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(D) Let the distance be $S 	ext{ km}$ and the usual speed be $u 	ext{ km/hr}$.According to the first condition:$\frac{S}{u} - \frac{S}{u+3} = \frac{

Vedclass · Accepted Answer

$40$

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(D) Let the distance be $S 	ext{ km}$ and the usual speed be $u 	ext{ km/hr}$.According to the first condition:$\frac{S}{u} - \frac{S}{u+3} = \frac{40}{60} = \frac{2}{3}$$\frac{3S}{u(u+3)} = \frac{2}{3}$$9S = 2u(u+3) \quad \dots(1)$According to the second condition:$\frac{S}{u-2} - \frac{S}{u} = \frac{40}{60} = \frac{2}{3}$$\frac{2S}{u(u-2)} = \frac{2}{3}$$3S = u(u-2) \quad \dots(2)$Dividing equation $(1)$ by equation $(2)$:$\frac{9S}{3S} = \frac{2u(u+3)}{u(u-2)}$$3 = \frac{2(u+3)}{u-2}$$3(u-2) = 2(u+3)$$3u - 6 = 2u + 6$$u = 12 	ext{ km/hr}$Substituting $u = 12$ in equation $(2)$:$3S = 12(12-2)$$3S = 12 	imes 10$$3S = 120$$S = 40 	ext{ km}$.

$A$ man covered a certain distance at some speed. Had he moved $3 \text{ km/hr}$ faster,he would have taken $40 \text{ minutes}$ less. If he had moved $2 \text{ km/hr}$ slower,he would have taken $40 \text{ minutes}$ more. The distance (in $\text{km}$) is:

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