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Question

Maths New.jpg

How do you work out this question?
The answer is 42 apparently and it does work if you use the information given and substitute the values into the answer.
Reply 1
Original post by Chittesh14
Maths New.jpg

How do you work out this question?
The answer is 42 apparently and it does work if you use the information given and substitute the values into the answer.

The number of yellow balls is an unknown so let's call that nn.

Then the number of red balls is n+6n+6

So the ratio of red balls to yellow balls is n+6:nn+6:n

Now work out how these expressions change when the balls are removed. And then write down the new ratio.

Please post all your working if you get stuck.
Reply 2
Original post by Chittesh14
...


Call the number of red balls xx, then:

Unparseable latex formula:

\displaystyle[br]\begin{equation*} \frac{x}{x-6} = \frac{x-4}{x-6-3} \end{equation*}



Solve this equation/quadratic/linear equation for xx, take away 6 for the number of yellow, then add those two numbers to get the total.
Reply 3
Original post by Zacken
Call the number of red balls xx, then:

Unparseable latex formula:

\displaystyle[br]\begin{equation*} \frac{x}{x-6} = \frac{x-4}{x-6-3} \end{equation*}



Solve this equation/quadratic/linear equation for xx, take away 6 for the number of yellow, then add those two numbers to get the total.


Thanks lol :P.
I don't know why I kept making mistakes, I obviously knew how to approach the question but for some reason I kept getting:

Unparseable latex formula:

\displaystyle[br]\begin{equation*} \frac{x}{x-6} = \frac{x-4}{x-1} \end{equation*}



or even:

Unparseable latex formula:

\displaystyle[br]\begin{equation*} \frac{x}{x-6} = \frac{x-4}{x-5} \end{equation*}



I kept doing this:

at the start red balls = x
yellow balls = x-6

Afterred balls = x-4
4 less red balls, so y = x - 2
then 3 yellow balls were removed so yellow balls = x - 5

Spoiler

(edited 7 years ago)
Reply 4
Original post by notnek
The number of yellow balls is an unknown so let's call that nn.

Then the number of red balls is n+6n+6

So the ratio of red balls to yellow balls is n+6:nn+6:n

Now work out how these expressions change when the balls are removed. And then write down the new ratio.

Please post all your working if you get stuck.


Thank you for your help. I've solved the question now.
But, I was just wondering, would this method be correct too?

Red balls = r

Unparseable latex formula:

\displaystyle[br]\begin{equation*} \frac{r}{r-6} = \frac{4}{3} \end{equation*}



Where, 4 = number of red balls removed and 3 = number of yellow balls removed.

Because, if 4 red balls were removed and 3 yellow balls were removed and the ratio at the start and after the balls were removed stays the same, then that must be the ratio 4:3.

Since,

Unparseable latex formula:

\displaystyle[br]\begin{equation*} \frac{r}{r-6} = \frac{4}{3} \end{equation*}



3r = 4r - 24
24 = r

number of yellow balls = number of red balls - 6 balls
y = r - 6

Then, substitute it in:

Total = t.
Total = number of red balls + number of yellow balls
Total = r + r - 6
Total = 24 + (24-6) = 24 + 18 = 42.

24 red balls
18 yellow balls
total = 42 balls

If you check it:

Unparseable latex formula:

\displaystyle[br]\begin{equation*} \frac{24}{18} = \frac{4}{3} \end{equation*}



24:18 = ratio 4:3

After 4 red balls are removed and 3 yellow balls are removed, 20:15 = 4:3.

Unparseable latex formula:

\displaystyle[br]\begin{equation*} \frac{20}{15} = \frac{4}{3} \end{equation*}



Ratio is same - answer is correct.

Method?
Reply 5
Oh darn, I didn't see Notnek's post! Sorry!
Reply 6
Original post by Chittesh14
Thank you for your help. I've solved the question now.
But, I was just wondering, would this method be correct too?


Yes your method is correct.

If you have a ratio x:yx:y and you add aa to xx and bb to yy to keep the ratio the same then a:ba:b = x:yx:y.


xy=x+ay+b\displaystyle \frac{x}{y} = \frac{x+a}{y+b}

xy+bx=xy+aybx=ay\displaystyle \Rightarrow xy + bx = xy + ay \Rightarrow bx = ay

xy=ab\displaystyle \Rightarrow \frac{x}{y} = \frac{a}{b}

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