Advertisements
Advertisements
प्रश्न
A dealer sells an article for Rs. 24 and gains as much percent as the cost price of the article. Find the cost price of the article.
उत्तर
et the cost price of article be Rs. x.
Then, gain percent = x
Therefore, the selling price of article
`=(x+x/100xx x)`
`=(x^2+100x)/100`
It is given that
`(x^2+100x)/100=24`
x2 + 100x = 2400
x2 + 100x - 2400 = 0
x2 + 120x - 20x - 2400 = 0
x(x + 120) - 20(x + 120) = 0
(x + 120)(x - 20) = 0
x + 120 = 0
x = -120
Or
x - 20 = 0
x = 20
Because x cannot be negative.
Thus, x = 20 is the require solution.
Therefore, the cost price of article be x = Rs. 20
APPEARS IN
संबंधित प्रश्न
Solve the following quadratic equation by Factorisation method: x2 + 7x + 10 = 0
Solve (i) x2 + 3x – 18 = 0
(ii) (x – 4) (5x + 2) = 0
(iii) 2x2 + ax – a2 = 0; where ‘a’ is a real number
If one of the equation ax2 + bx + c = 0 is three times times the other, then b2 : ac =
Solve the following equation: 3x2 + 25 x + 42 = 0
The hypotenuse of a right-angled triangle is 17cm. If the smaller side is multiplied by 5 and the larger side is doubled, the new hypotenuse will be 50 cm. Find the length of each side of the triangle.
In each of the following, determine whether the given values are solution of the given equation or not:
`a^2x^2 - 3abx + 2b^2 = 0; x = a/b, x = b/a`.
Solve the following equation by factorization
6p2+ 11p – 10 = 0
Find the values of x if p + 1 =0 and x2 + px – 6 = 0
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.
(x – 3) (x + 5) = 0 gives x equal to ______.