New Learning Composite Math Class 5 Chapter 8 Solutions
SK Gupta Anubhuti Gangal’s New Learning Composite Mathematics Class 5 Self Practice 8A, 8B, 8C, 8D, 8E & 8F Solutions is available here.
Self Practice 8A
Copy and multiply.
1)
a) 6/10 × 5/12
Solution: 6/10 × 5/12 = 1/2 × 1/2 = 1/4 |
b) 2/3 × 18/20
Solution: 2/3 × 18/20 = 3/5 |
c) 3/8 × 12/27
Solution: 3/8 × 12/27 = 1/6 |
d) 5/18 × 54/45
Solution: 5/18 × 54/45 = 1/3 |
2)
a) 4/10 × 2/6 × 25/16
Solution: 4/10 × 2/6 × 25/16 = 1 × 1/6 × 5/4 = 5/24 |
b) 7/11 × 33/49 × 14/44
Solution: 7/11 × 33/49 × 14/44 = 1 × 3 × 1/22 = 3/22 |
c) 42/15 × 3/7 × 5/36
Solution: 42/15 × 3/7 × 5/36 = 1/6 |
3)
a) {1+(7/8)} × {3+(1/5)}
Solution: {1+(7/8)} × {3+(1/5)} = 15/8 × 16/5 = 3 × 2 = 6 |
b) {9+(1/11)} × {10+(1/12)}
Solution: {9+(1/11)} × {10+(1/12)} = 100/11 × 121/12 = 25 × 11/3 = 275/3 = 91+(2/3) |
c) }2+(3/5)} × {1+(1/4)}
Solution: {2+(3/5)} × {1+(1/4)} = 13/5 × 5/4 = 13/4 = 3+(1/4) |
d) {2+(1/7)} × {3+(1/2)}
Solution: {2+(1/7)} × {3+(1/2)} = 15/7 × 7/2 = 15/2 = 7+(1/2) |
e) {3+(11/15)} × {1+(42/63)}
Solution: {3+(11/15)} × {1+(42/63)} = 56/15 × 105/63 = 56 × 1/9 = 56/9 = 6+(2/9) |
f) {1+(2/7)} × {5+(5/6)}
Solution: {1+(2/7)} × {5+(5/6)} = 9/7 × 35/6 = 3 × 5/2 = 15/2 = 7+(1/2) |
g) {4+(3/4)} × {1+(7/57)}
Solution: {4+(3/4)} × {1+(7/57)} = 19/4 × 64/57 = 16/3 = 5+(1/3) |
h) {12+(1/2)} × {3+(1/5)}
Solution: {12+(1/2)} × {3+(1/5)} = 25/2 × 16/5 = 5 × 8 = 40 |
4) Simplify:
a) {1+(8/17)} × {4+(4/5)} × {6+(3/8)}
Solution: {1+(8/17)} × {4+(4/5)} × {6+(3/8)} = 25/17 × 24/5 × 51/8 = 5 × 3 × 3 = 45 |
b) {1+(2/3)} × 7/8 × {3+(1/5)}
Solution: {1+(2/3)} × 7/8 × {3+(1/5)} = 5/3 × 7/8 × 16/5 = 1/3 × 7 × 2/1 = 14/3 = 4+(2/3) |
c) {2+(1/2)} × {3+(2/3)} × 6/11
Solution: {2+(1/2)} × {3+(2/3)} × 6/11 = 5/2 × 11/3 × 6/11 = 5 |
Self Practice 8B
Find the product and write in the simplest form.
1) 18 × 5/6
Solution: 18 × 5/6 = 3 × 5 = 15 |
2) 3 × 1/3
Solution: 3 × 1/3 = 1 |
3) 3 × 2/5
Solution: 3 × 2/5 = 6/5 = 1+(1/5) |
4) 10 × 1+(5/8)
Solution: 10 × 1+(5/8) = 10 × 13/8 = 65/4 = 16+(1/4) |
5) 10 × {1+(2/15)}
Solution: 10 × {1+(2/15)} = 10 × 17/15 = 34/3 = 11+(1/3) |
6) 34 × 10/68
Solution: = 34 × 10/68 = 5 |
7) 44 × 13/220
Solution: =44 × 13/220 = 13/5 = 2+(3/5) |
8) 75 × 19/100
Solution: = 75 × 19/100 = 57/4 = 14+(1/4) |
9) 12 × 7/20
Solution: = 12 × 7/20 = 21/5 = 4+(1/5) |
10) 18 × 4/33
Solution: = 18 × 4/33 = 24/11 = 2+(2/11) |
11) 60 × 11/105
Solution: = 60 × 11/105 = 44/7 = 6+(2/7) |
12) 220 × 15/66
Solution: = 220 × 15/66 = 10 × 5 = 50 |
Self Practice 8C
1) Neha took 5/9 L of orange juice to the gym. She drank 1/10 of it in the gym. How many litres of orange juice did she drink? How much was left?
Solution:
Neha drank orange juice = 1/10 of 5/9 L = 1/18 L
So, she drank 1/18 L of orange juice.
Orange juice left = 5/9 – 1/18 L = (5×2 – 1)/18 L = 1/2 L
2) One-fourth students of a class are in the animation club. 2/5 of these students are taking part in the movie making competition. What fraction of the class is taking part in the competition?
Solution:
Fraction of students of a class in the animation club = 1/4
Students taking part in the movie making competition is = 2/5 of 1/4 = 2/5 × 1/4 = 1/10
So, fraction of the class is taking part in the competition is 1/10.
3) Krish spent 8/11 of an hour on his Science project. He spent 3/4 of this time building his earth model. How much time did he spend building his earth model?
Solution:
Krish spend the time on his Science project = 8/11 hours
He spent the time on earth model = 3/4 of 8/11 hours = 3/4 × 8/11 hours = 6/11 hours
He spend 6/11 hours to building his earth project.
4) If 1 litre of oil costs ₹ 120+(2/5), how much money will be required to buy 8+(1/2) litres of oil?
Solution:
Cost of 1 litre of oil = ₹ 120+(2/5) = ₹ 602/5
Cost of 8+(1/2) L = 17/2 litres of oil = 17/2 × 602/5 = ₹ 5117/5 = ₹ 1023+(2/5)
Money will be required to buy 8+(1/2) litres of oil is ₹ 1023+(2/5).
5) One dress required 3+(3/4) m cloth. How much cloth will be needed for 16 such dresses?
Solution:
Cloth required for 1 dress = 3+(3/4) m = 15/4 m
Cloth required for 16 dresses = 15/4 × 16 m = 60 m
60 m of cloth will be required for 16 such dresses.
6) Ravi is 1+(1/4) m tall. Dinesh is 1+(1/5) times as tall as Ravi. What is Dinesh’s height?
Solution:
Ravi’s height = 1+(1/4) m = 5/4 m
So, Dinesh’s height = 5/4 × (1+1/5) m = 5/4 × 6/5 m = 3/2 m = 1+(1/2) m
7) Divyam had 2+(6/7) L of milk. He used 1+(1/4) of it to make custard. How much milk did he use?
Solution:
Diviyam had the quantity of milk = 2+(6/7) L = 20/7 L
He used = 1+(1/4) of 20/7 L = 5/4 ×20/7 L = 25/7 L = 3+(4/7) L
He used 3+(4/7) L of milk to make custard.
8) Everyday Ms Sharma spends 1+(1/3) hours teaching children at an orphanage. If she visited the orphanage for 20 days, what time did she spend in all?
Solution:
Per day Ms Sharma spends time for teaching children at an orphanage is = 1+(1/3) hour = 4/3 hour
For 20 days shespend = 20 × 4/3 h = 80/3 h = 26+(2/3) hour
So, Ms Sharma spend 26+(2/3) hour in allfor teaching children at an orphanage.
Self Practice 8D
Find the quotient in the simplest form.
1) 7 ÷ 1/5
Solution: 7 ÷ 1/5 = 7 × 5 = 35 |
2) 18 ÷ 3/5
Solution: 18 ÷ 3/5 = 18 × 5/3 = 30 |
3) 7/9 ÷ 7
Solution: 7/9 ÷ 7 = 7/9 × 1/7 = 1/9 |
4) 8/21 ÷ 7/3
Solution: 8/21 ÷ 7/3 = 8/21 × 3/7 = 8/49 |
5) 7/5 ÷ 7/5
Solution: 7/5 ÷ 7/5 = 7/5 × 5/7 = 1 |
6) 21/28 ÷ 14/3
Solution: 21/28 ÷ 14/3 = 21/28 × 3/14 = 9/56 |
7) 0 ÷ {5+(7/11)}
Solution: 0 ÷ {5+(7/11)} = 0 × 1/{5+(7/11)} = 0 |
8) {4+(1/2)} ÷ 18
Solution: {4+(1/2)} ÷ 18 = 9/2 × 1/18 = 1/4 |
9) 1 ÷ {3+(2/7)}
Solution: 1 ÷ {3+(2/7)} = 1 ÷ 23/7 = 1 × 7/23 = 7/23 |
10) {8+(4/7)} ÷ 4/21
Solution: {8+(4/7)} ÷ 4/21 = 60/7 × 21/4 = 45 |
11) {2+(3/4)} ÷ {1+(1/4)}
Solution: {2+(3/4)} ÷ {1+(1/4)} = 11/4 ÷ 5/4 = 11/4 × 4/5 = 11/5 = 2+(1/5) |
12) 1010 ÷ {3+(1/3)}
Solution: 1010 ÷ {3+(1/3)} = 1010 ÷ 10/3 = 1010 × 3/10 = 303 |
13) {2+(2/3)} ÷ {1+(2/3)}
Solution: {2+(2/3)} ÷ {1+(2/3)} = 8/3 ÷ 5/3 = 8/3 × 3/5 = 8/5 = 1+(3/5) |
14) {16+(1/2)} ÷ {2+(3/4)}
Solution: {16+(1/2)} ÷ {2+(3/4)} = 33/2 ÷ 11/4 = 33/2 × 4/11 = 6 |
15) 16 ÷ {2+(2/15)}
Solution: 16 ÷ {2+(2/15)} = 16 ÷ 32/15 = 16 × 15/32 = 15/2 = 7+(1/2) |
16) {1+(11/15)} ÷ {1+(4/35)}
Solution: {1+(11/15)} ÷ {1+(4/35)} = 26/15 ÷ 39/35 = 26/15 × 35/39 = 14/9 = 1+(5/9) |
17) {3+(3/17)} ÷ {2+(6/51)}
Solution: {3+(3/17)} ÷ {2+(6/51)} = 54/17 ÷ 108/51 = 54/17 × 51/108 = 3/2 = 1+(1/2) |
18) {3+(22/38)} ÷ {1+(18/57)}
Solution: {3+(22/38)} ÷ {1+(18/57)} = 136/38 ÷ 75/57 = 136/38 × 57/75 = 68/25 = 2+(18/25) |
Self Practice 8E
1) How many pieces each 3/5 m long may be cut from a rope 3+(4/5) m long?
Solution:
Length of a rope = 3+(4/5) m = 19/5 m
Length of each piece = 3/5 m
Number of pieces cut from the rope = 19/5 ÷ 3/5 = 19/5 × 5/3 = 19/3 = 6+(1/3)
2) Rita cut 2/3 of a ribbon equally into 6 pieces. What fraction of Rita’s ribbon is one piece?
Solution:
Fraction of Rita’s one piece ribbon is = 2/3 ÷ 6 = 2/3 × 1/6 = 1/9
3) A sweetshop uses 3+(1/7) L of milk everyday to prepare sweets. Find the number of days in which the shop will use 22 L of milk.
Solution:
Per day used of milk to prepare sweets = 3+(1/7) L = 22/7 L
Total used of milk by sweetshop = 22 L
Time required = 22 ÷ 22/7 day = 22 × 7/22 day = 7 day
4) Mansi did her homework in 3+(1/5) hours. She spent the same amount of time in each of her 4 subjects. What time did she spend on each subject?
Solution:
Mansi complete her homework in = 3+(1/5) h = 16/5 h
She spent the same amount of time in each of her 4 subjects
So, she spend the time on each subject is = 16/5 ÷ 4 h = 16/5 × 1/4 h = 4/5 h
5) Rahman plays tennis 6+(1/4) hours a week. If he plays 1+(1/4) hours each day, how many days does he play tennis each week?
Solution:
Rahman play tennis for a week is = 6+(1/4) h = 25/4 h
He plays each day = 1+(1/4) h = 5/4 h
So, Rahman play tennis each week for = 25/4 ÷ 5/4 days = 25/4 × 4/5 days = 5 days
6) Two-thirds of a tank can be filled in 17+(1/3) minutes. How many minutes will be required to fill the whole tank?
Solution:
Time required for filling 2/3 of a tank will be = 17+(1/3) m = 52/3 m
Time required for filling the whole tank = 52/3 ÷ 2/3 h = 52/3 × 3/2 m = 26 m
Self Practice 8F
1) Find:
a) 4/22 of 88
Solution: 4/22 of 88 = 4/22 × 88 = 16 |
b) 3/7 of 35
Solution: 3/7 of 35 = 3/7 × 35 = 15 |
c) 1/8 of 64
Solution: 1/8 of 64 = 1/8 × 64 = 8 |
d) 1/210 of 3000
Solution: 1/210 of 3000 = 1/210 × 3000 = 100/7 = 14+(2/7) |
2) Fill in the blanks.
a) 1/5 of one day = (_) hours.
Solution: 1/5 of one day = 1/5 × 1 day = 1/5 × 24 hour = 24/5 hours = 4+(4/5) hours |
b) 3/5 of 25 km = (_) km.
Solution: 3/5 of 25 km = 3/5 × 25 km = 15 km |
c) 1/6 of two hours = (_) minutes.
Solution: 1/6 of two hours = 1/6 of 2 hours = 1/6 × (2×60) minutes = 20 minutes |
d) 1/4 of dozen = (_) items.
Solution: 1/4 of dozen = 1/4 × 12 items = 3 items |
e) 5/12 of a minute = (_) seconds.
Solution: 5/12 of a minute = 5/12 of 1 minute = 5/12 × 60 seconds = 25 seconds |
f) 22/40 of a metre = (_) centimetres.
Solution: 22/40 of a metre = 22/40 of 1 metre = 22/40 × 100 centimetres = 55 centimetres |
3) Dinesh’s school is 4 km away. He walks 2/5 of the total distance and then boards the schools bus. Find out the distance he walks every day to school.
Solution:
Total distance of Dinesh’s school = 4 km
The distance he walks every day to go to school is = 2/5 of 4 km = 2/5 × 4 km = 8/5 km = 1+(3/5) km
4) There are 80 oranges in a box. Out of these 3/5 oranges are sold. How many oranges are left?
Solution:
Number of total oranges in a box = 80 oranges
Oranges sold = 3/5 of 80 oranges = 3/5 × 80 oranges = 48 oranges
So, the number of ranges left in the box = (80 – 48) oranges = 32 oranges
5) A library has in all 2500 books. Out of these 1/5 books are fictions and 7/10 books are autobiographies. How many more autobiographies are there in the library as compared to fictions books?
Solution:
A library has the total number of books = 2500
Number of books fictions are = 1/5 of 2500 = 1/5 × 2500 = 500
Number of books autobiographies are = 7/10 of 2500 = 7/10 × 2500 = 1750
Difference of fictions books and autobiographies books = 1750 – 500 = 1250
There are more than 1250 autobiographies books present in the library as compared to fictions books.
Also See: Composite Math Ex. 7C to 7E Solutions