Ration Question: £150 is divided into the ratio of 4x+10:2x+5:5x+3 How do I find the three amounts?

Any time you are adding parts of a thing, summing to the whole thing (whatever that "thing" might be), it may be possible to equate the sum of the parts to 1. (I haven't checked to see if that would be helpful in this particular context.)



...which is why I proposed summing the expressions for the three parts, and setting the result equal to the intended sum. Then one *can* solve for the value of [imath]x[/imath], and can then find the sizes of each of the three parts.



Going out on a limb... because that's the only way it works nicely...? ?
Whether the sum of parts can be equated to 1 depends on the meaning of "parts". If they are unitless fractions of the whole, then yes, they should add up to 1. This is not the case here.
The same idea applies to summing up to a measure (length, weight, amount of money). Again, this is not the case here - the given expressions are not amounts of money.
 
Dan,
It may look like the sum adds up to 1 (I don't see that) but in reality, it doesn't!.
Steven
Hmm, according to what I read in Dr Peterson's post, my post above is wrong. I really need to go to the corner and think about this one. I get fooled way too easily!
 
(4x+10) + (2x+5) + (5x+3) = 150
My guess: that's what the teacher is thinking
That's conceivable; but then they wouldn't really be talking about a ratio, but just saying the three amounts received by A, B, and C are (4x+10), (2x+5), and (5x+3).
A, B and C shared £150 in the ratio (4x+10) : (2x+5): (5x+3)

How much money did A, B and C get?

Ultimately, I think someone has to ask the teacher.
 
If we take
That's conceivable; but then they wouldn't really be talking about a ratio, but just saying the three amounts received by A, B, and C are (4x+10), (2x+5), and (5x+3).


Ultimately, I think someone has to ask the teacher.
I think the question is faulty. The amounts depend on what x is, and the ratio and the corresponding amounts will vary according to the value given to x.

BUT

If we take \(\displaystyle (4x + 10) + (2x + 5) + (5x + 3) = 150\) then we get \(\displaystyle x = 12\), which gives the amounts $58, $29 and $63.

I'd say this is probably what was intended. But ...
 
If we take

I think the question is faulty. The amounts depend on what x is, and the ratio and the corresponding amounts will vary according to the value given to x.

BUT

If we take \(\displaystyle (4x + 10) + (2x + 5) + (5x + 3) = 150\) then we get \(\displaystyle x = 12\), which gives the amounts $58, $29 and $63.

I'd say this is probably what was intended. But ...
Agreed. I'd tried solving for x in my head and didn't get a nice number (which is why I shouldn't do arithmetic in my head), so I didn't more strongly support this possibility.

It would be a fine problem if it weren't for the word "ratio".
 
Suppose the problem was 4x+10:2x+5:5x+12.5?
Then the ratio would be 2:1:2.5 no matter which value you pick for x.
The point is, just because there are x's involved doesn't mean that we need more information for the original problem to have an answer.
 
Suppose the problem was 4x+10:2x+5:5x+12.5?
Then the ratio would be 2:1:2.5 no matter which value you pick for x.
The point is, just because there are x's involved doesn't mean that we need more information for the original problem to have an answer.
But we do need more info, because the particular x's involved make the problem invalid.
 
The exact question: “Edmund, Susan and Peter shared £150 in the ratio (4x+10): (2x+5) : (5x+3)

how much money did each person get?”
 
Beer induced query follows.
The exact question: “Edmund, Susan and Peter shared £150 in the ratio (4x+10): (2x+5) : (5x+3)

how much money did each person get?”
Did you make that problem up yourself or did you get that problem from a textbook?
If that's from a textbook, what is the book's ISBN?
 
The exact question: “Edmund, Susan and Peter shared £150 in the ratio (4x+10): (2x+5) : (5x+3)

how much money did each person get?”
The solution (for one situation and the consequent set of numerical solution/s) has been discussed in response #26.
 
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