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Solved Example

Example-1
Find S.I. on Rs. 68000 at 16 2/3% per annum for 9 months ?
Solution
P = 68000 R = 50/3% p.a. T = 9/12 years = 3/4 years
S.I = (P x R x T) /100 = (68000 x (50/3) x (3/4) x (1/100) ) =Rs. 8500
 
 
Example-2

Find S.I. on Rs. 3000 at 18% per annum for the period from 4th Feb to 18th April 2009 ?

Solution

Time = (24 + 31 + 18) days = 73 days = 73/365 = 1/5 years
P = Rs. 3000
R = 18% p.a.
S.I. = (P × R × T) /100 = (3000 × 18 × 1/5 × 1/100) = Rs. 108
 

 

Note: The day on which money is deposited is not counted while the day on which money is withdrawn is counted.

 

Example-3
In how many years will a sum of money becomes triple at 10% per annum ?
Solution
Let principal = P
S.I. = 2P
S.I. = (P × T × R)/100
2P = (P × T × 10)/100
T = 20 years
 
 

Note:

  1. Total amount = Principal + S.I.
  2. If sum of money becomes double means Total amount or Sum = Principal + S.I. = P + P = 2P
 
Example-4

A sum at Simple interest at 13 1/2% per annum amounts to Rs. 2502.50 after 4 years. Find the sum?

Solution

Let Sum be S. then,
S.I. = (P × T × R)/100 = ((S × 4 × 27)/(100 × 2)) = 54S/100
Amount = (S + (54S)/100) = 77S/50
77S/50 = 2502.50
S = (2502.50 × 50)/77 S = 1625 Sum = Rs. 1625
 

 
Example-5
A sum of money becomes double of itself in 4 years, in 12 years it will become how many times at the same rate?
Solution
4 years = P
12 years = ?
(12/4) × P = 3P
Amount or Sum = P + 3P = 4 times
 
 
Example-6

A Sum was put at S.I. at a certain rate for 3 years Had it been put at 2% higher rate, it would have fetched Rs. 360 more. Find the Sum?

Solution

Let Sum = P
Original rate = R
T = 3 years
If 2% is more than the original rate, it would have fetched 360 more i.e., R + 2
(P × (R + 2) × 3/100) - (P × R × 3)/100 = 360 = >3PR + 6P - 3PR = 36000 = >6P = 36000 = >P = 6000
 

  
Example-7
Rs. 800 amounts to Rs. 920 in 3 years at S.I. If the interest rate is increased by 3%, it would amount to how much?
Solution
S.I. = 920 - 800 = 120
Rate = (100 × 120)/(800 × 3) = 5%
New Rate = 5 + 3 = 8%
Principal = 800
Time = 3 yrs
S.I. = (800 × 8 × 3)/100 = 192
New Amount = 800 + 192 = Rs. 992
 
 
Example-8

Prabhat took a certain amount as a loan from bank at the rate of 8% p.a. S.I. and gave the same amount to Ashish as a loan at the rate of 12% p.a.. If at the end of 12 yrs, he made a profit of Rs. 320 in the deal, What was the original amount?

Solution

Let the original amount be Rs. S.
T = 12
R1 = 8%
R2 = 12%
Profit = 320
P = S
(P × T × R2)/100 - (P × T × R1)/100 = 320
(S × 12 × 12)/100 - (S × 8 × 12)/100 = 320
S = 2000/3
S = Rs. 666.67

  
Example-9
Simple Interest on a certain sum at a certain rate is 9/16 of the sum. If the number representing rate percent and time in years be equal, then the rate is.
Solution
Let Sum = S
Then,
S.I. = 9S/16
Let time = n years & rate = n%
n = 100 × 9S/16 × 1/S × 1/n = >n × n = 900/16 = > n = 30/4 = 7 1/2%
 
  
Example-10

A certain sum of money amounts to 1680 in 3 years & it becomes 1920 in 7 years. What is the sum?

Solution

3 years - > 1680
7 years - > 1920
then, 4 years - > 240
1 yr - > ?
(1/4) × 240 = 60
S.I. in 3 years = 3 × 60 = 180
Sum = Amount - S.I = 1680 – 180 = 1500 we get the same amount if we take S.I. in 7 yrs
i.e., 7 × 60 = 420, Sum = Amount - S.I. = 1920 – 420 = 1500
 

  
Example-11
A Person takes a loan of Rs. 200 at 5% simple Interest. He returns Rs. 100 at the end of 1 yr. In order to clear his dues at the end of 2 yrs, he would pay:
Solution
Amount to be paid = Rs. (100 + (200 × 5 × 1)/100 + (100 × 5 × 1)/100) = Rs. 115
 
 
Example-12

A Man borrowed Rs. 24000 from two money lenders. For one loan, he paid 15% per annum and for other 18% per annum. At the end of one year, he paid Rs. 4050. How much did he borrow at each rate?

Solution

Let the Sum at 15% be Rs. A
& then at 18% be Rs. (24000 - A)
P1 = A, R1 = 15
P2 = (24000 - A), R2 = 18
At the end of ine year T = 1
(P1 × T × R1)/100 + (P2 × T × R2)/100 = 4050
(A × 1 × 15)/100 + ((24000 - A) × 1 × 18)/100 = 4050
15A + 432000 – 18A = 405000 = > A = 9000
Money borrowed at 15% = 9000
Money borrowed at 18% = (24000 - 9000) = 15000
 

  
Example-13
What annual instalment will discharge a debt of Rs. 1092 due in 3 years at 12% Simple Interest?
Solution
Let each instalment be Rs. A
(A + (A × 12 × 1)/100) + (A + (A × 12 × 2)/100) + A = 1092
= > 28A/25 + 31A/25 + A = 1092 => (28A + 31A + 25A) = (1092 × 25) = > 84A = 1092  25
= > A = (1092  25) /84 = 325 Each instalment = Rs. 325
 
 
Example-14

A Sum of Rs. 1550 was lent partly at 5% and partly at at 8% p.a. Simple interest. The total interest received after 3 years was Rs. 300. The ratio of the money lent at 5% to that lent at 8% is:

Solution

Let the Sum at 5% be Rs. A
at 8% be Rs. (1550 - A)
(A × 5 × 3)/100 + ((1500 - A) × 8 × 3)/100 = 300 = > 15A + 1500 × 24 – 24A = 30000 = > A = 800
Money at 5%/ Money at 8% = 800/(1550 - 800) = 800/750 = 16/15
 

 
Example-15
A Man invests a certain sum of money at 6% p.a. Simple interest and another sum at 7% p.a. Simple interest. His income from interest after 2 years was Rs. 354. One fourth of the first sum is equal to one fifth of the second sum. The total sum invested was?
Solution
Let the sums be A & B
R1 = 6, R2 = 7, T = 2
(P1 × R1 × T)/100 + (P2 × R2 × T )/100 = 354 = > (A × 6 × 2)/100 + (B × 7 × 2)/100 = 354 = > 6A + 7B = 17700
also one fourth of the first sum is equal to one fifth of the second sum
A/4 = B/5 = > 5A – 4B = 0
By solving the equations we get,
A = 1200, B = 1500
Total sum = Rs. (1200 + 1500) = Rs. 2700
 
  
Example-16

Rs. 2189 are divided into three parts such that their amounts after 1, 2 & 3 years respectively may be equal, the rate of S.I. being 4% p.a. in all cases. The Smallest part is?

Solution

Let these parts be A, B and [2189 - (A + B)] then, (A × 1 × 4)/100 = (B × 2 × 4)/100 = (2189 - (A + B)) × 3 × 4/100 = > 4A/100 = 8B/100 = > A = 2B
By substituting values
(2B × 1 × 4)/100 = (2189 - 3B) × 3 × 4/100 = > 44B = 2189 × 12 = > B = 597
Smallest Part = Rs. 597
 

  
Example-17
A man invested 1/3 of his capital at 7%, 1/4 at 8% and the remainder at 10%.If his annual income is Rs. 561. The capital is:
Solution
Let the capital be Rs. A. Then, (A/3 × 7/100 × 1) + (A/4 × 8/100 × 1) + (5A/12 × 10/100 × 1) = 561
7A/30 0 + A/50 + A/24 = 561
51A = 561 × 600 thus, A = Rs. 6600
 
 
Example-18

Find CI on Rs. 6250 at 16% per annum for 2 years, compounded annually?

Solution

Rate R = 16, n = 2, Principal = Rs. 6250
Amount = P[1 + (R/100)]n
= 6250[1 + (16/100)]2 = Rs. 8410
C.I. = Amount - P
= 8410 - 6250
= Rs. 2160

  
Example-19
Find C. I on Rs. 16000 at 20% per annum for 9 months compounded quarterly?
Solution
R = 20% then Rate for 12 months = 20%
Rate for 3 months = 20/4 = 5%
For 9 months, there are '3' 3months
Amount = P[1 + (R/100)]n = 160 00[1 + (5/100) 3 = Rs. 18522
C.I. = 18522 - 16000 = Rs. 2522
 
 
Example-20

The difference between C.I. and S.I. on a certain sum at 10% per annum for 2 years is Rs. 631. Find the sum?

Solution

Let the sum be Rs. P, Then
C.I. = P[1 + (10/100)]2 – P = 21P/100
S.I. = (P × 10 × 2) /100 = P/5
C.I. - S.I. = 21P/100 - P/5
= P/100
Given CI - SI = Rs. 631
= > P/100 = 631 = > P = Rs. 63100
 

  
Example-21
If C. I on a certain sum for 2 years at 12% per annum is Rs. 1590. What would be S.I?
Solution
C.I. = Amount - Principal and C.I. = P[1 + (12/100)]2 – P = > 784P/625 - P = 1590 = > P = Rs. 6250
S.I. = (6250 × 12 × 2)/100 = Rs. 1500
 




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