Quantity , 1: Original duration of flight. In | vedclass

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Question

(A) $Quantity \, 1:$
Let the original duration be $t$ hours and original speed be $s$ km/hr.
Given,$s 	imes t = 3000 \implies s = \frac{3000}{t}$.

Vedclass · Accepted Answer

Quantity $I >$ Quantity $II$

Vedclass · Answer

(A) $Quantity \, 1:$
Let the original duration be $t$ hours and original speed be $s$ km/hr.
Given,$s 	imes t = 3000 \implies s = \frac{3000}{t}$.
According to the problem,$(s - 100)(t + 1) = 3000$.
Substituting $s = \frac{3000}{t}$ into the equation:
$(\frac{3000}{t} - 100)(t + 1) = 3000$
$3000 + \frac{3000}{t} - 100t - 100 = 3000$
$\frac{3000}{t} - 100t - 100 = 0$
Dividing by $100$: $\frac{30}{t} - t - 1 = 0$
$30 - t^2 - t = 0 \implies t^2 + t - 30 = 0$
$(t + 6)(t - 5) = 0$. Since time cannot be negative,$t = 5 \, hours$.
$Quantity \, 2:$
Let the usual speed be $v$ and usual time be $T$.
New speed $= \frac{3}{4}v$. Since distance is constant,time is inversely proportional to speed.
New time $= \frac{4}{3}T$.
Given,$\frac{4}{3}T - T = 20 \, minutes$.
$\frac{1}{3}T = 20 \, minutes \implies T = 60 \, minutes = 1 \, hour$.
Comparing $Quantity \, 1$ $(5 \, hours)$ and $Quantity \, 2$ $(1 \, hour)$,we get $Quantity \, I > Quantity \, II$.

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