Math, asked by MAYAKASHYAP5101, 1 year ago

Hølã !!!

A train covered A certain distance at a uniform speed if the train would have been 10 km/h it would have taken 2hrs less than the scheduled time and if the train were slower by 10 km/h it would have taken 3 hours more than the scheduled time find the distance covered by the train ?

claßẞ => 10th

chäp => Pair of linear equations into two variables .

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Answers

Answered by porusvaid
0

speed = distance /time

distance = speed•time

distance = 50kmph • 12 hours

distance = 600Km

Attachments:
Answered by Anonymous
5
\huge \underline{\textbf {Solution ;}}

Let speed of train be x km/hr

Time taken = y hrs

We know that,

Distance = Speed × Time

So, Distance = xy....... (1)

______________________________

\underline{Case\:1:-}

If the train had been 10 km/hr faster :-

Speed = (x + 10) km/hr

Time = (y - 2) [As it would take 2 hours less time]

So, Distance = (x +10)(y - 2)

Putting this from equation 1 :-

=> xy = x (y - 2) + 10 (y - 2)

=> xy = xy - 2x + 10y - 20

=> xy - xy = 2x - 10y + 20

=> 2x - 10y + 20 = 0...... (2)

______________________________

\underline{Case\:2:-}

If the speed had been 10 km/hr slower,

Speed = (x - 10) km/hr

Time = (y + 3) hrs [As it is given that the time will increase by 3 hrs]

So, Distance = (x - 10)(y +3)

Putting this from equation 1:-

xy = (x - 10)(y +3)

=> xy = x (y +3)- 10(y +3)

=> xy = xy + 3x- 10y - 30

=> xy - xy = 3x - 10y - 30

=> 3x - 10y - 30 = 0....... (3)

Now,

From (2),

Taking 2 common,

2 ( x - 5y +10)= 0

=> x = 5y - 10........ (4)

Putting (4) in (3) :-

3 ( 5y - 10)10y - 30 = 0

=>15 y - 30 - 10 y - 30 = 0

=> 5y - 60 =0

=> y = 12

Putting y = 12 in (4)

X = 5 × 12 - 10 = 50

Hence, the answer will be :-

SPEED OF TRAIN = 50 km/hr

TIME TAKEN = 12 hrs

So, DISTANCE = 50 × 12

=  \huge\bold{600\: km}
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