Let x be the no. of 7's and y be the no. of 77's

x + y = n (n=Total numbers adding up to 350). We are asked to find the value of n.

350 = 7x + 77y

350 - 77y = 7x

Here, y should be less compared to x to satisfy the equation. Try plugging in y values,

y=1,
350 - 77 = 7x
273 = 7x
x = 39

Sub. x and y in this eqn,
x + y = n
39 + 1 = 40 (Choice C).

At this moment, you can stop and check this answer. If you wanna test other choices, you could do so.

hpgmat wrote:if each term in the sum a1+a2+....+an is either 7 or 77 and the sum equalls 350 , which of the following could be n?

38
39
40
41
42

I can think of a couple of approaches here: either backsolving plus reasoning OR pure reasoning.

Approach 1--Backsolving + Reasoning:

Reasoning part: We know there are a certain number of 7s and 77s that sum to 350. The question is asking for the number of 7s and 77s (whatever their ratio). If they were all 7s, you would have...50 7s (because 50*7=350). Because the answer choices run from 38-42, we know we have a lot of 7s, and that they can`t all be 7s.

Backsolving part: Start with answer choice C because it is clearly the easiest one to evaluate. 40*7=280. But, remember, they can`t all be 7s (because we would need 50 7s then). So if 39 of these 40 were 7s, we would have...one less 7 or 280-7=273. The other one would be 77 and 350-273 just happens to equal 77. Therefore, we have 39 7s and one 77. Done.

Approach 2--Pure Reasoning (and this is the approach I, personally, would have employed):

If 7 multiplied by a number equals another number whose units digits is zero, then the number being multiplied by 7 must be a multiple of 10. In other words, if 7*x= a number ending in zero, then x is a multiple of ten. Both 7 and 77 have 7 as their units digits. For a certain number of them to sum to 350, the number of them must be a multiple of ten. There is only one multiple of ten in the answer choices...choice C

I think a pure algebraic approach such as the one papgust adopted while technically correct is far more time-consuming than either of the approaches I outlined above (especially the second approach). Remember, GMAT wants to reward critical reasoning. It's not just about getting to the correct answer but also about how fast you get to the correct answer. Therefore, when reviewing, make sure you also review the questions you answered correctly and ask yourself "is there a quicker way I could have gotten to the correct answer? Could I have employed some reasoning or commonsense? Could I have used the answer choices or picked numbers?" Etc.

Thank you , but why not

Y= 3 and X =17 , therefore n=20

3(77)+17(7)=350

hpgmat wrote:Thank you , but why not

Y= 3 and X =17 , therefore n=20

3(77)+17(7)=350

there can be various combination of no. to get 350

but we need to get a pair which is given in option apart from C none other leads to 350

hpgmat wrote:Thank you , but why not

Y= 3 and X =17 , therefore n=20

3(77)+17(7)=350

Yes, n (n being the number of 7s and 77s in the set) COULD equal 20. I never said that n MUST equal 40. Instead, under my reasoning approach, we deduced that n MUST be a multiple of 10.

It is very important to note that the question stem said "could" and 20 does not present an answer choice. If both 20 and 40 presented as answer choices, then there would be 2 correct answers. The point is that among the five answer choices listed, 40 is the only value that n COULD take. It is entirely consistent with the wording of the question that there are other values not listed among the answer choices that n COULD also take. (And, under my reasoning approach, had 20 presented as an answer choice (and 40 did not), we would still be able to select the correct answer because 20 is a multiple of 10.)

Testluv wrote:

hpgmat wrote:Thank you , but why not

Y= 3 and X =17 , therefore n=20

3(77)+17(7)=350

Yes, n (n being the number of 7s and 77s in the set) COULD equal 20. I never said that n MUST equal 40. Instead, under my reasoning approach, we deduced that n MUST be a multiple of 10.

It is very important to note that the question stem said "could" and 20 does not present an answer choice. If both 20 and 40 presented as answer choices, then there would be 2 correct answers. The point is that among the five answer choices listed, 40 is the only value that n COULD take. It is entirely consistent with the wording of the question that there are other values not listed among the answer choices that n COULD also take. (And, under my reasoning approach, had 20 presented as an answer choice (and 40 did not), we would still be able to select the correct answer because 20 is a multiple of 10.)

I should have also mentioned that among the answer choices, there will be only one value that n can take. In other words, n can be the number in the correct answer and can't be any of the numbers in the incorrect answers. This is why, under any approach, the moment you find an answer choice that works ( ie, a value that n could be), you are done--absolutely no need to check the other four answer choices because there can be one and only one correct answer.

That's why in my first post, under both approaches, as soon as we found that n could be 40 in answer choice C, we were done. You can take this as a a general rule:

In PS, whenever the question says "could" or "possible value" as soon as you are confident that a certain answer choice "works", you are DONE.

Thank you so much . That clarifies