Tutor MathsGuru: Ask me for your burning Maths questions!
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Need help : Roy bought an equal number of pens and pencils. The pencils are sold at 5 for $3 and pens are sold at 4 for $5. He paid $65 more for pen than pencils. how many pen and pencils did he buy?
Thank you. -
silverhorse:
Lowest common multiple of 5 and 4 = 20Need help : Roy bought an equal number of pens and pencils. The pencils are sold at 5 for $3 and pens are sold at 4 for $5. He paid $65 more for pen than pencils. how many pen and pencils did he buy?
Thank you.
Let's say you buy 20 pens and 20 pencils,
(question says an equal number of pens and pencils)
Total cost of 20 pens = 20 / 4 x $5 = $25
Total cost of 20 pencils = 20 / 5 x $3 = $12
The 20 pens cost $25 - $12 = $13 more than the 20 pencils.
However, the question says that Roy paid $65 more for the pens than the pencils, not $13.
Therefore, $65 / $13 = 5 sets
Final answer is hence as follows:
Roy bought 5 x 20 = 100 pens and 100 pencils. -
Herbie:
A. |2x - 5| + 2 |10 - 4x| = 15Hi! Can advise how to solve this modulus function?
A. I2x - 5l + 2l10 -4xl =15
B. lx-3l/lxl=2
|2x - 5| + 4 |5 - 2x| = 15
|2x - 5| + 4 |-(2x-5)| = 15
|2x - 5| + 4 |2x-5| = 15
5 |2x - 5| = 15
|2x - 5| = 3
therefore, 2x - 5 = -3 or 2x - 5 = 3
solve for x = 1 or x = 4.
B. |x - 3| / |x| = 2
|(x - 3) / x| = 2
therefore, (x - 3) / x = -2 or (x - 3) / x = 2
solve for x = 1 or x = -3 -
ChewingPencilLine:
Thanks Chewing Pencil Line. Appreciate it.silverhorse:
Need help : Roy bought an equal number of pens and pencils. The pencils are sold at 5 for $3 and pens are sold at 4 for $5. He paid $65 more for pen than pencils. how many pen and pencils did he buy?
Thank you.
Lowest common multiple of 5 and 4 = 20
Let's say you buy 20 pens and 20 pencils,
(question says an equal number of pens and pencils)
Total cost of 20 pens = 20 / 4 x $5 = $25
Total cost of 20 pencils = 20 / 5 x $3 = $12
The 20 pens cost $25 - $12 = $13 more than the 20 pencils.
However, the question says that Roy paid $65 more for the pens than the pencils, not $13.
Therefore, $65 / $13 = 5 sets
Final answer is hence as follows:
Roy bought 5 x 20 = 100 pens and 100 pencils. -
No problem. Glad to be of help… As long as it doesn’t take up too much time, LOL.
Which is why I am not going to give a complete answer to the question "Find the possible equations of the circle which touches both coordinate axes and passes through (2,1)."
Basically, there should only be one possible circle right? In the first quadrant. It is not possible for a circle to touch the coordinate axes and pass through (2,1) if located in the other 3 quadrants.
The circle should have a radius of 1 unit (larger of 2, 1 divided by 2) and should be centered at (1,1). I trust you should be able to form the equation of the circle on your own by plugging in the values into your general equation for circles. -
ChewingPencilLine:
there are 2 possible circles ... http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=29&t=8315&start=690No problem. Glad to be of help... As long as it doesn't take up too much time, LOL.
Which is why I am not going to give a complete answer to the question \"Find the possible equations of the circle which touches both coordinate axes and passes through (2,1).\"
Basically, there should only be one possible circle right? In the first quadrant. It is not possible for a circle to touch the coordinate axes and pass through (2,1) if located in the other 3 quadrants.
The circle should have a radius of 1 unit (larger of 2, 1 divided by 2) and should be centered at (1,1). I trust you should be able to form the equation of the circle on your own by plugging in the values into your general equation for circles.
cheers. -
MathIzzzFun:
Sorry, but I am not too sure why? The question says \"touch the coordinate axes\", not \"cross\". I presumed that to mean 'tangential'. As such, the circle must be bounded within a quadrant. Therefore, my conclusion is that there can only be 1 such circle that passes through the point (2, 1). Can you please help to clarify?
there are 2 possible circles ... http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=29&t=8315&start=690ChewingPencilLine:
No problem. Glad to be of help... As long as it doesn't take up too much time, LOL.
Which is why I am not going to give a complete answer to the question \"Find the possible equations of the circle which touches both coordinate axes and passes through (2,1).\"
Basically, there should only be one possible circle right? In the first quadrant. It is not possible for a circle to touch the coordinate axes and pass through (2,1) if located in the other 3 quadrants.
The circle should have a radius of 1 unit (larger of 2, 1 divided by 2) and should be centered at (1,1). I trust you should be able to form the equation of the circle on your own by plugging in the values into your general equation for circles.
cheers. -
ChewingPencilLine:
Sorry, but I am not too sure why? The question says \"touch the coordinate axes\", not \"cross\". I presumed that to mean 'tangential'. As such, the circle must be bounded within a quadrant. Therefore, my conclusion is that there can only be 1 such circle that passes through the point (2, 1). Can you please help to clarify?[/quote]I believe you have not looked at the solution at the provided link.. look at diagram closely ... both the red circle and green circle touch x and y axes, not cross them.
there are 2 possible circles ... http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=29&t=8315&start=690MathIzzzFun:
[quote=\"ChewingPencilLine\"]No problem. Glad to be of help... As long as it doesn't take up too much time, LOL.
Which is why I am not going to give a complete answer to the question \"Find the possible equations of the circle which touches both coordinate axes and passes through (2,1).\"
Basically, there should only be one possible circle right? In the first quadrant. It is not possible for a circle to touch the coordinate axes and pass through (2,1) if located in the other 3 quadrants.
The circle should have a radius of 1 unit (larger of 2, 1 divided by 2) and should be centered at (1,1). I trust you should be able to form the equation of the circle on your own by plugging in the values into your general equation for circles.
cheers.
the two circles are centred at (1,1) and (5,5) respectively.
cheers. -
The diagram was too small for me. Sorry!
Actually, I still ain't too sure what you're talking about!
However, I think I understand now because I was thinking about it before I head off to bed last night (after finally finishing my project, omg) and I finally realized that there can be another circle in the same quadrant.
Thank you! -
Hi, MathIzzzFun. I read your PM. My apologies if you took offence to my claim that your diagram was too small for me to see the 2nd circle. Maybe it was due to the lousy colour contrast on my laptop. Maybe it was as you suggested in your PM title, that I looked at the wrong diagram (but I doubt so). In any case, I am sorry that I really could not see the diagram (as I was when I wrote that reply above ^). That is honestly not a lie, as you implied.
It was never my intention to start a ‘war of words’ on the forum, as you so put it. I wish to help people answer their questions while improving myself, that’s all. This is why I have decided to reply your PM on the forum; I apologize here to you, publicly and openly. My apologies if you found my comments offensive. Blame this on morning grumpiness, lol.
Anyway, I have not looked at your answers in entirety (ignored the numbers on the right) but I did look at your diagram for a substantial amount of time (5-10 minutes) because I was in a rush. In all frankness, I did feel rather offended by your accusations that I was provoking a ‘war of words’ and by your suggestions (by implications) that I was unwilling to admit my mistake, calling it ‘laughable’.
Did I not thank you for pointing out and correcting my mistake? My apologies once again if you did not find that sufficient. I am here to learn as much as I am here to help, really.
Also, I am sorry if it felt like someone hijacked your thread. I was bored, wanted a break from doing my homework and I wanted to help so I did the questions that were still unanswered. My apologies if that offended you.
Ultimately, however, I did figure out the solution on my own on my way to school earlier. My apologies if you found me rude for not giving your solution more attention and time as you would have deemed reasonable and right, for wanting to get my answer right on my own. I just thought that it would be more beneficial that way.
Basically, the center of the circle must lie on the line y = x since the circle must touch the coordinate axes. Also, the radius r is necessarily equal to h/k i.e. h = k = r. As such, you get (2 - r)^2 + (1 - r)^2 = r^2 which can be easily solved to obtain r = 1 or r = 5.
Moral of the story: don’t do math when half-awake.
Once again, my apologies if you found me carelessly rude with my innocent comment. Indeed, many others have looked at that diagram and found it sufficiently large. The problem lies with me, probably. Believe it or not however, I really could not see that 2nd circle even though I did look hard for it. I maintain that I was not lying and that I was not trying to provoke a ‘war of words’. Nevertheless, I apologize if it offended you.
Thank you.
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