill the blank.

We have to complete the table and then fill the blank in 8

lb = __________ oz.

So, we use the conversion 1 Pound =16 ounces and then complete the table by writing down values of ounces corresponding to the given pounds.

The table has values of pounds from 1 to 10 Using the conversion

3 pounds = 3 × 16

3 pounds = 48 ounces

4 pounds = 4 × 16

4 pounds = 64 ounces

5 pounds = 5 × 16

5 pounds=80 ounces

6 Pounds = 6 × 16

6 Pounds = 96 ounces

7 pounds = 7 × 16

7 pounds = 112 ounces

8 pounds = 8 × 16

8 pounds= 128 ounces

9 pounds = 9 × 16

9 pounds = 144 ounces

10 pounds = 10 × 16

10 pounds =160 ounces

Now the completed table is

From the table, we can see that 8 pounds is 128 ounces, so we can fill the blank as 8 lb = 128oz

For the given table

The completed table is

And the blank is filled as, 8 lb = 128 oz

Chapter 5 Exercise 5.1 Measures Workbook 4A Answers Page 155  Exercise 5.1  Problem 2

A table with pounds and ounces is given below

After that, we find the value from the table and then fill the blank.

We have to complete the table and then fill the blank in 80 oz = ______ lb.

So, we use the completed table from previous part and then fill the blank by locating values of pounds corresponding to the given ounces.

From problem 1 of Exercise 5.1, we have the completed table as below

From the table, we can see that corresponding to 80 ounces, we have 5 pounds, so we can fill the blank as 80 oz =5 lb

For the given table

The completed table is

And the blank is filled as 80 oz = 5 lb

Page 155  Exercise 5.1  Problem 3

A table with pounds and ounces is given as below

After that, we find the value from the table and then fill the blank.

We have to complete the table and then fill the blank in 6 lb 7 oz = ______ lb.

So, we use the completed table from previous part and then fill the blank by locating values of pound corresponding to the given ounces.

From problem 1 of Exercise 5.1, we have the completed table as below

From the table, we can see that corresponding to 6 pounds, we have 96 ounces.

And in the LHS, we have additional 7 ounces, so in order to represent entire LHS in ounces as per RHS, we add 96 and 7 to get 103 ounces.

So, we can fill the blank as 6 lb 7 oz = 10 oz

For the given table

The completed table is

And the blank is filled as 6lb 7 oz =103 oz

Workbook 4A Chapter 5 Exercise 5.1 Measures Solutions Page 155  Exercise 5.1  Problem 4

A table with pounds and ounces is given as below

After that, we find the value from the table and then fill the blank.

We have to complete the table and then fill the blank in 112 oz =_________ lb __________ oz.

So, we use the completed table from the previous part and then fill the blank by locating values of pound corresponding to the given ounces.

From problem 1 of Exercise 5.1, we have the completed table as below

From the table, we can see that corresponding to 7 pounds, we have 112 ounces.

The quantity on the LHS is 112 ounces and since we have a whole number as the conversion gives 7 pounds, the quantity of ounces on the RHS would be 0.

So, we can fill the blank as 112 oz = 7 lb 0 oz

For the given table

The completed table is

And the blank is filled as 112 oz = 7 lb 0 oz

Measures Exercise 5.1 Primary Mathematics Workbook 4A Explanation Page 155  Exercise 5.1  Problem  5

It is given to fill the blanks for 25m = _________ cm.

We know that 1m = 100cm.

So, to get 25min cm, we have to multiply 25 by 100.

Therefore, we get 25m  = 2500cm.

The given blank in 25cm = _____ cm has been filled using the conversion as 25cm =  2500 cm.

Common Core Workbook 4A Chapter 5 Measures Exercise 5.1 Help Page 155  Exercise 5.1  Problem  6

It is given to fill the blanks for 10ft ________ in.

We know that 1ft =  12 in.

So, to get 10ft in, we have to multiply 10 by 12.

Therefore, we get 10ft =  120 in.

The given blank in 10ft = _____ in has been filled using the conversion as 10ft = 120in

Measures Exercise 5.1 Workbook 4A Detailed Solutions Common Core Page 155  Exercise 5.1 Problem  7

It is given to fill in the blanks for 2gal = _________ qt.

We know that 1gal = 4qt

So, to get 2gal in qt we have to multiply 2 by 4

Therefore, we get 2 gal = 8qt

The given blank in 2gal= _____ qt has been filled using the conversion as 2gal =  8 qt.

Measures Exercise 5.1 Workbook 4A Detailed Solutions Common Core Page 155  Exercise 5.1 Problem  8

It is given to fill the blanks for 3km = _________ m.

We know that 1km = 1000m.

So, to get 3km in m, we have to multiply 3 by 1000.

Therefore, we get 3km = 3000m.

The given blank in 3km = _____ m has been filled using the conversion as 3km = 3000m.

It is given to fill the blanks for 5 lb= _________ oz.

We know that 1lb = 16 oz.

So, to get 5lb in oz, we have to multiply 5 by 16.

Therefore, we get 5 lb = 80 oz.

The given blank in 5lb = _____ oz has been filled using the conversion as 5 lb = 80 oz

Measures Exercise 5.1 Workbook 4A Detailed Solutions Common Core Page 155  Exercise 5.1  Problem  9

It is given to fill the blanks for 4kg= _________ g

We know that 1kg = 1000g

So, to get 4kg in g, we have to multiply 4 by 1000

Therefore, we get 4kg =  4000 g

The given blank in 4kg = _____ g has been filled using the conversion as 4kg = 4000 g.

Measures Exercise 5.1 Workbook 4A Detailed Solutions Common Core Page 155  Exercise 5.1 Problem  10

It is given to fill the blanks for 6L = _________ ml

We know that 1L = 1000ml

So, to get 6L in ml, we have to multiply 6 by 1000

Therefore, we get 6L = 6000ml.

The given blank in 6L = _____ ml has been filled using the conversion as 6L = 6000 ml.

Measures Exercise 5.1 Workbook 4A Detailed Solutions Common Core Page 155  Exercise 5.1 Problem  11

It is given to fill in the blanks for 11 days = _________ h

We know that 1 day = 24h.

So, to get 11 days in h, we have to multiply 11 by 24

Therefore, we get 11 days = 264h.

The given blank in 11days = _____ h has been filled using the conversion as 11days = 264 h

Primary Mathematics Workbook 4A Chapter 5 Measures Exercise 5.1 Problem Solving Page 156 Exercise 5. 1 Problem 12

It is required to fill the blanks in given 5 yr 6months =_______ months.

We first convert year to months and then add the quantities on the LHS such that it gets converted into single unit as required on the RHS.

So, we use 1yr  = 12 months to convert 5 yr to months.

Therefore, we get 5 yr = 5 × 12 months

5yr = 60months

Now, we have to add 60 months and 6 months and we get 66 months as the total quantity on the LHS.

Therefore, we can fill the blank on the RHS as 5yr 6 months = 66 months

The given blank in 5yr 6months = ____ months has been filled as 5yr 6months = 66months.

Page 156 Exercise 5. 1 Problem  13

It is required to fill the blanks in given 6km​​20m= _______ m.

We first convert kilometers to meters and then add the quantities on the LHS such that it gets converted into single unit as required on the RHS.

So, we use 1km = 1000m to convert 6km to m

Therefore, we get

6km = 6 × 1000m

6km = 6000m

Now, we have to add 6000 and 20 and we get 6020 as the total quantity on the LHS.

Therefore, we can fill the blank on the RHS as 6km​​20m = 6020m

The given blank in 6km20m= ____ m has been filled as 6km​​20m = 6020 m

Page 156 Exercise 5. 1 Problem  14

It is required to fill the blanks in given 8L​​ 100ml= _______ ml

We first convert liter to milliliter and then add the quantities on the LHS such that it gets converted into single unit as required on the RHS.

So, we use 1L = 1000ml to convert 8L to ml

Therefore, we get 8L = 8 × 1000ml

8L = 8000ml

Now, we have to add 8000 and 100 and we get 8100 as the total quantity on the LHS.

Therefore, we can fill the blank on the RHS

8L 100 ml =  8100 ml

The given blank in 8L100ml = ____ ml has been filled as 8L100ml = 8100ml

Page 156 Exercise 5. 1 Problem  15

It is required to fill the blanks in given 5ft​​3in _______ in.

We first convert feet to inches and then add the quantities on the LHS such that it gets converted into single unit as required on the RHS.

So, we use 1 ft = 12 in to convert 5 ft to in

Therefore, we get

5 ft = 5 × 12 in

5 ft = 60 in

Now, we have to add 60 and 3 and we get 63 as the total quantity on the LHS.

Therefore, we can fill the blank on the RHS as

5 ft 3 in =  63 in

The given blank in 5 ft 3 in = ____ in has been filled as 5 ft 3 in = 63 in

Page 156 Exercise 5. 1 Problem  16

It is required to fill the blanks in given 7lb​​15oz= _______ oz.

We first convert pounds to ounces and then add the quantities on the LHS such that it gets converted into single unit as required on the RHS.

So, we use 1lb = 16 oz to convert 7lb to  oz.

Therefore, we get

7 lb = 7 × 16 oz

7 lb = 112 oz

Now, we have to add 112 and 15 and we get 127 as the total quantity on the LHS.

Therefore, we can fill the blank on the RHS as 7lb​​15 oz = 127oz

The given blank in 7 lb​​15 oz = ____ oz has been filled as 7lb​​ 15 oz = 127 oz

Page 156 Exercise 5. 1 Problem  17

It is required to fill the blanks in given 1 qt 1 pt= _______ pt.

We first convert quarts to pints and then add the quantities on the LHS such that it gets converted into single unit as required on the RHS.

So, we use 1qt = 2pt

Now, we have to add 2 and 1 and we get 3 as the total quantity on the LHS.

Therefore we can fill the blank on the RHS as

1 qt = 2 pt

The given blank in 1 qt 1 pt = ______ pt has been filled as  1 qt 1pt = 3 pt 

Page 156 Exercise 5.1 Problem 18

It is required to fill in the blanks in the given 113 ft = _______ yd ______ ft.

The quantity on the LHS has to be written in terms of two units to fill the blanks on the RHS.

So, first, we find how many yards are there in 113 feet.

We know that

1 yd = 3 ft or 1 ft = \(\frac{1}{3}\)yd

So, 113 ft = \(\frac{1}{3}\) × 113 yd

Dividing it
​

So there are 37 yd and a remaining 2 ft.

Therefore, the blank can be filled as  113ft = 37 yd​​ 2 ft.

The given fill in the blanks in 113ft= ____ yd ____ ft can be filled as 113 ft = 37 yd​​ 2 ft

​Page 156  Exercise 5.1 Problem 19

It is required to fill in the blanks in the given 8030L = ____ L ____ ml

The quantity on the LHS has to be written in terms of two units to fill the blanks on the RHS.

We know that 1L = 1000ml

So, we can divide 8030 by 1000 to get how many liters and milliliters are there in it.

\(\frac{8030}{1000}\) = 8.030

This means that there are 8L and 30ml in it.

Therefore, the blank on the RHS can be filled as

The given fill in the blanks in 8030 L = ____ L ____ ml can be filled as 8030 L = 8L 30ml

Page 156 Exercise 5.1 Problem 20

It is required to fill in the blanks in the given 45 days = ____ weeks ____ days.

The quantity on the LHS has to be written in terms of two units to fill the blanks on the RHS.

So, first, we find how many weeks are there in 45 days.

We know that 1week = 7days

So, dividing 45 by 7

This means that there are 6 weeks and remaining 3 days in 45 days.

Therefore, the blanks on the RHS can be filled as 45days  = 6 weeks​​ 3 days

The given fill in blanks in 45 days = ____ weeks ____ days has been filled as 45 days =  6 weeks ​​3 days

Page 156 Exercise 5.1 Problem 21

It is required to fill in the blanks in the given 265qt = ____ gal _____ qt.

The quantity on the LHS has to be written in terms of two units to fill the blanks on the RHS.

So, first, we find how many gallons are there in 265qt.

We know that 1gal  = 4 qt.

So, divide 265 by 4 to get the number of gallons

So, it means that there are 66 gallons and the remaining 1quarts in 265 qt.

Therefore, the blanks on the RHS can be filled as 265 qt = 66 gal​​ 1 qt

The given fill in the blanks in 265qt = ____ gal ____ qt has been filled as 265qt = 66 gal​​ 1qt

Page 157 Exercise 5.2 Problem 1

It is given that 1L−385 ml =_____ ml.

To subtract them.

First, we convert the terms on the LHS into a single unit to be able to subtract them.

The unit to be converted would be same as required on the RHS.

Here, we require milliliters.

So, we know that 1L = 1000ml.

Thus the LHS becomes

1000ml − 385ml

Subtracting, we get
​
1 0 0 0

−3 8 5
______
6 1 5
​
Hence, the RHS will become 615ml and we get 1L−385ml = 615ml

The given quantities 1L-385ml have been subtracted and the result obtained is 1 L− 385ml = 615ml.

Page 157 Exercise 5.2 Problem 2

It is given that 1h−22 min =____ min.

To subtract them.

First, we convert the terms on the LHS into a single unit to be able to subtract them.

The unit to be converted would be same as required on the RHS.

Here, we require minutes.

So, we know that 1h = 60 min.

Thus the LHS becomes

60 min−22 min

Subtracting, we get
​
6 0

−2 2
______
4 4
​
Hence, the RHS will become 44min and we get

1h−22 min = 44 min

The given quantities 1h−22min have been subtracted and the result obtained is 1h−22 min = 44min.

Page 157  Exercise 5.2  Problem 3

It is given that 1 ft−4 in=____ in.

To subtract them.

First, we convert the terms on the LHS into a single unit to be able to subtract them.

The unit to be converted would be same as required on the RHS.

Here, we require inches.

So, we know that 1 ft = 12 in.

Thus the LHS becomes

12 in − 4 in

Subtracting, we get RHS as 8 in

Hence, we get 1 ft − 4 in = 8 in

The given quantities 1ft-4in have been subtracted and the result obtained is 1 ft−4 in = 8 in.

Page 157 Exercise 5.2 Problem 4

It is given that 1lb−7 oz= ____ oz

To subtract them.

First, we convert the terms on the LHS into a single unit to be able to subtract them.

The unit to be converted would be same as required on the RHS.

Here, we require ounces.

So, we know that 1lb = 16 oz.

Thus the LHS becomes

16 oz−7 oz

Subtracting, we get RHS as 9 oz

Hence, we get 1lb−7 oz = 9 oz

The given quantities 1lb−7 oz have been subtracted and the result obtained is 1 lb−7 oz = 9 oz

Page 157  Exercise 5.2  Problem 5

It is given that 1gal−1 pt =____ pt.

To subtract them.

First, we convert the terms on the LHS into a single unit to be able to subtract them.

The unit to be converted would be same as required on the RHS.

Here, we require pints.

So, we know that 1gal = 8 pt.

Thus the LHS becomes 8 pt−1 pt

Subtracting, we get RHS as 7 pt

Hence, we get 1gal−1 pt = 7 pt

The given quantities 1gal−1pt have been subtracted and the result obtained is 1gal−1 pt = 7 pt.

Page 157  Exercise 5.2  Problem 6

It is given that  3h 20min + 6h 45min = ____ h ____ min

To add them.

First, we add the smaller and bigger units together separately.

Then we get

3h + 6h = 9h

20min + 45min = 65min

Now we convert the smaller unit in terms of the bigger unit using the conversion 1h=60min.

65min will have 60min + 5min , which is equivalent to 1h + 5min.

So, now we get it as 1h and 5 min.

Next, we can add this along with the previously found value 9h and get 9h + 1h = 10h.

The remaining quantity is 5min

Therefore, we can write the result as  3h20min + 6h45min = 10h5min

The given quantities have been added and the result is obtained as 3h20min + 6h45min = 10h 5min.

Page  157 Exercise 5.2  Problem 7

It is given that 12kg10g − 10kg600g = ____ kg ____ g

To subtract them.

First, we convert each of the compound units into single unit using the conversion 1kg = 1000g.

Then we get

12kg10g = 12  ×  1000g + 10g

12kg10g ​​​​​​​​​​​​​​= 12000g + 10g

​12kg10g  = 12010g
​
​10kg600g = 10 × 1000g + 600g

​10kg600g = 10000g + 600g

​10kg600g =10600g
​
Now we can subtract the quantities and we get
​
​1 2 0 1 0

1 0 6 0 0
−
__________
1 4 1 0

​
So we got it as 1410g

Now we use the conversion 1kg = 1000g and convert it in terms of kg and g.

So, now we get it as \(\frac{1410}{1000}\)

= 1.410kg

This means that there is 1kg and 410g

Therefore, we can write the

12kg 10g− 10kg 600g  = 1kg 410g.

The given quantities have been subtracted and the result obtained is 12kg 10g −10kg 600g = 1kg 410g.

Page 157 Exercise 5.2 Problem 8

It is given that 17 ft 3 in−7 ft 4 in = ____ ft ____ in.

To subtract them.

First, we convert each of the compound units into single unit using the conversion 1 ft = 12 in.

Then we get

17 ft 3 in = 17 × 12 in + 3 in

17 ft 3 in = 204 in + 3 in

17 ft 3 in = 207 in
​
​7 ft 4 in = 7 × 12 in + 4 in

​7 ft 4 in = 84 in + 4 in

​7 ft 4 in = 88 in
​
Now we can subtract the quantities and we get
​​2 0 7

8 8
_
______
119

​So we got it as 119in.

Now we use the conversion 1 ft = 12 in and convert it in terms of feet and inches by

Dividing 119 by 12

This means that there are 9 feet and 11 inches in 119 inches.

Therefore, we get the result as

17 ft 3 in −  7 ft 4 in = 9 ft 11 in

The given quantities have been subtracted and the result obtained is 17 ft 3 in − 7 ft 4 in = 9ft 11 in.

Page 157 Exercise 5.2 Problem 9

It is given that 5gal 2 qt + 1gal 3 qt= ___ ga___ qt

To add them.

First, we add the smaller and bigger units together separately.

Then we get

5gal + 1gal = 6gal

2 qt + 3 qt = 5 qt

Now we convert the smaller unit in terms of the bigger unit using the conversion 1gal = 4 qt.

So, now we get it as 5 qt = 4 qt + 1 qt which is 1gal and 1qt.

Next, we can add this along with the previously found value of 6gal to get 7gal.

The remaining quantity is 1 qt.

Therefore, we can write the result as

5gal 2 qt + 1gal 3 qt = 7gal 1qt

The given quantities have been added and the result is obtained as 5gal 2 qt + 1gal 3 qt = 7gal 1 qt.

Page 158 Exercise 5.2  Problem 10

It is given that a bag of beans has a mass of 1kg 830g, a bag of flour is 340g heavier than a bag of beans, and 680g lighter than a bag of coffee.

We have to find the mass of a bag of coffee in kilograms and grams.

First, we interpret the given data and write it in mathematical terms.

So, a bag of flour is 340g heavier than bag of beans weighing 1kg830g, so we have to add these up to get the weight of bag of flour.

So, weight of a bag of flour = 1kg 830g + 340g.

Now, a bag of flour is 680g lighter than a bag of coffee.

It means that bag of coffee is 680g heavier than bag of flour.

⇒ Weight of bag of coffee = 1kg830g + 340g + 680g

Now adding all the grams, we get

8  3  0
3  4  0
6  8  0
+
________
1850

​
Converting this as kg and g using conversion 1kg = 1000g

= 1.850kgor1kg850g

We already have 1kg, so by adding it with the above result, we get  2kg 850g

∴  Weight of bag of coffee  = 2kg850g

For the given data that a bag of beans has a mass of 1kg830g , a bag of flour is 340g heavier than the bag of beans and 680g lighter than a bag of coffee, the mass of a bag of coffee has been computed as 2kg850g.

Page 158  Exercise 5.2  Problem 11

It is given that the backyard of a property is 130ft long and Mr.

Maxwell has 40 yd of fencing.

We have to find out how much more fencing he needs in yards and feet.

So, we first interpret given data and write it in mathematical form.

Total length of the backyard is 130 ft.

Available fencing is 40yd

⇒ More fencing required = 130 ft − 40 yd.

Now using conversion 1 yd = 3 ft

40 yd = 40 × 3 ft

40 yd = 120 ft

⇒  More fencing required = 130 ft−120 ft

More fencing required  = 10 ft

Dividing 10ft by 3

​We get 3 yd and 1ft

Therefore, Mr. Maxwell will need 3yd1ft more length of fencing.

For the given data that the backyard of a property is 130ft long and Mr. Maxwell has 40 yd of fencing, he will need 3yd1ft more fencing.

Page 159 Exercise 5.3 Problem 1

It is given to multiply 3m​​20cm × 4 = ____ m ____ cm

So, first we multiply the meters and then we multiply the centimeters by 4

Multiplying 3m by 4, we get

3m × 4 = 12m

Multiplying 20cm by 4, we get

20cm × 4 = 80cm

Therefore, we get the final result as 3m​​20cm × 4 = 12m​​ 80cm.

The multiplication has been done and the blanks can be filled as 3m​​20cm × 4 = 12m 80cm.

Page 159 Exercise 5.3 Problem 2

It is given to multiply 85cm×3=_____ cm and then express it in terms of meters and centimeters.

So, we multiply 85 and 3 to get 255cm

Now we use the conversion 1m = 100cm to convert into 255cm meters and centimeters.

Dividing 255by 100

\(\frac{255}{100}\) = 2.55 m

This means that there are 2m and 55cm.

So, we can write the final result as

​85cm × 3 = 255cm = 2m​​55cm
​
The multiplication has been done and the result obtained is  85cm × 4 = 255cm = 2m​​55cm
​

Page 159 Exercise 5.3 Problem 3

It is given to multiply 2m85cm × 3 = ____ m ____ cm.

So, first we multiply the meters and then we multiply the centimeters by 3

Multiplying 2m by 3, we get

2m × 3 = 6m

Multiplying 85cm by 3, we get

85cm × 3 = 255cm

Now we use conversion 1m = 100cm to convert 255cm into meters and centimeters.

So, dividing

\(\frac{255}{100}\) = 2.55 m

This means that it indicates 2m55cm.

Now adding it to previous result of 6m, we get 8m55cm.

Therefore, the final result can be filled in the blanks as

​2m85cm × 3 = 6m​​255cm

2m85cm × 3 = 8m​​55cm
​
The multiplication has been carried out and the blanks can be filled as2m85cm×3=6m​​255cm, 2m85cm × 3 = 8m​​55cm
​

Page 159 Exercise 5.3 Problem 4

It is given that 6 ft ​​2 in × 4 =____ ft ____ in.

We have to multiply and fill in the blanks.

To do that we will multiply each unit with the given integer separately.

Multiplying 6 ft by 4, we get

6 ft × 4 = 24 ft

Multiplying 2 in by 4, we get

2 in × 4 = 8 in

Therefore, we can fill the blanks as

6 ft​​ 2 in × 4 = 24 ft 8 in

The given fill-in-the-blanks has been filled using multiplication as 6 ft​​ 2 in × 4 = 24 ft​​ 8 in.

It is given that 9 in × 6= ____ in and we have to fill the blanks in terms of feet and inches as well.

First, we multiply with the given integer and then use the conversion to express it as feet and inches.

Multiplying 9 in by 6, we get 54 in.

Using conversion 1 ft = 12 in, we can convert 54 in as below.

Dividing

This means that it is 4 feet and 5 inches.

Therefore, we can fill the blanks as

​9 in × 6 = 54 in

9 in × 6 = 4 ft​​ 5 in
​
The given fill-in-the-blanks have been filled using multiplication as​ 9 in × 6 = 54 in , 9in × 6 = 4 ft​​ 5 in

Page 159 Exercise 5.3 Problem 5

It is given that 10 feet  and  ​​9 inches × 6 = ____ ft ____ in.

To fill the blanks.

First, we multiply each unit with the integer separately.

Multiplying 10 ft by 6, we get

10 ft × 6 = 60 ft

Multiplying 9 in by 6, we get

9 in × 6 = 54 in

Using conversion 1ft = 12in, we can convert 54in.

Be referring to Problem 4  Exercise 5.3, we get it as 4 ft 5 in.

So, adding it to the previous result of 60ft, we have 64 ft​​ 5 in.

Therefore, we can fill the blanks as 265 qt

​10 ft 9 in × 6 = 60 ft ​​54 in

10 ft 9 in × 6 = 64 ft  5 in
​

The given fill-in-the-blanks has been filled using multiplication as 10ft 9in × 6 = 60 ft ​​54 in, 10 ft 9 in × 6 = 64 ft  5 in

Page 160  Exercise 5.4  Problem 1

It is given that 4km​​250m ÷ 2= ____ km ____ m.

To fill the blanks with results obtained after division.

Dividing 4km by 2, we get 2km.

Dividing 250m by 2, we get 125m.

Therefore, the given blanks can be filled as 4km​​250m ÷ 2 = 2km​​125m.

The given blanks in the question 4km​​250m ÷ 2=___ km___ m can be filled as 4km​​250m ÷ 2 = 2km​​125m.

Page 160  Exercise 5.4  Problem 2

It is given that

1km200m ÷ 3 = 1200m ÷ 3 = ____ m.

We have to divide the given unit in meters by the given integer.

So, dividing 1200m by 3, we get 400m.

Therefore, the blanks can be filled as

1km200m ÷ 3 = 1200m ÷ 3

1km200m ÷ 3  =  400m

The given fill in the blanks has been filled after applying the division operation as 1km200m ÷ 3 = 1200m ÷ 3400 m

Page 160  Exercise 5.4  Problem 3

It is given that 4km​​200m ÷ 3 = ____ km ___ m.

We have to fill the blanks after dividing.

From the question, 4km can be written as  3km + 1km.

Adding 1km along with 200m and using the conversion 1km = 1000m , we can write it as 1200m

So, we get the LHS as 3km1200m ÷ 3

From Problem 2 Exercise 5.4, we have the result that 1200m ÷ 3 = 400m.

So, by dividing each unit separately by 3, we finally get it as 1km​​400m.

Therefore, we can write the blanks as

4km​​200m ÷ 3 = 1km​​400m

The given fill-in-the-blanks has been filled as 4km​​200m÷3= 1km​​400m using the division operation.

Page 161  Exercise 5.5  Problem 1

It is given that a bottle holds 1L500ml of water and a bucket holds 3 times as much water as the bottle.

We have to find out how much water the bucket holds.

So, first we interpret and write it is mathematical form.

Then, we use the multiplication operation for compound units and multiply each unit separately.

Then, we use the conversion that 1L = 1000ml.

Finally, we combine the results to get the final result.

The word “times” means multiplication.

So, 3 times would mean that quantity is being multiplied by 3.

Now we multiply 1L 500ml by 3.

Multiplying 1L by 3, we get 3L.

Multiplying 500ml by 3, we get 1500ml.

Now, we use the conversion that 1L=1000ml.

So, 1500ml would be \(\frac{1500}{1000}\)

= 1.5L​​ or ​​1L 500ml.

Adding it with previously found value of 3L, we get 4L​​500ml.

Therefore, the amount of water that the bucket can hold is 4L ​​500ml.

For the given situation when a bottle holds 1L 500ml of water and a bucket holds 3 times as much water as the bottle, the amount of water held by the bucket has been calculated as 4L 500ml.

Page 161 Exercise 5.5 Problem 2

It is given that a washing machine takes 1h40min to wash 1 load of laundry.

We have to find how long it takes to wash 4 loads of laundry.

So, we multiply the time taken for 1 load of laundry by 4.

We multiply hours and minutes separately.

Then use the conversion 1h = 60min to convert minutes in terms of hours and minutes.

Adding the hours together and writing it with the minutes will give us the required results.

We have time taken for 1 load of laundry as 1h40min.

Multiplying it with 4

1h × 4 = 4h and

40min × 4 = 160min

Now using conversion 1h = 60min

We divide 160 by 60

So, we get it as 2h​​ 40min.

Adding it along with 4h

6h 40min.

Hence, the time taken for 4 loads of laundry will be 6h 40min.

The time taken by the washing machine for 4 loads of laundry is 6h 40min when it is given that the washing machine takes 1h40min to complete 1 load of laundry.

Page 161  Exercise  5.5 Problem 3

It is given that a fruit seller packed all his oranges into 6 boxes and each box of oranges weighed 12 lb 12 oz.

We have to find total weight of the oranges.

So, we multiply the weight of each orange box by 6.

We multiply pounds and ounces separately.

Then use the conversion 1lb=16oz to convert ounces in terms of pounds and ounces.

Adding the pounds together and writing it with the ounces will give us the required results.

We have to multiply the weight of one box of orange by the number of boxes.

So, we have weight of one box of oranges as 12 lb 12 oz and the number of boxes is 6.

12 lb12 oz  ×  6

We get

12 lb × 6 = 72 lb

12 oz × 6 = 72 oz

We have to use conversion 1 lb = 16 oz to convert 72 oz.

Dividing 72 by 16

This gives 4lb8oz.

Adding it along with 72 lb, we get 76 lb 8oz.

Therefore, the weight of 6 boxes of oranges is 76 lb 8 oz.

For the given data that a box of oranges weighed 12lb​12oz and the fruit seller packed his entire oranges in 6 boxes, then the weight of 6 boxes of oranges is found out to be 76 lb ​​8 oz.

Page 162  Exercise 5.5  Problem 4

It is given that Meredith had 6lb12oz of mushrooms and she packed them equally into 9 boxes.

We have to find the weight of the mushrooms in each box.

It is said that “equally into 9 boxes” which means that we have to use the division operation.

First, convert into ounces using conversion 1lb = 16 oz

Divide that by 9.

Finally, reconvert back in terms of pounds and ounces.

We have the total weight of mushrooms as 6 lb​​12 oz.

The number of boxes is 9.

So, the weight of one box of mushrooms would be 6 lb ​​12oz ÷ 9.

Using 1 lb = 16 oz

​ 6 lb = 6 × 16 oz

​ 6 lb = 96 oz
​
Adding it with 12 oz,  108 oz

Now, divide 108 by 9

This gives 12oz.

Since it doesn’t exceed 16oz, we cannot convert to pounds.

So, the weight of one box of mushroom is 12 oz.

For the given situation where Meredith had  6 lb 12 oz of mushrooms and she packed them equally into 9 boxes, the weight of the mushrooms in each box has been calculated as 12 oz.

Page 162 Exercise 5.5 Problem 5

It is given that a box contains 5 identical books and it has a mass of 6kg850g.

The mass of the empty box is 600g.

We have to find the mass of each book.

Since mass of box + books is given, first deduct mass of empty box to get mass of books alone.

Use unit conversion 1kg =1000g to convert to grams and then subtract.

Then, divide the obtained mass by 5 to get mass of one book.

The mass of books and box is given as 6kg850g.

Using unit conversion, we get

6 × 1000g = 6000g

So, mass of books and box is 6000g+850g=6850g.

The mass of empty box is 600g.

The mass of books  = 6850g − 600g

The mass of books = 6250g

The mass of books is 6250g.

The number of books is 5.

So, the mass of one book is obtained as 6250g ÷ 5.

Dividing

This gives 1250g.

Using conversion 1kg = 1000g

\(\frac{1250}{100}\) = 1.25kg

Or we get 1kg​​250g.

Hence, the mass of one book is 1kg​​250g.

Therefore we got the mass of one book 1kg​​250g when we were given that mass of 5 identical books in a box is  6kg​​850g and the mass of the empty box is 600g.

Primary Mathematics  Chapter 5 Measures

Primary Mathematics Workbook 4A Common Core Solutions Chapter 5 Measures Exercise 5.9 Page 169  Exercise 5.9  Problem 1

It is given that 2km ​​634m =  ____

To fill – In the blanks.

First, we will convert the quantities in terms of meters.

Using conversion, we know that 1km=1000m

Hence, 2km can be written as 2000m

Adding it along with 634m, we get the final result as

2000m + 634m = 2634m

The blank is filled as 2km​​634m = 2634m

The given fill-in-the-blanks has been filled as 2km ​​634m = 2634m.

Chapter 5 Exercise 5.9 Measures Workbook 4A Answers Page 169  Exercise 5.9  Problem 2

It is given that 5kg​​17g = ___ g

To fill –  In the blanks.

First, we will convert the quantities in terms of grams.

Using conversion, we know that 1kg=1000g

Hence, 5kg can be written as 5000g

Adding it along with 17g, we get the final result as

5000g + 17g = 5017g

The blank is filled as

5kg​​17g =  5017 g

The given fill-in-the-blanks has been filled as  5kg​​17g = 5017g.

Workbook 4A Chapter 5 Exercise 5.9 Measures Solutions Page 169 Exercise 5.9  Problem 3

It is given that 3h​​4min = ___ min

To fill – in the blanks.

First, we will convert the quantities in terms of minutes.

Using conversion, we know that 1h = 60min

Hence, 3h can be written as 3×60 = 180min

Adding it along with 4min, we get the final result as 184min

The blank is filled as

3h​​4min = 184 min

The given fill-in-the-blanks has been filled as  3h​​4min = 184 min.

Page 169  Exercise 5.9  Problem 4

It is given that 6m​​5cm = ____ cm

To fill –  In the blanks.

First, we will convert the quantities in terms of centimeters.

Using conversion, we know that 1m = 100cm

Hence, 6m can be written as 600cm

Adding it along with 5cm, we get the final result as 605cm

The blank is filled as 6m​​5cm = 605cm

The given fill-in-the-blanks has been filled as 6m​​ 5cm = 605cm

Measures Exercise 5.9 Primary Mathematics Workbook 4A Explanation Page 169  Exercise 5.9  Problem 5

It is given that 8ft​​ 7in = _____ in

To fill  – In the blanks.

First, we will convert the quantities in terms of inches.

Using conversion, we know that 1ft = 12in

Hence, 8ft can be written as

8 × 12 = 96in

Adding it along with 7in, we get the final result as 103in

The blank is filled as

8ft ​​7in = 103 in

The given fill-in-the-blanks has been filled as  8ft ​​7in = 103 in

Page 169  Exercise 5.9  Problem 6

It is given that 260min= ___ h ___ min.

To fill  –  In the blanks.

First, we will convert the quantities in terms of hours.

Using conversion, we know that 1h=60min

Hence, 260min can be written as \(\frac{260}{60}\)h.

Dividing it, we get

This means that it is 4h and 20min The blank is filled as

260 min = 4h 20 min

The given fill-in-the-blanks has been filled as  260min = 4h ​​20 min.

Common Core Workbook 4A Chapter 5 Measures Exercise 5.9 Help Page 169  Exercise 5.9  Problem 7

It is given that 4​\(\frac{1}{4}\) h =___ h___ min

To fill –  In the blanks.

We use concept of compound units.

Here, the whole number is 4, so the bigger unit is 4h.

The improper fraction is ​\(\frac{1}{2}\)

The unit conversion 1h = 60min is to be used.

Multiplying ​\(\frac{1}{2}\) and  60

​\(\frac{1}{2}\) × 60 = ​\(\frac{60}{2}\)

Cancelling out common factors, we get 30

So, we have got 30min.

The blank has been filled as

4​​ \(\frac{1}{2}\) h = 4h​​30min

The given blanks have been filled as 4​​\(\frac{1}{2}\) h = 4h​​30min

Measures Exercise 5.9 Workbook 4A Detailed Solutions Common Core Page 169  Exercise 5.9  Problem 8 

It is given that 2\(\frac{3}{4}\) ft = ____ ft ____in.

To fill  – In the blanks.We use concept of compound units.

Here, the whole number is 2, so the bigger unit is 2ft

The improper fraction is \(\frac{3}{4}\)

The unit conversion 1ft=12in is to be used.

Multiplying \(\frac{3}{4}\).

Multiplying \(\frac{3}{4}\) and 12

\(\frac{3}{4}\) × 12 = \(\frac{3×12}{4}\)

Cancelling out common factors, 9.

So, we have got 9 in.

The blank has been filled as 2 \(\frac{3}{4}\)ft = 2ft​​ 9in

The given blanks have been filled as \(\frac{3}{4}\)ft = 2ft​​ 9in

Measures Exercise 5.9 Workbook 4A Detailed Solutions Common Core Page 169  Exercise 5.9  Problem 9

It is given that 3\(\frac{9}{10}\)m=___ m___ cm

To fill  – In the blanks.

We use concept of compound units.

Here, the whole number is 3, so the bigger unit is 3m

The improper fraction is 3\(\frac{9}{10}\).

The unit conversion 1m=100cm is to be used.

Multiplying 3\(\frac{9}{10}\)and 100

3\(\frac{9}{10}\)×100 = \(\frac{9×100}{10}\)

Cancelling out common factors, 90.

So, we have got 90cm.

The blank has been filled as

3\(\frac{9}{10}\)10m = 3m ​​90cm

The given blanks have been filled as   3\(\frac{9}{10}\)m = 3m ​​90cm

Page 169  Exercise 5.9  Problem 10

The given blanks have been filled as 3\(\frac{9}{10}\)10m = 3m​​90cm

It is given that 5 \(\frac{3}{8}\)kg =___ g

To fill –  In the blanks.

Here, the whole number is 5 so the bigger unit is 5kg.

The unit conversion 1kg = 1000g is to be used.

​5kg = 5 × 1000g = 5000g
​
The improper fraction is \(\frac{3}{8}\)

Again using conversion

\(\frac{3}{8}\) × 1000 = \(\frac{l3×1000}{8}\)

Cancelling out common factors,3 × 125

So, we have got 375g.

Now adding the results,5000g + 375g = 5375g

The blank has been filled as  5\(\frac{3}{8}\)kg = 5375g

The given blank has been filled as 5 \(\frac{3}{8}\)kg = 5375g.

Primary Mathematics Workbook 4A Chapter 5 Measures Exercise 5.9 Problem Solving Page 169  Exercise 5.9  Problem 11

It is given that 10 \(\frac{2}{3}\) yd= ____ ft.

To fill  – In the blanks.

Here, the whole number is 10, so the bigger unit is 10yd

The unit conversion 1yd =3ft is to be used.

​10yd = 10 × 3ft

​10yd = 30ft

​The improper fraction is \(\frac{2}{3}\).

Again using conversion

\(\frac{2}{3}\) × 3

Cancelling out common factors, we have got

2ft

Now adding the results

30 + 2 = 32ft

The blank has been filled as

10\(\frac{2}{3}\)yd = 32ft

Page 170  Exercise 5.9  Problem 12

It is given to multiply 2 yd​​ 2ft × 5 = ___ yd ____ ft

First, we will multiply each unit separately.

Multiplying 2yd by 5, we get 10 yds.

Multiplying 2ft by 5, we get 10ft.

Using the conversion 1yd = 3ft, we convert 10ft into yards and feet.

Dividing

This means that it has 3yd and 1ft

So, adding it to the previous result of 10yd, we get 13yd

Therefore, we can note down the final results and fill in the blanks as 2yd​​ 2ft × 5 = 13yd​​1ft.

The multiplication has been carried out and the given fill-in-the-blanks has been filled as 2yd​​ 2ft × 5 = 13yd ​​1ft.

Page 170 Exercise 5.9 Problem 13

It is given to multiply 6gal​​ 3qt × 6 = ___ gal ____ qt.

First, we will multiply each unit separately.

Multiplying 6gal by 6, we get 36gal.

Multiplying 3qt by 6, we get 18qt.

Using the conversion 1gal = 4qt, we convert 18qt into gallons and quarts.

Dividing

This means that it has 4gal and 2qt.

So, adding it to previous result of 36gal, we get 40gal.

Therefore, we can note down the final results and fill in the blanks as 6gal​​3qt×6= 40 gal​​ 2qt.

The multiplication has been carried out and the given fill-in-the-blanks has been filled as 6gal​​ 3qt × 6 = 40 gal​​ 2qt

Page 170  Exercise 5.9  Problem 14

It is given to multiply 4m​​25cm × 7 = ___ m ____ cm.

First, we will multiply each unit separately.

Multiplying 4m by 7, we get 28m

Multiplying 25cm by 7, we get 175cm.

Using the conversion 1m = 100cm, we convert into meters and centimeters.

Dividing

\(\frac{175}{100}\) = 1.75

This means that it has 1m and 75cm.

So, adding it to previous result of 28m, we get 29m

Therefore, we can note down the final results and fill in the blanks as

4m ​​25cm × 25 = 29m ​​75cm

The multiplication has been carried out and the given fill-in-the-blanks has been filled as 4m​​25cm × 25 = 29m​​75cm.

Page 170  Exercise 5.9  Problem 15

It is given to multiply 6h ​​40min × 4 =___ h ____ min.

First, we will multiply each units separately.

Multiplying 6h by 4, we get 24h.

Multiplying 40min by 4, we get 160min.

Using the conversion 1h = 60min, we convert 160min into hours and minutes.

Dividing

This means that it has 2h and 40min.

So, adding it to the previous result of 24h, we get 26h

Therefore, we can note down the final results and fill in the blanks as  6h​​40min × 4 = 26 h​​ 40min

The multiplication has been carried out and the given fill-in-the-blanks has been filled as 6h ​​40min × 4 = 26h​​ 40min.

Page 170  Exercise 5.9  Problem 16

It is given that 5lb ​​12oz ÷ 4 = ____ lb ____ oz

To fill – The blanks with results obtained after division.

So we can write 5lb = 4lb + 1lb.

Now if we add 1lb to 12oz and use the conversion 1lb = 12oz, then we get it as 12oz + 12oz = 24oz

Dividing 4lb by 4, we get 1lb

Dividing 24oz by 4, we get 6oz.

Therefore, the given blanks can be filled as 5lb​​ 12oz ÷ 4 = 1lb​​6oz

The division has been carried out and the given fill-in-the-blanks has been filled as 5lb​​ 12o z ÷ 4 = 1lb​​ 6oz.

Page 170  Exercise 5.9  Problem 17

It is given that 13ft ​​4in ÷ 5 = ____ ft____ in

To fill – The blanks with results obtained after division.

So we can write 13ft = 10ft + 3ft.

Now if we add 3ft to 4in and use the conversion 1ft=12in then we get it as

3 × 12in + 4in = 36in + 4in = 40in
​
Dividing 10ft by 5, we get 2ft.

Dividing 40in by 5, we get 8in.

Therefore, the given blanks can be filled as

13ft ​​4in ÷ 5= 2 ft​​ 8 in

The division has been carried out and the given fill-in-the-blanks has been filled as 13ft ​​4in ÷ 5 = 2ft​​ 8in

Page 170 Exercise 5.9 Problem 18

It is given that 7kg​​800g ÷ 6 = ____ kg ____ g

To fill – The blanks with results obtained after division.

So we can write 7kg = 6kg + 1kg.

Now if we add 1kg to 800g use the conversion 1kg = 1000g, then we get 1000g + 800g = 1800g

Dividing 6kg by 6, we get 1kg.

Dividing 1800g by 6, we get 300g.

Therefore, the given blanks can be filled as 7kg800g ÷ 6 = 1 kg​​ 300 g

The division has been carried out and the given fill-in-the-blanks has been filled as 7kg800g ÷ 6 = 1kg​​300g.

Page 170  Exercise 5.9  Problem 19

It is given that 10min​​21sec ÷ 9 = ____ min ____ sec

To fill  – The blanks with results obtained after division.

So we can write 10min = 9min + 1min.

Now we add 1min to 21sec and use the conversion 1min = 60 sec , then we get

60sec + 21sec = 81sec

Dividing 9min by 9, we get 1min

Dividing 81sec by 9, we get 9sec.

Therefore, the given blanks can be filled as  10min ​​21sec ÷ 9 = 1 min​​ 9sec

The division has been carried out and the given fill-in-the-blanks has been filled as 10min​​ 21sec ÷ 9 = 1 min​​ sec.

Page 170 Exercise 5.9  Problem 20

It is given that \(\frac{4}{5}\) m____ 80cm.

To compare and fill the blank.

First, we will convert the quantities on both sides of the blank into a single unit, i.e. centimeters.

Then we will compare them and use the appropriate sign to fill the blank.

Let us convert \(\frac{4}{5}\) m into centimeters.

Using conversion, we get it to be equivalent to

\(\frac{4}{5}\) × 100cm

Cancelling common terms, we get

4 × 20cm =  80cm

Now, we have \(\frac{4}{5}\) m = 80cm

For the given quantities, the comparison has been done and we get \(\frac{4}{5}\) m= ​​​ 80cm.

Page 170 Exercise 5.9  Problem  21

It is given that 3\(\frac{5}{6}\) yr ______ 3yr​​ 5months

To compare and fill the blank.

First, we will convert the quantities on both sides of the blank into a single unit, i.e. months.

Then we will compare them and use the appropriate sign to fill the blank.

Let us convert 3\(\frac{5}{6}\)yr into months.

Here, the whole number is 3yr.

Using conversion 1yr = 12months, we get it to be equivalent to

3 × 12 = 36 months

The improper fraction is \(\frac{5}{6}\)yr.

Using conversion, we get it to be equivalent to
​
\(\frac{5}{6}\)×12months

= 10months
​
Combining these by adding

36 + 10 = 46months

So, we have got 3\(\frac{5}{6}\) yr=46months.

Now let us convert 3yr​​ 5months into months.

The unit 3yr can be converted into months as

3 × 12 = 36months

Now adding it with 5 months, we get

36 + 5 = 41months

Finally, we have got that 3yr​​ 5months = 43months.

Now, we have 46months ____ 43months

Since 46 is greater than 43, we can say that

46months > 43months

Finally, we can fill the blank as 35 6yr>​​ 3yr ​​5months

For the given quantities, the comparison has been done and we get 3\(\frac{5}{6}\)6yr>​​3yr​​5months.

Page 170  Exercise 5.9  Problem  22

It is given that 2\(\frac{1}{10}\)kg ______ 2100g.

To compare and fill the blank.

First, we will convert the quantities on both sides of the blank into a single unit, i.e. grams.

Then we will compare them and use the appropriate sign to fill the blank.

Let us convert 2\(\frac{1}{10}\)kg _ into grams.

Here, the whole number is 2kg.

Using conversion 1kg = 1000g, we get it to be equivalent to 2000g.

The improper fraction is 1\(\frac{1}{10}\) kg.

Using conversion, we get it to be equivalent to

​\(\frac{1}{10}\) × 1000g

= 100g
​
Combining these by adding

2000g + 100g = 2100g

So, we have got 2100g.

Now, we have

2100g _____ 2100g

Since both are the same, we can say that

2100g​​ = 2100g

Finally, we can fill the blank as 2​\(\frac{1}{10}\) kg ​​= ​2100g

For the given quantities, the comparison has been done and we get 2​\(\frac{1}{10}\) kg​​=​2100g

Page 170  Exercise 5.9  Problem  23

It is given that 4\(\frac{1}{2}\)L______ 4L​​50ml.

To compare and fill the blank.

First, we will convert the quantities on both sides of the blank into a single unit, i.e. milliliters.

Then we will compare them and use the appropriate sign to fill the blank.

Let us convert 4 \(\frac{1}{2}\)L into meters.

Here, the whole number is 4L.

Using conversion 1L = 1000ml, we get it to be equivalent to 4000ml.

The improper fraction is 1\(\frac{1}{2}\)L.

Using conversion, we get it to be equivalent to \(\frac{1}{2}\) × 1000ml  = 500ml.

Combining these by adding

4000ml + 500ml

So, we have got 4500ml.

Now let us convert 4L ​​50 ml into months.

The unit 4L can be converted into milliliters as 4000ml.

Now adding it with 50ml, we get

4000ml + 50ml = 4050ml

Finally, we have got that 4050ml.

Now, we have 4500ml _____ 4050ml

Since 4500is greater than 4050, we can say that

4500ml > 4050ml

Finally, we can fill the blank as 4\(\frac{1}{2}\)L​​>4L​​50ml

For the given quantities, the comparison has been done and we get 4\(\frac{1}{2}\)L​​>4L​​50ml

Page 171  Exercise 5.9  Problem 24

It is given that 9 months.

To express it as a fraction of 2 years.

First, we will convert the quantities in terms of months.

Using conversion, we know that 1year = 12months.

So, 2 years = 24 months.

Hence, the fraction can be written as  \(\frac{9}{24}\)

Expressing it in simplest form by cancelling common factors  \(\frac{3}{8}\)

Therefore, 9 months has been expressed as a fraction of 2 years as \(\frac{3}{8}\).

Page 171  Exercise 5.9  Problem 25

It is given that 50min.

To express it as a fraction of 3h.

First, we will convert the quantities in terms of minutes.

Using conversion, we know that 1h = 60min.

3h = 3 × 60min = 180min

Hence, the fraction can be written as  \(\frac{150}{180}\)

Expressing it in simplest form by cancelling common factor \(\frac{15}{18}\)

Therefore, 50min has been expressed as a fraction of 3h as \(\frac{15}{18}\).

Page 171  Exercise 5.9  Problem 26

It is given that 500ml.

To express it as a fraction of 2L.

First, we will convert the quantities in terms of milliliters.

Using conversion, we know that 1L=1000ml.

So, we get 2L= 2000ml.

Hence, the fraction can be written as   \(\frac{500}{2000}\)

Expressing it in simplest form by cancelling common factor   is \(\frac{1}{4}\)

Therefore, 500ml has been expressed as a fraction of 2L as \(\frac{1}{4}\)

Page 171  Exercise 5.9  Problem 27

It is given that 3lb.

To find – what fraction of 3lb is 8oz

First, we will express it in mathematical terms.

If x is the fraction, then

3lb × x = 8oz

Convert the quantities in terms of ounces.

Using conversion, we know that 1lb = 16oz

So, we get

3lb = 3 × 16oz = 48oz

Hence, the fraction can be written as

x = \(\frac{8}{48}\)

Expressing it in simplest form by cancelling common factors

x = \(\frac{1}{6}\)

Therefore, 8oz is \(\frac{1}{6}\) fraction of 3lb.

Page 171  Exercise  5.9  Problem 28

It is given that Chris sleeps 8 hours a day.

To find –  what fraction of a day he sleeps.

So, we have to find what fraction of 1 day is 8 hours.

We know that 1day=24hours.

The fraction can be written as  \(\frac{8}{24}\)

Simplifying and expressing in simplest form  \(\frac{1}{3}\)

Hence, Chris sleeps \(\frac{1}{3}\) of a day.

Therefore, it has been found that the fraction of a day that Chris sleeps is \(\frac{1}{3}\), when it is given that he sleeps 8 hours in a day.

Page 172  Exercise  5.9  Problem 29

It is given that a carton contains 3L​​250ml of fruit juice.

We have to find how much fruit juice we can get from 6 such cartons.

Then express as liters and milliliters.

We use multiplication operation when quantity of one carton is given and we require quantity of six cartons.

So, we multiply the given quantity by 6

3L × 6 = 18L

250ml × 6 = 1500ml

Now using conversion 1L = 1000ml, we convert

1500ml = \(\frac{1500}{1000}\) L

1500ml = 1.5L
​
This means that it has got 1L and 500ml

Now adding it with previously found value

18L+1L = 19L

Therefore, we can get 19L​​500ml from 6 cartons of fruit juices.

We can get 19L​​500ml from 6 cartons of fruit juices when it is given that we get 3l​​250ml from 1 carton of fruit juice.

Page 173  Exercise  5.9  Problem 30

It is given that Jamie bought 6 identical dictionaries as prizes for some party games and the total weight of the dictionaries was 7lb​​ 8oz

We have to find the weight of each dictionary and express the answer in pounds and ounces.

Since total weight is given and we have to find weight of each dictionary, division operation has to be used.

So, we start by writing

Weight of one dictionary \(=\frac{\text { Total Weight of dictionary }}{6}\)

Weight of one dictionary = 7lb​​ 8oz ÷ 6

We can express 7lb = 6lb + 1lb.

Using conversion 1lb = 16oz.

Adding it along with 8oz to get 24oz.

So, we get 6lb ​​24oz ÷ 6.

Dividing

Weight of one dictionary = 1lb​​ 4oz.

Hence, got the answer.

So, we got the weight of one dictionary as 1lb ​​4oz when we know that Jamie bought 6 identical dictionaries as prizes for some party games and the total weight of the dictionaries was 7lb​​8oz.

Page 173   Exercise 5.9   Problem 31

It is given that Lindsey bought 8 bottles of orange juice.

Each bottle contained 1L​​275ml of orange juice.

She filled two 2L jugs with orange juice and poured the remaining juice into a barrel.

We have to find the amount of juice present in the barrel and express it in liters and milliliters.

So, first, we find the total amount of orange juice bought by Lindsey by multiplying 1L​​275ml by 8.

Then we convert it into milliliters and also convert 2L to milliliters using conversion 1L = 1000ml.

Now find the amount of orange juice filled in the jugs.

Then deduct this amount from the total amount to get the amount in barrel.

Again convert it in terms of liters and milliliters.

The amount of juice in each bottle is 1L​​275ml.

The number of bottles is 8.

So, the total amount of juice is 1L​​ 275ml × 8.

Multiplying we get

1L × 8 = 8L

275ml × 8 = 2200ml

Converting 8L into milliliters

8L × 1000 = 8000ml

Now the total amount of juice bought by Lindsey is

8000 + 2200 = 10200ml

The amount of juice that can be poured into a jug is 2L.

There are two such jugs.

So, total amount of juice poured in jugs is 2L ×  2 = 4L.

Using conversion, we get it as 4000ml.

The amount of remaining juice that is poured into a barrel would be

The total amount of juice Lindsey bought − Total amount of juice poured into jugs

​⇒ 10200ml − 4000ml

⇒ 6200ml
​
Now we have to express it as liters and milliliters.

Using conversion 1L =1000ml

​6200ml = \(\)\frac{6200]{1000}[\latex] L

​6200ml  = 6.2L

This gives us 6L and 200ml.

Therefore, when Lindsey bought 8 bottles of orange juice, each containing 1L​​ 275ml of orange juice, and filled two 2L jugs with orange juice, the remaining juice poured into a barrel is 6L​​ 200ml.

Primary Mathematics Chapter 4 Operations On Fractions

Primary Mathematics Workbook 4A Common Core Solutions Chapter 4 Operations On Fractions 4.10 Page 132  Exercise 4.10  Problem 1

Given:  90− page book is 50 pages.

What fraction of a 90− page book is 50 pages.

The fraction of a 90− page book in 50

pages= \(\frac{50}{90}\) or \(\frac{5}{9}\)

The fraction of a 90− page book in 50 pages are  \(\frac{5}{9}\).

Page 132  Exercise 4.10  Problem 2

Given: Cameron has 40 toy cars.15 are battery operated.

What fraction of the toy cars are battery-operated

A fraction of toys are in battery

Operated = \(\frac{15}{40}\) or \(\frac{3}{8}\)

The fraction of toy cars in battery-operated is \(\frac{3}{8}\).

Page 132  Exercise 4.10  Problem 3

Given: Jim bought a packet of 60 stamps.24 were Canadian stamps.

What fraction of stamps were Canadian stamps.

The fraction of the stamps in Canadian Stamps is = \(\frac{24}{60}\) or \(\frac{2}{5}\).

The fraction of stamps in Canadian stamps are  \(\frac{2}{5}\).

Chapter 4 Exercise 4.10 Operations On Fractions Workbook 4A Answers Page 133  Exercise 4.10  Problem 4

Given: Stewart had $25. He spent$5

What fraction of his money did he have left.

First of all, Stewart has $25 & He spent $5, then.

The money he left = $20

The fraction of his money he has left  \(\frac{20}{25}\) or  \(\frac{4}{5}\).

 The fraction of the money he has left is  \(\frac{4}{5}\)

Page 133  Exercise 4.10  Problem 5

Given: There are 100 children at a carnival,60 were boys.

Express the number of girls as a fraction of the total number of children at the carnival.

First of all, total children are 100 & boys are 60 , then girls are (100−60 = 40)

The fraction of number of girls are = \(\frac{240}{100}\)  or  \(\frac{2}{5}\).

The fraction of number of girls of the total number of children at the carnival is  \(\frac{2}{5}\).

Workbook 4A Chapter 4 Exercise 4.10 Operations On Fractions Solutions Page 134  Exercise 4.10   Problem 6

Given: Sara bought 40m of material.

She made 6 curtains from the material.

She used 2m to make each curtain.

What fraction of the material did she use for the 6 curtains.

First of all, if each curtain make 2m, then6 curtain make = 2 × 6

So,6 curtains make = 12m

The fraction of material for 6 curtains is

The fraction of material used for 6 curtains is \(\frac{1}{3}\)m.

Page 134  Exercise 4.10  Problem 7

In the question, we are asked how many mangoes Travis sold.

Given that, he had 160 mangoes and each mango costs $2

Use the equation mentioned in the Tip section and find the solution.

Given:

Each mango costs = $2

Total number of mangoes = 160

The total cost that he gained $ 240

We have

Total Cost = Number of mangoes×Cost of one mango

Number of mangoes \( =\frac{\text { Total Cost }}{\text { Cost of one mango }}\)

= \(\frac{240}{2}\)

Number of mangoes  = 120

Travis sells 120 mangoes for the cost of $240.

Operations On Fractions Exercise 4.10 Primary Mathematics Workbook 4A Explanation Page 134  Exercise 4.10  Problem 8

In the question, we are asked to find the fraction of his mangoes did he have left.

We found that how many mangoes he sold from exercise 8a.

So we can find how many mangoes left and can convert into fraction.

Number of mangoes that Travis sold =120

Number of mangoes left = 160 − 120

Number of mangoes left = 40

Fraction of mangoes that he has left

= \(\frac{40}{160}\)

Fraction of mangoes that he has left = \(\frac{1}{4}\)

Fraction of his mangoes he has left = \(\frac{1}{4}\)

Page 135  Exercise 4.11  Problem 1

In the question, asked to color by given fraction from the set.

Find how many number of elements in the set.

Then find the fraction and color accordingly.

Given set contains 8 tortoises.

So, we have to find the given fraction of 8.

Given:  Fraction =\(\frac{1}{2}\)

To find – \(\frac{1}{2}\) of 8

8 × \(\frac{1}{2}\) = 4

Therefore, color 4 Tortoises.

Color 4 tortoises.

Common Core Workbook 4A Chapter 4 Operations On Fractions Exercise 4.10 Help Page 135  Exercise 4.11  Problem 2

In the question, asked to color by given fraction from the set.

Find how many number of elements in the set.

Then find the fraction and color accordingly.

Given set contains 9 sandwiches.

So, we have to find given fraction of 9.

Given:  Fraction = \(\frac{1}{3}\)

To find – \(\frac{1}{3}\) of  9

9 × \(\frac{1}{3}\)  =  3

Therefore, color 3 sandwiches.

Color 3 Sandwiches.

Fractions Exercise 4.10 Workbook 4A Detailed Solutions Common Core Page 135  Exercise 4.11  Problem 3

In the question, asked to color by a given fraction from the set.

Find how many number of elements in the set.

Then find the fraction and color accordingly.

The given set contains 15 socks.

So we have to find the given fraction of 15

Given:  Fraction = \(\frac{2}{5}\)

To find – \(\frac{2}{5}\) of 15

15 × \(\frac{2}{5}\)

Therefore, Color 6 socks.

Color 6 socks.

Primary Mathematics Workbook 4A Chapter 4 Fractions Exercise 4.10 Problem Solving Page 135 Exercise 4.11 Problem 4

In the question, asked to color by a given fraction from the set.

Find how many number of elements in the set.

Then find the fraction and color accordingly.

The given set contains 18 cakes.

We have to find a given fraction of 18.

Given: Fraction =\(\frac{5}{9}\)

To find – \(\frac{5}{9}\) of 18

18 × \(\frac{5}{9}\)= 10

Therefore, color 10 cakes.

Color 10 cakes.

Page 135  Exercise 4.11  Problem 5

In the question, we are asked to find the value of given problem.

Multiply the number by the fraction to get the solution.

We have to find

\(\frac{1}{2}\) of 12 = 12 × \(\frac{1}{2}\)

Simplify

\(\frac{1}{2}\) × 12 = 6

\(\frac{1}{2}\) of 12 =  6

Page 135  Exercise 4.11  Problem 6

In the question, we are asked to find the value of the given problem.

Multiply the number by the fraction to get the solution.

We have to find

\(\frac{1}{4}\) of 12 = 12 × \(\frac{1}{4}\)

Simplify

12 × \(\frac{1}{4}\) = 3

\(\frac{1}{4}\) of 12 = 3

Page 135  Exercise 4.11  Problem 7

In the question, we are asked to find the value of the given problem.

Multiply the number by the fraction to get the solution.

We have to find

\(\frac{1}{3}\) of 24 = 24 × \(\frac{1}{3}\)

Simplify

24 × \(\frac{1}{3}\)  =  8

\(\frac{1}{3}\) of 24 =  8

Page 135  Exercise 4.11  Problem 8

In the question, we are asked to find the value of given problem.

Multiply the number by the fraction to get the solution.

We have to find

\(\frac{1}{6}\) of 36 = 36× \(\frac{1}{3}\)

Simplify

36× \(\frac{1}{6}\) = 6

\(\frac{1}{6}\) of 36 = 6

Page 136  Exercise 4.11  Problem 9

In the question, we are asked to find the value of given problem.

Multiply the number by the fraction to get the solution.

We have to find

\(\frac{1}{4}\) of 20 = 20 × \(\frac{1}{4}\)

= 5

To find –

\(\frac{3}{4}\) of 20 = 20× \(\frac{3}{4}\)

= 15

\(\frac{1}{4}\) of 20 = 5 , \(\frac{3}{4}\) of 20 = 15

Page 136  Exercise 4.11  Problem 10

In the question, we are asked to find the value of given problem.

Multiply the number by the fraction to get the solution.

We have to find

\(\frac{1}{5}\) of 25 = 25× \(\frac{1}{5}\)= 5

To find – 

\(\frac{3}{5}\) of 25 = 25 × \(\frac{3}{5}\) = 15

\(\frac{1}{5}\) of 25 = 5 , \(\frac{3}{5}\) of 25 = 15

Page 136  Exercise 4.11  Problem 11

In the question, we are asked to find the value of given problem.

Multiply the number by the fraction to get the solution.

We have to find

\(\frac{1}{3}\) of 21 = 21× \(\frac{1}{3}\)= 7

To find – 

\(\frac{2}{3}\) of 21 = 21 × \(\frac{2}{3}\) = 14

\(\frac{1}{3}\) of 21 = 7 , \(\frac{2}{3}\) of 21 = 14

Page 136  Exercise 4.11  Problem 12

In the question, we are asked to find the value of given problem.

Multiply the number by the fraction to get the solution.

We have to find

\(\frac{1}{10}\) of 30 = 30× \(\frac{1}{10}\)= 3

To find – 

\(\frac{7}{10}\) of 30 = 30 × \(\frac{7}{10}\) = 21

\(\frac{1}{10}\) of 30 = 3 , \(\frac{7}{10}\) of 30 = 21

Page 137  Exercise 4.12  Problem 1

In the question, we are asked to find the value of given problem.

Using the rule mentioned in the Tip multiply the numbers with the fraction to get the solution.

We have to find

\(\frac{1}{4}\) of 8 = \(\frac{1}{4}\) × \(\frac{8}{1}\)

1. Multiply the numerators.

2. Multiply the denominators.

\(\frac{1}{4}\) of 8 = \(\frac{1×8}{4×1}\)

\(\frac{1}{4}\) of 8  = \(\frac{8}{4}\)

Simplify

\(\frac{8}{4}\) = 2 , \(\frac{1}{4}\) of 8 = 2

Page 137  Exercise 4.12  Problem 2

In the question, we are asked to find the value of given problem.

Using the rule mentioned in the Tip multiply the numbers with the fraction to get the solution.

We have to find

\(\frac{1}{5}\) of 15 =\(\frac{1}{5}\) × \(\frac{15}{1}\)

1. Multiply the numerators.

2. Multiply the denominators.

\(\frac{1}{5}\) of 15 = \(\frac{1× 15}{5× 1}\)

\(\frac{1}{5}\) of 15 = \(\frac{15}{5}\)

Simplifying

\(\frac{15}{5}\) = 3 , \(\frac{1}{5}\) of 15 = 3

Page 137  Exercise 4.12  Problem 3

In the question, we are asked to find the value of given problem.

Using the rule mentioned in the Tip multiply the numbers with the fraction to get the solution.

We have to find

\(\frac{1}{3}\) of 12 = \(\frac{1}{3}\) × \(\frac{12}{1}\)

1. Multiply the numerators.

2. Multiply the denominators.

\(\frac{1}{3}\) of 12 =\(\frac{1×12}{3× 1}\)

\(\frac{1}{3}\) of 12  = \(\frac{12}{3}\)

Simplify

\(\frac{12}{3}\) = 4 , \(\frac{1}{3}\) of 12 = 4

Page 137  Exercise 4.12  Problem 4

In the question, we are asked to find the value of given problem.

Using the rule mentioned in the Tip multiply the numbers with the fraction to get the solution.

We have to find

\(\frac{1}{3}\) of 6 =\(\frac{6}{1}\) × \(\frac{1}{3}\)

1. Multiply the numerators.

2. Multiply the denominators.

\(\frac{1}{3}\) of 6 = \(\frac{6}{3}\)

Simplifying

\(\frac{6}{3}\) = 2

\(\frac{1}{3}\) of 6 = 2

Page 137  Exercise 4.12  Problem 5

In the question, we are asked to find the value of given problem.

Using the rule mentioned in the Tip multiply the numbers with the fraction to get the solution.

We have to find 

\(\frac{2}{3}\) of 6= \(\frac{1}{3}\) of 6 =\(\frac{6}{1}\) × \(\frac{2}{3}\)

= \(\frac{12}{3}\)

1. Multiply the numerators.

2. Multiply the denominators.

Simplifying

\(\frac{12}{3}\) = 4

\(\frac{2}{3}\) of 6 = 4

Page 138  Exercise 4.12  Problem 6

In the question, we are asked to find the value of the given problem.

Using the rule mentioned in the Tip multiply the numbers with the fraction to get the solution

We have to find

\(\frac{1}{2}\) of 8= \(\frac{8}{1}\) × \(\frac{1}{2}\)

1. Multiply the numerators.

2. Multiply the denominators.

\(\frac{1}{2}\) of 8 = \(\frac{8}{2}\)

Simplifying

\(\frac{8}{2}\) = 4

\(\frac{1}{2}\) of 8 = 4

Page 138  Exercise 4.12  Problem 7

In the question, we are asked to find the value of the given problem.

Using the rule mentioned in the Tip multiply the numbers with the fraction to get the solution.

We have to find

\(\frac{1}{3}\) of 15 = \(\frac{1}{3}\) × \(\frac{15}{1}\)

1. Multiply the numerators.

2. Multiply the denominators.

\(\frac{1}{3}\) of 15= \(\frac{15}{3}\)

Simplifying

\(\frac{15}{3}\) = 5

\(\frac{1}{3}\) of 15 = 5

Page 138  Exercise 4.12  Problem 8

In the question, we are asked to find the value of the given problem.

Using the rule mentioned in the Tip multiply the numbers with the fraction to get the solution.

We have to find

\(\frac{1}{4}\) of 20 = \(\frac{20}{1}\) × \(\frac{1}{4}\)

1. Multiply the numerators.

2. Multiply the denominators.

\(\frac{1}{4}\) of 20= \(\frac{20}{4}\)

Simplifying

\(\frac{20}{4}\) = 5

\(\frac{20}{4}\) of 20 = 5

Page 138  Exercise 4.12  Problem 9

In the question, we are asked to find the value of given problem.

Using the rule mentioned in the Tip multiply the numbers with the fraction to get the solution.

We have to find

\(\frac{1}{6}\) of 18 = \(\frac{18}{1}\) × \(\frac{1}{6}\)

1. Multiply the numerators.

2. Multiply the denominators.

\(\frac{1}{6}\) of 18 = \(\frac{18}{6}\)

Simplifying

\(\frac{18}{6}\) =  3

\(\frac{1}{6}\) of 18 = 3

Page 139  Exercise 4.12  Problem 10

Given:  The expression

\(\frac{2}{3}\) of 15 = \(\frac{2}{3}\) × 15

To find –  The value of the given expression.

We multiply numerator with numerator and denominator with denominator and simplify.

First, we will convert 15 into a fraction using 1 as the denominator.

So, it can be written as ⇒ \(\frac{1}{15}\) .

Now, multiply the numerators and denominators.

\(\frac{2}{3}\) × \(\frac{15}{1}\)= \(\frac{2×15}{3×1}\)

Simplifying we get

\(\frac{30}{3}\) = 10

Therefore, the value of  \(\frac{2}{3}\) of 15 is 10.

Page 139  Exercise 4.12  Problem 11

Given:  The expression \(\frac{3}{4}\) of 20.

To find – The value of the given expression.

We multiply numerator with numerator and denominator with denominator and simplify.

We know that \(\frac{3}{4}\) of 20 can be written as \(\frac{3}{4}\) × 20.

First, we will convert 20 into a fraction using 1 as the denominator.

So, it can be written as ⇒  \(\frac{20}{1}\)

Now, multiply the numerators and denominators.

\(\frac{3}{4}\) × \(\frac{20}{1}\)= \(\frac{3×20}{4×1}\)

Simplifying we get

= 15

Therefore, the value of \(\frac{3}{4}\) of 20 is 15.

Page 139  Exercise 4.12  Problem 12

Given: The expression \(\frac{4}{5}\) of 30.

To find – The value of the given expression.

We multiply numerator with numerator and denominator with denominator and simplify.

We know that \(\frac{4}{5}\) of 20 can be written as \(\frac{4}{5}\)× 30.

First, we will convert 20 into a fraction using 1 as the denominator.

So, it can be written as ⇒  \(\frac{30}{1}\)

Now, multiply the numerators and denominators.

\(\frac{4}{5}\) × \(\frac{30}{1}\)

= \(\frac{4×30}{5×1}\)

Simplifying we get

= 24

Therefore, the value of \(\frac{4}{5}\) of 30 is 24.

Page 139  Exercise 4.12  Problem 13

Given: The expression \(\frac{5}{6}\) of 36.

To find –  The value of the given expression.

We multiply numerator with numerator and denominator with denominator and simplify.

We know that \(\frac{5}{6}\) of 36 can be written as  \(\frac{5}{6}\) × 36.

First, we will convert 20 into a fraction using 1 as the denominator.

So, it can be written as  ⇒ \(\frac{36}{1}\)

Now, multiply the numerators and denominators.

\(\frac{5}{6}\) × \(\frac{36}{1}\) = \(\frac{5×36}{6×1}\)

Simplifying we get

= 30

Therefore, the value of \(\frac{5}{6}\) of 36 is 30.

Page 140  Exercise 4.13  Problem 1

Given:  Manfred had 25 picture cards. He gave \(\frac{2}{5}\) of them to his friends.

To find –  The number of cards Manfred gave to his friends.

We find \(\frac{2}{5}\) of 25 cards.

We are given that Manfred gave \(\frac{2}{5}\) of them to his friends that implies

\(\frac{2}{5}\) of 25 =  \(\frac{2}{5}\)×25

First, we will convert 25 into a fraction using 1 as the denominator.

So, it can be written as ⇒  \(\frac{25}{1}\)

Now, multiply the numerators and denominators.

\(\frac{2}{5}\)×\(\frac{25}{1}\) = \(\frac{2×25}{5×1}\)

Simplifying we get

\(\frac{50}{5}\) = 10

Therefore, Manfred gave 10 cards to his friends.

Page 140  Exercise 4.13  Problem 2

Given:  Manfred had 25 picture cards. He gave \(\frac{2}{5}\) of them to his friends

To find –  The number of cards left with Manfred.

We find \(\frac{2}{5}\) of 25 cards and subtract it from the total cards.

We are given that Manfred gave \(\frac{2}{5}\) of them to his friends that implies

\(\frac{2}{5}\) of 25 = \(\frac{2}{5}\) × 25

First, we will convert 25 into a fraction using 1 as the denominator.

So, it can be written as ⇒  \(\frac{25}{1}\)

Now, multiply the numerators and denominators.

\(\frac{2}{5}\)×\(\frac{25}{1}\) =\(\frac{2×25}{5×1}\)

Simplifying we get

\(\frac{50}{5}\) = 10

A number of remaining cards can be calculated by subtracting the cards he gave to his friends from the total number of cards he had.

25−10 = 15

Therefore, the number of cards left with Manfred is 15.

Page 140  Exercise 4.13 Problem 3

Given:  Sharon had 40 dollars. She spent \(\frac{3}{8}\) of the money on a storybook.

To find – the cost of the storybook.

We find the money spent on the storybook which will be the cost of a storybook.

We are given that Sharon spent \(\frac{3}{8}\) of the money on a storybook that implies

\(\frac{3}{8}\) of 40 = \(\frac{3}{8}\)× 40

First, we will convert 40 into a fraction using 1 as the denominator.

So, it can be written as \(\frac{40}{1}\).

Now, multiply the numerators and denominators.

\(\frac{3}{8}\)×\(\frac{40}{1}\) = \(\frac{3×40}{8×1}\)

Simplifying we get

\(\frac{120}{8}\) = 15

Hence, the money spent on storybooks is 15.

Also, it is known that the cost of the storybook is equal to the expenditure on the storybook which is 15.

Therefore, the cost of the storybook is 15.

Page 140  Exercise 4.13  Problem 4

Given:  Sharon had 40 dollars.

She spent \(\frac{3}{8}\) of the money on a storybook.

To find –  The amount of money left with Sharon.

We find the money spent on the storybook and subtract it from the total cost.

We are given that Sharon spent \(\frac{3}{8}\) of the money on a storybook that implies

\(\frac{3}{8}\) of 40 = \(\frac{3}{8}\) × 40

First, we will convert 40 into a fraction using 1 as the denominator.

So, it can be written as \(\frac{40}{1}\).

Now, multiply the numerators and denominators.

\(\frac{3}{8}\)×\(\frac{40}{1}\) =\(\frac{3×40}{8×1}\)

Simplifying we get

\(\frac{120}{8}\) = 15

The remaining amount left with Sharon can be calculated by subtracting the amount spent on the storybook from the total money he had.

40−15 = 25

Therefore, 25 dollars are left with Sharon.

Page 141  Exercise 4.13  Problem 5

Given:  There are 24 potatoes in a bag and Lily peeled \(\frac{2}{3}\)

To find – The number of potatoes Lily peeled.

We find the peeled potatoes by calculating \(\frac{2}{3}\) of 24 potatoes.

We are given that Lily peeled \(\frac{2}{3}\) of total potatoes that implies

\(\frac{2}{3}\) of = \(\frac{2}{3}\) × 24

First, we will convert 24 into a fraction using 1 as the denominator.

So, it can be written as ⇒ \(\frac{42}{1}\)

Now, multiply the numerators and denominators.

\(\frac{2}{3}\)×\(\frac{24}{1}\) =\(\frac{2×24}{3×1}\)

Simplifying we get

\(\frac{48}{3}\) = 16

Hence, the number of potatoes Lily peeled is 16.

Therefore, the number of potatoes Lily peeled is 16.

Page 141  Exercise 4.13  Problem 6

Given:  Lindsay had 48 dollars. She spent \(\frac{3}{8}\) of the money.

To find – The amount of money she had spent.

We find the money spent by calculating \(\frac{3}{8}\)of 48 dollars.

We are given that Lindsay spent \(\frac{3}{8}\) of the money on a storybook that implies

\(\frac{3}{8}\) of 48 = \(\frac{3}{8}\) ×48

First, we will convert 48 into a fraction using 1 as the denominator.

So, it can be written as  ⇒ \(\frac{48}{1}\)

Now, multiply the numerators and denominators.

\(\frac{3}{8}\)×\(\frac{48}{1}\)=\(\frac{3×48}{8×1}\)

Simplifying we get

\(\frac{144}{3}\) = 18

Hence, the money spent is 18.

Therefore, the amount of money Lindsay had spent is 18 dollars.

Page 142  Exercise 4.14  Problem 1

Given:  Mandy had 25 dollars. She spent \(\frac{1}{5}\) of the money and saved the rest.

To find –  The amount of money she saved.

We find the money spent and subtract it from the total amount.

We are given that Mandy spent \(\frac{1}{5}\) of the money that implies

\(\frac{1}{5}\) of 25 = \(\frac{1}{5}\)× 25

First, we will convert 25 into a fraction using 1 as the denominator.

So, it can be written as ⇒  \(\frac{25}{1}\).

Now, multiply the numerators and denominators.

\(\frac{1}{5}\)×\(\frac{25}{1}\)=\(\frac{1×25}{5×1}\)

Simplifying we get

\(\frac{25}{5}\)=5

Hence, the money spent is 5 dollars.

The remaining amount saved with Mandy can be calculated by subtracting the amount spent from the total money she had.

25−5 = 20

Therefore, the amount of money Mandy saved is 20 dollars.

Page 142  Exercise 4.14  Problem 2

Given:  Matthew bought 45 oranges and used \(\frac{3}{5}\) of them to make orange juice.

To find –  The number of oranges left with him.

We will subtract the used oranges from the total oranges.

We are given that Matthew used \(\frac{3}{5}\) of the total oranges that implies

\(\frac{3}{5}\) of 45 = \(\frac{3}{5}\)×45

First, we will convert 45 into a fraction using 1 as the denominator.

So, it can be written as  ⇒ \(\frac{45}{1}\).

Now, multiply the numerators and denominators.

\(\frac{3}{5}\)×\(\frac{45}{1}\)=

Simplifying we get

\(\frac{135}{5}\) = 27

Hence, the number of oranges used to make orange juice is 27.

The remaining oranges left with Matthew can be calculated by subtracting the number of oranges used from the total oranges he had.

45 − 27 = 18

Therefore, the number of oranges left with Matthew is 18 oranges.

Page 143  Exercise 4.14  Problem 3

Given:  Nellie had 48 dollars and spent \(\frac{1}{4}\) of it on a calculator.

She also bought a book for 14 dollars.

To find  –  The amount of money she spent all together.

We add the amount spent on calculator and book.

We are given that Nellie spent \(\frac{1}{4}\)

of money she had on a calculator that implies

\(\frac{1}{4}\) of 45 = \(\frac{1}{4}\)×48

First, we will convert 48 into a fraction using 1 as the denominator.

So, it can be written as ⇒  \(\frac{48}{1}\)

Now, multiply the numerators and denominators.

\(\frac{1}{4}\)×\(\frac{48}{1}\)=\(\frac{1×48}{4×1}\)

Simplifying we get

\(\frac{48}{4}\)=12

Hence, the amount of money spent on a calculator is 12 dollars.

Also, we know that the amount spent on books is 14 dollars.

So, the total amount spent will be the amount spent on calculator + amount spent on book.

That is

12 + 14 = 26

Therefore, the amount of money she spent altogether is 26 dollars.

Page 144  Exercise 4.14  Problem 4

Given: There were 96 people on a board ship and \(\frac{1}{4}\)×\(\frac{48}{1}\)= of them were females.

To find – The number of males on the board ship.

We subtract the number of females from the total number of people.

We are given that \(\frac{1}{4}\) of total people were females. This implies that

\(\frac{1}{4}\) of = \(\frac{1}{4}\) × 96

First, we will convert 96 into a fraction using 1 as the denominator.

So, it can be written as  ⇒ \(\frac{96}{1}\)

Now, multiply the numerators and denominators.

\(\frac{1}{4}\)×\(\frac{96}{1}\)=\(\frac{1×96}{4×1}\)

Simplifying we get

\(\frac{96}{4}\) = 24

Hence, the number of females is 24.

The number of males can be calculated by subtracting the number of females from the total number of people on the board ship.

This implies, Number of males

​= 96−24

= 72
​

Page 145 Exercise 4.15 Problem 1

Given: Lynn spent \(\frac{7}{10}\) of the money she had.

She spent 42 dollars.

To find – The amount of money Lynn had at first.

We find the amount of money for 10 units.

We are given that Lynn spent \(\frac{7}{10}\) of the money she had which is 42 dollars.

This implies

7 units = 42 dollars

1 unit = 6 dollars

Now, the amount for 10 units will be the total amount.

⇒ 10 units = 6 × 10 dollars

⇒ 10 units = 60 dollars

Therefore, the amount of money Lynn had at first is 60 dollars.

Page 145  Exercise 4.15  Problem 2

Given:  Lynn spent \(\frac{7}{10}\) of the money she had.

She spent 42 dollars.

To find – The amount of money Lynn saved.

We subtract the amount of money spent from the total money she had at first.

We are given that Lynn spent \(\frac{7}{10}\) of the money she had which is 42 dollars.

This implies

7 units  = 42 dollars

1 unit = 6 dollars

Now, the amount for 10 units will be the total amount.

⇒ 10 units = 6 × 10 dollars

⇒ 10 units = 60 dollars

Hence, the amount of money Lynn had at first is 60 dollars.

We find the money she saved by subtracting money spent from the total amount of money.

Amount of money saved = 60 − 42 which is 18 dollars.

Therefore, the amount of money Lynn saved is 18 dollars.

Page 145  Exercise 4.15  Problem 3

Given:   A group of children went for a school picnic and \(\frac{3}{7}\) of them were boys.

There were 18 boys

To find –  The number of children presents there.

We find a number of children for 7 units.

We are given that \(\frac{3}{7}\) of the group of children were boys.

This implies

7 units =18 children

1 unit = \(\frac{18}{7}\) children

Now, the number of children for 7 units will be the total number.

⇒ 7 units = \(\frac{18}{7}\) × 7

⇒ 7 units = 18 children

Hence, the total number of children who went to the school picnic is 18 children.

Therefore, 18 children went to the school picnic.

Page 145  Exercise 4.15  Problem 4

Given:   A group of children went for a school picnic and \(\frac{3}{7}\) of them were boys. There were 18 boys.

To find – The number of girls present there.

We find the number of girls by calculating the number of children for 4 units.

We are given that \(\frac{3}{7}\) of the group of children were boys.

This implies

3 units = 18 boys

1 unit = 6 boys

Now, the number of girls will be calculated by finding the number of children for 7−3=4 units.

⇒ 4 units =  6 × 4 girls

⇒ 7 units =  24 girls

Hence, the number of girls who went for the school picnic is 24 girls.

Therefore, the number of girls who went for the school picnic is 24 girls.

Page 146  Exercise 4.15  Problem 5

Given: Susan spent \(\frac{3}{10}\) of her money on a bag.

To find – The amount of money she had at first if the bag costs 9 dollars.

We find the total amount of money she had by calculating the amount for 10 units.

We are given that Susan spent \(\frac{3}{10}\) of her money on a bag which is 9 dollars.

This implies

3 units = 9 dollars

1 unit = 3 dollars

Now, the total amount of money she had is for 10 units.

⇒ 10 units =  10 × 3 dollars

⇒ 10 units = 30 dollars

Therefore, the amount of money she had at first if the bag costs 9 dollars is 30 dollars.

Page 147 Exercise 4.15  Problem 6

Given: John bought some stamps. He used \(\frac{3}{5}\)of them to make mail letters and had 12 stamps left.

To find  – The number of stamps that he used.

We find the number of stamps he bought and then find \(\frac{3}{5}\) of them.

We are given that John used of the stamps and had 12 stamps left.

Remaining units are 5−3 = 2 units.

This implies

2units =12 stamps

1 unit =6 stamps

Now, the total number of stamps that she had is for 5 units.

⇒ 5 units  = 6 × 5 stamps

⇒  5 units = 30 stamps

We know that she used \(\frac{3}{5}\) of the total stamps.

Thus, the used number of stamps is \(\frac{3}{5}\) of 30 stamps.

Thus, used number of stamps is

⇒ \(\frac{3}{5}\)× 30 = \(\frac{3}{5}\) \(\frac{30}{1}\)

⇒ \(\frac{3×30}{5×1}\)= \(\frac{90}{5}\) 18 stamps.

Therefore, the number of stamps that he used is 18 stamps.

Primary Mathematics Chapter 4 Operations On Fractions

Primary Mathematics Workbook 4A Common Core Solutions Chapter 4 Operations On Fractions 4.1  Page 112  Exercise 4.1  Problem 1

We are asked to find the missing numbers

Given: \(\frac{1}{3}+\frac{1}{12}\)

As we can see that the denominators are not equal let us do LCM and then solve to get the result

\(\frac{1}{3}+\frac{1}{12}\) = \(\frac{4+1}{12}\)

\(\frac{1}{3}+\frac{1}{12}\) = \(\frac{5}{12}\)

So therefore, the missing fraction is \(\frac{5}{12}\)

Therefore, we can say that the missing fraction is \(\frac{5}{12}\)

Primary Mathematics Workbook 4A Common Core Solutions Chapter 4 Operations On Fractions 4.1 Page 112  Exercise 4.1  Problem 2

We are asked to find the missing numbers

Given: \(\frac{3}{8}+\frac{1}{2}\)

As we can see that the denominators are not equal let us do LCM and then solve to get the result.

\(\frac{3}{8}+\frac{1}{2}\) = \(\frac{3+4}{8}\)

\(\frac{3}{8}+\frac{1}{2}\) = \(\frac{7}{8}\)

So therefore, the missing fraction is \(\frac{7}{8}\).

Therefore, we can say that the missing fraction is \(\frac{7}{8}\).

Chapter 4 Exercise 4.1 Operations On Fractions Workbook 4A Answers Page 112  Exercise 4.1  Problem 3

We are asked to find the missing numbers

Given: \(\frac{2}{5}+\frac{3}{10}\)

As we can see that the denominators are not equal let us do LCM and then solve to get the result.

\(\frac{2}{5}+\frac{3}{10}\) = \(\frac{4+3}{8}\)

\(\frac{2}{5}+\frac{3}{10}\) = \(\frac{7}{10}\)

So therefore, the missing fraction is \(\frac{7}{10}\)

Therefore, we can say that the missing fraction is \(\frac{7}{10}\)

Workbook 4A Chapter 4 Exercise 4.1 Operations On Fractions Solutions Page 113  Exercise 4.1  Problem 4

We are asked to write the answers in the simplest form.

So, by solving we get

\(\frac{1}{2}+\frac{1}{4}\) = \(\frac{3}{4}\)

\(\frac{1}{6}+\frac{2}{3}\) = \(\frac{1+4}{3}\)=\(\frac{5}{3}\)

\(\frac{2}{9}+\frac{2}{3}\) = \(\frac{2+6}{9}\)=\(\frac{9}{3}\)= 3

\(\frac{1}{9}+\frac{2}{3}\) = \(\frac{1+6}{3}\)=\(\frac{7}{3}\)

\(\frac{1}{5}+\frac{1}{10}\) = \(\frac{2+1}{10}\)=\(\frac{3}{10}\)

\(\frac{3}{10}+\frac{1}{5}\) = \(\frac{3+2}{10}\)=\(\frac{5}{10}\)= \(\frac{1}{2}\)

\(\frac{1}{8}+\frac{3}{4}\) = \(\frac{1+6}{8}\)=\(\frac{7}{8}\)

\(\frac{3}{8}+\frac{1}{4}\) = \(\frac{3+2}{4}\)=\(\frac{5}{4}\)

\(\frac{1}{8}+\frac{1}{4}\) = \(\frac{1+2}{8}\)=\(\frac{3}{8}\)

\(\frac{1}{4}+\frac{1}{12}\) = \(\frac{3+1}{12}\)=\(\frac{4}{12}\)= \(\frac{1}{3}\)

Therefore, we can say that the simplified fractions are \(\frac{3}{4}\),\(\frac{5}{3}\),3, \(\frac{7}{3}\),\(\frac{3}{10}\),\(\frac{1}{2}\),\(\frac{7}{8}\),\(\frac{5}{4}\) ,\(\frac{3}{8}\),\(\frac{1}{3}\)

Operations On Fractions Exercise 4.1 Primary Mathematics Workbook 4A Explanation Page 115  Exercise 4.2  Problem 1

We are asked to find the missing numbers.

Given: \(\frac{3}{4}- \frac{1}{2}\)

As we can see that the denominators are not the same.

Let us multiply the second fraction by 2 for the numerator and denominator, we get

⇒ \(\frac{3}{4}− \frac{1}{4}\)

= \(\frac{3−2}{4}\)

= \(\frac{1}{4}\)

Therefore, by solving the given fractions we get \(\frac{1}{4}\)

Therefore, we can say that by solving the given fractions we get \(\frac{1}{4}\)

Page 115 Exercise 4.2 Problem 2

We are asked to find the missing numbers

Given: \(\frac{5}{6}- \frac{2}{3}\)

As we can see the denominators are not the same.

Let us multiply the second fraction by 2 for the numerator and denominator, we get

⇒ \(\frac{5}{6}- \frac{4}{6}\)

= \(\frac{5-4}{6}\)

= \(\frac{1}{6}\)

Therefore, by solving the given fractions we get \(\frac{1}{6}\)

Therefore, we can say that by solving the given fractions we get \(\frac{1}{6}\)

Page 115  Exercise 4.2  Problem 3

We are asked to find the missing numbers.

Given: \(\frac{2}{3}- \frac{1}{12}\)

As we can see the denominators are not the same

Let us multiply the first fraction by 4 for the numerator and denominator, we get

⇒ \(\frac{8}{12}- \frac{1}{12}\)

= \(\frac{8- 1}{12}\)

​= \(\frac{7}{12}\)

Therefore, by solving the given fractions we get \(\frac{7}{12}\)

Therefore, we can say that by solving the given fractions we get \(\frac{7}{12}\)

Common Core Workbook 4A Chapter 4 Operations On Fractions Exercise 4.1 Help Page 116  Exercise 4.2  Problem 4

Given: Some subtraction to compute.

To calculate the given subtractions and make an appropriate word with the answers connecting with the letters, where the word should represent a four-sided figure.

Recall that, subtraction is the process of removing a number from a larger number.

Compute each subtraction row-wise and match the letters with the numbers given in the box.

Follow the steps given below.

Perform the subtractions in the first row.

The LCM of the number 2,6 is 6.

Therefore

⇒ \(\frac{1}{2}- \frac{1}{6}\)

=\(\frac{(6÷2)×1−(6÷6)×1}{6}\)

=\(\frac{3−1}{6}\)

=\(\frac{2}{6}\)

=\(\frac{1}{3}\) [Reduce the fraction]

Similarly, the LCM of the number 4,8 is 8.

Therefore

⇒ \(\frac{3}{4}- \frac{5}{8}\)

=\(\frac{3×2−1×5}{8}\)

= \(\frac{6−5}{8}\)

=\(\frac{1}{8}\).

Also, the LCM of the numbers 3,9 is 9.

Therefore \(\frac{2}{3}- \frac{2}{9}\) = \(\frac{2×3−2×1}{9}\)

= \(\frac{6−2}{9}\)

= \(\frac{4}{9}\).

Perform the subtractions in the second row.

The LCM of the numbers 4,12 is 12.

Therefore \(\frac{3}{4}- \frac{1}{12}\) = \(\frac{3×3−1×1}{12}\)

=\(\frac{9−1}{12}\)

= \(\frac{8}{12}\).

= \(\frac{2}{3}\).

Similarly, the LCM of the numbers 5,10 is 10.

⇒ \(\frac{2}{5}- \frac{1}{10}\) = \(\frac{2×2−1×1}{10}\)

=\(\frac{4−1}{10}\)

= \(\frac{3}{10}\).

Also, the LCM of the numbers 6,12 is 12.

Then

⇒ \(\frac{5}{6}- \frac{5}{12}\) = \(\frac{5×2−5×1}{12}\)

= \(\frac{10−5}{12}\)

= \(\frac{5}{12}\).

Perform the subtractions in the Third row.

The LCM of the numbers 5,10 is 10.

Therefore

⇒ \(\frac{4}{5}- \frac{3}{10}\) = \(\frac{4×2−3×1}{10}\)= \(\frac{8−1}{10}\)

= \(\frac{5}{10}\)

= \(\frac{1}{2}\). [Reduction the fraction]

Similarly, the LCM of the numbers 2,12 is 12.

\(\frac{1}{2}- \frac{5}{12}\) = \(\frac{1×6−5×1}{12}\)

= \(\frac{6−5}{12}\)

= \(\frac{1}{12}\).

Also, the LCM of the number 12,3 is 12.

Then

⇒ \(\frac{7}{2}- \frac{1}{3}\) = \(\frac{7−4}{12}\)

= \(\frac{3}{12}\)

= \(\frac{1}{4}\).

Match the answers with the letters to form a word that represents a four-sided figure.

The answers to the subtractions are given by

Thus, the letters that match the answers are given by

The letters that match the answers and the word that represents a four-sided figure are given by

Fractions Exercise 4.1 Workbook 4A Detailed Solutions Common Core Page 117  Exercise 4.3  Problem 1

We are given that Meredith brought a piece of cloth.

She used \(\frac{3}{8}\) of it to make a dress.

We are asked to find the remaining clothes that she has.

Let us consider the remaining cloth as x.

So, from the question

1 = x + \(\frac{3}{8}\)

Transferring \(\frac{3}{8}\) from RHS to LHS.
​
​x = 1 − \(\frac{3}{8}\)

x = \(\frac{5}{8}\)

Therefore, we can say that \(\frac{5}{8}\) the part of the cloth is left.

Therefore, we can say that Meredith has  \(\frac{5}{8}\) the part of cloth left with her.

Page 117  Exercise 4.3  Problem 2

Given: John spent \(\frac{1}{2}\) of his money on a toy car. He spent 1

⇒ \(\frac{1}{6}\) of his money on a pen.

To find the fraction of the money he spends altogether.

Here, add the fraction of the money he spends on a toy car and pen.

⇒ \(\frac{1}{2}\)+\(\frac{1}{6}\)

Multiply and divide the first fraction by 3, to get a common denominator.

⇒ \(\frac{3}{6}\)+\(\frac{1}{6}\)

Simplifying

⇒ \(\frac{3+1}{6}\)

⇒ \(\frac{4}{6}\)

⇒ \(\frac{2}{3}\)

Therefore, the fraction of money John spends altogether on his car is \(\frac{2}{3}\)

Primary Mathematics Workbook 4A Chapter 4 Fractions Exercise 4.1 Problem Solving Page 118  Exercise 4.3  Problem 3

Given: Mary drank \(\frac{7}{10}\) L of orange juice.

Jim drank  \(\frac{1}{5}\) L of orange juice less than Mary.

To find  – The fraction of orange juice they drank altogether.

Here, add the fraction of orange juice drank by Mary and Jim.

Jim drank \(\frac{1}{5}\) L of orange juice less than Mary.

Therefore, orange juice drank by Jim;
​
​⇒ \(\frac{7}{10}\) −\(\frac{1}{5}\)

⇒ \(\frac{7−2}{10}\)

⇒ \(\frac{5}{10}\)

⇒ \(\frac{1}{2}\)

​Add the fraction of orange juice drank by Mary and Jim
​
​⇒ \(\frac{7}{10}\)+\(\frac{1}{2}\)

⇒ \(\frac{7+5}{10}\)

⇒ \(\frac{12}{10}\)= \(\frac{6}{5}\)

Therefore, the fraction of orange juice Mary and Jim drank altogether is \(\frac{6}{5}\) L.

Page 118  Exercise 4.3  Problem 4

Given: Lily bought 1 yd of ribbon. She used \(\frac{1}{2}\) yd to tie a package and \(\frac{3}{10}\) to make a bow.

To find – How much ribbon she had left.

Here, add the fraction of ribbon used.

Then subtract it from the total ribbon brought.

She used \(\frac{1}{2}\)yd to tie a package and \(\frac{3}{10}\) to make a bow.

Therefore, the total ribbon used

​⇒ \(\frac{1}{2}\)+\(\frac{3}{10}\)

​⇒ \(\frac{5+3}{10}\)

⇒ \(\frac{8}{10}\)

​⇒ \(\frac{4}{5}\)

Subtract the fraction of ribbon used from the total ribbon brought

​⇒ 1 −  \(\frac{4}{5}\)

⇒ \(\frac{1}{5}\)

Therefore, the length of ribbon Lily left with is \(\frac{1}{5}\) yd.

Page 119  Exercise 4.4  Problem 1

Given: 1\(\frac{1}{12}\)+ 3\(\frac{1}{3}\)

To find  – The simplest form.

To add mixed numbers, we first add the whole numbers together, and then the fractions.

Adding the given mixed fraction:

⇒ \(\frac{1}{12}\) + 3\(\frac{1}{3}\)

Add the whole numbers together, and then the fractions

⇒  4\(\frac{1}{12}\)+ \(\frac{1}{3}\)

Multiple and divide the second fraction by 4
​
⇒ 4\(\frac{1}{12}\)+ \(\frac{1×4}{3×4}\)

⇒ 4\(\frac{1}{12}\)+\(\frac{4}{12}\)

⇒ 4\(\frac{5}{12}\)

The simplest form of the given expression 1\(\frac{1}{12}\)+ 3\(\frac{1}{3}\)= 4\(\frac{5}{12}\)

Page 119  Exercise 4.4  Problem 2

Given: 1\(\frac{3}{4}\)+ 1\(\frac{1}{8}\)

To find  – The simplest form.

To add mixed numbers, we first add the whole numbers together, and then the fractions.

Adding the given mixed fraction:

1\(\frac{3}{4}\)+ 1\(\frac{1}{8}\)

Add the whole numbers together, and then the fractions

⇒  2 \(\frac{3×2}{4×2}\)+ \(\frac{1}{8}\)

⇒  2 \(\frac{6}{8}\)+ \(\frac{1}{8}\)

⇒  2 \(\frac{7}{8}\)

The simplest  1\(\frac{3}{4}\) + 1\(\frac{1}{8}\) =  2 \(\frac{7}{8}\)

Page 119  Exercise 4.4  Problem 3

Given: 2\(\frac{3}{10}\)+ 2\(\frac{2}{5}\)

To find – The simplest form.

To add mixed numbers, we first add the whole numbers together, and then the fractions.

Adding the given mixed fraction:

⇒ 2\(\frac{3}{10}\)+ 2\(\frac{2}{5}\)

Add the whole numbers together, and then the fractions

⇒ 4\(\frac{3}{10}\) + \(\frac{2}{5}\)

Multiple and divide the second fraction by 2

⇒  4\(\frac{3}{10}\)+\(\frac{2×2}{5×2}\)

⇒ 4\(\frac{3}{10}\)+\(\frac{4}{10}\)

⇒  4\(\frac{7}{10}\)

The simplest form of the given expression  2\(\frac{3}{10}\)+ 2\(\frac{2}{5}\) = 4\(\frac{7}{10}\)

Page 119  Exercise 4.4  Problem 4

Given: 1\(\frac{2}{3}\)+ 5\(\frac{2}{15}\)

To find –  The simplest form.

To add mixed numbers, we first add the whole numbers together, and then the fractions.

Adding the given mixed fraction:

⇒ 1\(\frac{2}{3}\)+ 5\(\frac{2}{15}\)

Add the whole numbers together, and then the fractions

⇒ \(\frac{2}{3}\)+ 6\(\frac{2}{15}\)

Multiple and divide the first fraction by 5

⇒ \(\frac{2×5}{3×5}\) + 6\(\frac{2}{15}\)

⇒\(\frac{10}{15}\) +6\(\frac{2}{15}\)

⇒ 6\(\frac{12}{15}\)

⇒ 6\(\frac{4}{5}\)

The simplest form of the given expression 1\(\frac{2}{3}\)+ 5\(\frac{2}{15}\) = 6\(\frac{4}{5}\)

Page 120 Exercise 4.4 Problem  5

Given: 2\(\frac{1}{6}\) + 1\(\frac{2}{3}\)

To find –  The simplest form

To add mixed numbers, we first add the whole numbers together, and then the fractions.

Adding the given mixed fraction:

⇒ 2\(\frac{1}{6}\)+ 1\(\frac{2}{3}\)

Add the whole numbers together, and then the fractions

⇒ 3 \(\frac{1}{6}\)+ \(\frac{2}{3}\)

Multiple and divide the second fraction by 2
​
​⇒ 3 \(\frac{1}{6}\)+ \(\frac{2(2)}{3(2)}\)

⇒ 3 \(\frac{1}{6}\) + \(\frac{4}{6}\)

⇒ 3 \(\frac{5}{6}\)

⇒ 3 \(\frac{5}{6}\)

The simplest form of the given expression  2\(\frac{1}{6}\)+ 1\(\frac{2}{3}\)= 3 \(\frac{5}{6}\)\

Page 120  Exercise 4.4  Problem  6

Given: 2\(\frac{3}{8}\)+ 2\(\frac{3}{4}\)

To find  – The simplest form.

To add mixed numbers, we first add the whole numbers together, and then the fractions.

Adding the given mixed fraction:

⇒ 2\(\frac{3}{8}\)+ 2\(\frac{3}{4}\)

Add the whole numbers together, and then the fractions

⇒ 4\(\frac{3}{8}\)+ \(\frac{3}{4}\)

Multiple and divide the second fraction by 2

​⇒ 4\(\frac{3}{8}\)+\(\frac{3(2)}{4(2)}\)

⇒ 4\(\frac{3}{8}\)+ \(\frac{6}{8}\)

⇒ 4\(\frac{9}{8}\)

The simplest form of the given expression  2\(\frac{3}{8}\)+ 2\(\frac{3}{4}\)= 4\(\frac{9}{8}\)

Page 120  Exercise 4.4  Problem  7

Given:

3\(\frac{1}{3}\)+ 2\(\frac{7}{9}\)

To find – The simplest form.

To add mixed numbers, we first add the whole numbers together, and then the fractions.

​Adding the given mixed fraction:

⇒ 3\(\frac{1}{3}\)+ 2\(\frac{7}{9}\)

Add the whole numbers together, and then the fractions

⇒ 5\(\frac{1}{3}\)+\(\frac{7}{9}\)

Multiple and divide the first fraction by 3
​
​⇒ 5 \(\frac{1(3)}{3(3)}\)+ \(\frac{7}{9}\)

⇒ 5 \(\frac{3}{9}\)+\(\frac{7}{9}\)

⇒ 5\(\frac{10}{9}\)

​The simplest form of the given expression  3\(\frac{1}{3}\)+ 2\(\frac{7}{9}\)= 5\(\frac{10}{9}\)

Page 120  Exercise 4.4  Problem  8

Given:

2\(\frac{3}{12}\)+ 2 \(\frac{1}{6}\)

To find  – The simplest form.

To add mixed numbers, we first add the whole numbers together, and then the fractions.

Adding the given mixed fraction:

⇒ 2\(\frac{3}{12}\)+ 2 \(\frac{1}{6}\)

Add the whole numbers together, and then the fractions

⇒ 4\(\frac{3}{12}\)+ \(\frac{1}{6}\)

Multiple and divide the second fraction by 2

⇒ 4\(\frac{3}{12}\)+ \(\frac{1(2)}{6(2)}\)

⇒ 4\(\frac{3}{12}\)+ \(\frac{2}{12}\)

⇒ 4\(\frac{5}{12}\)

The simplest form of the given expression  2\(\frac{3}{12}\)+ 2 \(\frac{1}{6}\)= 4\(\frac{5}{12}\)

Page 121  Exercise 4.5 Problem 1

Given: 4\(\frac{7}{8}\)−1\(\frac{1}{2}\)

To find – The simplest form.

To add mixed numbers, we first add the whole numbers together, and then the fractions

Adding the given mixed fraction:

⇒ 4\(\frac{7}{8}\)−1\(\frac{1}{2}\)

Add the whole numbers together, and then the fractions

⇒ 3\(\frac{7}{8}\)−\(\frac{1}{2}\)

Multiple and divide the second fraction by 4

⇒ 3\(\frac{7}{8}\)− \(\frac{1(4)}{2(4)}\)

⇒ 3\(\frac{7}{8}\)− \(\frac{4}{8}\)
​
⇒ 3\(\frac{3}{8}\)

The simplest form of the given equation  4\(\frac{7}{8}\)−1\(\frac{1}{2}\) = 3\(\frac{3}{8}\)

Page 121  Exercise 4.5  Problem 2

Given: 7\(\frac{4}{5}\)−3\(\frac{1}{10}\)

To find –  The simplest form.

To add mixed numbers, we first add the whole numbers together, and then the fractions.

Adding the given mixed fraction:

⇒ 7\(\frac{4}{5}\)−3\(\frac{1}{10}\)

Add the whole numbers together, and then the fractions

⇒ 4\(\frac{4}{5}\)−\(\frac{1}{10}\)

Multiple and divide the first fraction by 2

⇒ 4\(\frac{4(2)}{5(2)}\)−\(\frac{1}{10}\)

⇒ 4\(\frac{8}{10}\)−\(\frac{1}{10}\)

⇒ 4\(\frac{7}{10}\)

The simplest form of the given equation  7\(\frac{4}{5}\)−3\(\frac{1}{10}\) = 4\(\frac{8}{10}\)

Page 121  Exercise 4.5  Problem 3

Given: 2\(\frac{5}{12}\)−1\(\frac{1}{4}\)

To find –  The simplest form.

To add mixed numbers, we first add the whole numbers together, and then the fractions.

Adding the given mixed fraction:

⇒ 2\(\frac{5}{12}\)−1\(\frac{1}{4}\)

Add the whole numbers together, and then the fractions

⇒ 1\(\frac{5}{12}\)−\(\frac{1}{4}\)

Multiple and divide the second fraction by 3

​⇒ 1\(\frac{5}{12}\)−\(\frac{1(3)}{4(3)}\)

⇒ 1\(\frac{5}{12}\)−\(\frac{3}{12}\)
​
⇒ 1\(\frac{2}{12}\)= 1\(\frac{1}{6}\)

The simplest form of the given equation  2\(\frac{5}{12}\)−1\(\frac{1}{4}\)= 1\(\frac{1}{6}\)

Page 121   Exercise 4.5 Problem 4

Given:

3\(\frac{2}{3}\)−2\(\frac{1}{9}\)

To find – The simplest form.

To add mixed numbers, we first add the whole numbers together, and then the fractions.

Adding the given mixed fraction:

⇒3\(\frac{2}{3}\)−2\(\frac{1}{9}\)

Add the whole numbers together, and then the fractions

⇒ 1\(\frac{2}{3}\)−\(\frac{1}{9}\)

Multiple and divide the first fraction by 3
​
​⇒ 1\(\frac{2(3)}{3(3)}\)−\(\frac{1}{9}\)

​​⇒ 1\(\frac{6}{9}\)−\(\frac{1}{9}\)

​⇒ 1\(\frac{5}{9}\)

The simplest form of the given equation  3\(\frac{2}{3}\)−2\(\frac{1}{9}\)= 1\(\frac{5}{9}\)

Page 122  Exercise 4.5  Problem 5

Given:

4\(\frac{1}{3}\)−1\(\frac{2}{9}\)

To find – The simplest form.

To add mixed numbers, we first add the whole numbers together, and then the fractions.

Adding the given mixed fraction:

⇒ 4\(\frac{1}{3}\)−1\(\frac{2}{9}\)

Add the whole numbers together, and then the fractions

⇒ 3\(\frac{1}{3}\)−\(\frac{2}{9}\)

Multiple and divide the first fraction by 3

⇒ 3\(\frac{1(3)}{3(3)}\)−\(\frac{2}{9}\)

⇒ 3\(\frac{3}{9}\)−\(\frac{2}{9}\)

⇒ 3\(\frac{3}{9}\)

The simplest form of the given equation   4\(\frac{1}{3}\)−1\(\frac{2}{9}\)= 3\(\frac{3}{9}\)

Page 121  Exercise 4.5  Problem 6

Given:

3\(\frac{3}{4}\)−\(\frac{1}{12}\)

To find  – The simplest form.

To add mixed numbers, we first add the whole numbers together, and then the fractions.

Adding the given mixed fraction:

⇒  3\(\frac{3}{4}\)−\(\frac{1}{12}\)

Add the whole numbers together, and then the fractions

⇒  3\(\frac{3}{4}\)−\(\frac{1}{12}\)

Multiple and divide the first fraction by 3

⇒ 3\(\frac{3(3)}{4(3)}\)−\(\frac{1}{12}\)

⇒ 3\(\frac{9}{12}\)−\(\frac{1}{12}\)

⇒ 3\(\frac{8}{12}\)= 3\(\frac{2}{3}\)

The simplest form of the given equation  3\(\frac{3}{4}\)−\(\frac{1}{12}\)= 3\(\frac{2}{3}\)

Page 121  Exercise 4.5  Problem 7

Given:

3\(\frac{5}{9}\)−1\(\frac{1}{3}\)

To find – The simplest form.

To add mixed numbers, we first add the whole numbers together, and then the fractions.

Adding the given mixed fraction:

⇒3\(\frac{5}{9}\)−1\(\frac{1}{3}\)

Add the whole numbers together, and then the fractions

⇒ 2\(\frac{5}{9}\)−\(\frac{1}{3}\)

Multiple and divide the second fraction by 3
​
⇒ 2\(\frac{5}{9}\)−\(\frac{1(3)}{3(3)}\)

⇒ 2\(\frac{5}{9}\)−\(\frac{3}{9}\)

⇒ 2\(\frac{2}{9}\)

The simplest form of the given equation  3\(\frac{5}{9}\)−1\(\frac{1}{3}\)= 2\(\frac{2}{9}\)

Page 121  Exercise 4.5  Problem 8

Given:

4\(\frac{7}{8}\)−2\(\frac{5}{16}\)

To find –  The simplest form.

To add mixed numbers, we first add the whole numbers together, and then the fractions.

Adding the given mixed fraction:

​​⇒ 4\(\frac{7}{8}\)−2\(\frac{5}{16}\)

Add the whole numbers together, and then the fractions

⇒ 2\(\frac{7}{8}\)−\(\frac{5}{16}\)

Multiple and divide the first fraction by 2

​​⇒  2\(\frac{7(2)}{8(2)}\)−\(\frac{5}{16}\)

​⇒  2\(\frac{14}{16}\)−\(\frac{5}{16}\)

​⇒  2\(\frac{9}{16}\)

The simplest form of the given equation   4\(\frac{7}{8}\)−2\(\frac{5}{16}\)= 2\(\frac{9}{16}\)

Page 123  Exercise 4.6  Problem 1

Given: John used 3\(\frac{5}{6}\)m of wire to create flower pot hangers, leaving him with   1\(\frac{7}{12}\) m. J

To find – The length of wire he had at first.

Here, add the length of wire used to make the flower pot and the length of the wire

Adding the length of wire used to make the flower pot and the length of the wire left.

⇒ 3\(\frac{5}{6}\)+1\(\frac{7}{12}\) m. J

Add the whole numbers together, and then the fractions

⇒ 4\(\frac{5}{6}\)+\(\frac{7}{12}\)

Multiple and divide the first fraction by 2

​⇒ 4\(\frac{5(2)}{6(2)}\)+\(\frac{7}{12}\)
​
⇒ 4\(\frac{10}{12}\)+\(\frac{7}{12}\)
​
⇒ 4\(\frac{3}{12}\)= 4\(\frac{1}{4}\)

The length of wire he had first is 4\(\frac{1}{4}\).

Page 123  Exercise 4.6  Problem 2

Given: The capacity of a container is 4\(\frac{4}{5}\) gal.

It contains  1\(\frac{3}{10}\) gal of water.

To find the water needed to fill the container

Here, subtract the amount of water in the container from the capacity of the container.

Subtract the amount of water in the container from the capacity of the container.

⇒ 4\(\frac{4}{5}\) − 1\(\frac{3}{10}\)

Add the whole numbers together, and then the fractions

⇒ 3\(\frac{4}{5}\) – \(\frac{3}{10}\)

Multiple and divide the first fraction by 2​

​⇒ 3\(\frac{4(2)}{5(2)}\) – \(\frac{3}{10}\)
​
​⇒ 3\(\frac{8}{10}\) – \(\frac{3}{10}\)

⇒ 3\(\frac{5}{10}\) = 3\(\frac{1}{2}\)

Water required to fill the container is 3\(\frac{1}{2}\).

Page 124  Exercise 4.6  Problem 3

Given: Mrs. Lopez bought  3\(\frac{3}{4}\) kg of beans, 1\(\frac{1}{2}\) kg of lettuce, and 1\(\frac{3}{4}\)kg of carrots.

To find  – The total kilograms of vegetables she brought all together.

Here, add all the amount of vegetables brought.

Add all the amount of vegetables brought.

⇒ 3\(\frac{3}{4}\) + 1\(\frac{1}{2}\) + 1\(\frac{3}{4}\)

Add the whole numbers together, and then the fractions;

⇒ 5\(\frac{3}{4}\) + \(\frac{1}{2}\) + \(\frac{3}{4}\)

Multiple and divide the second fraction by 2 

⇒ 5\(\frac{3}{4}\) + \(\frac{1(2)}{2(2)}\) + \(\frac{3}{4}\)

⇒ 5\(\frac{3}{4}\) + \(\frac{2}{4}\) + \(\frac{3}{4}\)

⇒ 5\(\frac{8}{4}\) = 5(2) = 10

The total amount of vegetables she brought is 10kg

Page 124  Exercise 4.6 Problem 4

Given: Lauren purchased 7\(\frac{1}{2}\) lb of flour.

She baked several banana cakes using 2\(\frac{2}{5}\) Ib of flour.

She baked several chocolate cakes with another 3\(\frac{3}{10}\) lb of flour.

To find – The total amount of flour used.

Here, add the amount of flour used for banana cakes and chocolate cakes.

Add the amount of flour used for banana cakes and chocolate cakes.

⇒ 2\(\frac{2}{5}\) + 3\(\frac{3}{10}\)

Add the whole numbers together, and then the fractions

⇒ 5\(\frac{2}{5}\) + \(\frac{3}{10}\)

Multiple and divide the first fraction by 2

​⇒ 5\(\frac{2(2)}{5(2)}\) + \(\frac{3}{10}\)

⇒ 5\(\frac{4}{10}\) + \(\frac{3}{10}\)
​
⇒ 5\(\frac{7}{10}\)

The total amount of flour used is 5\(\frac{7}{10}\)

Page 124   Exercise 4.6  Problem 5

Given: Lauren purchased  \(\frac{1}{2}\) lb of flour.

She baked several banana cakes using  2\(\frac{2}{5}\) Ib of flour.

She baked several chocolate cakes with another 3\(\frac{3}{10}\) lb of flour.

To find – The total amount of flour used.

Here, subtract the amount of flour used for banana cakes and chocolate cakes from the total amount of flour brought

Subtract the amount of flour used for banana cakes and chocolate cakes from the total amount of flour brought.

⇒ 7\(\frac{1}{2}\) – 5\(\frac{7}{10}\)

Add the whole numbers together, and then the fractions

⇒ 2\(\frac{1}{2}\) – \(\frac{7}{10}\)

Multiple and divide the first fraction by 5
​
⇒   2 \(\frac{1(5)}{2(5)}\). 10

⇒  2\(\frac{5}{10}\). 10

⇒  2\(\frac{-2}{10}\) = − 2\(\frac{1}{5}\)

⇒ 1\(\frac{4}{5}\)

The total amount of flour left is 1\(\frac{4}{5}\)

Page 126  Exercise 4.7  Problem 1

Given: \(\frac{1}{3}\) × 9

The answer should be in the simplest form

Therefore

⇒ \(\frac{1}{3}\)× 9 = \(\frac{9}{3}\)

= 3

Simplest form of  \(\frac{1}{3}\) × 9 = 3

Page 126  Exercise 4.7  Problem 2

Given: \(\frac{1}{2}\)× 12 =

The answer should be in the simplest form

Therefore

⇒ \(\frac{1}{2}\)× 12 = \(\frac{12}{2}\)

= 6

Simplest form of   \(\frac{1}{2}\)×12 = 6

Page 126  Exercise 4.7  Problem 3

Given: \(\frac{1}{4}\)× 14 =

The answer should be in the simplest form

Therefore

⇒ \(\frac{1}{4}\)× 14 = \(\frac{14}{4}\)

= \(\frac{7}{2}\)

= 3\(\frac{1}{2}\)

Simplest form of  \(\frac{1}{4}\) × 14= 3\(\frac{1}{2}\)

Page 126  Exercise 4.7  Problem 4

Given: \(\frac{1}{6}\) × 5 =

Multiply write in simplest forms.

The expression  is \(\frac{1}{6}\)×5= \(\frac{5}{6}\)

The simplest form is \(\frac{5}{6}\)

Page 127  Exercise 4.8  Problem 1

Given: 4 by \(\frac{1}{3}\)

Multiply

The expression is

4×  \(\frac{1}{3}\)

The solution of expression is \(\frac{4}{3}\)

Page 127  Exercise 4.8  Problem 2

Given : 5 by \(\frac{3}{4}\)

Multiply

The expression is

5×  \(\frac{3}{4}\)= \(\frac{15}{4}\)

The solution of expression is \(\frac{15}{4}\)

Page 128 Exercise 4.8   Problem 3

Given: 8 × \(\frac{1}{3}\) =

Multiply and write in the simplest form

The expression is 8×\(\frac{8}{3}\)

The simplest form is \(\frac{8}{3}\)

Page 128  Exercise 4.8  Problem 4

Given: 12×\(\frac{1}{2}\)=

Multiply and write in simplest form.

The expression

12×\(\frac{1}{2}\)=\(\frac{6}{1}\)

The simplest form = \(\frac{6}{1}\)

Page 128  Exercise 4.8   Problem 5

Given: 14×\(\frac{1}{4}\)=

Multiply and write in simplest form.

The expression is

14× \(\frac{1}{4}\)=\(\frac{7}{2}\)

The simplest form is \(\frac{7}{2}\)

Page 128  Exercise 4.8 Problem 6

 Given: 5× \(\frac{1}{6}\)

Multiply and write in simplest form.

The Expression is

5× \(\frac{1}{6}\)

= \(\frac{5}{6}\)

The simplest form is \(\frac{5}{6}\)

Page 129  Exercise 4.9  Problem 1

We are given a set of fruits.

We are asked to divide the whole set into two equal parts.

From the given set, by counting the number of fruits we get the total count as 16.

Now, we have to divide the whole set into two parts with an equal number of fruits.

So, the number of fruits in each set is given as

Now, dividing the given set into two sets with 8 fruits in each we get

Therefore, the given set is divided into two equal parts which are given as

There are 8 fruits in each set.

Page 129  Exercise 4.9  Problem 2

We are given a set of ice – creams.

We are asked to divide the whole set into three equal parts.

From the given set, by counting the number of ice creams we get the total count as 18.

Now, we have to divide the whole set into three parts with an equal number of ice creams.

So, the number of ice creams in each set is given as

\(\frac{18}{3}\)  = 6

Now, dividing the given set into three sets with 6 ice creams in each we get

Therefore, the given set is divided into three equal parts which is given as

There are 6 ice creams in each set.

Page 129  Exercise 4.9  Problem 3

Given:  A picture with flowers shown.

We are asked to represent what fraction of flowers are shaded in their simplest form.

In the figure, it is clear that the total number of flowers is 7 and the shaded flowers are.

So, the fraction of each set is shaded =\(\frac{2}{7}\)

As we cannot divide either numerator or denominator with a common number we can say that it is in its simplest form.

Therefore, the fraction of flowers that are shaded, expressed in simplest form is \(\frac{2}{7}\).

Page 129  Exercise 4.9  Problem 4

Given: Apple pictures are shown.

What fraction of each set is shaded

Write the simplest form.The total apples = 12 and

Shaded apples = 8

Then fraction of set is shaded =\(\frac{8}{12}\)

= \(\frac{1}{2}\)

The fraction of shaded = \(\frac{1}{2}\)

Page 129  Exercise 4.9   Problem 5

Given: Octopus pictures are shown.

What fraction of each set is shaded.

Each answer is in simplest form.The total octopus =12

The total octopus =12

Shaded Octopus = 9

And fraction of set is shaded = \(\frac{9}{12}\)

= \({3}{4}\)

The fraction of set shaded = \(\frac{3}{4}\)

Page 129   Exercise 4.9  Problem 6

Given: Some key pictures are shown.

Write fraction of each set is shaded and written in simplest form.

The total keys =21

And

Shaded keys = 9 and

Fraction of each set is shaded= \(\frac{9}{21}\)

= \(\frac{3}{7}\)

The fraction of each set is shaded = \(\frac{3}{7}\)

Page 130  Exercise 4.9   Problem 7

Given: A circle shape is shown.

Write the fraction of each set that is shaded and write the simplest form.

The total circle shape =8 and shaded shape of circle = 4

Then, the fraction of set is shaded = \(\frac{4}{8}\)

= \(\frac{1}{2}\)

The fraction of set is shaded = \(\frac{1}{2}\)

Page 129  Exercise 4.9  Problem 8

Given: A circle shape is shown.

Write the fraction of each set that is shaded and write the simplest form.

The total circle shape = 12 and

The shaded shape of the circle =10 then

The fraction of the set is shaded = \(\frac{5}{6}\)

The fraction of the set is shaded=\(\frac{5}{6}\)

Page 129  Exercise 4.9  Problem 9

Given: A triangle shape is shown.

Write the fraction of each set that is shaded and write the simplest form.

The total triangle shape=16 and

The shaded shape of the triangle = 4, then

The fraction of the set is shaded = \(\frac{1}{4}\)

Page 129  Exercise 4.9  Problem 10

Given: A triangle shape is shown.

Write the fraction of each set that is shaded and write the simplest form.

The total triangle shape =16 and

The shaded shape of the triangle = 6

The fraction of the set is shaded = \(\frac{6}{16}\)

The fraction of the set is shaded = \(\frac{3}{8}\)

The fraction of the set is shaded =\(\frac{3}{8}\)

 Page 131  Exercise 4.9 Problem 11

Given: Some Apple’s pictures are shown.

Write the fraction of each set that is shaded and write the simplest form.

The total Apple’s are =15 , then

The fraction of green Apple’s=\(\frac{6}{15}\)

The fraction of green Apple’s=\(\frac{2}{5}\)

The fraction of green apples is \(\frac{2}{5}\)

Page 131  Exercise 4.9  Problem 12

Given: Some triangle, circle & square pictures are shown.

Write the fraction of each set is shaded & write the simplest form.

The total shapes are = 24 , then

The fraction of shapes in a circle = \(\frac{12}{24}\)

The fraction of shapes in a circle = \(\frac{1}{2}\), then

The fraction of shapes in a triangle = \(\frac{4}{24}\) or \(\frac{1}{6}\)

The fraction of shapes in squares = \(\frac{8}{24}\) or \(\frac{1}{3}\)

The fraction of the circle is \(\frac{1}{2}\) & triangles is \(\frac{1}{6}\) & square is \(\frac{1}{3}\)

Page 131  Exercise 4.9   Problem 13

Given: Some bead pictures are shown.

Write fractions and write each answer in the simplest form.

The total beads are20 and the fraction of black beads = \(\frac{10}{20}\) or \(\frac{1}{2}\)

The fraction of black beads is \(\frac{1}{2}\).

Primary Mathematics  Chapter 4 Operations On Fractions

Primary Mathematics Workbook 4A Common Core Solutions Chapter 3 Fractions Exercise 3.4  Page 148  Exercise 4.16  Problem 1

Given:  The expression \(\frac{5}{8}\) +\(\frac{3}{4}\)

Numerator and Denominator of \(\frac{3}{5}\) by 10

⇒ \(\frac{9 \times 5}{10 \times 5}-\frac{3 \times 10}{5 \times 10}\)

⇒ \(\frac{45}{50}-\frac{30}{50}\)

Since the Denominators are same, subtract the numerator

⇒ \(\frac{9 \times 5}{10 \times 5}\) – \(\frac{3 \times 10}{5 \times 10}\)

⇒ \(\frac{45}{50}\) – \(\frac{30}{50}\)

⇒ \(\frac{15}{50}\)

Now, simplify  \(\frac{15 \div 5}{50 \div 5}=\frac{3}{10}\)

Hence, the solution of \(\frac{9}{10}-\frac{3}{5} \text { is } \frac{3}{10} \text {. }\)

Chapter 3 Fractions Exercise 3.4 Answers Workbook 4A Common Core Page 148  Exercise 4.16   Problem 2

Given: \(\frac{5}{8}\)−\(\frac{3}{4}\)

First, we, find the LCM of \(\frac{5}{8}\) and \(\frac{3}{4}\)

LCM of \(\frac{5}{8}\) is \(\frac{3}{4}\)

Now, we make the denominators of both fractions equal.

⇒ \(\frac{3}{4} \times \frac{2}{2}=\frac{6}{8}\)

Simplify the expression we have

⇒ \(\frac{5}{8}+ \frac{3}{4}=\frac{5}{8}+\frac{6}{8}\)

⇒ \(\frac{5+6}{8}=\frac{11}{8}\)

Therefore the solution for  \(\frac{5}{8}+\frac{3}{4}=\frac{11}{8}\)

Chapter 3 Fractions Exercise 3.4 Answers Workbook 4A Common Core Page 148  Exercise 4.16  Problem 3

Given: 4\(\frac{5}{6}\)+ 3\(\frac{2}{3}\)

To Find –  Find 4\(\frac{5}{6}\)+ 3\(\frac{2}{3}\)

To add fractions with unlike denominators, first, find the least common multiple of the two denominators and then add and simplify.

The LCM of 6 and 3 is 6. So, we need to find fractions equivalent to 4\(\frac{5}{6}\) and 3\(\frac{2}{3}\) which have 18 in the denominator.

Multiply the numerator and denominator of 4\(\frac{5}{6}\) by 3 , and multiply the numerator and denominator of 3\(\frac{2}{3}\) by 6.
​
⇒ 4\(\frac{5}{6}\)

⇒ \(\frac{6×4+5}{6}\)

⇒ \(\frac{29}{6}\)

​Similarly 3\(\frac{2}{3}\)

⇒ \(\frac{3×3+2}{3}\)

⇒ \(\frac{11}{3}\)

Now, add the fractions
​
4\(\frac{5}{6}\) + 3\(\frac{2}{3}\)

⇒ \(\frac{29}{6}\) + \(\frac{11}{3}\)

⇒ \(\frac{29×3}{6×3}+\frac{11×6}{3×6}\)
​
⇒ \(\frac{87}{18}\) + \(\frac{66}{18}\)

Since the denominators are the same, add the numerators.

⇒ \(\frac{87}{18}\)+\(\frac{66}{18}\)

= \(\frac{87+66}{18}\)

= \(\frac{153}{18}\)

Now, simplify
​
⇒ \(\frac{153}{18}\)

= \(\frac{153÷9}{18÷9}\)

= \(\frac{17}{2}\)

Hence, the solution of  4\(\frac{5}{6}\) + 3\(\frac{2}{3}\)  is  \(\frac{17}{2}\).

Primary Mathematics 4A Fractions Exercise 3.4 Step-By-Step Solutions Page 148  Exercise 4.16  Problem 4

Given:  5\(\frac{1}{4}\) – 2\(\frac{5}{12}\) =

To Find –  Find 5\(\frac{1}{4}\) – 2\(\frac{5}{12}\)
​
To subtract fractions with unlike denominators, first, find the least common multiple of the two denominators and then subtract and simplify.

The LCM of 4 and 12 is 12. So, we need to find fractions equivalent to  5\(\frac{1}{4}\)  and  2\(\frac{5}{12}\) which have 48 in the denominator.

Multiply the numerator and denominator of 5\(\frac{1}{4}\) by 12, and multiply the numerator and denominator of  2\(\frac{5}{12}\) by, 4.

⇒ 5\(\frac{1}{4}\)

= \(\frac{4×5+1}{4}\)

= \(\frac{21}{4}\)

Similarly

⇒ 2\(\frac{5}{12}\)

= \(\frac{12×2+5}{12}\)

​= \(\frac{29}{12}\)
​
Now, add the fractions

⇒ 5\(\frac{1}{4}\)– 2\(\frac{5}{12}\)

= \(\frac{21}{4}\) – \(\frac{29}{12}\)
​
= \(\frac{21×12}{4×12}\) – \(\frac{29×4}{12×4}\)

= \(\frac{252}{48}\) – \(\frac{116}{48}\)

Since the denominators are the same, subtract the numerators.
​
⇒ \(\frac{252}{48}\) – \(\frac{116}{48}\)

​= \(\frac{252-116}{48}\)

= \(\frac{136}{48}\)

Now, simplify

⇒ \(\frac{136}{48}\)

= \(\frac{136÷8}{48÷8}\)

= \(\frac{17}{6}\)

Hence, the solution of   5\(\frac{1}{4}\)− 2\(\frac{5}{12}\)  is \(\frac{17}{6}\).

Fractions Exercise 3.4 Explanation Common Core Workbook 4A Page 149  Exercise 4.16  Problem 5

Given: ______− 6\(\frac{3}{4}\) = \(\frac{1}{2}\)

Take \(\frac{3}{4}\) to the other side and add to \(\frac{1}{2}\)

⇒ \(\frac{1}{2}+6 \frac{3}{4}\)

⇒ \(\frac{1}{2}+\frac{27}{4}\)

Take LCM of 2 and 4 to simplify

⇒ \(\frac{1}{2}+\frac{27}{4}\)

⇒ \(\frac{2+27}{4}\)

⇒ \(\frac{29}{4}\)

Hence the solution is \(\frac{29}{4}-6 \frac{3}{4}=\frac{1}{2}\)

​Page 149  Exercise 4.16   Problem  6

Given: 3\(\frac{7}{8}\) +_____ =  5 \(\frac{1}{8}\)

Take 3\(\frac{7}{8}\) to the other side and subtract from 5 \(\frac{1}{8}\)

⇒ \(5 \frac{1}{8}-3 \frac{7}{8}\)

⇒ \(\frac{8 \times 5+1}{8}-\frac{8 \times 3+7}{8}\)

⇒ \(\frac{41}{8}-\frac{31}{8}\)

As the denominator is the same, simplify

⇒ \(\frac{41}{8}-\frac{31}{8}\)

⇒ \(\frac{10}{8} \text { or } \frac{5}{4}\)

Hence solution is [latex3 \frac{7}{8}+\frac{5}{4}=5 \frac{1}{8}][/latex]

Exercise 3.4 Fractions Detailed Explanation Workbook 4A Page 150  Exercise 4.16  Problem 7

Given: Rope A is 10\(\frac{2}{3}\)ft long and rope B is 8\(\frac{5}{6}\) ft long.

To Find – Find the sum of the length of the two ropes.

By adding the integers rule, find the sum of the length of the two ropes.

Simplify 10\(\frac{2}{3}\)ft

⇒ 10\(\frac{2}{3}\)

=  \(\frac{10×3+2}{3}\)

=  \(\frac{32}{3}\)

Similarly for 8\(\frac{5}{6}\) ft.

⇒ 8\(\frac{5}{6}\)

= \(\frac{6×8+5}{6}\)

= \(\frac{53}{6}\)

Add the length of the two ropes

⇒ \(\frac{32}{3}\)+\(\frac{53}{6}\)

Take LCM of 3 and 6, then add
​
⇒ \(\frac{32}{3}\)+\(\frac{53}{6}\)

​= \(\frac{64}{6}\)+\(\frac{53}{6}\)

​=  \(\frac{64+53}{6}\)

= \(\frac{117}{6}\)

Or , \(\frac{39}{2}\) ft

Hence, the sum of the length of the two ropes is \(\frac{39}{2}\) ft.

Page 150 Exercise 4.16   Problem 8

Given: Rope A is  10\(\frac{2}{3}\) ft  long and rope B is  8\(\frac{5}{6}\)

To Find –  Find the difference in length of the two ropes.

By subtracting the integers rule, find the difference in length of the two ropes.

Simplify 10\(\frac{2}{3}\)ft

⇒ 10\(\frac{2}{3}\)

=  \(\frac{10×3+2}{3}\)

=  \(\frac{32}{3}\)

Similarly for 8\(\frac{5}{6}\) ft.

⇒ 8\(\frac{5}{6}\)

=  \(\frac{6×8+5}{6}\)

=  \(\frac{53}{6}\)

Subtract the length of the two ropes

⇒ \(\frac{32}{3}\) −  \(\frac{53}{6}\)

Take LCM of 3 and 6, then subtract
​
⇒ \(\frac{32}{3}\)− \(\frac{53}{6}\)

=  \(\frac{64}{6}\)−\(\frac{53}{6}\)

​=  \(\frac{64−53}{6}\)

=  \(\frac{11}{6}\)

Hence, the difference in the length of the two ropes is \(\frac{11}{6}\).

Common Core Primary Mathematics Workbook 4A Fractions Help Chapter 3 Page 150  Exercise 4.16  Problem 9

Given: Brandy bought 3 yd of raffia. She used \(\frac{5}{6}\) yd to make a doll.

To Find –  Find the length of the raffia left.

By subtracting the integers rule, find the length of the raffia left.

Given, Brandy bought 3 yd of raffia. She used \(\frac{5}{6}\) yd to make a doll.

Total raffia is 3yd . Out of which, \(\frac{5}{6}\) yd is used to make a doll. So, the length of the raffia left

⇒ (3-\(\frac{5}{6}\)) yd

Solve (3-\(\frac{5}{6}\)) yd

⇒ (3-\(\frac{5}{6}\)) yd

= (\(\frac{18-5}{6}\)) yd

= \(\frac{13}{6}\)

Hence, the length of the raffia left is \(\frac{13}{6}\).

Primary Mathematics 4A Exercise 3.4 Common Core Fraction Problems Page 151 Exercise 4.16   Problem 10

Given: Total 10 number of fruits, out of which 3 number of apples, 1 number of pears, and 6 number of oranges.

To Find – What fraction of the fruits are oranges?

The total number of fruits is 10. Out of which, 6 are oranges.

So, a fraction of the fruits are oranges is  \(\frac{6}{10}\)  =  \(\frac{3}{5}\)

Hence, \(\frac{3}{5}\) of the fruits are oranges.

Page 151 Exercise 4.16   Problem 11

Given: In a test, Matthew answered 32 out of 40 questions correctly.

To Find –  What fraction of the questions did he answer correctly?

In a test, Matthew answered 32 out of 40 questions correctly.

So, a fraction of the questions did he answer correctly  \(\frac{32}{40}\)

=  \(\frac{4}{5}\)

Hence, a fraction of the questions he answered correctly is  \(\frac{4}{5}\).

Fractions Exercise 3.4 Primary Mathematics Workbook 4A Worksheet Help Page 151 Exercise 4.16   Problem 12

Given: Mrs. Reed bought 30 eggs.

She used 5 of them to bake a cake.

To Find –  What fraction of the eggs did she have left?

Mrs. Reed bought 30 eggs.

She used 5 of them to bake a cake.

So, the remaining eggs she left were 30 − 5 = 2 eggs.

Thus, the fraction of the eggs she has left is  \(\frac{25}{30}\) = \(\frac{5}{6}\).

Hence, the fraction of the eggs left with her is  \(\frac{5}{6}\).

Page 152 Exercise 4.16   Problem 13

Given: Taylor made 98 sugar buns and 42 plain buns.

To Find – What fraction of the buns were plain buns?

The total number of buns Taylor has is 98 + 42 = 140 buns.

Thus, the fraction of the buns that were plain buns is  \(\frac{42}{140}\) =  \(\frac{21}{70}\).

Hence, \(\frac{21}{70}\) of the buns were plain buns.

Page 152  Exercise 4.16   Problem 14

Given: Rosa wants to tie 6 packages. She needs \(\frac{3}{5}\) yd of string for each package.

To Find – How many yards of the string must she buy?

Given, Rosa wants to tie 6 packages.

For each package, she needs \(\frac{3}{5}\) yd of string.

Hence , for 6 packages she needs 6 × \(\frac{3}{5}\)

= \(\frac{18}{5}\)  yd of string.

Hence, she must buy  \(\frac{18}{5}\) yd  of string for 6 packages.

Page 152  Exercise 4.16  Problem 15

Given: The job had $24

He used  \(\frac{7}{8}\)  of it to buy a book.

To Find – What was the cost of the book?

Given, that Job had $24. He used \(\frac{7}{8}\) of $24 to buy a book.

So, the cost book is \(\frac{7}{8}\) × $24  =  $21.

Hence, the cost of the book was $21.

Page 153  Exercise 4.16  Problem 16

Given: There are 60 roses. \(\frac{7}{10}\) of them are red roses.

The rest are yellow roses.

To Find – How many yellow roses are there?

Given

There are 60 roses. \(\frac{7}{10}\) of them are red roses.

So, red roses are

⇒ \(\frac{7}{10}\) × 60 = 42

Total of 60 roses, 42 roses are red.

Hence remaining roses are 60 − 42 = 18.

Therefore, there are 18 yellow roses.

Hence, there are 18 yellow roses.

Page 153  Exercise 4.16   Problem 17

Given: Ben had $35.

He spent \(\frac{2}{7}\) $35 on a pair of shoes.

To Find – How much money did he have left?

Given, Ben had $35. He spent \(\frac{2}{7}\) $35 on a pair of shoes.

Hence, money he had spent  \(\frac{2}{7}\) × $35 = $10.

Therefore, he had left  35 −10 = $25.

Hence, Ben has $25 left.

Page 154  Exercise 4.16  Problem 18

Given: After spending \(\frac{3}{5}\) of his money on a tennis racket, Sean had $14 left.

To Find – How much did the tennis racket cost?

First, assume total money x and then find the value of x, for which he left $14

Then, substitute the value of x, in money he had spent that is \(\frac{3x}{5}\), and find the cost.

Let’s assume, Sean had total money x.

He spent  \(\frac{3}{5}\)  of his money x on a tennis racket.

So, he spent  \(\frac{3x}{5}\)  and left $14.

Therefore, the total money he had
​
⇒ ​x−\(\frac{3x}{5}\) =14

⇒ \(\frac{5x−3x}{5}\) = 14

⇒ \(\frac{2x}{5}\) = 14

⇒ 2x = 70

Or , x = 35

Hence, the total money he had was $35. He spent on tennis

⇒ \(\frac{3x}{5}\)
​
=\(\frac{3×35}{5}\)

= 21

Hence, the tennis racket cost $21.

Page 154  Exercise 4.16  Problem 19

Given: A computer costs $2290.

An oven costs \(\frac{1}{5}\)  the cost of the computer.

To Find –  How much more does the computer cost than the oven?

A computer costs $2290. An oven costs  \(\frac{1}{5}\) the cost of the computer.

Hence, the oven cost  \(\frac{1}{5}\) ×2290  = $458.

Therefore, the computer costs more than the oven is

$2290 − $458 = $1832

Hence, the computer costs $1832 more than the oven.

Primary Mathematics  Chapter 3 Fractions

Primary Mathematics Workbook 4A Common Core Edition Solutions Page 107  Exercise 3.10 Problem 1

We are given a figure and asked to convert it into an improper fraction.

Draw a long line between the figures and then we can see 8 triangles and in that 2 triangles are shaded region.

Therefore we can say that the improper fraction will be \(\frac{2}{8}\)

Therefore, we can say that the fraction for the given figure is \(\frac{2}{8}\)

Page 107   Exercise 3.10    Problem 2

We are asked to find the missing numerators and denominators. As we can see the given are equivalent fractions.

Given:  First fraction as \(\frac{2}{3}\)

The second fraction has a numerator which is 3 times 2

So, the denominator also has to be 3 times 3

Which means \(\frac{6}{9}\)

In the third fraction, it is 4 times the first fraction i.e. \(\frac{8}{12}\)

In the fourth fraction, it is 5 times the first fraction i.e.  \(\frac{10}{12}\)

In the fifth fraction, it is 7 times the first fraction i.e.  \(\frac{14}{21}\)

Therefore, we can say that the completed fractions are

\(\frac{2}{3}\) = \(\frac{6}{9}\)

⇒ \(\frac{8}{12}\)

⇒ \(\frac{10}{12}\)

⇒ \(\frac{14}{21}\)

Chapter 3 Fractions Exercise 3.10 Solutions Page 107  Exercise 3.10   Problem 3

We are asked to write any two fractions that are equivalent to \(\frac{1}{4}\)

As we know equivalent fraction means multiple of the given fraction.

Therefore the equivalent fraction for the given fraction is

⇒ \(\frac{1×2}{4×2}\) = \(\frac{2}{8}\)

⇒ \(\frac{1×3}{2×3}\) =  \(\frac{3}{6}\)

Therefore, equivalent fractions are \(\frac{2}{8}\), \(\frac{3}{6}\)

Page 107  Exercise 3.10  Problem 4

We are asked to find the fractions that the number line represents.

In the given number it is divided by 10 parts.

So, we can see that the P is the 2nd part of the 10 parts.

So, the fraction will be  3\(\frac{6}{10}\)

Q is in the 6th part. So, the fraction will be 3\(\frac{6}{10}\)= 3\(\frac{3}{5}\).

R is the second part between 4 and 5. So, fraction will be 4\(\frac{2}{10}\)

= 4\(\frac{1}{5}\).

Therefore, the values of P, Q,R are 3\(\frac{3}{5}\) and 4\(\frac{1}{5}\)

Page 108  Exercise 3.10   Problem 5

We are asked to find the fractions that in missing in the number line.

In the given number line it is divided by 10 parts between the numbers.

As we can see the first blank is 5 the blank of 8 blanks.

So, the fraction is 1 \(\frac{5}{8}\)

The second blank is at the 7 position in 8 parts.

So, the fraction is 1\(\frac{6}{8}\)

The last blank is the first part between 2 and 3

So, 2\(\frac{1}{8}\) is the fraction.

Therefore, the values of the blanks are 1 \(\frac{5}{8}\), 1\(\frac{6}{8}\), 2\(\frac{1}{8}\) respectively.

Page 108  Exercise 3.10  Problem 6

We are asked to find which of the given fractions is small.

Given: \(\frac{1}{2}, \frac{2}{5}, \frac{3}{4}, \frac{4}{7} .\)

Let’s find the decimal values of the given fraction.

⇒ \(\frac{1}{2}\) =  0.5

⇒ \(\frac{2}{5}\) =  0.4

⇒ \(\frac{3}{4}\) =  0.75

⇒ \(\frac{4}{7}\) = 0.5714

So,\(\frac{2}{5}\) is the smallest faction.

Therefore, we can say that a given fraction is the smallest in given fractions.

Primary Mathematics 4A Chapter 3 Step-By-Step Solutions For Exercise 3.10 Page 108  Exercise 3.10  Problem 7

We are asked to find which of the given fractions is in its simplest form.

Given: \(\frac{3}{9}, \frac{2}{10}, \frac{5}{7}\)

In \(\frac{3}{9}\) we can see that 9 is a multiple of 3 so it is not in its simplest form.

In \(\frac{2}{10}\)we can see that 10 is a multiple of 2 so it is not in its simplest form.

In \(\frac{5}{7}\)we can see that 5 is not a multiple of 7 so it is not in its simplest form.

Therefore, we can say that  \(\frac{5}{7}\) is in its simplest form.

Page 108  Exercise 3.10   Problem 8

We are asked to find which of the given fractions is smaller than\(\frac{1}{2}\)

Let us find the decimal of all the given fractions, we get

⇒ \(\frac{7}{8}\) = 0.875

⇒ \(\frac{5}{9}\) =  0.55

⇒ \(\frac{2}{5}\) = 0.4

Therefore we can say that \(\frac{2}{5}\) is smaller than  \(\frac{1}{2}\)  is \(\frac{1}{2}\)

Therefore, we can say that \(\frac{2}{5}\) is the fraction that is smaller than  is \(\frac{1}{2}\)

Page 108  Exercise 3.10  Problem 9

We are asked to find which of the given fractions is nearest to 1

Let us find the decimal of all the given fractions, we get

⇒ \(\frac{4}{5}\) = 0.8

⇒ \(\frac{6}{7}\) = 0.85

⇒ \(\frac{11}{12}\) = 0.91

⇒ \(\frac{8}{9}\) = 0.88

Therefore we can say that  \(\frac{11}{12}\)  is nearest to 1

Therefore we can say that \(\frac{11}{12}\) is nearest to 1

Fractions Exercise 3.10 Primary Mathematics Workbook Answers Page 108  Exercise 3.10  Problem 10

We are asked to find which of the given fractions is greater.

Given:

⇒ \(\frac{3}{8}\) or \(\frac{7}{12}\)

Let us find the decimal of all the given fractions, we get

⇒ \(\frac{3}{8}\) = 0.37

⇒ \(\frac{7}{12}\) = 0.5

Therefore we can say that \(\frac{7}{12}\) is greater than 3\(\frac{3}{8}\)

Therefore we can say that \(\frac{7}{12}\) is greater than 3\(\frac{3}{8}\)

Page 109 Exercise 3.10  Problem 11

We are asked to arrange fractions in increasing order.

Given:

⇒ \(\frac{3}{4}, \frac{7}{6}, \frac{5}{12}\), 1

Given: \(\frac{3}{4}, \frac{7}{6}, \frac{5}{12}, 1\)

Let us find the decimal of all the given fractions, we get

⇒ \(\frac{3}{4}=0.75\)

⇒ \(\frac{7}{6}=1.16\)

⇒ \(\frac{5}{2}=0.415\)

Therefore we can say that the increasing order will be

⇒ \(\frac{5}{12}<\frac{3}{4}<1<\frac{7}{6}\)

Therefore we can say that the \(\frac{5}{12}<\frac{3}{4}<1<\frac{7}{6}\)

Page 109  Exercise 3.10  Problem 12

We are asked to arrange fractions in decreasing order.

Given:  \(\frac{2}{9}, \frac{2}{7}, \frac{9}{7}, \frac{2}{3}\)

⇒ \(\frac{2}{9}\) = 0.22

⇒ \(\frac{2}{7}\) = 0.28

⇒ \(\frac{9}{7}\) = 1.2

⇒ \(\frac{2}{3}\) = 0.6

Therefore we can say that the decreasing order will be.

⇒ \(\frac{9}{7}\)>\(\frac{2}{3}\)>\(\frac{2}{7}\)>\(\frac{2}{9}\).

Therefore we can say that the decreasing order will be. \(\frac{9}{7}\)>\(\frac{2}{3}\)>\(\frac{2}{7}\)>\(\frac{2}{9}\).

Page 109 Exercise 3.10  Problem 13

We are asked to find the missing fraction that makes the equation true.

Given:  ____ + \(\frac{3}{8}\)=1

Let us transfer \(\frac{3}{8}\) from LHS to RHS, we get

_______= 1− \(\frac{3}{8}\)

By solving the RHS, we get

______ = \(\frac{8-3}{8}\)

______ = \(\frac{5}{8}\)

Therefore, given blank fraction value is \(\frac{5}{8}\)

Therefore, we can say that the value of the given blank is \(\frac{5}{8}\)

Common Core Primary Mathematics 4A Chapter 3 Solved Examples For 3.10 Page 109  Exercise 3.10  Problem 14

We are asked to find the missing fraction that makes the equation true.

Given: 1−______ = \(\frac{5}{12}\)

Let us transfer 1 from LHS to RHS and multiply both sides with a negative sign, we get

⇒ 1−______ = \(\frac{5}{12}\)

⇒ _____ \(1-\frac{5}{12}\)

By solving the RHS, we get

______ \(\frac{12-5}{12}\)

______ \(\frac{7}{12}\)

Therefore, given a blank fraction value is \(\frac{7}{12}\)

Therefore we say that the value of a given blank is \(\frac{7}{12}\)

Solutions For Fractions Exercise 3.10 In Primary Mathematics 4A Page 109 Exercise 3.10  Problem 15

We are asked to write an improper fraction for each of the following.

Given:  9 –  Eighths.

As we know eighth means \(\frac{1}{8}\)

So, 9 eighths is 9(\(\frac{1}{8}\)). i.e. \(\frac{9}{8}\)

Therefore, we can say that the value of nine-eighths means  \(\frac{9}{8}\)

Page 109  Exercise 3.10  Problem 16

We are asked to write an improper fraction for each of the following.

Given:  11 –  quarters.

As we know quarter means\(\frac{1}{4}\)

So, 11 eighths is 11(\(\frac{1}{8}\)) i.e \(\frac{11}{8}\)

Therefore, we can say that the value of nine-eighths means  \(\frac{11}{8}\).

Page 109  Exercise 3.10  Problem 17

We are asked to write an improper fraction for each of the following.

Given: 7-fifths.

As we know fifths means \(\frac{1}{5}\)

So, 7 fifths is 7(\(\frac{1}{5}\)) i.e \(\frac{5}{7}\)

Therefore, we can say that the value of nine-eighths means \(\frac{5}{7}\)

Detailed Solutions For Exercise 3.10 Fractions In 4A Workbook Page 109  Exercise 3.10  Problem 18

We are asked to write an improper fraction for each of the following.

Given:  15 – sixths.

As we know sixths means \(\frac{1}{6}\)

So, 15 fifths is 15 (\(\frac{1}{6}\)) i.e \(\frac{15}{6}\)

Therefore, we can say that the value of 15 sixths means \(\frac{15}{6}\)

Page 110  Exercise 3.10  Problem 19

We are asked to express   2\(\frac{3}{5}\) as an improper fraction.

As we know mixed fraction contains a whole number and a fraction part we have to multiply the denominator with the whole number and we have to add this to the numerator to get the numerator of the improper fraction.

So, we get

2\(\frac{3}{5}\) = \(\frac{2×5+3}{5}\)

= \(\frac{13}{5}\)

Therefore, we can say that by converting a given mixed fraction into an improper fraction we get  \(\frac{13}{5}\).

Page 110  Exercise 3.10   Problem 20

We are asked to find how many thirds are there in 3

Let us write an equivalent fraction for \(\frac{3}{1}\)

Multiply the numerator and denominator by 3, and we get

= \(\frac{9}{3}\)

So, we can say that we have 9 thirds in three.

Therefore, we can say that we have 9 thirds in three.

Step-By-Step Guide For Fractions Exercise 3.10 In 4A Workbook Page 110  Exercise 3.10  Problem 21

We are asked to find how many sixths are there in 2\(\frac{1}{6}\)

Let us write an equivalent fraction for \(\frac{2}{1}\)

Multiply numerator and denominator by 6 , we get = \(\frac{12}{6}\)

So, we can say that we have 12 sixths in three.

Therefore, we can say that we have 12 sixths in two.

Page 110 Exercise 3.10  Problem 22

We are given to convert the given improper fraction into a mixed fraction.

Given: \(\frac{27}{6}\)

To convert an improper fraction into a mixed fraction let us divide the given fraction and then we have to write the divisor as the whole number and quotient as the numerator and the remainder as the numerator.

So, for 27÷6 we get 4 as quotient and 3 as remainder.

So, Mixed fraction = 4 \(\frac{3}{6}\)

Therefore, we can say that the mixed fraction for the given improper fraction is  4 \(\frac{3}{6}\)

Page 110  Exercise 3.10  Problem 23

We are asked to find which will match \(\frac{4}{5}\).

As from \(\frac{4}{5}\)we can say that 5 is diving 4.

So, therefore we can say that the correct option is 4÷5.

Therefore, we can say that the correct option is  4÷5.

Page 110  Exercise 3.10   Problem 24

We are asked to write a fraction of 8 girls sharing 2 cakes equally.

So, we can say that 2 cakes should be shared with 8 girls equally.

It implies \(\frac{2}{8}\)

⇒ \(\frac{1}{4}\)

So, each girl gets \(\frac{1}{4}\)  portion from the 2 cakes.

Therefore, we can the fraction that represents the part of each girl that the get is  \(\frac{1}{4}\)

Primary Mathematics Workbook 4A exercise 3.10 Fractions Page 110  Exercise 3.10  Problem 25

We are asked to add the given fractions.

Given:

⇒ \(\frac{2}{7}+\frac{4}{7}\)

As we can see that the denominators are the same we can add them directly, and we get

⇒ \(\frac{2}{7}+\frac{4}{7}\) = \(\frac{6}{7}\)

Therefore, by adding  \(\frac{2}{7}+\frac{4}{7}\) we get\(\frac{6}{7}\).

Page 110  Exercise 3.10  Problem 26

We are asked to subtract the given fractions.

Given:

⇒ \(\frac{7}{12}-\frac{3}{12}\)

As we can see that the denominators are the same we subtract and add them directly, and we get

⇒ \(\frac{7}{12}-\frac{3}{12}\)=\(\frac{3}{12}\)

⇒ \(\frac{1}{3}\)

Therefore, by subtracting  \(\frac{7}{12}-\frac{3}{12}\)=\(\frac{3}{12}\) we get\(\frac{1}{3}\)

Page 110  Exercise 3.10  Problem 27

We are asked to add the given fractions.

Given:

⇒ \(\frac{6}{9}+\frac{3}{9}\).

As we can see that the denominators are the same we add them directly, we get

⇒ \(\frac{6}{9}+\frac{3}{9}\)=\(\frac{9}{9}\)

= 1

Therefore, by adding  \(\frac{6}{9}+\frac{3}{9}\) we get 1

Chapter 3 Fractions Exercise 3.10 Breakdown With Solutions Page 110  Exercise 3.10  Problem 28

We are asked to subtract the given fractions.

Given: 2-\(\frac{6}{11}\)

As we can see the denominators are not the same, so let us find the LCM of the two denominators, and then by solving we get

⇒ \(\frac{22−6}{11}\)= \(\frac{6}{11}\)

Therefore, by subtracting  \(\frac{22−6}{11}\)  we get  \(\frac{6}{11}\)

Page 110  Exercise 3.10   Problem 29

We are asked to add the given fractions.

Given: \(\frac{2}{8}+\frac{3}{8}+\frac{1}{8}\)

As we can see that the denominators are the same.

So we can add them directly

⇒ \(\frac{2}{8}+\frac{3}{8}+\frac{1}{8}\)

⇒ \(\frac{2+3+1}{8}\)

⇒ \(\frac{6}{8}\)

⇒ \(\frac{3}{4}\)

There fore by adding \(\frac{2}{8}+\frac{3}{8}+\frac{1}{8} \text { we get } \frac{3}{4}\)

Common Core 4A Chapter 3 Exercise 3.10 Solutions Page 110  Exercise 3.10   Problem 30

We are asked to subtract the given fractions.

Given: 1− \(\frac{3}{10}-\frac{5}{10}\)

As we can see that the denominators are not the same.

So we have to do LCM and then we have to solve them.

By doing LCM and solving we get

⇒ \(\frac{10-3-5}{10}\)= \(\frac{2}{10}\)

⇒ \(\frac{1}{5}\)

Therefore, by subtracting 1-\(\frac{3}{10}-\frac{5}{10}\) we get \(\frac{1}{5}\)

Page 111  Exercise 3.10   Problem 31

Given: The total weight of two bags is 2 lb.

One of them is \(\frac{1}{4}\)lb.

We are asked to find the weight of the other bag.

Let us consider the other bag weight as x.

So, from the question x+\(\frac{1}{4}\) =2

Transferring \(\frac{1}{4}\) from LHS to RHS, we get
​
​x = 2−\(\frac{1}{4}\)

x = \(\frac{8-1}{4}\)

x = \(\frac{7}{4}\)

​Therefore, we can say that the other bag weight is \(\frac{7}{4}\) lb.

Page 111  Exercise 3.10  Problem 32

Given: She used \(\frac{5}{8}\) meters rope and she has 7 meters left.

The total rope is the sum of the remaining rope and the used rope

Total rope is \(\)

⇒ \(\frac{5}{8}+7\)

⇒ \(\frac{5}{8}+\frac{7}{1}\)

⇒ \( \frac{5+7(8)}{8}\)

⇒ \(\frac{5+56}{8}\)

⇒ \(\frac{61}{8}\)

Therefore we say that the total rope is \(\frac{61}{8}\)