Compound interest - Practice

Important Question on Compound Interest:

1.  The simple and compound interests on a sum of money for two years are Rs 2880 and Rs 3052.80 respectively. The Rate of interest per annum is..

(1) 10%                                

(2)12%                              

(3)15%                                 

(4)18%             

   

2.  What annual instalment would clear a debt of Rs 4.24 lakhs in two years at the rate of 12% p.a. compound interest if interest is paid at end of year.

(1) 250880                        

(2)276440         

(3)284220                          

(4)292660

3.  Find the annual payment that will discharge a debt of Rs 15260 due in 4 years at 6% per annum simple interest( assume that each annual payment is made at the end of the year.)

(1) 3600                              

(2)3500              

(3)3450                               

(4)3400

4.   A person took a loan of Rs 6000 for 3 years, at 5% per annum compound interest. He repaid Rs 2100 in each of the first 2 years.The amount he should pay at the end of third year to clear all his debts is.

(1) 2425.5                          

(2)2552.5           

(3)2635.5                            

(4)2745.5

5.   If the difference between simple interest and compound interest on a certain sum of money for 3 years at 10% per annum is Rs 31, The sum is.

(1) 500                                 

(2)750 

(3)1000                               

(4)1250

Answers with Explanation:

1.       (2) Since, CI/SI = (200+R)/200

                3052.8/2880 = (1+R/200)

                R = 12%

2.       (1) 4.24 x 100000 = x/ (1+r/100) + x / (1+r/100)^2

                = x/(1+12/100) + x/(1+12/100)^2

                = 25x/28 + 25² x/28² = 1325x/784

                X = 250880

3.       (2) Let annual payment 100 Rs.

                100 +(100+R) + (100 +2*R) + (100 +3*R) = 15260

                100+ 106 + 112 + 118 = 15260

                436 = 15260

                100 = 15260x100 / 436 = 3500

4.       (1)  P = 6000

                For 1sr year CI = 5% of 6000 = 300

                Amount = 6000 + 300 = 6300

                P for 2^nd year = 63000 – 2100 = 4200

                CI for 2^nd year = 5% of 4200 = 210

                Amount 2^nd year = 4200 + 210 = 4410

                P for 3^rd year = 4410 – 2100 = 2310

                CI for 3^rd year = 5% of 2310 = 115.5

                Required amount = 2310 + 115.5 = 2425.5

5.       (3) Since difference = PR² (300+R)

                31 = Px10² (300+10)/ 100³

                P = 1000