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

J :)
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Re: Ellen can purchase a certain computer at a local store at th [#permalink]
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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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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Ellen can purchase a certain computer at a local store at th [#permalink]
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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 25359 times ]

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

The question asks: is 1.06p>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.

Answer: C.

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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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.

The question asks: is 1.06p>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.

Answer: C.

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

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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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]
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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Re: Ellen can purchase a certain computer at a local store at th [#permalink]
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Video solution from Quant Reasoning starts at 9:58 here:
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Re: Ellen can purchase a certain computer at a local store at th [#permalink]
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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


Store Price = 1.06p
Catalog Price = q

(1) Value of q and p is unknown. For example, if p=100 q=100, then 100-106=-6, which less than 50 and the answer is yes, but if q=130 and p=100, then 130-106=26, which is less than 50 and the answer is no. INSUFFICIENT.

(2) No information of P. INSUFFICIENT.

Using both information.
1150-p<50
p>1100
1.06p>1100*1.06
10.6p>1166
Sufficient
The Answer is C
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Re: Ellen can purchase a certain computer at a local store at th [#permalink]
PC1=p$+0.06p PC2=q$

Is PC1>PC2?

(1) q-p<50
Nothing about q and p
NOT SUFF

(2) q=1150
NOT SUFF

(1) and (2)
If q=1150 and p=1101
PC1=1167.06
so PC1>PC2
So, if 1101<p<1149
PC1>PC2
SUFF

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