Class 6 Math Solutions WBBSE Chapter 9 Percentage Exercise 9
Question 1. Let’s convert the following percentages into proper fractions:
1. 10%
Solution:
= \(\frac{10}{100}=\frac{1}{10}\)
2. 70%
Solution:
= \(\frac{70}{100}=\frac{7}{10}\)
3. 15%
Solution:
= \(\frac{15}{100}=\frac{3}{20}\)
4. 257 %
Solution:
= \(\frac{257}{100}\)
5. \(33 \frac{1}{3} \%\)
Solution:
= \(\frac{100}{3} \times \frac{1}{100}=\frac{1}{3}\)
Question 2. Let’s convert the following percentages into proper fractions:
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1. 61%
Solution:
= \(\frac{61}{100}=0.61\)
2. 3%
Solution:
= \(\frac{3}{100}=0.03\)
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3. 105%
Solution:
= \(\frac{105}{100}=1.05\)
4. 1.26%
Solution:
= \(\frac{126}{100 \times 100}=0.0126\)
5. 0.07%
Solution:
= \(\frac{7}{100} \times \frac{1}{100}\)
= \(\frac{7}{10000}=0.0007\)
Question 3. Let us express the following in percentages and arrange them in ascending order
1. \(\frac{1}{2}, \frac{1}{4}, \frac{3}{5}\)
Solution:
⇒ \(\frac{1}{2}=\frac{1}{2} \times 100 \%=50 \%\)
⇒ \(\frac{1}{4}=\frac{1}{4} \times 100 \%=25 \%\)
⇒ \(\frac{3}{5}=\frac{3}{5} \times 100 \%=60 \%\)
Ascending order: 25%; 50% ; 60%
∴ \(\frac{1}{4}, \frac{1}{2}, \frac{3}{5}\)
2. \(\frac{2}{5} ; \frac{13}{25} ; \frac{7}{10}\)
Solution:
⇒ \(\frac{2}{5}=\frac{2}{5} \times 100 \%\) = 40%
⇒ \(\frac{13}{25}=\frac{13}{25} \times 100 \%\) = 52%
⇒ \(\frac{7}{10}=\frac{7}{10} \times 100 \%\)= 70
Ascending order: 40% ; 52% ; 70%
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i.e., \(\frac{2}{5} ; \frac{13}{25} ; \frac{7}{10}\)
3. \(1 \frac{2}{5}, 1 \frac{1}{2}, 1 \frac{9}{10}\)
Solution:
⇒ \(1 \frac{2}{5}\)
= \(\frac{7}{5}=\frac{7}{5} \times 100 \%\) =140%
⇒ 1 \(\frac{1}{2}\)
= \(\frac{3}{2}=\frac{3}{2} \times 100 \%\) =150%
⇒ 1 \(\frac{9}{10}\)
= \(\frac{19}{10}=\frac{19}{10} \times 100 \%\) =190%
Ascending order: 140% ; 150% ; 190%
i.e., \(\frac{7}{5} ; \frac{3}{2} ; \frac{19}{10}\)
i.e., \(1 \frac{2}{5}, 1 \frac{1}{2}, 1 \frac{9}{10}\)
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4. 0.02,0.15,0.6
Solution:
0.02 = \(\frac{2}{100} \times 100 \%=2 \%\)
0.15 = \(\frac{15}{100} \times 100 \%=15 \%\)
0.6 = \(\frac{6}{10} \times 100 \%=60 \%\)
Ascending order: 2%; 15%; 60%
i.e., 0.02; 0.15; 0.6
Question 4. 1. If 3 out of 5 candies were taken, let’s find the percentage of candies taken.
Solution:
3 out of 5 candies are taken.
∴ Percentage of candies taken = \(\frac{3}{5}\) x 20% = 60%.
2. If 6 out of 24 berries are rotten. Let’s find the percentage of berries that have rotten.
Solution:
6 out of 24 berries are rotten.
∴ Percentage of rotten berries = \(\frac{6}{24}\) x 100% =25%.
WBBSE Math Solutions Class 6
3. Today 7 students of our class are absent. There are 35 students in our class. Let us find the percentage of students present today in our class.
Solution:
7 students out of 35 students are absent.
∴ 35 -7 = 28 students are present.
∴ Percentage of present students = \(\frac{28}{35}\) x100% = 80%.
4. A bamboo is of length 55 m. 11 m of it is under muddy water. Let’s find what percentage of the bamboo is above the muddy water.
Solution:
Total length of bamboo = 55 m.
Under muddy water = 11 m.
Length of bamboo above muddy water = (55-11)m = 44m.
Percentage of the bamboo above the muddy water
= \(\frac{44}{55}\) 100% = 80 %
WBBSE Math Solutions Class 6
Question 5. There are 2100 story books in our local library. If 30% more storybooks are bought, what will be the total number of storybooks in the library now also find out the additional number of storybooks bought.
Solution:
Total number of story books in our library = 2100.
30% of 2100 story books = \(\frac{30}{100}\) x 2100 = 630.
∴ Total number of books now = 2100 + 630 = 2730.
Question 6. It’s raining heavily since morning. Hence only 20% of students were present in Alam’s school. Let’s find how many students were actually present in school if the total number of students in Alam’s school is 1230.
Solution:
Total number of students in Alam’s school = 1230.
Number of students present today = 20% of 1230
= \(\frac{20}{100}\) X 1230 = 246.
Question 7. 10. Today I shall prepare orange juice, ‘sherbet’ (drink) myself. For preparing 300 ml of ‘sherbet’ I added 18% orange juice to it. Let’s find how many milliliters of orange juice I added to the ‘sherbet’.
Solution:
Quantity of sherbet = 300 ml.
Orange juice added = 18% of 300ml
= \(\frac{18}{100}\) x 300 ml = 54 ml.
WBBSE Math Solutions Class 6
Question 8. Shobhan used organic fertilizer this year in his land This increased the production of paddy by 25% than last year. If last year’s production of paddy is 12 quintals, let’s find the production of oaddv this year.
Solution:
Last year production of paddy = 12 Quintals.
This year production increased by 25%
= \(\frac{25}{100}\) x 12 Quintals = 3 Quintals.
∴ Total production = (12 + 3) Quintals = 15 Quintals.
Question 9. In Rasulpur village, the population has increased by 12% than the previous year. Till last year, the population of the village was 775. Let’s find the present population of the village.
Answer:
Till last year, the population of the village was 775.
The population increased by 12%
= 12/100 x 775 = 993
∴ Present population of the village = 775 + 93 = 868.
Question 10. 80 students appeared for the Madhyamik Examination this year from our school. If 65% of students have passed, let us find how many students have actually passed.
Answer:
Number of students appeared for the Madhyamik Examination this year = 80.
Percentage of passed = 65%.
∴ Number of students passed = \(\frac{65}{100}\) x 80 = 52.
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Question 11. A certain alloy contains 70% popper and rest zinc. In 20 kg of such alloy let’s find how many kg of zinc is required.
Solution:
Total weight of alloy = 20 kg.
Percentage of copper present = 70%
= \(\frac{70}{100}\) X 20 kg = 14 kg.
∴ Quantity of zinc required = 20 – 14 = 6 kg.
Question 12. Due to the rise in the price of sugar we decided to reduce the consumption of sugar by 4%. Presently, we consume 625 gm of sugar every day. Let’s find how many grams the consumption of sugar is reduced and also the amount of sugar consumed each day now.
Solution:
At present we consume sugar every day = 625 gm.
Consumption of sugar reduced by 4% = \(\frac{4}{100}\) x 625 = 25gm
∴ Amount of sugar consumption today = 625gm – 252gm = 600gm
Question 13. Anilbabu pays 22% of his salary for house rent. If he pays Rs. 1870 per month for rent, let’s find his monthly salary.
Solution:
Monthly rent
22
1870
Monthly salary
100
?
When Rs. 22 is the rent, monthly salary = Rs. 100.
When Rs. 1 is the rent, monthly salary = Rs. \(\frac{100}{22}\)
When Rs. 1870 is the rent, monthly salary = Rs. \(\frac{100}{22}\) x 1870 = Rs. 8500.
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Question 14. Yasina Khatoon cultivates jute in 55% of her agricultural land. If she cultivates jute in 11 bighas of land, let’s find her total agricultural land.
Solution:
If Yasina cultivated 55 bighas of land, she had 100 bighas of agricultural land.
Cultivated land
55
11
Total agricultural land
100
?
If 55 bighas of cultivated land, total agricultural land = 100 bighas.
If 11 bighas of cultivated land, total agricultural land = 100 bighas = \(\frac{100}{55}\) x 11 bighas = 20 bighas.
Question 15. Out of total monthly expenses of our family, Rs. 4750 is spent on food and 5900 for all other expenses. If expenses on food is increased by 10% and other expenses is decreased by 16%, then let’s calculate whether the total monthly expenses will increase or decrease.
Solution:
Expenses for food = Rs. 4750
and other expenses = Rs. 5900
Expenses for food increased by 10%
= Rs. 4750 x \(\frac{10}{100}\)
= Rs. 475
Now expenses for food = Rs. 4750 + Rs. 475 = Rs. 5225
Expenses for other decreased by 16%
= Rs. 5900 x \(\frac{16}{100}\) = Rs. 944
Now other expenses = Rs. 5900 – Rs. 944 = Rs. 4956
Previously total expenses were = Rs. 4750 + Rs. 5900 = Rs. 10650
Presently total expenses are = Rs. (5225 + 4956) = Rs. 10181
Expenses Decrease in = Rs. (10650 -10181) = Rs. 469
Class 6 WBBSE Math Solutions
Question 16. The present population of the city is 26250. If the population increases at the rate of 4% every year, then let’s find what will the population of the city be next year. Also, let’s find what the population will be after two years.
Solution:
The present population of a city = 26250.
Population increases by 4%
= \( x 26250 = 1050.
∴ The population of the city in 1st year = 26250 + 1050 = 27300.
Next year population again increased by 4%
= [latex]\frac{4}{100}\) x 27300 = 1092
∴ The population of the city in 2nd year = 27300 + 1092 = 28392.
WBBSE Class 6 Maths Solution for Question 17. During harvesting the price of paddy was Rs. 1080 per quintal. In monsoon the price increased by 15%, let’s find out how much more the farmer who sold 12 quintals of paddy earlier could have earned, had he sold the same amount during monsoon.
Solution:
During harvesting, the price of paddy per quintal was Rs. 1080.
In monsoon, the price increased by 15%
= \(\) x Rs. 1080
= Rs. 162.
∴ During monsoon, he sold paddy at
= Rs. (1080 + 162) per quintal.
= Rs. 1242 per quintal
During monsoon, he will earn by selling 12 quintals
= Rs. 162×12
= Rs. 1944 more.
