Question

Rahul purchased a scooter at $(\frac{13}{15}{)}^{th}$ of its selling price and sold it at $12\%$ more than its selling price. His gain is _____.

• A. $20$
• B. $29\frac{3}{13}$
• C. $30$
• D. $38\frac{1}{13}$

Answer 1

Answer: (B)

Let, the initial selling price of scooter be $Rs.x$

$\Rightarrow$ Cost price (C.P) $=\frac{13x}{15}$ ---(1)

It is given that, the scooter again sold by $12\%$ more than its selling price.

i.e., Selling price of scooter at second time,

S.P $=x+12\%\text{of}x$

$\Rightarrow$ S.P $=x+\frac{12x}{100}$

$\Rightarrow$ S.P $=\frac{100x+12x}{100}$

$\Rightarrow$ S.P $=\frac{112x}{100}$ ---(2)

From (1) and (2),

Gain $=$ S.P $-$ C.P

$=\frac{112x}{100}-\frac{13x}{15}$

$=\frac{28x}{25}-\frac{13x}{15}$

LCM of $25,15$ is $75$

$=(\frac{28x}{25}\times \frac{3}{3})-(\frac{13x}{15}\times \frac{5}{5})$

$=\frac{84x}{75}-\frac{65x}{75}$

$=\frac{84x-65x}{75}$

$=\frac{19x}{75}$

$\Rightarrow$ Gain $=\frac{19x}{75}$ --(3)

Now, The Gain percent $=\frac{\text{Gain}}{\text{C.P}}\times 100$

$=\frac{\frac{19x}{75}}{\frac{13x}{15}}\times 100$

$=\frac{19x}{75}\times \frac{15}{13x}\times 100$

$=\frac{19}{1}\times \frac{1}{13}\times 20$

$=\frac{380}{13}$

$=29\frac{3}{13}$

$\Rightarrow$ Gain percent $=29\frac{3}{13}$

i.e., In whole transaction, his gain is $29\frac{3}{13}\%$ .

Hence, Option B is correct.