Question
Q and L have chocolates in the ratio of 7: 8 and B and N
have chocolates in the ratio of 2: 5. Q has 20 more chocolates than B and sum of chocolates of B and N is 3 times of number of chocolates Q have, then what is the average number of chocolates Q, L, B and N have?Solution
Let Q = 7x and L = 8x and B = 2y and N = 5y
Given, Q = 20 + B
=> 7x - 2y = 20 ---(1)
And (B + N) = 3Q
=> 2y + 5y = 3 * 7x
=> y = 3x ---(2)
Putting value of y in equation (1), we get
7x - 2 * 3x = 20
=> x = 20
Thus, y = 60
So, required average = (15x + 7y)/4 = (15 * 20 + 7 * 60)/4 = 180
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