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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

Solution

✅ Correct Option: 3

Let original speed = ss kmph

Distance = DD km

Original time = tt hours

For original journey:

Ds=t\frac{D}{s} = t ...(1)

For faster speed (+4 kmph):

Ds+4=t−12\frac{D}{s+4} = t - \frac{1}{2} ...(2)

For slower speed (-2 kmph):

Ds−2=t+13\frac{D}{s-2} = t + \frac{1}{3} ...(3)

From (1): D=stD = st

Substituting in (2):

sts+4=t−12\frac{st}{s+4} = t - \frac{1}{2}

st=(s+4)(t−12)st = (s+4)(t - \frac{1}{2})

st=st+4t−s2−2st = st + 4t - \frac{s}{2} - 2

s2+2=4t\frac{s}{2} + 2 = 4t ...(4)

Substituting in (3):

sts−2=t+13\frac{st}{s-2} = t + \frac{1}{3}

st=(s−2)(t+13)st = (s-2)(t + \frac{1}{3})

st=st+s3−2t−23st = st + \frac{s}{3} - 2t - \frac{2}{3}

2t+23=s32t + \frac{2}{3} = \frac{s}{3} ...(5)

From (4): s=8t−4s = 8t - 4

From (5): s=6t+2s = 6t + 2

Equating:

8t−4=6t+28t - 4 = 6t + 2

2t=62t = 6

t=3t = 3

Substituting t=3t = 3 in s=6t+2s = 6t + 2:

s=6(3)+2s = 6(3) + 2

s=18+2s = 18 + 2

s=20s = 20 kmph

Therefore:

Distance = st=20×3=60st = 20 × 3 = 60 km

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