Question

A boy travels a distance of $25 \mathrm{~km}$ in 4 hours partly on foot at rate $3.5 \mathrm{~km} / \mathrm{h}$ and partly on cycle at $9 \mathrm{~km} / \mathrm{h}$. Find the distance on foot and on cycle.

• A. $7 \mathrm{~km}$
• B. $2 \mathrm{~km}$
• C. $6 \mathrm{~km}$
• D. $18 \mathrm{~km}$

Answer 1

Answer: (A), (D)

Let distance by foot is $\mathrm{X} \mathrm{km}$ in $t$ hours By cycle is $25-\mathrm{X} \mathrm{km}$ in $4-\mathrm{t}$ hours
$ \text { Distance }=\text { speed } \times \text { time } $
$\Rightarrow \mathrm{X}=3.5 \times t $
$\Rightarrow 25-x=9 \times(1-t) $
$\Rightarrow 25-3.5 t=36-9 t $
$\Rightarrow 5.5 \mathrm{t}=11 $
$ t=2 $
$X=3.5 \times 2=7 \mathrm{~km} \text { (Distance by foot) } $
$ \text { By cycle }=25-x \text { in } 4-\mathrm{t} \text { hours } $
$ \Rightarrow 4-2=2 \text { hours } $
$ \Rightarrow 25-x=25-7$
$=18 \mathrm{~km} \text { (Distance by cycle) } $