Question

A train cover certain distance, if a train could have been 10km/h faster, it would take 2 hour less than the Schedule times and if train were slower by 10km/h, it would be taken 3 hour more. Find the distance cover by the train.​

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that

• Speed of the train be x km per hour

• Time taken to covered the distance be y hours

So,

• Distance covered by the train = xy km

Case :- 1

If a train could have been 10km/h faster, it would take 2 hour less than the schedule time.

So,

➢ Speed of the train = x + 10 km per hour

➢ Time taken = y - 2 hours

Thus,

$\rm :\longmapsto\:(x + 10)(y - 2) = xy$

$\rm :\longmapsto\:xy - 2x + 10y - 20= xy$

$\rm :\longmapsto\: - 2x + 10y - 20= 0$

$\rm :\longmapsto\: - 2(x - 5y + 10)= 0$

$\rm :\longmapsto\:x - 5y + 10= 0 - - - - (1)$

Case - 2

If train were slower by 10km/h, it would be taken 3 hour more than the schedule time.

So,

➢ Speed of the train = x - 10 km per hour

➢ Time taken = y + 2 hours

Thus,

$\rm :\longmapsto\:(x - 10)(y + 3) = xy$

$\rm :\longmapsto\:xy + 3x - 10y - 30 = xy$

$\rm :\longmapsto\:3x - 10y - 30 =0 - - - - (2)$

Now, multiply equation (1) by 3, we get

$\rm :\longmapsto\:3x - 15y + 30= 0 - - - - (3)$

On Subtracting equation (3) from equation (2), we get

$\rm :\longmapsto\:5y - 60 = 0$

$\rm :\longmapsto\:5y = 60$

$\bf\implies \:y = 12$

On substituting the value of y in equation (1), we get

$\rm :\longmapsto\:x - 60 + 10 = 0$

$\rm :\longmapsto\:x - 50 = 0$

$\bf\implies \:x = 50$

Thus,

➢ Speed of the train = 50 km per hour

➢ Time taken = 12 hours

➢ Thus, Distance covered = xy = 12 × 50 = 600 km

Answer 2

Let assume that

Speed of the train be x km per hour

Time taken to covered the distance be y hours

So,

Distance covered by the train = xy km

Case :- 1

If a train could have been 10km/h faster, it would take 2 hour less than the schedule time.

So,

➢ Speed of the train = x + 10 km per hour

➢ Time taken = y - 2 hours

Thus,

:⟼(x + 10)(y - 2) = xy

:⟼ (x+10)(y−2)=xy

:⟼xy - 2x + 10y-20=xy

:⟼xy−2x+10y−20=xy

:⟼ - 2x + 10y - 20= 0

:⟼−2x+10y−20=0

:⟼ - 2(x - 5y + 10)= 0

:⟼−2(x−5y+10)=0

:⟼x - 5y + 10= 0 - - - - (1)

:⟼x−5y+10=0−−−−(1)

Case - 2

If train were slower by 10km/h, it would be taken 3 hour more than the schedule time.

So,

➢ Speed of the train = x - 10 km per hour

➢ Time taken = y + 2 hours

Thus,

:⟼(x−10)(y+3)=xy

:⟼xy+3x−10y−30=xy

:⟼3x−10y−30=0−−−−(2)

Now, multiply equation (1) by 3, we get

:⟼3x−15y+30=0−−−−(3)

On Subtracting equation (3) from equation (2), we get

⟼5y−60=0

⟼5y=60

y = 12⟹y=12

On substituting the value of y in equation (1), we get

x - 60 + 10 = 0:⟼x−60+10=0

:⟼x−50=0

⟹x=50

Thus,

➢ Speed of the train = 50 km per hour

➢ Time taken = 12 hours

➢ Thus, Distance covered = xy = 12 × 50 = 600 km