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

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Re: A store sells a certain product at a fixed price per unit. A [#permalink]

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New post 30 Dec 2014, 02:04
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. A [#permalink]

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Re: A store sells a certain product at a fixed price per unit. A [#permalink]

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New post 09 Mar 2016, 19:50
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


my approach:
pq=300
(p-5)(q+2n)=300
pq+2pn-5q-10n=300
now we know that pq=300, deduct from both sides:
2n-5q-10n=0

second equation
(p+5)(q-n)=300
pq-pn+5q-5n=300 - same thing - deduct 300
5q-pn-5n=0

now..take both new equations and add them:
2n-5q-10n + 5q-pn-5n= 0
we get np-15n=0
factor n:
n(p-15)=0
either n=0, or p=15
but n can't be 0, because otherwise (P-5)*Q or (P+5)*Q would not be the same as PQ, so it MUST be true that P=15
in this case Q=20.

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Re: A store sells a certain product at a fixed price per unit. A [#permalink]

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New post 20 Mar 2017, 15:43
first I jot down the equations:
x=price

x*q=300
(x-5)(q+2n)=300
(x+5)(q-n)=300

MNGMT advises to start from B and based on the move
B wasn't fine
so I tried C=>ok

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Re: A store sells a certain product at a fixed price per unit. A [#permalink]

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New post 21 Mar 2017, 03:33
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


price per unit = 300/q

300/q - 5 = 300/(q+2n) ... (1)

300/q + 5 = 300/(q-n) .... (2)

solving (1) and (2), we get
q = 20 n = 5

Option C

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Re: A store sells a certain product at a fixed price per unit. A [#permalink]

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New post 23 May 2017, 21:10
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


This problem requires a great deal of space on a yellow pad. So it's going to likely need two quadrants on a GMAT folio. That said, the best approach is to set up a table with the values P | PQ = 300 | Q | (P-5)(Q+2N) = 300 | (P+5)(Q-N) = 300

Then at this point it's about plugging and chugging through the answers. The criterion by which to eliminate is the variable N which should be consistent within a given row in your table (e.g., for the correct answer, it's equal to 5).

Q = 20, P = 15.
(10)*(Q+2N) = 300
Q + 2N = 30
2N = 30 - Q = 30 -20
2N = 10
N = 5

(20)*(20-5) = 20*15 = 300

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Re: A store sells a certain product at a fixed price per unit. A [#permalink]

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New post 02 Sep 2017, 10:30
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


A very simple solution to this question can be framed which will take approx 2 minutes by following method.
1st case : price is lowered by $5
So Total price lowered = 5q
This lead to increase in number of units by 2n
So, 5q = 2n(300/q - 5)
5q/2n = 300/q - 5 ...........................(i)

2nd case : price is increased by $5
So, total price increased = 5q
This lead to decrease in number of units by n.
So, 5q = n(300/q +5)
5q/n = 300/q + 5.............................(ii)

From (i) * 2 - (ii), we get
0 = 600/q - 10 - 300/q - 5
15 = 300/q
q = 20

Answer C

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Re: A store sells a certain product at a fixed price per unit. A   [#permalink] 02 Sep 2017, 10:30

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