**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 = (325 × 100)13

⇒ x = 32500/13

⇒ x = 2500

∴ Salary = Rs. 2500

Number wrongly read as = 3.265

Error = 3.625 – 3.265

= 0.360

% Error = 0.360/3.625 × 100

= 360/3625 × 100

⇒ 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 × 10**^{3}**is wrongly written as 5.87 × 10**^{3}**; find the percentage error**

**Solution****(i)**Correct number = 3.625Number 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%

Number wrongly written as = 5.87 × 10

Error = 5.87 × 10

= 0.09 × 10

% Error = (0.09 × 10

= 0.09/5.78 × 100

= 9/578 × 100

= 900/578%

= 1.56%

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 = 18,33,600/16

⇒ 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 polled = 1,14,600

Votes secured by winning candidate = 66,468

Votes secured by losing candidate = 48,132

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 polled = 2,06,100

Votes secured by winning candidate = 1,09,233

Present cost of scooter = Rs. 8000

The cost of scooter depreciates by 15% every year

= 85/100 × 8000

= Rs 6800

= 85/100 × 6800

= Rs. 5780

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

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

Total questions = 150

John got correct answers = 80%

Mohan got correct answers = 64%

Numbers of correct answers got by Mohan = 64/100 × 150

= 64/4 × 6

= 96

= 4/5 × 100

= 4 × 20

= 80%

The resulting number = The original number × (1 + 20/100) × (1 – 20/100)

= 8000 × 120/100 × 80/100

= 7,680

The resulting = The original number × (1 – 25/100) × (1 + 25/100)

= 12000 × 75/100 × 125/100

= 11,250

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

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

= 9.93%

**(ii)**Correct number = 5.78 × 10^{3}Number wrongly written as = 5.87 × 10

^{3}Error = 5.87 × 10

^{3}– 5.78 × 10^{3}= 0.09 × 10

^{3}% Error = (0.09 × 10

^{3})/(5.78 × 10^{3}) × 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 = 18,33,600/16

⇒ 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 polled = 1,14,600

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 polled = 2,06,100

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

**Cost of scooter after 2 years = (100 – 15)/100 × 6800**

(ii)(ii)

= 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?****(ii) What percent is Mohan’s correct answers to John’s correct answers?**

**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 = 120Numbers 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%