Advertisements
Advertisements
Question
A man spent Rs. 2800 on buying a number of plants priced at Rs x each. Because of the number involved, the supplier reduced the price of each plant by Rupee 1.The man finally paid Rs. 2730 and received 10 more plants. Find x.
Advertisements
Solution
Amount spent = Rs. 2800
Price of each plant = Rs. x
Reduced price = Rs. (x – 1)
No. of plants in first case = `(2800)/x`
No. of plants received in second case = `(2800)/x + 10`
Amount paid = Rs. 2730
According to the condition,
`(2800/x + 10)(x - 1)` = 2730
⇒ `((2800 + 10x)(x - 1))/x` = 2730
⇒ (2800 + 10)(x – 1) = 2730x
⇒ 2800x – 2800 + 10x2 – 10x – 2730 = 0
⇒ 10x2 + 2800x – 10x – 2730x – 2800 = 0
⇒ 10x2 + 60x – 2800 = 0
⇒ x2 + 60x – 280 = 0 ...(Dividing by 10)
⇒ x2 + 20x – 14x – 280 = 0
⇒ x(x + 20) – 14(x + 20) = 0
⇒ (x + 20)(x – 14) = 0
Either x + 20 = 0,
then x = –20,
but it is not possible as it is in negative.
or
x – 14 = 0,
then x = 14.
APPEARS IN
RELATED QUESTIONS
Solve for x :
`1/(x + 1) + 3/(5x + 1) = 5/(x + 4), x != -1, -1/5, -4`
Solve the following quadratic equations by factorization:
`4sqrt3x^2+5x-2sqrt3=0`
Solve the following quadratic equations by factorization:
`x^2-(sqrt3+1)x+sqrt3=0`
Solve the following quadratic equations by factorization:
`(x+3)/(x+2)=(3x-7)/(2x-3)`
Three consecutive positive integers are such that the sum of the square of the first and the product of other two is 46, find the integers.
Solve the following quadratic equation by factorisation.
2y2 + 27y + 13 = 0
Write the set of value of 'a' for which the equation x2 + ax − 1 = 0 has real roots.
A two digit number is such that its product of its digit is 18. When 63 is subtracted from the number, the digits interchange their places. Find the number.
Forty years hence, Mr. Pratap’s age will be the square of what it was 32 years ago. Find his present age.
Find the roots of the following quadratic equation by the factorisation method:
`2/5x^2 - x - 3/5 = 0`
