Aditya covered a certain distance at some spe | vedclass

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(A) Let the distance be $d 	ext{ km}$ and the usual speed be $s 	ext{ km/h}$.According to the first condition:$\frac{d}{s} - \frac{d}{s+3} = \frac{4

Vedclass · Accepted Answer

$40$

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(A) Let the distance be $d 	ext{ km}$ and the usual speed be $s 	ext{ km/h}$.According to the first condition:$\frac{d}{s} - \frac{d}{s+3} = \frac{40}{60} = \frac{2}{3}$$\frac{d(s+3) - ds}{s(s+3)} = \frac{2}{3} \implies \frac{3d}{s(s+3)} = \frac{2}{3} \implies 9d = 2s(s+3)$ .....$(i)$According to the second condition:$\frac{d}{s-2} - \frac{d}{s} = \frac{40}{60} = \frac{2}{3}$$\frac{ds - d(s-2)}{s(s-2)} = \frac{2}{3} \implies \frac{2d}{s(s-2)} = \frac{2}{3} \implies 3d = s(s-2)$ .....$(ii)$From $(ii)$, $d = \frac{s(s-2)}{3}$. Substituting this into $(i)$:$9 \left( \frac{s(s-2)}{3} ight) = 2s(s+3)$$3s(s-2) = 2s(s+3)$Since $s 
eq 0$, divide by $s$:$3s - 6 = 2s + 6$$s = 12 	ext{ km/h}$Now, substitute $s = 12$ into $(ii)$:$d = \frac{12(12-2)}{3} = \frac{12 	imes 10}{3} = 40 	ext{ km}$.

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

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