A certain sum amount to $1725$ at $15$ $%$ per annum at simple interest and $1800$ in the same time at $20$ $%$ per annum at simple interest. Find the sum.

1700

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1500

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1650

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1200

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Solution

The correct option is B $1700$
$A = P (1 + \frac{R T}{100})$
$1725 = P (1 + \frac{15 T}{100}) 1800 = P (1 + \frac{20 T}{100})$
$\frac{1725}{P} = 1 + \frac{3 T}{20} \frac{1800}{P} = 1 + \frac{4 T}{20}$
Dividing both equations, we get
$\frac{1 + \frac{3 T}{20}}{1 + \frac{4 T}{20}} = \frac{1725}{1800} = \frac{69}{72} = \frac{23}{24}$
$\frac{20 + 3 T}{20 + 4 T} = \frac{23}{24} \Rightarrow 1 - \frac{20 + 3 T}{20 + 4 T} = 1 - \frac{23}{24} = \frac{1}{24}$
$\frac{T}{20 + 4 T} = \frac{1}{24} \Rightarrow 24 T = 4 T + 20 \Rightarrow T = 1$
$\frac{1800}{P} = 1 + \frac{4}{20} = 1 + \frac{1}{5} = \frac{6}{5}$
$\Rightarrow P = 1800 \times \frac{5}{6} = 1500$

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A certain sum amount to 1725 at 15 % per annum at simple interest and 1800 in the same time at 20 % per annum at simple interest. Find the sum.

The correct option is B 1700A=P(1+RT100)1725=P(1+15T100)1800=P(1+20T100)1725P=1+3T201800P=1+4T20Dividing both equations, we get1+3T201+4T20=17251800=6972=232420+3T20+4T=2324⇒1−20+3T20+4T=1−2324=124T20+4T=124⇒24T=4T+20⇒T=11800P=1+420=1+15=65⇒P=1800×56=1500

A certain sum amount to $1725$ at $15$ $%$ per annum at simple interest and $1800$ in the same time at $20$ $%$ per annum at simple interest. Find the sum.