# Selina Concise Solutions for Chapter 7 Percent and Percentage Class 8 ICSE Mathematics

## ### Exercise 7A

1. Evaluate:

(i) 55% of 160 + 24% of 50 – 36% of 150
(ii) 9.3% of 500 – 4.8% of 250 – 2.5% of 240
Solution
(i) 55% of 160 + 24% of 50 – 36% of 150
= (55 × 160)/100 + (24 × 50)/100 – (36 × 150)/100
= 11 × 8 + 12 – 18 × 3 = 88 + 12 – 54 = 46

(ii) 9.3 % of 500 – 4.8% of 250 – 2.5% of 240
= (9.3 × 500)/100 – (4.8 × 250)/100 – (2.5 × 240)/100
9.3 × 5 – 1.2 × 10 – 0.5 × 12
= 46.5 – 12 – 6
= 46. 5 – 8
= 28.5

2. (i) A number is increased from 125 to 150; find the percentage increase.
(ii) A number is decreased from 125 to 100; find the percentage decrease.
Solution
(i) Original value = 125,  New value = 150
Increase = (150 – 125) = 25
Increase % = 25/125 × 100 = 20%

(ii) Original number = 125, New value = 100,
Decrease = (125 – 100) = 25
Decrease % = 25/125 × 100 = 200%

3. Find:
(i) 45 is what percent of 54?
(ii) 2.7 is what percent of 18?
Solution
(i) Let 45 = x percent of 54 = (54 × x)/100
⇒ x = (45 × 100)/54 = (5 × 100)/6
= 250/3 = 83 1/3%
∴ Reqd. percentage = 83 1/3%

(ii) Let 2.7 = x percent of 18 = (18 × x)/100
∴ x = (2.7 × 100)/18 = 270/18 = 30/2 = 15
∴  Reqd.  percentage = 15%

4. (i) 252 is 35% of a certain number, find the number.
(ii) If 14% of a number is 315; find the number.
Solution
(i) Let the number be x
By the given condition,
252 = (x × 35)/100 = (x × 7)/20
∴ x = (252 × 20)/7 = 36 × 20 = 720
Hence reqd. number = 720

(ii) Let the number be x
By the given condition,
315 = (x × 14)/100
∴ x = (315 × 100)/14
= (45 × 100)/2
= 45 × 50
= 2250
Hence, reqd. number = 2250

5. Find the percentage change, when a number is changed from :
(i) 80 to 100
(ii) 100 to 80
(iii) 6.25 to 7.50
Solution
(i) Original number = 80
New number = 100,
Change = (100 – 80) = 20
∴ Percentage change = (increase)
= 20/80 × 100
= 25%

(ii) Original number = 100
New number = 80
Change (100 – 80) = 20
∴ Percentage change(decrease) = 20/100 × 100
= 20%

(iii) Original number = 6.25,
New number = 7.50
Change (Increase) = (7.50 – 6.25) = 1.25
∴ Increase = 1.25/6.25 × 100
= 20%

6. An auctioneer charges 8% for selling a house. If a house is sold for Rs. 2,30,500; find the charges of the auctioneer.
Solution
Selling price of the house = Rs 2,30,500
Rate of charges of the auctioneer = 8% of selling price
∴ Charges of the auctioneer =8% of 2,30,500
= 8/100 × 2,30,500
= Rs. 18,440

7. Out of 800 oranges, 50 are rotten. Find the percentage of good oranges.
Solution
Total number of oranges = 800
Number of good oranges = 800 – 50
= 750
Percentage of good oranges = 750/850 × 100
= 750/8 = 375/4 = 93 ¾%

8. A cistern contains 5 thousand litres of water. If 6% water is leaked. Find how many litres of water are left in the cistern.
Solution
Water in the cistern = 5000 Liters
Quantity of water leaked = 6/100 × 5000
= 300 litres
Quantity of water left in the cistern = (5000 – 300) = 4700 litres

9. A man spends 87% of his salary. If he saves Rs. 325; find his salary.
Solution
Let salary = Rs x
∴ Expenditure = 87/100 of x
= Rs 87x/100
Saving = Rs. 325
∴ x - 87x/100 = 325
⇒ (100x – 87x)/100 = 325
⇒ 13x/100 = 325
⇒ x = (325 × 100)13
⇒ x = 32500/13
⇒ x = 2500
∴  Salary = Rs. 2500

10. (i) A number 3.625 is wrongly read as 3.265; find the percentage error.
(ii) A number 5.78 × 103 is wrongly written as 5.87 × 103; find the percentage error
Solution
(i) Correct number = 3.625
Number wrongly read as = 3.265
Error = 3.625 – 3.265
= 0.360
% Error = 0.360/3.625 × 100
= 360/3625 × 100
= 36000/3625
= 9.93%

(ii) Correct number = 5.78 × 103
Number wrongly written as = 5.87 × 103
Error = 5.87 × 103 – 5.78 × 103
= 0.09 × 103
% Error = (0.09 × 103)/(5.78 × 103) × 100
= 0.09/5.78 × 100
= 9/578 × 100
= 900/578%
= 1.56%

11. In an election between two candidates, one candidate secured 58% of the votes polled and won the election by 18, 336 votes. Find the total number of votes polled and the votes secured by each candidate.
Solution
Since, winning candidate secured 58% of the votes polled.
∴ Losing candidate secured = (100 – 58)% of the votes polled
= 42% of the votes polled
Difference of votes = 58 – 42
= 16% of the votes polled
We are given :
16% of votes polled = 18, 336
⇒ 16/100 of votes polled = 18, 336
⇒ Votes polled = 18,336 × 100/16
⇒ Votes Polled = 1, 14,600
∴ Votes secured by winning candidate = 58/100 × 1,14, 600
= 66,468
Votes secured by losing candidate = 42/100 × 1,14,600
= 48,132
Votes secured by winning candidate = 66,468
Votes secured by losing candidate = 48,132

12. In an election between two candidates, one candidate secured 47% of votes polled and lost the election by 12, 366 votes. Find the total votes polled and die votes secured by the winning candidate.
Solution
Since, the losing candidate secured 47% of the votes polled
Winning candidate secures votes = (100 – 47)% of the votes polled
= 53% of the votes polled
Difference of votes = 53 – 47
= 6% of the votes polled
We are given:
6% of the votes polled = 12,366
⇒ 6/100 of the votes polled = 12,366
⇒ Votes polled = 12,366 × 100/6
⇒ 1236600/6
⇒ 2,06,100
Votes secured by winning candidate = 53/100 × 2,06,100 = 1,09233
Votes secured by winning candidate = 1,09,233

13. The cost of a scooter depreciates every year by 15% of its value at the beginning of the year. If the present cost of the scooter is
₹ 8,000; find its cost:
(i) after one year
(ii) after 2 years
Solution
Present cost of scooter = Rs. 8000
The cost of scooter depreciates by 15% every year
(i) Cost of scooter after one year = (100 – 15)/100 × 8000
= 85/100 × 8000
= Rs 6800

(ii)
Cost of scooter after 2 years = (100 – 15)/100 × 6800
= 85/100 × 6800
= Rs. 5780

14. In an examination, the pass mark is 40%. If a candidate gets 65 marks and fails by 3 marks; find the maximum marks.
Solution
Marks obtained by the candidate = 65
Fails by = 3marks
Pass marks = 65 + 3 = 68
% of Pass marks = 40%
∴ Required maximum marks = 100/40 × 68
= 10 × 17
= 170

15. In an examination, a candidate secured 125 marks and failed by 15 marks. If the pass percentage was 35 %; find the maximum marks.
Solution
Total marks secured = 125
Failed by 15 marks
∴ Pass marks = 125 + 15 = 140
Let Maximum marks = x
∴ (x × 35)/100 = 140
⇒ x = (140 × 100)/35 = 4 × 100 = 400
Hence maximum marks = 400

16. In an objective type paper of 150 questions; John got 80% correct answers and Mohan got 64% correct answers.
(i) How many correct answers did each get?
Solution
Total questions = 150
John got correct answers = 80%
Mohan got correct answers = 64%

(i) Number of correct answers got by John = 80/100 × 150 = 120
Numbers of correct answers got by  Mohan = 64/100 × 150
= 64/4 × 6
= 96

(ii) % of Mohan’s correct answers to John’s correct answers = 96/120 × 100
= 4/5 × 100
= 4 × 20
= 80%

17 The number 8,000 is first increased by 20% and then decreased by 20%. Find the resulting number.
Solution
The resulting number = The original number × (1 + 20/100) × (1 – 20/100)
= 8000 × 120/100 × 80/100
= 7,680

18. The number 12,000 is first decreased by 25% and then increased by 25%. Find the resulting number.
Solution
The resulting = The original number × (1 – 25/100) × (1 + 25/100)
= 12000 × 75/100  × 125/100
= 11,250

19. The cost of an article is first increased by 20% and then decreased by 30%, find the percentage change in the cost of the article.
Solution
Let the original cost = ₹ 100
Increased by 20%
∴  New cost = 100 + 20 = ₹120
Decreased by 30% = (120 × 30)/100
= ₹ 36
∴ New cost = 120 – 36  = ₹ 84
Overall change = 100 – 84 = ₹ 16
Required percentage = 16/100 × 100
= 16% decrease

20. The cost of an article is first decreased by 25% and then further decreased by 40%. Find the percentage change in the cost of the article.
Solution
Let the original cost = ₹ 100
Decreased by 25%
∴ New cost = 100 – 25 = ₹75
Decreased by 40%  = (75 × 40)/100 = ₹30
∴ New cost = ₹75 - 30 = ₹45
Overall change = 100 – 45 = ₹55
Required percentage = 55/100 × 100
= 55% decrease

### Exercise 7 B

1. A man bought a certain number of oranges; out of which 13 percent were found rotten. He gave 75% of the remaining in charity and still has 522 oranges left. Find how many had he bought?
Solution
Suppose number of oranges bought = 100
Number of Rotten oranges = 13/100 × 100
= 13
Remaining oranges = 87
Oranges given in charity = 75/100 × 87
= 3 × 87/4
= 261/4
Net balance of oranges = 87 – 261/4
= (348 – 261)/4
= 87/4
If the balance is 87/4, then number of oranges bought = 100
If the balance is 1 then number of oranges bought = 100 × 4/87
If the balance is 522 then number of oranges bought = 100 × 4/87 × 522
= (100 × 4 × 522)/87
= 100 × 4 × 6
= 2400

2. 5% pupil in a town died due to some diseases and 3% of the remaining left the town. If 2,76,450 pupil are still in the town; find the original number of pupil in the town.
Solution
Let original number of pupil in the town = 100
Number of pupil did due to disease = 5/100 × 100 = 5
Remaining pupil = 100 – 5 = 95
Number of pupil who left the town = 3/100 × 95
= (3 × 95)/100
= 57/20
Actual remaining pupil = 95 – 57/20
= (1900 – 57)/20
= 1843/20
If the remaining pupil in the town are 1843/20, then original number of pupil = 100
If the remaining pupil in the town is 1, then original number of pupil = 100 × 20/1843
If the remaining pupil in the town are 276450, then original number of pupil =  100 × 20/1843 × 276450
= (100 × 20 × 276450)/1843
= 100 × 20 × 150
= 300000

3. In a combined test in English and Physics ; 36% candidates failed in English ; 28% failed in Physics and 12% in both ; find:
(i) the percentage of passed candidates
(ii) the total number of candidates appeared, if 208 candidates have failed.
Solution
Candidates failed only in English = 36% - 12% = 24%
Candidates failed only in Physics = 28% - 12% = 16%
Candidates failed in both subjects = 12%
Total failed candidates = 24% + 16% + 12%
= 52%
(i) Percentage of passed candidates = 100% - 52% = 48%
(ii) If failed candidates are 52, then total candidates appeared = 100
If failed candidate is 1, then total candidates appeared = 100/52
If failed candidate are 208, then total candidates appeared = 100/52 × 208
= 100 × 4 = 400

4. In a combined test in Maths and Chemistry; 84% candidates passed in Maths; 76% in Chemistry and 8% failed in both. Find :
(i) the percentage of failed candidates ;
(ii) if 340 candidates passed in the test; then how many appeared?
Solution
Since, candidates passed  in Maths = 84%
∵ Candidates failed in Maths = 100% - 84%
= 16%
Again, candidates passed in Chemistry = 76%
∴ Candidates failed in Chemistry = 100% - 76%
= 24%
Candidates failed in both = 8%
∴ Candidates failed in only Maths = 16% - 8%
= 8%
Candidates failed in only Chemistry = 24% - 8% = 16%
Total failed in candidates = 8% + 16% + 8%
= 32%
(i) Percentage of failed candidates = 32%
(ii) Passed candidates = 100% - 32% = 68%
If passed candidates are 68, then total candidates appeared = 100
If passed candidate appeared is 1 then total candidates appeared = 100/68
(ii) If passed candidates are 340 total candidates appeared = 100/68 × 340
= (100 × 340)/68
= 500

5. A’s income is 25% more than B’s. Find B’s income is how much percent less than A’s.
Solution
Let B’s income = Rs.100
Then A’s income = 100 + 25
= Rs. 125
Now, difference of income of A and B = Rs. (125 – 100)
= Rs. 25
If A’s income is Rs. 125, then B’s income less than A = Rs. 25
If A’s income is Re, 1, then B’s income less than A = Rs. 25/125
If A’s income is Rs. 100, then B’s income less than A = Rs. (25/125 × 100)
= 1/5 × 100
= Rs 20
∴ B’s income is less than A’s income = 20%

6. Mona is 20% younger than Neetu. How much percent is Neetu older than Mona ?
Solution
Let Neetu’s age = 100 years
Then, Mona’s age = 100 – 20 = 80 years
Difference of ages = 100 – 80 = 20 years
If Mona is 80 years, then Neetu is older than Mona by = 20 years
If Mona is 1 year, then Neetu is older than Mona by = 20/80 years
If Mona is 100 years, then Neetu is older than Mona by = 20/80 × 100 years = (20 × 100)/80
= 25%

7. If the price of sugar is increased by 25% today; by what percent should it be decreased tomorrow to bring the price back to the original ?
Solution
Let original price of sugar = Rs 100
∴ Price of sugar for today = Rs 100 + Rs 25
= Rs 125
In order to bring  down the price to original i.e., Rs. 100, its price should be decreased by = Rs. 125 – Rs. 100
= Rs. 25
∴ On Rs 125, the price should be decreased by = Rs. 25
On Re. 1, the price should be decreased  by = Rs. 25/125
On Rs. 100, the price should be decreased by = Rs. 25/125 × 100
= Rs. 1/5 × 100
= Rs. 20
∴ Price should be decreased by 20%

8. A number increased by 15% becomes 391. Find the number.
Solution
Let the required number = x
∴ According to the statement,
15% of x + x = 391
⇒ 15/100 × x + x = 391
⇒ x[15/100 + 1] = 391
⇒ x[(15 + 100)/100] = 391  ⇒ x × 115/100 = 391
⇒ x = 391 × 100/115  ⇒ x = (391 × 100)/115
⇒ (17 × 100)/5  ⇒ x = 340
∴ required number = 340

9. A number decreased by 23% becomes 539. Find the number.
Solution
Let the number = x
According to the statement,
x – 23% of x = 539
⇒ x – 23/100 × x = 539
⇒ x[1 – 23/100] = 539
⇒ x[(100 – 23)/100] = 539
⇒ x × 77/100 = 539  ⇒ x = 539 × 100/77
⇒ x = (539 × 100)/77  = 7 × 100 ⇒  x = 700
∴ Required number = 700

10. Two numbers are respectively 20 percent and 50 percent more than a third number. What percent is the second of the first?
Solution 10
Let the third number = x
∴ First number = x + 20/100x
= (100x + 20x)/100
= 120x/100
Second number = x + 50/100x
= (100x + 50x)/100
= 150x/100
Required % = (150x/100)/(120x/100)
= 150x/100 × 100/120x ×100
= (150 × 100)/120
= 1500/12
= 125%

11. Two numbers are respectively 20 percent and 50 percent of a third number. What percent is the second of the first?
Solution
Let the third number be 100
∴ The first number = 20% of 100
= 20/100 × 100
= 20
And the second number = 50% of 100
= 50/100 × 100
= 50
∴  The second no. as the percent of the first = 50/20 × 100%
= 250%

12. Two numbers are respectively 30 percent and 40 percent less than a third number. What percent is the second of the first ?
Solution
Let the third number = x
∴ First number = x – 30x/100
= (100x – 30x)100
= 70x/100
= 7x/10
Second number = x – 40x/100
= (100x – 40x)/100  = 60x/100
= 6x/10
∴ Required % = (6x/10)/(7x/10) × 100
= 6x/10 × 10/7x × 100
= 600/7
= 85 5/7%