Word Problem

thechicinnovation

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The denominator of a proper fraction is 4 times bigger than the numerator. If we increase the numerator by 6, and the denominator by 5, than the recieved fraction will be .5 bigger than the original. Find the original fraction.

I know that D=4N, so
N+6/D+5= something... I don't understand what I should put here...
 
The denominator of a proper fraction is 4 times bigger than the numerator. If we increase the numerator by 6, and the denominator by 5, than the recieved fraction will be .5 bigger than the original. Find the original fraction.

I know that D=4N, so
N+6/D+5= something... I don't understand what I should put here...

1. Let x denote the value of the numerator.

2. Then the original fraction is \(\displaystyle \displaystyle{\frac x{4x}}\). The modified fraction is \(\displaystyle \displaystyle{\frac{x+6}{4x+5}}\)

3. If you add \(\displaystyle \displaystyle{.5=\frac12}\) to the original fraction you'll get the modified fraction:

\(\displaystyle \displaystyle{\frac x{4x} + \frac12=\frac{x+6}{4x+5}}\)

4. Solve for x.
 
Thanks

1. Let x denote the value of the numerator.

2. Then the original fraction is \(\displaystyle \displaystyle{\frac x{4x}}\). The modified fraction is \(\displaystyle \displaystyle{\frac{x+6}{4x+5}}\)

3. If you add \(\displaystyle \displaystyle{.5=\frac12}\) to the original fraction you'll get the modified fraction:

\(\displaystyle \displaystyle{\frac x{4x} + \frac12=\frac{x+6}{4x+5}}\)

4. Solve for x.
I got 1/4. Is that right? Oh, and thank you so much!
 
I think there is something wrong with the problem statement.

Following the instruction we get \(\displaystyle x \ = \ \frac{9}{8}\)

then the original fraction becomes: (9/8)/(9/2)

check → (9/8 + 6)/(9/2 + 5) = (57/8)/(19/2) = 3/4 = 1/4 + 0.5 → checks

However, 1/4 does not satisfy the given condition.

Question is then is the answer sought for (9/8)/(9/2) - is that a proper fraction?
 
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Hello, thechicinnovation!

What a weird problem . . . I really don't like it.


The denominator of a proper ** fraction is 4 times bigger than the numerator.
If we increase the numerator by 6, and the denominator by 5,
. . then the resulting fraction will be 1/2 bigger than the original.
Find the original fraction.
Redundant!
If the denominator is greater than the numerator, it is, by definition. a "proper" fraction.

The original fraction is: .\(\displaystyle \dfrac{N}{D} \:=\:\dfrac{1}{4} \quad\Rightarrow\quad D \,=\,4N\) .[1]


The new fraction is: .\(\displaystyle \dfrac{N+6}{D+5} \:=\:\dfrac{3}{8}\)

. . \(\displaystyle 8N + 48 \:=\:3D + 15 \quad\Rightarrow\quad3D - 8N \:=\:33\)


Substitute [1]: .\(\displaystyle 3(4N) - 8N \:=\:33 \quad\Rightarrow\quad 4N \:=\:33\)

Hence: .\(\displaystyle \begin{Bmatrix}N &=& \frac{33}{4} \\ D &=& 33 \end{Bmatrix}\)

Therefore, the original fraction is: .\(\displaystyle \boxed{\dfrac{\frac{33}{4}}{33}}\)


Check:

\(\displaystyle \dfrac{\frac{33}{4} + 6}{33+ 5} \:=\:\dfrac{\frac{57}{4}}{38} \:=\:\dfrac{57}{4}\cdot\dfrac{1}{38} \:=\:\dfrac{3}{8}\)


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


Note: the original fraction is not \(\displaystyle \dfrac{1}{4}.\)

. . . . .Because \(\displaystyle \dfrac{1+6}{4+5}\) is not equal to \(\displaystyle \dfrac{3}{8}.\)
 
Hello, thechicinnovation!

What a weird problem . . . I really don't like it.



The original fraction is: .\(\displaystyle \dfrac{N}{D} \:=\:\dfrac{1}{4} \quad\Rightarrow\quad D \,=\,4N\) .[1]


The new fraction is: .\(\displaystyle \dfrac{N+6}{D+5} \:=\:\dfrac{3}{8}\)

. . \(\displaystyle 8N + 48 \:=\:3D + 15 \quad\Rightarrow\quad3D - 8N \:=\:33\)


Substitute [1]: .\(\displaystyle 3(4N) - 8N \:=\:33 \quad\Rightarrow\quad 4N \:=\:33\)

Hence: .\(\displaystyle \begin{Bmatrix}N &=& \frac{33}{4} \\ D &=& 33 \end{Bmatrix}\)

Therefore, the original fraction is: .\(\displaystyle \boxed{\dfrac{\frac{33}{4}}{33}}\)


Check:

\(\displaystyle \dfrac{\frac{33}{4} + 6}{33+ 5} \:=\:\dfrac{\frac{57}{4}}{38} \:=\:\dfrac{57}{4}\cdot\dfrac{1}{38} \:=\:\dfrac{3}{8}\)


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


Note: the original fraction is not \(\displaystyle \dfrac{1}{4}.\)

. . . . .Because \(\displaystyle \dfrac{1+6}{4+5}\) is not equal to \(\displaystyle \dfrac{3}{8}.\)

Hello Soroban,

I don't want to pick at you but I really don't understand how you got this equation:

\(\displaystyle \dfrac{N+6}{D+5} \:=\:\dfrac{3}{8}\)

Especially: Where does the\(\displaystyle \displaystyle{ \frac38} \)
come from?
 
I don't want to pick at you but I really don't understand how you got this equation:
\(\displaystyle \dfrac{N+6}{D+5} \:=\:\dfrac{3}{8}\)
Especially: Where does the\(\displaystyle \displaystyle{ \frac38} \)
come from?
In your reply #2 you misread the phrase "will be .5 bigger than the original."
You see three is .5 bigger than two. It means \(\displaystyle \frac32\) the original.
 
If we increase the numerator by 6, and the denominator by 5,
than the recieved fraction will be > > .5 bigger than < < the original.
Find the original fraction.

What are we to make of this phrase?

A synonym of "bigger" is "larger."


For instance, 1 bigger than x would be 1 + x, not 1x.


Then, .5 bigger than the original would be .5 + the original.


If the problem poser intends .5 of the original, then the phrasing should

just be ".5 of the original," or just ".5 the original."



Edit 1:

This problem is flawed. If the fraction is to be a proper fraction, then it wouldn't

have a form of \(\displaystyle \dfrac{\frac{33}{4}}{33}, \ \ \ which \ \ is \ \ a \ \ complex \ \ fraction.\)


For a proper fraction, neither the numerator nor the denominator would be
allowed to be a fraction.


--------------------------------------------------------------------------------------------


Edit 2:

But, had the phrase been "50% bigger than the original," or "50% more than the original,"
or "50% larger than the original," then it would make sense to me to mean 3/2
multiplied by the original.



Then I find the phrase ".5 bigger than" to be ambiguous.


 
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What are we to make of this phrase?
A synonym of "bigger" is "larger."
For instance, 1 bigger than x would be 1 + x, not 1x.
Then, .5 bigger than the original would be .5 + the original.
I disagree. The common reading of .5 bigger is \(\displaystyle 50\%\text{ bigger }\)
Or \(\displaystyle \text{Larger by }50\%\) or even half again as large.
 
I disagree. The common reading of .5 bigger is \(\displaystyle 50\%\text{ bigger }\)
Or \(\displaystyle \text{Larger by }50\%\) or even half again as large.

The whole phrase is ".5 bigger than the original."

This suggests .5 larger than the original, or .5 in addition to the original.


Example:

.5 bigger than 3 =

.5 larger than 3 =

.5 greater than 3 =

3 + .5 =

3.5



Then, again, I would classify that phrase, ".5 bigger than"

as ambiguous/meaningless.


As I see it (in my example), 3 is a number that is increased by another number
("another number" here being .5).


But 50% is a ratio.

So 3 + 50% is meaningless.


The common reading of "50% bigger than the original" should mean "50% larger than the original,"

but .5 and 50% are not interchangeable, as the former is (sometimes) a number to the latter,
which is a ratio and also a non-number.



Here's another example:

1 bigger than 3 =

1 more than 3 =

3 + 1 =

4


But,

100% bigger than 3 =

larger by 100% of 3 =

3 + (100%)(3) =

3 + 3 =

6
 
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This seems to disagree also.

In response to that example for that WolframAlpha page:

3 + 50% is meaningless.



In contrast, 3 + 50%(3) does equal

3 + .5(3) =

3 + 1.5 =

4.5



-------------------------


Edit:

When I typed "3 + (50%)(3)" into WolphramAlpha,

it came back with 4.5 (again).


WolphramAlpha program, you can't have it both ways.
 
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Hello, pappus!


I really don't understand how you got this equation: .\(\displaystyle \dfrac{N+6}{D+5} \:=\:\dfrac{3}{8}\)

Where does the \(\displaystyle \frac38\)
come from?
If TCI meant "one-half more than the original fraction",
. . I assumed that he would have said so.

So I concluded he meant "50% more than the original fraction".

Hence: .\(\displaystyle \frac{1}{4} + (50\%\text{ of }\frac{1}{4}) \:=\:\frac{3}{8}\)
 
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