Question:
Two numbers differ by 40. When each number is increased by 8, the bigger becomes thrice the lesser number. If one number is x, then the
other number is (40 – x).
Solution:
False
Given, one number $=x$
and other number $=40-x$
Let $(40-x)>x$
Then, according to the question,
$40-x+8=3(x+8)$
$\Rightarrow \quad 48-x=3 x+24$
$\Rightarrow \quad-x-3 x=24-48 \quad$ [transposing $3 x$ to LHS and 48 to RHS]
$\Rightarrow$ $-4 x=-24$
$\Rightarrow$ $x=-24 \times\left(-\frac{1}{4}\right)$
$\therefore$ $x=6$
Hence, one number $=x=6$
and other number $=40-x=40-6=34$
Now, difference between numbers $=34-6=28 \neq 40$ which is not satisfy the condition given in question.