Tutor MathsGuru: Ask me for your burning Maths questions!
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Mathizzzfun, tq so much. U made it so easy ! Really appreciate our reply
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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!
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