For each month of a given year except December,

[shohet.mark]

For each month of a given year except December, a worker earned the same monthly salary and donated one-tenth of that salary to charity. In December, the worker earned N times his usual monthly salary and donated one-fifth of his earnings to charity. If the worker's charitable contributions totaled one-eighth of his earnings for the entire year, what is the value of N?


8/5

5/2

3

11/3

4


The answer says that this can be done algebraically or by picking numbers. I understand both approaches, but I don't understand why picking a number for S (monthly salary) is okay, because I was always under the impression that if the answer choices contain a definite number rather than a ratio or variables, numbers cannot be picked. Am I wrong about this--what makes it okay to pick a number for S? Thank you.

[RonPurewal]

If it's not possible to solve for a quantity, then the problem MUST work for arbitrary values of that quantity. (Otherwise, the problem could well be impossible to solve!)

If you look over this problem for a bit, you'll note that there is no indication whatsoever that it's possible to solve for the salary itself. So, just pick a number for it.

[RonPurewal]

Further thoughts:

- If it is possible to solve for the salary"”but you pick a random value instead"”then you'll probably just get an answer that doesn't agree with any of the choices. In that case, there's no damage done except the loss of a small amount of time.

"- More importantly"”The GMAT doesn't have "trick" questions.
If a multiple-choice problem doesn't mention solving for a particular quantity, then you can be pretty sure that you won't have to"”and probably can't"”solve for that quantity.

[RonPurewal]

Finally"”

If you are confused by why it's not necessary to solve for the salary here, then it's worth your time to sit for a moment and think about WHY you don't need to (and, in fact, can't) solve for the salary.

Here's where I'm going with this:
- In this problem, everything is just a fraction of a larger (monthly or annual) amount.
"- Fractions are relative. They grow along with the absolute quantities.
- So, if you are only thinking about fractions"”i.e., there's no mention of absolute numbers in the problem"”then those absolute sizes are irrelevant.

This is actually the whole point of using fractions/ratios/percentages"”to MAKE absolute numbers irrelevant.
That's a hugely important principle.
Even more so on data sufficiency, which is primarily a test of focus. DS is only a "math test" in the superficial sense; in reality, it's more of a "solve for THIS, not THAT"”are you sure you're actually solving the right problem, sir/ma'am?" test.
I.e., if a DS problem asks you for a fraction/ratio/percentage and you try to solve for absolute numbers, you're going to be wrong pretty much every single time.

[shohet.mark]

Thanks Ron; this is helpful. Using your explanation, I think I can prove to myself in two ways that the value of S is irrelevant:

1) If you plug in S=10, the end of the calculation yields 3NS = 11S and the S's just cancel out, yielding N = 11/3. So S does not matter.

2) If you plug the N= 11/3 back into the formula (11)(S/10) + (NS/5) = (1/8)(11S + NS), you end up with 11S = 11S. So S does not matter.

So two questions:

1) Does the above "proof" make sense, and if so, how do I identify this (that a variable is irrelevant and can be randomly assigned) on the exam ahead of time without calculating it out?

2) You say "If a multiple-choice problem doesn't mention solving for a particular quantity, then you can be pretty sure that you won't have to"”and probably can't"”solve for that quantity."

Does this generally mean that if ever given a variable in a problem that does not appear in the answer choices, I can always just assign it a random value (i.e. in this case, S is mentioned but the problem asks for N, so I can just assign S a random value)? I believe this explains why we can usually just assigned Distance = 100 to rate problems?

[RonPurewal]

1) If you plug in S=10, the end of the calculation yields 3NS = 11S and the S's just cancel out, yielding N = 11/3. So S does not matter.

I'm not quite following this. If you plug in a numerical value for S, you can't still have the letter S present in the equation.
Where S would have appeared, there should just be numbers.

Are you trying to say that, by seeing "30" and "110", you realize that the S drops out?

[RonPurewal]

1) Does the above "proof" make sense, and if so, how do I identify this (that a variable is irrelevant and can be randomly assigned) on the exam ahead of time without calculating it out?

This is mostly a matter of developing a relevant intuition.

More importantly, though, you're overvaluing the whole "without calculating it" thing.
How long did it take you to do the above figuring? like 10-15 seconds?

Don't be afraid to pick up the proverbial shovel and dig.

[RonPurewal]

Does this generally mean that if ever given a variable in a problem that does not appear in the answer choices, I can always just assign it a random value (i.e. in this case, S is mentioned but the problem asks for N, so I can just assign S a random value)? I believe this explains why we can usually just assigned Distance = 100 to rate problems?

"Always" is a dangerous word. If you have two quantities that depend on each other, then you may not be able to assign one of them a random value.
For instance, in OG13 multiple-choice #167, Jack's and Bill's ages are tied together. The problem can only ask for one of them"”mostly just since a multiple-choice problem can't ask for two different things"”but, in that problem, you can't randomly pick the other one.

As I emphasized above, the whole point of fractions/percentages/proportions is to make absolute sizes irrelevant. So, in such problems, your being able to pick the "whole" amount is a pretty safe bet.
In other situations, though, you should just "pick up the shovel and dig" if you are not 100.0000000% sure that a given shortcut will work.

[beakdas]

In reference to the question posted above my solution was based on picking the options directly & backsolving to see if the premise in the question is satiated.


Now, as a rule of thumb ( or maybe not) I saw that my options were arranged in an ascending order & were all numerical in nature. I picked option 'C' as it is the middle value in terms of magnitude & placement.

On using N=3 & taking Salary = $10, my total earnings were coming out to 17.5 whereas my charitable contribution was 17.

Since, 'C' gave me a value > 17 I assumed values after C would give me an even larger answer ( since data in choices is arranged from smallest-largest).But, in doing so I ended up picking 'B' & landed up with an incorrect answer.


My question here is does the method of picking 'C' work to gauge the nature of answer only when the choices are Integral & not fractional/decimal.

Regards,
Abheek Das

[RonPurewal]

i am confused by your numbers.

i'm going to assume that $10 is a monthly figure, since all of the calculations in the problem are based on monthly figures. (if it isn't, then maybe we've found the issue right there.)

thus the worker's usual donation (in every month except december) is $1 (= one-tenth of $10).

Jan.-Nov. (11 months)
salary: 11 x $10 = $110
donation: 11 x $1 = $11

Dec. using choice C
salary: 3 x $10 = $30
donation: 1/5 x $30 = $6

total
salary: $140
donation: $17

$17 is LESS than one-eighth of $140.

the whole point of the problem, of course, is that the worker is giving relatively MORE to charity in december. so, if charitable donations are too low, then you have underestimated Dec. income.

bye-bye A, B, C.

what did you do that i didn't?

[RonPurewal]

oooooooohhhhh okay, i see, your $17.50 is $140/8.

in this case, you just got the significance of the answer backward.

you got an excessively LOW figure for charitable donations.

remember, december is the month that raises the fraction of money given to charity.
just think about it—the worker gives 1/5 that month, instead of the usual 1/10.

so, if the charity figure is too low, then you have to move the december income UP.
somehow you managed to get the bold-face thing backward.

[RonPurewal]

now, here's the question for YOU:

when you decided (incorrectly) to eliminate C/D/E,
were you...
... 100.0000000000 per cent sure?
... 99.9999999999 per cent sure or less?

i'm betting that your answer is 'i was 99.9999999999 per cent sure or less.'

if you are not good at quantifying this sort of thing, here is a(n excessively melodramatic) situation:
• someone is holding a gun to your head.
• if you DON'T eliminate these choices, the person does not shoot.
• if you eliminate these choices correctly, the person does not shoot, and you get 50 cents for your trouble.
• if you eliminate these choices incorrectly, the person shoots you dead on the spot.
do you still eliminate the choices?

if the answer is 'no' then DO NOT eliminate the choices on the real exam. just take an extra twenty seconds of your life and plug in random choices, until one of them makes the problem work.

[RonPurewal]

...and now i will address your more general questions.

I don't understand why picking a number for S (monthly salary) is okay, because I was always under the impression that if the answer choices contain a definite number rather than a ratio or variables, numbers cannot be picked.

pro tip:
if you are confused about the process (as opposed to 'i'm confused about this specific math thing'), MAKE A SUPER-EASY EXAMPLE, and then the issue should become self-explanatory.

here is a super-easy example:

X and Y are nonzero numbers.
X is 2 times Y.
Therefore, Y is __ times X.
a 1/4
b 1/2
c 1
d 2
e 4

hopefully this is entirely self-explanatory, but these are the highlights:
• yes, the correct answer is a number.
• BUT the correct answer IS NOT 'x' or 'y'.
• instead, the correct answer is a RATIO.
• it is perfectly possible to have a fixed ratio of unknown quantities. (if this were not true, then the concept of 'ratio' would not even exist.)
• so, you can pick the quantities.

now, if you generalize from this easy example, you'll understand why this problem allows 'smart numbers'.

[RonPurewal]

ok, i am officially stupid—i think i just responded to a much older post.
...but, my response (MAKE EASY EXAMPLES to understand the process better) is sufficiently important that i'll leave it up. (this is the single most effective way to iron out the kinks in ANY conceptual process. literally, any conceptual process at all, whatsoever.)

My question here is does the method of picking 'C' work to gauge the nature of answer only when the choices are Integral & not fractional/decimal.

nope.

remember—algebra CANNOT distinguish integers from non-integers.

thus, if a problem can be solved algebraically, 'integer vs. non-integer' CANNOT possibly affect the outcome of that problem.

(corollary:
if it's clear that 'integer vs. non-integer' DOES matter on some problem, then you will have to use NON-ALGEBRAIC methods to separate the integers from the non-integers.)