There are 28 steps in a temple. In the time A, initially at the 28th step, comes down two steps, B, initially at 1st step, goes one step up. If they start simultaneously and keep their speed uniform, then at which step from the bottom will they meet ?
8<sup>th</sup>
9<sup>th</sup>
10<sup>th</sup>
not possible to meet
Answer is Right!
Answer is Wrong!
This question was previously asked in
UPSC CAPF – 2014
A’s position after time $t$ (from the bottom) is $P_A(t) = 28 – 2t$.
B’s position after time $t$ (from the bottom) is $P_B(t) = 1 + 1t$.
They meet when their positions are equal: $P_A(t) = P_B(t)$.
$28 – 2t = 1 + t$
$28 – 1 = t + 2t$
$27 = 3t$
$t = 9$.
So, they meet after 9 units of time. Their position at this time is:
$P_B(9) = 1 + 9 = 10$.
$P_A(9) = 28 – 2(9) = 28 – 18 = 10$.
They meet at the 10th step from the bottom.