**1. Mohan invested a certain sum in a simple interest bond whose value grew to Rs. 300 at the end of 3 yr and to Rs. 400 at the end of another 5 yr. The rate of interest was**

**2. Suman lent some money to Rohan at 6% p.a. simple interest. Rohan lent the entire amount to Mira on the same day at 8.5 % p.a. In this transaction Rohan earned a profit of Rs. 250. Find the sum lent by Suman to Rohan.**

**3. On retirement, a person gets 1.53 lakhs of his provident find which he invests in a scheme at 20% p.a. His monthly income from this scheme will be**

**4.If Rs. 450 amount to Rs. 540 in 4 years, what will it amount to in 6 years at the same rate %?**

**5.In what time will Rs. 64,000 invested at 5% p.a. fetch an interest of Rs. 4,921, the interest being compounded half yearly.**

**6.The population of a city two years ago was 1,25,000. Due to migration from cities, it decreases every year at the rate of 4% per annum. Find its present population. How many person have migrated in last two years?**

**7.The population of a town in the year 2001 was 4 lakhs. Due to migration from cities, it decreases every year at the rate of 40 per thousand. Find the population of the town in lakhs during 2003.**

**8.A sum of money at compound interest amounts to thrice itself in three years. In how many years will it be nine times itself?**

**9.The difference between simple and compound interest on a certain sum of money for 3 years at 5 per cent per annum is Rs. 325. The sum of money is :**

**10.The difference between simple and compound interest on a certain sum of money for 2 years at 4 per cent per annum is Rs. 12. The sum of money is :**

**Answers and Solution:**

1.(c)

(Pr*3)/100 +P = 300…….(1)

(Pr*8)/100 +P = 400…….(2)

subtract (1) from (2)

p*r = 2000 ………..(3)

put (3) in (1)

p = 240, then r = 8.33%

2.(c)

3.(c)

Use, SI = PRT/100 = (153000*20)/100*1/12 = Rs. 2550

4.(b)

SI per year = (540-450)/4 = 22.5

so after 6 year = 22.5 x 6 = 135

So Amount 450+135 = Rs.585

5.(d)

6.(b)

Population 2 year ago = 125000 = p

for two years,rate of decrease = 4% p.a.

population after 2 year = present population = p(1-R/100)^2

Present population = 125000(1-4/100)^2

= 115200

So number of persons migrated during last 2 years= 125000-115200 = 9800

7.(d)

8.(c)

9.(a)

Difference of CI and SI for 3 years = pr^2(300+r)/(100)^3

325 = (p*25*305)/1000000

so p = 42623

10.(3)

Difference of CI and SI for 2 years = pr^2/(100)^2

12 = (p*16)/10000

so p = 7500

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**to see Concepts and Tricks on Compound Interest and simple Interest**