Advertisements
Advertisements
प्रश्न
If the list price of a toy is reduced by Rs. 2, a person can buy 2 toys more for Rs. 360. Find the original price of the toy.
उत्तर
Let the original list price of the toy be Rs. x .
Then, the number of toys brought for Rs.360 `=360/x`
According to question, reduced list price of the toys = Rs. (x - 2).
Therefore, the number of toys brought for Rs.360 `=360/(x-2)`
It is given that
`360/(x-2)-360/x=2`
`360x-360(x-2)/((x-2)x)=2`
`(360x-360x+720)/(x^2-2x)=2`
`720/(x^2-2x)=2`
`720/2=x^2-2x`
360 = x2 - 2x
x2 - 2x - 360 = 0
x2 + 18x - 20x - 360 = 0
x(x + 18) - 20(x + 18) = 0
(x + 18)(x - 20) = 0
x + 18 = 0
x = -18
Or
x - 20 = 0
x = 20
Because x cannot be negative.
Thus, x = 20 is the require solution.
Therefore, the original list price of the toy be x = Rs. 20
APPEARS IN
संबंधित प्रश्न
The sum of the squares of three consecutive natural numbers as 149. Find the numbers
Solve each of the following equations by factorization:
`X^2 – 10x – 24 = 0 `
Solve the following quadratic equations by factorization:
`4/(x+2)-1/(x+3)=4/(2x+1)`
Divide 57 into two parts whose product is 680.
Solve the following quadratic equations by factorization:
\[16x - \frac{10}{x} = 27\]
Solve the following quadratic equations by factorization:
\[9 x^2 - 6 b^2 x - \left( a^4 - b^4 \right) = 0\]
Solve the following quadratic equations by factorization: \[\frac{1}{2a + b + 2x} = \frac{1}{2a} + \frac{1}{b} + \frac{1}{2x}\]
If the equation ax2 + 2x + a = 0 has two distinct roots, if
Solve the following equation :
`sqrt 2 "x"^2 - 3"x" - 2 sqrt 2 = 0`
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`.