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A store sells a certain product at a fixed price per unit. A [#permalink]
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A store sells a certain product at a fixed price per unit. At the product's current price, q units cost a total of exactly $300. If the price were lowered by $5 from its current value, then q + 2n units would cost exactly $300; if the price were raised by $5, then q – n units would cost exactly $300. What is the value of q? A. 10 B. 15 C. 20 D. 25 E. 30 Is there any easy way to solve this problem? Ans is the real exam going to have these sort of problems? Cheers
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Jp27 wrote: A store sells a certain product at a fixed price per unit. At the product's current price, q units cost a total of exactly $300. If the price were lowered by $5 from its current value, then q + 2n units would cost exactly $300; if the price were raised by $5, then q – n units would cost exactly $300. What is the value of q? A. 10 B. 15 C. 20 D. 25 E. 30
Is there any easy way to solve this problem? And is the real exam going to have these sort of problems? Dear Jp27, I'm happy to help with this. This problem is very hard  definitely an upper 700 level question. If you are doing very well on the Quant section, the CAT could feed you a question like this. This question is a very challenging if you take an algebraic approach, but it's remarkably simple if you backsolve. Let's start with (C). Assume q = 20 units If we can buy 20 units, they must cost a price of 300/20 = $15 Lower the price $5 to a new price of $10  then we could buy 300/10 = 30 units (10 more than the original case) Raise the price $5 to a new price of $20  then we could buy 300/20 = 15 units (5 fewer than the original case) This is the case for which we are looking  the increase in units from a $5 decrease in price is twice as much as the decrease in units from a $5 increase in price. Does this make sense? Mike
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Re: A store sells a certain product at a fixed price per unit. A [#permalink]
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04 Oct 2012, 14:09
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It got too complicated when I used algebra. Using plugging in, it was quite fast.
Price Quantity total value p q pq = 300 p5 q+2n (p5)(q+2n) = 300 p+5 qn (p+5)(qn) = 300
Solving three equations for three unknowns. Tough!!
Plugging in, I always start with C. C was the answer here, so saved calculation!
Putting values in above equations:
Price Quantity total value 15 20 300 10 20+2n 300 > 10(20 + 2n)=300 > 200 +20n = 300 > 20n = 100 > n =5 20 15 300
So q = 20 satisfies all equations!!
What is the source?



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Re: A store sells a certain product at a fixed price per unit. A [#permalink]
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As it is said for GMAT "whenever you see yourself dealing with some extra long equations or calculations trust!! there is an easy way out" here in this case back solving makes this question way too easy.
B is the answer



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Re: A store sells a certain product at a fixed price per unit. A [#permalink]
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06 Oct 2012, 20:52
mikemcgarry wrote: Jp27 wrote: A store sells a certain product at a fixed price per unit. At the product's current price, q units cost a total of exactly $300. If the price were lowered by $5 from its current value, then q + 2n units would cost exactly $300; if the price were raised by $5, then q – n units would cost exactly $300. What is the value of q? A. 10 B. 15 C. 20 D. 25 E. 30
Is there any easy way to solve this problem? And is the real exam going to have these sort of problems? Dear Jp27, I'm happy to help with this. This problem is very hard  definitely an upper 700 level question. If you are doing very well on the Quant section, the CAT could feed you a question like this. This question is a very challenging if you take an algebraic approach, but it's remarkably simple if you backsolve. Let's start with (C). Assume q = 20 units If we can buy 20 units, they must cost a price of 300/20 = $15 Lower the price $5 to a new price of $10  then we could buy 300/10 = 30 units (10 more than the original case) Raise the price $5 to a new price of $20  then we could buy 300/20 = 15 units (5 fewer than the original case) This is the case for which we are looking  the increase in units from a $5 decrease in price is twice as much as the decrease in units from a $5 increase in price. Does this make sense? Mike Thanks Mike. It does, especially the way you have circumvented calculating the variable N. I guess all the thinking goes before even touching the pen! cheers



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Re: A store sells a certain product at a fixed price per unit. A [#permalink]
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07 Oct 2012, 07:57
piyatiwari wrote: It got too complicated when I used algebra. Using plugging in, it was quite fast.
Price Quantity total value p q pq = 300 p5 q+2n (p5)(q+2n) = 300 p+5 qn (p+5)(qn) = 300
Solving three equations for three unknowns. Tough!!
Plugging in, I always start with C. C was the answer here, so saved calculation!
Putting values in above equations:
Price Quantity total value 15 20 300 10 20+2n 300 > 10(20 + 2n)=300 > 200 +20n = 300 > 20n = 100 > n =5 20 15 300
So q = 20 satisfies all equations!!
What is the source? Is there any reason why you start with C while using POE? What if the numbers are not given in a particular order (ascending here)? Do you use the same strategy in other question types?
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Re: A store sells a certain product at a fixed price per unit. A [#permalink]
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07 Oct 2012, 08:53
closed271 wrote: Is there any reason why you start with C while using POE? What if the numbers are not given in a particular order (ascending here)? Do you use the same strategy in other question types? No reason. I always start with option C while using plugging in. MGMAT advanced math book has some nice techniques about ruling out option choices. But here, I didn't use any of those.



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Re: A store sells a certain product at a fixed price per unit. A [#permalink]
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navigator123 wrote: Jp27 wrote: A store sells a certain product at a fixed price per unit. At the product's current price, q units cost a total of exactly $300. If the price were lowered by $5 from its current value, then q + 2n units would cost exactly $300; if the price were raised by $5, then q – n units would cost exactly $300. What is the value of q?
A. 10 B. 15 C. 20 D. 25 E. 30
Is there any easy way to solve this problem? Ans is the real exam going to have these sort of problems?
Cheers Since from the options is of pure chance, one would end up losing lots of time. Any other solution for this question? Back to the basics. Write down the prices and corresponding quantities using the given answers. You will get a small table: (Q, P) (10, 30) (15, 20) (20, 15) (25, 12) (30, 10) From the given information, about raising/reducing the price, you can conclude the following about the prices and corresponding quantities: Q+2N P5 Q P QN P+5 The corresponding prices for the quantities Q+2N, Q, and QN are three consecutive multiples of 5: P5, P, and P+5. From the table, the prices should be 10, 15, and 20, and thus the quantity should be 20. Answer C. Remark: this is my corrected post, as in my previous one, I mixed up prices and quantities. If somebody saw it, please, just forget about it.
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Re: A store sells a certain product at a fixed price per unit. A [#permalink]
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08 Oct 2012, 12:00
closed271 wrote: Is there any reason why you start with C while using POE? What if the numbers are not given in a particular order (ascending here)? Do you use the same strategy in other question types? On official GMAT problems, if the answer choices are all single numbers, those numbers will be in numerical order. This is true on all official material and all high quality material, and therefore it's an excellent test of how authentic a given prep source is  if you see a number of questions with the answers out of order, that's a red flag  you should question whether that source is trustworthy. Some GMAT prep sources are excellent, and some are not worth the paper on which they are printed. The reason we start with the middle answer (i.e. (C) on official material) is so that we know which way to go if our first choice is not right. Consider this hypothetical question: Frank started with X money. He bought blah blah, blah blah, blah % of blah, etc etc and was left with $41.50 in cash after those purchases. What was Frank's original starting amount? (A) $100 (B) $120 (C) $150 (D) $180 (E) $200I will start with (C) 150  if by chance I'm right, that's great. If I wind up with more leftover cash than $41.50, I know I started with too much  I can eliminate (C) & (D) & (E). If I wind up with less leftover cash than $41.50, I know I started with too little  I can eliminate (A) & (B) & (C). Does that make sense? Mike
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Re: A store sells a certain product at a fixed price per unit. A [#permalink]
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10 Nov 2012, 14:00
Even with plugging in this is a humdinger problem!
I tried the algebraic approach first and got stuck. Finally, took a guess.
With plugging in, I will try to time myself but it will take exactly two minutes still.
Start with C (so we know which direction to go in)
Case 1: (x) = Price of the unit (q) = # of units Let q = 20 Therefore, x = 15 because qx = 300 Case 2: x5 = 10 Given: q + 2n = 30; therefore: 20 + 2n = 30 2n = 10 n = 5
Case 3: x+5 = 20 Given : q  n = 15 20  5 = 15 Done. D is the answer because the equation checks out!



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Re: A store sells a certain product at a fixed price per unit. A [#permalink]
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Hello !
I was reviewing some questions and I think this one is not as difficult as it seems. I solved it in less than 2 minutes using equations solving so I felt like it could be useful to other people :
We can write the following equations :
(1) pq = 300 (2) (q+2n)(p5) = 300 (3) (qn)(p+5) = 300
Distribute (2) and (3) :
(2) qp 5q +2np 10n = 300 (3) qp +5q np  5n = 300
Add them together :
(2) + (3) : 2qp +np 15n = 600 Thanks to (1) you get : np15n = 0 or n(p15) = 0
Since n can't be 0 ( it would mean that a lower or higher price doesn't affect the total price and quantities ) you have p15 = 0 and p =15. With (1), you get q = 300/15 = 20
I know it seems time consuming but it worked for me. And I usually don't like picking numbers..



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Re: A store sells a certain product at a fixed price per unit. A [#permalink]
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28 Mar 2013, 03:56
pancakeFR wrote: Hello !
I was reviewing some questions and I think this one is not as difficult as it seems. I solved it in less than 2 minutes using equations solving so I felt like it could be useful to other people :
We can write the following equations :
(1) pq = 300 (2) (q+2n)(p5) = 300 (3) (qn)(p+5) = 300
Distribute (2) and (3) :
(2) qp 5q +2np 10n = 300 (3) qp +5q np  5n = 300
Add them together :
(2) + (3) : 2qp +np 15n = 600 Thanks to (1) you get : np15n = 0 or n(p15) = 0
Since n can't be 0 ( it would mean that a lower or higher price doesn't affect the total price and quantities ) you have p15 = 0 and p =15. With (1), you get q = 300/15 = 20
I know it seems time consuming but it worked for me. And I usually don't like picking numbers.. Hi this also works for me.Most of the times i can`t pick the number.1 kudos for u



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Re: A store sells a certain product at a fixed price per unit. A [#permalink]
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Let p be price per unit.
From the first condition we get pq = 300.
From the other conditions we get, (p5)(q+2n) = pq => pq 5q + 2pn  10n = pq => 5q = 2pn  10n
(p+5)(qn) = pq => pq + 5q  pn 5n = pq => 5q = pn + 5n (3)
so, 2pn  10n = pn + 5n => pn = 15n => p = 15 Then q = 300/15 = 20.
Answer : Option C.  Please press KUDOS if you like my post.



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Re: A store sells a certain product at a fixed price per unit. A [#permalink]
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22 Oct 2013, 10:48
mikemcgarry wrote: Jp27 wrote: A store sells a certain product at a fixed price per unit. At the product's current price, q units cost a total of exactly $300. If the price were lowered by $5 from its current value, then q + 2n units would cost exactly $300; if the price were raised by $5, then q – n units would cost exactly $300. What is the value of q? A. 10 B. 15 C. 20 D. 25 E. 30
Is there any easy way to solve this problem? And is the real exam going to have these sort of problems? Dear Jp27, I'm happy to help with this. This problem is very hard  definitely an upper 700 level question. If you are doing very well on the Quant section, the CAT could feed you a question like this. This question is a very challenging if you take an algebraic approach, but it's remarkably simple if you backsolve. Let's start with (C). Assume q = 20 units If we can buy 20 units, they must cost a price of 300/20 = $15 Lower the price $5 to a new price of $10  then we could buy 300/10 = 30 units (10 more than the original case) Raise the price $5 to a new price of $20  then we could buy 300/20 = 15 units (5 fewer than the original case) This is the case for which we are looking  the increase in units from a $5 decrease in price is twice as much as the decrease in units from a $5 increase in price. Does this make sense? Mike I don't get this at all; based on the first part: 300/(20+2n)=10 (since q=20, the original price was 15 per unit, minus 5 gives you 10), then the 2nd piece gives you 300/(20n)=20...and I get stuck. What do I do with those two facts. Also, the part highlighted in red, how do you you know you're looking for that case?



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Re: A store sells a certain product at a fixed price per unit. A [#permalink]
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22 Oct 2013, 11:02
AccipiterQ wrote: A store sells a certain product at a fixed price per unit. At the product's current price, q units cost a total of exactly $300. If the price were lowered by $5 from its current value, then q + 2n units would cost exactly $300; if the price were raised by $5, then q – n units would cost exactly $300. What is the value of q? A. 10 B. 15 C. 20 D. 25 E. 30
This is the case for which we are looking  the increase in units from a $5 decrease in price is twice as much as the decrease in units from a $5 increase in price.
I don't get this at all;
based on the first part: 300/(20+2n)=10 (since q=20, the original price was 15 per unit, minus 5 gives you 10), then the 2nd piece gives you 300/(20n)=20...and I get stuck. What do I do with those two facts. Also, the part highlighted in red, how do you you know you're looking for that case? Dear AccipiterQ, You seem to be doing the problem with a mix of algebra and backsolving together, which is very very confusing. Do one or the other. For this problem, I believe backsolving is much easier. As always, start with (C)  suppose q = 20. That means original price is $300/20 = $15. Now, don't do algebra here  just follow the numerical consequences. If the price is lowered $5 to $10, then we could buy $300/$10 = 30 items, which is q + 10 If the price is raised $5 to $20, then we could buy $300/$20 = 15 items, which is q  5 The difference between (q + 2p) and (q  p) is that the distance between q and the higher item number (at the lower price) is TWICE the distance between q and the lower item number (at the higher price). Well, here, 10 is twice 5, so we know we have met that condition. Does all this make sense? Mike
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Re: A store sells a certain product at a fixed price per unit. A [#permalink]
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22 Oct 2013, 12:51
mikemcgarry wrote: AccipiterQ wrote: A store sells a certain product at a fixed price per unit. At the product's current price, q units cost a total of exactly $300. If the price were lowered by $5 from its current value, then q + 2n units would cost exactly $300; if the price were raised by $5, then q – n units would cost exactly $300. What is the value of q? A. 10 B. 15 C. 20 D. 25 E. 30
This is the case for which we are looking  the increase in units from a $5 decrease in price is twice as much as the decrease in units from a $5 increase in price.
I don't get this at all;
based on the first part: 300/(20+2n)=10 (since q=20, the original price was 15 per unit, minus 5 gives you 10), then the 2nd piece gives you 300/(20n)=20...and I get stuck. What do I do with those two facts. Also, the part highlighted in red, how do you you know you're looking for that case? Dear AccipiterQ, You seem to be doing the problem with a mix of algebra and backsolving together, which is very very confusing. Do one or the other. For this problem, I believe backsolving is much easier. As always, start with (C)  suppose q = 20. That means original price is $300/20 = $15. Now, don't do algebra here  just follow the numerical consequences. If the price is lowered $5 to $10, then we could buy $300/$10 = 30 items, which is q + 10 If the price is raised $5 to $20, then we could buy $300/$20 = 15 items, which is q  5 The difference between (q + 2p) and (q  p) is that the distance between q and the higher item number (at the lower price) is TWICE the distance between q and the lower item number (at the higher price). Well, here, 10 is twice 5, so we know we have met that condition. Does all this make sense? Mike aaaah I get it now, I was creating a stew of algebra & backsolving. Makes perfect sense, thanks!



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A store sells a certain product at a fixed price per unit. At the product's current price, q units cost a total of exactly $300. If the price were lowered by $5 from its current value, then q + 2n units would cost exactly $300; if the price were raised by $5, then q – n units would cost exactly $300. What is the value of q?
a: 10 b: 15 c: 20 d: 25 e: 30
in the solution, they say, this problem is to hard to solve with algebra, so they just put in the answer choices and figure out what ist right... That is ok for me, but I don't get how you can se if a problem of this kind is solvable quickly or if it is better just to go with the answer options. Has anybody an advice about that? The problem above didn't look that hard for me so I spent 2 min with it and got the wrong answer:)



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Re: A store sells a certain product at a fixed price per unit. [#permalink]
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Am not a big fan of substitution so am posting a solution using algebra (doesn't look like a tough equation to solve) We just need to eliminate the variables which we don't really need to compute If we notice the three equations then we can get that "n" is the variable which is making the equations complex for us. So, let eliminate n first Let price be p pq = 300 (p5)*(q+2n) = 300 => q+2n = 300/(p5) ...(1) (p+5)*(qn) = 300 => qn = 300/(p+5) => 2q  2n = 600/(p+5) ..(2) Adding (1) and (2), we get 3q = 300/(p5) + 600/(p5) Now, we just have two equations in p and q. (Solving for p as the equation looks easy to me) put q = 300/p 900/p = 300/(p5) + 600/(p+5) 3/p = (p+5 +2p  10)/(p^225) simplifying we get 3p^2  75  3p^2 + 5p =0 p =15 => q = 20 edsafari wrote: A store sells a certain product at a fixed price per unit. At the product's current price, q units cost a total of exactly $300. If the price were lowered by $5 from its current value, then q + 2n units would cost exactly $300; if the price were raised by $5, then q – n units would cost exactly $300. What is the value of q?
a: 10 b: 15 c: 20 d: 25 e: 30
in the solution, they say, this problem is to hard to solve with algebra, so they just put in the answer choices and figure out what ist right... That is ok for me, but I don't get how you can se if a problem of this kind is solvable quickly or if it is better just to go with the answer options. Has anybody an advice about that? The problem above didn't look that hard for me so I spent 2 min with it and got the wrong answer:)
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Re: A store sells a certain product at a fixed price per unit. [#permalink]
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have three equations
300/q=p
300/q+2n=p5
300/qn=p+5
one option is solving algebraically but it is too long for three variables
started backsolve from q=20 (C)
find that p=15, n=5 and all is OK in this case
C




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