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(B) Let the distance be $d \, km$ and the initial speed be $s \, km/h$.Time taken at speed $s$ is $t = \frac{d}{s} \, hours$.Case $1$: If speed is inc

Vedclass · Accepted Answer

$420$

Vedclass · Answer

(B) Let the distance be $d \, km$ and the initial speed be $s \, km/h$.Time taken at speed $s$ is $t = \frac{d}{s} \, hours$.Case $1$: If speed is increased by $10 \, km/h$,time taken is $t - 1$.$\frac{d}{s+10} = \frac{d}{s} - 1 \implies \frac{d}{s} - \frac{d}{s+10} = 1 \implies \frac{10d}{s(s+10)} = 1 \implies 10d = s(s+10) \dots (i)$.Case $2$: If speed is increased by $20 \, km/h$ (total increase),time taken is $t - 1 - \frac{45}{60} = t - 1.75 = t - \frac{7}{4}$.$\frac{d}{s+20} = \frac{d}{s} - \frac{7}{4} \implies \frac{d}{s} - \frac{d}{s+20} = \frac{7}{4} \implies \frac{20d}{s(s+20)} = \frac{7}{4} \implies 80d = 7s(s+20) \dots (ii)$.From $(i)$,$d = \frac{s(s+10)}{10}$. Substituting into $(ii)$:$80 \left( \frac{s(s+10)}{10} ight) = 7s(s+20) \implies 8s(s+10) = 7s(s+20)$.Since $s 
eq 0$,$8s + 80 = 7s + 140 \implies s = 60 \, km/h$.Substituting $s = 60$ into $(i)$:$10d = 60(60+10) = 60 	imes 70 = 4200 \implies d = 420 \, km$.

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