IIFT 2013 Question Paper Question 122

Question 122

Two alloys of aluminium have different percentages of aluminium in them. The first one weighs 8 kg and the second one weighs 16 kg. One piece each of equal weight was cut off from both the alloys and first piece was alloyed with the second alloy and the second piece alloyed with the first one. As a result, the percentage of aluminium became the same in the resulting two new alloys. What was the weight of each cut-off piece?

Solution

Let x and y be the percentage of aluminum in the 8 kg and 16 kg piece.
Let us take a kg away from both.
Thus, we will have xa and ya amount of aluminum in both the pieces.
In the remaining 8-a and 16-a pieces we will have (8-a)x and (16-a)y amount of aluminum.
Given that, $$\dfrac{(8-a)x+ay}{8} = \dfrac{(16-a)y + ax}{16}$$
Thus, $$2*(8x-ax+ay) = 16y-ay+ax$$
=> $$16x-2ax+2ay = 16y-ay+ax$$
Thus, $$16(x-y) = 3a(x-y)$$
Thus, $$a = \dfrac{16}{3} = 5.33$$
Hence, option C is the correct answer.



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