Mathematics Part II Solutions Solutions for Class 6 Math Chapter 1 Decimal System are provided here with simple step-by-step explanations. These solutions for Decimal System are extremely popular among Class 6 students for Math Decimal System Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part II Solutions Book of Class 6 Math Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Mathematics Part II Solutions Solutions. All Mathematics Part II Solutions Solutions for class Class 6 Math are prepared by experts and are 100% accurate.
Page No 88:
Question 1:
Can you write the following decimals as fractions?
(i) 0.7
(ii) 0.27
(iii) 0.90
(iv) 0.900
(v) 0.09
(vi) 0.009
(vii) 1.23
(viii) 2.731
(ix) 2.30
(x) 3.05
(xi) 7.003
Answer:
(i) Since there is only one digit after the decimal, the denominator of the equivalent fraction will be 10.
Therefore, 0.7 can be written in fraction as
(ii) Since there are two digits after the decimal, the denominator of the equivalent fraction will be 100.
Therefore, 0.27 can be written in fraction as
(iii) Since there are two digits after the decimal, the denominator of the equivalent fraction will be 100.
Therefore, 0.90 can be written in fraction as
Note: It can also be written as_{ }
(iv) Since there are three digits after the decimal, the denominator of the equivalent fraction will be 1000.
Therefore, 0.900 can be written in fraction as
Note: It can also be written as_{ }
(v) Since there are two digits after the decimal, the denominator of the equivalent fraction will be 100.
Therefore, 0.09 can be written in fraction as
(vi) Since there are three digits after the decimal, the denominator of the equivalent fraction will be 1000.
Therefore, 0.009 can be written in fraction as
(vii) Since there are two digits after the decimal, the denominator of the equivalent fraction will be 100.
Therefore, 1.23 can be written in fraction as
(viii) Since there are three digits after the decimal, the denominator of the equivalent fraction will be 1000.
Therefore, 2.731 can be written in fraction as
(ix) Since there are two digits after the decimal, the denominator of the equivalent fraction will be 100.
Therefore, 2.30 can be written in fraction as
Note: It can also be written as
(x) Since there are two digits after the decimal, the denominator of the equivalent fraction will be 100.
Therefore, 3.05 can be written in fraction as
(xi) Since there are three digits after the decimal, the denominator of the equivalent fraction will be 1000.
Therefore, 7.003 can be written in fraction as
Page No 90:
Question 1:
Three sides of a triangle are of lengths 12.4 centimetres, 8.3 centimetres and 15.9 centimetres. Find the perimeter.
Answer:
The lengths of the sides of the triangle are 12.4 cm, 8.3 cm and 15.9 cm.
The lengths can also be written as:
12.4 cm = 12 cm + 4 mm
8.3 cm = 8 cm + 3 mm
15.9 cm = 15 cm + 9 mm
We know that the perimeter of a triangle is equal to the sum of the lengths of all its sides.
âˆ´Perimeter of the given triangle = 12.4 cm + 8.3 cm + 15.9 cm
The addition can be shown as follows:
cm |
mm |
||||
12 |
4 |
||||
8 |
3 |
||||
+ |
15 |
9 |
|||
36 |
6 |
Therefore, the perimeter of the triangle is 36 cm and 6 mm, i.e. 36.6 cm.
Page No 90:
Question 2:
24.375 kilometres of road in a panchayat needs tarring. Of this, 18.43 kilimetres has been tarred. How much more remains to be tarred?
Answer:
Length of the panchayat road that needs tarring = 24.375 km
Length of the panchayat road that has been tarred = 18.43 km
These lengths can also be written as:
24.375 km = 24 km + 375 m
18.43 km = 18 km + 43 m
Length of the panchayat road that remains to be tarred = Length of the panchayat road that needs tarring âˆ’ Length of the panchayat road that has already been tarred
âˆ´ Length of the panchayat road that remains to be tarred = (24 km + 375 m) âˆ’ (18 km + 43 m)
km |
m |
||||
24 |
375 |
||||
âˆ’ |
18 |
430 |
|||
05 |
945 |
The length of the panchayat road that remains to be tarred is 5 km and 945 m, i.e. 5.945 km.
Page No 90:
Question 3:
The perimeter of a triangle is 29.6 centimetre and two of the sides are of length 11.8 centimetres and 9.4 centimetres. Find the length of the third side.
Answer:
Perimeter of the triangle = 29.6 cm
The lengths of two sides of the triangle are 11.8 cm and 9.4 cm.
These lengths can also be written as:
29.6 cm = 29 cm + 6 mm
11.8 cm = 11 cm + 8 mm
9.4 cm = 9 cm + 4 mm
We know that the perimeter of a triangle is equal to the sum of the lengths of all its sides.
Hence, the length of the third side can be calculated by subtracting the sum of the lengths of the other two sides from the perimeter of the triangle.
âˆ´ Length of the third side = (29 cm + 6 mm) âˆ’ {(11 cm + 8 mm) âˆ’ (9 cm + 4 mm)}
The addition of the sides:
cm |
mm |
||||
11 |
8 |
||||
+ |
9 |
4 |
|||
21 |
2 |
The subtraction of the sum of the lengths of the other two sides from the perimeter:
cm |
mm |
||||
29 |
06 |
||||
âˆ’ |
21 |
02 |
|||
08 |
04 |
Therefore, the length of the third side is 8 cm and 4 mm, i.e. 8.4 cm.
Page No 93:
Question 1:
Babu bought 1.5 metres of cloth for a shirt, at Rs 42.50 rupees a metre. How much did he pay for it?
Answer:
Length of the cloth bought by Babu = 1.5 m
Cost of 1 m length of the cloth = Rs 42.50
Total amount paid by Babu = Cost of 1 m length of the cloth Ã— Length of the cloth bought
= Rs 42.50 Ã— 1.5
= Rs
= Rs
= Rs
= Rs 63.75
Page No 93:
Question 2:
One kilogram of plastic rope is 11.5 metre long. A man bought 3.5 kilograms of it. What is the total length of the rope?
Answer:
Length of 1 kilogram of plastic rope = 11.5 m
Total weight of rope bought by the man = 3.5 kg
Total length of the rope = Length of 1 kilogram of the rope Ã— Total weight of the rope bought
= 11.5 m Ã— 3.5
= m
= m
= m
= 40.25 m
Page No 93:
Question 3:
The price of one litre of petrol is Rs 42.40. A man paid Rs 100 for 2.5 litre of petrol for his bike. Is it more than the required amount or less? By how much?
Answer:
Price of 1 litre of petrol = Rs 42.40
Money paid for 2.5 litres of petrol = Rs 100
Total cost of petrol = Price of 1 litre of petrol Ã— Total litres of petrol bought
= Rs 42.40 Ã— 2.5
= Rs
= Rs
= Rs
= Rs 106
The amount paid by the man is less than the required amount.
Required difference = Rs 106 âˆ’ Rs 100 = Rs 6
Thus, the man paid Rs 6 less than the required amount.
Page No 93:
Question 4:
Using 236 Ã— 43 = 10148, can you find 2.36 Ã— 4.3?
Answer:
Yes, we can find the value of 2.36 Ã— 4.3 from 236 Ã— 43.
Given: 236 Ã— 43 = 10148 â€¦(1)
Now, 2.36 Ã— 4.3
Page No 93:
Question 5:
From 45 Ã— 738 = 33210, can you find 0.45 Ã— 7.38?
Answer:
Yes, we can find the value of 0.45 Ã— 7.38 from 45 Ã— 738.
Given: 45 Ã— 738 = 33210 â€¦(1)
Now, 0.45 Ã— 7.38 =
= 3.3210
Page No 93:
Question 6:
Can you find 847 Ã— 0.312 from 8.47 Ã— 31.2 = 264.264?
Answer:
Yes, we can find the value of 847 Ã— 0.312 from 8.47 Ã— 31.2.
Given: 8.47 Ã— 31.2 = 264.264
â‡’= 264.264
â‡’= 264.264
â‡’ = 264.264 â€¦ (1)
Now, 847 Ã— 0.312 =
_{= 264.264 (Using (1)}
Page No 93:
Question 7:
Which of the following is equal to 12.3 Ã— 7.89?
(i) 1.23 Ã— 7.89
(ii) 0.123 Ã— 789
(iii) 1.23 Ã— 78.9
(iv) 123 Ã— 7.89
(v) 12.3 Ã— 0.789
Answer:
12.3 Ã— 7.89 =
=
=
(i) 1.23 Ã— 7.89 =
It can be observed that the obtained fraction has 10000 as the denominator while the fraction obtained from the question has 1000 as the denominator.
Therefore, 1.23 Ã— 7.89 is not equal to 12.3 Ã— 7.89.
(ii) 0.123 Ã— 789 =
It can be observed that the obtained fraction has 1000 as the denominator, which is same as the denominator of the fraction obtained from the question. Also, the numerators of both the fractions are same.
Therefore, 0.123 Ã— 789 is equal to 12.3 Ã— 7.89.
(iii) 1.23 Ã— 78.9 =
It can be observed that the obtained fraction has 1000 as the denominator, which is same as the denominator of the fraction obtained from the question Also, the numerators of both the fractions are same.
Therefore, 1.23 Ã— 78.9 is equal to 12.3 Ã— 7.89.
(iv) 123 Ã— 7.89 =
It can be observed that the obtained fraction has 100 as the denominator while the fraction obtained from the question has 1000 as the denominator.
Therefore, 123 Ã— 7.89 is not equal to 12.3 Ã— 7.89.
(v) 12.3 Ã— 0.789 =
It can be observed that the obtained fraction has 10000 as the denominator while the fraction obtained from the question has 1000 as the denominator.
Therefore, 12.3 Ã— 0.789 is not equal to 12.3 Ã— 7.89.
Page No 95:
Question 1:
The cost of 5 pens is 42.50 rupees. What is the cost of one pen?
Answer:
Cost of five pens = Rs 42.50
âˆ´ Cost one pen = Rs 42.50 Ã· 5
The division can be done as follows:
Thus, the cost of one pen is Rs 8.50.
Page No 95:
Question 2:
If a rod of length 71.2 metres is cut into 8 equal pieces, what is the length of each piece?
Answer:
Length of the rod = 71.2 m
Total number of equal divisions of the rod = 8
Length of each piece = 71.2 m Ã· 8
The division can be done as follows:
Thus, the length of each piece is 8.9 m.
Page No 96:
Question 1:
Try the following.
(i) 43.6 Ã· 8
(ii) 49.4 Ã· 4
(iii) 7.1 Ã· 2
(iv) 100.31 Ã· 7
(v) 53.28 Ã· 4
(vi) 207.42 Ã· 6
(vii) 9.256 Ã· 8
(viii) 75.276 Ã· 9
Answer:
(i) 43.6 Ã· 8
The division of 43.6 by 8 can be done as follows:
âˆ´ 43.6 Ã· 8 = 5.45
(ii) 49.4 Ã· 4
The division of 49.4 by 4 can be done as follows:
âˆ´ 49.4 Ã· 4 = 12.35
(iii) 7.1 Ã· 2
The division of 7.1 by 2 can be done as follows:
âˆ´ 7.1 Ã· 2 = 3.55
(iv) 100.31 Ã· 7
The division of 100.31 by 7 can be done as follows:
âˆ´ 100.31 Ã· 7 = 14.33
(v) 53.28 Ã· 4
The division of 53.28 by 4 can be done as follows:
âˆ´ 53.28 Ã· 4 = 13.32
(vi)
207.42 Ã· 6
The division of 207.42 by 6 can be done as follows:
âˆ´ 207.42 Ã· 6 = 34.57
(vii)
9.256 Ã· 8
The division of 9.256 by 8 can be done as follows:
âˆ´ 9.256 Ã· 8 = 1.157
(viii)
75.276 Ã· 9
The division of 75.276 by 9 can be done as follows:
âˆ´ 75.276 Ã· 9 = 8.364
Page No 98:
Question 1:
Try the following problems.
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
Answer:
(i)
(ii)
The division of 745 by 11 can be done as follows:
âˆ´ = 67.727â€¦.
(iii)
The division of 642.84 by 12 can be done as follows:
âˆ´ = 53.57
(iv)
The division of 3890 by 2 can be done as follows:
âˆ´ = 1945
(v)
The division of 546 by 13 can be done as follows:
âˆ´ = 42
(vi)
The division of 627.3 by 5 can be done as follows:
âˆ´= 125.46
(vii)
Page No 98:
Question 2:
Express the following fractions in the decimal form.
(i)
(ii)
(iii)
(iv)
(v)
(vi)
Answer:
(i)
Since has 10 as the denominator, can be written as 0.4.
(ii)
The division of 25 by 2 can be done as follows:
(iii)
The division of 75 by 2 can be done as follows:
(iv)
The division of 125 by 2 can be done as follows:
(v)
Since has 100 as the denominator, can be written as 0.04.
(vi)
The division of 5 by 8 can be done as follows:
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