Is the answer B ?yvonne12 wrote:Marta bought several pencils, if each pencil was either a 23 cent pencil or a 21 cent pencil, how many 23 cent pencils did Marta buy?
1. Marta bought a total of 6 pencils
2. The total value of the pencils Marta bought was 130 cents
please explain how you would approack this question
jayhawk2001 wrote:Is the answer B ?yvonne12 wrote:Marta bought several pencils, if each pencil was either a 23 cent pencil or a 21 cent pencil, how many 23 cent pencils did Marta buy?
1. Marta bought a total of 6 pencils
2. The total value of the pencils Marta bought was 130 cents
please explain how you would approack this question
1 - insufficient. Out of the 6, any number of them can be 21 or 23.
2 - Sufficient. You can write down multiples of 23 and 21
alongside and see if there's more than 1 pair that adds up to 130
23 -- 23, 46, 69, 92, 115
21 -- 21, 42, 63, 84, 105, 126
There's only 1 unique value of 23c and 21c pencils that
add up to 130 i.e. 46 + 84. Hence sufficient.
your heart's in the right place, but the issue here actually has nothing to do with prime numbers -- this is a fundamental issue with all of these problems, in which your answer is restricted to whole numbers.logitech wrote:Once again , if you see PRIME numbers in these kind of DS questions , beware of C trap! You might need only either A or B!
B winsyvonne12 wrote:Marta bought several pencils, if each pencil was either a 23 cent pencil or a 21 cent pencil, how many 23 cent pencils did Marta buy?
1. Marta bought a total of 6 pencils
2. The total value of the pencils Marta bought was 130 cents
please explain how you would approack this question
There are always shortcuts in these questions that can save a ton of time. In the question above, the prices of the pencils are very close in value, which suggests that if we know the total cost, we'll be able to determine the total number of pencils. If Marta buys 5 pencils, the total cost is at most 5*0.23 = $1.15, and if she buys 7 pencils, the total cost is at least 7*0.21 = $1.47. So if she spent $1.30, she bought six pencils. If we know the total cost and the number of pencils, there can clearly be only one value for the number of 23 cent pencils, since the more 23 cent pencils she buys, the more she'll spend (alternatively, you have two distinct linear equations in two unknowns, so we can solve). So we don't actually need to test a single number here to see that the answer is B.lunarpower wrote: if you have a linear equation(s) to which your answer must be in WHOLE NUMBERS, then the only reliable way to solve the equation is to TEST VALUES.
yeah, i thought about this sort of thing. and, yes, you are certainly correct -- with this reasoning, you can solve the whole problem without firing a single shot, so to speak.Ian Stewart wrote:There are always shortcuts in these questions that can save a ton of time. In the question above, the prices of the pencils are very close in value, which suggests that if we know the total cost, we'll be able to determine the total number of pencils. If Marta buys 5 pencils, the total cost is at most 5*0.23 = $1.15, and if she buys 7 pencils, the total cost is at least 7*0.21 = $1.47. So if she spent $1.30, she bought six pencils. If we know the total cost and the number of pencils, there can clearly be only one value for the number of 23 cent pencils, since the more 23 cent pencils she buys, the more she'll spend (alternatively, you have two distinct linear equations in two unknowns, so we can solve). So we don't actually need to test a single number here to see that the answer is B.lunarpower wrote: if you have a linear equation(s) to which your answer must be in WHOLE NUMBERS, then the only reliable way to solve the equation is to TEST VALUES.
I think you're just explaining why your 5-cent/7-cent example couldn't appear on the GMAT. The test will never (or almost never) require you to test six or more different sets of numbers for a question, so there needs to be a shortcut available. On a question like this, one in which you get a linear equation with positive integer unknowns which has only one solution, there are only two shortcuts that could, in theory, be available; the one I described here and the one (using divisibility) that I described in the post I linked to.lunarpower wrote:in fact, just to underscore the dangers of recommending such shortcuts -- the above shortcut doesn't even cut it on my two previously cited examples:
I certainly agree that there's no value in learning shortcuts which are unique to certain problem setups. If there's no useful general takeaway that might be applied to a variety of future problems, then there's no reason to bother learning a shortcut. There are even entire questions in the OG that I think are pointless to study since their setup is so idiosyncratic, the test taker is simply never going to see something similar again - the question late in the DS section about Carl's credit card balance is one example.lunarpower wrote: 1) these shortcuts are almost never going to be consistent from problem to problem, while the general recommendation (plug in numbers) is highly generalizable.
remember that the student is never going to see this problem again, and so we should judge the different methods on the probability that the student could actually be able to use them again. (i.e., students already have tons of stuff to study, so i don't want to emphasize the "low-percentage shots")
I've worked with a lot of people who are capable of at least sometimes seeing and applying shortcuts on questions like this one. The methods that should be 'put forward' really depend on the capabilities of the student. While I agree a certain mathematical sophistication is required to recognize when to use shortcuts, that sophistication certainly is not beyond the abilities of a 40-51 scoring test taker. Of course, the more time-consuming alternative methods - number plugging - should also be offered for those who won't see, under time pressure, a faster method; even a high level test taker won't see shortcuts much of the time. But if a test taker can spot a quick shortcut on even two or three questions during the test, he or she can save so much time that pacing may no longer be an issue. It's for that reason that I think these kinds of approaches are worthwhile for the higher level test taker - I teach methods like these not with the expectation that the test taker will see some super-fast technique for all 37 questions on the test and finish the Quant section in fifteen minutes, but rather with the goal of saving them 4-6 minutes on 2-3 questions somewhere in the test. That is, it's specifically because of the time pressure on the test that these shortcuts can be invaluable.lunarpower wrote: your post seems to have a certain subtext that students should preferentially use methods like this one. for those blessed students to whom this sort of intuition comes naturally, i would agree, but, since 100% of students should be able to understand and execute the number plugging -- and because number plugging will be universally applicable on these problems -- that's the method we should put forward as #1.
