Simple and Compound Interest ICSE Class-8th Concise Selina Mathematics  Solutions Chapter-9 . We provide step by step Solutions of Exercise / lesson-9 Simple and Compound Interest forICSE Class-8 Concise Selina Mathematics. Our Solutions contain all type Questions with Exe-9 A,  Exe-9 B and Exe-9 C to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-8.

## Simple and Compound Interest ICSE Class-8th Concise Selina Mathematics  Solutions Chapter-9

–: Select Topics :–

Exe-9 A,

Exe-9 B,

Exe-9 C,

### Exercise – 9 A  Simple and Compound Interest

#### Question 1 :-

Find the interest and the amount on:
(i) ₹ 750 in 3 years 4 months at 10% per annum.
(ii) ₹ 5,000 at 8% per year from 23rd December 2011 to 29th July 2012.
(iii) ₹ 2,600 in 2 years 3 months at 1% per month.
(iv) ₹ 4,000 in 1$\frac { 1 }{ 3 }$ years at 2 paisa per rupee per month.

#### Question 2 :-

Rohit borrowed Rs. 24,000 at 7.5 percent per year. How much money will he pay at the end of 4th years to clear his debt?

Principal (P) = Rs.24,000

Rate (R) = 7.5% P.A.

Time (T) = 4 years

= Rs. 240 × 4 × 7.5

= 240 × 30

= Rs. 7200

Amount needed to clear the debt at the end of 4th year

= Rs. 24000 + Rs. 7200 =  Rs. 3,1200

#### Question 3 :-

The interest on a certain sum of money is Rs. 1,480 in 2 years and at 10 percent per year. Find the sum of money.

Let P = Rs. x

Time (T) = 2 years

Rate (R) = 10%

x/5= Rs. 14810 ….(Given)

∴ x = 1480 × 5 = Rs. 7400

Hence the money Rs. 7400

#### Question 4 :-

On what principal will the simple interest be Rs. 7,008 in 6 years 3 months at 5% per year?

Let Principal = Rs. P

Time (T) = 6 years 3 months = 6 year +3/12

year =75/12=25/4 year =6 (1/4) years

Rate (R) = 5%

= Rs. 22425.60

#### Question 5 :-

Find the principal which will amount to Rs. 4,000 in 4 years at 6.25% Per annum.

Let Principal = Rs. P, Time (T) = 4 years

= 4 × 800

⇒ P = Rs. 3200

Hence principal = Rs. 3200

#### Question 6 :-

(i) At what rate per cent per annum will Rs. 630 produce an interest of Rs. 126 in 4 years ?
(ii) At what rate per cent per year will a sum double itself in 6$\frac { 1 }{ 4 }$ years ?

(i)

P = Rs. 630, I = Rs. 126, T = 4 years

(ii)

Let P = Rs. 100

∴ Amount = 2 × Rs. 100 = Rs. 200

Interest = A − P

= Rs. 200 − Rs. 100

= Rs. 100

= 16%

#### Question 7 :-

(i) In how many years will Rs.950 produce Rs.399 as simple interest at 7%?
(ii) Find the time in which Rs.1200 will amount to Rs.1536 at 3.5% per year.

(i)

P = Rs.950

S.I. = Rs.399

R = 7%

We know that :

(ii)

A = Rs.1536

P = Rs.1200

I = A − P

= Rs.1536 − Rs.1200

= Rs.336

We know that :

#### Question 8 :-

The simple interest on a certain sum of money is $\frac { 3 }{ 8 }$ of the sum in 6$\frac { 1 }{ 4 }$ years. Find the rate percent charged.

Let P = Rs.8

S.I. = Rs. (3/8)×8

= Rs.3

T =6 (1/4) years = 25/4 years

We know that :

= 6%

#### Question 9 :-

What sum of money borrowed on 24th May will amount to Rs. 10210.20 on 17th October of the same year at 5 percent per annum simple interest.

A = Rs. 10210.20
R = 5% P.A.
T=May + June + July + August + Sept.+ Oct.

= 7 + 30 + 31 + 31 + 30 + 17

= 146/365 days = 2/5 year

We know that :

P + I = A

⇒ P = Rs.10010

∴ Money to be borrowed = Rs.10010

#### Question 10 :-

In what time will the interest on a certain sum of money at 6% be $\frac { 5 }{ 8 }$ of itself?

Let P = Rs.8

Interest = Rs.8 × 5/8 = Rs.5

R = 6%

= 10 years 5 months

∴ Time = 10 years 5 months

#### Question 11 :-

Ashok lent out Rs.7000 at 6% and Rs.9500 at 5%. Find his total income from the interest in 3 years.

case I  :

P = Rs.7000

R = 6%

T = 3 years

= Rs.1260

case II  :

P = Rs.9500

R = 5%

T = 3 years

= Rs.1425

Total income from the interest

= Rs.1260 + Rs.1425

= Rs.2685

#### Question 12 :-

Raj borrows Rs.8,000; out of which Rs. 4500 at 5% and the remainder at 6%. Find the total interest paid by him in 4 years.

Total sum borrowed by Raj = Rs.8000

In First Case :

P = Rs.4500

R = 5%

T =4 years

= Rs.900

In Second Case :

p = Rs.8000 − Rs.4500

= Rs.3500

R = 6%

T = 4 years

= 35 × 6 × 4 = Rs.840

Total interest paid by Raj

= Rs.900 + Rs.840

= Rs.1740

#### Question 13 :-

Mohan lends Rs.4800 to John for 4$\frac { 1 }{ 2 }$ years and Rs.2500 to Shy am for 6 years and receives a total sum of Rs.2196 as interest. Find the rate percent per annum, it being the same in both the cases.

In first case :

P = Rs.4800

R = x% (Suppose)

= Rs.216x

In second case :

P = Rs.2500

R = x%

T = 6 years

= Rs.150x

According to statement,

Interest in first case + Interest in second case

= Rs.2196

∴ Rs.216x + Rs.150x = Rs.2196

⇒ Rs.366x = Rs.2196

⇒ x =2196/366

⇒ x = 6

∴ Rate = 6%

#### Question 14 :-

John lent Rs. 2550 to Mohan at 7.5 percent per annum. If Mohan discharges the debt after 8 months by giving an old black and white television and Rs. 1422.50; find the price of the television.

### Exercise – 9 B ICSE Class-8th Concise Selina Mathematics

#### Question 1 :-

The interest on a certain sum of money is 0.24 times of itself in 3 years. Find the rate of interest.

#### Question 2 –

If ₹ 3,750 amount to ₹ 4,620 in 3 years at simple interest. Find:
(i) the rate of interest
(ii) the amount of Rs. 7,500 in 5$\frac { 1 }{ 2 }$ years at the same rate of interest

(i) A = Rs.4620

P = Rs.3750

I = A − P = Rs.4620 − Rs.3750 = Rs.870

T = 3 years

(ii) P = Rs.7500

R = 116/15 %

T =5 (1/2) years = 11/2 years

= 10 × 29 × 11 = 290 × 11 = Rs.3190

Amount = Rs.7500 + 3190 = Rs.10,690

#### Question 3 :-

A sum of money, lent out at simple interest, doubles itself in 8 years. Find :
(i) the rate of interest
(ii) in how many years will the sum become triple (three times) of itself at the same rate percent?

Let P = Rs.100, A = Rs.200

I = Rs.200 − Rs.100 = Rs.100, T = 8 years

Now again P = Rs.100

A = Rs.300

I = Rs.300 − Rs.100 = Rs.200

= 16 Years

So the given sum of money will become triple in 16 years.

#### Question 4 :-

Rupees 4000 amount to Rs.5000 in 8 years ; in what time will Rs.2100 amount to Rs.2800 at the same rate ?

First case :

A = Rs.5000

P = Rs.4000

I = A − P

= Rs.5000 − Rs.4000

= Rs.1000

T = 8 years

Second Case :

A = Rs.2800

P = Rs.2100

I = Rs.2800 − Rs.2100 = Rs.700

= 10 years 8 months

#### Question 5 :-

What sum of money lent at 6.5% per annum will produce the same interest in 4 years as Rs.7500 produce in 6 years at 5% per annum?

First case :

P = Rs.7500

R = 5%

T = 6 years

= Rs.75 × 5 × 6

= Rs.2250

Second case:

According to the statement, interest = Rs.2250

R = 6.5% P.A.

T = 4 years

= Rs.8653.85

Required principal = Rs.8653.85

#### Question 6 :-

A certain sum amounts to Rs.3825 in 4 years and to Rs.4050 in 6 years. Find the rate percent and the sum.

In 6 years sum amounts to = Rs.4050

In 4 years sum amounts to = Rs.3825

∴ Interest of 2 years = Rs.4050 − Rs.3825 = Rs.225

The interest of 4 years = Rs.(225/2)×4

= Rs.450 …(∵ Rs.225 is interest for 2 years)

Now P = A − I

= Rs.3825 − Rs.450

= Rs.3375

I = Rs.450

T = 4 years

P = Rs.3375

#### Question 7 :-

At what rate percent of simple interest will the interest on Rs.3750 be one-fifth of itself in 4 years? To what will it amount in 15 years?

P = Rs.3750

I = Rs.3750×1/5 = Rs.750

T = 4 years

= 5%

Again, P = Rs.3750

The interest of 4 years = Rs.750

= Rs.2812.50

Amount in 15 years will be = Rs.3750 + Rs.2812.50

= Rs.6562.50

∴ Rate = 5%

The amount in 15 years will be = Rs.6562.50

#### Question 8 :-

On what date will ₹ 1950 lent on 5th January 2011 amount to ₹ 2125.50 at 5 percent per annum simple interest?

P = Rs.1950

A = Rs.2125.50

R = 5% P.a.

I = A − P

= Rs.2125.50 − Rs,1950

= Rs.175.50

= 1 years 292 days

∵ 4/5 years

=(4/5)×365 days = 292 days

Jan. + Feb. + March + April + May + June + July +Aug. + Sept. + Oct.

(31 − 5) + 29 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 23

= 292 days

∴ Required date = 23rd October 2012

#### Question 9 :-

If the interest on Rs.2400 is more than the interest on Rs.2000 by Rs.60 in 3 years at the same rate percent; find the rate.

First case :

P = Rs.2400

R = x% (Assume)

T = 3 years

= Rs.72x

Second case :

P = Rs.2000

R = x% ….(Rate same as in first case)

T = 3 years

= Rs.60x

According to the statement,

72x = 60x + 60

⇒ 72x − 60x = 60

⇒ 12x = 60

⇒ x =60/12

⇒ x = 5

∴ Rate = 5%

#### Question 10 :-

Divide Rs. 15,600 into two parts such that the interest on one at 5 percent for 5 years may be equal to that on the other at 4$\frac { 1 }{ 2 }$ per cent for 6 years.

Let one part = Rs. x

∴ Second part = Rs. (15,600 − x)

By the given condition

⇒ 25x = 27 × 15,600 − 27x

⇒ 25x + 27x = 27 × 15,600

⇒ 52x = 27 × 15,600

= 27 × 300

= 8100

Hence one part = Rs.8100 and second part Rs. (15,600 − 8,100) = Rs.7,500

### Exercise – 9 C Simple and Compound Interest ICSE Mathematics Class-8th Concise Selina Solutions

#### Question 1 :-

A sum of Rs. 8,000 is invested for 2 years at 10% per annum compound interest. Calculate:
(i) interest for the first year.
(ii) principal for the second year.
(iii) interest for the second year.
(iv) final amount at the end of second year
(v) compound interest earned in 2 years.

(i) Here Principal (P) = Rs. 8,000
Rate of interest = 10%

= Rs.800

(ii) ∴ Amount = Rs.8,000 + Rs.800 = Rs.8,800

Thus Principal for the second year = Rs.8,800

(iii) Interest for the second year

= Rs.880

(iv) Amount at the end of second year = Rs.8,800 + Rs.880 = Rs.9,680

(v) Hence compound interest earned in 2 years

= Rs.9,680 − Rs.8,000 = Rs.1680

#### Question 2 :-

A man borrowed Rs. 20,000 for 2 years at 8% per year compound interest. Calculate :
(i) the interest of the first year.
(ii) the interest of the second year.
(iii) the final amount at the end of second year.
(iv) the compound interest of two years.

Here Principal (P) =Rs.20,000, Time = 1 year

Rate = 8%

(i) ∴ The interest of the first year =(20,000×8×1)/100

= Rs.1600

(ii) ∴ Amount after one year

i.e. Principal for second year = Rs.20,000 + Rs.1,600 = Rs.21,600

∴ Interest for second year =(21,600×8×1)/100

= 216 × 8

= Rs.1728

(iii) Final amount at the end of second year

= Rs. (21,600 + 1728) = Rs.23,328

(iv) Interest of two years = Rs.23,328 − Rs.20,000 = Rs.3,328

#### Question 3 :-

Calculate the amount and the compound interest on Rs. 12,000 in 2 years and at 10% per year.

In Ist year

Principal (P) = Rs.12,000

Rate (R) = 10%

Time (T) = 1 year

I = Interest =(12,000×10×1)/100

= 120 × 10

= Rs.1200

Amount = P + I = Rs.12,000 + Rs.1200 = Rs.13,200

In IInd year

P = Rs.13,200, R = 10%, Time (T) = 1 year

∴ Interest =(13,200×10×1)/100 = 132 × 10

= Rs.1320

∴ Amount in 2 years = Rs. (13,200) + (1320)

= Rs.14520

Compound interest in 2 years = Rs.1200 + Rs.1320 = Rs.2520

[or directly = Rs.14520 − Rs.12000 = Rs.2520]

#### Question 4 :-

Calculate the amount and the compound interest on Rs. 10,000 in 3 years at 8% per annum.

In I st year

Principal (P) = Rs.10,000,

Rate (R) = 8%

Time (T) = 1 year

∴ Interest = (10,000×8×1)/100

= 100 × 8

= Rs.800

In II nd year

P = Rs.10,000 + Rs.800 = Rs.10,800

Rate (R) = 8%,

Time (T) = 1 year

∴ Interest =(10,800×8×1)/100

= 108 × 8

= Rs.864

In III rd year

∴ P = Rs.10,800 + Rs.864 = Rs.11664,

R = 8%, T = 1 year

∴ Interest =(11664×8×1)/100 = (11664×2)/25

= Rs.933.12

∴ Amount = Rs.11664+ 933.12 = Rs.12597.12

Hence required amount = Rs.12597.12

∴ Compound interest = Rs.12597.12 − 10000

= Rs.2597.12

#### Question 5 :-

Calculate the compound interest on Rs. 5,000 in 2 years; if the rates of interest for successive years be 10% and 12% respectively.

In I st year

Principal (P) = Rs.5,000, Rate (R) = 10%, Time (T) = 1 year

∴ Interest =(5,000×10×1)/100 = 50 × 10 = Rs.500

∴ Amount at the end of 1st year = Rs. (5000 + 500) = Rs.5500

In II nd year

P = Rs.5550, Rate = 12%, T = 1 year

∴ Interest =(5500×12×1)/100 = 55 × 12 = Rs.660

∴ Amount at the end of 2nd year = Rs.5500 + Rs.660 = Rs.6160

Hence compound interest =Rs.6160 − Rs.5000 = Rs.1160

#### Question 6 :-

Calculate the compound interest on Rs. 15,000 in 3 years; if the rates of interest for successive years be 6%, 8% and 10% respectively.

In I st year

Principal (P) = Rs.15,000, Rate (R) = 6%, Time (T) = 1 year

∴ Interest =(15,000×6×1)/100 = 150 × 6 = Rs.900

∴ Amount at the end of 1nd year

= Rs.15,000 + Rs.900

= Rs.15900

In II nd year

P = Rs.15900, R = 8%, T = 1 year

∴ Interest =(15,900×8×1)/100 = 159 × 8 = Rs.1272

∴ Amount at the end 2nd year

= Rs. (15900 + 1272)

= Rs.17172

In III rd year

P = Rs.17172, R = 10%, T = 1 year

∴ Interest =(17172×10×1)/100 = Rs.1717.20

∴ Amount at the end of 3rd year

= Rs. (17172 + 1717.20)

= Rs.18889.20

∴ Compound interest = 18889.20 − 15,000

= Rs.3889.20

#### Question 7 :-

Mohan borrowed Rs. 16,000 for 3 years at 5% per annum compound interest. Calculate the amount that Mohan will pay at the end of 3 years.

In I st year

Principal (P) = Rs.16,000, Rate (R) = 5%, Time (T) = 1 year

∴ Interest =(16000×5×1)/100= 160 × 5 = Rs.800

∴ Amount at the end of 1st year = Rs. (16,000 + 800) = Rs.16,800

In II nd year

P = Rs.16,800, R = 5%, T = 1 year

∴ Interest =(16,800×5×1)/100 = 168 × 5 = Rs.840

∴ Amount at the end of 2nd year = Rs. (16,800 + 840) = Rs.17640

In III rd year

P = 17640, R = 5%, T = 1 year

∴ Interest =(17640×5×1)/100=17642 = Rs.882

∴ Amount at the end of 3rd year = Rs. (17640 + 882) = Rs.18522

Hence reqd. amount = Rs.18522

#### Question 8 :-

Rekha borrowed Rs. 40,000 for 3 years at 10% per annum compound interest. Calculate the interest paid by her for the second year.

In I st year

Principal = Rs.40,000, Rate = 10%, Time = 1 year

∴ Interest =(40,000×10×1)/100

= 400 × 10

= Rs.4000

∴ Amount at the end of 1st year = Rs. (40,000 + 4000) = Rs.44,000

In II nd year

P = Rs.44,000, R = 10%, T = 1 year

∴ Interest = Rs.(44,000×10×1)/100

= 440 × 10

= Rs.4400

Thus interest earned in the second year = Rs.4400

#### Question 9 :-

Calculate the compound interest for the second year on Rs. 15000 invested for 5 years at 6% per annum.

Principal (P) = Rs.15000

Rate (R) = 6% p.a.

Period (n) = 5 years

Interest for the first year =PRT/100

=(15000×6×1)/100

= Rs.900

∴ Amount for the first year = Rs.15000 + 900

= Rs.15900

Principal for the second year= Rs.15900

Interest for the second year =(15900×6×1)/100

= 159 x 6

= Rs. 954

#### Question 10:-

A man invests Rs. 9600 at 10% per annum compound interest for 3 years. Calculate :
(i) the interest for the first year.
(ii) the amount at the end of the first year.
(iii) the interest for the second year.
(iv) the interest for the third year.

Principal (P) = Rs.9600

Rate (R) = 10% p.a.

Period (n) = 3 years

(i) ∴ Interest for the first year =PRT/100

= (9600×10×1)/100

= Rs.960

(ii) Amount at the end of first year

= P + S.I.

= Rs.9600 + 960

= Rs.10560

(iii) Principal for the second year = Rs.10560

Interest for the second year = (10560×10×1)/100

= Rs.1056

∴ Amount after second year = Rs.10560 + 1056 = Rs.11616

(iv) Principal for the third year = Rs.11616

Interest for the third year = (11616×10×1)/100

= 116.16 × 10

= Rs.1161.60

#### Question 11 :-

A person invests Rs. 5,000 for two years at a certain rate of interest compounded annually. At the end of one year, this sum amounts to Rs. 5,600. Calculate :
(i) the rate of interest per year.
(ii) the amount at the end of the second year.

Principal (P) = Rs.5000

Period (T) = 2 years

Amount at the end of one year = Rs.5600

∴ Interest for the first year = A − P

= Rs.5600 − 5000

= Rs.600

(i) ∴ Rate of interest =(S.I.×100)/(P×T)

=(600×100)/(5000×1)

= 12% p.a.

(ii) Principal for the second year = Rs.5600

Interest for the second year = (5600×12×1)/100

= ₹672

∴ Amount at the end of the second year

= P + S.I.

= 5600 + 672

= ₹ 6272

#### Question 12 :-

Calculate the difference between the compound interest and the simple interest on ₹ 7,500 in two years and at 8% per annum.

Principal (P) = ₹7500

Rate (R) =  8% p.a.

Period (T) = 2 years

∴ Simple interest =(PRT/100) = (7500×8×2)/100

= ₹1200

Interest for the first year = (7500×8×1)/100

= ₹600

∴ Amount at the end of first year = P + S.I.

= ₹7500 + ₹600

= ₹8100

Principal for the second year = ₹8100

∴ Interest for the second year = (8100×8×1)/100

= ₹648

∴ Total C.I. for 2 years = ₹600 + ₹648

= ₹1248

∴ Difference between C.I. and S.I. for 2 years

= ₹1248 − ₹1200

= ₹48

#### Question 13 :-

Calculate the difference between the compound interest and the simple interest on ₹ 8,000 in three years and at 10% per annum.

Principal (P) = ₹8000

Rate (R) = 10% p.a.

Period (T) = 3 years

∴ S.I. for 3 years = (PRT/100) = (8000×10×3)/100

= ₹2400

Now, S.I. for 1st year = ₹(8000×10×1)/100

= 80 × 10 × 1

= ₹800

Amount for the first year = P + S.I.

= ₹8000 + ₹800

= ₹8800

Principal for the second year = ₹8800

Interest for the second year = (8800×10×1)/100

= ₹880

∴ Amount after second year = ₹8800 + ₹880

= ₹9680

Principal for the third year = ₹9680

Interest for the third year

= ₹ (9680×10×1)/100

= ₹968

∴  C.I. for 3 year = ₹800 + ₹880 + ₹968

= ₹2648

∴ Difference between C.I. and S.I. for 3 year

= ₹ 2648 − ₹2400

= ₹ 248

#### Question 14 :-

Rohit borrowed ₹ 40,000 for 2 years at 10% per annum C.I. and Manish borrowed the same sum for the same time at 10.5% per annum simple interest. Which of these two gets less interest and by how much?

Sum borrowed (P) = ₹40000

Rate (R) = 10% p.a. compounded annually

Time (T) = 2 years

∴ Interest for the first year =PRT/100

= ₹ (40000×10×1)/100

= ₹ 4000

Amount after one year = ₹40000 + 4000

= ₹ 44000

Principal for the second year = ₹44000

∴ Interest for the second year

= (44000×10×1)/100

= ₹4400

∴ Compound Interest for 2 years = ₹4000 + 4400

= ₹8400

In the second case,

Principal (P) = ₹40000

Rate (R) = 10.5% p.a.

Time (T) = 2 years

∴ Simple Interest = PRT/100 = (40000×10.5×2)/100

= ₹ (40000×105×2)/100×10

= ₹8400

In both the cases, interest is same.

#### Question 15 :-

Mr. Sharma borrowed ₹ 24,000 at 13% p.a. simple interest and an equal sum at 12% p.a. compound interest. Find the total interest earned by Mr. Sharma in 2 years.

Sum borrowed (P) = ₹24000

Rate (R) = 13% p.a.

Time (T) = 2 years

In case of simple interest,

Simple interest for 2 years =PRT/100

= ₹ (24000×13×2)/100

= ₹6240

In the case of compound interest,

Interest for the first year = (24000×12×1)/100

= ₹ 2880

Amount after the first year

= ₹ 24000 + 2880

= ₹ 26880

Interest for the second year = ₹ (26880×12×1)/100

= ₹ (322560/100)

= ₹3225.60

∴ C.I. for 2 years = ₹2880 + 3225.60

= ₹6105.60

Total interest = ₹6240 + 6105.60

= ₹12345.60

#### Question 16 :-

Peter borrows ₹ 12,000 for 2 years at 10% p.a. compound interest. He repays ₹ 8,000 at the end of first year. Find:
(i) the amount at the end of first year, before making the repayment.
(ii) the amount at the end of first year, after making the repayment.
(iii) the principal for the second year.
(iv) the amount to be paid at the end of second year, to clear the account.

Sum borrowed = ₹12000

Rate (R) = 10% p.a. compound annually

Time (T) = 2 years

Interest for the first year =PRT/100

=(12000×100×1)/100

= ₹1200

(i) Amount = ₹12000 + 1200 = ₹13200

Amount paid = ₹8000

(ii) balance amount = ₹13200 − 8000 = ₹5200

(iii) ∴ Principal for the second year = ₹5200

(iv) Interest for the second year = (5200×10×1)/100

= ₹520

∴ Amount = ₹5200 + 520 = ₹5720

#### Question 17 :-

Gautam takes a loan of ₹ 16,000 for 2 years at 15% p.a. compound interest. He repays ₹ 9,000 at the end of first year. How much must he pay at the end of second year to clear the debt?

Loan taken (P) = ₹ 16000
Rate (R) = 15% p.a.
Time (T) = 2 years

∴ Interest for the first year

= PRT/100 = (16000×15×1)/100

= ₹2400

Amount after one year = ₹16000 + 2400

= ₹18400

At the end of one year amount paid back = ₹9000

Balance amount = ₹18400 − 9000

= ₹9400

Interest for the second year = (9400×15×1)/100

= ₹1410

Amount after second year = ₹9400 + 1410

= ₹10810

#### Question 18 :-

A certain sum of money, invested for 5 years at 8% p.a. simple interest earns an interest of ₹ 12,000. Find:
(i) the sum of money.
(ii) the compound interest earned by this money in two years and at 10% p.a. compound interest.

Rate (R) = 8% p.a.

Period (T) = 5 years

Interest (I) = ₹12000

(i) ∴ Sum = (I×100)/(R×T)

= ₹ (12000×100)/8×5

= ₹30000

(ii) Rate (R) = 10% p.a.

Time (T) = 2 years

Principal (P) = ₹30000

Interest for the first year = PRT/100

= ₹ (30000×10×1)/100

= ₹3000

∴ Amount after one year = ₹30000 + 3000

= ₹33000

Principal for the second year = ₹33000

Interest for the second year = (33000×10×1)/100

= ₹3300

∴ Compound Interest for two years

= ₹3000 + 3300

= ₹6300

#### Question 19 :-

Find the amount and the C.I. on ₹ 12,000 at 10% per annum compounded half-yearly.

Principal (P) = ₹12,000

Rate (r) = 10%

Time (t) = 1 years

= ₹13,230

C.I. = Amount − Principal

= ₹13230 − ₹12000

= ₹1230

#### Question 20 :-

Find the amount and the C.I. on ₹ 8,000 in 1$\frac { 1 }{ 2 }$ years at 20% per year compounded half- yearly.

Principal (P) = ₹8000

Rate = 20%

= ₹10648

C.I. = Amount − Principal

= ₹10648 − ₹8000

= ₹2648

#### Question 21 :-

Find the amount and the compound interest on ₹ 24,000 for 2 years at 10% per annum compounded yearly.

Principal (P) = ₹24,000

Time (t) = 2 years

Rate (r) = 10%

= ₹29,172

C.I. = Amount − Principal

= ₹29,172 − ₹24,000

= ₹5,172

#### Question 22 :-

Find the amount and the compound interest on ₹ 16,000 for 3 years at 5% per annum compounded annually.

Principal (P) = ₹16,000

Time (t) = 3 years

Rate (r) = 5%

= ₹18,555

C.I. = Amount − Principal

= ₹18,555 − ₹16,000

= ₹2555

#### Question 23 :-

Find the amount and the compound interest on ₹ 20,000 for 1$\frac { 1 }{ 2 }$ years at 10% per annum compounded half-yearly.

Principal (P) = ₹20,000

= ₹23,152.50

C.I. = Amount − Principal

= ₹23,152.50 − ₹20,000

= ₹3,152.50

#### Question 24 :-

Find the amount and the compound interest on ₹ 32,000 for 1 year at 20% per annum compounded half-yearly.

Principal (P) = ₹32,000

Time (t) = 1 year

Rate (r) = 20%

= ₹38,720

C.I. = Amount − Principal

= ₹38,720 − ₹32,000

= ₹6,720

#### Question 25 :-

Find the amount and the compound interest on ₹ 4,000 in 2 years, if the rate of interest for first year is 10% and for the second year is 15%.

Principal (P) = ₹4,000

Time (t) = 2 years

Rate (R1) = 10% and rate (R2) = 15%

= ₹ 5060

C.I. = Amount − Principal

= ₹ 5060 − ₹4000

= ₹ 1060

#### Question 26 :-

Find the amount and the compound interest on ₹ 10,000 in 3 years, if the rates of interest for the successive years are 10%, 15% and 20% respectively.

Principal (P) = ₹10,000

Time (t) = 3 years

Rate (R1) = 10%

Rate (R2) = 15%

Rate (R3) = 20%

= ₹15,180

C.I. = Amount − Principal

= ₹15,180 − ₹10,000

= ₹5180

— End of Simple and Compound Interest Solutions :–

Thanks