Question

A sum of money lent out at simple interest amounts to 2200 rupees in one year and to 2800 rupees in four years. Find the sum of money and the rate of interest.

Answer 1

Sum = P lent out

after one year, Rs 2, 200 = P (1 + r /100 )

After 4 years  Rs 2, 800 =  P (1 + 4 r /100)

Subtract equation 1 from equation 2 :  Rs 2, 800 - RS 2, 200 = 3 P r /100
                                     P r = Rs 20,000

Substituting the value of P r in equation 1 :
           2, 200  = P + 200            So P = Rs 2,000.

             r = 20,000 / 2,000 = 10

Answer 2

Let  the  principle be  P  and rate  of  interest  r.

For  one  year,

Time (t) = 1 year

rate  of interest = r

principle  = P

Simple  interest (I₁) = A - P = 2200 - P

Amount  = Rs. 2200

We know that,

$S.I.=\frac{p\times r\times t}{100}\\\;\\I_1=\frac{P\times r\times1}{100}\\\;\\A-P=\frac{Pr}{100}\\\;\\2200-P=\frac{Pr}{100}\;\;\;................i)$

For  four   year,

Time (t) = 4 year

rate  of interest = r

principle  = P

Simple  interest (I₂) = A - P = 2800 - P

Amount  =  Rs.2800

We know that,

$S.I.=\frac{p\times r\times t}{100}\\\;\\I_2=\frac{P\times r\times4}{100}\\\;\\A-P=\frac{4Pr}{100}\;\;........................ii)$

Comparing  Eq. i) and  ii)

$2200-P=\frac{2800-P}{4}\\\;\\2200-P=\frac{2800}{4}-\frac{P}{4}\\\;\\2200-P=700-\frac{P}{4}\\\;\\2200-700=P-\frac{P}{4}\\\;\\1500=\frac{4P-P}{4}\\\;\\6000=3P\\\;\\P=2000\\\;\\\;\\\text{Putting P=2000 in eq. ii)}\\\;\\\frac{2800-P}{4}=\frac{Pr}{100}\\\;\\\frac{2800-2000}{4}=\frac{2000\times r}{100}\\\;\\\frac{800}{4}\\\;\\200=20r\\\;\\r=10\%$