Question

On selling a TV at 5% gain and a fridge at 10% gain, a shopkeeper gains ₹3250. But, if he sells the TV at 10% gain and the fridge at 5% loss, he gains ₹1500. Find the actual price of TV and that of the fridge.​

Answer 1

Let the cost price of the TV be ₹x and that of the fridge be ₹y.

Then,

$\sf \dashrightarrow Total \: CP \: of \: TV \: and \: fridge = (x + y)$

Selling price in first case is,

$\sf \dashrightarrow \bigg( \dfrac{105x}{100} + \dfrac{110y}{100} \bigg) = \bigg( \dfrac{21x}{20} + \dfrac{11y}{10} \bigg)$

So, the gain in this case is,

$\sf \dashrightarrow \bigg\{ \bigg( \dfrac{21x}{20} + \dfrac{11y}{10} \bigg) - (x + y) \bigg\} = \bigg( \dfrac{x}{20} + \dfrac{y}{10} \bigg)$

So,

$\sf \dashrightarrow \dfrac{x}{20} + \dfrac{y}{10} = 3250$

$\sf \dashrightarrow x + 2y = 65000 - - - (i)$

SP in second case,

$\sf \dashrightarrow \bigg( \dfrac{110x}{100} + \dfrac{95y}{100} \bigg) = \bigg( \dfrac{11x}{10} + \dfrac{19y}{20} \bigg)$

So, gain in this case,

$\sf \dashrightarrow \bigg\{ \bigg( \dfrac{11x}{10} + \dfrac{19y}{20} \bigg) - (x + y) \bigg\} = \bigg( \dfrac{x}{10} + \dfrac{y}{20} \bigg)$

So,

$\sf \dashrightarrow \dfrac{x}{10} - \dfrac{y}{20} = 1500$

$\sf \dashrightarrow 2x - y = 30000 - - - (ii)$

Multiplying (ii) by two and adding the result with (i), we get

$\sf \dashrightarrow 5x = 60000 + 65000$

$\sf \dashrightarrow 5x = 125000$

$\sf \dashrightarrow x = 25000$

Putting x = 25000 in (i), we get

$\sf \dashrightarrow 25000 - 2y = 65000$

$\sf \dashrightarrow 2y = 65000 - 25000$

$\sf \dashrightarrow 2y = 40000$

$\sf \dashrightarrow y = 20000$

$\sf \therefore x = 25000 \: \: and \: \: y = 20000$

Hence, the CP of the TV set is ₹25000 and that of the fridge is ₹20000.