Two cars start out simultaneously from a point in the same direction, one of them going at a speed of $50$ $k m / h r$ and the other at $40$ $k m / h r$ . In half an hour a third car starts out from the same point and overtakes the first car $1.5$ hours after catching up with the second car. The speed of the third car is

55 k m / h r

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90 k m / h r

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72 k m / h r

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60 k m / h r

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Solution

The correct option is D $60 k m / h r$
Positions of the cars when third car starts is shown.
Let third car catches $2^{n d}$ car after time $T$ it starts out.
$v T = 40 T + 20 \dots (i)$
Now third car catches $1^{s t}$ car after time $T + 1.5.$
$So v [T + 1.5] = 50 [T + 1.5] + 25 \dots (i i)$
Put the value of T from equation $(i)$ into equation $(i i)$ , we get
$1.5 v^{2} - 140 v + 3000 = 0$
$3 v^{2} - 280 v + 6000 = 0$
$v = \frac{280 \pm \sqrt{(280)^{2} - 72000}}{6}$
$\Rightarrow v = \frac{280 \pm 80}{6}$
$= 60 k m / h r, \frac{100}{3}$ $k m / h r$
As $v$ cannot be less than $50 k m / h r$
Final Answer: (b)

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