Question

A man purchases two clocks A and B at a total cost of Rs 650. He sells A at 20% profit and B at a loss of 25% and gets the same selling price for both the clocks. What are the purchasing prices of A and B respectively?

• A. Rs. 225, Rs. 425
• B. Rs. 275, Rs. 375
• C. Rs. 250, Rs. 400
• D. Rs. 300, Rs. 350

Answer 1

Answer: (C)

Rs. 250, Rs. 400

Let the cost price (C.P) of A be Rs x

and that of B be Rs y 3

Total cost $=Rs650$ - [Given]

$\therefore x+y=650$

For A

Profit $=20\%$

Profit $\%=\frac{Profit}{C.P}\times 100$

$\therefore 20=\frac{sellingpriceof(S.P)-C.PofA}{C.PofA}\times 100$

$\therefore \frac{1}{5}=\frac{S.P-x}{x}$

$\therefore S.P.$ of $A=\frac{6x}{5}$

For B

Loss $=25\%$

Loss% $\frac{Loss}{C.P}\times 100$

$25=\frac{C.PofB-sellingprice(S.P)ofB}{C.PofB}\times 100$

$\frac{1}{4}=\frac{y-S.P}{y}$

$\frac{y}{4}=y-S.P.$ of B

$\therefore$ SP of B $=\frac{3y}{4}$

But, according to the given concoction

S.P of $A=S.P$ of B

$\frac{6x}{5}=\frac{3y}{4}$

$\frac{6x}{5}-\frac{3y}{4}=0$

$8x-5y=0...\left(2\right)$

Multiplying equation (1) by 5

$\therefore 5x+5y=\mathrm{3250...}\left(3\right)$

Adding equations (2) and (3)

$\therefore 8x+5y+5x+5y=0+3250$

$13x=3250$

$x=Rs250$

Substituting the above value in equation (1)

$\therefore 250+y=650$

$\therefore y=Rs400$

$\therefore$ Cost price of $A=Rs250$

$\therefore$ Cost price of $B=Rs400$