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
[spoiler]The OA to this question is Option D. Could you please analyze my solution and let me know whats wrong?
Let the cost of cheap pairs be =x
Cost of Expensive pair=(2x-1)
Number of cheap socks=(a)
Number of Expensive socks=(a-3)
Therefore ax+(a-3)(2x-1)=18
which implies a=(6x+15)/(3x-1)
if we substitute x as 6, a would be equal to 3. Thus I arrived at A as the answer.[/spoiler]
Regards,
Amit
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Hi Amit,
This question could be solved rather easily by TESTing THE ANSWERS. Since you're asking about the algebraic approach you used, you made a small error in your calculations. $18 can be used to buy cheap socks OR 3 fewer pairs of expensive socks. In your calculation, you assume that BOTH are purchased for $18, but that's not the case. If you change your $18 to $36, then you can continue with the algebra and get the correct answer.
GMAT assassins aren't born, they're made,
Rich
This question could be solved rather easily by TESTing THE ANSWERS. Since you're asking about the algebraic approach you used, you made a small error in your calculations. $18 can be used to buy cheap socks OR 3 fewer pairs of expensive socks. In your calculation, you assume that BOTH are purchased for $18, but that's not the case. If you change your $18 to $36, then you can continue with the algebra and get the correct answer.
GMAT assassins aren't born, they're made,
Rich
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gmat_for_life
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Thanks a lot Rich!
That was so silly of me indeed! Should have noticed the wordings of the question properly.
Regards,
Amit
That was so silly of me indeed! Should have noticed the wordings of the question properly.
Regards,
Amit
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Algebraically you'd have
(2x - 1) * (a - 3) = 18
and
x * a = 18
The first equation is really
2ax - a - 6x + 3 = 18
We want to solve for a, so we'll sub out a*x = 18 and x = 18/a, giving
2*18 - a - 6*(18/a) + 3 = 18
0 = a - 108/a - 21
0 = a² - 21a - 108
0 = (a - 12) * (a - 9)
So a = 9 or a = 12, and we're set.
(2x - 1) * (a - 3) = 18
and
x * a = 18
The first equation is really
2ax - a - 6x + 3 = 18
We want to solve for a, so we'll sub out a*x = 18 and x = 18/a, giving
2*18 - a - 6*(18/a) + 3 = 18
0 = a - 108/a - 21
0 = a² - 21a - 108
0 = (a - 12) * (a - 9)
So a = 9 or a = 12, and we're set.
