Mathematics Solutions Solutions for Class 6 Maths Chapter 5 Decimal Fractions are provided here with simple step-by-step explanations. These solutions for Decimal Fractions are extremely popular among class 6 students for Maths Decimal Fractions Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Solutions Book of class 6 Maths Chapter 5 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Solutions Solutions. All Mathematics Solutions Solutions for class 6 Maths are prepared by experts and are 100% accurate.
Page No 30:
Question 1:
In the table below, write the place value of each of the digits in the number 378.025.
Place | Hundreds | Tens | Units | Tenths | Hundredths | Thousandths |
100 | 10 | 1 | ||||
Digit | 3 | 7 | 8 | 0 | 2 | 5 |
Place value | 300 |
Answer:
Place | Hundreds | Tens | Units | Tenths | Hundredths | Thousandths |
100 | 10 | 1 | ||||
Digit | 3 | 7 | 8 | 0 | 2 | 5 |
Place value | 300 | 7 × 10 = 70 | 8 × 1 = 8 |
Page No 30:
Question 2:
Solve.
(1) 905.5 + 27.197
(2) 39 + 700.65
(3) 40 + 27.7 + 2.451
Answer:
(i)
∴ 905.5 + 27.197 = 932.697
(ii)
∴ 39 + 700.65 = 739.65
(iii)
∴ 40 + 27.7 + 2.451 = 70.151
Page No 30:
Question 3:
Subtract.
(1) 85.96 – 2.345
(2) 632.24 – 97.45
(3) 200.005 – 17.186
Answer:
(i)
∴ 85.96 − 2.345 = 82.615
(ii)
∴ 632.24 − 97.45 = 534.79
(iii)
∴ 200.005 − 17.186 = 182.819
Page No 31:
Question 4:
Avinash travelled 42 km 365 m by bus, 12 km 460 m by car and walked 640 m. How many kilometres did he travel altogether? (Write your answer in decimal fractions.)
Answer:
Distance travelled by bus = 42 km 365 m = 42 km + 365 m = 42 km + km = 42 km + 0.365 km = 42.365 km (1 km = 1000 m)
Distance travelled by car = 12 km 460 m = 12 km + 460 m = 12 km + km = 12 km + 0.460 km = 12.460 km
Distance travelled by walking = 640 m = km = 0.640 km
∴ Total distance travelled altogether
= Distance travelled by bus + Distance travelled by car + Distance travelled by walking
= 42.365 km + 12.460 km + 0.640 km
= 55.465 km
Thus, the total distance travelled by Avinash altogether is 55.465 km.
Page No 31:
Question 5:
Ayesha bought 1.80 m of cloth for her salwaar and 2.25 m for her kurta. If the cloth costs 120 rupees per metre, how much must she pay the shopkeeper?
Answer:
Length of cloth bought for salwaar = 1.80 m
Length of cloth bought for kurta = 2.25 m
∴ Total length of cloth bought
= Length of cloth bought for salwaar + Length of cloth bought for kurta
= 1.80 m + 2.25 m
= 4.05 m
Rate of cloth = Rs 120/m
∴ Amount paid to the shopkeeper
= Total length of cloth bought × Rate of cloth
= 4.05 m × â¹ 120/m
= â¹ 486
Thus, Ayesha paid â¹ 486 to the shopkeeper.
Page No 31:
Question 6:
Sujata bought a watermelon weighing 4.25 kg and gave 1 kg 750g to the children in her neighbourhood. How much of it does she have left?
Answer:
Total weight of the watermelon = 4.25 kg
Weight of the watermelon given to the children = 1 kg 750 g = 1 kg + 750 g = 1 kg + kg = 1 kg + 0.75 kg = 1.75 kg
∴ Weight of the watermelon left with her
= Total weight of the watermelon − Weight of the watermelon given to the children
= 4.25 kg − 1.75 kg
= 2.5 kg
Thus, the weight of watermelon left with Sujata is 2.5 kg.
Page No 31:
Question 7:
Anita was driving at a speed of 85.6 km per hour. The road had a speed limit of 55 km per hour. By how much should she reduce her speed to be within the speed limit?
Answer:
Original driving speed of Anita = 85.6 km/h
Speed limit for driving = 55 km/h
∴ Speed reduced by her to be within the speed limit
= Original driving speed of Anita − Speed limit for driving
= 85.6 km/h − 55 km/h
= 30.6 km/h
Thus, Anita needs to reduce her speed by 30.6 km/h to be within the speed limit.
Page No 32:
Question 1:
Write the proper number in the empty boxes.
(1)
(2)
(3)
(4)
Answer:
(1)
(2)
(3)
(4)
Page No 32:
Question 2:
Convert the common fractions into decimal fractions.
(1)
(2)
(3)
(4)
(5)
(6)
(7)
Answer:
(1)
(2)
(3)
(4)
(5)
(6)
(7)
Page No 32:
Question 3:
Convert the decimal fractions into common fractions.
(1) 27.5
(2) 0.007
(3) 90.8
(4) 39.15
(5) 3.12
(6) 70.400
Answer:
(1) 27.5 =
(2) 0.007 =
(3) 90.8 =
(4) 39.15 =
(5) 3.12 =
(6) 70.400 = 70.4 =
Page No 33:
Question 1:
If, 317 × 45 = 14265, then 3.17 × 4.5 = ?
Answer:
Page No 33:
Question 2:
If, 503 × 217 = 109151, then 5.03 × 2.17 = ?
Answer:
Page No 33:
Question 3:
Multiply.
(1) 2.7 × 1.4
(2) 6.17 × 3.9
(3) 0.57 × 2
(4) 5.04 × 0.7
Answer:
(i)
(ii)
(iii)
(iv)
Page No 34:
Question 4:
Virendra bought 18 bags of rice, each bag weighing 5.250 kg. How much rice did he buy altogether? If the rice costs 42 rupees per kg, how much did he pay for it?
Answer:
Weight of each bag of rice = 5.250 kg
Number of bags of rice bought = 18
∴ Weight of rice bought altogether = Weight of each bag of rice × Number of bags of rice bought
= 5.250 kg × 18
=
=
=
= 94.5 kg
Rate of rice = â¹ 42/kg
∴ Total amount paid for the rice = Weight of rice bought altogether × Rate of rice
= 94.5 kg × â¹ 42/kg
=
=
=
= â¹ 3,969
Thus, Virendra bought 94.5 kg of rice altogether and the total amount paid by him is â¹ 3,969.
Page No 34:
Question 5:
Vedika has 23.50 metres of cloth. She used it to make 5 curtains of equal size. If each curtain required 4 metres 25 cm to make, how much cloth is left over?
Answer:
Total length of the cloth = 23.50 m
Length of cloth required to make each curtain = 4 m 25 cm = 4 m + 25 cm = 4 m + m = 4 m + 0.25 m = 4.25 m (1 m = 100 cm)
Number of curtains made of equal size = 5
∴ Length of cloth required to make 5 curtains = Length of cloth required to make each curtain × 5
= 4.25 m × 5
=
=
=
= 21.25 m
∴ Amount of cloth left = Total length of the cloth − Length of cloth required to make 5 curtains
= 23.50 m − 21.25 m
= 2.25 m
Thus, the length of the cloth left over is 2.25 m.
Page No 34:
Question 1:
Carry out the following divisions.
(1) 4.8 ÷ 2
(2) 17.5 ÷ 5
(3) 20.6 ÷ 2
(4) 32.5 ÷ 25
Answer:
(1) 4.8 ÷ 2 =
(2) 17.5 ÷ 5 =
(3) 20.6 ÷ 2 =
(4) 32.5 ÷ 25 =
Page No 34:
Question 2:
A road is 4 km 800 m long. If trees are planted on both its sides at intervals of 9.6 m, how many trees were planted?
Answer:
Total length of the road = 4 km 800 m = 4 km + 800 m = 4000 m + 800 m = 4800 m (1 km = 1000 m)
Distance between two plants = 9.6 m
∴ Number of trees planted on each side of the road = (Total length of the road ÷ Distance between two plants) + 1
= (4800 m ÷ 9.6 m) + 1
= + 1
= + 1
= 500 + 1
= 501
∴ Total number of trees planted on both sides of the road = Number of trees planted on each side of the road × 2 = 501 × 2 = 1002
Thus, the number of trees planted on both sides of the road is 1002.
Page No 34:
Question 3:
Pradnya exercises regularly by walking along a circular path on a field. If she walks a distance of 3.825 km in 9 rounds of the path, how much does she walk in one round?
Answer:
Total distance walked by Pradnya in 9 rounds of the path = 3.825 km
∴ Distance walked by Pradnya in one round of the path = Total distance walked by Pradnya in 9 rounds of the path ÷ 9
= 3.825 km ÷ 9
=
=
=
= 0.425 km
Thus, the distance walked by Pradnya in one round is 0.425 km.
Page No 34:
Question 4:
A pharmaceutical manufacturer bought 0.25 quintal of hirada, a medicinal plant, for 9500 rupees. What is the cost per quintal of hirada ? (1quintal = 100 kg)
Answer:
Amount paid for 0.25 quintal of hirada = â¹ 9,500
∴ Cost per quintal of hirada = Amount paid for 0.25 quintal of hirada ÷ 0.25
= â¹ 9,500 ÷ 0.25
=
=
= 380 × 100
= â¹ 38,000
Thus, the cost per quintal of hirada is â¹ 38,000.
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