8. Chapter 8. Percentage and Its Applications NIOS Class 10 Mathematics Textbook Solutions
Chapter 8. Percentage and Its Applications NIOS Class 10 Mathematics Textbook Solutions for Exam Preparation
Chapter 8. PERCENTAGE AND ITS APPLICATIONS
CHECK YOUR PROGRESS 8.1
1. Convert each of the following fractions into a percent:
(a) 
(b) 
(c) 
(d) 
(e) 
(f) 
(g) 
(h) 
(i) 
(j) 
Solution: (a) 
(b) 
(c) 
(d) 
(e) 
(f) 
(g) 
(h) 
(i) 
(j) 
2. Write each of the following percents as a fraction:
(a) 53% (b) 85% (c) 
(d) 3.425% (e) 6.25% (f) 70% (g) 
(h) 0.0025% (i) 47.35% (j) 0.525%
[Note : 
]
Solution: (a) 
(b) 
(c) 
(d) 
(e) 
(f) 
(g) 
(h) 
(i) 
(j) 
3. Write each of the following decimals as a percent:
(a) 0.97 (b) 0.735 (c) 0.03 (d) 2.07 (e) 0.8 (f) 1.75 (g) 0.0250 (h) 3.2575 (i) 0.152 (j) 3.0015
Solution: (a) 
(b) 
(c) 
(d) 
(e) 
(f) 
(g) 
(h) 
(i) 
(j) 
4. Write each of the following percents as a decimal:
(a) 72% (b) 41% (c) 4% (d) 125% (e) 9% (f) 410% (g) 350% (h) 102.5% (i) 0.025% (j) 10.25%
Solution: (a) 
(b) 
(c) 
(d) 
(e) 
(f) 
(g) 
(h) 
(i) 
(j) 
5. Gurpreet got half the answers correct, in an examination. What percent of her answers were correct?
Solution: The number of correct answers is 
of the total answers.
Therefore, the percentage of Gurpreet's answers that were correct 
6. Prakhar obtained 18 marks in a test of total 20 marks. What was his percentage of marks?
Solution: Total marks of the test = 20
Marks obtained by Prakhar = 18
Marks obtained in percent 
Therefore, Prakhar scored 90% in the test.
7. Harish saves Rs 900 out of a total monthly salary of Rs 14400. Find his percentage of saving.
Solution: Total monthly salary of Harish = Rs 14,400
Amount saved by Harish = Rs 900
Percentage of saving 
8. A candidate got 47500 votes in an election and was defeated by his opponent by a margin of 5000 votes. If there were only two candidates and no votes were declared invalid, find the percentage of votes obtained by the winning candidate.
Solution: Votes obtained by the losing candidate = 47,500
Margin of defeat = 5,000 votes
Votes obtained by winning candidate = 47500 + 5000 = 52500
Total votes = 47500 + 52500 = 100000
Therefore, the percentage of votes obtained by the winning candidate 
9. In the word PERCENTAGE, what percent of the letters are E’s?
Solution: Total number of letters = 10
Number of E’s = 3
∴ Percent of E’s 
10. In a class of 40 students, 10 secured first division, 15 secured second division and 13 just qualified. What percent of students failed.
Solution: Total number of students in the class = 40
Number of students who secured first division = 10
Number of students who secured second division = 15
Number of students who just qualified = 13
Number of students who passed=10+15+13=38
Number of students who failed=40−38=2
Percentage of failures 
Therefore, the percentage of students who failed is 5%.
CHECK YOUR PROGRESS 8.2
1. Find: (i) 16% of 1250 (ii) 47% of 1200
Solution: (i) 
(ii) 
2. A family spends 35% of its monthly budget of Rs 7500 on food. How much does the family spend on food?
Solution: Total monthly budget of the family = Rs 7,500
Percentage spent on food = 35%
Amount spent on food 
Therefore, the family spends Rs 2,625 on food.
3. In a garden, there are 500 plants of which 35% are trees, 20% are shrubs, 25% are herbs and the rest are creepers. Find out the number of each type of plants.
Solution: Total number of plants in the garden = 500 .
Percentage of trees = 35%
Percentage of shrubs = 20%
Percentage of herbs = 25%
The number of trees 
The number of shrubs 
The number of herbs 
Percentage of creepers = 100%−(35%+20%+25%) = 100% – 80% = 20%
The number of creepers 
4. 60 is reduced to 45. What percent is the reduction?
Solution: Here, reduction = 60 – 45 = 15
∴ Reduction percent 
5. If 80 is increased to 125, what is the increase percent?
Solution: Here, Increase = 125 – 80 = 45
∴ Increase percent 
6. Raman has to score a minimum 40% marks for passing the examination. He gets 178 marks and fails by 22 marks. Find the maximum marks.
Solution: Marks obtained by Raman = 178
Passing marks=178+22=200
Let the maximum marks be x.
A/Q, 
7. It takes me 45 minutes to go to school and I spend 80% of the time travelling by bus. How long does the bus journey last?
Solution: Total time taken to go to school = 45 minutes
Percentage of time spent travelling by bus = 80%
Time on bus 
Therefore, the bus journey lasts 36 minutes.
8. In an election, between 2 candidates 25% voters did not cast their votes. A candidate scored 40% of the votes polled and was defeated by 900 votes. Find the total number of voters.
Solution: Let the total number of voters be x.
25% of voters did not cast their votes, so the percentage of voters who polled their votes = 100% − 25% = 75%
Therefore, votes polled = 75% of total voters
Votes polled 
The losing candidate scored 40% of the votes polled.
Votes of losing candidate=40% of votes polled 
Votes of losing candidate 
A/Q, Votes of winning candidate−Votes of losing candidate = 900
Therefore, the total number of voters is 6000 .
9. A rise of 25% in the price of sugar compels a person to buy 1.5 kg of sugar less for Rs 240. Find the increased price as well as the original price per kg of sugar.
Solution: Let the original price of sugar be x rupees per kg.
Original quantity 
Increased price 
New quantity 
A/Q, Original quantity − New quantity = 1.5
Increased price 
Therefore, the original price is Rs 32 per kg and the increased price is Rs 40 per kg.
10. A number is first increased by 20% and then decreased by 20%. What is the net increase or decrease percent?
Solution: Let the original number be x
Increased number 
Final number 
Net decrease 
Net decrease percent 
Therefore, there is a net decrease of 4%.
11. ‘A’ scored 12 marks, while B scored 10 marks, in the first terminal examination. If in the second terminal examination (with same total number of marks) ‘A’ scored 14 marks and ‘B’ scored 12 marks, which student showed more improvement?
Solution: For Student A: Marks in first examination = 12 , Marks in second examination = 14
Increase in marks=14−12=2
Percentage increase 
For Student B: Marks in first examination = 10 , Marks in second examination = 12
Increase in marks=12−10=2
Percentage increase 
Therefore, Student B showed more improvement.
12. 30,000 students appeared in a contest. Of them 40% were girls and the remaining boys. If 10% boys and 12% girls won the contest with prizes, find the percentage of students who won prizes.
Solution: Total number of students = 30,000
Percentage of girls = 40%
Percentage of boys = 100% – 40% = 60%
Number of girls 
Number of boys 
The number of boys who won prizes 
The total number of students who won prizes 
The total number of students who won prizes = 1800 + 1440 = 3240
The percentage of students who won prizes 
Therefore, 10.8% of the students won prizes.
13. Sunil earns 10% more than Shailesh and Shailesh earns 20% more than Swami. If Swami earns Rs 3200 less than Sunil, find the earnings of each.
Solution: Let Swami's earnings be x rupees.
Shailesh’s earnings 
Sunil’s earnings 
A/Q, Sunil’s earnings − Swami’s earnings = 3200
Swami earns Rs 10,000.
Shailesh’s earnings 
Sunil’s earnings 
CHECK YOUR PROGRESS 8.3
1. A shopkeeper bought an almirah from a wholesale dealer for Rs 4500 and sold it for Rs 6000. Find his profit or loss percent.
Solution: Here C.P. = Rs. 4500 and S.P. = Rs. 6000
Since C.P. < S.P., ∴ there is a loss.
profit = S.P. – C.P. = Rs (6000 – 4500) = Rs. 1500
Gain% 
2. A retailer buys a cooler for Rs 3800 but had to spend Rs 200 on its transport and repair. If he sells the cooler for Rs 4400, determine, his profit percent.
Solution: Here C.P. = Cost price + Overhead charges = Rs (3800 + 200) = Rs 4000
S.P. = Rs 4400
Since, S.P. > C.P., ∴ Gain = Rs (4400 – 4000) = Rs 400
Gain% 
3. A vendor buys lemons at the rate of 5 for Rs 7 and sells them at Rs 1.75 per lemon. Find his gain percent.
Solution: Cost Price of 1 lemon 
Selling Price of 1 lemon = Rs 1.75
Gain per lemon = SP - CP = Rs 1.75 - Rs 1.40 = Rs 0.35
Gain % 
4. A man purchased a certain number of oranges at the rate of 2 for Rs 5 and sold them at the rate of 3 for Rs 8. In the process, he gained Rs 20. Find the number of oranges he bought.
Solution: The cost price of 1 orange 
The selling price of 1 orange 
Profit per orange = Selling price − Cost price 
Total profit = Rs 20.
Let, x be the number of oranges .
A/Q, 
5. By selling a bi-cycle for Rs 2024, the shopkeeper loses 12%. If he wishes to make again of 12% what should be the selling price of the bi-cycle?
Solution: Selling Price = 100% – 12% = 88%
So, 
The cost Price = Rs 2300.
Selling Price 
Therefore, the selling price = Rs 2576 .
6. By selling 45 oranges for Rs 160, a woman loses 20%. How many oranges should she sell for Rs 112 to gain 20% on the whole?
[ Note: 
and 
]
Solution: Given, Selling Price (S.P.) of 45 oranges = Rs 160 and Loss = 20%
So, the C.P. of 45 oranges = Rs 200.
C.P. of each orange 
So, to gain 20%, S.P. of 1 orange 
Number of oranges 
She should sell 21 oranges for Rs 112 to gain 20% on the whole.
7. A dealer sold two machines at Rs 2400 each. On selling one machine, he gained 20% and on selling the other, he lost 20%. Find the dealer’s net gain or loss percent.
Solution: Given, Selling Price (S.P.) of each machine = Rs 2400
Gain on first machine = 20%
Loss on second machine = 20%
Total C.P. = 2000 + 3000=Rs 5000
Total S.P. =2400+2400=Rs 4800
Total C.P. (Rs 5000) > Total S.P. (Rs 4800) .
Loss = Total C.P. − Total S.P. = 5000 – 4800 =Rs 200
Therefore, the dealer has a net loss of 4% on the whole transaction.
8. Harish bought a table for Rs 960 and sold it to Raman at a profit of 5%. Raman sold it to Mukul at a profit of 10%. Find the money paid by Mukul for the table.
Solution: Given, Harish's Cost Price (C.P.) = Rs 960 , Harish's profit = 5% and Raman's profit = 10%
Again, 
Therefore, the money paid by Mukul for the table is Rs 1108.80 .
9. A man buys bananas at 6 for Rs 5 and an equal number at Rs 15 per dozen. He mixes the two lots and sells them at Rs14 per dozen. Find his gain or loss percent, in the transaction.
Solution: For first lot :
6 bananas cost Rs 5
C.P. of 1 banana 
Cost of 12 bananas 
For second lot :
12 bananas cost Rs 15
C.P. of 1 banana 
Cost of 12 bananas 
Total bananas bought = 12 + 12 = 24 bananas
Total C.P. = 10 + 15 = Rs 25
So, C.P. of 24 bananas = Rs 25.
S.P. of 12 bananas = Rs 14
S.P. of 24 bananas = 14×2=Rs 28
S.P. (Rs 28) > C.P. (Rs 25)
Gain = S.P. − C.P. = 28 − 25 = Rs 3
Therefore, the man gains 12% in the whole transaction.
10. If the selling price of 20 articles is equal to the cost price of 23 articles, find the loss or gain percent.
Solution: Given, Selling Price (S.P.) of 20 articles = Cost Price (C.P.) of 23 articles.
C.P. of 23 articles = 23 × 1 = Rs 23
C.P. of 20 articles = 20 × 1 = Rs 20
A/Q, S.P. of 20 articles = C.P. of 23 articles = Rs 23
Total C.P. (for 20 articles) = Rs 20
Total S.P. (for 20 articles) = Rs 23
S.P. (Rs 23) > C.P. (Rs 20)
Gain = S.P. − C.P. = 23 − 20 = Rs 3
The dealer gains 15% in the transaction.
CHECK YOUR PROGRESS 8.4
1. A shirt with marked price Rs 375/- is sold at a discount of 15%. Find its net selling price.
Solution: Here, Marked Price (M.P.) of the shirt = Rs 375
Discount = 15%
Net selling price (S.P.) = M.P. – Discount
Thus, the net selling price of shirt is Rs 318.75 .
2. A pair of socks marked at Rs 60 is being offered for Rs 48. Find the discount percent being offered.
Solution: Here, Marked Price (M.P.) = Rs 60
Net selling price (S.P.) = Rs 48
Discount offered = Rs (60 – 48) = Rs 12
3. A washing machine is sold at a discount of 10% on its marked price. A further discount of 5% is offered for cash payment. Find the selling price of the washing machine if its marked price is Rs 18000.
Solution: Marked Price (M.P.) = Rs 18000
First Discount = 10% and Second Discount = 5%
Price after 1st discount 
Price after 2nd discount 
The selling price of the washing machine is Rs 15390.
4. A man pays Rs 2100 for a machine listed at Rs 2800. Find the rate of discount offered.
Solution: Marked Price = Rs 2800
Selling Price = Rs 2100
Discount = Rs 2800 – Rs 2100 = Rs 700
5. The list price of a table fan is Rs 840 and it is available to a retailer at a discount of 25%. For how much should the retailer sell it to earn a profit of 15%.
Solution: Marked Price = Rs 840
Discount for the retailer = 25%
Desired Profit = 15%
Cost Price (C.P.) 
Selling Price (S.P.) 
Therefore, the retailer should sell the table fan for Rs 724.50 to earn a profit of 15%.
6. A shopkeeper marks his goods 50% more than their cost price and allows a discount of 40%, find his gain or loss percent.
Solution: Let the C.P. of an article = Rs 100
∴ Marked Price (M.P.) 
Discount offered = 40%
∴ Net selling Price 
∴ Loss = Rs (100 – 90 ) = Rs 10
7. A dealer buys a table listed at Rs 2500 and gets a discount of 28%. He spends Rs 100 on transportation and sells it at a profit of 15%. Find the selling price of the table.
Solution: The dealer gets a 28% discount on the listed price of Rs 2,500. 
Total C.P. = Price after discount + Transportation cost
= 1800 + 100 = Rs 1900
Profit = 15%
Selling Price (S.P.) 
8. A retailer buys shirts from a manufacturer at the rate of Rs 175 per shirt and marked them at Rs 250 each. He allows some discount and earns a profit of 28% on the cost price. What percent discount does he allow to his customers ?
Solution: Here, Cost Price (C.P.) of 1 shirt = Rs 175
Marked Price (M.P.) of 1 shirt = Rs 250
Profit = 28% on cost price
Discount = Marked Price − Selling Price = 250 – 224 = Rs 26
CHECK YOUR PROGRESS 8.5
1. Rama borrowed Rs 14000 from her friend at 8% per annum simple interest. She returned the money after 2 years. How much did she pay back altogether?
Solution: Here, P = Rs 14000 , R = 8% and T 2yrs
We have, Simple Interest (SI) 
Total Amount (A) = P + I = 14000+ 2240 = 16240
Therefore, Rama paid back Rs 16,240 altogether.
2. Ramesh deposited Rs 15600 in a financial company, which pays simple interest at 8% per annum. Find the interest he will receive at the end of 3 years.
Solution: Here, Principal (P) = Rs 15600 , Rate (R) = 8% per annum and Time (T) = 3 years
Simple Intersect (SI) 
Therefore, Ramesh will receive Rs 3,744 as interest at the end of 3 years.
3. Naveen lent Rs 25000 to his two friends. He gave Rs 10,000 at 10% per annum to one of his friend and the remaining to other at 12% per annum. How much interest did he receive after 2 years.
Solution: Total lent = Rs 25,000
For first friend: P = 10000 ,R=10%, T=2 years
For second friend: P =25000 − 10000 = 15000, R =12%, T=2 years
Total interest (SI)=2000+3600=5600
Therefore, Naveen received Rs 5600 as total interest after 2 years.
4. Shalini deposited Rs 29000 in a finance company for 3 years and received Rs 38570 in all. What was the rate of simple interest per annum?
Solution: Here, P = 29000 , T = 3 yrs , A = 38570
We have , 
⇒R=11% 
Therefore, the rate of simple interest per annum is 11% .
5. In how much time will simple interest on a sum of money be 2/5th of the sum, at the rate of 10% per annum.
Solution: Let, the principal = P .
So, 
Rate R=10% per annum
We have, 
Thus, the simple interest will become 
of the principal in 4 years at 10% per annum.
6. At what rate of interest will simple interest be half the principal in 5 years.
Solution: Here, T = 5 yrs
Let, P be the principal .
Simple interest
We have,
%
The required rate of interest is 10% per annum.
7. A sum of money amounts to Rs 1265 in 3 years and to Rs 1430 in 6 years, at simple interest. Find the sum and the rate percent.
Solution: Amount after 3 years = Rs 1,265
Amount after 6 years = Rs 1,430
Time difference = 6 − 3 = 3 years
Interest for 3 years (SI) = 1430 – 1265 = 165
Principal (P) = 1265 – 165 = 1100
We have, 
So,the sum is Rs 1,100 and the rate is 5% per annum.
8. Out of Rs 75000 to invest for one year, a man invested Rs 30000 at 5% per annum and Rs 24000 at 4% per annum. At what percent per annum, should he invest the remaining money to get 6% interest on the whole money.
Solution: Total investment (P) = Rs 75,000
Desired overall rate (R) = 6% per annum , T = 1 yrs
Total Interest Required 
First: Rs 30,000 at 5% for 1 year
Second: Rs 24,000 at 4% for 1 year
Total interest from these two = 1500+960 =2460
Remaining interest needed = 4500−2460=2040
Remaining money to invest = 75000−(30000+24000)=75000−54000 = Rs 21000
Let the required rate be R% .
A/Q, 
He should invest the remaining Rs 21,000 at about 9.7% per annum to get an overall 6% return on the total Rs 75,000.
9. A certain sum of money doubles itself in 8 years. In how much time will it become 4 times of itself at the same rate of interest?
Solution:Solution: Let the principal = P
Amount after 8 years = 2P
So, interest earned in 8 years = 2P−P=P
We have, 
Again, Amount = 4P
Interest needed = 4P−P = 3P
We have,
Thus, the sum will become 4 times itself in 24 years at the same rate.
10. In which case, is the interest earned more:
(a) Rs 5000 deposited for 5 years at 4% per annum, or
(b) Rs 4000 deposited for 6 years at 5% per annum?
Solution: (a) Here, P = 5000 , R = 4% , T = 5 yrs
We have, 
(a) Here, P = 4000 , R = 5% , T = 6 yrs
We have, 
The interest earned is more in case (b): Rs 4,000 at 5% for 6 years.
CHECK YOUR PROGRESS 8.6
1. Calculate the compound interest on Rs 15625 for 3 years at 4% per annum, compounded annually.
Solution: Here, P = 15625 , r = 4% , n = 3
We have, 
2. Calculate the compound interest on Rs 15625 for 
years at 8% per annum, compounded semi-annually.
Solution: Here, P = 15625 , r = 8% per annum = 4% per half year ,
We have, 
3. Calculate the compound interest on Rs 16000 for 9 months at 20% per annum, compounded quarterly
Solution: Here, Principal (P) = Rs 16,000
Time = 9 months = 9/3=3 quarters
Rate = 20% per annum = 20/4 = 5% quarters
We have, 
4. Find the sum of money which will amount to Rs 27783 in 3 years at 5% per annum, the interest being compounded annually.
Solution: Here, A = 27783 , n= 3 yrs , r = 5%
We have, 
5. Find the difference between simple interest and compound interest for 3 years at 10% per annum, when the interest is compounded annually on Rs 30,000.
Solution: For Simple Interest: Here, P = Rs 30,000 ,T = 3 yrs R = 10%
We have, 
For compound interest : Here, P = 30000 , n= 3 yrs , r = 10%
We have,
Therefore, the difference between compound interest and simple interest = 9930 – 9000 = 930 .
6. The difference between simple interest and compound interest for a certain sum of money at 8% per annum for years, when the interest is compounded half-yearly is Rs 228. Find the sum.
Solution: Here, Rate (r)= 8% per annum = 4% half yrs
Time (n)= 1 1/2 years = 3/2 years = 3 half yrs
Difference between CI and SI = Rs 228
Let, P be the principal .
Rate per half-year = 8/2 =4%
We have, Simple Interest (SI) 
and 
A/Q, 
So, the sum of money is Rs 46,875.
7. A sum of money is invested at compound interest for 9 months at 20% per annum, when the interest is compounded half yearly. If the interest were compounded quarterly, it would have fetched Rs 210 more than in the previous case. Find the sum.
Solution: Let the principal sum be Rs x .
Rate = 20% per annum = 10% per half-year.
Time = 9 months
For first 6 months ,
Interest for 3 months ,
Here, R = 20% annually = 20/4 = 5% per quarter .
Time (n) = 9 months = 9/3 = 3 quarters.
We have, 
A/Q, 
Therefore, the principal sum is Rs 80000 .
8. A sum of Rs15625 amounts to Rs 17576 at 8% per annum, compounded semi-annually. Find the time.
Solution: Here, Principal (P) = Rs 15,625 , Amount (A) = Rs 17,576
Rate = 8% per annum = 4% semi-annum , n= ?
We have, 
Therefore, the time 
9. Find the rate at which Rs 4000 will give Rs 630.50 as compound interest in 9 months, interest being compounded quarterly.
Solution: Here, Principal (P) = Rs 4,000
Compound Interest (CI) = Rs 630.50
Time = 9 months = 9/3 = 3 quarters , r = ?
We have,
10. A sum of money becomes Rs17640 in two years and Rs18522 in 3 years at the same rate of interest, compounded annually. Find the sum and the rate of interest per annum.
Solution: Let the principal be P and rate be r% per annum.
We know that,
For 2 years : Here, A = 17640 , n = 2 yrs
For 3 years : Here, A= 18522 , n=3 yrs
From (i) , we get 
Therefore, the sum is Rs 16000 and the rate of interest is 5% per annum
CHECK YOUR PROGRESS 8.7
1. The population of a town is 281250. What will be its population after 3 years, if the rate of growth of population is 4% per year?
Solution: Here, 
We have, 
2. The cost of a car was Rs 4,36,000 in January 2005. Its value depreciates at the rate of 15% in the first year and then at the rate of 10% in the subsequent years. Find the value of the car in January 2008.
Solution: Here, 
, r= 15% , n= 2008 – 2005 =3 yrs
We have, 
For the first year (2005-2006), the rate is 15% .
For the subsequent years (2006-2007), the rate is 10%
For the subsequent years (2007-2008), the rate is 10%
3. The cost of machinery is Rs 360000 today. In the first year the value depreciates by 12% and subsequently, the value depreciates by 8% each year. By how much, the value of machinery has depreciated at the end of 3 years?
Solution: Given, Initial cost of machinery, 
Depreciation in the 1st year = 12%
Depreciation in the 2nd and 3rd years = 8% each year
We have,
For the first year, the rate is 12% .
Depreciation in the 2nd years = 8% each year
Depreciation in the 3rd years = 8% each year
4. The application of manure increases the output of a crop by 10% in the first year, 5% in the second year and 4% in the third year. If the production of crop in the year 2005 was 3.5 tons per hectare, find the production of crop per hectare in 2008.
Solution: Here, 
tons per hectare
Time period = 3 years (from 2005 to 2008)
We have,
Increase in 1st year (2005–2006) = 10%
Increase in 2nd year (2006–2007) = 5%
Increase in 3rd year (2007–2008) = 4%
Therefore, the production of crop per hectare in 2008 is 4.2042 tons.
5. The virus of a culture decreases at the rate of 4% per hour due to a medicine. If the virus count in the culture at 9.00 AM was
, find the virus count at 11.00 AM on the same day.
Solution: Here, 
Rate of decrease (depreciation) = 4% per hour
Time duration = From 9:00 AM to 11:00 AM = 2 hours
We have, 
Therefore, the virus count at 11:00 AM is 
.
6. Three years back, the population of a village was 50000. After that, in the first year, the rate of growth of population was 5%. In the second year, due to some epidemic, the population decreased by 10% and in the third year, the population growth rate was noticed as 4%. Find the population of the town now.
Solution: Here, 
Time period = 3 years
We have, 
For 1st year (growth) = +5%
For 2nd year (decrease ) = 10%
For 3rd (growth) = 4%
Therefore, the population of the village is 49140.
TERMINAL EXERCISE
1. Write each of the following as a percent
(a) 9/20 (b) 7/10 (c) 0.34 (d) 0.06
Solution: (a) 
(b) 
(c) 
(d) 
2. Write each of the following as a decimal:
(a) 36% (b) 410% (c) 2% (d) 0.35%
Solution: (a) 
(b) 
(c) 
(d) 
3. Write each of the following as fraction:
(a) 0.12% (b) 2.5% (c) 25.5% (d) 255%
Solution: (a) 
(b) 
(c) 
(d) 
4. Find each of the following:
(a) 23% of 500 (b) 2.5% of 800 (c) 0.4% of 1000 (d) 115% of 400
Solution: (a) 23% of 500
(b) 2.5% of 800
(c) 0.4% of 1000
(d) 115% of 400
5. What percent of 700 is 294?
Solution: We have, 
6. By what percent is 60 more than 45?
Solution: Increase = 60 – 45 =15
Increase% 
7. What number increased by 10% of itself is 352?
Solution: Let the required number be y .
A/Q, 
8. Find the number whose 15% is 270.
Solution: Let the required number be p .
A/Q, 
9. What number decreased by 7% of itself is 16.74?
Solution: Let the required number be k .
A/Q, 
10. If three fourth of the students of a class wear glasses, what percent of the students of the class do not wear glasses?
Solution: The students wear glasses 
The fraction of students who do not wear glasses 
Therefore, the percent of students who do not wear glasses
11. There are 20 eggs in a fridge and 6 of them are brown. What percent of eggs are not brown?
Solution: Total eggs = 20 , Brown eggs = 6
Not brown = 20−6=14
Therefore, the percent of eggs are not brown 
12. 44% of the students of a class are girls. If the number of girls is 6 less than the number of boys, how many students are there in the class?
Solution: Let the total number of students be y .
Girls 
Boys 
A/Q, 
13. During an election, 70% of the population voted. If 70,000 people cast their votes, what is the population of the town?
Solution: Let P be the total population .
A/Q, 
14. A man donated 5% of his monthly income to a charity and deposited 12% of the rest in a Bank. If he has Rs. 11704 left with him, what is his monthly income?
Solution: Let x be the man's monthly income .
Income left after donation 
Bank deposit 
Amount left 
A/Q, 
15. Ratan stores has a sale of Rs 12000 on Saturday, while Seema stores had a sale of Rs 15000 on that day. Next day, they had respective sales of Rs 15000 and Rs 17500. Which store showed more improvement in Sales?
Solution: Ratan Stores: Sales on Saturday = Rs. 12,000
Sales on the next day = Rs. 15,000
Increase in sales = 15,000−12,000 = 3,000
Percentage increase 
Seema Stores: Sales on Saturday = Rs. 15,000
Sales on the next day = Rs. 17,500
Increase in sales = 17,500−15,000=2,500
Percentage increase 
Since 25% > 16.67%
Therefore, Ratan Stores showed more improvement in sales.
16. A candidate has to secure 45% marks in aggregate of three papers of 100 marks each to get through. He got 35% marks in the first paper and 50% marks in the second paper. At least how many marks should he get in third paper to pass the examination?
Solution: Total marks=3×100=300
Passing marks
For first paper:
For second paper:
Total marks first two paper = 35+50=85
Therefore, marks required in the third paper = 135 – 85 = 50
17. The price of sugar rises by 25%. By how much percent should a householder reduce his consumption of sugar, so as not to increase his expenditure on sugar?
Solution: Let, the original price of sugar is Rs. 100 per kg.
The new price = 100+25 = 125 per kg.
Percentage reduction in consumption 
Therefore, the householder should reduce his consumption of sugar by 20%.
18. By selling 90 ball pens for Rs 160, a person loses 20%. How many ball pens should he sell for Rs. 96, so as to have a gain of 20%?
Solution: At 20% loss,
For a 20% gain, Selling Price= (100%+20%) of Cost Price
The number of pens 
Therefore, the person must sell 36 ball pens for Rs. 96 to achieve a 20% gain .
19. A vendor bought bananas at 6 for 5 rupees and sold them at 4 for 3 rupees. Find his gain or loss percent.
Solution: Cost Price (CP) per banana 
Selling Price (SP) per banana
Since, C.P. > S.P.
Loss% 
20. A man bought two consignments of eggs, first at Rs 18 per dozen and an equal number at Rs20 per dozen. He sold the mixed egges at Rs 23.75 per dozen. Find his gain percent.
Solution: Cost of first consignment = Rs 18
Cost of second consignment = Rs 20
Total C.P.= 18 + 20 = Rs 38
Total dozens = 1 + 1 = 2 dozen
Total S.P.= 2 × 23.75 = Rs 47.50
Gain = S.P.− C.P.= 47.50 – 38 = Rs 9.50
Gain% 
21. A man sells an article at a gain of 10%. If he had bought it for 10% less and sold it for Rs 10 more, he would have gained 25%. Find the cost price of the article.
Solution: Let, x be the cost price (C.P.) of the article .
Selling price (S.P.) 
New C.P. 
A/Q, New S.P. = New C.P. + 25% of new C.P.
Therefore, the cost price of the article is Rs. 400.
22. A pair of socks is marked at Rs 80 and is being offered at Rs 64. Find the discount percent being offered.
Solution: Here, Marked price (M.P.) = Rs 80
Selling price (S.P.) = Rs 64
Discount=M.P.−S.P.= 80−64=Rs16
Therefore, the discount percent being offered is 20%.
23. A dealer buys a table listed at Rs 1800 and gets a discount of 25%. He spends Rs 150 on transportation and sells it at a profit of 10%. Find the selling price of the table.
Solution: Here, Marked price (M.P.) = Rs 1800
Discount = 25%
Transportation cost = Rs 150
Profit = 10%
Purchase price=1800 − 450 = Rs 1350
Total C.P. = 1350 + 150 = Rs 1500
S.P. = 1500 + 150 = Rs1650
Therefore, the selling price of the table is Rs 1650.
24. A T.V. set was purchased by paying Rs 18750. If the discount offered by the dealer was 25%, what was the marked price of the TV set?
Solution: Here, Selling Price = 18750 ,
We have, S.P.= M.P.− Discount
25. A certain sum of money was deposited for 5 years. Simple interest at the rate of 12% was paid. Calculate the sum deposited if the simple interest received by the depositor is Rs 1200.
Solution: Here, T = 5 years , R = 12% , SI = Rs 1200
Let, P be the Principal .
We have, 
Therefore, the sum deposited was Rs 2000.
26. Simple intrest on a sum of money is 1/3rd of the sum itself and the number of years is thrice the rate percent. Find the rate of interest.
Solution: Let, P be the principal .
Simple Interest 
, Time (T) = 3 × R = 3R
We have,
Therefore, the rate of interest is 
per annum .
27. In what time will Rs 2700 yield the same interest at 4% per annum as Rs 2250 in 4 years at 3% per annum?
Solution: Here, P = Rs 2250 , R = 3% per annum , T = 4 years
We have , 
Again, P = 2700 , R = 4% , Let, the time be T ,
Therefore, the required time is 2.5 years (or 2 years 6 months).
28. The difference between simple interest on a sum of money for 3 years and for 2 years at 10% per annum is Rs 300. Find the sum.
Solution: Let, P be the principal .
Here, R = 10% per annum
Time difference (T) = 3 years − 2 years = 1 year
We have , 
Therefore, the required sum is Rs 3,000.
29. Find the sum which when invested at 4% per annum for 3 years will becomes Rs 70304, when the interest is compounded annually.
Solution: Let, P be the Principal .
Here, A= 70304 , r = 4% , n= 3
We have,
Therefore, the Principal (sum) is Rs 62500 .
30. The difference between compound interest and simple interest at 10% per annum in 2 years (compounded annually) is Rs 50. Find the sum.
Solution: Let, P be the principal .
Here, Rate (r) = 10% per annum , Time (n) = 2 years
And 
A/Q, 
Therefore, the sum is Rs 5,000.
31. A sum of money becomes Rs 18522 in three years and Rs 19448.10 in 4 years at the same rate of interest, compounded annually. Find the sum and the rate of interest per annum.
Solution: Let the principal be P and rate be r% per annum.
We know that,
For 3 years : Here, A = 18522 , n = yrs
For 4 years : Here, A= 19448.10 , n=4 yrs
From (i) , we get 
Therefore, the sum of money is Rs 16000 and the rate is 5% .
32. Find the sum of money which will amount to Rs 26460 in six months at 20% per annum, when the interest is compounded quarterly.
Solution: Let P be the Principal .
Here, A = 26460 ,
r = 20% per annum = 20/4 = 5% per quarters
n= 6 months = ½ yrs = 4 × ½ =2 quarters
We have,
Therefore, the required sum of money is Rs 24,000.
33. At what rate percent per annum will a sum of Rs 12000 amount to Rs 15972 in three years, when the interest is compounded annually?
Solution: Here, P = Rs 12,000 , A = Rs 15,972 , n = 3 years
We have,
Therefore, the required rate of interest is 10% per annum.
34. The price of a scooter depreciates at the rate of 20% in the first year, 15% in the second year and 10% afterwards, what will be the value of a scooter now costing Rs 25000, after 3 years.
Solution: Here, 
We have, 
35. The population of a village was 20,000, two years ago. It increased by 10% during first year but decreased by 10% in the second year. Find the population at the end of 2 years.
Solution: Given, 
Population after the first year (10% increase) : Here , n = 1
We have, 
Population after the second year (10% decrease) : Here, n = 1
We have , 
Therefore , the population at the end of 2 years is 19,800.
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