One aeroplane started 30 , text{minutes} late | vedclass

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(A) Let the scheduled time be $T$ hours and the original speed be $v \, 	ext{km/hr}$.Distance $D = 1500 \, 	ext{km}$.So,$v = \frac{1500}{T}$,which i

Vedclass · Accepted Answer

$750$

Vedclass · Answer

(A) Let the scheduled time be $T$ hours and the original speed be $v \, 	ext{km/hr}$.Distance $D = 1500 \, 	ext{km}$.So,$v = \frac{1500}{T}$,which implies $T = \frac{1500}{v}$.Since the plane started $30 \, 	ext{minutes}$ $(0.5 \, 	ext{hours})$ late,the time taken in the second case is $(T - 0.5)$ hours at a speed of $(v + 250) \, 	ext{km/hr}$.$1500 = (v + 250)(T - 0.5)$.Substituting $T = \frac{1500}{v}$:$1500 = (v + 250)(\frac{1500}{v} - 0.5)$.$1500 = 1500 - 0.5v + \frac{375000}{v} - 125$.$0.5v + 125 - \frac{375000}{v} = 0$.Multiplying by $2v$: $v^2 + 250v - 750000 = 0$.$(v + 1000)(v - 750) = 0$.Since speed cannot be negative,$v = 750 \, 	ext{km/hr}$.

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