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Two Variables, One Equation, and a lot of headache!! [#permalink]
17 May 2012, 17:42
Guys,
As you know in some GMAT problems, GMAC loves placing implicit conditions on variables we are working with. The most notable conditions are that the variables be both positive and non-negative (either 0 or positive). For example ( made this question up):
John purchased x amount of apples and y pears and paid $19, how many apples and pears did John buy? 1) Each apple costs $3 and pears $4 2) 6x+5y=12
Now from statement 1) we can say 3x+4y=19, Now from this simple statement, we know x and y have to be nonnengative and integer (thats an implicit condition we can't buy a negative amounts of apples and pears and we cant buy half an apple). But this statement is sufficient because only one set of values satisfies this equality and that is x=1 and y=4. We don't need the typical two equations, two variables to solve this problem.
However if the question had asked:
John purchased an x amount of apples and y pears and paid $13, how many apples and pears did John buy? 1) Apples cost $3 and pears $4 2) Statement 2 (just ignore)
Statement 1 would be insufficient as {x=3 and y=1} or {x=1 or y=3}.
My question is there a way of determining that ax+by=c (If x and y are both non-negative integers, a,b,c are constants.) without trial and error by plugging? There just has to be a more judious method for determining whether equation 3x+4y=19 or 3x+4y=13 has more than one solution set. If we factor the number on the RHS can that tell us something?
Re: Two Variables, One Equation, and a lot of headache!! [#permalink]
17 May 2012, 20:01
alphabeta1234 wrote:
Guys,
As you know in some GMAT problems, GMAC loves placing implicit conditions on variables we are working with. The most notable conditions are that the variables be both positive and non-negative (either 0 or positive). For example ( made this question up):
John purchased x amount of apples and y pears and paid $19, how many apples and pears did John buy? 1) Each apple costs $3 and pears $4 2) 6x+5y=12
Now from statement 1) we can say 3x+4y=19, Now from this simple statement, we know x and y have to be nonnengative and integer (thats an implicit condition we can't buy a negative amounts of apples and pears and we cant buy half an apple). But this statement is sufficient because only one set of values satisfies this equality and that is x=1 and y=4. We don't need the typical two equations, two variables to solve this problem.
However if the question had asked:
John purchased an x amount of apples and y pears and paid $13, how many apples and pears did John buy? 1) Apples cost $3 and pears $4 2) Statement 2 (just ignore)
Statement 1 would be insufficient as {x=3 and y=1} or {x=1 or y=3}.
My question is there a way of determining that ax+by=c (If x and y are both non-negative integers, a,b,c are constants.) without trial and error by plugging? There just has to be a more judious method for determining whether equation 3x+4y=19 or 3x+4y=13 has more than one solution set. If we factor the number on the RHS can that tell us something?
Hi
I understand your concern, but still i think that subsctituion/trail & error is still the best & quickest way to solve such problmes especially in DS...
That's all DS is all about after all.. _________________
Best Vaibhav
If you found my contribution helpful, please click the +1 Kudos button on the left, Thanks
Re: Two Variables, One Equation, and a lot of headache!! [#permalink]
19 May 2012, 18:10
Bunuel, please help. For example, here is a problem you recently did that shows the problem.
1. A citrus fruit grower receives $15 for each crate of oranges shipped and $18 for each crate of grapefruit shipped. How many crates of oranges did the grower ship last week?
a. Last week the number of crates of oranges that the grower shipped was 20 more than twice the number of crates of grapefruit shipped. b. Last week the grower received a total of $38,700 from the crates of oranges and grapefruit shipped.
A citrus fruit grower receives $15 for each crate of oranges shipped and $18 for each crate of grapefruit shipped. How many crates of oranges did the grower ship last week?
Let \(x\) be the # of oranges and \(y\) the # of grapefruits. Note that \(x\) and \(y\) must be an integers. Q: \(x=?\)
(1) Last week the number of crates of oranges that the grower shipped was 20 more than twice the number of crates of grapefruit shipped --> \(x=2y+20\). Not sufficient to calculate \(x\)
(2) Last week the grower received a total of $38,700 from the crates of oranges and grapefruit shipped --> \(15x+18y=38700\) --> \(5x+6y=12900\). Multiple values are possible, for istance: \(x=180\) and \(y=2000\) OR \(x=60\) and \(y=2100\).
(1)+(2) Two unknowns, two different linear equations --> We can calculate unique value of \(x\). Sufficient.
Re: Two Variables, One Equation, and a lot of headache!! [#permalink]
22 May 2012, 08:44
alphabeta1234 wrote:
Bunuel, please help. For example, here is a problem you recently did that shows the problem.
1. A citrus fruit grower receives $15 for each crate of oranges shipped and $18 for each crate of grapefruit shipped. How many crates of oranges did the grower ship last week?
a. Last week the number of crates of oranges that the grower shipped was 20 more than twice the number of crates of grapefruit shipped. b. Last week the grower received a total of $38,700 from the crates of oranges and grapefruit shipped.
A citrus fruit grower receives $15 for each crate of oranges shipped and $18 for each crate of grapefruit shipped. How many crates of oranges did the grower ship last week?
Let \(x\) be the # of oranges and \(y\) the # of grapefruits. Note that \(x\) and \(y\) must be an integers. Q: \(x=?\)
(1) Last week the number of crates of oranges that the grower shipped was 20 more than twice the number of crates of grapefruit shipped --> \(x=2y+20\). Not sufficient to calculate \(x\)
(2) Last week the grower received a total of $38,700 from the crates of oranges and grapefruit shipped --> \(15x+18y=38700\) --> \(5x+6y=12900\). Multiple values are possible, for istance: \(x=180\) and \(y=2000\) OR \(x=60\) and \(y=2100\).
(1)+(2) Two unknowns, two different linear equations --> We can calculate unique value of \(x\). Sufficient.
Answer: C.
Hi alphabeta1234,
I agree with narangvaibhav. But how does x = 1 and y = 3 satisfy 3x + 4y = 13? Just thought of pointing this out.
It actually depends on the kind of question. The expression 3x + 4y = 19 can be easily converted into x = (19 - 4y)/3.
So, you see you have only four values of y that we need to check. Hence, here substitution is the best way forward. You will rarely find problems that would be very difficult to deduce.
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