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Martha bought several pencils. If each pencil was either a 23 cent pencil or a 21 pencil, how many 23 cent pencils did martha buy?

a) Martha bought a total of 6 pencils

b) The total value of the pencils Martha bought was 130 cents.

This is C-trap question. C-trap question is the question which is obviously sufficient if we take statements together. When we see such questions we should become very suspicious.

Martha bought several pencils. if each pencil was either 23 cents pencil or a 21 cents pencil. How many 23 cents pencils did Marta buy?

Let \(x\) be the # of 23 cent pencils and \(y\) be the # of 21 cent pencils. Note that \(x\) and \(y\) must be an integers. Q: \(x=?\)

(1) Martha bought a total of 6 pencils --> \(x+y=6\). Clearly not sufficient.

(2) The total value of the pencils Martha bought was 130 cents --> \(23x+21y=130\). Now x and y must be an integers (as they represent the # of pencils). The only integer solution for \(23x+21y=130\) is when \(x=2\) and \(y=4\) (trial and errors). Sufficient.

Yes, but only if they the range of possible numbers is small.. as is the case here.

One good way to attack the statements in DS, is to prove insufficiency by getting two different values which meet the preimposed conditions

e.g. in this question

Quote:

Statement 1 x+y = 6

we have two that are possible ..x=1,y=5 x=2,y=4 ... Therefore insufficient

Quote:

Statement 2 23x+21y = 130

a. to get an even sum(130), either both x and y have to be even or both have to be odd b. to get units digit of 0, 3x + y should have units digit of 0

--> take possible values of x from 1 to 5.. x=1 --> 3 .. y has to be odd x=2 --> 6 .. y has to be even x=3 --> 9 .. y has to be odd x=4 --> 12 .. y has to be even x=5 --> 15 .. y has to be odd

the only value that fits that for x =2,y=4; On substituting in the equation 23x+21y, we get 130 --> Statement 2 alone is sufficient coz we get only one possible value for x and y

Hope this helps
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The way I thought about this one that probably will take less time than trying to actually write out formulas, is;

1) This doesn't tell us anything about the total amount spent on the pencils so it could be 5 of 23 cent pencil and 1 of the 21 cent pencils, or 4 of the 23 cent pencils and 2 of the 21 cent pencils etc... These have different totals so 1) is sufficient.

2) 21 and 23 cents are fairly large portions of the 130 total cents spent and 23 is a prime number so I concluded that there would only be one solution to this and we could figure out how much she spend on each. This by itself would be enough to conclude B, 2) is sufficient, but to validate I used the first digits of 23 and 21, which according to 2) must add to a multiple of ten. If you write out the two equations you can deduce from this it because readily evident that you can figure out this problem with just 2)

A rental car agency purchases fleet vehicles in two sizes: a full-size car costs $10,000, and a compact costs $9,000. How many compact cars does the agency own? (1) The agency owns 7 total cars. (2) The agency paid $66,000 for its cars.

This is classic C-trap question. C-trap questions are the questions which are obviously sufficient if we take statements together. When we see such questions we should become very suspicious.

Let # of full-size car be F and # of compact cars be C. Question: C=?

(1) The agency owns 7 total cars --> F+C=7. Clearly insufficient to get C.

(2) The agency paid $66,000 for its cars --> 10,000F+9,000C=66,000 --> 10F+9C=66. Here comes the trap: generally such kind of linear equations (ax+by=c) have infinitely many solutions for x and y, and we can not get single numerical values for the variables. But since F and C represent # of cars then they must be non-negative integers and in this case 10F+9C=66 is no longer simple linear equation it's 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 (F and C in out case) possible to satisfy it. When you encounter such kind of problems you must always check by trial and error whether it's the case.

Now, it's quite easy to check whether 10F+9C=66 has one or more solutions. 9C=66-10F so 66 minus multiple of 10 must be multiple of 9: 66 is not multiple of 9; 56 is not; 46 is not; 36 IS MULTIPLE OF 9 (F=3 and C=4); 26 is not; 16 is not and 6 is not. So only one combination of F and C satisfies equation 10F+9C=66, namely F=3 and C=4. Sufficient.

How can this problem be solved without 2 equations. How do you in fact spot such cases?

To solve for two variables, you need two equations. But if there are constraints on the solutions (e.g. x and y should be positive integers), sometimes, one equation is enough. When you have real world examples, where they talk about number of pens, pencils etc which cannot be negative or a fraction, you need to ascertain whether one equation is enough.

Re: Martha bought several pencils. If each pencil was either a [#permalink]

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04 Jul 2013, 01:02

1

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Is this question below 600 difficulty level? There are complex questions, which test similar concept and can take a while to solve....
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I am not sure if I am missing something, but it sounds a bit silly to just try out numbers. Of course it is logical that a combination could work, but having to look for it is not really reasonable - even if it is not hard.

Apart from this, I have a question. Let's say that it wasn't possible to find the needed combination for the equation in B. Could we use A and B together to figure it out?

So, [1] says that x+y = 6, so x = 6-y and [2] says that 23x+24y = 130 and then replace x with 6-y. Because in this way you indeed result in x=2 and y=4.

For this reason, I chose C, as it sounded more reasonable to use them in combination that to have to find a random combination or numbers. Any thoughts?

As it is very obvious that C will give you 2 equations and 2 variables, it is thus called a "C-trap" question in which people mark C without realising that 1 equation with 2 equations can be solved for both the variables as the additional condition will come from the fact that the number of pencils MUST be an integer. You can not have 0.3 or 1.5 or 3.7 pencils.

Usually, you need 2 equations for uniquely obtaining values for 2 variables, but in some cases (such as the one above) 1 equation can provide you sufficient information on its own.

As for your question, yes, if there were no integer values (you do not need to go on for hours to see how many values satisfy the given equation such that x, y are both integers.) then you could have used both the equations and come up with C as the correct answer as you have 2 equations and 2 variables.

For the given question,

23x+24y = 130 ---> x = (130 -24y) / (23) , the most you will have to check is for y = 0 to 5 as x MUST be \(\geq\)0 (24 X 6 = 144 > 130) and (130-24y) should be a multiple of 23
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I am not sure if I am missing something, but it sounds a bit silly to just try out numbers. Of course it is logical that a combination could work, but having to look for it is not really reasonable - even if it is not hard.

Apart from this, I have a question. Let's say that it wasn't possible to find the needed combination for the equation in B. Could we use A and B together to figure it out?

So, [1] says that x+y = 6, so x = 6-y and [2] says that 23x+24y = 130 and then replace x with 6-y. Because in this way you indeed result in x=2 and y=4.

For this reason, I chose C, as it sounded more reasonable to use them in combination that to have to find a random combination or numbers. Any thoughts?

There is logic associated with this and you do not need to just try random numbers. That is why the answer here is not (C). The link that I have given in my post above discusses the logic.
_________________

Yes, but only if they the range of possible numbers is small.. as is the case here.

One good way to attack the statements in DS, is to prove insufficiency by getting two different values which meet the preimposed conditions

e.g. in this question

Quote:

Statement 1 x+y = 6

we have two that are possible ..x=1,y=5 x=2,y=4 ... Therefore insufficient

Quote:

Statement 2 23x+21y = 130

a. to get an even sum(130), either both x and y have to be even or both have to be odd b. to get units digit of 0, 3x + y should have units digit of 0

--> take possible values of x from 1 to 5.. x=1 --> 3 .. y has to be odd x=2 --> 6 .. y has to be even x=3 --> 9 .. y has to be odd x=4 --> 12 .. y has to be even x=5 --> 15 .. y has to be odd

the only value that fits that for x =2,y=4; On substituting in the equation 23x+21y, we get 130 --> Statement 2 alone is sufficient coz we get only one possible value for x and y

Hope this helps

Yes, but the odd + odd or even + even formula is just to check if the equation has te aptitud to deliver a result that is even, what they are really asking is what is the number or amount of each unit that was bought in order to get that result. and that should be accomplished by a formula and not by trial and error, since in that case there are many other problems in the gmat that could be solve by trial and error, thus, leaving space to get the answers right without the need of knowing the formulas that the gmat want or should be testing. In this case despite the fact that can be solve that way, the gmat team got confused.

Re: Martha bought several pencils. If each pencil was either a [#permalink]

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19 Aug 2012, 10:56

I see a nice discussion on this topic and would like to add my bits here. Trying to find a general method of solving such questions where the constraints are 1) linear equation in two variables 2) the solution is only non negative integers . Just by combining the ways used by community members for this particular question,I can say a shortcut method for such questions can be

1) express in form of a linear equation 21x + 23y = 130 . . find y

2) find max value of required variables y = 130/23 . therefore y< 6 and x <7

3) plug in values of y , 1,2,3,4,5.

4) calculate only units digit . .y = 1 => 23y ( units digit is 3) 2 (6) 3(9) 4 (2) 5(5)

5) check if an integer value of x does exist for the set of values of y, by plugging in values and calculating only units digit, the solution should follow the required constraint i.e. x<7

for y=1 not possible (units digit of 21x need to be 7) for y = 2 units digit of 21x should be 4 possible and thats one solution for y = 3 21x to end in 1 possible but not a solution because does not follow 21x + 23y = 130 ( no need to calculate can rule out as ans being to small) for y = 4 21x should end in 8 , not possible for y = 5 21x should end in 5 possible but does not follow 21x + 23y = 130 ( again no calculation required can rule out as ans being too large)

can you please elaborate this sarathy : b. to get units digit of 0, 3x + y should have units digit of 0. . your equation is 23x + 21y = 130 . . this is along same lines to calculate units digit only but i guess we cannot make an equation like this because 4x + 2y can also have units digit of 0 and thats infact our solution to this problem. .
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Re: Martha bought several pencils. If each pencil was either a [#permalink]

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19 Aug 2012, 11:03

An addition to above procedure by referring to Kasishma's link . . after finding one unique solution in case the coefficients a and b in the equation ax + by = z are too small than z, then we can take lcm of a and b and can find more solutions by plugging in numbers for x and y in steps of lcm(a,b)/a or lcm(a,b)/b . . for b and a respectively. before doing this do ensure to reduce equation such that a and b are co-prime and also that the equation is in form of ax + by = z and not ax-by = z, that form will have infinite many solutions. . . sincere apologies to form an abstract summary of a great article by respected Karishma maam
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Re: Martha bought several pencils. If each pencil was either a [#permalink]

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16 Aug 2014, 10:49

Bunuel wrote:

Bull78 wrote:

Martha bought several pencils. If each pencil was either a 23 cent pencil or a 21 pencil, how many 23 cent pencils did martha buy?

a) Martha bought a total of 6 pencils

b) The total value of the pencils Martha bought was 130 cents.

This is C-trap question. C-trap question is the question which is obviously sufficient if we take statements together. When we see such questions we should become very suspicious.

Martha bought several pencils. if each pencil was either 23 cents pencil or a 21 cents pencil. How many 23 cents pencils did Marta buy?

Let \(x\) be the # of 23 cent pencils and \(y\) be the # of 21 cent pencils. Note that \(x\) and \(y\) must be an integers. Q: \(x=?\)

(1) Martha bought a total of 6 pencils --> \(x+y=6\). Clearly not sufficient.

(2) The total value of the pencils Martha bought was 130 cents --> \(23x+21y=130\). Now x and y must be an integers (as they represent the # of pencils). The only integer solution for \(23x+21y=130\) is when \(x=2\) and \(y=4\) (trial and errors). Sufficient.

Question for you -- I totally fell for the CTrap. How would I have know that this was a C trap vs. the investment question, which wasn't a C trap. They are worded almost identically?

Question for you -- I totally fell for the CTrap. How would I have know that this was a C trap vs. the investment question, which wasn't a C trap. They are worded almost identically?

Though the question is addressed to Bunuel, I would like to add my two cents here, hoping Bunuel will not mind!

The question in this thread is based on the concept of "integral solutions of an equation in two variables". In these equations, it seems that you need more information, but actually you can often get a unique solution due to constraints on values that x and y can take. Such questions are often based on real life situations involving number of objects which cannot be a fraction. Hence the test maker does not need to specifically write that x and y must be non negative/positive integers but you need to logically assume it. Whenever you get such questions where you need to make an equation in two variables based on physical objects, you need to be alert of the possibility of C-Traps. In the investment question, the amount invested/rate of interest needn't be integral and hence multiple values are possible.

Obviously, there are many other questions where people could fall for C-Trap but be aware of integral solutions questions.

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