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Peter went to the store to buy paint. Small cans cost $30 and larger cans cost $80. How many small cans of paint did he buy? Statement #1: Peter spent $220 on paint. Statement #2: Peter bought four cans of paint in total.

Some problems on the GMAT Quant section cannot be solved with formulas or with algebra. For some problems, the GMAT makes you wrestle with the number properties themselves. For a discussion of difficult numerical reasoning questions, as well as the full solution to this question, see: http://magoosh.com/gmat/2013/difficult- ... questions/

Re: Peter went to the store to buy paint. Small cans cost $30 [#permalink]

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23 Dec 2013, 14:45

Not really a 700 question but nevertheless an interesting one:

Choice A: If you buy one can at 80, than there are no way to reach the $220. If you buy zero also. And if you buy three than you are already at 240. Than you need two cans at $80 and two at $30. CORRECT

Re: Peter went to the store to buy paint. Small cans cost $30 [#permalink]

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07 Jan 2014, 09:09

mikemcgarry wrote:

Peter went to the store to buy paint. Small cans cost $30 and larger cans cost $80. How many small cans of paint did he buy? Statement #1: Peter spent $220 on paint. Statement #2: Peter bought four cans of paint in total.

Some problems on the GMAT Quant section cannot be solved with formulas or with algebra. For some problems, the GMAT makes you wrestle with the number properties themselves. For a discussion of difficult numerical reasoning questions, as well as the full solution to this question, see: http://magoosh.com/gmat/2013/difficult- ... questions/

Mike

Not really a 700 question. Not really a Number Property question, this should be tagged as a Word Problem

Integer constraints are commonly tested on the GMAT

Not really a 700 question. Not really a Number Property question, this should be tagged as a Word Problem

Integer constraints are commonly tested on the GMAT

This problem is meant to deal with exactly that

I suggest always simplifying first

3x + 8y = 22

And then x = 22 - 8y / 3

Only 2 works here

Cheers! J

Dear jlgdr, By your statement "Only 2 works here", do you mean statement #2? In other words, do you mean to imply that (B) is the answer? If not, it's not clear to me what you think the answer is. Furthermore, I would argue that introducing algebra into this problem doesn't necessarily facilitate the solution, and for some students, may confuse them. Mike
_________________

Mike McGarry Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)

Re: Peter went to the store to buy paint. Small cans cost $30 and larger [#permalink]

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19 Apr 2017, 18:57

(A) Statement 1 - total spending 220$ We know that number of cans will be a natural number. Let x be number of small cans and y be number of large cans If y=1 then 30x=140 => no solution If y=2 then 30x=60=> x=2 Then y cannot be 3 otherwise sum will be greater than 220. Statement 1 is sufficient.

Statement 2 : Total 4 cans Possible cases (0,4) (1,3) (2,2) (3,1) (4,0)

Let's see the statements.. I. It gives us equation 30s+80b=220... ONLY possibility is s=b=2 Sufficient

The moment I see two variables I immediately concluded that I need a second linear equation. How to avoid this trap?

Generally such kind of linear equations (ax + by = c) have infinitely many solutions for x and y, and we cannot get single numerical values for the variables. But, here in 30x + 80y = 220, x and y must represent the # of cans, so they must be non-negative integers and in this case 30x + 80y = 220 is no longer simple linear equation, it's a Diophantine equation (equations whose solutions must be integers only) and for such kind on equations there might be only one combination of x and y possible to satisfy it. When you encounter such kind of problems you must always check by trial and error whether it's the case.