Class 6 Math WBBSE Solutions Chapter 7 Multiplication And Division Of Decimal Fraction By Whole Number And Decimal Fraction Exercise 7
Question 1. Let’s draw and find the product of 0.4 x 0.7 and 0.2 x 0.8 by taking a square of white paper.
Solution:
1. 0.4 x 0.7
2. 0.2 x 0.8
1. 0.4 x 0.7
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2. 0.2 x 0.8
Question 2. The side of a square is 5.2 cm. Let’s find its perimeter and area.
Solution:
The side of a square = 5.2 cm.
1. Perimeter of the square = 4 x one side
= 4 x 5.2 cm
= 20. 8 cm
2. Area of the square = side x side
= 5.2 cm x 5.2 cm
= \(\frac{52}{10} \times \frac{52}{10} \mathrm{sq} \cdot \mathrm{cm}\)
= \(\frac{2704}{100} \mathrm{sq} . \mathrm{cm}\)
= 27.04 sq. cm.
Class 6 Math WBBSE Solutions
Question 3. Let’s find the values of the following:
1. 6.2 x 3.1
Solution:
= \(\frac{62}{10} \times \frac{31}{10}=\frac{1922}{100}\)
= 19.22
2. 11.3 x 2.5
Solution:
= \(\frac{113}{10} \times \frac{25}{10}=\frac{2825}{100}\)
= 282.5
3. 11.25 x 7.3
Solution:
= \(\frac{1125}{100} \times \frac{73}{10}=\frac{82125}{1000}\)
= 82.125
4. 0.03 x 0.3
Solution:
= \(\frac{3}{100} \times \frac{3}{10}=\frac{9}{1000}\)
= 0.009
5. 0.04 x 0.04.
Solution:
= \(\frac{4}{100} \times \frac{4}{100}=\frac{16}{10000}\)
= 0.0016
= 19
Class 6 Math Solutions WBBSE Multiplication And Division Of Decimal Fraction By Whole Number And Decimal Fraction Exercise 7.1
Question 1. On a white square paper let’s draw 10×10 small squares. By using different colours let’s find the value of 0.5 x 0.9, 0.2 x 0.8.
Solution:
1. 0.5 x 0.9
2. 0.2 x 0.8
Question 2. Mithu wants to buy 4 exercise books. If an exercise book costs Rs. 12.75, let’s find how much money Mithu will need.
Solution:
Price of each exercise book = Rs. 12.75.
∴ Price of 4 exercise books = Rs. 12.75 x 4 = Rs. 51.
∴ Mithu will need = Rs. 51.
Class 6 Math Solutions WBBSE
Question 3. Rajinabibi built a house in 0.35 part of her land. She cultivated flowers in 0.2 part of the remaining land. Let’s find in what part of her land she cultivated flowers.
Solution:
Rajinabibi built a house in 0.35 part of her land.
Remaining land = 1 – 0.35 = 0.65 part.
She cultivated flowers on 0.2 part of the remaining land
= \(0.2 \times 0.65=\frac{2}{10} \times \frac{65}{100}=\frac{13}{100}=0.13 \mathrm{part}\)
Question 4. 1 have Rs. 150. With 0.3 part of my money, I bought exercise books and with 0.4 part I bought a storybook. Let me find out what amount of money is left with me.
Solution:
I have Rs. 150.
I bought an exercise book with 0.3 part and bought a story book with 0.4 part.
∴ Total expenditure (0.3 + 0.4) = 0.7 part
= Rs. 150 x 0.7
= Rs. 150 x \(\frac{7}{10}\) = Rs. 105
∴ Remaining amount = Rs. (150 – 105) = Rs. 45.
Class 6 Math Solutions WBBSE
Question 5. Today we shall travel a distance of 94.5 km. If 0.078 liter of petrol is used per kilometer, let’s calculate the total amount of petrol required.
Solution:
To travel 1 km petrol required 0.078 litre.
∴ To travel 94.5 km petrol required = 0.078 x 94.5 litre = 7.37 litre.
Question 6. Alisha’s brother took 1.4 hours to reach Shibpur Launch pier (ghat) from his house in a cycle. If the speed of his cycle is 11.5 km per hour, then let us calculate the distance of Shibpur Launch pier from Alisha’s house.
Solution:
Speed of the cycle is 11.5 km per hour, i.e., in 1 hour he goes 11.5 km.
∴ In 1.4 hour he goes 1.4 x 11.5 km = 16.1 km.
∴ Required distance = 16.1 km.
Question 7. My mother asked me to buy 2.5 kg of pulse. The cost of 1 kg of pulse is Rs. 62.50. Let me calculate how much money I must carry to the shop.
Solution:
The cost of 1 kg of pulse is Rs. 62.50.
∴ The cost of 2.5 kg of pulse = Rs. 62.50 x 2.5.
∴ I must carry to the shop Rs. 156.25 = Rs. 156.25.
Class 6 Math Solutions WBBSE
Question 8. The perimeter of an equilateral triangle is 10.4 cm. Let’s find the length of the side of the equilateral triangle in decimal fractions.
Solution:
The perimeter of an equilateral triangle = 14.4 cm.
∴ Length of each side of this triangle = \(\frac{14.4}{3}\) cm = 4.8 cm.
Question 9. Let’s multiply
1. 0.7 x 0.9
Solution: = 0.63
2. 0.6 x 0.5
Solution: = 0.30
3. 0.02 x 0.2
Solution: = 0.004,
4. 0.67 x 0.39
Solution:
= \(\frac{67}{100} \times \frac{39}{100}=\frac{2613}{10000}=0.2613\)
5. 0.52 x 0.43
Solution:
= \(\frac{52}{100} \times \frac{43}{100}=\frac{2236}{10000}\) = 0.2236
6. 0.07 x 0.97
Solution:
= \(\frac{7}{100} \times \frac{97}{100}=\frac{679}{10000}\) = 0.0679
WBBSE Math Solutions Class 6
7. 6.23 x 2.51
Solution:
= \(\frac{623}{100} \times \frac{251}{100}=\frac{156373}{10000}\) = 15.6373
8. 5.77 x 2.93
Solution:
= \(\frac{577}{100} \times \frac{293}{100}=\frac{169061}{10000}\) = 16.
9. 8.23 x 0.3
Solution:
= \(\frac{823}{100} \times \frac{3}{10}=\frac{2469}{1000}\) = 2.469
10. 82.03 x 0.06
Solution:
= \(\frac{8203}{100} \times \frac{6}{100}=\frac{49218}{10000}\) = 4.9218
11. 85.29 x 3.92
Solution:
= \(\frac{8529}{100} \times \frac{392}{100}=\frac{3343368}{10000}\) = 334.3368
12. 72.2 x 2.65
Solution:
= \(\frac{722}{10} \times \frac{265}{100}=\frac{191330}{1000}\) =191.33
WBBSE Math Solutions Class 6
13. 72.04 x 0.05
Solution:
= \(\frac{7204}{100} \times \frac{5}{100}=\frac{36020}{10000}\)
= 3.602
14. 72.156 x 12.16
Solution:
= \(\frac{72156}{1000} \times \frac{1216}{100}\)=\(\frac{87741696}{100000}\)
= 877.41696
15. 0.6 x 0.3 x 0.2
Solution:
= \(\frac{6}{10} \times \frac{3}{10} \times \frac{2}{10}=\frac{36}{1000}\)
= 0.036
16. 0.2 x 0.06 x .03
Solution:
= \(\frac{2}{10} \times \frac{6}{100} \times \frac{3}{100}=\frac{36}{100000}\)
= 0.00036
17. 2. 14 x 0.4 x 0.9
Solution:
= \(\frac{214}{100} \times \frac{4}{10} \times \frac{9}{10}=\frac{7704}{10000}\)
= 0.7704
WBBSE Math Solutions Class 6
18. 1.21 x0.5 x 5.2
Solution:
= \(\frac{121}{100} \times \frac{5}{10} \times \frac{52}{10}=\frac{31460}{10000}\)
= 3.146
19. 3.06 x 100
Solution:
= \(\frac{306}{100} \times 100\)
= 306
20. 7.92 x 1000
Solution:
= \(\frac{792}{100} \times 1000\)
= 7920
Question 10. Arrange the following in descending order (bigger to smaller) of their values:
1. 0.5 x 0.3, 0.5, 0.3
Solution:
= 0.15; 0.5; 0.3
WBBSE Math Solutions Class 6
In descending order = 0.5; 0.3; 0.5 x 0.3
2. 0.6 x 0.7, 0.6, 0.7
Solution:
= 0.42 ; 0.6; 0.7
In descending order = 0.7; 0.6 x 3; 0.6 x 0.7
3. 0.9 x 0.2, 0.9, 0.2
Solution:
= 0.18; 0.9; 0.2
In descending order = 0.9; 0.2 ; 0.9 x 0.2
4. 0.4 x 0.8, 0.4, 0.8
Solution:
= 0.32; 0.4; 0.8
In descending order = 0.8; 0.4; 0.4 x 0.8
5. 1.2 X 1.5,1.2,1.5
Solution:
= 1.8; 1.2; 1.5
In descending order = 1.2 x 1.5; 1.5; 1.2
6. 2.3 x 2.4, 2.3, 2.4
Solution:
= 5.52; 2.3; 2.4
In descending order = 2.3 x 2.4; 2.4; 2.3
7. 6.7 x 7.2, 6.7, 7.2
Solution:
= 48.24; 6.7; 7.2
In descending order = 6.7 x 7.2 ; 7.2 ; 6.7
8. 8.2 x 1.9, 8.2,1.9
Solution:
= 15.58; 8.2; 1.9
In descending order = 8.2 x 1.9; 8.2; 1.9
Question 11. Let’s find values in decimal numbers
1. 0.625÷ 5
Solution:
⇒ \(\frac{625}{1000} \times \frac{1}{5}=\frac{125}{1000}=0.125\)
2. \(0.627 \div 3\)
Solution:
⇒ \(\frac{627}{1000} \times \frac{1}{3}=\frac{209}{1000}\)
= 0.209
Class 6 WBBSE Math Solutions
3. \(0.343 \div 7\)
Solution:
⇒ \(\frac{343}{1000} \times \frac{1}{7}\)=\(\frac{49}{1000}\)
= 0.049
4. \(651.2 \div 4\)
Solution:
⇒ \(\frac{6512}{10} \times \frac{1}{4}=\frac{1628}{10}\).
= 162.8
5. \(7 \div 3.5\)
Solution:
⇒ \(7 \div \frac{35}{10}=\frac{7 \times 10}{35}=\frac{70}{35}\)
= 2
6. \(18 \div 0.2\)
Solution:
⇒ \(18 \div \frac{2}{10}=\frac{18 \times 10}{2}\)
= 90
7. \(28.8 \div 1.2\)
Solution:
⇒ \(\frac{288}{10} \div \frac{12}{10}\)
= \(\frac{288}{10} \times \frac{10}{12}\)
= 24
8. \(11.7 \div 1.3\)
Solution:
⇒ \(\frac{117}{10} \times \frac{10}{13}\)
= 9
Class 6 WBBSE Math Solutions
9. \(1.35 \div 1.5\)
Solution:
⇒ \(\frac{135}{100} \div \frac{15}{10}\)
= \(\frac{135}{100} \times \frac{10}{15}\)
= \(\frac{9}{10}\)
= 0.9
10. \(0.65 \div 0.5\)
Solution:
⇒ \(\frac{65}{100} \div \frac{5}{10}\)
= \(\frac{65}{100} \times \frac{10}{5}\)
= \(\frac{13}{10}\)
= 1.3
11. \(0.07 \div 0.5\)
Solution:
⇒ \(\frac{7}{100} \div \frac{5}{10}\)
= \(\frac{7}{100} \times \frac{10}{5}\)
= \(\frac{14}{100}\)
= 0.14
Class 6 WBBSE Math Solutions
Question 12. Let’s make a story and then work out:
1. 2.50 x 5 + 3.25 x 6
Solution:
= 12.50 +19.50 = 32
2. 3.75 x 8 – 2.50 x 3
Solution:
Find the difference of prices of 8 pens (each cost Rs. 3.75) and 3 pencils (each cost Rs. 2.50)
= Rs. 3.75 x 8 – Rs. 2.50 x 3
= Rs. (30 – 7.50) = Rs. 22.50
3. 22.50 x 3.50
Solution:
If the cost price of 1 kg rice is Rs. 22.50, then the cost price of 3.50 kg rice
= Rs. 22.50 x 3.50 = Rs. 78.75
4. 13.75 x 2 + 12.50 x 3
Solution:
The total cost price color boxes (the price of each Rs. 13.75) and 3 pencils (price of each Rs. 12.50)
Rs. 13.75 x 2 + Rs. 12.50 x 3
= Rs. 27.50 + Rs. 37.50 = Rs. 65.
Question 13. Let’s find the values of the following simplification:
1. 13.28 – 4.07 + 2.7 x 0.02
Solution:
13.28-4.07 + 0.054 = 13.334-4.07
= 9.264
2. {45.85 – (6.29 + 15.06)} ÷ 5
Solution:
{45.85 -21.35} ÷ 5
= 24.5-5
= 4.9
Class 6 WBBSE Math Solutions
3. (7.8 – 7.8 x 0.2) ÷ 1.2
Solution:
= (7.8-1.56) ÷ 1.2
= 6.24 ÷ 1.2
= 5.2
4. 0.35 x 0.35 + 0.15 x 0.15 + 2 x 0.35 x 0.15
Solution:
0.1225 + 0.0225 + 0.105
= 0.25
5. {(4 – 2.07) x 2.5} -1.93
Solution:
= (1.93 x 2.5) -1.93
= 4.825 – 1.93
= 2.5
