Question

A certain sum in certain time becomes Rs 560 at the rate of 6% per annum at simple interest and the same sum amount to Rs 320 at the rate of 2% in same duration. Find the sum and time​

Answer 1

$Let \: the \: Principal = Rs \: P$

$and \: Time = T \: years$

$i) Amount ( A_{1} ) = Rs \: 560$

$Rate \: of \: interest (R_{1}) = 6\% \: p.a$

$A_{1} = P\Big( 1 + \frac{TR_{1}}{100}\Big)$

$\implies 560 = P\Big( 1 + \frac{6T}{100}\Big) \: --(1)$

$ii) Amount ( A_{2} ) = Rs \: 320$

$Rate \: of \: interest (R_{2}) = 2\% \: p.a$

$A_{2} = P\Big( 1 + \frac{TR_{2}}{100}\Big)$

$\implies 320 = P\Big( 1 + \frac{2T}{100}\Big) \: --(2)$

$\red { Do \: (1) \div (2) }, we \:get$

$\implies \frac{560}{320} = \frac{ P\Big( 1 + \frac{6T}{100}\Big)}{P\Big( 1 + \frac{2T}{100}\Big)}$

$\implies \frac{7}{4} = \frac{ \Big( \frac{100+6T}{100}\Big)}{\Big( \frac{100+2T}{100}\Big)}$

$\implies \frac{7}{4} = \frac{100+6T}{100+2T}$

$\implies 7(100+2T) = 4(100+6T)$

$\implies 700+14T= 400+24T$

$\implies 700 - 400= 24T - 14T$

$\implies 10T = 300$

$\implies T = 30 \: years$

/* Put T = 30 in equation (1) , we get */

$560 = P \Big ( 1 + \frac{6 \times 30 }{100} \Big)$

$\implies 560 = P\Big ( \frac{(100+180)}{100} \Big)$

$\implies 560 = P\Big ( \frac{280}{100} \Big)$

$\implies P = \frac{560 \times 100}{280}$

$\implies P = Rs \: 200$

Therefore.,

$\green { Required \: sum (P) = Rs \: 200 } \\\green { and \: Time = 30 \: years }$

•••♪

Answer 2

Given:

$a_1=560\\r_1=6\%\\a_2=320\\r_2=2\%$

To find:

t,p=?

Solution:

Formula:

$\bold{A=P(1+r)^t}$

Let p=P and t= T

put the given value in the above formula:

$\to 560=P(1+\frac{6T}{100}).............(i)$

$\to 320=P(1+\frac{2T}{100}).............(ii)$

Divide the equation (i) from the equation (ii):

$\Rightarrow \frac{560}{320} =\frac{P(1+\frac{6T}{100})}{P(1+\frac{2T}{100})}\\\\\Rightarrow \frac{56}{32} =\frac{P(1+\frac{6T}{100})}{P(1+\frac{2T}{100})}\\\\\Rightarrow \frac{7}{4} =\frac{(\frac{100+6T}{100})}{(\frac{100+2T}{100})}\\\\\Rightarrow \frac{7}{4} =(\frac{100+6T}{100} \times \frac {100}{100+2T})\\\\\Rightarrow \frac{7}{4} =(\frac{100+6T}{1} \times \frac {1}{100+2T})\\\\\Rightarrow \frac{7}{4} =(\frac{100+6T}{100+2T})\\\\\Rightarrow 7 \times (100+2T) = 4\times (100+6T)$

$\Rightarrow 700+14T = 400+24T\\\\\Rightarrow 700- 400=24T-14T \\\\\Rightarrow 300=10T\\\\\Rightarrow T= \frac{300}{10}\\\\\Rightarrow T= 30$

put the value of the T in equation (i) and (ii):

$\ Equation (i): \\\\ \to 560=P(1+\frac{6T}{100})\\\\\to 560=P(1+\frac{6\times 30}{100})\\\\\to 560=P(1+\frac{180}{100})\\\\\to 560=P(1+\frac{18}{10})\\\\\to 560=P(\frac{10+18}{10})\\\\\to 560=P(\frac{28}{10})\\\\\to P= \frac{560 \times 10}{28}\\\\\to P= 20 \times 10\\\\\to P=200$

$\ Equation (ii): \\\\ \to 320=P(1+\frac{2T}{100})\\\\\to 320=P(1+\frac{2\times 30}{100})\\\\\to 320=P(1+\frac{60}{100})\\\\\to 320=P(1+\frac{6}{10})\\\\\to 320=P(\frac{10+6}{10})\\\\\to 320=P(\frac{16}{10})\\\\\to P= \frac{320 \times 10}{16}\\\\\to P= 20 \times 10\\\\\to P=200$

The final value of P and T is 200 and 30