A shopkeeper mixed $10$ kg of rice costing Rs. $10$ per kg with a certain quantity of another variety of rice costing Rs. $15$ per kg. He sold the mixture at the rate of Rs. $14$ per kg. What was the quantity of the second variety of rice, if he gained by $5$ $%$ ?

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Solution

The correct option is A $20$ kg
Let the quantity of second verity of rice $= x$ kg
Then C.P. of the next mixture $=$ $10 \times 10 + 15 \times x =$ Rs. $(100 + 15 x)$
S.P. of the mixture $=$ $(10 + x) \times$ Rs. $14 =$ Rs. $(140 + 14 x)$

Gain $= (140 + 14 x) - (100 + 15 x) = 40 - x$

According to the question
$∴$ Gain $% =$ $(\frac{40 - x}{100 + 15 x}) \times 100 = 5$
$\Rightarrow$ $4000 - 100 x = 500 + 75 x$

$\Rightarrow$ $175 x = 3500$

$\Rightarrow x = \frac{3500}{175} = 20$ kg

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A shopkeeper mixed 10 kg of rice costing Rs. 10 per kg with a certain quantity of another variety of rice costing Rs. 15 per kg. He sold the mixture at the rate of Rs. 14 per kg. What was the quantity of the second variety of rice, if he gained by 5 %?

A shopkeeper mixed $10$ kg of rice costing Rs. $10$ per kg with a certain quantity of another variety of rice costing Rs. $15$ per kg. He sold the mixture at the rate of Rs. $14$ per kg. What was the quantity of the second variety of rice, if he gained by $5$ $%$ ?