The Student Room Group

Algebraic Fractions question

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 :biggrin:

Posted from TSR Mobile
(edited 10 years ago)

Scroll to see replies

Reply 1
Avatar for kara1
kara1
11
start by writing down what you know in some algebraic form. assign letters to the painter/his apprentice and form some equations
Reply 2
Avatar for interstitial
interstitial
OP
18
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 :colondollar:

Posted from TSR Mobile
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 :colondollar:

Posted from TSR Mobile


Start by writing down their work-rates (i.e. rooms/hour), then find their combined work-rate. These questions are always kind of confusing. :tongue:
(edited 10 years ago)
Reply 4
Avatar for interstitial
interstitial
OP
18
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. :tongue:


So they can do 1 room in 90 minutes. I think this gives

\dfrac{x}{90} + \dfrac{x+4}{90} = 1 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 :colondollar:

Posted from TSR Mobile
Reply 5
Avatar for Old_Simon
Old_Simon
12
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.
Reply 6
Avatar for interstitial
interstitial
OP
18
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?

Posted from TSR Mobile
Ignore this, just read the entire question again.
(edited 10 years ago)
Reply 8
Avatar for interstitial
interstitial
OP
18
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 :tongue:

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

Posted from TSR Mobile
Original post by majmuh24
So they can do 1 room in 90 minutes. I think this gives

\dfrac{x}{90} + \dfrac{x+4}{90} = 1


Posted from TSR Mobile

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

(edited 10 years ago)
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 :biggrin:

I think I get it now, so it would be

\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 :colondollar:)

Posted from TSR Mobile
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 :redface:

You have to use algebraic fractions if that helps

Posted from TSR Mobile
Original post by majmuh24
Thanks for that, that explains a lot :biggrin:

I think I get it now, so it would be

\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 :colondollar:)

Posted from TSR Mobile


No problem. :smile: But it'd be equal to 3/2, btw, not 90 (hours, not minutes, as the x and x + 4 are in hours).
Original post by StrangeBanana
No problem. :smile: 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 :wink: I always make stupid mistakes like this :tongue:

Anyway, are you doing the maths challenge next week?

Posted from TSR Mobile
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?

Posted from TSR Mobile


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
(edited 10 years ago)
Original post by majmuh24
Still no, cos this would give a negative time which is just nonsensical :redface:

You have to use algebraic fractions if that helps

Posted from TSR Mobile


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
Original post by majmuh24
Ah right, thanks for that :wink: I always make stupid mistakes like this :tongue:

Anyway, are you doing the maths challenge next week?

Posted from TSR Mobile


Me too, they're so annoying. xD

The UKMT? Yeah, you? :biggrin: Have you done it in previous years?
Original post by StrangeBanana
Me too, they're so annoying. xD

The UKMT? Yeah, you? :biggrin: Have you done it in previous years?


It's always the worst when you have the right method but make a stupid numerical error :getmecoat:

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 :ninja: Have you done any prep yet?

Posted from TSR Mobile
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 :biggrin:

Posted from TSR Mobile
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 :ninja: Have you done any prep yet?


Aww I'm only in year 11 now. :frown: 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?

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you