Question

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 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 for 5 bananas, his total collection would have been Rs. 460. Find the total number of bananas he had.
A. 200
B. 500
C. 600
D. 900

Answer 1

Hint: We had to assume the number of bananas of each lot and then form two linear equations for the two different cases and after solving both of the equations we will get the number of bananas in each lot.

Complete step-by-step answer:
Let the number of bananas in the first lot is x.
Let the number of bananas in the second lot be y.
So, the total number of bananas will be equal to x + y.
So, as we know that in first case,
Vijay sold the first lot at the rate of Rs. 2 for 3 bananas.
And Rs. 1 per banana for the second lot.
So, the cost of one banana of the first lot in this case will be equal to Rs. \[\dfrac{2}{3}\].
And the cost of x bananas of the first lot in this case will be equal to Rs. \[\dfrac{{2x}}{3}\].
Cost of y bananas of the second lot in this case will be equal to Rs. y.
So, the total cost of all the bananas in this case will be equal to the sum of the cost of bananas of each lot.
So, according to the question
\[\dfrac{{2x}}{3}\] + y = 400
Now taking LCM on the LHS of the above equation and then cross multiplying the above equation. We get,
2x + 3y = 1200 ---- (1)
Now in another case Vijay sold the bananas of the first lot at a rate of Rs. 1 per banana and the bananas of the second lot at a rate of Rs. 4 for 5 bananas.
And the cost of x bananas of first lot in this case will be equal to Rs. x
So, the cost of one banana of the second lot in this case will be equal to Rs. \[\dfrac{4}{5}\].
And the cost of my bananas of the second lot in this case will be equal to Rs. \[\dfrac{{4y}}{5}\].
So, the total cost of all the bananas in this case will be equal to the sum of the cost of bananas of each lot.
So, according to the question
\[x + \dfrac{{4y}}{5} = 460\]
Now taking LCM on the LHS of the above equation and then cross multiplying the above equation. We get,
5x + 4y = 2300 ---- (2)
Now we have to solve equation 1 and equation 2, to find the value of x and y.
Now for solving equation 1 and equation 2.
Multiply both sides of the equation 1 by 5 and both sides of the equation 2 by 2. And then subtracting equation 2 from 1. We get,
10x + 15y – 10x – 8y = 6000 – 4600
7y = 1400
Now dividing both sides of the above equation by 7. We get,
y = 200
Now putting the value of y in equation 1. We get,
2x + 600 = 1200
2x = 600
On dividing both sides of the above equation by 2. We get,
x = 300
So, the total number of bananas will be equal to x + y = 300 + 200 = 500.
Hence, the correct option will be B.

Note: Whenever we come up with this type of problem then first we have to assume the number of both of the lots equal to x and y. After that we will find the cost of each banana of both the lots in each case. And then we will find the cost of x bananas of the first lot and the cost of y bananas of the second lot in both cases. After that we had to form two equations using the cost of all of the bananas given in each of the cases. On solving these two equations we will get the value of x and y. And the total number of bananas will be equal to x + y. We can solve these equations using the substitution method as well.