Advertisements
Advertisements
Question
A passenger train takes one hour less for a journey of 150 km if its speed is increased by 5 km/hr from its usual speed. Find the usual speed of the train.
Solution
Let the usual speed of train be x km/hr then
Increased speed of the train = (x + 5)km/hr
Time taken by the train under usual speed to cover 150km = `150/x`hr
Time taken by the train under increased speed to cover 150km = `150/(x + 5)`hr
Therefore,
`150/x-150/(x+5)=1`
`(150(x+5)-150x)/(x(x+5))=1`
`(150x+750-150)/(x^2+5x)=1`
`750/(x^2+5x)=1`
750 = x2 + 5x
x2 + 5x - 750 = 0
x2 - 25x + 30x - 750 = 0
x(x - 25) + 30(x - 25) = 0
(x - 25)(x + 30) = 0
So, either
x - 25 = 0
x = 25
Or
x + 30 = 0
x = -30
But, the speed of the train can never be negative.
Hence, the usual speed of train is x = 25km/hr
APPEARS IN
RELATED QUESTIONS
Solve the following quadratic equation by factorization:
`(x-5)(x-6)=25/(24)^2`
Solve the following quadratic equations by factorization:
a2b2x2 + b2x - a2x - 1 = 0
The product of Ramu's age (in years) five years ago and his age (in years) nice years later is 15. Determine Ramu's present age.
`4x^2+4sqrt3x+3=0`
One of the roots of equation 5m2 + 2m + k = 0 is `(-7)/5` Complete the following activity to find the value of 'k'.
Solve the following quadratic equations by factorization: \[\frac{3}{x + 1} - \frac{1}{2} = \frac{2}{3x - 1}, x \neq - 1, \frac{1}{3}\]
If sin α and cos α are the roots of the equations ax2 + bx + c = 0, then b2 =
The product of a girl's age five years ago and her age 3 years later is 105. Find her present age.
Solve: x(x + 1) (x + 3) (x + 4) = 180.
A boat can cover 10 km up the stream and 5 km down the stream in 6 hours. If the speed of the stream is 1.5 km/hr. find the speed of the boat in still water.