**Calculate the amount and compound interest for the following:**

**(a) Rs12,000 for 2 years at 10% compounded annually.**

**(b) Rs 20,000 for 3 years at 8% compounded annually.**

**(c) Rs 5,000 for 1 year at 4% compounded semi-annually.**

**(d) Rs 10,000 for 1 ^{1}/_{2} years at 5% compounded half – yearly.**

**(e) Rs 500 for 1year at 2% compounded quarterly.**

Solution:

(a) Rs12,000 for 2 years at 10% compounded annually.

P = 12,000;

R = 10%

n = 2

A = P [1 + ^{R}/_{100}]^{n}

=12,000[1 + ^{10}/_{100}]^{2}

=12,000[1+0.1]^{2}

=12,000[1.1]^{2}

= 12,000[1.21]

=14,520.

Compound interest = Amount – Principal

= 14,520 – 12,000

= 2520

(b) Rs 20,000 for 3 years at 8% compounded annually.

P = 20,000;

R = 8%

n = 3

A = P [1 + ^{R}/_{100}]^{n}

= 20000[1 + ^{8}/_{100}]^{3}

= 20,000 [1.08]^{3}

= 20,000[1.259712]

=25,194.24

Compound interest = Amount – Principal

= 25,194.24 – 20,000

= 5194.24

(c) Rs 5,000 for 1 year at 4% compounded semi-annually.

P = 5,000

n = 1

R = 4

A = P(1 + ^{R}/_{2×100})^{2n}

A = 5,000x(1 + ^{4}/_{2×100})^{2×1}

=5000x(1.02)^{2}

= 5,202

Compound interest = Amount – Principal

=5,202 – 5,000

= 202

(d) Rs 10,000 for 1^{1}/_{2} years at 5% compounded half – yearly.

P = 10,000

n = 1^{1}/_{2}

R = 5

A = P(1 + ^{R}/_{2×100})^{2n}

= 10,000(1 + ^{5}/_{2×100})^{2×1.5}

= 10,000(^{41}/_{40})^{3}

= 10,000(1.02)^{3}

= 10,768.91

Compound interest = Amount – Principal

=10,768.91 – 10,000

= 768.91

=769.00

(e) Rs 500 for 1year at 2% compounded quarterly.

P = 500

n = 1

R = 2

A = P(1 + ^{R}/_{4×100})^{4n}

=500(1 + ^{2}/_{4×100})^{4×1}

= 500(1 + ^{1}/_{200})^{4}

= 500(^{201}/_{200})^{4}

=510.07

Compound interest = Amount – Principal

= 510.07 – 500

= 10.07

**A man invests Rs 5000 for 2 years at compound interest. After one year his money amount to Rs 5150. Find the interest for second year.**

Solution:

Principal for the first year = P_{1} = 5000.00

Amount at the end of first year = A_{1} = 5150.00

Interest = A_{1} – P_{1} = 5150 – 5000 =150

Rate of interest = R = ^{100xI}/_{PT} = ^{100×150}/_{5000} = ^{15000}/_{5000} = 3

P = 5000 + 150 = 5150

Interest in the second year = ^{PTR}/_{100} = ^{5150x1x3}/_{100} = 154.5

**Find the amount on 10000 after 2 years if the rate of interest are 3% and 4% doe successive years.**

Solution:

P_{1} = 10000.00

R_{1} = 3%

T = 1

I_{1} = ^{PTR}/_{100} = ^{10000x1x3}/_{100} = 300

Principal for the second year P_{2} = 10000 + 300 = 10300

R_{2} = 4%

I_{2} = ^{PTR}/_{100} = ^{10300x1x4}/_{100} = 412

Amount at the end of second year P_{2} = 10300 + 412 = 10712

**Pralhad invests a sum of money in a bank and gets Rs 3307.5 Rs 3472.87 in 2nd and 3rd year respectively. Find the sum he invested.**

Solution:

Amount at the end of second year A_{2} = RS 3307.50

Amount at the end of third year A_{3} = Rs 3472.87

Interest on Rs 3307.50 for a year = 3472.87 – 3307.50 = 165.37

Rate of interest = ^{100xI}/_{PT} = ^{100*165.37}/_{3307.50*1} = 5%

Let P be the initial investment

Amount after 2 years A = P(1+ ^{R}/_{100})^{n}

3307.50 = P(1+^{5}/_{100})^{2}

3307.50 = P(^{21}/_{20})^{2}

P = 3020.00

- O
**n what sum of money will be difference between the simple interest and compound interest for 2 years at 4% per annum will be equal to RS 100? (Hint : Assuming the principal to be RS 100 first calculate the SI and CI, then proceed)**

Solution:

Let the principal be Rs. 100.00

R = 4

T = 2 years

S.I = ^{PTR}/_{100} = ^{100*2*4}/_{100} = Rs. 8.00

Amount at the end of 2 years if C.

I is calculated = 100(1 + ^{4}/_{100})^{2}

= 100*^{26}/_{25}*^{26}/_{25}=108.16

C.I = 108.16 – 100.00 = 8.16

Difference between SI and CI = 8.16 – 8.00 = 0.16

When the difference is Rs. 0.16

The amount invested = Rs. 100.00

The amount invested when the difference between SI and CI is Rs. 100.00 = ^{100*100}/_{0.16} = Rs. 62500.00

**A sum of money is invested at compound interest payable annually. The interest in two successive year are RS 275.00 and Rs 300.00. Find the rate of interest.**

Solution:

Interest at the end of 1 year = 275

Interest at the end of II year = 300

Interest per year = 300 -275 = RS. 25

Interest on RS 100 per year = 25

I = ^{PTR}/_{100}

25 = ^{275*1*R}/_{100} = ^{25*100}/_{275} = ^{2500}/_{275} = 9.09

Rate of interest = 9.09

**The difference between compound interest and simple interest on a certain sum for 2 year at 7 ½% per annum is RS. 360. Find the sum and verify answer.**

Solution:

To find the simple interest and compound interest let us assume that the principal be Rs. 100

Time = 2 years

Rate =7 ½ % = ^{15}/_{2}%

SI = ^{PTR}/_{100} = ^{100*2*15}/_{100*2} = 15

CI = P[(1 + ^{R}/_{100})^{n} – 1]

= 100[(1 + ^{15}/_{2×100})^{2} – 1]

= 100[(^{215}/_{200})^{2} – 1]

= 100[^{46225}/_{40000} – 1 ]

= 100[^{46225-40000}/_{40000}]

_{= 100*}^{6225}/_{40000} = ^{6225}/_{400}

= 15.5625

CI – SI = 15.5625 – 15 = 0.5625

When P = 100 ; Difference in CI and SI = 0.5625

If the difference is Rs. 360 then principal = ^{100}/_{0.5625} x 360 = 64000

Thus, the principal is Rs. 64000

__Verification:__

S.I = PTR/100 = ^{64000x2x15}/_{2×100 } = 9600

CI = P[(1 + ^{R}/_{100})^{n} – 1] = 64000x[(1+^{15}/_{2×100})^{2} – 1] = 9960

SI – CI = 9960 – 9600 = 360

**Teju invests Rs. 12,000 at 5% interest compounded annually. IF he receives an amount Rs. 13320 at the end find the period.**

Solution:

P = 12,000

R = 5%

A = 13230

A = P(1+^{R}/_{100})^{n}

13230 = 12000(1+ ^{5}/_{100})^{n}

^{13230}/_{12000} = (1+^{5}/_{100})^{n}

^{21}/_{20} = (^{21}/_{20})^{n}

n = 2

**The present population of a village is Rs . 18000. It is estimated that the population of the village grows by 3% per year . Find the population of the village after 4 years.**

Solution:

P = 18000

R = 3%

n = 4

A = P(1+^{R}/_{100})^{n}

= 18000(1 + ^{3}/_{100})^{4}

= ^{18000x103x103x103x103}/_{100x100x100x100}

= 20256.15858

= 20259

Population after 4 years = 20259

**Jeshu purchased a bike by paying Rs. 52000. If the value depreciates by 2% every year. Find the value of the bike after 3 years .**

Solution:

Original price of the bike =rs.52,000

Rate of depreciation =2%

Price after 3 years = P(1 – ^{R}/_{100})^{ n}

= 52000 x (1 – ^{2}/_{100})^{3}

= 52000 x (^{98}/_{100})^{3}

= 48941.98 ≈ 48942

**Using the ready reckoner find the compound interest in the following:**

**(a). Principal Rs 15000 for 4 years at 6.5% p.a.**

**(b). Principal Rs 22000 for 5 years at 12% p.a.**

Solution:

(a) Principal Rs 15000 for 4 years at 6.5% p.a.

P = 15000

T = 4

R = 6.5%

From the table the interest on Rs. 1 t 6.5% for 4 years is Rs. 0.2865

CI on Rs. 15000 = 0.2865*15000 = Rs. 4297.5

(b)P = Rs. 22000

R = 12%

n = 5years

From the table the interest on Rs. 1 t 12% for 5 years is Rs. 0.7623

CI on Rs. 22000 for 5 years

= 0.7623*22000 = Rs. 16770.60 ≈ 16771

**Using the ready reckoner find the period of interest in the following :**

**(a) Compound interest Rs 4026 at 7% p.a. principal Rs 10000**

**(b) Amount RS 16939.2 principal RS 12000 at 9% p.a.**

Solution:

(a) P = 30000 ; R = 7% ; CI = Rs. 4026

Thus, Rs. 4026 is the compound interest for the principal of Rs. 10000 at 7%.

Now, we have to find compound interest for Re. 1. For Rs. 10000, th compound interest is Rs. 4026

For the principal Re. 1, the compound interest is = ^{4026}/_{10000} = 0.4026

From the ready recknors 0.4026 corresponds to 5 years.

Therefore, n = 5.

(b) Amount RS 16939.2 principal RS 12000 at 9% p.a.

We have A = Rs. 16939.2

P = 12000

R = 9

Compound interest = A – P = 16939.2 – 12000 = 4939.2

Thus, Rs. 4939.2 is the compound interest for the principal of Rs. 12000 at 9%.

Now, we have to find compound interest for Re. 1. For Rs. 12000, the compound interest is Rs. 4939.2

For the principal Re. 1, the compound interest is = ^{4939.2}/_{12000} =0.4116

From the ready recknors 0.4116 corresponds to 4 years.

Therefore, n = 4.

**Using the ready recknors ,Find the rate of interest in the following :**

**(a) Compound interest RS 1733.6 for 3 years on principal of RS 11000**

**(b) Amount RS 35246 principal SR 20000 for 5 years .**

Solution :

(a) Compound interest = Rs. 1733.6

n = 3

P = Rs. 11000

Thus, Rs. 4939.2 is the compound interest for the principal of Rs. 12000 at 9%.

Now, we have to find compound interest for Re. 1. For Rs. 11000, the compound interest is Rs. 1733.6

For the principal Re. 1, the compound interest is = ^{1733.6} /_{11000} = 0.1576

From the ready recknors 0.1576 corresponds to 5%

Therefore the rate of interest is 5%.

(b)

Amount = Rs. 35246

P = Rs. 20000

n = 5

Compound interest = A – P = 35246 – 20000 = 15246

Now, we have to find compound interest for Re. 1. For Rs. 20000, the compound interest is Rs. 15246

For the principal Re. 1, the compound interest is = ^{15246} /_{20000} = 0.7623

From the ready recknors 0.7623 corresponds to 12%

Therefore the rate of interest is 12%.

## 1 thought on “Compound Interest Exercise 2.2.5 – Class IX”

Comments are closed.