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If a certain company purchased computers at $2000 each and printers $

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If a certain company purchased computers at $2000 each and printers $  [#permalink]

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New post Updated on: 06 Sep 2015, 03:28
5
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A
B
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E

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  45% (medium)

Question Stats:

69% (01:19) correct 31% (01:22) wrong based on 160 sessions

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If a certain company purchased computers at $2000 each and printers $300 each , how many computers did it purchase?

(1) More than three printers were purchased
(2) The total amount for the purchase of the computers and printers was $15,000

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Originally posted by sunita123 on 05 Sep 2015, 22:08.
Last edited by Bunuel on 06 Sep 2015, 03:28, edited 1 time in total.
Renamed the topic and edited the question.
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Re: If a certain company purchased computers at $2000 each and printers $  [#permalink]

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New post 05 Sep 2015, 23:45
Yes E is correct. I will explain how.

Statement 1 is clearly not sufficient.

Statement 2 says That the total was 15,000.

So (c)2000 + (p)300 = 15,000
where p is for printers and c is for computers.

when p=10 then the remaining amount left for computers is 12000 which means 6 computers.

but when p=30 the then the remaining amount left is 15,000 - 30(300) = 6000 which means 3 computers.
Different values means the statement is insufficient

Combining:

St. 1 says p>3 St. 2 says (c)2000 + (p)300 = 15,000.

But we know that the number of printers has to be a minimum of 10 for the remaining amount to be a multiple of 2000. ( from ST. 2)
Combining does not help because p>10 from statement 2. and statement 1 says p>3. p>10 is the more limiting inequality so we use that but it will not get us the answer.

So the answer is E.


The easiest approach according to me:
Start subtracting multiples of 300 from 15000 till you get a multiple of 2000.

15,000-300 = 14,700( not a multiple of 2000)
15,000- 600 = 14,400(not a multiple of 2000)
till you get to 15,000 - 3000 = 12,000 (Bingo you found the multiple)

Now you have to check if any other value of p satisfies the equation.
15,000- 9,000 = 6000( multiple of 2000)

So answer is E.

Hope this helps!
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If a certain company purchased computers at $2000 each and printers $  [#permalink]

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New post 06 Sep 2015, 19:42
Forget conventional ways of solving math questions. In DS, Variable approach is

the easiest and quickest way to find the answer without actually solving the

problem. Remember equal number of variables and equations ensures a solution.



If a certain company purchased computers at $2000 each and printers $300 each , how many computers did it purchase?

1) more than three printers were purchased
2) The total amount for the purchase of the computers and printers was $15,000

Transforming the original condition by variable approach method, we have 2 variables (number of computers;x, number of printers:y) and since we need to match the number of variables and equations, we need 2 equations. Since there is 1 each in 1) and 2), C is likely the answer. Using both 1) & 2) together, 2000x+300y=15000 ==> 20x+3y=150. then we have pair (x,y)=(3,30), (6,10), and since the answer is Not unique the conditions are not sufficient. Therefore the answer is E.
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If a certain company purchased computers at $2000 each and printers $  [#permalink]

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New post 18 Jan 2016, 09:19
VeritasPrepKarishma and other experts:
The answer is not B as the LCM of 2000 and 300 is less than 15,000. This is the reason the equation will not form a Diophantine equation and will not result in a unique solution.

I think my logic is sound. But can you pls comment on my logic and if I can use it reliably on similar questions, in which, it ultimately comes down to unicity or plurality of solution to an equation, without testing numbers? thanks
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Re: If a certain company purchased computers at $2000 each and printers $  [#permalink]

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New post 18 Jan 2016, 09:47
NoHalfMeasures wrote:
VeritasPrepKarishma and other experts:
The answer is not B as the LCM of 2000 and 300 is less than 15,000. This is the reason the equation will not form a Diophantine equation and will not result in a unique solution.

I think my logic is sound. But can you pls comment on my logic and if I can use it reliably on similar questions, in which, it ultimately comes down to unicity or plurality of solution to an equation, without testing numbers? thanks


Dont know about the application of diophantine equation to other similar questions but this is how I would solve this question.

You are given that each computer costs 2000 while each printer costs 300. You need to find the # of computers = c.

Per statement 1, p>3. Clearly not sufficient.

Per statement 2, 2000c+300p=15000 ---> 20c+3p=150. Solution sets for (c,p) can be (6,10) or (3,30). Thus no unique solution. Not sufficient.

Combining the 2 statements, still the 2 solution sets above (6,10) or (3,30) are applicable and hence E is the correct answer as you do not get a unique solution for the number of computers.

Hope this helps.
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Re: If a certain company purchased computers at $2000 each and printers $  [#permalink]

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New post 18 Jan 2016, 10:09
Thanks for your reply Engr2012. However, I know how to solve it using conventional methods. Im more interested in conceptual understanding of my logic and its applicability to similar questions. thanks again
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Re: If a certain company purchased computers at $2000 each and printers $  [#permalink]

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New post 18 Jan 2016, 21:10
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If a certain company purchased computers at $2000 each and printers $300 each , how many computers did it purchase?

(1) More than three printers were purchased
(2) The total amount for the purchase of the computers and printers was $15,000

When you modify the original condition and the question, it is frequently given on GMAT Math test which is "2 by 2" que.
Attachment:
GCDS  sunita123   If a certain company purchased (20160119).jpg
GCDS sunita123 If a certain company purchased (20160119).jpg [ 25.83 KiB | Viewed 2233 times ]

On the table, there are 2 variables(a,b), which should match with the number of equations. So you need 2 equations. For 1) 1 equation, for 2) 1 equation, which is likely to make C the answer. When 1) & 2), 2000a+200b=15000 -> 20a+3b=150 -> (a,b)=(3,30),(6,10), which is not unique and not sufficient. Therefore, the answer is E.


 For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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Re: If a certain company purchased computers at $2000 each and printers $  [#permalink]

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New post 12 Feb 2017, 21:15
Statement (1);
Says nothing about computers

Statement (2):
From the stem we know that the total purchase of computers and printers can be expressed as:

2000c + 300p

From this statement, we know that sum will be $15,000, so:

15000 = 2000c + 300p

This is a single expression with TWO variables so it is unlikely to give us a unique (integer) value for both of our variables, but we can see what pairs would work:

First we check the high priced item, Computers.

0 comp = 0 (15,000 left for printers = 50 printers)
1 comp = 2,000 (13,000 left for printers = NOT WHOLE # of printers)
2 comps = 4,000 (11,000 left for printers = NOT WHOLE # of printers)
3 comps = 6,000 (9,000 left for printers = 30 printers)
4 comps = 8,000 (7,000 left for printers = NOT WHOLE # of printers)
5 comps = 10,000 (5,000 left for printers = NOT WHOLE # of printers)
6 comps = 12,000 (3,000 left for printers = 10 printers)
7 comps = 14,000 (1,000 left for printers = NOT WHOLE # of printers)
8 comps = TOO MUCH

So statement (2) gives 3 possible outcomes:
- 0 computers and 50 printers
- 3 computers and 30 printers
- 6 computers and 10 printers


Statement (1+2)
We have not eliminated any of the possible scenarios from statement (1) because in ALL cases more than 3 printers were purchased.

The correct answer is E
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Re: If a certain company purchased computers at $2000 each and printers $  [#permalink]

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New post 17 Jul 2017, 17:17
Engr2012 wrote:
NoHalfMeasures wrote:
VeritasPrepKarishma and other experts:
The answer is not B as the LCM of 2000 and 300 is less than 15,000. This is the reason the equation will not form a Diophantine equation and will not result in a unique solution.

I think my logic is sound. But can you pls comment on my logic and if I can use it reliably on similar questions, in which, it ultimately comes down to unicity or plurality of solution to an equation, without testing numbers? thanks


Dont know about the application of diophantine equation to other similar questions but this is how I would solve this question.

You are given that each computer costs 2000 while each printer costs 300. You need to find the # of computers = c.

Per statement 1, p>3. Clearly not sufficient.

Per statement 2, 2000c+300p=15000 ---> 20c+3p=150. Solution sets for (c,p) can be (6,10) or (3,30). Thus no unique solution. Not sufficient.

Combining the 2 statements, still the 2 solution sets above (6,10) or (3,30) are applicable and hence E is the correct answer as you do not get a unique solution for the number of computers.

Hope this helps.


Hello Engr2012, Bunuel,
What is a good way to reach the two solutions (6,10) or (3,30), from the equation 20c + 3p = 150?
looks like a lot of these DS word problems hinge on determining whether there are 1 or 1+ solutions to these equations.
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Re: If a certain company purchased computers at $2000 each and printers $  [#permalink]

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