A man bought two mobiles phones for Rs 50,000. He sold one of them at a profit of 15% and the other at a loss of 20%,

A man bought two mobiles phones for Rs 50,000. He sold one of them at a profit of 15% and the other at a loss of 20%, If the selling price of each mobile is the same, what is the approximate cost price (to the nearest rupee) of the mobile that was sold at a loss?

Options
a) 27,368
b) 25,465
c) 29,487
d) 20,513

Solution

As per question
CP (1) + CP (2) = 50,000 Rs.
P = 15%
L = 20%
SP = $CP \times \frac {100+Profit}{100}$
SP = $CP \times \frac {100-Loss}{100}$
SP = $CP \times \frac{100+15}{100}$
SP = $CP \times \frac{100-20}{100}$
x + y = 50,000 Rs.

$x \times \frac {100+15}{100} = y \times \frac {100-20}{100}$
$x \times \frac {115}{100} = y \times \frac {80}{100}$
$x \times \frac {23}{20} = y \times \frac {4}{5}$
$\frac {x}{y} = \dfrac {\frac {4}{5}} {\frac{23}{20}}$
$\frac {x}{y} = \frac {80}{115}$
$\frac {x}{y} = \frac {16}{23}$
x= 16
y = 23
Sum of x + y = 16 + 23 = 39
Cost price of the mobile that was sold at a loss
y = $50,000 \times \frac {23}{39}$
y = 1282.05 x 23 = 29487.15
Nearest rupee= 29487 Rs.