Simple interest

Black Holes of Aptitude-Simple Interest
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Simple Interest
 Interest: (I) Interest is money paid to the lender by the borrower for using his money for
a specified period of time. Various terms and their general representation are as follows:
 Principal(P): The original sum borrowed
 Time (t): Time for which money is borrowed.
 Rate of Interest(r): Rate at which interest is calculated on the original sum.
 Amount (A): Sum of Principal and Interest. (P+I).
 Simple Interest (SI): When interest is calculated every year (or every time period) on
the original Principal, such interest is called Simple Interest.
SI =
P × r × t
100
A=P+I
A=P+𝑡 × 𝑥
Type I:
Exp 1: A Sum of Rs. 1600 gives a SI of Rs. 252 in 2 years and 3 months. The rate of interest is:
Solution:
P=1600
SI=252
t= 2 years 3 months=2+
3
12
=
9
4
r=?
SI =
P × r × t
100
252 =
1600 × r × 9
100 × 4
=7%

Type II: Times
Part A: Simple Interest becomes x times of Principal
Take
P=1
SI=x
Exp2: In what time will the SI be
2
5
of the principal at 8 ppa.
Solution:
SI =
P × r × t
100
2
5
=
1 × 8 × t
100
t=5 years
Part B: A sum of money becomes x times of itself.
Take
P=1
SI=x-1
Exp3: A sum of money becomes
7
6
times of itself in 3 years at a certain rate of SI .The rate ppa
(percent per annum) is:
Solution:
P=1
SI=
7
6
-1 =
1
6
SI =
P × r × t
100
1
6
=
1 × r × 3
100
r=5
5
9
%

Part C: A sum of money becomes x times of Amount.
A=P+I
1=P+x×1
P=1-x
Take
SI=x
P=1-x
Exp 4: At what rate ppa will the SI on a sum of money be
2
5
of the amount in 10 years.
Solution:
SI=
2
5
P=1-
2
5
=
3
5
SI =
P × r × t
100
2
5
=
3
5
× r × 10
100
r=6
2
3
%
Part D: x times in t1 years
y times in t2 years
𝑥 − 1
𝑦 − 1
=
t1
t2
Exp 5: A certain sum of money becomes 3 times of itself in 20 years at SI. In How many years
does it become double of itself at SI?
Solution:
𝑥 − 1
𝑦 − 1
=
t1
t2

3 − 1
2 − 1
=
20
t2
t2 = 10 years
Type III
Part A: A sum of money P amounts to A1 in t years, it will amount to A2 if it was put at r %
higher.
A2= A1+
P×r×t
100
Exp 6: A sum of 800 amounts to 920 in 3 years at SI. If the rate is increased by 3ppa .What will
be the sum amount to in the same period.
A2= 920+
800×3×3
100
=992
Part B: A sum was put at SI at a certain rate for t years .Had it been put at r ppa higher, it would
have fetched D Rs. more find the sum:
P=
D×100
𝑡×𝑟
Exp 7: A sum of money was lent at SI at certain rate for 3 years .Had it been lent at 2.5 % higher
rate, it would have fetched 540 more. The money lent was:
P=
D×100
𝑡×𝑟
P=
540×100
3×2.5
=7200
Type IV: The SI on certain sum of money at r1 ppa for t1 years is Rs. D more than the interest
on the same sum for t2 years at r2 ppa.
P =
D×100
(r1×t1−r2×t2)

Type V: Annual Payment
Annual Payment=
P×100
𝑛×100+
𝑛(𝑛−1)×𝑟
2
Exp 8: What annual payment will discharge a debt of Rs. 848 in 4 years at 4ppa.
Annual Payment=
P×100
𝑛×100+
𝑛(𝑛−1)×𝑟
2
Annual Payment=
848×100
4×100+
4(4−1)×4
2
=200
Type VI: If certain money amounts to A1 in t1 years, A2 in t2 years. Find Principal and rate of
interest:
Solution:
A1= P+ t1× x…………………(i)
A2= P+ t2× x………………..(ii)
(ii)-(i)
X=
A2−A1
t2−t1
……………………(iii)
( iii) in (i)
P =
A1t2−A2t1
t2−t1
From
SI =
P × r × t
100
R =
100×(A2−A1)
(A1t2−A2t1)
Exp 9: A certain sum of money amounts to 756 in 2 years and 873 in 3.5 yeas at a certain rate
find P and SI:

Solution:
756=P+2×x……..(i)
873=P+3.5×x………(ii)
X=78…………(iii)
(iii)in (i) or (ii)
P=600
from
SI =
P × r × t
100
R=13%
Type VII Partly at x % partly at y %,avg. is r %
Ratio of principal=
(𝑦−𝑟)
(𝑟−𝑥)
Exp10: A sum of Rs. 10000 lent partly at 8 % and remaining at 10ppa.If the yearly interest on
avg. is 9.2%, the two parts are:
Solution: Ratio of principal=
(𝑦−𝑟)
(𝑟−𝑥)
Ratio of principal=
(10−9.2)
(9.2−8)
=2:3
Parts will be 4000 and 6000

Simple interest

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