IBPS RRB Assistant 9 Sep 2012

Question 152

The product of two consecutive even numbers is 7568. What is 150% of the sum of the two numbers?

Solution

Let even numbers be '2n' and '2n+2'.
According to question,
$$2n\times(2n+2)=7568$$.
$$4n(n+1)=7568$$.
$$n(n+1)=1892$$.
$$n=43$$.
Hence, numbers are 86 and 88 respectively.
Sum of numbers=86+88.
=174.
150% of sum of numbers=$$1.5\times174$$.
$$=261$$.
Hence, Option B is correct.


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