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A train takes 4/3 hours less for a journey of 528km . If its speed is increased by 11/2 km/hour from its usual speed . The usual speed of the train is ( in km/hour )
a. 44
b. 48
c. 52
d. 55
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Let Speed of train= v km/hour

Time taken to complete journey= t hour

$Time=\frac{Distance}{Speed}$

$v \times t =Distance$

$v \times t =528 \dots\dots(1)$

Given that if speed increased by $\frac{11}{2} km /hour$ from its usual speed then train takes $\frac{4}{3}$ hour less.

$(v+\frac{11}{2}) \times (t-\frac{4}{3}) =528 \dots\dots(2)$

$(vt-\frac{4}{3}v+\frac{11}{2}t-\frac{44}{6}) =528$

From (1)

$(528-\frac{4}{3}v+\frac{11}{2}t-\frac{44}{6}) =528$

$(-\frac{4}{3}v+\frac{11}{2}t-\frac{44}{6}) =0$

$33t-8v=44$

$33\times \frac{528}{v}-8v=44$

$8v^{2}+44v-17424$

v=44 , $-\frac{99}{2}$

Hence,Option (a)44 is the correct choice.
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