**1.A man saves Rs 200 at the end of each year and lends the money at 5% compound interest. How much will it become at the end of 3 years.**

**2.Find the compound interest on Rs.16,000 at 20% per annum for 9 months, compounded quarterly**

**3.At what rate of compound interest per annum will a sum of Rs. 1200 become Rs. 1348.32 in 2 years**

**4.The least number of complete years in which a sum of money put out at 20% compound interest will be more than doubled is**

**5.Simple interest on a certain sum of money for 3 years at 8% per annum is half the compound interest on Rs. 4000 for 2 years at 10% per annum. The sum placed on simple interest is**

**6.What will be the difference between simple and compound interest @ 10% per annum on the sum of Rs 1000 after 4 years**

**7.The difference between simple and compound interests compounded annually on a certain sum of money for 2 years at 4% per annum is Rs 1. Find the sum**

**8.On a sum of money, simple interest for 2 years is Rs 660 and compound interest is Rs 696.30, the rate of interest being the same in both cases.**

**9.If the simple interest on a sum of money for 2 years at 5% per annum is Rs. 50, what is the compound interest on the same at the same rate and for the same time?**

**10.The difference between compound interest and simple interest on an amount of Rs. 15,000 for 2 years is Rs. 96. What is the rate of interest per annum?**

**Answers:**

**1. (c) **

Amount = Rs. [200 X (1+ 5/100)^3 + 200(1 +( 5/100))^2 + 200(1 + 5/100)

= Rs. [200 X (21/20) X (21/20) X (21/20) +200 X (21/20) X (21/20) + 200 X 21/20 ].

= Rs. [200 X (21/20)[(21/20) X (21/20) + (21/20) +1]

= Rs. 662.02

**2.C**

Principal= Rs.16,000;

Time=9 months = 3 quarters;

Rate = 20%, it will be 20/4 = 5%

So lets solve this question now

Amount=16000(1+5/100)^3=18522

C.I=18522-16000=2522

**3.D**

Let Rate will be R

1200(1+R/100)^2=134832/100

(1+R/100)^2=134832/120000

(1+R/100)^2=11236/10000

(1+R/100)=106/100

R=6%

**4.A**

As per question we need something like following

P(1+R/100)^^n>2P

(1+20/100)^n>2

(6/5)^n>2

6/5�6/5�6/5�6/5>2

So answer is 4 years

**5.B**

C.I.=(4000�(1+10/100)^2-4000)

=4000*11/10*11/10-4000=840

So S.I. = 840/2=420

So Sum = (S.I.*100)/(R*T)

=(420*100)/(3*8)

=Rs1750

**6.C**

S.I.=(1000*10*4)/100=400

C.I.=[1000(1+10/100)^4-1000]=464.10

So difference between simple interest and compound interest will be

464.10 – 400 = 64.10

**7.B**

**8.D**

Difference in CI and SI for 2 years=Rs(696.30-660)=Rs 36.30

SI for one year = Rs 330.

therefore SI on Rs 330 for 1 year = Rs 36.30

therefore,Rate= (100*36.30/330*1)% = 11%

**9.A**

**10.A**

C.I.-S.I.=P(R/100)*(R/100)

96=15000(R/100)(R/100)

R*R=960/15=64

R=8