Question

Two brothers John and Robert started from school to reach their home through the same path. John started running at 5 km/hr and reached home in 1 hr. If Robert was running at 6 km/hr, by how much time John arrived late than Robert?

• A. $\frac{1}{6}$ hr
• B. $\frac{5}{6}$ hr
• C. $\frac{7}{6}$ hr
• D. $\frac{6}{5}$ hr

Answer 1

The correct option is A $\frac{1}{6}$ hr

Let the speed be $S$ and the time taken to reach home be $T$.
We know,
$\text{Distance}=\text{Speed}\times \text{Time}$

Distance is the same for John and Robert, so it can be considered as constant($k$).

Hence, $\text{S}\times \text{T}=\text{k}$
If Product of two variable is constant, then they are inversely proportional to each other.
or $S\propto \frac{1}{T}$

$\Rightarrow S=\frac{k}{T}$ ; where k is a constant

According to question,
John started running at 5 km/hr and reached home in 1 hr
i.e. $S=5,T=1$
So,
$\Rightarrow 5=\frac{K}{1}\Rightarrow k=5$
Also,
Robert ran at 6 km/hr speed and reached home in T time
i.e. $S=6,T=?$

$\Rightarrow 6=\frac{5}{T}$
$\Rightarrow T=\frac{5}{6}\text{hr}$
$\text{Hence, John arrived late by}=1-\frac{5}{6}$
$=\frac{1}{6}\text{hr}$