Worked Out Problem Simple Interest and Compound Interest
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.
- 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%
- 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
- 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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- 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
- 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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- 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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- 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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- 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
- 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(1 + r/100)4
2 = (1 + r/100)4
When amount = 8x
8x = x(1 + r/100)t
8 = (1 + r/100)t
23 = (1 + r/100)t
(1 + r/.100)12 = (1 + r/100)t
T = 12 years
- 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
- 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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