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# Ellen can purchase a certain computer at a local store at th

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Ellen can purchase a certain computer at a local store at th  [#permalink]

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Updated on: 18 Jan 2015, 03:05
2
23
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Difficulty:

75% (hard)

Question Stats:

61% (02:23) correct 39% (02:24) wrong based on 536 sessions

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Ellen can purchase a certain computer at a local store at the price of p dollars and pay a 6 percent sales tax. Alternatively, Ellen can purchase the same computer from a catalog for a total of q dollars, including all taxes and shipping costs. Will it cost more for Ellen to purchase the computer from the local store than from the catalog?

(1) q - p < 50
(2) q = 1,150

Originally posted by SUNGMAT710 on 11 Sep 2013, 03:16.
Last edited by Bunuel on 18 Jan 2015, 03:05, edited 2 times in total.
Edited the question.
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Re: Ellen can purchase a certain computer at a local store at th  [#permalink]

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11 Sep 2013, 03:38
12
7
Ellen can purchase a certain computer at a local store at the price of p dollars and pay a 6 percent sales tax. Alternatively, Ellen can purchase the same computer from a catalog for a total of q dollars, including all taxes and shipping costs. Will it cost more for Ellen to purchase the computer from the local store than from the catalog?

The price from the store = p+0.06p=1.06p.
The price from the catalog = q.

(1) q - p < 50. If q=p, then the answer is YES (1.06p>q) but if q=120 and p=100, then the answer is NO (1.06p<q). Not sufficient.

(2) q = 1,150. The question becomes: is 1.06p>1,150. No info about p. Not sufficient.

(1)+(2) Since q = 1,150, then from (1) we have that 1,150-p<50 --> p>1,100 --> 1.06p>1,100*1.06>1,150. Sufficient.

Hope it's clear.
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Re: Ellen can purchase a certain computer at a local store at th  [#permalink]

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11 Sep 2013, 03:25
2
SUNGMAT710 wrote:
Ellen can purchase a certain computer at a local store at the price of p dollars and pay a 6 percent sales tax. Alternatively, Ellen can purchase the same computer from a catalog for a total of q dollars, including all taxes and shipping costs. Will it cost more for Ellen to purchase the computer from the local store than from the catalog?

(1) q - p < 50
(2) q = 1,150

For F.S 1, assume p=100,q=120. Thus, the total cost for purchasing from the local store : 100+6% = 106, and here 106<120.
Again, for p=1000,q=1001, the cost from local store = 1000+6% = 1060, and 1060>1001.Insufficient.

From F.S 2, we just know the value of q, no information about p. Insufficient.

Taking both together, we know that 1150-p<50 = p>1100. Thus, the minimum cost by purchasing at the local store, assuming p=1100 = 1100+6% = 1166 and here we clearly see that p>q. Sufficient.
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Re: Ellen can purchase a certain computer at a local store at th  [#permalink]

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16 Jan 2014, 16:39
1
SUNGMAT710 wrote:
Ellen can purchase a certain computer at a local store at the price of p dollars and pay a 6 percent sales tax. Alternatively, Ellen can purchase the same computer from a catalog for a total of q dollars, including all taxes and shipping costs. Will it cost more for Ellen to purchase the computer from the local store than from the catalog?

(1) q - p < 50
(2) q = 1,150

Or one could rephrase as follows

Is 1.06p>q?

(1) q-p<50
50+p>q

Is 1.06p>50+p?

We don't know

(2) q = 1150

Not enough

(1) and (2)

Is 1.06p>1150?

Is p more than a little less than 1150?

50+p> 1150

So the price 'p' which is obviously positive is greater than 1150

Therefore we have an answer and its gonna be C

Hope its clear

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Re: Ellen can purchase a certain computer at a local store at th  [#permalink]

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24 Feb 2015, 18:28
guys..

for (1) q - p < 50, can we take q = 100 and p = 110 ? (where q - p = -10 )
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Re: Ellen can purchase a certain computer at a local store at th  [#permalink]

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24 Feb 2015, 19:07
1
suhasancd wrote:
guys..

for (1) q - p < 50, can we take q = 100 and p = 110 ? (where q - p = -10 )

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Yes.
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Re: Ellen can purchase a certain computer at a local store at th  [#permalink]

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24 Feb 2015, 22:26
3
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Hi All,

Certain types of DS questions will appear on Test Day to measure the thoroughness of your thinking. You can see from the statistics above this prompt that almost half of the people who attempted it got it wrong. My guess is that most (if not all) of them COULD have gotten it correct, but they just didn't do enough work to PROVE what the correct answer is.

The prompt tells us about two ways that Ellen could purchase a computer:
1) From a store - for $P + 6% tax 2) From a catalog - for a total price of$Q

We're asked if it will cost Ellen MORE to buy the computer from the store than from the catalog. In real simple terms, it's asking "Is 1.06(P) > Q?" This is a YES/NO question.

Fact 1: Q - P < 50

From this, we don't know if Q or P is bigger, but we can still TEST VALUES to prove the inconsistency....

IF...
Q = 50
P = 1
1.06(1) is NOT > 50. The answer to the question is NO.

IF....
Q = 50
P = 53
1.06(53) IS > 50. The answer to the question is YES.
Fact 1 is INSUFFICIENT.

Fact 2: Q = 1150

This tells us nothing about the value of P, so we cannot figure out the value of 1.06(P).
Fact 2 is INSUFFICIENT

Combined, we know....
Q - P < 50
Q = 1150

At this point, many Test Takers would assume that this was insufficient, without gathering ANY proof. It's always interesting when Quant questions include specific numbers - those numbers were CHOSEN for a reason. They are NOT by accident. So it's interesting that Q = 1150 EXACTLY. Maybe there's a reason....maybe there's a pattern here....

Substituting in the value of Q, we have...

1150 - P < 50
1100 < P

Now this provides a lower limit to what P can be, but we MUST factor in the 6% tax too....

At the 'low end', P = just over $1100. With a 6% tax, (.06)(1100) =$66, so 1.06(P), at the minimum = a little more than 1100 + 66 = \$1166. Every other possible price for P is GREATER than this. By extension, EVERY possible value of 1.06(P) is greater than Q.
Combined, SUFFICIENT.

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Ellen can purchase a certain computer at a local store at th  [#permalink]

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04 Aug 2016, 19:31
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Attached is a visual that should help. For the final step, use estimation. 6 percent of 1000 is 60, so $$(>1100) + (>60)$$ is always $$> 1,150$$.
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Screen Shot 2016-08-04 at 7.31.57 PM.png [ 921.55 KiB | Viewed 10552 times ]

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Re: Ellen can purchase a certain computer at a local store at th  [#permalink]

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07 Aug 2016, 00:19
Bunuel wrote:
Ellen can purchase a certain computer at a local store at the price of p dollars and pay a 6 percent sales tax. Alternatively, Ellen can purchase the same computer from a catalog for a total of q dollars, including all taxes and shipping costs. Will it cost more for Ellen to purchase the computer from the local store than from the catalog?

The price from the store = p+0.06p=1.06p.
The price from the catalog = q.

(1) q - p < 50. If q=p, then the answer is YES (1.06p>q) but if q=120 and p=100, then the answer is NO (1.06p<q). Not sufficient.

(2) q = 1,150. The question becomes: is 1.06p>1,150. No info about p. Not sufficient.

(1)+(2) Since q = 1,150, then from (1) we have that 1,150-p<50 --> p>1,100 --> 1.06p>1,100*1.06>1,150. Sufficient.

Hope it's clear.

Hi when you club option 1& 2, you get 1150-p<50. But concluding p>1100 is not correct because we still don't know whether (1150-p) is greater than zero or less than zero. Hence i propose my answer as E. Kindly correct me

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Re: Ellen can purchase a certain computer at a local store at th  [#permalink]

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07 Aug 2016, 01:02
ramkumarsm@live.com wrote:
Bunuel wrote:
Ellen can purchase a certain computer at a local store at the price of p dollars and pay a 6 percent sales tax. Alternatively, Ellen can purchase the same computer from a catalog for a total of q dollars, including all taxes and shipping costs. Will it cost more for Ellen to purchase the computer from the local store than from the catalog?

The price from the store = p+0.06p=1.06p.
The price from the catalog = q.

(1) q - p < 50. If q=p, then the answer is YES (1.06p>q) but if q=120 and p=100, then the answer is NO (1.06p<q). Not sufficient.

(2) q = 1,150. The question becomes: is 1.06p>1,150. No info about p. Not sufficient.

(1)+(2) Since q = 1,150, then from (1) we have that 1,150-p<50 --> p>1,100 --> 1.06p>1,100*1.06>1,150. Sufficient.

Hope it's clear.

Hi when you club option 1& 2, you get 1150-p<50. But concluding p>1100 is not correct because we still don't know whether (1150-p) is greater than zero or less than zero. Hence i propose my answer as E. Kindly correct me

Sent from my Lenovo P1ma40 using GMAT Club Forum mobile app

We have to find the difference between 1.06p and 1150.

So, even if you take p=1100, you will get 1.06p = 1166 which is greater than 1150.

Hence, 1.06p will always be greater than q.

Hence C.
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Re: Ellen can purchase a certain computer at a local store at th  [#permalink]

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02 Dec 2016, 13:18
For me it is more intuitive if you consider min/max from the differences

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Re: Ellen can purchase a certain computer at a local store at th  [#permalink]

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01 Jan 2018, 11:42
Hello Everyone,
I have a question regarding this post.
for (1) q - p < 50, can we take q = 100 and p = 110 ? (where q - p = -10 )
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01 Jan 2018, 12:13
Hi dawnemily6,

Yes, the prompt never states that Q or P is larger than the other, so with Fact 1 you can certainly TEST a larger value for P.

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Re: Ellen can purchase a certain computer at a local store at th  [#permalink]

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16 Jul 2019, 01:01
guys.. :D

for (1) q - p < 50, can we take q = 100 and p = 110 ? (where q - p = -10 )
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16 Jul 2019, 10:21
Hi havip,

To answer your immediate question: YES, with the information in Fact 1, the value of P could be greater than the value of Q. Knowing that, were you able to correctly answer the question?

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Ellen can purchase a certain computer at a local store at th  [#permalink]

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15 Oct 2019, 02:00
Stem: $$106P>Q?$$

I) $$Q<50+P \implies 106P>50+P \implies 21P>10? \implies Not \ Sufficient$$

II) $$Q=1150 \implies P>1110 \implies Not \ Sufficient$$

I) & II) $$21(1100)>10 \implies Sufficient$$
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Ellen can purchase a certain computer at a local store at th   [#permalink] 15 Oct 2019, 02:00
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