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The product of two consecutive even numbers is 7568. What is 150% of the sum of the two numbers?
- A204
- B261
- C304
- D198
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CCommunity
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.
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.