Decimals (Exercise 7.9) RD Sharma Solutions | Mathematics (Maths) Class 6 PDF Download

 Page 1


 
 
 
 
 
 
Exercise 7.9                                                                             page: 7.31 
1. Subtract: 
 
Solution: 
 
(i) We know that 
46.23 – 37.5 = 8.73 
 
(ii) We know that  
128.4 – 53.05 = 75.35 
 
(iii) We know that 
45.03 – 27.8 = 17.23 
 
(iv) We know that  
23.93 – 5.946 = 17.984 
 
2. Find the value of: 
(i) 9.756 – 6.28 
(ii) 21.05 – 15.27 
(iii) 18.5 – 6.79 
(iv) 48.1 – 0.37 
(v) 108.032 – 86.8 
(vi) 91.001 – 72.9 
(vii) 32.7 – 25.86 
(viii) 100 – 26.32 
Solution: 
 
(i) 9.756 – 6.28 
We know that 
9.756 – 6.280 = 3.476 
 
(ii) 21.05 – 15.27 
We know that 
21.05 – 15.27 = 5.78 
 
(iii) 18.5 – 6.79 
We know that 
18.50 – 6.79 = 11.71 
 
(iv) 48.1 – 0.37 
We know that 
48.10 – 0.37 = 47.73 
 
(v) 108.032 – 86.8 
We know that 
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Exercise 7.9                                                                             page: 7.31 
1. Subtract: 
 
Solution: 
 
(i) We know that 
46.23 – 37.5 = 8.73 
 
(ii) We know that  
128.4 – 53.05 = 75.35 
 
(iii) We know that 
45.03 – 27.8 = 17.23 
 
(iv) We know that  
23.93 – 5.946 = 17.984 
 
2. Find the value of: 
(i) 9.756 – 6.28 
(ii) 21.05 – 15.27 
(iii) 18.5 – 6.79 
(iv) 48.1 – 0.37 
(v) 108.032 – 86.8 
(vi) 91.001 – 72.9 
(vii) 32.7 – 25.86 
(viii) 100 – 26.32 
Solution: 
 
(i) 9.756 – 6.28 
We know that 
9.756 – 6.280 = 3.476 
 
(ii) 21.05 – 15.27 
We know that 
21.05 – 15.27 = 5.78 
 
(iii) 18.5 – 6.79 
We know that 
18.50 – 6.79 = 11.71 
 
(iv) 48.1 – 0.37 
We know that 
48.10 – 0.37 = 47.73 
 
(v) 108.032 – 86.8 
We know that 
 
 
 
 
 
 
108.032 – 86.800 = 21.232 
  
(vi) 91.001 – 72.9 
We know that 
91.001 – 72.900 = 18.101 
 
(vii) 32.7 – 25.86 
We know that 
32.70 – 25.86 = 6.84 
 
(viii) 100 – 26.32 
We know that 
100 – 26.32 = 73.68 
 
3. The sum of two numbers is 100. If one of them is 78.01, find the other. 
Solution: 
 
One of the number = 78.01 
Sum of two numbers = 100 
Consider x as the other number 
It can be written as 
78.01 + x = 100 
On further calculation 
x = 100 – 78.01 
By subtraction 
x = 21.99 
 
Hence, the other number is 21.99. 
 
4. Waheeda’s school is at a distance of 5 km 350 m from her house. She travels 1 km 70 m on foot and the 
rest she travels by bus. How much distance does she travel by bus? 
Solution: 
 
Distance of school from house = 5 km 350 m = 5.350 km 
Distance travelled on foot = 1 km 70 m = 1.070 km 
Consider x km as the distance travelled by bus 
It can be written as 
1.070 + x = 5.350 
On further calculation 
x = 5.350 – 1.070 
So we get 
x = 4.280 km 
 
Hence, the distance travelled by bus is 4.280 km. 
 
5. Raju bought a book for Rs 35.65. He gave Rs 50 to the shopkeeper. How much money did he get back 
from the shopkeeper? 
Solution: 
 
Cost of book = Rs 35.65 
Page 3


 
 
 
 
 
 
Exercise 7.9                                                                             page: 7.31 
1. Subtract: 
 
Solution: 
 
(i) We know that 
46.23 – 37.5 = 8.73 
 
(ii) We know that  
128.4 – 53.05 = 75.35 
 
(iii) We know that 
45.03 – 27.8 = 17.23 
 
(iv) We know that  
23.93 – 5.946 = 17.984 
 
2. Find the value of: 
(i) 9.756 – 6.28 
(ii) 21.05 – 15.27 
(iii) 18.5 – 6.79 
(iv) 48.1 – 0.37 
(v) 108.032 – 86.8 
(vi) 91.001 – 72.9 
(vii) 32.7 – 25.86 
(viii) 100 – 26.32 
Solution: 
 
(i) 9.756 – 6.28 
We know that 
9.756 – 6.280 = 3.476 
 
(ii) 21.05 – 15.27 
We know that 
21.05 – 15.27 = 5.78 
 
(iii) 18.5 – 6.79 
We know that 
18.50 – 6.79 = 11.71 
 
(iv) 48.1 – 0.37 
We know that 
48.10 – 0.37 = 47.73 
 
(v) 108.032 – 86.8 
We know that 
 
 
 
 
 
 
108.032 – 86.800 = 21.232 
  
(vi) 91.001 – 72.9 
We know that 
91.001 – 72.900 = 18.101 
 
(vii) 32.7 – 25.86 
We know that 
32.70 – 25.86 = 6.84 
 
(viii) 100 – 26.32 
We know that 
100 – 26.32 = 73.68 
 
3. The sum of two numbers is 100. If one of them is 78.01, find the other. 
Solution: 
 
One of the number = 78.01 
Sum of two numbers = 100 
Consider x as the other number 
It can be written as 
78.01 + x = 100 
On further calculation 
x = 100 – 78.01 
By subtraction 
x = 21.99 
 
Hence, the other number is 21.99. 
 
4. Waheeda’s school is at a distance of 5 km 350 m from her house. She travels 1 km 70 m on foot and the 
rest she travels by bus. How much distance does she travel by bus? 
Solution: 
 
Distance of school from house = 5 km 350 m = 5.350 km 
Distance travelled on foot = 1 km 70 m = 1.070 km 
Consider x km as the distance travelled by bus 
It can be written as 
1.070 + x = 5.350 
On further calculation 
x = 5.350 – 1.070 
So we get 
x = 4.280 km 
 
Hence, the distance travelled by bus is 4.280 km. 
 
5. Raju bought a book for Rs 35.65. He gave Rs 50 to the shopkeeper. How much money did he get back 
from the shopkeeper? 
Solution: 
 
Cost of book = Rs 35.65 
 
 
 
 
 
 
Amount given = Rs 50 
So the balance returned = 50 – 35.65 = Rs 14.35 
 
Hence, the balance returned by the shopkeeper is Rs 14.35. 
 
6. Ruby bought a watermelon weighing 5 kg 200 g. Out of this she gave 2 kg 750 g to her neighbour. What 
is the weight of the watermelon left with Ruby? 
Solution: 
 
Weight of watermelon bought by Ruby = 5 kg 200 g = 5.200 kg 
Weight of watermelon Ruby gave to neighbour = 2 kg 750 g = 2.750 kg 
So the weight of watermelon left with Ruby = Weight of watermelon bought by Ruby – Weight of watermelon 
Ruby gave to neighbour 
We get 
Weight of watermelon left with Ruby = 5.200 – 2.750 = 2.450 kg 
 
Hence, the weight of watermelon left with Ruby is 2.450 kg. 
 
7. Victor drove 89.050 km on Saturday and 73.9 km on Sunday. How many kilometres more did he drive 
on Sunday? 
Solution: 
 
Distance travelled by Victor on Saturday = 89.050 km 
Distance travelled by Victor on Sunday = 73.9 km 
So the distance travelled more by Victor on Saturday = 89.050 – 73.9 = 15.15 km 
 
Hence, Victor drove 15.15 km more on Saturday. 
 
8. Raju bought a book for Rs 35.65. He gave Rs 50 to the shopkeeper. How much money did he get back 
from the shopkeeper? 
Solution: 
 
Cost of the book = Rs 35.65 
Amount given = Rs 50 
So the balance retuned = 50 – 35.65 = Rs 14.35 
 
Hence, the shopkeeper returned back Rs 14.35. 
 
9. Gopal travelled 125.5 km by bus, 14.25 km by pony and the rest of distance to Kedarnath on foot. If he 
covered a total distance of 15 km, how much did he travel on foot? 
Solution: 
 
Distance travelled by Gopal by bus = 125.5 km 
Distance travelled by Gopal by pony = 14.25km 
Consider x km as the distance travelled on foot 
We know that 
Total distance = Distance travelled by Gopal by bus + Distance travelled by Gopal by pony + Distance travelled 
by Gopal on foot 
By substituting the values 
150 = 125.5 + 14.25 + x 
Page 4


 
 
 
 
 
 
Exercise 7.9                                                                             page: 7.31 
1. Subtract: 
 
Solution: 
 
(i) We know that 
46.23 – 37.5 = 8.73 
 
(ii) We know that  
128.4 – 53.05 = 75.35 
 
(iii) We know that 
45.03 – 27.8 = 17.23 
 
(iv) We know that  
23.93 – 5.946 = 17.984 
 
2. Find the value of: 
(i) 9.756 – 6.28 
(ii) 21.05 – 15.27 
(iii) 18.5 – 6.79 
(iv) 48.1 – 0.37 
(v) 108.032 – 86.8 
(vi) 91.001 – 72.9 
(vii) 32.7 – 25.86 
(viii) 100 – 26.32 
Solution: 
 
(i) 9.756 – 6.28 
We know that 
9.756 – 6.280 = 3.476 
 
(ii) 21.05 – 15.27 
We know that 
21.05 – 15.27 = 5.78 
 
(iii) 18.5 – 6.79 
We know that 
18.50 – 6.79 = 11.71 
 
(iv) 48.1 – 0.37 
We know that 
48.10 – 0.37 = 47.73 
 
(v) 108.032 – 86.8 
We know that 
 
 
 
 
 
 
108.032 – 86.800 = 21.232 
  
(vi) 91.001 – 72.9 
We know that 
91.001 – 72.900 = 18.101 
 
(vii) 32.7 – 25.86 
We know that 
32.70 – 25.86 = 6.84 
 
(viii) 100 – 26.32 
We know that 
100 – 26.32 = 73.68 
 
3. The sum of two numbers is 100. If one of them is 78.01, find the other. 
Solution: 
 
One of the number = 78.01 
Sum of two numbers = 100 
Consider x as the other number 
It can be written as 
78.01 + x = 100 
On further calculation 
x = 100 – 78.01 
By subtraction 
x = 21.99 
 
Hence, the other number is 21.99. 
 
4. Waheeda’s school is at a distance of 5 km 350 m from her house. She travels 1 km 70 m on foot and the 
rest she travels by bus. How much distance does she travel by bus? 
Solution: 
 
Distance of school from house = 5 km 350 m = 5.350 km 
Distance travelled on foot = 1 km 70 m = 1.070 km 
Consider x km as the distance travelled by bus 
It can be written as 
1.070 + x = 5.350 
On further calculation 
x = 5.350 – 1.070 
So we get 
x = 4.280 km 
 
Hence, the distance travelled by bus is 4.280 km. 
 
5. Raju bought a book for Rs 35.65. He gave Rs 50 to the shopkeeper. How much money did he get back 
from the shopkeeper? 
Solution: 
 
Cost of book = Rs 35.65 
 
 
 
 
 
 
Amount given = Rs 50 
So the balance returned = 50 – 35.65 = Rs 14.35 
 
Hence, the balance returned by the shopkeeper is Rs 14.35. 
 
6. Ruby bought a watermelon weighing 5 kg 200 g. Out of this she gave 2 kg 750 g to her neighbour. What 
is the weight of the watermelon left with Ruby? 
Solution: 
 
Weight of watermelon bought by Ruby = 5 kg 200 g = 5.200 kg 
Weight of watermelon Ruby gave to neighbour = 2 kg 750 g = 2.750 kg 
So the weight of watermelon left with Ruby = Weight of watermelon bought by Ruby – Weight of watermelon 
Ruby gave to neighbour 
We get 
Weight of watermelon left with Ruby = 5.200 – 2.750 = 2.450 kg 
 
Hence, the weight of watermelon left with Ruby is 2.450 kg. 
 
7. Victor drove 89.050 km on Saturday and 73.9 km on Sunday. How many kilometres more did he drive 
on Sunday? 
Solution: 
 
Distance travelled by Victor on Saturday = 89.050 km 
Distance travelled by Victor on Sunday = 73.9 km 
So the distance travelled more by Victor on Saturday = 89.050 – 73.9 = 15.15 km 
 
Hence, Victor drove 15.15 km more on Saturday. 
 
8. Raju bought a book for Rs 35.65. He gave Rs 50 to the shopkeeper. How much money did he get back 
from the shopkeeper? 
Solution: 
 
Cost of the book = Rs 35.65 
Amount given = Rs 50 
So the balance retuned = 50 – 35.65 = Rs 14.35 
 
Hence, the shopkeeper returned back Rs 14.35. 
 
9. Gopal travelled 125.5 km by bus, 14.25 km by pony and the rest of distance to Kedarnath on foot. If he 
covered a total distance of 15 km, how much did he travel on foot? 
Solution: 
 
Distance travelled by Gopal by bus = 125.5 km 
Distance travelled by Gopal by pony = 14.25km 
Consider x km as the distance travelled on foot 
We know that 
Total distance = Distance travelled by Gopal by bus + Distance travelled by Gopal by pony + Distance travelled 
by Gopal on foot 
By substituting the values 
150 = 125.5 + 14.25 + x 
 
 
 
 
 
 
On further calculation 
x = 150 – 125.5 – 14.25 
We get 
x = 10.25 km 
 
Hence, the distance travelled by Gopal on foot is 10.25 km. 
 
10. Tina had 20 m 5 cm long cloth. She cuts 4 m 50 cm length of cloth from this for making a curtain. How 
much cloth is left with her? 
Solution: 
 
Length of cloth Tina had = 20 m 5 cm = 20.05 m 
Length of cloth cut to make curtain = 4m 50 cm = 4.50 m 
So the length of cloth left with her = Length of cloth Tina had - Length of cloth cut to make curtain 
By substituting values 
The length of cloth left with her = 20.05 – 4.50 = 15.55 m 
 
Hence, the length of cloth left with Tina is 15.55 m. 
 
11. Vineeta bought a book for Rs 18.90, a pen for Rs 8.50 and some papers for Rs 5.05. She gave fifty rupee 
to the shopkeeper. How much balance did she get back? 
Solution: 
 
Cost of book = Rs 18.90 
Cost of pen = Rs 8.50 
Cost of papers = Rs 5.05 
So the total cost = 18.90 + 8.50 + 5.05 = Rs 32.45 
Amount given to the shopkeeper = Rs 50 
So the balance returned back = 50 – 32.45 = Rs 17.55 
 
Hence, the shopkeeper returned back Rs 17.55. 
 
12. Tanuj walked 8.62 km on Monday, 7.05 km on Tuesday and some distance on Wednesday. If he walked 
21.01 km in three days, how much distance did he walk on Wednesday? 
Solution: 
 
Distance walked by Tanuj on Monday = 8.62 km  
Distance walked by Tanuj on Tuesday = 7.05 km 
Consider x km as the distance walked by Tanuj on Wednesday 
It can be written as 
Total distance = Distance walked by Tanuj on Monday + Distance walked by Tanuj on Tuesday + Distance 
walked by Tanuj on Wednesday 
By substituting the values 
21.01 = 8.62 + 7.05 + x 
On further calculation 
x = 21.01 – 8.62 – 7.05 
So we get 
x = 5.34 km 
 
Hence, the distance walked by Tanuj on Wednesday is 5.34 km.