Question 21
Three numbers A, B and C are such that A is 40% less than B, and C is 40% of the sum of A and B. The difference between A and B is what percentage of C?
Solution
Given, A is 40% less than B
So, $$A=\left(1-0.4\right)B=0.6B$$
Also, C is 40% of sum of A and B
So, $$C=0.4\left(A+B\right)$$
or, $$C=0.4\left(0.6B+B\right)$$
or, $$C=0.4\times\ 1.6B$$
or, $$C=0.64\ B$$
Now, $$B-A=B-0.6B=0.4B$$
So, required percentage = $$\dfrac{0.4B}{C}\times\ 100=\dfrac{0.4B}{0.64B}\times\ 100\ \%\ =62.5\ \%\ $$
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