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

[cdwashin]

Hello, I am having problems understanding how to complete this problem. I get a totally different answer.

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 the fuel oil bill for February had been $40 more, the corresponding ratio would have been 5/3. How much was Alex's fuel oil bill for January?

The answer should be $240

Can someone please provide a thorough explanation.

[RonPurewal]

well, with a problem like this, one of the easiest ways to get the solution is by PLUGGING IN THE ANSWER CHOICES and WORKING BACKWARDS.

since the only answer choice you supplied is the correct answer, we'll work with that one, and show you how the solution proceeds.

you take the given value for january and input it into the given ratio of 3 : 2 to find the bill for february. either by using a proportion (x/240 = 3/2) or just by inspection ("hey, they're multiplying by 120"), you find that the bill for february is $360.
now you check the hypothetical: add $40 to the february bill, producing $400. CHECK this ratio against what's supposed to happen: 400/240 = 5/3, as required. this means that you have the correct answer; if you'd chosen one of the incorrect answers, then you'd get something other than the desired 5/3 ratio once you'd worked out the whole problem.

--

algebraically:

the openers for this problem are AUTOMATIC; you should have no trouble generating them. if you've practiced ratios enough, this sort of problem should be completely routine.

OPENER #1: IF YOU HAVE A RATIO OF TWO TOTALLY UNKNOWN QUANTITIES, USE THE "X" MULTIPLIER:
the first thing you're told is that february and january are in a ratio of 3:2. therefore, call them "3x" (february) and "2x" (january). do not use two variables, as getting rid of one of them will simply be a waste of time.

OPENER #2: IF YOU ALREADY HAVE ALGEBRAIC EXPRESSIONS, SET UP A PROPORTION:
once you have the 3x and the 2x, do what you're told, and add $40 to the 3x (the hypothetical situation described in the problem).
the new, hypothetical february quantity is 3x + 40, while january is still 2x.
therefore, set up the proportion:
(3x + 40) / 2x = 5 / 3
cross-multiply: 9x + 120 = 10x
120 = x
january = 2x = 240

[cdwashin]

Thank you very much. I really like to algebraic approach. I simply forgot to ad the X to the bottom of the equations. This was very helpful.

Thank you.

[esledge]

Thank you very much. I really like to algebraic approach. I simply forgot to ad the X to the bottom of the equations. This was very helpful.

Thank you.

(speaking for Ron) You are welcome.

[munique2]

This is simply a typo, right?
"you find that the bill for january is $360."
It should say February instead, shouldn't it?

[RonPurewal]

This is simply a typo, right?
"you find that the bill for january is $360."
It should say February instead, shouldn't it?

fixed. thanks.

how weird, that it took almost 3 years for someone to notice that.

[claire.b.public]

Here's how I solved it - slightly more time consuming:
(1) Ratio F/J = 3/2 → F = (3J)/2
(2) Ration (F +40)/J = 5/3
Plug (1) into (2):
((3J)/2) + 40) / J = 5/3
(3J + 80) / 2J) = 5/3
3(3J + 80) = 5 x 2J
9J + 240 = 10J
J = 240

[RonPurewal]

Here's how I solved it - slightly more time consuming:
(1) Ratio F/J = 3/2 → F = (3J)/2
(2) Ration (F +40)/J = 5/3
Plug (1) into (2):
((3J)/2) + 40) / J = 5/3
(3J + 80) / 2J) = 5/3
3(3J + 80) = 5 x 2J
9J + 240 = 10J
J = 240

the more approaches the better!

ideally, you should try not to use unnecessary variables -- in other words, it's always better to solve a problem with one variable than with two, if it's possible to do so -- but the most important thing is to learn as many different solution methods as possible.

[jackson.b.allan]

Just watched this problem on TWR archive.

When I calculated it, I approached it in the following way;

The increase in the ratio from the real to hypothetical situation, expressed as a fraction is;

5/3-3/2 = 10/6-9/6 = 1/6

Therefore the increase was in a ratio of 1 part increase : 6 parts original, said another way, the original ratio was increased by a 6th. So 6*$40=$240

Is this just coincidental, or a valid way to approach this problem. Just wondering if approaching in this manner might lead me to problems down the track with other problems.

[RonPurewal]

Yes. That is a thing.

Remember that a ratio involving only two quantities can also be written as a multiple. E.g., if two quantities are in the ratio 3/2, then the first is 3/2 times the second.

So, if J is the dollar amount of the January bill, then the February bill is 3J/2, and adding $40 would bring it to 5J/3. It should thus be pretty clear that subtraction will work.

[RonPurewal]

On the other hand"”for the same reason"”this idea does NOT extend to changes made to the second quantity in the ratio (= the bottom of the fraction), since you can't just add or subtract denominators.

In that case, if you already understand the principle you've described, you're best off inverting all the ratios, so that once again the changes occur in the numerator.