Advertisements
Advertisements
प्रश्न
An aeroplane flying with a wind of 30 km/hr takes 40 minutes less to fly 3600 km, than what it would have taken to fly against the same wind. Find the planes speed of flying in still air.
Advertisements
उत्तर
Let the speed of the plane in still air = x km/hr
Speed of wind = 30km/hr
Distance = 3600km
∴ Time taken with the wind = `(3600)/(x + 30)`
and time taken against the wind = `(3600)/(x - 30)`
According to the condition,
`(3600)/(x - 30) - (3600)/(x + 30) = 40"mnutes" = (2)/(3)"hour"`
⇒ `3600((1)/(x - 30) - (1)/(x + 30)) = (2)/(3)`
⇒ `3600((x + 30 - x + 30)/((x - 30)(x + 30))) = (2)/(3)`
⇒ `(3600 xx 60)/(x^2 - 900) = (2)/(3)`
⇒ 2x2 - 1800 = 3 x 3600 x 60
⇒ 2x2 - 1800 = 648000
⇒ 2x2 - 1800 - 648000 = 0
⇒ 2x2 - 649800 = 0
⇒ x2 - 324900 = 0 ..(Dividing by 2)
⇒ x2 - (570)2 = 0
⇒ (x + 570)(x - 570) = 0
Either x + 570 = 0,
then x = -570
which is not possible as it is negative
or
x - 570 = 0,
then x = 570
Hence speed of plane in still air = 570km/hr.
APPEARS IN
संबंधित प्रश्न
Find the roots of the following quadratic equation by factorisation:
`2x^2 – x + 1/8 = 0`
Solve the following quadratic equations by factorization:
(x − 4) (x + 2) = 0
Solve the following equation: 2x2 - x - 6 = 0
The present age of the mother is square of her daughter's present age. 4 years hence, she will be 4 times as old as her daughter. Find their present ages.
Solve the following quadratic equation using factorization method:
`"x"^2-11"x"+24=0`
Two pipes flowing together can fill a cistern in 6 minutes. If one pipe takes 5 minutes more than the other to fill the cistern, find the time in which each pipe would fill the cistern.
Solve the following quadratic equation by factorisation:
2x2 + ax - a2 = 0 where a ∈ R.
A shopkeeper buys a certain number of books for Rs 960. If the cost per book was Rs 8 less, the number of books that could be bought for Rs 960 would be 4 more. Taking the original cost of each book to be Rs x, write an equation in x and solve it to find the original cost of each book.
The age of a man is twice the square of the age of his son. Eight years hence, the age of the man will be 4 years more than three times the age of his son. Find the present age.
Solve the following equation by factorisation :
2x2 + ax – a2= 0
