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

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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?

(1) Martha bought a total of 6 pencils

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

For more on this type of questions check:
eunice-sold-several-cakes-if-each-cake-sold-for-either-109602.html
martha-bought-several-pencils-if-each-pencil-was-either-a-100204.html
a-rental-car-agency-purchases-fleet-vehicles-in-two-sizes-a-105682.html
joe-bought-only-twenty-cent-stamps-and-thirty-cent-stamps-106212.html
a-certain-fruit-stand-sold-apples-for-0-70-each-and-bananas-101966.html
joanna-bought-only-0-15-stamps-and-0-29-stamps-how-many-101743.html
at-an-amusement-park-tom-bought-a-number-of-red-tokens-and-126814.html
collections-confused-need-a-help-81062.html

Hope it helps.
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amitjash wrote:
Does GMAT ask trial and error questions?

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

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General Discussion
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Does GMAT ask trial and error questions?
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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)

3x+1y=10
x+y=6

Hope this helps.

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

As for your question, see the solution for another similar question (discussed here: car-dealer-data-sufficiency-105682.html?hilit=diophantine#p826606):

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.

Similar problems: gmat-prep2-92785.html?hilit=linear%20type

Hope it helps.
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shash wrote:
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.

Check out this blog post where I have discussed in detail how to solve such problems:
http://gmatquant.blogspot.com/2010/11/integral-solutions-of-ax-by-c.html
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Martha bought several pencils. If each pencil was either a  [#permalink]

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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?

(1) Martha bought a total of 6 pencils

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

You can also look at this as follows: if she bought 7 pencils, she would spend at least 7*0.21 = $1.49. If she bought 5 pencils, she would spend at most 5*0.23 =$1.15. So if she spent $1.30, as we learn from Statement 2 alone, she must have bought exactly 6 pencils. So Statement 2 is sufficient alone (once we know that she bought 6 pencils, there can only be one combination of$0.23 and $0.21 pencils that give a total cost of$1.30, since the more \$0.23 pencils she buys, the greater the cost will be).
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Re: Martha bought several pencils. If each pencil was either a  [#permalink]

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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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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.

For more on this type of questions check:
eunice-sold-several-cakes-if-each-cake-sold-for-either-109602.html
martha-bought-several-pencils-if-each-pencil-was-either-a-100204.html
a-rental-car-agency-purchases-fleet-vehicles-in-two-sizes-a-105682.html
joe-bought-only-twenty-cent-stamps-and-thirty-cent-stamps-106212.html
a-certain-fruit-stand-sold-apples-for-0-70-each-and-bananas-101966.html
joanna-bought-only-0-15-stamps-and-0-29-stamps-how-many-101743.html
at-an-amusement-park-tom-bought-a-number-of-red-tokens-and-126814.html
collections-confused-need-a-help-81062.html

Hope it helps.

Hi Bunuel,

Interesting question. I landed on this after solving this a-total-of-60-000-was-invested-for-one-year-part-of-this-126655.html#p1035647:

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

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russ9 wrote:
Hi Bunuel,

Interesting question. I landed on this after solving this a-total-of-60-000-was-invested-for-one-year-part-of-this-126655.html#p1035647:

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.

Here is a post on integral solutions: http://www.veritasprep.com/blog/2011/06 ... -of-thumb/
Look at question 2.
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Re: Martha bought several pencils. If each pencil was either a  [#permalink]

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I am approaching the problem from divisibility angle.

Since 23x+21y=130 and 130 is not divisible by 23 or 21. Martha can not only buy one type of pencil.
Now, If she buys both I need to know how many of 23 and 21 number pencil she bought. i.e. 23+21 = 44 => 130/44 = 2*44 + 42. So she can buy 2 -2 pencil of each number.

Now, reminder 42 is divisible by only 21 i.e. 21*2. so she bought additional two 21 number pencil.

Net we found the values that satisfies the conditions. the statement is sufficient - The answer is B.
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Re: Martha bought several pencils. If each pencil was either a  [#permalink]

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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,  says that x+y = 6, so x = 6-y and  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?
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Martha bought several pencils. If each pencil was either a  [#permalink]

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pacifist85 wrote:
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,  says that x+y = 6, so x = 6-y and  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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Re: Martha bought several pencils. If each pencil was either a  [#permalink]

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pacifist85 wrote:
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,  says that x+y = 6, so x = 6-y and  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.
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Re: Martha bought several pencils. If each pencil was either a  [#permalink]

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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?

(1) Martha bought a total of 6 pencils

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

Target question: How many 23 cent pencils did Martha buy?

Given: Martha bought several pencils. Each pencil was either a 23-cent pencil or a 21-cent pencil.

Statement 1: Martha bought a total of 6 pencils
Clearly, statement 1 is NOT SUFFICIENT

Statement 2: The total value of the pencils Martha bought was 130 cents.
This statement APPEARS to be insufficient. However, since the pencils cost 21 cents and 23 cents, we can't buy many for 130 cents.
So, let's LIST THE POSSIBLE CASES

Case a: Martha bought ZERO 23-cent pencils (for a total of 0 cents).
This means she spent all 130 cents on the 21-cent pencils. However, 21 cents does not evenly divide into 130 cents
So, it cannot be the case that Martha bought ZERO 23-cent pencils

Case b: Martha bought ONE 23-cent pencil.
This means she spent the remaining 107 cents on the 21-cent pencils. However, 21 cents DOES NOT evenly divide into 107 cents.
So, it cannot be the case that Martha bought ONE 23-cent pencil

Case c: Martha bought TWO 23-cent pencils (for a total of 46 cents).
This means she spent the remaining 84 cents on the 21-cent pencils. 21 cents divides into 84 cents exactly FOUR TIMES.
So, it's possible that Martha bought TWO 23-cent pencils and FOUR 21-cent pencils
In this case, the answer to the target question is Martha bought TWO 23-cent pencils

Case d: Martha bought THREE 23-cent pencils (for a total of 69 cents).
This means she spent the remaining 61 cents on the 21-cent pencils. However, 21 cents DOES NOT evenly divide into 61 cents.
So, it cannot be the case that Martha bought THREE 23-cent pencils

Case e: Martha bought FOUR 23-cent pencils (for a total of 92 cents).
This means she spent the remaining 38 cents on the 21-cent pencils. However, 21 cents DOES NOT evenly divide into 38 cents.
So, it cannot be the case that Martha bought FOUR 23-cent pencils

Case f: Martha bought FIVE 23-cent pencils (for a total of 15 cents).
This means she spent the remaining 15 cents on the 21-cent pencils. However, 21 cents DOES NOT evenly divide into 15 cents.
So, it cannot be the case that Martha bought FIVE 23-cent pencils

Now that we've examined all of the cases, the only possible solution is the one described in case c
So, it MUST be the case that Martha bought TWO 23-cent pencils
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Cheers,
Brent
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