Advertisements
Advertisements
Question
The time taken by a person to cover 150 km was 2.5 hrs more than the time taken in the return journey. If he returned at a speed of 10 km/hr more than the speed of going, what was the speed per hour in each direction?
Advertisements
Solution
Let the ongoing speed of person be x km/hr. Then,
Returning speed of the person is = (x + 10)km/hr.
Time taken by the person in going direction to cover 150km = `150/x`hr
Time taken by the person in returning direction to cover 150km = `150/(x+10)`hr
Therefore,
`150/x-150/(x+10)=5/2`
`(150(x+10)-150x)/(x(x+10))=5/2`
`(150x+1500-150x)/(x^2+10x)=5/2`
`1500/(x^2+10)=5/2`
1500(2)=5(x2+10x)
3000 = 5x2 + 50x
5x2 + 50x - 3000 = 0
5(x2 + 10x - 600) = 0
x2 + 10x - 600 = 0
x2 - 20x + 30x - 600 = 0
x(x - 20) + 30(x - 20) = 0
(x - 20)(x + 30) = 0
So, either
x - 20 = 0
x = 20
Or
x + 30 = 0
x = -30
But, the speed of the train can never be negative.
Thus, when x = 20 then
= x + 10
= 20 + 10
= 30
Hence, ongoing speed of person is x = 20 km/hr
and returning speed of the person is x = 30 km/hr respectively.
APPEARS IN
RELATED QUESTIONS
Solve the following quadratic equation by Factorisation method: x2 + 7x + 10 = 0
Solve for x : `(x+1)/(x-1)+(x-1)/(x+2)=4-(2x+3)/(x-2);x!=1,-2,2`
Solve the following quadratic equations by factorization:
ax2 + (4a2 − 3b)x − 12ab = 0
Solve the following quadratic equations by factorization:
`(x+3)/(x+2)=(3x-7)/(2x-3)`
Solve the following quadratic equations by factorization:
`(x-a)/(x-b)+(x-b)/(x-a)=a/b+b/a`
Solve the following quadratic equation by factorization:
`(x-5)(x-6)=25/(24)^2`
The hypotenuse of a right triangle is 25 cm. The difference between the lengths of the other two sides of the triangle is 5 cm. Find the lengths of these sides.
A takes 10 days less than the time taken by B to finish a piece of work. If both A and B together can finish the work in 12 days, find the time taken by B to finish the work.
Solve:
`(x/(x + 2))^2 - 7(x/(x + 2)) + 12 = 0; x != -2`
Solve the following quadratic equations by factorization: \[\frac{3}{x + 1} - \frac{1}{2} = \frac{2}{3x - 1}, x \neq - 1, \frac{1}{3}\]
Find the values of k for which the roots are real and equal in each of the following equation:
\[4 x^2 - 2\left( k + 1 \right)x + \left( k + 1 \right) = 0\]
The sum of the square of two numbers is 233. If one of the numbers is 3 less than twice the other number. Find the numbers.
Solve the following by reducing them to quadratic form:
`sqrt(y + 1) + sqrt(2y - 5) = 3, y ∈ "R".`
Solve the equation:
`6(x^2 + (1)/x^2) -25 (x - 1/x) + 12 = 0`.
In each of the following determine whether the given values are solutions of the equation or not
2x2 - 6x + 3 = 0; x = `(1)/(2)`
Solve the following equation by factorization
x2 – (p + q)x + pq = 0
Solve the following equation by factorization
`4sqrt(3)x^2 + 5x - 2sqrt(3)` = 0
Solve the following equation by factorization
`(8)/(x + 3) - (3)/(2 - x)` = 2
Find the values of x if p + 1 =0 and x2 + px – 6 = 0
Solve the quadratic equation by factorisation method:
x2 – 15x + 54 = 0
