# Algebraic Fractions question

Thread starter 8 years ago
#1
So, my teacher gave me a question that goes like this:

A painter's apprentice takes 4 hours longer than a painter to pain a room on his own, but working together, they can paint the room in 1 and a half hours. How long does it take the painter to paint the room on his own?

I understand that this question involves setting up an equation involving algebraic fractions, but what would this equation be and how would you get there. From there I can solve it myself.

Thanks a lot guys

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8 years ago
#2
start by writing down what you know in some algebraic form. assign letters to the painter/his apprentice and form some equations
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Thread starter 8 years ago
#3
(Original post by kara1)
start by writing down what you know in some algebraic form. assign letters to the painter/his apprentice and form some equations
That's the problem, I have that the painter time is x and apprentice time is x+4 but where would you go from there? I just don't know which equations to form

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8 years ago
#4
(Original post by majmuh24)
That's the problem, I have that the painter time is x and apprentice time is x+4 but where would you go from there? I just don't know which equations to form

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Start by writing down their work-rates (i.e. rooms/hour), then find their combined work-rate. These questions are always kind of confusing.
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Thread starter 8 years ago
#5
(Original post by StrangeBanana)
Start by writing down their work-rates. (i.e. rooms/hour), then find their combined work-rate. These questions are always kind of confusing.
So they can do 1 room in 90 minutes. I think this gives

and it's simply a matter of solving from there? I'm normally pretty good at this kind of thing, it's just that this question confused me a little

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8 years ago
#6
Focus on this thought: Ultimately what we are being asked is the speed with which the painter paints the room - hence how long it takes. From the info given that is what we need to calculate. Currently we only know the apprentice's speed and the combined speed.
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Thread starter 8 years ago
#7
(Original post by Old_Simon)
Focus on this thought: Ultimately what we are being asked is the speed with which the painter paints the room - hence how long it takes. From the info given that is what we need to calculate. Currently we only know the apprentice's speed and the combined speed.
Do you really know the apprentice's speed though? All the information you have is that it's 4 hours longer than the painter's time, so you don't have an actual value for it, only the combined time.

I think I've set up an equation above, so would that be right?

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8 years ago
#8
Ignore this, just read the entire question again.
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Thread starter 8 years ago
#9
(Original post by MathsNerd1)
Wouldn't the two equations just be x+y=1.5, or 90 minutes but because you already know the painter apprentice takes 4 times longer than the painter, you've got x (the apprentice)= 4y, then just solve for y? Where y is the painter himself
I don't think so, as you only have their speeds, not just the time it takes. Anyway, this method would give that x = 18, which implies that he can work quicker on his own than with someone else which is kind of weird

Also, he doesn't take 4 times longer, but 4 hours longer. Thanks for trying though

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8 years ago
#10
(Original post by majmuh24)
So they can do 1 room in 90 minutes. I think this gives

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But you have defined x and x + 4 as the lengths of time the painter and his apprentice take to paint a room, so when you do x over 90, you are doing a time divided by a time, which doesn't give you a "rate"; a rate or speed would be in Rooms per Time.

Spoiler:
Show

Reciprocals...
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Thread starter 8 years ago
#11
(Original post by StrangeBanana)
But you have defined x and x + 4 as the lengths of time the painter and his apprentice take to paint a room, so when you do x over 90, you are doing a time divided by a time, which doesn't give you a "rate", which would be in Rooms per Time.
Thanks for that, that explains a lot

I think I get it now, so it would be

[Latex]\dfrac{1}{x} + \dfrac{1}{x+4} = 90[\latex]

Thanks for the help though, I understand this type of question much more now (I always used to find stuff like this kind of confusing )

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Thread starter 8 years ago
#12
(Original post by MathsNerd1)
I know, I realised the mistake as I sent it, what you should have then is just x+y=1.5, like before but say y=x+4 like you've suggested so you arrange for x to get you an answer? I'm pretty sure this is the correct route to take.
Still no, cos this would give a negative time which is just nonsensical

You have to use algebraic fractions if that helps

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8 years ago
#13
(Original post by majmuh24)
Thanks for that, that explains a lot

I think I get it now, so it would be

[Latex]\dfrac{1}{x} + \dfrac{1}{x+4} = 90[\latex]

Thanks for the help though, I understand this type of question much more now (I always used to find stuff like this kind of confusing )

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No problem. But it'd be equal to 3/2, btw, not 90 (hours, not minutes, as the x and x + 4 are in hours).
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Thread starter 8 years ago
#14
(Original post by StrangeBanana)
No problem. But it'd be equal to 3/2, btw, not 90 (hours, not minutes, as the x and x + 4 are in hours).
Ah right, thanks for that I always make stupid mistakes like this

Anyway, are you doing the maths challenge next week?

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8 years ago
#15
(Original post by majmuh24)
Do you really know the apprentice's speed though? All the information you have is that it's 4 hours longer than the painter's time, so you don't have an actual value for it, only the combined time.

I think I've set up an equation above, so would that be right?

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We do not know the apprentices speed but we need to calculate it first.

Painters rate = joint rate - apprentices rate expressed as area per minute
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8 years ago
#16
(Original post by majmuh24)
Still no, cos this would give a negative time which is just nonsensical

You have to use algebraic fractions if that helps

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I'm not being much help today, so I'll stop as I can see someone else is pointing you in the correct direction I believe
1
8 years ago
#17
(Original post by majmuh24)
Ah right, thanks for that I always make stupid mistakes like this

Anyway, are you doing the maths challenge next week?

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Me too, they're so annoying. xD

The UKMT? Yeah, you? Have you done it in previous years?
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Thread starter 8 years ago
#18
(Original post by StrangeBanana)
Me too, they're so annoying. xD

The UKMT? Yeah, you? Have you done it in previous years?
It's always the worst when you have the right method but make a stupid numerical error

Yeah, I've done it every year since Year 11 and did the Senior one this year, but that kind of failed since I only got Kangaroo Have you done any prep yet?

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Thread starter 8 years ago
#19
(Original post by Old_Simon)
We do not know the apprentices speed but we need to calculate it first.

Painters rate = joint rate - apprentices rate expressed as area per minute
I've got a new equation above so would that be right?

Thanks for the help BTW

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8 years ago
#20
(Original post by majmuh24)
Yeah, I've done it every year since Year 11 and did the Senior one this year, but that kind of failed since I only got Kangaroo Have you done any prep yet?
Aww I'm only in year 11 now. Y u do dis?

And nah, I haven't really done any prep; might do some past questions tomorrow. The questions are always fun. ^^ And you?
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