Question

A TRAIN COVERD CERTAIN DISTANCE AT A UNIFORM SPEED. IF THAE TRAIN WOULD HAVE BEEN 10KM/H FASTER, IT WOULD HAVE TAKEN 2 HOURS LESS THEN THE SCHEDULE TIME AND IF THE TRAIN WERE SLOWER BY 10KM/H IT WOULD HAVE TAKEN 3 HOURS MORE THEN THE SCHEDULE TIME FIND THE DISTANCE COVERED BY THE TRAIN

Answer 1

Hey there ☺

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SOLUTION -:

Let speed of train = x km/h
& Time taken = y hours

We know that,

Speed = Distance/Time
Distance = Speed × Time
Distance = xy ...... (1)

If the train would have been 10km/h faster i.e.,
Speed = x + 10
It would have taken 2 hours less i.e.,
Time = y - 2

Now,

Distance = Speed × Time
Distance = (x + 10) (y -2)
Putting distance = xy from equation (1)

$xy = (x + 10)(y - 2) \\ xy = x(y - 2) + 10(y - 2) \\ xy = xy - 2x + 10y - 20 \\ 2x - 10y + 20 = xy - xy \\ 2x - 10y + 20 = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: ............(2)$
Also, If the train were slower by 10km/h
Speed = x - 10
It would have been 3 hours more
Time = y + 3

Now,

Distance = Speed × Time
Distance = ( x-10) (y+3)
Putting Distance = xy from equation (1)

$xy = (x - 10)(y + 3) \\ xy = x(y + 3) - 10y - 30 \\ xy = xy + 3x - 10y - 30 \\ xy - xy = 3x - 10y - 30 \\ 3x - 10y - 30 = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: ............(3)$
Hence, the equations are

$2x - 10y + 20 = 0 \: \: \: \: \: \: \: \: ....(2) \\ 3x - 10y - 30 = 0 \: \: \: \: \: \: \: .....(3)$
From equation (2)

$2x - 10y + 20 = 0 \\ 2(x - 5y + 10) = 0 \\ x - 5y + 10 = 0 \\ x = 5y - 10 \: \: \: \: \: \: \: \: ....(4)$
Putting (4) in equation (3)

$3x - 10y - 30 = 0 \\ 3(5y - 10) - 10y - 30 = 0 \\ 15y - 30 - 10y - 30 = 0 \\ 5y - 60 = 0 \\ 5y = 60 \\ y = \frac{60}{5} \\ y = 12$
Putting y = 12 in equation (4)

$x = 5y - 10 \\ x = (5 \times 12) - 10 \\ x = 60 - 10 \\ x = 50$
Thus,

Speed of train = x = 50km/h
& Time taken by train = y = 12 hours



Now,

Distance = Speed × Time
Distance = 50 × 12
Distance = 600km


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Hope it helps :)

Answer 2

Answer:

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