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# Worked Out Problem Simple Interest and Compound Interest

Worked Out Problem Simple Interest and Compound Interest for SSC,SBI PO,IBPS PO and government exam.

1. A  sum of Rs. 1600 gives a simple interest of Rs. 252 in 2 years and 3 months. the rate of interest per annum is

Solution: I =P × R × T/100

252 = 1600 × R × 9/4 /100

R = 7%

1. What sum of money will amount to Rs. 520 in 5 years and to Rs. 568 in 7 years at simple Interest ?

Solution: principal + interest for 5 years = 520                                                (1)

Principal + interest for 7 years = 568                                                                    (2)

Subtracting (1) from (2)

Interest for two years = 568 – 520 = 48

Interest for 1 years = 24

Interest for 5 years = 120

Principal = 520 – 120 = 400

1. What should be the least number of years in which the simple interest on Rs. 2600 at 6 2/3 % will be an exact number of rupees ?

Solution : Interest for 1 year = 26002031100 = 520/3

If we multiply by 3 in interest for 1 year, we get the exact number of rupees

Minimum time = 3 years

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1. Ratio of the principal and the amount after 1 year is 10 :12 . then the rate of interest  per  annum is

Solution: P:A = 10 :12

P : I = 10 :2

P/I = 10/2 = 5

I = P × R × T/100

I = 5I × R × T/100

R= 20%

## Worked Out Problem Simple Interest and Compound Interest

1. A person invests money in three different schemes for 6 years, 10 years and 12 years at 10 percent,12 percent and 15 percent respectively. At the completion of each scheme, he gets the same interest. The ratio of his investment is

Solution: let the principal be P1,P2 and P3 for 6 , 10 and 12 years respectively

P1 = 100 × I/6 × 10

P2 = 100 × I/10 × 12

P3 = 100 × I/12 × 15

P1:P2:P3 = 1/60:1/120:1/180
= 6:3:2

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1. The sum of money that yields a compound interest of Rs. 420 during the second year at  5% p.a. Is

Solution: C.I. = P [(1 + R/100)t-1] – PR/100

420 = P [(1+ 5/100)2-1] – P5/100

420 = 41P/400 – 5P/100

420 = 21P/400

P = 8000

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1. A man saves Rs. 2000 at the end of each year and invests the money at 5 Compound interest. At the end of 3 years he will have

Solution: he saves 2000 at the end of 1st year

Interest in the  second year = 2000 × 5 × 1/100 = 100

Amount = 2000 + 100 = 2100

He saves another 2000 at the end of 2nd year

Principal for the 3rd year = 2100 + 2000 = 4100

Interest in the 3rd year = 4100 × 5 × 1/100 = 205

Amount = 4100 + 205 = 4305

But he saves 2000 at the end of 3rd year

So total amount = 4305 + 2000 = 6305

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1. The difference between the compound interest and simple interest for the amount Rs. 5000 in 2 years is Rs. 32. The rate of interest is

Solution: simple interest for 1st year = 5000R1/100 = 50R

Simple interest for 2nd year = 50R

Compound interest for 2nd year = 50R +50RR/100

Difference between C.I. and S.I. = 32

50R2/100 = 32

R2 = 64

R = 8%

Alternative

Difference between C.I. and S.I. = P R2/100100

32 = 5000 R2/10000

R2= 64

R = 8

1. A sum of money doubles itself in 4 years at compound interest. It will amount to 8 Times it self at the same rate of interest in :

Solution: let the principal be x

Amount = 2x in 4 years

2x = x(1r/100)4

2 = (1 + r/100)4

When amount = 8x

8x =  x(1 + r/100)t

8 =  (1r/100)t

23 = (1 + r/100)t

(1 + r/.100)12 = (1 + r/100)t

T = 12 years

1. If the compound interest on a certain sum for 2 years at 3% per annum is Rs. 101.50 , then the simple interest on the same sum at the same rate and for the Same time will be

Solution: The compound interest for 2 years in terms of interest is

C.I. = I +  I × r/100*2

101.5 = y + 3y/200

203y/200 = 101.5

Y = 20300/203
Y = 100

1.  An amount of money appreciates to Rs. 7000 after 4 years and to Rs. 10000 After 8 years at a certain compound interest compounded annually. The Initial amount of money was

Solution: After 4 years the amount 7000 will be the principal for the next 4 years

10000 = 7000 (1 + r/100)4

10/7 = (1 + r/100)4

Now, when amount =  7000, T= 4 years,

7000 = P(1 + r/100)4

7000 = P × 10/7

P = 4900

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