Primary Mathematics Chapter 5 Measures
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
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.
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.
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.
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
= 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.
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
= 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
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.
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.
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
= 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.