# Samacheer Kalvi Class 7 Maths Term 3 Chapter 2 Solutions

## Samacheer Kalvi Class 7 Maths Term 3 Chapter 2 Percentage and Simple Interest Solutions

Welcome to NCTB Solutions. Here with this post we are going to help 7th class students by providing Solutions for Samacheer Kalvi Class 7 Maths Term 3 chapter 2 Percentage and Simple Interest. Here students can easily find all the solutions for Percentage and Simple Interest Exercise 2.1, 2.2, 2.3, 2.4 and 2.5. Also here our Expert Maths Teacher’s solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 2 solutions. Here all the solutions are based on Tamil Nadu State Board latest syllabus. #### Percentage and Simple Interest Exercise 2.1 Solutions :

(2) A picture of chess board is given.

(i) Find the percentage of the white coloured squares.

(ii) Find the percentage of grey coloured squares.

(iii) Find the percentage of the squares that have the pieces and

(iv) The squares that do not have the pieces.

Solution :

(i) Total squares = 64

white colored squares = 32

32/64 = 32/64 × 100/100

= 32/64 × 100%

= 3200/64

= 50%

(ii) Total squares = 64

Grey colored squares = 32

Thus, 32/64 = 32/64 × 100/100

= 32/64 × 100%

= 3200/64

= 50%

(iii) Total squares = 64

Squares have the pieces = 20

Thus, 20/64 = 20/64 × 100/100

= 20/64 ×100%

= 2000/64

= 31.25 %

(iv) The squares that do not have the pieces.

Total squares = 64

Squares have not the pieces = 44

Thus, 44/64 = 44/64 × 100/100

= 44/64 × 100%

= 4400/64

= 68.75%

(4) Write each of the following fraction as percentage

(i) 36/50

(ii) 81/30

(iii) 42/56

(iv) 2 1/4

(v) 1 3/5

Solution :

(i) 36/50 = 36/50 × 100/100

= 36/50 × 100%

= 3600/50%

= 72 %

(ii) 81/30 = 81/30 × 100/100

= 81/30 × 100%

= 8100/30%

= 270%

(iii) 42/56 = 42/56 × 100/100

= 42/56 × 100 %

= 4200/56

= 75%

(iv) 2 1/4 = 9/4

= 9/4 × 100/100

= 9/4 ×100%

= 900/4%

= 225%

(v) 1 3/5 = 8/5

= 8/5 × 100/100

= 8/5 ×100%

= 800/5%

= 160%

(5) Anbu scored 436 marks out of 500 in his exams. What was the percentage he scored?

Solution :

Given, Anbu scored 436 marks out of 500 in his exams.

Thus, 436/500 = 436/500 × 100/100

= 436/500 × 100%

= 436/5%

= 87.2 %

Therefore, Anbu will scored is 87.2%

(6) Write each of the following percentage as fraction.

(i) 21%

(ii) 93.1%

(iii) 151%

(iv) 65%

(v) 0.64%

Solution :

(i) 21%

= 21/100

(ii) 93.1%

= 93.1/100

= 931/1000

(iii) 151%

= 151/100

(iv) 65%

= 65/100

= 13/20

(v) 0.64%

= 0.64/100

= 4/625

(7) Iniyan bought 5 dozen eggs. Out of that 5 dozen eggs, 10 eggs are rotten. Express the number of good eggs as percentage.

Solution :

Given, Iniyan bought 5 dozen eggs.

We know, 1 dozen = 12

Thus, 5 dozen eggs = 60 eggs

Since, 10 eggs rotten

Thus, good eggs = 60 – 10 = 50 eggs

Now, 50/60 = 50/60 × 100/100

= 50/60 × 100%

= 5000/60%

= 83.33 %

Hence, the percentage of good eggs 83.33%

(8) In an election, Candidate X secured 48% of votes. What fraction will represent his votes?

Solution :

Given, in an election, Candidate X secured 48% of votes.

Thus, 48%

= 48/100

= 12/25

Hence, fraction of votes is 12/25.

(9) Ranjith’s total income was ₹ 7,500. He saved 25% of his total income. Find the amount saved by him.

Solution :

As per the question,

Save out of 100 = 25

So, save out of 7500 = x

Therefore,

x = 7500 × 25/100

= 75 × 25

= 1875

Therefore, Ranjith’s save the amount is ₹1875.

(10) Thendral saved one fourth of her salary. her savings percentage is

(i) 3/4%

(ii) 1/4 %

(iii) 25%

(iv) 1%

Solution :

Correct Option → (iii)

Thendral saved one fourth of her salary. her savings percentage is 25%.

(11) Kavin scored 15 out of 25 in a test. The percentage of his marks is

(i) 60%

(ii) 15%

(iii) 25%

(iv) 15/25

Solution :

Correct Option → (i)

The percentage of his marks will be 60%

(12) 0.07% is

(i) 7/10

(ii) 7/100

(iii) 7/1000

(iv) 7/10000

Solution :

Correct Option → (i)

0.07% is 7/10000

#### Percentage and Simple Interest Exercise 2.2 Solutions :

(1) Write each of the following percentage as decimal.

(i) 21%

(ii) 93.1%

(iii) 151%

(iv) 65%

(v) 0.64%.

Solution :

(i) 21% = 21/100

= 0.21

(ii) 93.1% = 93.1/100

= 931/1000

= 0.931

(iii) 151% = 151/100

= 1.51

(iv) 65% = 65/100

= 0.65

(v) 0.64% = 0.64/100

= 64/10000

= 0.0064

(2) Convert each of the following decimal as percentage

(i) 0.282

(ii) 1.51

(iii) 1.09

(iv) 0.71

(v) 0.858

Solution :

(i) 0.282

= 282/1000 × 100%

= 282/10 %

= 28.2%

(ii) 1.51

= 151/100 ×100%

= 151 %

(iii) 1.09

= 109/100 × 100%

= 109%

(iv) 0.71

= 71/100 × 100%

= 71%

(v) 0.858

= 858/1000 × 100%

= 858/10 %

= 85.8%

(3) In an examination a student scored 75% of marks. Represent the given percentage in decimal form?

Solution :

Given, in an examination a student scored 75% of marks.

Thus, 75%

= 75/100

= 0.75

(4) In a village 70.5% people are literate. Express it as a decimal.

Solution :

Given, in a village 70.5% people are literate.

Thus, 70.5%

= 70.5/100

= 0.705

(5) Scoring rate of a batsman is 86%. Write his strike rate as decimal.

Solution :

Given, scoring rate of a batsman is 86%

Thus, 86%

= 86/100

= 0.86

(6) The height of a flag pole in a school is 6.75m. Express it as percentage.

Solution :

Given, the height of a flag pole in a school is 6.75 m.

Thus, 6.75 = 675/ 100

= 675/100 × 100%

= 675%

(7) The weights of two chemical substances are 20.34 g and 18.78 g. Write the difference in percentage.

Solution :

Difference,

= 20.34 – 18.78

= 1.56 g

Thus, 1.56

= 156/100 × 100%

= 156%

(8) Find the percentage of shaded region in the following figure.

Solution :

The fraction in the Fig, that is shaded = 1/4

So, the percentage of shaded portion

= 1/4 × 100%

= 25%

Therefore, 25% of the fig. is shaded.

Objective Type Question Solution :

(9) Decimal value of 142.5% is

(i) 1.425

(ii) 0.1425

(iii) 142.5

(iv) 14.25

Solution :

Correct Option → (i)

Decimal value of 142.5% is 1.425.

(10) The percentage of 0.005 is

(i) 0.005%

(ii) 5%

(iii) 0.5%

(iv) 0.05%

Solution :

Correct Option → (iii)

The percentage of 0.005 is 0.5%

(11) The percentage of 4.7 is

(i) 0.47%

(ii) 4.7%

(iii) 47%

(iv) 470%

Solution :

Correct Option → (iv)

The percentage of 4.7 is 470%

#### Percentage and Simple Interest Exercise 2.3 Solutions :

(1) 14 out of the 70 magazines at the bookstore are comedy magazines. What percentage of the magazines at the bookstore are comedy magazines?

Solution :

Given, 14 out of the 70 magazines at the bookstore are comedy magazines.

Let X% of comedy magazines

So, X% of 70 = 14

X/100 × 70 = 14

X = 14 × 100/70

X = 1400/70

X = 20 %

Therefore, 20% of comedy magazines in the bookstore.

(2) A tank can hold 50 litres of water. At present, it is only 30 % full. How many litres of water will fill the tank, so that it is 50 % full?

Solution :

Given, A tank can hold 50 litres of water.

Let x litres water required to 50% full tank

Thus, 50% of 50

= 50/100 × 50

= 2500/100

= 25 litres

Therefore, 25 litres water required to 50% full tank

Since, at present, it is only 30% full.

Thus, 30 % of 50

= 30/100 × 50

= 1500/100

= 15 litres

Now, 25 – 15

= 10 litres

Hence, 10 litres water required.

(4) An agent of an insurance company gets a commission of 5% on the basic premium he collects. What will be the commission earned by him if he collects ₹ 4800?

Solution :

5% of ₹4800

= 5/100 × 4800

= 5 × 48

= ₹240

Therefore, the agent will earned ₹240 commission on ₹4800.

(5) A biology class examined some flowers in a local Grass land. Out of the 40 flowers they saw, 30 were perennials. What percentage of the flowers were perennials?

Solution :

Since, out of the 40 flowers they saw, 30 were perennials.

Thus, 30/40

= 30/40 × 100%

= 75%

Thus, the percentage of the flowers were perennials is 75%

Solution :

Thus, 15/50

= 15/50 × 100%

= 15 × 2

= 30%

Therefore, the percentage of brown beads is 30%

(7) Ramu scored 20 out of 25 marks in English, 30 out of 40 marks in Science and 68 out of 80 marks in mathematics. In which subject his percentage of marks is best?

Solution :

Given, Ramu scored

In English out of 25 = 20

% in English

= 20/25 × 100%

= 20×4

= 80%

In Science out of 40 = 30

% is Science

= 30/40 × 100%

= 75%

In Mathematics out of 80 = 68

% in Mathematics

= 68/80 × 100

= 85%

Therefore, in Mathematics subject Ramu percentage is best.

(8) Peter requires 50% to pass. If he gets 280 marks and falls short by 20 marks, what would have been the maximum marks of the exam?

Solution :

Since, he gets 280 marks and falls short by 20 marks

Thus, 280 + 20 = 300 marks

Let x marks have maximum marks.

Thus, maximum Marks,

= 300/50 × 100

= 300 × 2

= 600 marks

Thus, maximum marks will be 600 marks.

(9) Kayal scored 225 marks out of 500 in revision test 1 and 265 out of 500 marks in revision test 2. Find the percentage of increase in her score.

Solution :

Given, For revision test 1

Kayal scored out of 500 = 225 marks

For revision test 2

Kayal scored out of 500 = 265 marks

Thus, 265 – 225 = 40

Let x% of increase in her score

Thus,

x% = 40×100/500

= 40/5

= 8%

Therefore, the 8% of increase in her score.

(10) Roja earned ₹ 18,000 per month. She utilized her salary in the ratio 2 : 1 : 3 for education, savings and other expenses respectively. Express her usage of income in percentage.

Solution :

Given, Roja earned ₹ 18,000 per month.

Let ₹2x , ₹x and ₹3x for education, savings and other expenses respectively.

Thus, 2x + x + 3x = 18000

6x = 18000

x = 18000/6

= 3000

Therefore, for education

= 2x = 2 × 3000

= ₹6000

% for education

= 6000/18000 × 100

= 33.33%

Now, for saving = x = 3000

%for saving

= 3000/18000 × 100

= 16.66 %

Also, for other expenses

= 3x = 3 × 3000

= 9000

% for other expenses

= 9000/18000 × 100

= 50%

Hence, her usage of income in percentage is 33.33%,16.66% and 50% for education, savings and other expenses respectively.

#### Percentage and Simple Interest Exercise 2.4 Solutions :

(1) Find the simple interest on ₹ 35,000 at 9% per annum for 2 years?

Solution :

Given, Principal (P) = ₹35000

r = 9 % , n = 2 years

We know simple interest (I) = Prn/100

= 35000 × 9 × 2/100

= 350×18

= ₹6300

Therefore, the simple interest is ₹6300

(2) Aravind borrowed a sum of ₹ 8,000 from Akash at 7% per annum. Find the interest and amount to be paid at the end of two years.

Solution :

Given, P = ₹ 8,000 r = 7% and n = 2 years

We know, I = Prn/100

= 8000 × 7 × 2/100

= 80×14

= ₹ 1120

Now, Amount = P + I

= 8000 + 1120

= ₹ 9120

Therefore, Arvind paid ₹ 9120 at the end of 2 years.

(3) Sheela has paid simple interest on a certain sum for 4 years at 9.5% per annum is ₹ 21,280. Find the sum.

Solution :

Given, I = ₹ 21,280, r = 9.5 % and n = 4 years

To find P = ?

We know, I = Prn/100

P = I × 100/rn

P = 21280 × 100/9.5 × 4

P = 2128000/38

P = ₹ 56,000

Therefore, the sum will be ₹56,000

(4) Basha borrowed ₹ 8,500 from a bank at a particular rate of simple interest. After 3 years, he paid ₹ 11,050 to settle his debt. At what rate of interest he borrowed the money?

Solution :

Given, P = ₹8,500 , n = 3 , A = ₹11,050

Thus, I = A – P = 11050 – 8500 = ₹2550

To find r = ?

We know, I = Pnr/100

r = I × 100/Pn

r = 2550 × 100/8500 × 3

r = 2550/255

r = 10%

Therefore, the rate of interest will be 10%

(5) In What time will ₹ 16,500 amount to ₹ 22,935 at 13% per annum?

Solution :

Given, P = ₹ 16,500 , A = ₹ 22,935 and r = 13 %

Thus, I = A – P = 22935 – 16500 = ₹6435

We know, I = Prn /100

n = I × 100/Pr

n = 6435 × 100/16500 × 13

n = 643500/214500

n = 3

Therefore, the time is 3 years

(6) In what time will ₹ 17800 amount to ₹ 19936 at 6% per annum?

Solution :

Given, P = ₹ 17800 , A = ₹19936 and r = 6 %

Thus, I = A – P = 19936 – 17800 = ₹2136

We know, I = Prn /100

n = I × 100/Pr

n = 2136 × 100/17800 × 6

n = 213600/106800

n = 2

Therefore, the time is 2 years.

(8) A principal becomes ₹ 17,000 at the rate of 12% in 3 years. Find the principal.

Solution :

Given, A = ₹17,000 , r = 12% and n = 3 years

We know, I = Pnr/100

A = P + I

Thus, A = P + Pnr/100

A = P(1 + nr/100)

A = P ( 1 + 3 × 12/100)

A = P (1 + 36/100)

A = P(136/100)

A = P × 1.36

P = A/1.36

P = 17000/1.36

P = ₹ 12500

Therefore, the principal will be ₹ 12,500

Objective Type Question Solution :

(9) The interest for a principle of ₹ 4,500 which gives an amount of ₹ 5,000 at end of certain period is

(i) ₹ 500

(ii) ₹ 200

(iii) 20%

(iv) 15%

Solution :

Correct Option → (i)

At end of certain period is Rs 500.

(10) Which among the following is the simple interest for the principle of ₹ 1,000 for one year at the rate of 10% interest per annum?

(i) ₹ 200

(ii) ₹ 10

(iii) ₹ 100

(iv) ₹ 1,000

Solution :

Correct Option → (iii)

₹ 100 is the correct simple interest.

(11) Which among the following rate of interest yields an interest of ₹ 200 for the principle of ₹ 2,000 for one year.

(i) 10%

(ii) 20%

(iii) 5%

(iv) 15%

Solution :

Correct Option → (iii)

10% is the correct rate of interest.

#### Percentage and Simple Interest Exercise 2.5 Solutions :

(1) When Mathi was buying her flat she had to put down a deposit of 1/10 of the value of the flat. what percentage was this?

Solution :

1/10 = 1/10 × 100%

= 10 %

Hence, the percentage of the deposit is 10%.

(2) Yazhini scored 15 out of 25 in a test. Express the marks scored by her in percentage.

Solution :

Given, Yazhini scored 15 out of 25 in a test.

Thus, 15/25 = 15/25 × 100%

= 15 × 4 %

= 60 %

Thus, her marks percentage is 60%

(3) Out of total 120 teachers of a school 70 were male. Express the number of male teachers as percentage.

Solution :

Given, out of total 120 teachers of a school 70 were male.

Thus, 70/120 = 7/12 ×100%

= 700/12

= 58.33 %

Therefore, the percentage of male teacher is 58.33%

(4) A cricket team won 70 matches during a year and lost 28 matches and no results for two matches. Find the percentage of matches they won.

Solution :

Thus, total match

= 70 + (28 + 2)

= 70 – 30

= 100

Therefore, 70/100

= 70/100 × 100 %

= 70%

Hence, the percentage of won matches is 70%

(5) There are 500 students in a rural school. If 370 of them can swim, what percentage of them can swim and what percentage cannot?

Solution :

Given, there are 500 students in a rural school.

Can swim students = 370

Can’t swim students

= 500 – 370

= 130

Thus, 370/500

= 370/500 × 100 %

= 370/5

= 74 %

Also, 130/500

= 130/500 × 100 %

= 130/5

= 26%

Therefore, percentage of them can swim and what percentage cannot is 74% and 26% respectively.

(7) A salesman is on a commission rate of 5%. How much commission does he make on sales worth ₹ 1,500?

Solution :

Given, a salesman is on a commission rate of 5%.

Since, total sales = ₹1500

So, 5% of ₹1500

= 5/100 × 1500

= 5 × 15

= ₹75

Thus, a salesman commission on ₹1500 is ₹75.

(8) In the year 2015 ticket to the world cup cricket match was ₹ 1,500. This year the price has been increased by 18%. What is the price of a ticket this year?

Solution :

Since, This year the price has been increased by 18%.

So, 18% of ₹1500

= 18/100 × 1500

= 18 ×15

= ₹270

Thus, 1500 + 270 = 1770

Hence, the price of ticket in this year is ₹1770.

(9) 2 is what percentage of 50?

Solution :

2/50 = 2/50 × 100%

= 2 × 2%

= 4%

(10) What percentage of 8 is 64?

Solution :

64/8 = 64/8 ×100%

= 6400/8%

= 800%

(11) Stephen invested ₹ 10,000 in a savings bank account that earned 2% simple interest. Find the interest earned if the amount was kept in the bank for 4 years.

Solution :

Given, P = ₹10,000, r = 2% and n = 4 years

We know, I = Prn/100

I = 10000 × 2 × 4/100

I = 100 × 2 × 4

I = 100 × 8

I = ₹ 800

Therefore, he earned interest is ₹800

(12) Riya bought ₹ 15,000 from a bank to buy a car at 10% simple interest. If she paid ₹ 9,000 as interest while clearing the loan, find the time for which the loan was given.

Solution :

Given, P = ₹ 15,000 , r = 10% and I = ₹9000

To find n = ?

We know

I = Prn/100

n = I × 100/Pr

n = 9000 × 100/ 15000 × 10

n = 90/15

n = 6

Hence, the time is 6 years.

(13) In how much time will the simple interest on ₹ 3,000 at the rate of 8% per annum be the same as simple interest on ₹ 4,000 at 12% per annum for 4 years?

Solution :

Given, P1 = ₹3000 , r1 = 8%

P2 = ₹4000, r2 = 12 % and n2 =4 years

Since, I1 = I2

We know, I = Prn/100

Thus, I1 = I2 = P2 × r2 × n2/100

= 4000 × 12 × 4/100

= 40 × 12 × 4

= ₹1920

Thus, I1 = ₹1920

Now, I1 = P1 × r1 × n1/100

n1 = I1 × 100/P1 × r1

= 1920 × 0100/3000 × 8

= 192/24

= 8 years

Hence, time is 8 years for the simple interest on ₹ 3,000 at the rate of 8%

Challenge Problem Solutions :

(14) A man travelled 80 km by car and 320 km by train to reach his destination. Find what percent of total journey did he travel by car and what per cent by train?

Solution :

Given, a man travelled 80 km by car and 320 km by train to reach his destination.

Total distance = 80 + 320 = 400 km

Thus, % travel by car

= 80/400 × 100%

= 80/4

= 20 %

Also, % travel by train

= 320/400 ×100 %

= 320/4

= 80 %

Hence, percent of total journey did he travel by car and by train is 20% and 80 % resp.

(15) Lalitha took a math test and got 35 correct and 10 incorrect answers. What was the percentage of correct answers?

Solution :

Given, Lalitha took a math test and got 35 correct and 10 incorrect Solution were.

Total Solution were = 35 + 10 = 45

Thus, % of correct Solution were

= 35/45 × 100 %

= 3500/45

= 700/9

= 77.77 %

Hence, the percentage of correct Solution were is 77.77 %.

(16) Kumaran worked 7 months out of the year. What percentage of the year did he work?

Solution :

Given, Kumaran worked 7 months out of the year.

Thus, 7/12

= 7/12 × 100%

= 700/12

= 58.33 %

Hence, the percentage of the year did he work is 58.33%

(17) The population of a village is 8000. Out of these, 80% are literate and of these literate people, 40% are women. Find the percentage of literate women to the total population?

Solution :

Given, the population of a village is 8000.

Out of these, 80% are literate

So, 80% of 8000

= 80/100 × 8000

= 80 × 80

= 6400

Thus, there are 6400 people literate.

Since, 40% of women literate out of 6400.

Thus, 40% of 6400

= 40/100 × 6400

= 40×64

= 2560

Now, 2560/8000 = 2560/8000 × 100 %

= 2560/80

= 256/8

= 32%

Hence, the percentage of literate women to the total population is 32%

(18) A student earned a grade of 80% on a math test that had 20 problems. How many problems on this test did the student answer correctly?

Solution :

Given, a student earned a grade of 80% on a math test that had 20 problems.

Thus, 80% of 20

= 80/100 × 20

= 8 × 2

= 16

Hence, the student Solution were correctly is 16 problems.

(19) A metal bar weighs 8.5 kg. 85% of the bar is silver. How many kilograms of silver are in the bar?

Solution :

Given a metal bar weighs 8.5 kg. 85% of the bar is silver.

Thus, 85% of 8.5

= 85/100 × 8.5

= 7.225 kg

Hence, 7.225 kg of silver are in the bar

(20) Concession card holders pay ₹ 120 for a train ticket. Full fare is ₹ 230. What is the percentage of discount for concession card holders?

Solution :

Given, Concession card holders pay ₹ 120 for a train ticket.

Since, full fare is ₹ 230

Thus, Concession = ₹ 230 – ₹120 = ₹110

Now, 110/230 = 110/230 × 100%

= 1100/23 %

= 47.82%

Hence, the percentage of discount for concession card holders is 47.82%.

(21) A tank can hold 200 litres of water. At present, it is only 40 % full. How many litres of water to fill in the tank, so that it is 75 % full?

Solution :

Given, a tank can hold 200 litres of water.

Now, 75% of 200

= 75/100 × 200

= 75 × 2

= 150 litres

Since, at present, it is only 40 % full.

So, 40% of 200

= 40/100 × 200

= 40 × 2

= 80 litres.

Thus, 150 litres – 80 litres. = 70 litres

Hence, there are 70 litres water required.

(22) Which is greater 16 2/3 or 2/5 or 0.17

Solution :

16 2/3 = 50/3 = 12.06

2/5 = 0.4 and 0.17 = 0.17

Since, 0.17 < 0.4 < 12.06

Hence, 16 2/3 is greater.

(25) A sum of 46900 was lent out at simple interest and at the end of 2 years, the total amount was 53466. find the rate of interest per year.

Solution :

Given, P = ₹ 46,900, n = 2 years and A = ₹ 53,466

We know, I = A – P

I = ₹ 53,466 – ₹ 46,900

I = ₹ 6566

To find r = ?

Now, I = Prn/100

r = I × 100/Pn

r = 6566×100/46900 × 2

r = 6566/938

r = 7%

Therefore, the rate of interest per year is 7%.

(26) Arun lent 5000 to Balaji for 2 years and 3000 to Charles for 4 years on simple interest at the same rate of interest and received 2200 in all from both of them as interest. find the rate of interest per year

Solution :

Given, P1 = ₹ 5,000 , n1 = 2 years, P2 = ₹ 3,000 , n2 = 4 years and r1=r2 = r

Since, I1 +I2 = ₹2200

We know, I = Prn/100

Thus, I1 + I2 = P1 × r × n1/100 + P2 × r × n2/100

2200 = 1/100(5000 × r × 2 + 3000 × r × 4)

2200×100 = r × (10000 + 12000)

220000 = r × 22000

r = 220000 /22000

r = 10%

Hence, the rate of interest per year is 10%.

(27) If a principal is getting doubled after 4 years, then calculate the rate of interest

Solution :

Let P = ₹ 100

As a principal is getting doubled after 4 years

Thus, A = ₹200

Here, n = 4 years and I = A – P = 200 – 100 = ₹100

We know, I = Prn/100

r = I × 100/Pn

r = 100 × 100 /100 × 4

r = 100/4

r = 25 %

Hence, the rate of interest is 25 %.

Next Chapter Solution :

👉 Algebra

Updated: July 28, 2023 — 4:00 pm