Question

From the same place at $7$ am, A started walking in the north at the speed of $5km/hr$. After $1$ hour, $B$ started cycling in the east at a speed of $16km/hr$. At what time (in am), they will be at a distance of $52km$ apart from each other?

• A. $11$
• B. $9$
• C. $12$
• D. $10$

Answer 1

Answer: (A)

$11$

Let the time taken by each be $t$ hours to have a distance of $52kms$ between them.
After $t$ hrs, $A$ will be at $5t$ distance from the point, and $B$ will be at $16(t-1)kms$ from the point.
Since, they are moving perpendicularly, the square sum root of the distance should be equal to $52$.
Hence, $(5t{)}^2+(16(t-1){)}^2={52}^2$
$25t^2+256(t^2-2t+1)=2704$
$281t^2-512t+256=2704$
$281t^2-512t-2448=0$
$t=\frac{512\pm \sqrt{(512{)}^2-4(281\left)\right(-2448)}}{2\times 281}$
$t=\frac{512\pm 1736}{2\times 281}$
$t=\frac{512+1736}{562},\frac{512-1736}{562}$
$t=4,\frac{-612}{281}$
Neglecting the negative value, $t=4$ hrs
Thus, the distance between them is $52kms$ after $4$ hours i.e. at $11a.m.$