The ratio of the amount of Alex's fuel oil bill for the mont

Ok. I figured it out. Was having trouble with the setup of the equation.

Two ratios. 3/2 : 5/3

you are add $40 to the 3 (in 3/2); this is what i setup:
(3x+40)/2x = 5/3
multiply across ---
10x=9x+120 (subtract 9x)
x=120

Plug back into the original equation to find out the amount for january:
2(120)= $240

It depends a lot on your level, and on how comfortable you are with algebra and other more conceptual approaches (no algebra is necessary to solve the question posted here from start to finish, with no answer choices, if you understand ratios well, incidentally). If your level is average or below, so you'll be seeing a lot of easy or medium questions on a test, you'll see a few questions where testing answers is an available strategy. It's still usually more time-consuming than solving the question some other way (because for each answer choice you test, you're usually solving the whole problem once, and you often need to test two or more answers), but of course that's only true if you're able to solve the question properly. If you aren't comfortable with algebra or conceptual approaches, you need to get right answers somehow, and testing answers can be a good way to do that on lower level questions, and on those questions it's often practical to do within two minutes.

But if you're a high level test taker, and therefore will mostly be seeing harder questions on a real test, then testing answers is almost worthless as a test taking strategy. I've collected a lot of data about this, using high-level official questions. On about 2/3 of harder real PS problems, testing answers isn't even theoretically possible (you can't, for example, backsolve a probability question), and on the 1/3 of questions where it is possible, more than 95% of the time it takes anywhere from slightly longer to vastly longer than solving in some more direct way. So if you're a high-level test taker, testing answers is something you should only be thinking about as a last resort fallback strategy if you get stuck, but even then you won't often be able to use it. Of course I'm not counting questions where you're obligated to test answers -- questions that might ask "Which of the following is largest?" followed by five answer choices, say. Everyone has to test the answers on those questions. And I'm not talking about taking advantage of answer choices to bypass some steps, which can be useful on harder questions. If answers are spread out, you might be able to estimate, or you might be able to use units digits or divisibility to pick a right answer from five choices. But that's not what I'd call "testing answers" - I'm talking about inserting an answer choice back into a problem and seeing if it agrees with the given data.

In general, you should be making strategy choices with your own skills and ability level in mind. There is not one set of strategies that is best for every test taker. Incidentally, if you're evaluating study material, one of the easiest ways to identify what ability level the material is aimed at is just by seeing how much emphasis the material places on backsolving / testing answer choices. If it places a lot of emphasis on that strategy, it might still be good material for a lot of test takers, but it's not going to be much help to a Q44 test taker aiming for a Q50. A test taker at that level needs material that actually teaches the content of the test.
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A. f/j = 3/2
or f = 3j/2

(f+40)/j = 5/3
or 3f+120 = 5j
or 3 (3j/2) + 120 = 5j
or 9j + 240 = 10j
or f = 240

F/J=3/2
F=3J/2

(F+40)/J=5/3
3F+120=5J
120= 5J-9J/2
J=240

A.

F/J=3/2

F+40/J=5/3

Two equations two unknowns, solving for J leads you to 240.

I went at it in a different way and got a wrong result.
Can someone see where I went wrong?

2F=3J
3F = 5J-120

6F = 9J
6F = 15J-360

9J = 15J-360

360 = 6J

J = 120.....


anyone?

Should be: 3F = 5J - 120 --> 6F = 10J - 240 (multiplied by 2).

Hope it helps.
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Plug the answers and you will be able to solve it faster.

Another way to look at this is to calculate the difference that the $40 makes as follows:

5/3j - 3/2j = 40
10/6j - 9/6j = 40
1/6j = 40
j = 240

Thoughts?

Something is wrong with my method???

3/2 -> Feb is (3/5)x and Jan is (2/5)x

If we add 40 -> (3/5)(x+40) = 5/8x because of the new ratio
3/5x+120/5 = 5/8x
120/5 = 5/8x - 3/5x -> 120/5 = 1/40x
x = 960
2/5(960) = 384????

The issue is the text marked in red above.

For the 2nd scenario, the additional 40 should be added as follows:

(3/5)x + 40 = (5/8) (x+40) ---> x = 600 ....(you multiply x+40 to (5/8)x to account for the fact that x was the total when the original ratio stood. The 'new' total should thus be = old total + 40)

---> Jan = (2/5) *600 = 240. A is the answer.

The way you wrote it above, (3/5)(x+40) means that Feb's contribution is 3/5ths of the total and this equals to 5/8ths as well. This is where you made a mistake.

Hope this helps.
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F=3/2J
3/2J+40=5/3J
40=5/3J-3/2J
40=10/6J-9/6J
40=1/6J
240=J

A. 240.

We are given that the ratio of the amount of Alex's fuel oil bill for the month of February to the amount of his fuel oil bill for the month of January was 3/2. If F = the amount of the Febuary fuel bill and J = the amount of the January fuel bill, we can create the following ratio:

F : J = 3x : 2x

We are also given that if the fuel bill for February had been $40 more, the ratio would have been 5/3. We can create the following equation to determine x:

(3x + 40)/2x = 5/3

3(3x + 40) = 10x

9x + 120 = 10x

120 = x

Thus, the fuel bill in January was 2x = 2(120) = 240 dollars.

Answer: A
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The algebraic solution is straightforward, but we can also test the answers in case one is stuck

Let f = February and j = January
Start by testing B: January = $300
So \(\frac{f}{$300}\) = \(\frac{3}{2}\) ==> f = $450 ==> \(\frac{$450+$40}{300}\) ==> \(\frac{$490}{$300}\), which is slightly less than \(\frac{5}{3}\), so this not our answer. What one can learn from this step is that the answer is close to $300, so you can conveniently eliminate D & E.

Let us try A: January = $240
So \(\frac{f}{$240}\) = \(\frac{3}{2}\) ==> f = $360 ==> \(\frac{$360+$40}{$300}\) ==> \(\frac{$400}{$240}\) = \(\frac{5}{3}\), therefore our answer is A.

The more I practice official questions, the more I get to see how GMAC sets straps for test takers. I know my answer is close to 300, so I have to decide between testing A or C next (notice that they are both $60 from the first answer I tested). I decided to test A only because I already solved the question algebraically.

I need experts' opinions here: which method will you suggest a test taker goes with? Do the algebra or test the answers? Bunuel ScottTargetTestPrep chetan2u EMPOWERgmatRichC VeritasKarishma

Hi bebs,

TESTing THE ANSWERS works perfectly on this question - and all you had to do was a little Arithmetic to prove what the correct answer was. There is a math 'pattern' that you could save you a little time on this question - and under similar circumstances (especially if you weren't sure whether to TEST Answer A or C next).

Once you determined that Answer B was not a match for what we were looking for (49/30 is clearly less than 50/30 = 5/3, so it's TOO SMALL), we need the extra $40 that's added to February to have a larger 'impact' on the calculation than it did with Answer B. By decreasing the denominator (in this case, going from 300 to 240) we increase the impact that the $40 increase has on the calculation - meaning that the resulting ratio would INCREASE. If you were low on time - and/or you were certain about the pattern - then you could stop working after just TESTing Answer B.

GMAT assassins aren't born, they're made,
Rich

You can test answers, but honestly, you should be able to solve this question without doing so.
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