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Vijay had some bananas, and he divided them into two lots A and B. He sold the first lot at the rate of Rs 2 for 3 bananas and the second lot at the rate of Re 1 per banana, and got a total of Rs 400 . If he had sold the first lot at the rate of Re 1 per banana, and the second lot at the rate of Rs 4 for 5 bananas, his total collection would have been Rs 460 . Find the total number of bananas he had.
Given:
Vijay had some bananas, and he divided them into two lots A and B. He sold first lot at the rate of Rs. 2 for 3 bananas and the second lot at the rate of Rs. 1 per banana and got a total of Rs. 400. If he had sold the first lot at the rate of Rs. 1 per banana and the second lot at the rate of Rs. 4 per five bananas, his total collection would have been Rs. 460.
To do:
We have to find the total number of bananas he had.
Solution:Let the number of bananas in lot A and the number of bananas in lot B be $x$ and $y$ respectively.
The total number of bananas $=x+y$.
According to the question,
$x\times\frac{2}{3} + y\times1 = 400$
$3(\frac{2x}{3}+y)=3(400)$ (Multiplying both sides by 3)
$2x+3y=1200$.....(i)
$x\times 1 + y\times \frac{4}{5} = 460$
$5(x+\frac{4y}{5})=5(460)$ (Multiplying both sides by 5)
$5x+4y=2300$.....(ii)
Adding equations (i) and (ii), we get,
$2x+3y+5x+4y=1200+2300$
$7x+7y=3500$
$7(x+y)=7(500)$
$x+y=500$
Therefore, Vijay had 500 bananas.
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