A store sells a certain product at a fixed price per unit. At the prod

It got too complicated when I used algebra. Using plugging in, it was quite fast.

Price Quantity total value
p q pq = 300
p-5 q+2n (p-5)(q+2n) = 300
p+5 q-n (p+5)(q-n) = 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?

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

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?

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 P-5
Q P
Q-N P+5

The corresponding prices for the quantities Q+2N, Q, and Q-N are three consecutive multiples of 5: P-5, 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.

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) $200

I 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

Let p be price per unit.

From the first condition we get pq = 300.

From the other conditions we get,
(p-5)(q+2n) = pq
=> pq -5q + 2pn - 10n = pq
=> 5q = 2pn - 10n

(p+5)(q-n) = 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.
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Please press KUDOS if you like my post.

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/(20-n)=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

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:)

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
(p-5)*(q+2n) = 300 => q+2n = 300/(p-5) ...(1)
(p+5)*(q-n) = 300 => q-n = 300/(p+5) => 2q - 2n = 600/(p+5) ..(2)
Adding (1) and (2), we get
3q = 300/(p-5) + 600/(p-5)

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/(p-5) + 600/(p+5)
3/p = (p+5 +2p - 10)/(p^2-25)
simplifying we get
3p^2 - 75 - 3p^2 + 5p =0
p =15
=> q = 20

have three equations

300/q=p

300/q+2n=p-5

300/q-n=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

Tough question. Need to deal with 3 equations.

I will try to explain it in a simple way:

Let's assume the article price is "P".

As per Question stem, Pq= $300..... equation 1

Also, if price is reduced by $5, quantity will increase by 2n; (c-5) * (q+2n) = $300 ..... equation 2
Again, if price is increased by $5, quantity will decrease by n; (c-5) * (q-n) = $300 ..... equation 3

Equaiton 2= Equation 3; we will get 2cn-5q-10n= 0 (by solving eq 2) & cn-5q+5n=0 9by solving eq 3)

Subtract the above two, cn-15n=0, n(c-15)=0, Therefore c=15.

From equation 1, q=300/15=20, option C

I agree with Mike that Plugging In The Answers is a great way to go on this question and others for which the instinct is to start setting up complex algebraic equations. One tweak (an important one, I think): rather than starting with C, I suggest starting with B or D. Here's why.

If A is correct: The C-first approach requires you to test two answer choices, C and then either A or B. The B-first approach only requires you to test one answer choice, B and then there is only left that is smaller. B-first approach wins.

If B is correct: The C-first approach requires you to test two answer choices, C and then either A or B. The B-first approach only requires you to test one answer choice. B-first approach wins.

If C is correct: The C-first approach requires you to test one answer choice. The B-first approach requires you to test two answer choices. You do B first and find that you need something larger. You're down to C, D, and E. If you try D and need something smaller, C is the only option available. C-first approach wins.

If D is correct: Both approaches require you to test two answer choices. No winner.

If E is correct: Both approaches require you to test two answer choices. No winner.

And an additional benefit of B-and-D is that sometimes you'll test an answer choice, it won't be right, and you won't be positive whether you need something smaller or larger. In that case, the C-first approach could mean having to test ass many as three answer choices (if you go the wrong direction in choosing which to try second and are therefore still left with two). B-first still only requires a maximum of two. If you need something between B and D, it's C. If you went the wrong direction in trying D after B, it's A. If you got closer, but not all the way there, it's E.

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