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???
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.
: 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]