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A Car Covers a Distance of 400 Km at a Certain Speed. Had the Speed Been 12 Km/Hr More, the Time Taken for the Journey Would Have Been 1 Hour 40 Minutes Less. Find the Original Speed of the Car. - Mathematics

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Question

A car covers a distance of 400 km at a certain speed. Had the speed been 12 km/hr more, the time taken for the journey would have been 1 hour 40 minutes less. Find the original speed of the car.

Sum

Solution 1

Let the original speed of the car be x
km/hr,
so, Time taken by car = `(400)/x"hrs"`.

Again, Speed = (x + 12) km/hr

Time taken by car = `(400)/(x + 12)`

so, `(400)/x - (400)/(x + 12) = 1 "hr" + (40)/(60)`

`400((x + 12 - x))/(x (x + 12)) = (5)/(3)`

`(4800)/(x^2 + 12x) = (5)/(3)`

⇒ 5 (x2 + 12x) = 14,400
⇒ x2 + 12x - 2,880 = 0
⇒ x2 + 60x - 48x - 2,880 = 0
⇒ x (x + 60) - 48 (x + 60) = 0
⇒ (x + 60) (x - 48) = 0
Either, x + 60 = 0
x = -60   ...(Neglect, Speed can't be negative)
or
x -  48 = 0
x = 48
⇒ Original speed of the car is 48 km/hr.

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Solution 2

Let the original speed of the car = x km/h.
Distance covered = 400km

Time taken to cover 400km = `(400)/x"h"`

In second case,
Speed of car = (x + 12)km/h

New time taken to cover 400km = `(400)/(x + 12)"h"`
According to the condition

`(400)/x - (400)/(x + 12)`

= `1(40)/(60)` 

= `1(2)/(3)`

= `(5)/(3)`.

⇒ `400((x + 12 - x)/(x(x + 12))) = (5)/(3)`

⇒ `(400 xx 12)/(x^2 + 12x) = (5)/(3)`

400x 12 x 3 = 5x2 + 60x
⇒ 1400 = 5x2 + 60x
⇒ 5x2 + 60x - 14400 = 0
⇒ x2 + 12x - 2880 = 0   ...(dividing both side by 5)
⇒ x2 + 60x - 48x - 2880 = 0
⇒ x(x + 60) - 48(x + 60) = 0
⇒ (x + 60)(x - 48) = 0
⇒ x = 48
or
⇒ x = -60
⇒ x = 48  ...(Rejecting x = -60, being speed)
Hence, orginal speed of the car = 48km/h.

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Chapter 6: Quadratic Equation - Exercise 1

APPEARS IN

ICSE Mathematics [English] Class 10
Chapter 6 Quadratic Equation
Exercise 1 | Q 51
ML Aggarwal Understanding ICSE Mathematics [English] Class 10
Chapter 5 Quadratic Equations in One Variable
Exercise 5.5 | Q 27
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