Primary Mathematics Chapter 5 Measures
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 2km634m = 2634m
The given fill-in-the-blanks has been filled as 2km 634m = 2634m.
Page 169 Exercise 5.9 Problem 2
It is given that 5kg17g = ___ 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
5kg17g = 5017 g
The given fill-in-the-blanks has been filled as 5kg17g = 5017g.
Page 169 Exercise 5.9 Problem 3
It is given that 3h4min = ___ 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
3h4min = 184 min
The given fill-in-the-blanks has been filled as 3h4min = 184 min.
Page 169 Exercise 5.9 Problem 4
It is given that 6m5cm = ____ 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 6m5cm = 605cm
The given fill-in-the-blanks has been filled as 6m 5cm = 605cm
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.
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 = 4h30min
The given blanks have been filled as 4\(\frac{1}{2}\) h = 4h30min
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
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 = 3m90cm
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.
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
= 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 = 13yd1ft.
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 6gal3qt×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 4m25cm × 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 4m25cm × 25 = 29m75cm.
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 6h40min × 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 = 1lb6oz
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 7kg800g ÷ 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 = 1kg300g.
Page 170 Exercise 5.9 Problem 19
It is given that 10min21sec ÷ 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>3yr5months.
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______ 4L50ml.
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>4L50ml
For the given quantities, the comparison has been done and we get 4\(\frac{1}{2}\)L>4L50ml
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 3L250ml 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
= 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 19L500ml from 6 cartons of fruit juices.
We can get 19L500ml from 6 cartons of fruit juices when it is given that we get 3l250ml 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 7lb8oz.
Page 173 Exercise 5.9 Problem 31
It is given that Lindsey bought 8 bottles of orange juice.
Each bottle contained 1L275ml 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 1L275ml 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 1L275ml.
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
= 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.