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The cost of 22 CDs, 25 pens & 7 erasers is $ 142. What is the cost of [#permalink]
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27 Jun 2010, 01:33
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The cost of 22 CDs, 25 pens & 7 erasers is $ 142. What is the cost of 1 CD, 1 pen & 1 eraser? (1) The cost of 7 CDs & 6 erasers is $ 23. (2) The cost of 5 CDs & 6 pens is $ 31.
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Re: The cost of 22 CDs, 25 pens & 7 erasers is $ 142. What is the cost of [#permalink]
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27 Jun 2010, 05:37
mikioso wrote: the answer is c please analyse the question carefully..... I have put it under 700+ level questions...



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Re: The cost of 22 CDs, 25 pens & 7 erasers is $ 142. What is the cost of [#permalink]
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27 Jun 2010, 14:35
The answer is C. Given in question: \(22x + 25y + 7z = 142\) Asked to find: \(x+y+z\) Statement 1: \(7x+6z = 23\) Clearly this is insufficient. There are many values of x and y which can satisfy this equation. So we are down to B C or E for answer choices Statement 2: \(5x+6y = 31\) This is also insufficient by itself. So now putting both together and using initial equation from question. \(22x + 25y + 7z = 142\) \(7x+6z = 23\) \(5x+6y = 31\) Three equations and three variables. Solvable. Hence the answer is C. apex1umt wrote: Question: The cost of 22 CDs, 25 pens & 7 erasers is $ 142. What is the cost of 1 CD, 1 pen & 1 eraser?
1. The cost of 7 CDs & 6 erasers is $ 23. 2. The cost of 5 CDs & 6 pens is $ 31.



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Re: The cost of 22 CDs, 25 pens & 7 erasers is $ 142. What is the cost of [#permalink]
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27 Jun 2010, 23:36
I still insist, please have a careful look at the question...
Moreover, it's NOT C



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Re: The cost of 22 CDs, 25 pens & 7 erasers is $ 142. What is the cost of [#permalink]
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28 Jun 2010, 20:32
The answer is B 5c + 6p =31 so 15c+ 18p=93..(1) 22c+25p+7e=142..(2) (2)(1) 7c+7p+7e=49 c+p+e=7



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Re: The cost of 22 CDs, 25 pens & 7 erasers is $ 142. What is the cost of [#permalink]
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30 Jun 2010, 23:13
apex1umt wrote: Question: The cost of 22 CDs, 25 pens & 7 erasers is $ 142. What is the cost of 1 CD, 1 pen & 1 eraser?
1. The cost of 7 CDs & 6 erasers is $ 23. 2. The cost of 5 CDs & 6 pens is $ 31. Verry tricky. It must be B 22C + 25P+7E= 7(C+P+E)+ 15C+18P= 7(C+P+E)+3(5C+6P)=7(C+P+E)+ 93 SO C+P+E= 7 22C + 25P+7E= 25(C+P+E)3C18E=25(C+P+E)3(C+6E) INSUF I need 3 more Kudos, Kudo me if OA is B



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Re: The cost of 22 CDs, 25 pens & 7 erasers is $ 142. What is the cost of [#permalink]
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01 Jul 2010, 02:13
B, seems to be the answer as per hemantbedi's explanation. However can someone give me a strong fundamental concept for a three variable equation, which tells when the three variable equation can be solved and when it cannot be solved. I know that a 3variable equation can be solved for sure if 3 equations are available. But when 2 equations are available it can sometimes be solved as shown here or sometimes cannot be solved. SO HOW DO i FIGURE THIS OUT ??? Please help.
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Re: The cost of 22 CDs, 25 pens & 7 erasers is $ 142. What is the cost of [#permalink]
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01 Jul 2010, 06:24
Good solution David Archuleta. I looked for tricks to manipulate, but completely forgot that I could commonly factorize the 7 from this equation. Good job, +1 from me.
Devashish: I think only practice can help you solve the equations. I think the easiest way to start practicing for these problems is to get the given equation in terms of what you want. In this case:
You want \(x+y+z\) and you're given \(22x+25y+7z\) The only way you can split this into something of the form \(a(x+y+z)\) is by taking the 7 out common and breaking it down as suggested.
I'm afraid there's no real shortcut guarantee to getting this type of problem, only practice helps.



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Re: The cost of 22 CDs, 25 pens & 7 erasers is $ 142. What is the cost of [#permalink]
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01 Jul 2010, 10:12
devashish wrote: B, seems to be the answer as per hemantbedi's explanation.
I know that a 3variable equation can be solved for sure if 3 equations are available. But when 2 equations are available it can sometimes be solved as shown here or sometimes cannot be solved.
We must have 3 equations to get 3 variables but this question doesn't ask to determine 3 vars, it just asks sum of 3 vars



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Re: The cost of 22 CDs, 25 pens & 7 erasers is $ 142. What is the cost of [#permalink]
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01 Jul 2010, 11:04
@David: Thanks a very obvious point, and still missed all the time. @whiplash2411: Thanks, it makes sense and should be the starting point the next time I start solving a 3variable equation.
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Re: The cost of 22 CDs, 25 pens & 7 erasers is $ 142. What is the cost of [#permalink]
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01 Jul 2010, 14:28
devashish wrote: B, seems to be the answer as per hemantbedi's explanation.
However can someone give me a strong fundamental concept for a three variable equation, which tells when the three variable equation can be solved and when it cannot be solved. I know that a 3variable equation can be solved for sure if 3 equations are available. But when 2 equations are available it can sometimes be solved as shown here or sometimes cannot be solved.
SO HOW DO i FIGURE THIS OUT ???
Please help. Hi, here's the rule: To solve for a system of n variables, one requires n distinct, linear, equations.Accordingly, if you have n distinct linear equations for n variables, you can answer any question about the system. However, some questions can potentially be answered with fewer than n distinct linear equations. First, if a question just asks about part of the system, you may not need the full number of equations. For example: Quote: If \(x + y + z = 25\), what's the value of \(x\)?
(1) \(y + z = 12\) Even though we only have 2 equations for our 3 unknowns, since statement (1) eliminates both y and z from the first equation, it's sufficient to solve for x. Second, if a question asks about a relationship among variables, rather than the value of variables, you may not need the full number of equations. For example: Quote: What's the value of \(3x  2y + 9z\)?
(1) \(6x + 18z  12 = 4y\) Here we only have 1 equation and 3 variables, but since we can rearrange (1) to: \(6x  4y + 18z = 12\) and since we can then divide by 2 to get: \(3x  2y + 9z = 6\) statement (1) is sufficient alone to answer the question. The key here was to rearrange the equation in (1) to make it easy to compare to the question and then to note that the equation in (1) is just a multiple of the expression in the question. The "number of equations vs number of unknowns" rule is, undoubtedly, THE most powerful rule for data sufficiency. The better you understand the rule, its exceptions and its application the less time it will take you to confidently answer many DS questions.



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Re: The cost of 22 CDs, 25 pens & 7 erasers is $ 142. What is the cost of [#permalink]
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27 Jul 2010, 05:48
The answer is B. If x, y, and z respectively the unit price of CD, pen, and eraser, we have: 22x+25y+7z=142$ We would like to know the price of 1CD, 1 pen, and 1 eraser, in other words x+y+z ? The first equation can be written as 7*(x+y+z)+(15x+18y)=142$ So if we know how much 15x+18y is, the question can be answered. On the other hand, statement 2 give us 5x+6y=31, or 15x+18y=93. Therefore statement 2 is sufficient.
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Re: The cost of 22 CDs, 25 pens & 7 erasers is $ 142. What is the cost of [#permalink]
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27 Jul 2010, 07:45
Even I thought of C but after reading the following comment: apex1umt wrote: I still insist, please have a careful look at the question...
Moreover, it's NOT C I looked the question in a different way and came up with B. Nice question and nice comment tooo ....
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Re: The cost of 22 CDs, 25 pens & 7 erasers is $ 142. What is the cost of [#permalink]
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28 Jul 2010, 02:20
Hi.. Guys... I think you have got the exact funda of the question.........
Moreover I am recently done with my GMAT and have scored a 710 with 51 (Quant) & 34(verbal) and 5 in AWA... Though 34 seems or rather is low, but I have no more patience to give the exam again.... I would rather better the profile then appear again..
Best of luck to you guys....



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