Looking for an Algebraic Solution

[robert]

I apologize in advance if this particular question has been addressed on this forum (i could not find it in the search).

I am curious to see if there is an algebraic solution to the following problem. The CAT Explanation only provided a "plugging in answer choices" solution. Thank You.

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?

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[tim]

yes there is. try it out and let us know if you get stuck..

[robert]

yes there is. try it out and let us know if you get stuck..

I have. I can get to: PQ=300 and (P-5) (Q+2N)=(P+5) (Q-N) ....but I cant simplify from there to get only Q.

[vikash.121186]

I apologize in advance if this particular question has been addressed on this forum (i could not find it in the search).

I am curious to see if there is an algebraic solution to the following problem. The CAT Explanation only provided a "plugging in answer choices" solution. Thank You.

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?

10
15
20
25
30

The following three equations can be framed with the given conditions.

PQ=300...... (i)

(Q+2N)(P-5)=300...... (ii)

(Q-N)(P+5)=300.........(iii)

Equate LHS of equation(i) and equation(ii) to get equation (iv).

Equate LHS equation(i) and equation (iii) to get equation (v).

Equate LHS equation(ii) and equation(iii) to get equation (vi).


I think after doing all this you will start to see the solution. Q should come out to be 20. Correspondingly, P will come out to be 15. I hope my solution is correct.

However, i believe that there could be a better solution to this problem but it can't think of it. Will post if i could think of a better solution to this.

Regards

[robert]

The following three equations can be framed with the given conditions.

PQ=300...... (i)

(Q+2N)(P-5)=300...... (ii)

(Q-N)(P+5)=300.........(iii)

Equate LHS of equation(i) and equation(ii) to get equation (iv).

Equate LHS equation(i) and equation (iii) to get equation (v).

Equate LHS equation(ii) and equation(iii) to get equation (vi).


I think after doing all this you will start to see the solution. Q should come out to be 20. Correspondingly, P will come out to be 15. I hope my solution is correct.

However, i believe that there could be a better solution to this problem but it can't think of it. Will post if i could think of a better solution to this.

Regards[/quote]

Do you think you could write that math out for me? I am still having a difficult time. This question is driving me nuts.....maybe the algebra is too time consuming to even consider as an option....i just thought i could learn something from seeing it all written out....thanks

[tim]

PQ=300...... (i)

(Q+2N)(P-5)=300...... (ii)

(Q-N)(P+5)=300.........(iii)

expand the last two equations:

PQ - 5Q + 2NP -10N = 300
PQ + 5Q - NP - 5N = 300

subtract 300 from both sides of each equation (remember PQ is 300):

-5Q + 2NP - 10N = 0
5Q - NP - 5N = 0

now add these two equations:

NP - 15N = 0

divide both sides by N:

P - 15 = 0
P = 15

from this you can plug P=15 into PQ=300 to get a value for Q..

[robert]

Thanks. It was the last step of subtracting the equations from each other that i was missing.

[jlucero]

Glad that it helped.

[saintforlife]

PQ=300...... (i)

(Q+2N)(P-5)=300...... (ii)

(Q-N)(P+5)=300.........(iii)

I got to this point and then for some reason I expanded and made both equations equal instead of solving them simultaneously. This got me nowhere.

PQ - 5Q + 2NP -10N = PQ + 5Q - NP - 5N

Then I got stuck at this equation and didn't know how to proceed after this:

10Q + 3NP - 5N = 0.

[RonPurewal]

PQ=300...... (i)

(Q+2N)(P-5)=300...... (ii)

(Q-N)(P+5)=300.........(iii)

I got to this point and then for some reason I expanded and made both equations equal instead of solving them simultaneously. This got me nowhere.

PQ - 5Q + 2NP -10N = PQ + 5Q - NP - 5N

Then I got stuck at this equation and didn't know how to proceed after this:

10Q + 3NP - 5N = 0.

If you scroll up three posts, you'll see a complete solution process, posted by Tim.

Here, you've lost the fact that both expressions are equal to 300. That fact is essential to solving the system; if you lose it, you can't solve the system.

[NL]

I'll give up on this question. Too much calculation or algebra manipulation. Not much sexy.

I read the explanation. It seems Ron's voice. How did you create such convoluted question?

[RonPurewal]

I honestly don't remember whether I wrote this question. Don't really remember the ones I wrote more than, say, half a year ago.

I'll give up on this question. Too much calculation or algebra manipulation. Not much sexy.

Backsolving is not very much work here.

[lsyang1212]

If you scroll up three posts, you'll see a complete solution process, posted by Tim.

Here, you've lost the fact that both expressions are equal to 300. That fact is essential to solving the system; if you lose it, you can't solve the system.

Why is it that in this specific problem, you cannot set both expressions equal to each other to solve? Why is it essential to keep the $300 on the right side of both equations? Is there a general rule for when we should stack and add, vs set equal to each other?


Thanks.

[RonPurewal]

Why is it that in this specific problem, you cannot set both expressions equal to each other to solve? Why is it essential to keep the $300 on the right side of both equations?

For the same reason why "I'm 6 feet tall, and my brother is also 6 feet tall" is not the same statement as "I'm exactly as tall as my brother".

[RonPurewal]

If "combining" equations causes you to lose information (like the actual height of 6 feet here), it's essential that you keep track of that information, as you'll almost certainly have to bring it back into play later.

Think about the typical "two equations, two unknowns" problems from first-year high-school algebra.
After you combine the equations (by substitution/adding/subtracting/whatever) to find one variable, you're not done until you go back and plug back in to an EARLIER equation. That's the information you "lost" by combining the equations, so now you have to go back and fetch it.