Simple Interest & Compound Interest Questions are asked
in all competition exams. If you want to ask us any CI & SI problems Short
Tricks, How to Solve Compound Interest Questions Easily, How to Solve Simple
Interest Questions in Less Time, Best Method to Solve CI and SI Questions
please write your query in comment box we will give you easy method to solve SI
and CI Questions. If you want to crack Competition Exam stay connect our blog.
In this blog we will provide Simple Method to Solve Compound Interest Questions,
Simple Interest (SI) Solved Examples, CI Solved Examples, Compound Interest
Questions with Answer, SI Questions with Solution, Top Method to Solve CI &
SI Problems.
Q.1 The SI Occurred
on a sum of money at the rate of Interest 5% per annum for two years is 410 Rs.
The Compound Interest is the same as S.I as occurred on the sum of money at the
same rate of Interest and the same time. Find the difference B/W the two sum of
money ( Principle)
Solution
From SI, P =
(410×100)/(5×2) = Rs 4100
From CI , P = 410/(5/50+ (5/100)2) =
(410 ×400)/41
= 4000
Q. 2 RS 2100 is divided into two parts such that the simple
interest on the one part at 4.5% for 3.5 years be the same as that on the other
at 5.25% for 4 years. Find Out the second Part?
Solution
When a money is divided in such two part so that SI on 1st
part at X% for T1 equal to SI on II part at X2 for T2 then the
ration of moneyto be divided = 1st :
2nd
=
X2T2 : X1T1
Ratio = 5.25% *
4 :
4.4 * 3.5
= 4 : 3
Therefore money of IInd part
= 2100 × 3/7
= Rs 900 Ans.
Q. 3 Rahul Borrowed an amount of Rs 15000 at the simple
interest rate of 12 P.C.P.A and another amount at the simple Interest rate of
15 P.C.P.A for a period of two years each. He Paid amount of Rs 9000 as the
total interest, What is the total amount borrowed?
Let the amount borrowed
at 15% p. a is X
15000 ×(2 ×12)/100
+ (15 ×X×2)/100 = 9000
3600 + 3X/10 = 9000
3X/10 = 5400
X = 18000
Total amount borrowed is
= 15000 + 18000
= Rs 33000 Ans
.
Q. 4 A sum of Money Placed at compound Interest doubles
itself in 5 years and Rs. 4624 in 2 years and 4913 in 3 years, the sum of Money
is?
Solution
A = P ( 1 + R/100)n = 2P
=> ( 1 + R/1000)5
( 1 + R/1000)5
= 2
Let P = ( 1 + R/100)n = 8
(2)3 = ( 1 +
R/100)n
=> {( 1 + R/1000)5 }3 = (
1+ R/100)n
=> n = 15 Years {1+10/100) 3
Q.5 there is a 60% increase in an amount in 6 Years at the
rate of simple Interest, what will be the compound Interest of Rs 12000 after 3 years at the same rate?
Let P = Rs 100 Then S.I Rs 60 & T = 6 Years
R = 100* 60(100
×60)/(100×6) = 10 %p.a
Now P = Rs 12000, T = 3 years & R = 10 % p.a.
C. I = Rs [ 12000 ×{1+ 10/100)n- 1 }]
= 12000 ×331/1000 = Rs 3972 Ans
Q. 6 The simple Interest on a certain sum of money for 3
years at 8% per annum is half of the compound interest on Rs 4000 for 2 years
at 10% per annum. The Sum placed on simple Interest is?
Solution
C. I = Rs [ 4000 × (1
+10/100)2 - 4000]
= 4000 ×11/10 ×
11/10 -4000 = Rs 840
Sum = (420 ×100)/(3×8) = Rs 1750 Ans
Q. 7 The Compound interest on a certain sum of money for 2
years at 8% per annum is Rs. 525. The Simple Interest on the same sum for
double the time at the half the rate percent per annum is?
Solution
Let the sum be Rs P Then
[P ((1 +10/100)2
- P ] = 525
=> P [(11/10)2
- 1] = 525
P = (525 ×100)/21 = 2500
Sum = 2500 Rs
So SI = Rs ((2500
×5*4)/100 ) = Rs 500 Ans.
Q. 8 If the compound Interest on a sum for 2 years at 12 1/2
% per Annum is Rs 510, The simple Interest on the same sum at the same rate for
the same period time is?
Solution
[ P ( 1 + 25/(2 ×100))2 - P ]
= 510
Or [ P ( 9/8)2 - 1] = 510
Or P = (510×64)/17 = 1920
So S.I = Rs
(1920×25×2)/(2×100) = Rs 480 Ans.
Q. 9 In how many years will a sum of Rs 800 at 10% per annum
compounded half yearly become rs 926.10?
Solution
Let the time be N years
Then 800 *( 1 + 5/100)2n = 926.10
( 1 + 5/100)2n
= 9261/8000
Or ( 21/20)2n
= (21/20)3
2n = 3
N = 3/2 Years Ans.
Q. 10 The Principle that amounts to Rs 4913 in 3 years at 6
1/4 % per annum compound interest compounded annually is?
Solution
P = [ 4913/█((1+25/4×100)
)3
= 4913 * 16/17 *
16/17 * 16/17 = 4091 Ans.
Q. 11 The difference between simple interest and compound
interest on Rs 1200 for one year at 10% per annum reckoned yearly is?
Solution
N = 1 year = 2 half years
R = 10/2 = 5 %
S. I = (1200× 5×1)/100
= Rs 60
C. I = Rs [ 1200 ×(1+ 5/100)2 - 1200
Rs = 63
Difference = 63 – 60
= Rs 3 Ans.
Q. 12 The Compound Interest on Rs 30,000 at 7% per annum is
Rs 4347. The Period in Years is?
Solution
Amount = 30000 +
4347 = Rs 34347
Let time be n years
then
30000 ((1+ 7/100)n = 34347
(107/100)n
= 34347/30000 = (107/100)2
N = 2 Years Ans
Q. 13 At what rate of Interest per Annum will a sum of Rs
1200 become Rs 1343.32 in 2 Years?
Solution
Let the Rate be R % p.a
then
1200 ×(1+ R/100)2
= 1343.32
(1+ R/100) 2
= 1343.32/ 120000
(1+ R/100) = 106/100
R = 6 %
Q. 14 Rs 800 becomes Rs 956 in 3 years at a certain rate of
simple Interest. If the rate of interest is increased by 4%, what amount will rs 800 become in 3
Years?
Solution
S. I = Rs ( 956 –
800) = Rs 156
Rate = (100×156)/(800×3)
= 6 1/2
New rate is = (6 ½
+ 4)% = 10 1/2 %
New S. I =
(800*21*3)/(100*2) = Rs 252
New amount = ( 800 +
252) = 1052 Ans.
Q. 15 The Simple Interest on a sum of money will be Rs
600after 10 Years. If the principle is tripled after 5 Years, what will be the
total interest at the end of the tenth year?
Solution
Let the sum be Rs , Now S.I
is given = Rs 600 and Time = 10 Years
Then Rate = (100 *
600 / X * 100)% = (6000/X)%
S.I for first five years
= ( X * 5 * 6000/ X * 100) = Rs 300
S. I for last Five Years
= (3 X * 5 * 6000/ X * 100) = Rs 900
Then Total Interest is
= Rs 1200 Ans
If the difference between the compound interest and simple interest on a sum of money at 5% p.a for 2 years is Rs. 16, then find the simple interest.
ReplyDelete