WBBSE Solutions For Class 6 Maths Arithmetic Chapter 8 Percentage

Arithmetic Chapter 8 Percentage

Question 1: Convert the following fractions the percentages:


1. 9/10;

Solution:

9/10 = (9/10 x 100)%

= 90%.

9/10 = 90%.


2. 43/50;

Solution:

43/50 = (43/50 x 100)%

= 86%

43/50 = 86%


3.1 2/5;

Solution:

1 2/5 = 7/5

= 7/5 x 100 %

= 140

1 2/5 = 140


4. 4 3/8.

Solution:

4 3/8 = 35/8

= (35/8 x 100)%

= 437.5%

4 3/8 = 437.5%

 

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Question 2. Convert the following percentage into proper fractions:

1. 10%;

Solution:

10% = 10/100

= 1/10;

10% = 1/10


2. 70%;

Solution:

70% = 70/100

7/10;

70% = 7/10


3. 257%;

Solution:

257% = 257/100

= 2 57/100;

257% = 2 57/100


4. 33 1/3%;

Solution:

33 1/3% = 100/3 %

= 100/3 x 1/100

= 1/3.

33 1 = 1/3.

 

Question 3. Convert the following decimal fractions into percentages:

1. 0.6;

Solution:

0.6 = 6/10

= (6/10 x 100)%

= 60%

0.6 = 60%


2. 0.02

Solution:

0.02 = 2/100

= (2/100 x 100)%

= 2%

0.02 = 2%


3. 1.21

Solution:

1.21 = 121/100

=(121/100 x 100)%

= 121%

1.21 = 121%


4. 0.003

Solution:

0.003 = 3/1000

=(3/1000 x 100)%

= 3/10%

0.003 = 3/10%

 

Question 4. Convert the following percentages into decimal fractions:


1. 61%

Solution:

61% = 61/100

= 0.61;

61% = 0.61;


2. 105%

Solution:

105 = 105/100

= 1.05

105 = 1.05


3. 1.26% 

Solution:

1.26% = 1.26/100

= 0.0126

1.26% = 0.0126


4. 0.07%

Solution:

0.07% = 0.07/100

3= 0.0007

0.07% 0.0007

 

Question 5. Express the following vulgar fractions and decimal fractions in percentages and arrange them in ascending order:

1. 2/5 , 13/25 , 7/10

Solution:

2/5,

2/5 = (2/5 x 100)%

= 40%;

2/5 = 40%


13/25,

13/25 = (13/25 x 100)%

= 52%;

13/25 = 52%


7/10

7/10 = (7/10 x 100)%

=70%

7/10 =70%

∴ Arranging in ascending order of magnitudes, we get,

2/5 ,13/25,7/10


2. 1 2/5, 1 1/2, 1 9/10

Solution:


1 2/5,

1 2/5 = 7/5

= (7/5 x 100)%

= 140%;

1 2/5 = 140%;


1 1/2,

1 1/2 = 3/2

= (3/2 x 100)%

= 150%

1 1/2 = 150%


1 9/10,

1 9/10= 19/10

= (19/10 x 100)%

190%

1 9/10 = 190%

∴ Arranging in ascending order, we get,

1 2/5, 1 1/2, 1 9/10


3. 0.02, 0.15, 0.6

Solution:


0.02,

= 02/100

= (2/100 x 100)%

= 2%

0.02 = 2%


0.15

0.15 = 15/100

= (15/100 x 100)%

= 15%

0.15 = 15%


0.6

0.6 = 6/10

=(6/10 x 100)%

= 60%

0.6 = 60%

∴ Arranging in ascending order of magnitude, we get,

0.02, 0.15, 0.6.

 

Question 6. Today out of 35 students 7 are absent from class of Riya. What is the percentage of the total students present today in that class?

Solution: Total number of students in Riya’s class = 35

Number of students absent today = 7

∴ Total number of students present today = 35 – 7

= 28

∴ Percentage of students present today = (28/35 x 100)

= 80

So today 80% of the total students are present.

Alternative method :

The number of students present today = 35 – 7 = 28

∴ Out of 35 students, the present number of students = 28

∴ Out of 1 student part of the students present = 28/35

∴ Out of 100 students, the present number of students = 28/35 x 100 = 80

∴ Today present student = 80%.

 

Question 7. There are 2100 storybooks in the library of Palpara. If 30% of it, more storybooks are purchased, then what will be the total number of storybooks in the library now, and what number of additional storybooks are purchased?

Solution: Total number of storybooks in the library = 2100. 30% of it is purchased.

30% of 2100 = 2100 x 30/100 = 630.

∴ An additional number of story books that are purchased = 630.

The total present number of storybooks in the library after purchasing = is 2100 + 630 = 2730.

∴ There will be 2730 story books in the library after purchasing.

 

Question 8. Abanibabu pays 22% of his monthly salary as house rent. If he pays Rs. 1870 as house rent every month, then what is his monthly salary?

Solution: Abanibabu pays as house rent = 22% of his monthly salary and which is ₹ 1870.

∴ 22% of monthly salary = ₹ 1870

∴ 1% of monthly salary = ₹ 1870/22

∴ 100% of monthly salary

 

WBBSE Solutions For Class 6 Maths Arithmetic Chapter 8 Percentage Question 8

 

= ₹ 8500

∴ Abanibabu’s monthly salary = ₹ 8500.

 

Question 9: The present population in a village is 26250. If the population increases at the rate of 4% every year, then what will be the total population in the next year? What will be the total population after two years?

Solution: Total number of present population in the village = 26250.

It increases every year by 4%.

∴ 4% of 26250 = 4/100 x 26250

= 1050

∴ The total population will be in the next year = (26250 + 1050) = 27300.

Again 4% of 27300

 

WBBSE Solutions For Class 6 Maths Arithmetic Chapter 8 Percentage Question 9

 

= 1092

∴ After 2 years, the total population village will be

(27300 + 1092) or 28392.

 

Question 10. Out of the total monthly expenses of the family of Reba, ₹ 4750 is spent on food and for all other expenses ₹ 5900 is spent. If the expenses on food are increased by 10% and other expenses are decreased by 16%, then calculate whether the total monthly expenses will increase or decrease and by how much.

Solution: The expenses on food is increased by 10%.

∴ 10% of ₹ 4750 = 10/100 x ₹ 4750 = ₹ 475.

The expenses for food will be increased by ₹ 475.

So the expenses for food will be after increment = ₹ (4750 + 475) = ₹ 5225.

Again the expenses for all other items are decreased by 16%.

∴ 16% of ₹ 5900 = 16/100 x Rs. 5900

= ₹ 944

∴ Other expenses will be decreased by = ₹ 944.

For all other expenses total amount to be spent = is ₹ (5900 – 944)

= ₹ 4956.

Total previous expenditure = ₹ (4750 + 5900)

= ₹ 10650.

Total present expenditure = ₹ (5225 + 4956)

= ₹ 10181.

As Rs. 10650 > ₹ 10181, the total expenditure will decrease by

₹ (10650 – 10181)

= ₹ 469.

 

Question 11. During harvesting, the price of rice was ₹ 1080 per quintal. In monsoon, the price of rice is increased by 15%. How much more money will be earned by a farmer who sold 12 acquittals of rice earlier if he will sell the same amount of rice during monsoon? Find also the total money he will receive by selling the rice during monsoon.

Solution: During monsoon, the price of rice is increased by 15%, whose earlier price was ₹ 1080 per quintal.

∴ 15% of ₹ 1080 = 15/100 x ₹ 1080 = ₹ 162.

∴ The price of rice per quintal during monsoon = ₹ (1080 + 162)

= ₹ 1242.

∴ The farmer will earn ₹ 162 more by selling per quintal of rice during monsoon.

∴ The farmer will earn more by selling 12 quintals of rice in monsoon than in earlier = ₹ (162 x 12)

= ₹ 1944.

The total amount of money will be received by the farmer by selling 12 quintals of rice in monsoon = ₹ (1242 x 12)

= ₹ 14904.

 

Question 12. 80 students have appeared for the Madhyamika examination this year from the school of Geeta. If 65% of them have passed, how many students have failed this year?

Solution: Total number of students who appeared for the Madhyamika examination = 80.

65% of 80 = 65/100 x 80 = 52.

∴ 52 students have passed the examination this year.

∴ (80 – 52) or 28 students have failed in Madhyamika examination this year.

 

Question 13. Due to the rise in the price of sugar, a family decides that the consumption of sugar will be reduced by 4%. If the family consumed 625 gm of sugar every day previously, how much gm of sugar consumption will be reduced, and also how much gm of sugar will be consumed each day now?

Solution: Previous consumption of sugar each day = 625 gm.

It is reduced by 4%.

∴ 4% of 625 gm

= 4/100 x 625

= 25 gm.

∴ The family will reduce sugar consumption each day = to 25 gm.

Again (625 – 25) = 600 gm

∴ The family will consumes sugar each day now = 600 gm.

 

Question 14. In a special kind of brass, copper is 70% and the rest is zinc. How many kg of copper and how many kg of zinc will be required to prepare 20 kg of this type of brass?

Solution: In the special kind of brass, copper is 70% and the rest is zinc.

∴ 70% of 20 kg = (70/100 x20) kg = 14 kg.

∴ Copper = 14 kg. So zinc = (20 – 14) = 6 kg.

∴ To prepare 20 kg of brass 14 kg of copper and 6 kg of zinc will be required.

 

 

 

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