Martha bought several pencils. If each pencil was either a

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


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


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

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

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:
https://anaprep.com/algebra-integer-sol ... variables/

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).

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. .

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.

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.

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

Answer: B

Cheers,
Brent

Bunuel VeritasKarishma

Hi, I read the below on another similar c trap question:



As per this, statement 2 would be enough to give a unique answer as it satisfies all three conditions. Would you say this approach is correct and I can use it for all similar questions?

Kritisood

Usually such instruction are given by a particular GC handle but I won't name that

I won't suggest making the approach such mechanical. We don't get extra credit to save time so use all 62 minutes to answer 31 questions.

Although to answer your question, yes these instructions are CORRECT.

Hi Bunuel
Nice Explantion Again.

I did the trial and error which you mentioned for st 2, but it took me 3 mins 2 secs.
So how can i work it with speed?
Than you.

Posted from my mobile device

Check out how to solve questions of this type (link given below). Understand the logic. Even when C > LCM of A and B, sometimes you may have a single solution only. Plus, how many such conditions will you learn for how many distinct cases?

https://anaprep.com/algebra-integer-sol ... variables/

Try 3p + 5e = 16

I don't remember the exact source but i learnt the way to solve these questions and it is relatively easy once we know a few basic things. The following equation Ax + By = C is called a diophantine Equation.

The first thing to know is how many solutions are possible i.e how many combination of number of 23 cents pencils and 21 cent pencils are possible.

How do we figure that out?

Ax + By = C
If A*B > C (There is at most 1 solution)
If A*B < C (There is at least 1 solution)
To know how many solutions are possible we need to know by how many times can we divide C from A*B when A*B <C Eg. (Digressing from the question) Lets assume C is 25 and A is 2 and B is 3. Now A*B = 6. If we divide C by A*B, the number is between 4 and 5 (4.16). Hence the number of solution will be either 4 or 5.

Now to solve the question given
Statement 1. Does not tell us the total cost which may lead to multiple values for x and y. Not sufficient

Statement 2:The total value of the pencils Martha bought was 130 cents.
From the information given above
23x + 21y = 130 (This is a diophantine equation with A = 23 and B =21 and C = 130)

Looking at the question we realize that A*B >130 (This would mean there is at most 1 solution i.e could be 0 or 1. Sneakily if we use the 1st equation we know its not 0. However we do not need to officially use that )

Hence answer is B

Kudos if this helps please

P.S: Found the source. Its by Anthony Ritz. Here is his video https://www.youtube.com/watch?v=SCf-HrtYUMg