# Frank Solutions for Chapter 1 Compound Interest Class 10 ICSE Mathematics

### Exercise 1.2

1. Calculate the amount and the compound interest for each of the following:

(a) ₹ 7,500 at 12% p.a. in 3 years.

(b) ₹ 13,500 at 10% p.a. in 2 years.

(c) ₹ 17,500 at 12% p.a. in 3 years.

(d) ₹ 23,750 at 12% p.a. in 2½ years.

(e) ₹ 30,000 at 8% p.a. in 2½ years.

(f) ₹ 10,000 at 8% p.a. in 2¼ years.

(g) ₹ 20,000 at 9% p.a. in 2.1/3 years.

(h) ₹ 25,000 at 8.2/5% p.a. in 1.1/3  years.

(a) From the question it is given that,

Principal, P = ₹ 7,500, Rate, r = 12% p.a., Time, t = 3 years

For the first year, t = 1 year

We know that, S.I. = (P × r × t)/100

= (7,500 × 12 × 1)/100

= ₹ 900

Then, A = P + S.I.

= 7,500 + 900

= ₹ 8,400

Therefore, new principal is ₹ 8,400.

Now, for the second year, t = 1 year, p = ₹ 8,400

S.I. = (P × r × t)/100

= (8,400 × 12 × 1)/100

= ₹ 1,008

Then, A = P + S.I.

= 8,400 + 1,008

= ₹ 9,408

Therefore, new principal is ₹ 9,408.

Now, for the second year, t = 1 year, p = ₹ 9,408

S.I. = (P × r × t)/100

= (9,408 × 12 × 1)/100

= ₹ 1,128.96

Then, A = P + S.I.

= 9,408 + 1,128.96

= ₹ 10,536.96

We know that,

C.I. = Interest in first year + interest in second year + interest in third year

= ₹ (900 + 1,008 + 1,128.96)

= ₹ 3,036.96

(b) ₹ 13,500 at 10% p.a. in 2 years.

From the question it is given that,

Principal, P = ₹ 13,500, Rate, r = 10% p.a., Time, t = 2 years

For the first year, t = 1 year

We know that, S.I. = (P × r × t)/100

= (13,500 × 10 × 1)/100

= ₹ 1,350

Then, A = P + S.I.

= 13,500 + 1,350

= ₹ 14,850

Therefore, new principal is ₹ 14,850.

Now, for the second year, t = 1 year, p = ₹ 14,850

S.I. = (P × r × t)/100

= (14,850 × 10 × 1)/100

= ₹ 1,485

Then, A = P + S.I.

= 14,850 + 1,485

= ₹ 16,335

We know that,

C.I. = Interest in first year + interest in second year

= ₹ (1,350 + 1,485)

= ₹ 2,835

(c) ₹ 17,500 at 12% p.a. in 3 years.

From the question it is given that,

Principal, P = ₹ 17,500, Rate, r = 12% p.a., Time, t = 3 years

For the first year, t = 1 year

We know that, S.I. = (P × r × t)/100

= (17,500 × 12 × 1)/100

= ₹ 2,100

Then, A = P + S.I.

= 17,500 + 2,100

= ₹ 19,600

Therefore, new principal is ₹ 19,600.

Now, for the second year, t = 1 year, p = ₹ 19,600

S.I. = (P × r × t)/100

= (8,400 × 12 × 1)/100

= ₹ 1,008

Then, A = P + S.I.

= 19,600 + 2,352

= ₹ 21,952

Therefore, new principal is ₹ 21,952.

Now, for the second year, t = 1 year, p = ₹ 21,952

S.I. = (P × r × t)/100

= (21,952 × 12 × 1)/100

= ₹ 2,634.24

Then, A = P + S.I.

= 21,952 + 2,634.24

= ₹ 7,086.24

We know that,

C.I. = Interest in first year + interest in second year + interest in third year

= ₹ (2,100 + 2,352 + 2,634.24)

= ₹ 7,086.24

(d) ₹ 23,750 at 12% p.a. in 2½ years.

From the question it is given that,

Principal, P = ₹ 23,750, Rate, r = 12% p.a., Time, t = 2½ years

For the first year, t = 1 year

We know that, S.I. = (P × r × t)/100

= (23,750 × 12 × 1)/100

= ₹ 2,850

Then, A = P + S.I.

= 23,750 + 2,850

= ₹ 26,600

Therefore, new principal is ₹ 26,600.

Now, for the second year, t = 1 year, p = ₹ 26,600

S.I. = (P × r × t)/100

= (26,600 × 12 × 1)/100

= ₹ 3,192

Then, A = P + S.I.

= 26,600 + 3,192

= ₹ 29,792

Therefore, new principal is ₹ 29,792.

Now, for the third year, t = ½ year, p = ₹ 29,792.

S.I. = (P × r × t)/100

= (29,792 × 12 × 1)/(100 × 2)

= ₹ 1,787.52

Then, A = P + S.I.

= 29,792 + 1,787.52

= ₹ 31,579.52

We know that,

C.I. = Interest in first year + interest in second year + interest in third year

= ₹ (2,850 + 3,192 + 1,787.52)

= ₹ 7,829.52

(e) ₹ 30,000 at 8% p.a. in 2½ years.

From the question it is given that,

Principal, P = ₹ 30,000, Rate, r = 8% p.a., Time, t = 2½ years

For the first year, t = 1 year

We know that, S.I. = (P × r × t)/100

= (30,000 × 8 × 1)/100

= ₹ 2,400

Then, A = P + S.I.

= 30,000 + 2,400

= ₹ 32,400

Therefore, new principal is ₹ 32,400.

Now, for the second year, t = 1 year, p = ₹ 32,400

S.I. = (P × r × t)/100

= (32,400 × 8 × 1)/100

= ₹ 2,592

Then, A = P + S.I.

= 32,400 + 2,592

= ₹ 34,992

Therefore, new principal is ₹ 34,992.

Now, for the third year, t = ½ year, p = ₹ 34,992.

S.I. = (P × r × t)/100

= (34,992 × 8 × 1)/(100 × 2)

= ₹ 1,399.68

Then, A = P + S.I.

= 34,992 + 1,399.68

= ₹ 36,391.68

We know that,

C.I. = Interest in first year + interest in second year + interest in third year

= ₹ (2,400 + 2,592 + 1,399.68)

= ₹ 6,391.68

(f) ₹ 10,000 at 8% p.a. in 2¼ years.

From the question it is given that,

Principal, P = ₹ 10,000, Rate, r = 8% p.a., Time, t = 2¼ years

For the first year, t = 1 year

We know that, S.I. = (P × r × t)/100

= (10,000 × 8 × 1)/100

= ₹ 800

Then, A = P + S.I.

= 10,000 + 800

= ₹ 10,800

Therefore, new principal is ₹ 10,800.

Now, for the second year, t = 1 year, p = ₹ 10,800

S.I. = (P × r × t)/100

= (10,800 × 8 × 1)/100

= ₹ 864

Then, A = P + S.I.

= 10,800 + 864

= ₹ 11,664

Therefore, new principal is ₹ 11,664.

Now, for the third year, t = ½ year, p = ₹ 11,664.

S.I. = (P × r × t)/100

= (11,664 × 8 × 1)/(100 × 4)

= ₹ 233.28

Then, A = P + S.I.

= 11,664 + 233.28

= ₹ 11,897.28

We know that,

C.I. = Interest in first year + interest in second year + interest in third year

= ₹ (800 + 864 + 233.28)

= ₹ 1,897.28

(g) ₹ 20,000 at 9% p.a. in 2.1/3 years.

From the question it is given that,

Principal, P = ₹ 10,000, Rate, r = 8% p.a., Time, t =2.1/3
years

For the first year, t = 1 year

We know that, S.I. = (P × r × t)/100

= (20,000 × 9 × 1)/100

= ₹ 1,800

Then, A = P + S.I.

= 20,000 + 1,800

= ₹ 21,800

Therefore, new principal is ₹ 21,800.

Now, for the second year, t = 1 year, p = ₹ 21,800

S.I. = (P × r × t)/100

= (21,800 × 9 × 1)/100

= ₹ 1,962

Then, A = P + S.I.

= 21,800 + 1,962

= ₹ 23,762

Therefore, new principal is ₹ 23,762.

Now, for the third year, t = 1/3 year, p = ₹ 23,762.

S.I. = (P × r × t)/100

= (23,762 × 9 × 1)/(100 × 3)

= ₹ 712.86

Then, A = P + S.I.

= 23,762 + 712.86

= ₹ 24,474.86

We know that,

C.I. = Interest in first year + interest in second year + interest in third year

= ₹ (1,800 + 1,962 + 712.86)

= ₹ 4,474.86

(h) ₹ 25,000 at 8.2/5% p.a. in 1.1/3  years.

From the question it is given that,

Principal, P = ₹ 25,000, Rate, r = 8.2/5 % p.a. = 42/5, Time, t = 1.1/3years

For the first year, t = 1 year

We know that, S.I. = (P × r × t)/100

= (25,000 × 42 × 1)/(100 × 5)

= ₹ 2,100

Then, A = P + S.I.

= 25,000 + 2,100

= ₹ 27,100

Therefore, new principal is ₹ 27,100.

Now, for the second year, t = 1/3 year, p = ₹ 27,100

S.I. = (P × r × t)/100

= (27,100 × 42 × 1)/(100 × 5 × 3)

= ₹ 758.80

Then, A = P + S.I.

= 27,100 + 758.80

= ₹ 27,858.80

We know that,

C.I. = Interest in first year + interest in second year

= ₹ (2,100 + 758.80)

= ₹ 2,858.80

(i) ₹ 40,000 at 5¼ p.a. in 1.1/3 years.

From the question it is given that,

Principal, P = ₹ 25,000, Rate, r = 5¼ p.a. = 21/4 %, Time, t = 1.1/3 years

For the first year, t = 1 year

We know that, S.I. = (P × r × t)/100

= (40,000 × 21 × 1)/(100 × 4)

= ₹ 2,100

Then, A = P + S.I.

= 40,000 + 2,100

= ₹ 42,100

Therefore, new principal is ₹ 42,100.

Now, for the second year, t = 1/3 year, p = ₹ 42,100

S.I. = (P × r × t)/100

= (42,100 × 21 × 1)/(100 × 4 × 3)

= ₹ 736.75

Then, A = P + S.I.

= 42,100 + 736.75

= ₹ 42,836.75

We know that,

C.I. = Interest in first year + interest in second year

= ₹ (2,100 + 736.75)

= ₹ 2,836.75

(j) ₹ 76,000 at 10% p.a. in 2½ years.

From the question it is given that,

Principal, P = ₹ 10,000, Rate, r = 8% p.a., Time, t = 2½ years

For the first year, t = 1 year

We know that, S.I. = (P × r × t)/100

= (76,000 × 10 × 1)/100

= ₹ 7,600

Then, A = P + S.I.

= 76,000 + 7,600

= ₹ 83,600

Therefore, new principal is ₹ 83,600.

Now, for the second year, t = 1 year, p = ₹ 83,600

S.I. = (P × r × t)/100

= (83,600 × 10 × 1)/100

= ₹ 8,360

Then, A = P + S.I.

= 83,600 + 8,360

= ₹ 91,960

Therefore, new principal is ₹ 91,960.

Now, for the third year, t = ½ year, p = ₹ 91,960.

S.I. = (P × r × t)/100

= (91,960 × 10 × 1)/(100 × 2)

= ₹ 4,598

Then, A = P + S.I.

= 91,960 + 4,598

= ₹ 96,558

We know that,

C.I. = Interest in first year + interest in second year + interest in third year

= ₹ (7,600 + 8,360 + 4,598)

= ₹ 96,558

(k) ₹ 22,500 at 12% p.a. in 1¾ years.

From the question it is given that,

Principal, P = ₹ 22,500, Rate, r = 12% p.a., Time, t = 1¾ years

For the first year, t = 1 year

We know that, S.I. = (P × r × t)/100

= (22,500 × 12 × 1)/100

= ₹ 2,700

Then, A = P + S.I.

= 22,500 + 2,700

= ₹ 25,200

Therefore, new principal is ₹ 25,200.

Now, for the second year, t = ¾ year, p = ₹ 25,200

S.I. = (P × r × t)/100

= (25,200 × 12 × 3)/(100 × 4)

= ₹ 2,268

Then, A = P + S.I.

= 25,200 + 2,268

= ₹ 27,468

We know that,

C.I. = Interest in first year + interest in second year

= ₹ (2,700 + 2,268)

= ₹ 4,968

(l) ₹ 16,000 at 15% p.a. in 2.2/3 years.

From the question it is given that,

Principal, P = ₹ 16,000, Rate, r = 15% p.a., Time, t = 2.2/3
years

For the first year, t = 1 year

We know that, S.I. = (P × r × t)/100

= (16,000 × 15 × 1)/100

= ₹ 2,400

Then, A = P + S.I.

= 16,000 + 2,400

= ₹ 18,400

Therefore, new principal is ₹ 18,400.

Now, for the second year, t = 1 year, p = ₹ 18,400

S.I. = (P × r × t)/100

= (18,400 × 15 × 1)/100

= ₹ 2,760

Then, A = P + S.I.

= 18,400 + 2,760

= ₹ 21,160

Therefore, new principal is ₹ 21,160.

Now, for the third year, t = 1/3 year, p = ₹ 21,160.

S.I. = (P × r × t)/100

= (21,160 × 15 × 2)/(100 × 3)

= ₹ 2116

Then, A = P + S.I.

= 21,160 + 2116

= ₹ 23,276

We know that,

C.I. = Interest in first year + interest in second year + interest in third year

= ₹ (2,400 + 2760 + 2116)

= ₹ 7,276

2. A sum of ₹ 65,000 is invested for 3 years at 8% p.a. compound interest.

(i) Find the sum due at the end of the first year.

(ii) Find the sum due at the end of the second year.

(iii) Find the compound interest earned in the first two years.

(iv) Find the compound interest earned in the last year.

From the question it is given that,

Principal, P = ₹ 65,000, Rate, r = 8% p.a., Time, t = 3 years

(i) C1 = (P × r × t)/100

= (65,000 × 8 × 1)/100

= ₹ 5,200

Then, P1 = 5200 + 65000

= ₹ 70,200

(ii) C2 = (P × r × t)/100

= (70,200 × 8 × 1)/100

= ₹ 5,616

Then, P2 = 70,200 + 5,616

= ₹ 75,816

(iii) C1 + C2 = 5,200 + 5,616

= ₹ 10,816

(iv) C3 = (P × r × t)/100

= (75,816 × 8 × 1)/100

= ₹ 6,065.28

3. Alisha invested ₹ 75,000 for 4 years at 8% p.a. compound interest,

(i) Find the amount at the end of the second year.

(ii) Find the amount at the end of third year.

(iii) Find the interest earned in the third year.

(iv) Calculate the interest for the fourth year.

From the question it is given that,

Alisha invested ₹ 75,000 for 4 years at 8% p.a.

Principal, P = ₹ 75,000, Rate, r = 8% p.a., Time, t = 4 years

(i) C1 = (P × r × t)/100

= (75,000 × 8 × 1)/100

= ₹ 6,000

Then, P1 = 75,000 + 6,000

= ₹ 81,000

C2 = (P × r × t)/100

= (81,000 × 8 × 1)/100

= ₹ 6,480

Then, P2 = 81,000 + 6,480

= ₹ 87,480

(ii) C3 = (P × r × t)/100

= (87,480 × 8 × 1)/100

= ₹ 6,998.4

Then, P3 = 6,998.4 + 87,480

= ₹ 94478.4

(iii) C3 = (P × r × t)/100

= (87,480 × 8 × 1)/100

= ₹ 6,998.4

(iv) C4 = (P × r × t)/100

= (9,4478.4 × 8 × 1)/100

= ₹ 7,558.272

4. Aryan borrowed a sum of ₹ 36,000 for 1½ years at 10% p.a. compound interest

(i) Find the total interest paid by him.

(ii) Find the amount he needs to return to clear the debt.

From the question it is given that,

Aryan borrowed a sum of ₹ 36,000 for 1½ years at 10% p.a

Principal, P = ₹ 36,000, Rate, r = 10 % p.a., Time, t = 1½ years

(i) C1 = (P × r × t)/100

= (36,000 × 10 × 1)/100

= ₹ 3,600

Then, P1 = 36,000 + 3,600

= ₹ 39,600

(ii) C2 = (P × r × t)/100

= (39,600 × 10 × 1)/200

= ₹ 1,980

Then, P2 = 36,000 + 3,600

= ₹ 41,580

5. Ameesha loaned ₹ 24,000 to a friend for 2½ at 10% p.a. compound interest.

(i) Calculate the interest earned by Ameesha.

(ii) calculate the amount by her at the end of time period.

From the question it is given that,

Ameesha loaned ₹ 24,000 to a friend for 2½ at 10% p.a.

Principal, P = ₹ 24,000, Rate, r = 10 % p.a., Time, t = 1½ years

(i) C1 = (P × r × t)/100

= (24,000 × 10 × 1)/100

= ₹ 2,400

Then, P1 = 24,000 + 2,400

= ₹ 26,400

C2 = (P × r × t)/100

= (26,400 × 10 × 1)/100

= ₹ 2,640

Then, P2 = 26,400 + 2,640

= ₹ 29,040

C3 = (P × r × t)/100

= (29,040 × 10 × 1)/200

= ₹ 2,904

Then, P3 = 29,040 + 2,904

= ₹ 31,944

(ii) From above,

The total interest = 2,400 + 2,640 + 2,904

= ₹ 7944

6. Harjyot deposited ₹ 27,500 in a deposit scheme paying 12% p.a. compound interest. If the duration of the deposit is 3 years, calculate:

(i) The amount received by him at the end of three years.

(ii) The compound interest received by him.

(iii) The amount received by him had he chosen the duration of the deposit to be 2 years.

From the question it is given that,

Harjyot deposited ₹ 27,500 in a deposit scheme paying 12% p.a.

Time, t = 3 years

(i) C1 = (P × r × t)/100

= (27,500 × 12 × 1)/100

= ₹ 3,300

Then, P1 = 27,500 + 3,300

= ₹ 30,800

C2 = (P × r × t)/100

= (30,800 × 12 × 1)/100

= ₹ 3,696

Then, P2 = 30,800 + 3,696

= ₹ 34,496

C3 = (P × r × t)/100

= (34,496 × 12 × 1)/100

= ₹ 4139.52

Then, P3 = 4,139.52 + 34,496

= ₹ 38,636

(ii) Then, the compound interest received by him = ₹ 3,300 + ₹ 3,696 + ₹ 4,139.52

= ₹ 11,135.52

(iii) The amount received by him had he chosen the duration of the deposit to be 2 years, P2 = 34,496

7. Natasha gave ₹ 60,000 to Nimisha for 3 years at 15% p.a. compound interest. Calculate to the nearest rupee:

(i) The amount Natasha receives at the end of 3 years.

(ii) The compound interest paid by Nimisha

(iii) The amount saved by Nimisha had he cleared the debt in 2 years.

From the question it is given that,

Natasha gave ₹ 60,000 to Nimisha for 3 years at 15% p.a.

(i) C1 = (P × r × t)/100

= (60,000 × 15 × 1)/100

= ₹ 9,000

Then, P1 = 60,000 + 9,000

= ₹ 69,000

C2 = (P × r × t)/100

= (69,000 × 15 × 1)/100

= ₹ 10,350

Then, P2 = 69,000 + 10,350

= ₹ 79,350

C3 = (P × r × t)/100

= (79,350 × 15 × 1)/100

= ₹ 1190.25

Then, P3 = 79,350 + 1,190.25

= ₹ 91,252.5

(ii) The compound interest paid by Nimisha,

Ctotal = C1 + C2 + C3

= 9,000 + 10,350 + 1,190.25

= ₹ 20,541

8Gayatri invested ₹ 25,000 for 3 years and 6 months in a bank which paid 10% p.a. compound interest. Calculate the amount, to the nearest Ts.10, that she received at the end of the period.

From the question it is given that,

Gayatri invested ₹ 25,000 for 3 years and 6 months in a bank which paid 10% p.a.

C1 = (P × r × t)/100

= (25,000 × 10 × 1)/100

= ₹ 2,500

Then, P1 = 25,000 + 2,500

= ₹ 27,500

C2 = (P × r × t)/100

= (27,500 × 10 × 1)/100

= ₹ 2,750

Then, P2 = 27,500 + 2,750

= ₹ 30,250

C3 = (P × r × t)/100

= (30,250 × 10 × 1)/100

= ₹ 3,025

Then, P3 = 30,250 + 3,025

= ₹ 33,275

C4 = (P × r × t)/100

= (33,275 × 10 × 1)/100

= ₹ 1,663.75

Then, P4 = 33,275 + 1,663.75

= ₹ 34,940

9. Prerna borrowed ₹ 16,000 from a friend at 15% p.a. Compound interest. Find the amount, to the nearest rupees, that she needs to return at the end of 2.4 years to clear the debt.

From the question it is given that,

Prerna borrowed ₹ 16,000 from a friend at 15% p.a.

C1 = (P × r × t)/100

= (16,000 × 15 × 1)/100

= ₹ 2,400

Then, P1 = 16,000 + 2,400

= ₹ 18,400

C2 = (P × r × t)/100

= (18,400 × 15 × 1)/100

= ₹ 2,760

Then, P2 = 18,400 + 2,760

= ₹ 21,160

C3 = (P × r × t)/100

= (21,160 × 15 × 1)/400

= ₹ 7,935

Then, P3 = 21,160 + 7,935

= ₹ 29,095

10. Shekhar had a fixed deposit of ₹ 24,000 for 3 years. If he received interest at 10% p.a. compounded annually, find the amount received by him at the time of maturity.

From the question it is given that,

Shekhar had a fixed deposit of ₹ 24,000 for 3 years.

Where, P = ₹ 24,000, t = 3 years, r = 10 % p.a.

Amount = P(1 + r/100)t

Amount = 24,000 (1 + (10/100))3

= ₹ 31,944

Hence, Shekhar received ₹ 31,944 at the time of maturity.

11. Neha loaned ₹ 27,500 to a friend for 1¾ years at 8% p.a. compound interest. Find the interest earned by her.

From the question it is given that,

Neha loaned ₹ 27,500 to a friend for 1¾ years at 8% p.a.

Where, P = ₹ 27,500, t = 1¾ years = 1.75 years, r = 8% p.a.

Amount = P(1 + r/100)t

Amount = 27,000 (1 + (10/100))1.75

= ₹ 3,982

Hence, shekhar received ₹ 31,944 at the time of maturity.

12Prashant borrowed ₹ 35,000 at 12% p.a. compounded semi-annually. Find the amount he needs to pay back at the end of 1 ½ years.

From the question it is given that,

Prashant borrowed ₹ 35,000 at 12% p.a.

Where, p = ₹ 35,000, t = 1 ½ years = 1.5 years, r = 12% p.a.

Amount = P(1 + r/100)2t

Amount = 35,000 (1 + (12/200))3

= ₹ 41,685.56

Hence, Prashant has to pay back ₹ 41,685.56 at the end of 1 ½ years.

13. Amita wanted to start a business for which she needed ₹ 40,000. She borrowed this from Dolly at 10% p.a. compounded semi-annually. Find the extra amount that she needs to pay at the end of two years to clear her debt.

From the question it is given that,

Amita needed = ₹ 40,000

Where, p = ₹ 40,000, t = 1 ½ years = 2 years, r = 10% p.a.

Amount = P(1 + r/100)2t

Amount = 40,000 (1 + (10/200))4

= ₹ 48,620.25

Hence, Amita has to pay ₹ 48,620.25 at the end of two years to clear her debt.

14. Pradeep gave ₹ 16,000 to a friend for 1.5 years at 15% p.a. compounded semi-annually. Find the interest earned by him at the end of 1.5 years.

From the question it is given that,

Pradeep gave ₹ 16,000 to a friend for 1.5 years at 15% p.a.

Where, p = ₹ 16,000, t = 1.5 years, r = 15% p.a.

Amount = P(1 + r/100)2t

Amount = 16,000 (1 + (15/200))3

= ₹ 19,876.75

Then, C = 19,876 – 16,000

= ₹ 3,876.75

Therefore, the interest earned by Pradeep at the end of 1.5 years is ₹ 3,876.75.

15. Mr. Mohan invested ₹ 12,500 at 16% p.a. compounded annually. If the duration of the deposit was 1.5 years, find the amount Mr. Mohan received at the end of 1.5 years.

From the question it is given that,

Mr. Mohan invested ₹ 12,500 at 16% p.a.

Where, p = ₹ 12,500, t = 1.5 years, r = 16% p.a.

Amount = P(1 + r/100)t

Amount = 12,500 (1 + (16/100))1.5

= ₹ 15,660

Therefore, Mr. Mohan received ₹ 15,660 at the end of 1.5 years

### Exercise 1.2

1.Calculate the amount and the compound interest for each of the following :

(a) Rs 12, 500 for 2 years at 8% for the first year and 10% for the second:

(b) Rs 15,000 for 2 years at 6% for the first year and 7% for the second:

(c) Rs 12, 5000 for 3 years at 12% for the first the year, 15% for the second year and 17% for the third year.

(d) Rs 20, 000 for 3 years at 7.1/2% for the first year, 8% for the second year and 10% for the third year.

(a) Rs 12,500 for 2 years at 8% for the first year and 10% for the second year.

(b) Rs 15,000 for 2 years at 6% for the first year and 7% for the second year.

(c) Rs 12,500 for 3 years at 12% for the first year, 15% for the second year and 17% for the third year.

(d) Rs 20,000 for 3 years at 7.1/2% for the first year, 8% for the second year and 10% for the third year.

2. Mohan borrowed Rs 25,000 at 10% p.a. compound interest. If he pays back Rs 7,500 every year, find the amount of loan outstanding at the begining of the fourth year.

P = Rs 25,000, R = 10% p.a.

3. Rajan borrowed Rs 90,000 at 15% p.a. compound interest. If he repays Rs 35,000 at the end of each year, find the amount of loan outstanding at the beginning of the fourth year.

P = Rs 90,000, R = 15% p.a.

4. Pooja borrowed Rs 15,000 from Sonali at 11% p.a compound interest. If she repays Rs 7,550 at the end of first year and Rs 6,101 at the end of second year, find the amount Pooja needs to give to Sonali at the end of third year to clear her debt.

P = Rs 15,000, R = 11% p.a.

5. Archana borrowed Rs 18,000 from Ritu at 12% p.a. compound interest. If at the end of the 1st, 2nd, and 3rd years, Archana returned Rs 5,250, Rs 5,875 and Rs 6,875 respectively, find the amount Archana has to pay Ritu at the end of the 4th year to clear her debt.

P = Rs 18,000, R = 12% p.a.

6. Rajeev borrowed Rs 15,000 from Sanjay at 12% p.a. compound interest. After 2 years Rajeev gave Rs 7,500 and a scooter to clear the account. Find the cost of scooter.

Here, P = Rs 15,000; r = 12% p.a.; t = 2 years

7. Manoj borrowed Rs 25,000 from Sohan at 8.4 % p.a. compound interest. After 2 years Manoj cleared Rs 17,500 and a motorcycle. Find the cost of the motorcycle.

Here, P = Rs 25,000; r = 8.4% p.a.; t = 2 years

8. Prakash borrowed Rs10,000 from Rajesh for 2 years at 6% and 8% p.a. compound interest for successive years. If Prakash returns Rs 5,600 at the end of the first year, how much does he have to give to Rajesh at the end of the second year to clear the loan?

P = Rs 10,000, R = 6% p.a.

9. Meera borrowed Rs 12,500 on compound interest from Rajeev for 2 years when the rates of interest for successive years were 8% and 10%. If Meera returned Rs 7,500 at the end of the first year, find the amount she has to return at the end of second year.

10. Mr. Chatterjee borrowed Rs 50,000 in compound interest from Mr. Patel for 2 years when the rates of interest for the successive years were 7.1/2 % and 9.1/4%. If Mr. Chatterjee returned Rs 27,750 at the end of the first year, find the amount he needs to return at the end of the second year to clear the loan.

### Exercise 1.3

1. What sum of money will amount to Rs 9,447.84 in 3 years at 8% p.a. compound interest?

Here, P = x; A = Rs 9, 447.84; t = 3 years; r = 8% p.a.

2. What sum of money will amount to Rs 16,637.50 in 3 years at 10% p.a. compound interest?

Here P = x; A = Rs 16,637.50; t = 3 years; r = 10% p.a.

3. What sum of money will amount to Rs 7,128 in 2 years at compound interest, if the rates of interest are 8% and 10% for successive years?

For the second year

4. What sum of money will amount to Rs 3,326.40 in 3 years at compound interest, if the rates of interest are 8%, 10% and 12% for the successive years?

For the third year,

5. What sum of money will amount to Rs 13,675.20 in 3 years at compound interest, if the rates of interest are 10%, 11% and 12% for the successive years?

For the third year

6. Ramesh saves Rs 4,000 every year and invests it at 10% p.a. compound interest. Calculate his savings at the end of the third year.

P = Rs 4,000; R = 10% p.a.; T = 3 years

7. Manoj saves Rs 5,000 every year and invests it at 12% p.a. compound interest. Calculate his savings at the end of the third year.

P = Rs 5,000; R = 12% p.a.; T = 3 years

8. A man’s savings increases by Rs 50 every year. If he saves Rs 500 in the first year and puts it at 10% compound interest, find his savings at the end of the third year.

P = Rs 500; R = 10% p.a.; T = 3 years

9. Neena’s savings increased by Rs 1,000 every year. If she saves Rs 4,000 in the first year and invests it 15% compound interest, find her total savings at the end of the third year.

P = Rs 4,000; R = 15% p.a.; T = 3 years

10. The value of car depreciated by 10% in the first 2 years and by 8% in the third year. Express the total depreciation of the car as a single per cent during the three years.

Let value of car be Rs x.

11. The value of a machine depreciates by 10%, 12% and 15% in the first 3 years. Express the total depreciation of the machine as a single percent during the three years.

Let value of machine be Rs x.

12. The value of a scooter depreciates by 12% of its value at the beginning of the year. Find the original value of the scooter if it depreciates by Rs 2,640 in the second year.

Let value of the scooter be Rs x.

13. The value of a refrigerator depreciates by 8% of its value at the beginning of the year. Find the original value of the refrigerator if it depreciated by Rs 2,392 in the second year.

Let value of the refrigerator be Rs x.

14. The value of a machine depreciates by 15% in the first year and by 12% in the second year. Find the value of the machine if its depreciation in the second year was 1,632.

Let value of the machine be Rs x.

15. The value of a ‘Honda’ bike depreciated by 16% in the first year and by 13% in the second year. Find the value of the bike if it depreciates by Rs 7,098 in the second year.

Let value of the bike be Rs x.

### Exercise 1.4

1. Mohan invested a certain sum at compound interest, compound annually. If the interests for two successive years were Rs 600 and Rs 648, calculate the rate of interest and the sum invested.

For the second year,

2. The interests on two successive years for a sum invested at compound interest compounded annually are Rs 840 and Rs 940.80. Calculate the rate of interest and the sum invested.

For the second year,

3. The simple interest on a certain sum in 2 years in Rs 1,300, whereas the compound interest on the same sum at the same rate and for the same time in Rs 1,365. Find the rate percent and the sum.

The extra interest earned = C.I. – S.I. = Rs (1,365 – 1,300) = Rs 65.

4. The simple interest and the compound interest on a certain sum of money for 2 years at the same rate of interest are Rs 8,000 and Rs 8,640 respectively. Calculate the rate of interest.

The extra interest earned = C.I. – S.I.

= Rs (8,640 – 8,000)

= Rs 640

5. A certain sum of money invested at compound interest compounded annually amounted to Rs 5,082 after 2 years and to Rs 5,590.20 after 3 years. Calculate the rate of interest and the sum invested.

Here, r = ? P = x (say)

6.A certain sum of money invested at compound interest compounded annually amounted to Rs 26,450 in 2 years and to Rs 30,417.50 in 3 years. Calculate the rate of interest and the sum invested.

Here, r = ? P = x (say)

7. Find the difference between the compound interest and the simple interest in 2 years on Rs 5,000 at 8% p.a. compounded annually.

Here, P = Rs 5,000; r = 8%, t = 2 years

8. Find the difference between the compound interest and the simple interest in 3 years on Rs 15,000 at 8% p.a. compounded yearly.

Here, P = Rs 15,000; r = 8% ; t = 3 years

9. Anand borrows Rs 20,000 at 9% p.a. simple interest for 3 years. He immediately gave it to Prakash at 8.1/2 p.a. compound interest compounded annually. Find Anand’s loss or gain.

Here, P = Rs 20,000; t = 3 years

10. Meera borrowed Rs 35,000 at 12.5% p.a. simple interest for 3 years. She immediately gave it to Archana at 12% p.a. compound interest compounded annually. Find Meera’s loss or gain at thr end of 3 years.

Here, P = Rs 35,000; t = 3 years

11. The cost of a scooter depreciated by Rs 5,100 during the second year and by Rs 4,335 during the year. Calculate:

(a) the rate of depreciation.

(b) the original cost of the scooter.

(c) the cost of the scooter at the end of the third year.

(a) the rate of depreciation,

12. The cost of a machine depreciated by Rs 2,592 during the third year and by Rs 2,332.80 during the fourth year. Calculate:

(a) The rate of depreciation.

(b) The original cost.

(c) The cost at the end of the fourth year.

(a) The rate of depreciation

### Exercise 1.5

1. Ramesh borrowed Rs 12,000 at 15% compound interest for 2 years. At the end of first year he returned some amount and on paying Rs 9,200 at the end of the second year, he cleared the loan. Calculate the amount of money Ramesh returned at the end of the first year.

Interest for first year:

2. Rajan borrowed Rs 32,000 at 12% compound interest for 2 years. At the end of the first year he returned some amount and on paying Rs 17,920 at the end of the second year, he cleared the loan. Calculate the amount Rajan paid at the end of the first year.

Interest for first year:

3. Find the sum invested at 8% p.a. compound interest on which the interest for the third year exceeds that of the first year by Rs 166.40.

Let the sum be P

4. Find the sum invested at 12.1/2% p.a. compound interest for the third year exceeds that of the first year by Rs 531.25.

Let the sum be P

5. The simple interest on an amount for 2 years at 8% is Rs 320. Calculate the compound interest on the same amount at the same rate for 1 year if the interest is compounded half-yearly.

Here, P = ?, t = 2 years; r = 8% p.a.

### Exercise 1.6

1. Calculate the amount and compound interest for each of the following, when compounded annually:

(a) Rs 12,000 for 3 years at 15% p.a.

(b) Rs 25,000 for 3 years at 8% p.a.

(c) Rs 16,000 for 3 years at 7.1/2% p.a.

(d) Rs 20,000 for 2 years at 12.1/2% p.a.

(e) Rs 8,000 for 1.1/2 years at 12% p.a.

(f) Rs 7,500 for 2.1/2 years; r = 16% p.a.

(a) Rs 12,000 for 3 years at 15% p.a.

2. Calculate the amount and the compound interest for each of the following:

(a) Rs 6,000 for 1.1/2 years at 10% p.a.

(b) Rs 25,000 for 1.1/2 years at 12%.

(a) Rs 6,000 for 1.1/2 years at 10% p.a.

3. Calculate the amount and the compound interest for each of the following :

(a) Rs 9,125 for 2 years if the rates of interest are 12% and 14% for the

(b) Rs 20,000 for 2 years if the rates of interest are 12.1/4% and 5.1/2% for the successive years.

(c) Rs 12,500 for 3 years if the rates for the successive years are 8%, 9% and 10% respectively.

(d) Rs 10,000 for 3 years if the rates of interest are 10%, 11% and 12% for the successive years.

(a) Rs 9,125 for 2 years if the rates of interest are 12% and 14% for the successive years.

4. Find the difference between the compound interest compounded yearly and half-yearly for each of the following:

(a) Rs 15,000 for 1.1/2 years at 12% p.a.

(b) Rs 20,000 for 1.1/2 years at 16% p.a.

(a) Rs 15,000 for 1.1/2 years at 12% p.a.

5. What sum of money will amount to Rs 8,073 in 2 years at compound interest if the rates of interest for the successive years are 15% and 17%?

6. Find the principal which will amount to Rs 22,344 in 2 years at compound interest if the rates of interest for the successive years are 12% and 14%?

Here P = ?; t = 2 years; r = 12% and 14% successively; A = Rs 22,344

7. What sum of money will amount to Rs 10,256 in 3 years at compound interest if the rates of interest for the successive years are 10%, 11% and 12%?

Here P = ?; t = 3 years; r = 10%, 11% and 12% successively; A = Rs 10,256.40

8. What sum of money will amount to Rs 18,792 in 1.1/2 years at 16% p.a. compounded yearly?

P = ? ; A = Rs 18,792; t = 1.1/2 years; r = 16%

9. What sum of money will amount to Rs 15,746.40 at 16% p.a. compounded half-yearly?

P = ?; A = Rs 15,746.40; t = ½ years; r = 16%

10. On what sum of money will the compound interest for 2 years 8% per annum amount to Rs 1399.68 ?

P = x; t = 2 years; r = 8%; A = Rs (x + 1399.68)

11. On what sum of money will the compound interest for 2.1/2 years at 12% per annum amount to Rs 8,241.60?

P = x; t = 2.1/2 years; r = 12%; A = Rs (x + 8,241.60)

12. On what sum of money will the compound interest for 2.1/2 years at 12.1/2% per annum amount to Rs 82,734.37?

P = x; t = 2.1/2 years; r = 12.1/2% = 25/2%; A = Rs (x + 82,734.37)

13. On what sum of money will the compound interest for 1.1/2 years at 16% p.a. compounded half-yearly amount to Rs 649.28?

P = x; t = 1.1/2 years = 3 × 6 months; r = 16% compounded half-yearly = 16/2% = 8%; A = Rs (x + 649.28)

14. On what sum of money will the compound interest for 2 years at 10 p.a. compounded half-yearly amount to Rs 3,448.10?

P = x; t = 2 years = 4 × 6 months; r = 10% compounded half-yearly = 10/2% = 5%; A = Rs (x + 3,448.10)

15. Calculate the rate per cent at which Rs 12,250 will yield Rs 3,116.40 as compound interest in 2 years.

P = Rs 12,250; A = Rs (12,250 + 3,116.40) = Rs 15,366.40; t = 2 years;

16. Calculate the rate percent at which Rs 15,000 will yield Rs 8,413.44 as compound interest in 3 years.

P = Rs 15,000; A = Rs (15,000 + 8,413.44) = Rs 23,413.44; t = 3 years; r = ?

17. Calculate the rate percent at which Rs 16,000 will yield Rs 3,876.75 as compound interest in 3 years.

P = Rs 16,000; A = Rs (16,000 + 3,876.75) = Rs 19,876.75; t = 3 years; r = ?

18. In what time will Rs 8,000 amount to Rs 12,167 at 15% per annum compounded annually?

P = Rs 8,000; A = Rs 12,167; r = 15%; t = ?

19. In what time will Rs 50,000 yield an interest of Rs 32,151.60 at 18% per annum interest compounded annually?

P = Rs 50,000; A = Rs (50,000 + 32,151.60) = Rs 82,151.60; r = 18%; t = ?

20. On what sum will be difference between compound interest and the simple interest for 2 years at 7.1/2% be Rs 22.50?

P = x. t = 2 years; r = 7.1/2% = 15/2%

21. On what sum will the difference between compound interest and the simple interest for 3 years at 12% be Rs 1,123.20?

P = x; t = 3 years; r = 12%

22. A sum of money placed at compound interest compounded annually amounts to Rs 47,610 in 2 years and to Rs 54,751.50 in 3 years. Calculate the rate of interest and sum.

P = x; r = ?; t = 2 and 3 years; A = Rs 47,610 (2 years) 54,751(3 years)

23. A sum of money placed at compound interest compounded annually amounts to Rs 31,360 in 2 years and to Rs 35,123.20 in 3 years. Calculate the rate of interest and the sum.

P = x; r = ?; t = 2 and 3 years; A = Rs 31, 360 (2 years) and Rs 35,123.20 (3 years)

24. A sum of money placed at compound interest compounded annually amounts to Rs 26,460 in 2 years and to Rs 29,172.15 in 4 years. Calculate the rate of interest and the sum.

P = x; r = ?; t = 2 and 4 years; A = Rs 26,460 (2 years) and Rs 29,172.15 (4 years)

25. The compound interest on a certain sum of money at 5% p.a. for 2 years is Rs 512.50. What will be the simple interest on the same sum for 3 years at 6% p.a. ?

P = x; t = 2 years; r = 5%; A = Rs (x + 512.50)

26. The compound interest on a certain sum of money at 10% p.a. for 3 years is Rs 4,965. What will be the simple interest on the same sum for 3 years at 11% p.a. ?

P = x; t = 3 years; r = 10%; A = Rs (x + 4,965)

### Exercise 1.7

1. The population of a town in the year 2005 was 4,25,000. Find its population in the year 2007 if the rate of annual increase is 4% per year.

Vn = ?; V0 = 4,25,000; r = 4%; t = 2 years

2. The population of a city is 1,25,000. If the annual birth rate and death rate are 5.5% and 3.5% respectively, calculate the population of the city after 3 years.

Vn = ?; V0 = 1,25,000; r = 5.5% (birth) and 3.5% (death); t = 3 years

3. The population of a village increase at the rate of 50 per thousand its population after 2 years will be 22,050. Find the present population.

Rate of increase =

4. In a factory the production of scooters rose to 46,305 from 40,000 in 3 years. Find the annual rate of growth of the production of scooters.

Vn = 46,305; V0 = 40,000; r = ? ; t = 3 years

5. The present population of a town is 1,15,200. If it increases at the rate of 6.2/3 % per annum, find

(i) its population after 2 years

(ii) Its population 2 years

Vn = ? ; V0 = 1,15,200; r = 6.2/3% = 20/3% ; t = 2 years

6. A machine was purchased 2 years ago. Its value decreases by 10% every year. Its present value is Rs 19,083.60. For how much money was the machine purchased?

Vn = Rs 19,083.60 ; V0 = ? ; r = 10% ; t = 2 years

7. The population of city is 24,000. In the next 3 years it will be 27,783. Find the rate of growth of the population.

Vn = 27,783; V0 = 24,000; r = ? ; t = 3 years

8. Under the electrification programme of villages, the number of villages with electricity rose to 27,040 from 25,000 in 2 years. Find the rate of growth in the number of villages with electricity.

Vn = 27,040; V0 = 25,000; r = ? ; t = 2 years

9. A new car is purchased for Rs 4,00,000. Its value depreciates at the rate of 10% per annum. What will be its value after 4 years?

Vn = ? ; V0 = Rs 4,00,000; r = 10% ; t = 4 years

10. The value of a property decreases every year at the rate of 5%. If its value at the end of 3 year be Rs 44,540, what was the original value at the beginning of these 3 years?

Vn = Rs 44.540; V0 = ? ; r = 5% ; t = 3 years

11. The value of the refrigerator which was purchased 2 years ago, depreciates at 12% per annum. If its present value is Rs 9,680. For how much was it purchased?

Vn = Rs 9,680; V0 = ? ; r = 12% ; t = 2 years

12. A building worth Rs 1,33, 100 is constructed on a plot of land worth Rs 72,900. After how many years will the values of both be same, if the land appreciates at 10% p.a. and the building depreciates at 10% p.a. ?

For the building,

13. The cost of a T.V. was quoted Rs 17,000 at the beginning of the the year 1999. In the beginning of 2000 the price was hiked by 5%. Because of decrease in demand the cost was reducted by 4% in the beginning of 2001. What is the cost of the T.V. in 2001?