3 years from now will be 360. We would like I THU
(iv) A train travels a distance of 480 km at a uniform speed. If the speed

3 years from now will be 360. We would like I THU
(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been
8 km/h less, then it would have taken 3 hours more to cover the same distance. We
need to find the speed of the train. a quadratic equation for this??? ​

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2 thoughts on “3 years from now will be 360. We would like I THU<br />(iv) A train travels a distance of 480 km at a uniform speed. If the speed”

  1. Answer

    Let the speed of the train be x km/hr. Then

    Time taken to travel a distance of 480km=

    x

    480

    hr

    Time taken by the train to travel a distance of 480km with the speed (x−8)km/hr=

    x−8

    480

    hr

    It is given that if the speed had been 8km/hr less, then the train would have taken 3 hours more to cover the same distance

    x−8

    480

    =

    x

    480

    +3

    x−8

    480

    x

    480

    =3⇒

    x(x−8)

    480(x−x+8)

    =3⇒

    x(x−8)

    480×8

    =3

    ⇒3x(x−8)=480×8⇒x(x−8)=160×8⇒x

    2

    −8x−1280=0

    This is the required quadratic equation.

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  2. : ANSWER

    The speed of train is = 40 km /hr

    : GIVEN

    Distance covered = 480 km

    if the speed has been 8 km / hr less

    : FIND

    Speed of the train

    : SOLUTION

    [tex] \looparrowright[/tex]Let the normal speed be x km \hr

    [tex] \looparrowright[/tex] We know that

    [tex] \looparrowright[/tex] Time = Distance/ speed

    [tex]\mapsto[/tex] [tex] \frac{480}{x – 8} – \frac{480}{x} = 3[/tex][tex]\mapsto[/tex] Cross multiplying

    [tex] \frac{480x – 480(x – 8)}{x(x – 8)} = 3[/tex][tex]\mapsto[/tex] 480x – 480x + 3840=3(x2- 8x)

    [tex]\mapsto[/tex] [tex]3x {}^{2} – 24x – 3840 = 0[/tex]

    [tex] \looparrowright[/tex] Factories by splitting the middle term

    [tex]\mapsto[/tex] [tex]x {}^{2} – 40x + 32x – 1280 = 0[/tex]

    [tex]\mapsto[/tex] [tex]x(x – 40) + 32(x – 40) = 0[/tex]

    [tex]\mapsto[/tex] [tex](x – 40)(x + 32) = 0[/tex]

    [tex]\mapsto[/tex] [tex]x = – 32 \: x = 40[/tex]

    [tex] \looparrowright[/tex] Here speed can’t be negative

    [tex] \implies[/tex] So, the train speed is 40 km /hr

    [tex] \sf \boxed{speed \: of \: the \: train \: is = 40 \: km \ \:hr}[/tex]

    Reply

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