General Aptitude | Practice Question | Unacademy

Question

In a car race, car X beats car Y by 36 km, car Y beats car Z by 72 km, and car X beats car Z by 90 km. The distance (in m) over which the race has been conducted is:

• A. 88000
• B. 108000
• C. 56000
• D. 144000

Please explain your answer.

 

Comments

Suppose, race has been conducted for distance $S$ Km and speed of cars $X,Y,Z$ are $v_1,v_2,v_3$ respectively in km per unit time.

“car X beats car Y by 36 km” means for the same time, X is ahead of Y by $36$ km. Similar meaning is for other two statements. So, we will be having three equations as:

$\frac{S}{v_1} = \frac{S- 36}{v_2}$   $(1)$

$\frac{S}{v_2} = \frac{S- 72}{v_3}$   $(2)$

$\frac{S}{v_1} = \frac{S- 90}{v_3}$   $(1)$

From $(1),$  $\frac{v_2}{v_1} = \frac{S- 36}{S}$

From $(2),$ $\frac{v_2}{v_3} = \frac{S}{S – 72}$

After dividing both equations, we get, $\frac{v_3}{v_1} = \frac{(S- 36)(S – 72)}{S^2}$ $(3)$

From $(3),$ $\frac{v_3}{v_1} = \frac{S- 90}{S}$  $(4)$

From $(3)$ and  $(4),$ $\frac{S- 90}{S} = \frac{(S- 36)(S – 72)}{S^2}$, $S \neq 0$

$S= 144$ km = $144000 m$

Answer 1

Let the total distance of the race is $d$ km .

$X$ beats $Y$ by 36 Km which means when $X$ completed the $d$ km distance $Y$ was at distance $d-36$ km.

Let $X$ completed the race in $t$ sec.

So, Speed of $X$ is $\large S_{X}=\frac{d}{t}$

       Speed of $Y$ is $\large S_{Y}=\frac{d-36}{t}$

So, ratio of $\large \frac{S_{X}}{S_{Y}}=\frac{d}{d-36}$……………………..(1)

$Y$ beats $Z$ by 72 Km which means when $Y$ completed the $d$ km distance $Z$ was at distance $d-72$ km.

Let $Y$ completed the race in $t_{1}$ sec.

So, Speed of $Y$ is $\large S_{X}=\frac{d}{t_{1}}$

       Speed of $Z$ is $\large S_{Y}=\frac{d-72}{t_{1}}$

So, ratio of $\large \frac{S_{Y}}{S_{Z}}=\frac{d}{d-72}$……………………..(2)

$X$ beats $Z$ by 90 Km which means when $X$ completed the $d$ km distance $Z$ was at distance $d-90$ km.

Let $Y$ completed the race in $t_{2}$ sec.

So, Speed of $Y$ is $\large S_{X}=\frac{d}{t_{2}}$

       Speed of $Z$ is $\large S_{Y}=\frac{d-90}{t_{2}}$

So, ratio of $\large \frac{S_{X}}{S_{Z}}=\frac{d}{d-90}$……………………..(3)

Dividing (3) by (1) we get ,

$\large \frac{\frac{S_{X}}{S_{Z}}}{\frac{S_{X}}{S_{Y}}}=\frac{\frac{d}{d-90}}{\frac{d}{d-36}}$

=>$\large \frac{S_{Y}}{S_{Z}}=\frac{d-36}{d-90}$ ….…….……...(4)

Now (4) and (2) are equal so equating them,

$\large \frac{d}{d-72}=\frac{d-36}{d-90}$

=>$\large d(d-90)=(d-36)(d-72)$

=>$\large d=144$km.

So correct answer is $144000$ $m$.