JIPMAT 2021 (QA) - A boy covers a certain distance on a toy train. If the train moved 4 kmph faster, it would take 30 minutes less. If it moved 2 kmph slower, it would have taken 20 minutes more. The distance is | PYQs + Solutions

JIPMAT 2021

Arithmetic

Time, Speed & Distance

Conceptual

A boy covers a certain distance on a toy train. If the train moved 4 kmph faster, it would take 30 minutes less. If it moved 2 kmph slower, it would have taken 20 minutes more. The distance is

40 km

50 km

60 km

80 km

Correct Option: 3

Let original speed =

s

kmph

\newline

Distance =

D

\newline

Original time =

t

hours

For original journey:

\newline

\frac{D}{s} = t

...(1)

For faster speed (+4 kmph):

\frac{D}{s+4} = t - \frac{1}{2}

...(2)

For slower speed (-2 kmph):

\frac{D}{s-2} = t + \frac{1}{3}

...(3)

From (1):

D = st

Substituting in (2):

\frac{st}{s+4} = t - \frac{1}{2}

st = (s+4)(t - \frac{1}{2})

st = st + 4t - \frac{s}{2} - 2

\frac{s}{2} + 2 = 4t

...(4)

Substituting in (3):

\frac{st}{s-2} = t + \frac{1}{3}

st = (s-2)(t + \frac{1}{3})

st = st + \frac{s}{3} - 2t - \frac{2}{3}

2t + \frac{2}{3} = \frac{s}{3}

...(5)

From (4):

s = 8t - 4

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From (5):

s = 6t + 2

Equating:

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8t - 4 = 6t + 2

\newline

2t = 6

\newline

t = 3

Substituting

t = 3

s = 6t + 2

\newline

s = 6(3) + 2

\newline

s = 18 + 2

\newline

s = 20

kmph

Therefore:

\newline

Distance =

st = 20 × 3 = 60