Frank Solutions for Class 10 Maths Chapter 1 Compound Interest are available here. These exercises are formulated by our subject experts in order to assist you with your exam preparation to attain good marks in Maths exams. Students who wish to score good marks in Maths should practise Frank Solutions for Class 10 Maths Chapter 1. This book is one of the top materials when it comes to providing a question bank to practise.
Chapter 1 – Compound Interest explains that it is the interest on a loan, which is calculated based on both the initial principal and the accumulated interest from previous periods. In the Frank Solutions for Class 10 Maths Chapter 1, students learn how to calculate interest.
Download the PDF of Frank Solutions for Class 10 Maths Chapter 1 Compound Interest
Access Answers to Frank Solutions for Class 10 Maths Chapter 1 Compound Interest
1. Calculate the amount and the compound interest for each of the following:
(a) ₹ 7,500 at 12% p.a. in 3 years.
Solution:-
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.
Solution:-
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.
Solution:-
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.
Solution:-
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.
Solution:-
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.
Solution:-
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 
years.
Solution:-
From the question it is given that,
Principal, P = ₹ 10,000, Rate, r = 8% p.a., Time, t = 
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
% p.a. in
years.
Solution:-
From the question it is given that,
Principal, P = ₹ 25,000, Rate, r =
% p.a. = 42/5, Time, t =
years
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
years.
Solution:-
From the question it is given that,
Principal, P = ₹ 25,000, Rate, r = 5¼ p.a. = 21/4 %, Time, t =
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.
Solution:-
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.
Solution:-
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
(i) ₹ 16,000 at 15% p.a. in
years.
Solution:-
From the question it is given that,
Principal, P = ₹ 16,000, Rate, r = 15% p.a., Time, t =
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.
Solution:-
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.
Solution:-
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.
Solution:-
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.
Solution:-
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.
Solution:-
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.
Solution:-
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
8. Gayatri 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.
Solution:-
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.
Solution:-
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.
Solution:-
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.
Solution:-
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.
12. Prashant 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.
Solution:-
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.
Solution:-
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.
Solution:-
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.
Solution:-
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
Comments