Chapter 1 Simplification Decimal Fraction Examples
Question 1. Express the following numbers in decimal fractions:
1. 2 + \(\frac{3}{10}\)
Class 6 West Bengal Board Math Solution :
2 + \(\frac{3}{10}\) = 2 + 0.3
= 2.3
Class 6 West Bengal Board Math Solution
2. 10 + 7 + \(\frac{8}{1000}\)
Solution:
10 + 7 + \(\frac{8}{1000}\)
= 17 +.008
= 17.008
WBBSE Class 6 Decimal Fraction Simplification Notes
3. 6 one-tenths
Solution:
6 one-tenths 0.6
Wbbse Class 6 Maths Solutions
4. 9 one-hundredths
Solution:
9 one-hundreths =0.09
Read And Learn More: WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Solved Problems
5. Four one-thousandths
Class 6 West Bengal Board Math Solution:
Four one-thousandths = 0.004
6. Two hundred three decimal four five
Solution:
Two hundred three decimal four five = 203.45
7. Four thousand two units five one-thousandths
Class 6 West Bengal Board Math Solution:
Four thousand two units five one-thousandth = 4002.005
8. 400 + 50 + \(\frac{9}{100}\) + \(\frac{1}{1000}\)
Solution:
400 + 50 + \(\frac{9}{100}\) + \(\frac{1}{1000}\)
= 450 +0.09 + 0·001 = 450.091
9. Two lac two units four 100 1000 one-thousandths
Solution:
Two lac two units four one-thousandths = 20002.004
10. Six hundred twenty-nine decimal zero five.
Class 6 West Bengal Board Math Solution :
Six hundred twenty-nine decimal zero five = 629.005
Short Questions on Simplifying Decimal Fractions
Question 2. Put the following decimal numbers in the respective places of the place value table and then express them in words:
1. 27.9
2. 1.28
3. 65.134
4. 42.009
5. 38.205
6. 4003.08
7. 712.5
8. 45.06
Solution:
Expression in Words:
1. 27.9 = Twenty-seven decimal nine or Twenty-seven nine one-tenths.
2. 128 = One unit two one-tenths eight one-hundred.
3. 65.134 = Sixty-five one one-tenths three one-hundredths four one-thousandths.
4. 42.009 = Forty-two nine one-thousandths or forty-two decimal zero nine.
5. 38.205 = Thirty-eight two one-tenths five one-thousandths
6. 4003.08 = Four thousand three eight one-hundredths.
7. 712.5 = Seven hundred twelve five one-tenths.
8. 45.06 = Forty-five six one-hundredths or forty-five decimal zero six.
Common Questions About Decimal Fraction Simplification
Question 3. Complete the following tables:
Class 6 West Bengal Board Math Solution:
Question 4. Convert the following decimal fractions to vulgar fractions:
1. 0.3
Solution:
0.3 = \(\frac{3}{10}\)
2. 0.039
Solution:
0.039 = \(\frac{39}{1000}\)
Question 5. Put >, = or < in the blank spaces of the following:
1. 5.0 0.5
Solution:
The left-hand number is 5.0, and its integral (or whole part) part is 5.
The right-hand number is 0.5, its integral part is 0.
∴ 5.0 0.5.
2. 72.1 72.10
Solution:
The whole or integral parts of both the left-hand number and right-hand number are the same (each equal to 72) and also in the decimal part both the numbers have the same one-tenths and so both the numbers are equal.
∴ 72.1 72.10.
3. 68.5 68.52
Solution:
The integral part of both the numbers is the same; in the decimal part both the numbers have the same one-tenths (each equal to 5) but the one-hundredth part of the left-hand number is 0 and in the right-hand number, the one-hundredth part is 2.
So the right-hand number is greater than the left.
∴ 68.5 68.52.
Practice Problems on Simplifying Decimal Fractions
4. 72.93 729.3
Solution:
The integral part of the left-hand number is 72 and that of the right-hand number is 729.
∴ 72.97 729.3.
5. 42.6 42.600
Solution:
Both numbers have same the same integral part as well as the decimal part.
So the numbers are the same.
∴ 42.6 42.600.
6. 2.33 3.22
Solution:
The integral part of the left-hand number is 2 and that in the right-hand number is 3.
∴ 2.33 3.22.
7. 92.4 924.00
Solution:
924 = 924.0 = 924.00
∴ 924 924.00.
8. 10.01 10.10
Solution:
Since both the numbers have the same integral parts and the one-tenths of the left-hand number is 0 but the right-hand number is 1.
∴ 10.01 10.10.
Question 6. Arrange the following numbers in ascending order of magnitude:
1. 0.534, 0.52, 5.34, 0.513
Class 6 West Bengal Board Math Solution:
Given
0.534, 0.52, 5.34, 0.513
The integral parts of the given numbers are 0, 0, 5, and 0.
∴ 5.34 is the greatest number.
Again,
0.534 = 0.534
0.52 = 0.520
0.513 = 0.513
As, 513 < 520 < 534
∴ 513 < 520 < 534.
So arranging the given numbers in ascending order of magnitude, we get,
0.513, 0.52, 0.534; 5.34.
2. 0.536, 0.335, 0.3354,. 0.52.
Solution:
Given
0.536, 0.335, 0.3354,. 0.52
The integral parts of the given numbers are the same, we have,
0.536 = 0.5360
0.335 = 0.3350
0.3354 = 0.3354
0.52 = 0.5200
∴ As, 3350 < 3354 < 5200 < 5360
0.0335 < 0.3354 < 0·5200 < 0.536
or, 0.335 < 0.3354 <0.52 < 0.536
So arranging the given numbers in ascending order of magnitude we get,
0.335, 0.3354, 0.52, 0:536
Question 7. Arrange the following numbers in descending order of magnitude:
1. 13.3, 11.3, 1.33, 2.31;
Class 6 West Bengal Board Math Solution
13 \cdot 3=13 \cdot 30 \\
11 \cdot 3=11 \cdot 30 \\
1 \cdot 33=1 \cdot 33 \\
2 \cdot 31=2 \cdot 31
\end{array}\right\}\)
The integral parts of the given numbers are
13, 11, 1, 2.
∵ 13 > 11 2 > 1
∴ 13.30 11.30 > 2.31> 1-33
Arranging the numbers in descending order of magnitudes, we get,
13.3, 11.3, 2.31, 1.33.
2. 3.007, 3.07, 37.30, 7.13
Class 6 West Bengal Board Math Solution:
\(\left.\begin{array}{l}3 \cdot 007=3 \cdot 007 \\
3 \cdot 07=3.070 \\
37 \cdot 30=37 \cdot 300 \\
7 \cdot 13=7 \cdot 130
\end{array}\right\}\)
3.007, 3.07, 37.30, 7.13
We get, 37.300 > 7.130
The integral parts of the given numbers are
3, 3, 37, 7
∵ 37 > 7 > 3
We get,
∴ 37.300 > 7.130 > 3.070 > 3.007 ( ∵ 3.070> 3.007)
or, 37.30 > 7.13 > 3.07 > 3.007.
So arranging the given numbers in descending order of magnitudes, we get 37.30, 7.13, 3.07, 3.007.
Important Definitions Related to Decimal Fractions
Question 8. Expand the following numbers according to the place value of the digits:
1. 101.153
Class 6 West Bengal Board Math Solution:
101-153 = 101 + \(\frac{1}{10}\) + \(\frac{5}{100}\) + \(\frac{3}{1000}\)
= 100 + 1 + \(\frac{1}{10}\) + \(\frac{5}{100}\) + \(\frac{3}{1000}\)
2. 57.031.
Solution:
57.031 = 57 + \(\frac{3}{100}\) + \(\frac{1}{1000}\)
= 50 + 7 + \(\frac{3}{100}\) + \(\frac{1}{1000}\)
Wbbse Class 6 Maths Solutions
Question 9. Find the values of the following:
1. 0.07 + 0.09
Class 6 West Bengal Board Math Solution :
∴ 0.07 + 0.09 = 0.16.
2. 4.11 + 1.6
Solution:
4.11 = 4.11; 1.6 = 1.60
∴ 4.11 + 1.6 = 4.11 + 1.60
= 5.71
3. 312.61+ 276.72
Class 6 West Bengal Board Math Solution:
312.61+ 276.72
∴ 312.61 + 276.72 = 586.33
Wbbse Class 6 Maths Solutions
4. 5 – 0.555
Solution:
5 = 5.000; 5.000 = 0.555
∴ 5 – 0.555 = 5.000 – 0.555 = 4.445.
5. 27.56+14.69
Examples of Real-Life Applications of Decimal Fractions
27.56 + 14.69 = 42.25.
6. 4.3 +36.4
Solution:
4.3 = 4.3; 3 = 3.0; 6.4 = 6.4
∴ 4.3 + 3 – 6.4 = 4.3 + 3.0 – 6.4 = 0.9.
7. 3.36 – 4.62 + 2.18
Conceptual Questions on Operations with Decimal Fractions
3.36 4.62 + 2.18 = 3.36+ 2.184.62.
∴ 3.36 = 4.62 + 2.18
= 3.36 + 2.18 – 4.62 = 0.92.
8. 2.67 – 3.727 + 4.2
Class 6 West Bengal Board Math Solution:
2.67 – 3.727 + 4.2
∴ 2.67 – 3.727 + 4.2
= 2.670 + 4·200 – 3.727 = 3.143
Question 10. For an occasion at our house, Father bought rice for 200, pulses for 125.50, and fish for what was spent by Father? 242.50. How much total amount of money
Solution :
Father spent 568.00 as the total amount of money for the occasion.
Question 11. Your exercise book is rectangular in shape and its length is 24.25 cm and its breadth is 10.75 cm. What is the perimeter of your exercise book?
Solution:
The perimeter of the exercise book = 2 (Length + Breadth)
= 2 (24.25 + 10.75) cm (2 x 35.00) cm 70 cm.
∴ The perimeter of the exercise book = is 70 cm.
Question 12. You had 5. You bought a pen for 3.50. How much money is left with you at present?
Solution :
₹ 5 = ₹ 5.00
∴ 1.50 is left with you at present.
Question 13. What must be added to 2.75 to get 3?
Solution
Now 3 = 3.00
∴ 0.25 must be added.
Question 14. Taniya cuts off a length of 8.5 cm of string from a string of length 12.5 cm. What is the length of the string left?
Solution:
The total length of the string = is 12.5 cm.
From it, a length of 8.5 cm is cut off.
∴ The length of the string is still left = 4 cm.
Question 15. What must be added to 2.172 to get 5?
Solution:
5 = 5.000
∴ 2.828 is to be added.
Real-Life Scenarios Involving Measurements and Money
Question 16. 2.647 is subtracted from 4.15. How much is to be added to the result of subtraction to get 10?
Solution:
4.15 = 4.150
∴ The result of subtraction = 1.503.
Now we have to find how much is to be added to 1-503 to get 10.
Now 10 = 10.000
∴ 8.497 must be added.
Question 17. In a long-jump competition, Susmita jumped 179.25 cm and Sagarika jumped 182.88 cm. Who and how much more length did jump?
Solution: Susmita jumped 179.25 cm and Sagarika jumped 182.88 cm.
As 179.25 cm < 182.88 cm, therefore Sagarika jumped more length than Susmita.
∴ Sagarika jumped 3.63 cm more in length than Susmita.
Question 18. Simplify :
1. \(\frac{10 \cdot 573+2 \cdot 227-1 \cdot 8}{2 \cdot 347-4 \cdot 32+12 \cdot 973}\)
Solution:
The given expression = \(\frac{10 \cdot 573+2 \cdot 227-1 \cdot 8}{2 \cdot 347-4 \cdot 32+12 \cdot 973}\)
2. \(\frac{4 \cdot 5-6 \cdot 12+7 \cdot 432-1 \cdot 1}{5 \cdot 234+10 \cdot 2-2 \cdot 33-8 \cdot 392}\)
Solution:
The given expression = \(\frac{4 \cdot 5-6 \cdot 12+7 \cdot 432-1 \cdot 1}{5 \cdot 234+10 \cdot 2-2 \cdot 33-8 \cdot 392}\)
Question 19. A man gave 0-5 part of his own property to his wife, 0.2 part to his son, and 0.25 part to his daughter. If he had still the property whose value is ₹ 5000. What is the value of the whole property? What is the value of the property of his wife?
Solution:
Let the whole property = 1 unit.
∴ The total part of the property that his wife, son, and daughter got (0.502+0.25) = 0.95 part
∴ The remaining part of the property = (1 – 0.95) = 0.05 part
∴ Value of 0.05 part of the property = ₹ 5000
∴ Value of the whole property = ₹ (5000 ÷ 0.05) = ₹ 100000.
∴ The value of his wife’s property = ₹ 100000 × 0.5 = ₹ 50000
So the value of the whole property = ₹ 100000. and the value of the wife’s property
= ₹ 50,000.