Question

Product of two consecutive number is 56. Find them?

• A. 5, 6
• B. 8, 7
• C. 28, 2
• D. 4, 4

Answer 1

Answer: (B)

8, 7

Let two consecutive numbers be $n$ and $n+1$

Given:

$n(n+1)=56\Rightarrow n^2+n-56=0\Rightarrow n^2+8n-7n-56=0\Rightarrow (n+8)(n-7)=0$

$\therefore n=-8$ or$7$

$\therefore$ Possible cases are $-8,-7$ or $7,8$

Answer:-$B$