Advertisements
Advertisements
प्रश्न
The time taken by a person to cover 150kms was 2.5 hrs more than the time taken in the return j ourney. If he returned at a speed of 10km/ h more than the speed when going, find his speed per hour in each direction.
उत्तर
Let the speed of the person be S, Hence return speed of the person = S +10.
D=150 Km, Time = Distance / Speed. Time difference= 2.5 Hours
Hence, in these two conditions,
`150/"S" - 150/("S" + 10) = 2.5`
⇒ 150 x (S+10) - 150 x S= S x (S+10) x 2.5
⇒ 2.5 s2 + 25 s - 1500 = 0
⇒ 25 s2+ 250 s - 15000 = 0
⇒ S2 + 10 S - 600 = 0
⇒ S2 + 30S - 20S - 600 = 0
⇒ S (S+30) - 20 (S + 30) = 0
⇒ (S+30) (S - 20) = 0
As the speed can't be negative, S = 20km/ hr and while return, its 30km/ hr
APPEARS IN
संबंधित प्रश्न
Check whether the following are the quadratic equation:
(x + 1)2 = 2(x - 3)
Solve the following quadratic equation by using formula method :
2x2 - 3x = 2
Solve `(1 + 1/(x + 1))(1-1/(x - 1)) = 7/8`
Solve the following equation using the formula:
2x2 + 7x + 5 = 0
`2x^2+ax-a^2=0`
`1/x-1-1/(x+5)=6/7,x≠1,-5`
`1/(x+1)+2/(x+2)=5/(x+4),x≠-1,-2,-4`
A fast train takes 3 hours less than a slow train for a journey of 600kms. If the speed of the slow train is 1 Okm/ hr less than the fast train, find the speed of the fast train.
A shopkeeper buys a number of books for Rs 840. If he had bought 5 more books for the same amount, each book would have cost Rs 4 less. How many books did he buy?
In each of the following find the values of k of which the given value is a solution of the given equation:
7x2 + kx -3 = 0; x = `(2)/(3)`