Comment
Share
Q)

# Ajay covers a certain distance in a certain times. If $\large\frac{1}{4}$ of the distance is covered in $\large\frac{3}{4}$ of the time with speed $V_1$ and the rest of the $\large\frac{3}{4}$ distance in $\large\frac{1}{4}$ of the time with speed $V_2$ then find the ratio $V_1:V_2$

$\begin{array}{1 1} 1:4 \\ 1:9 \\ 9:1 \\ 3:1 \end{array}$

Comment
A)
Solution :
Let the distance be - x km
time be - y sec
$\large\frac{1}{4}$$=> \large\frac{3}{4}$$(y)=> V_1$
$\large\frac{3}{4}$$=> \large\frac{1}{4}$$(y)=> V_2$
Speed $= \large\frac{distance}{time}$
$V_1=\large\frac{\Large\frac{1}{4}x}{\Large\frac{3}{4}y}$
$\qquad= \large\frac{1}{4} $$x \times \large\frac{4}{3}$$y$
$\qquad= \large\frac{1}{3} $$xy V_2=\large\frac{\Large\frac{3}{4}x}{\Large\frac{1}{4}y} \qquad= \large\frac{3}{4}$$x \times \large\frac{4}{1}$$(y)$
$\qquad= 3xy$
$\large\frac{V_1}{V_2}=\frac{\Large\frac{1}{3}xy}{\Large\frac{3}{1}xy}$
$\qquad= \large\frac{1}{3} \times \frac{1}{3}$
$\qquad= \large\frac{1}{9}$
$V_1 :V_2 =1:9$