Psc maths -simple interest

*Simple Interest,S.I=PNR/100
Where P=Principal
N=Time in years and R=Rate of interest.

*Amount,A=Principal+interest=P+PNR/100

=P+[1+NR/100]

*If a sum becomes 𝒳 times in N years at simple interest ,then NR=❲𝒳-1)100

Solved questoins

1.A person borrowed Rs 500 at the rate of 5% per annum simple Interest.What amount will he pay to clear the debt after 4 years?

S.I=PNR/100=(500𝗑4𝗑5)/100=100

Amount=Principal+S.I=500+100=RS 600.

(Or Amount,A=P[1+NR/100]  =500[1+4𝗑5/100]=500𝗑120/1O0=600).


2.A certain sum of money lent a S.I .amounts to Rs 690 in 3 years and to Rs 750 in 5 years. The sum lent is :

Principal+S.I .for 5 years =750 ⇒A

Principal+S.I for 3 years=690  ⇒B

(A)-(B)→ S.I for 2 years =750-690=60

So S.I for 5 years =(60/2)𝗑5=150

From (A)Principal+150=750

⇒Principal =750-150=600.

3.In how many years will the simple interest on a sum of money be equal to principal at the ratio of 16  2/3%  per annum?

Here P= I

So ...I=PNR/100=P

N=P𝗑100/P𝗑R=100/R

(100)/50÷3=100𝗑3/50=6

4.At what rate percent of simple interest will a sum of money double itself in 12 years?
A)8 1/3%
B)8  1/2% 
C)8  1/4%
D)9  1/2%

If a sum becomes 𝒳 times in N years at S.I ,Then

NR=(𝒳-1)100

ie,12𝗑R=(2-1)100;

R=1x100/12=8 1/3.

Answer:A

5.Rs 1750 is divided into two parts such that the annual interest on first parts at 8% simple interest and that on second part at 6% simple interest are equal.The interest on each part is ?

A)60

B)65

C)70

D)40

Let  𝒳 be the first part .Then the second part =1750−𝒳

Then 𝒳𝗑1𝗑8/100=(1750−𝒳)𝗑1𝗑6/100

⇒8𝒳 = 6𝗑1750 − 6𝒳;

⇒𝒳=10,500/14=750  ⇒ Second parts =1000.

So interest =750𝗑8/100  or 1000𝗑6/100=60

Answer:A

6.If a sum of money at simple interest doubles itself in 6 years .it will become 4 times in?

A)14 years

B)12years

C)18 years

D)16 years

NR=(𝒳−1)100 

So    6𝗑R=(2−1)100

R=100/6

Now ,N𝗑100/6=(4−1)100

N=3𝗑100𝗑 6/100 =18

Answer:C