Advertisements
Advertisements
Question
An aeroplane takes 1 hour less for a journey of 1200km, if its speed is increased by 100km/ hrfrom its usual speed. Find the usual speed.
Solution
Let the Usual Speed of the aircraft be S, and time taken be t. Distance = 1200km
Time = Distance/ Speed.
Time taken is reduced by 1 hr when speed increases by 100 km/ hr
`1200/"S" = 1200/("S" + 100) + 1`
`=> (1200 + "S" + 100)/("S" + 100) = 1200/"S"`
⇒ 1300S + S2 = 1200 S + 120000
⇒ S2 + 100 S - 120000 = 0
⇒ S2 + 400 S - 300 S - 120000 = 0
⇒ S (S + 400) - 300 (S + 400) = 0
⇒ (S + 400) (S - 300) = 0
⇒ S= -400 ,300
Speed cannot be negative .
Hence, S=300 km/ kr
APPEARS IN
RELATED QUESTIONS
A shopkeeper purchases a certain number of books for Rs. 960. If the cost per book was Rs. 8 less, the number of books that could be purchased for Rs. 960 would be 4 more. Write an equation, taking the original cost of each book to be Rs. x, and solve it to find the original cost of the books.
Check whether the following is quadratic equation or not.
x2 + 6x − 4 = 0
In the following, find the value of k for which the given value is a solution of the given equation:
x2 + 3ax + k = 0, x = -a
Solve for x using the quadratic formula. Write your answer correct to two significant figures.
(x – 1)2 – 3x + 4 = 0
Solve:
(x2 + 5x + 4)(x2 + 5x + 6) = 120
`x^2-4ax-b^2+4a^2=0`
`x/(x-1)+x-1/4=4 1/4, x≠ 0,1`
`1/(x+1)+2/(x+2)=5/(x+4),x≠-1,-2,-4`
Solve the following equation by using formula :
`(2)/(x + 2) - (1)/(x + 1) = (4)/(x + 4) - (3)/(x + 3)`
From the following equations, which one is the quadratic equation?
