PS : In a certain clothing store

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PS : In a certain clothing store

by u1983 » Sat Sep 29, 2018 8:20 am

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In a certain clothing store, the most expensive pair of socks sells for one dollar less than twice the price of the cheapest pair of socks. A customer notices that for exactly $18, she can buy three fewer pairs of the most expensive socks than the cheapest socks. What could be the number of pairs of the cheapest socks she could have purchased?

(A) 3
(B) 5
(C) 6
(D) 12
(E) 36
D

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by Brent@GMATPrepNow » Sun Sep 30, 2018 9:10 am
u1983 wrote:In a certain clothing store, the most expensive pair of socks sells for one dollar less than twice the price of the cheapest pair of socks. A customer notices that for exactly $18, she can buy three fewer pairs of the most expensive socks than the cheapest socks. What could be the number of pairs of the cheapest socks she could have purchased?

(A) 3
(B) 5
(C) 6
(D) 12
(E) 36
D
Let C = price of one pair of the CHEAPEST socks
So, 2C - 1 = price of one pair of the MOST EXPENSIVE socks

A customer notices that for exactly $18, she can buy three fewer pairs of the most expensive socks than the cheapest socks.
As a word equation, we can write: (# of pairs of EXPENSIVE socks purchased with $18) = (# of pairs of CHEAP socks purchased with $18) - 3

With $18, the number of pairs of CHEAP socks we can purchase = 18/C
With $18, the number of pairs of EXPENSIVE socks we can purchase = 18/(2C - 1)

So, we can write: (18/(2C - 1)) = (18/C) - 3
Multiply both sides by C to get: 18C/(2C - 1) = 18 - 3C
Multiply both sides by (2C - 1) to get: 18C = 18(2C - 1) - 3C(2C - 1)
Simplify: 18C = 36C - 18 - 6C² + 3C
Simplify: 18C = 39C - 18 - 6C²
Rearrange to get: 6C² - 21C + 18 = 0
Divide both sides by 3 to get: 2C² - 7C + 6 = 0
Factor to get: (2C - 3)(C - 2) = 0
So, EITHER 2C - 3 = 0 OR C - 2 = 0
In other words, EITHER C = 1.5 OR C = 2

Let's check each answer choice.
If C = $1.5 (the price of one pair of the CHEAPEST socks), then 2C - 1 = $2 (the price of one pair of the MOST EXPENSIVE socks)
So, the number of CHEAP socks purchased for $18 = 18/$1.5 = 12
And the number of EXPENSIVE socks purchased for $18 = 18/$2 = 9
Perfect, this meets the condition that the customer can buy three fewer pairs of the most expensive socks than the cheapest socks.

If C = $2 (the price of one pair of the CHEAPEST socks), then 2C - 1 = $3 (the price of one pair of the MOST EXPENSIVE socks)
So, the number of CHEAP socks purchased for $18 = 18/$2 = 9
And the number of EXPENSIVE socks purchased for $18 = 18/$3 = 6
This ALSO meets the condition that the customer can buy three fewer pairs of the most expensive socks than the cheapest socks.

So, the customer bought EITHER 12 pairs of cheap socks or 9 pairs of cheap socks.
Check the answer choices, we have 12

Answer: D

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by swerve » Sun Sep 30, 2018 9:45 am
Let the price Cheapest pair is x. Hence the price of the expensive pair is 2x-1.

For $18, we get n cheapest pair & N expensive pair. Hence, n = 18/x & N = 18/(2x-1)

As per the Q steam: n - N = 3.
i.e., 18/x - 18/(2x-1) =3
or, 6/x - 6/(2x-1) =1

or, 6(2x-1)-6x = x(2x-1)
or, 2x^2 - 7x + 6 = 0
or, (2x-3)(x-2) = 0 =======> x=2, 1.5

When x=2, 2x-1=3 , n=9 & N=6
When x=1.5, 2x-1=2 , n=12 & N=9.......................Hence the ans is 12.

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by deloitte247 » Sun Sep 30, 2018 10:34 am
Let the price of cheapest pair = x
Price of expensive pair = 2x - 1
For $ 18
A customer gets n cheapest pair and N expensive pair.
$$Therefore,\ n\ =\ \frac{$18}{x}\ and$$
$$\ N\ =\ \frac{$18}{2x\ -\ 1}$$
The customer notices that she can buy 3 fewer pairs of the most expensive socks than the cheapest socks.
n - N = 3
$$Therefore,\ \frac{18}{x\ }-\frac{18}{2x-1}=3$$
Dividing through with 3 (BOTH LHS & RHS)
$$\frac{6}{x}-\frac{6}{2x-1}=1$$
6(2x-1) - 6x = x (2x - 1)
$$2x^2-7x\ +\ 6\ =\ 0$$
x = 2 or x = 1.5
When x = 2
2x - 1 = 3 $$N\ =\ \frac{18}{3}=\ 6\ and\ n\ =\ \frac{18}{2}\ =\ 9$$
$$When\ x\ =\ 1.5;\ 2x\ -\ 1\ =\ 2$$
$$N\ =\ \frac{18}{2}=\ 9\ and\ n\ =\ \frac{18}{1.5}=\ 12$$
Option D is CORRECT.

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by Scott@TargetTestPrep » Thu Apr 18, 2019 5:43 pm
u1983 wrote:In a certain clothing store, the most expensive pair of socks sells for one dollar less than twice the price of the cheapest pair of socks. A customer notices that for exactly $18, she can buy three fewer pairs of the most expensive socks than the cheapest socks. What could be the number of pairs of the cheapest socks she could have purchased?

(A) 3
(B) 5
(C) 6
(D) 12
(E) 36
D

We can let c = the price of the cheapest socks and e = the price of the expensive socks, and thus:

e = 2c - 1

Letting n = the number of pairs of the cheapest socks that she could buy, we have:

nc = 18

and

(n - 3)e = 18

Equating the second and the third equations, we have:

nc = (n - 3)e

nc = (n - 3)(2c - 1)

nc = 2nc - n - 6c + 3

18 = 2(18) - n - 6(18/n) + 3

18 = 36 - n - 108/n + 3

n - 21 + 108/n = 0

n^2 - 21n + 108 = 0

(n - 9)(n - 12) = 0

n = 9 or n = 12

We see that n is either 12 or 9, but since only 12 is one of the answer choices. 12 is the correct answer.

Answer: D

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