Primary Mathematics Workbook 4A Common Core Edition Solutions Chapter 5 Measures Exercises 5.1

Primary Mathematics  Chapter 5 Measures

 

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

= 48 ounces

4 pounds = 4 × 16

= 64 ounces

5 pounds = 5 × 16

=80 ounces

6 Pounds = 6 × 16

= 96 ounces

7 pounds = 7 × 16

= 112 ounces

8 pounds = 8 × 16

= 128 ounces

9 pounds = 9 × 16

= 144 ounces

10 pounds = 10 × 16

=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

 

Page 155  Exercise 5.1  Problem 2

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 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

 

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 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

 

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.

 

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

 

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.

 

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

 

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.

 

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.

 

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

 

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

​​​​​​​​​​​​​​= 12000g + 10g

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

= 10000g + 600g

=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

= 204 in + 3 in

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

= 84 in + 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

\(\frac{1850}{1000}\)

= 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

= 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

=  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

= 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

= 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.