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 be2

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=55

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=62

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

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Simple Interest