Question

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

Answer 1

Answer:

Let the no of toys be x and price of each toy be y

xy = 360

x=360/y  {eg -1}

2nd condition

(x+2)(y-2)=360

xy + 2y - 2x - 4 = 360

2y - 2(360/y) - 4 = 0  { since xy=360 & from eq 1}

2y² - 4y -720 = 0

y² - 2y - 360 =0

y² - 20y +18y -360 =0

y(y-20) +18(y-20) = 0

(y+18)(y-20)=0

y= -18 or y = 20

in this case y=20 since price of quantity cannot e in negative

Original price of toy is 20 rupees.

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Answer 2

SOLUTION:-

Given:

•The list price of a toy is reduced by Rs.2

•A person can buy 2 toys more for Rs.360.

To find:

The original price of the toy.

Explanation:

•Assume the list price of the toy be Rs.R

•Total amount of toys that are bought for Rs.360= 360/x

•The list price of toy is reduced by Rs.2

•New list price of toy=Rs.(R-2).

&

The number of toys that can be bought for Rs.360,

$= > \frac{360}{R - 2}$

According to the question:

$= > \frac{360}{R - 2} - \frac{360}{R} = 2 \\ \\ = > \frac{1}{R - 2} - \frac{1}{R} = \frac{2}{360} \\ \\ = > \frac{R - R + 2}{ {R}^{2} - 2R } = \frac{1}{180} \\ \\ = > \frac{2}{ {R}^{2} - 2R} = \frac{1}{180} \\ \\ = > {R}^{2} - 2R = 360 \\ \\ = > {R}^{2} - 2R- 360 = 0 \\ \\ = > {R}^{2} - 20R+ 18R - 360 = 0 \\ \\ = > R(R - 20) + 18(R - 20) = 0 \\ \\ = > (R - 20)(R + 18) = 0 \\ \\ = > R - 20 = 0 \: \: \: \: or \: \: \: \: R + 18 = 0 \\ \\ = > R = 20 \: \: \: or \: \: \: R = - 18$

Price of the toy can't be negative.

So,

R= 20 rupees.

Thus,

The list price is Rs.20.

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