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Basic math problems(not homework & not ,but would like help anyway)

  1. Aug 27, 2009 #1
    Basic math problems(not homework & not urgent,but would like help anyway)

    Let me start by saying,I don't do math for my A/Ls(instead I chose bio :yuck:)and I haven't done any math for a very long loooong time,but I've always loved math.
    I have some time off,so I got this set of miscellaneous questions I enjoyed solving and was hoping you-all could help me out with some of the problems I've been having.

    Here's just a few

    Q1) The 5 tyres of a car (four road tyres and a spare) were each used equally on a car that had travelled 20,000km.The number of kilometers of use of each tyre was
    (a)4000 (b) 5000 (c) 16000 (d) 20000 (e)100000

    It's definitely not (a),(b),(d) or (e),which leaves me with only one choice, (c),but I'm not very convinced if this is how we find the answer,
    (20,000*4)/5 = 16000 ?


    The following questions I have no idea how to do.I'm not expecting a complete solution,but any hints on how i could solve them would be much appreciated

    Q2) The largest area of a right triangle having positive integer side lengths and a circle of radius 2 inscribed is
    (a)24 (b)30 (c) 54 (d) 60 (e)48

    drew a triangle and the circle & tried everything i could think of to solve it,still no luck .


    Q3) 3 distinct corners of a cube of volume 1m3 are P,Q and R.What could be the area of the triangle PQR in square metres ?
    I think the answer is 1/2 ,but it doesn't feel correct?


    Q4) If a sequence of numbers a1,a2,a3,...is given by an+1 = 1/(1-an ) for n>1 and a1 = 1/4 ,then the value of a2005 is ...?
    (a)1/4 (b) 4/3 (c) -3 (d) -1/2 (e) 2/3
    I have no idea how to do this ?


    Q5) Which is the smallest number in
    [tex]\sqrt[3]{7}[/tex] - [tex]\sqrt[3]{6}[/tex] , [tex]\sqrt[3]{8}[/tex] - [tex]\sqrt[3]{7}[/tex] , [tex]\sqrt[3]{9}[/tex] - [tex]\sqrt[3]{8}[/tex] , [tex]\sqrt[3]{10}[/tex] - [tex]\sqrt[3]{9}[/tex] , [tex]\sqrt[3]{11}[/tex] - [tex]\sqrt[3]{10}[/tex] ?

    Is there a way of doing this without actually having to use a calculator to find each value?:blushing:
     
    Last edited: Aug 28, 2009
  2. jcsd
  3. Aug 27, 2009 #2

    Mark44

    Staff: Mentor

    Re: Basic math problems(not homework & not urgent,but would like help anyway)

    Q3 - After not much thinking about it, I believe you have two possibilities: 1) P and Q are on one face of the cube and R is also on the same face, and 2) P and Q are as before, but R is on the opposite face. These two scenarios give you two different sizes of triangles.
    Q5 - Draw a graph of y = [itex]\sqrt[3]{x}[/itex]. The graph is concave down, meaning that a one-unit change in x for smaller numbers gives you a larger increment in y values than does the same change for larger numbers.
     
  4. Aug 27, 2009 #3
    Re: Basic math problems(not homework & not urgent,but would like help anyway)

    Q1) Imagine the driver has to rotate the tires so that each tire spends the same amount of time at each corner of the car and in the trunk. That's five positions. So 20,000/5 = 4,000 km in each position. Since 4 of them are on the road, each tire experiences 16,000 km of wear.

    (Or at least that was the fastest intuitive way I came up with the answer).

    Q2) The area of a right triangle is greatest when the legs are of equal length and gets less as the sides become more uneven (Calculus can proves this). Unfortunately, an isosceles right triangle cannot have all integer sides. I believe the closest is a 3-4-5 tringle. You just need to find which one of these has a circle inscribed of radius 2 (by similarity). There is the possibility that this doesn't have integer sides then you'd need to find the next Pythagorean triple and work from there.

    Q3) There are three distinct types of triangles that can be formed from 3 corners of a cube: One where all corners are on one face, one where exaclty two are on the same edge, and the other when no corners share an edge. I believe the possible areas are: [itex]1/2, \sqrt{(2)}/3, \sqrt{(3)}/2[/itex].

    Q4) The sequence is cyclic of order 4.

    Q5) The sequence is decreasing.

    I hope this helps.

    --Elucidus
     
  5. Aug 27, 2009 #4
    Re: Basic math problems(not homework & not urgent,but would like help anyway)

    Q2) You have to first note that the you're essentially asked to find a pythagorean triple, i.e., right triangles with integer sides. The most common ones have side lengths 3-4-5, 8-15-17, and 7-24-25, and it's easy to see that right triangles with side lengths that are integer multiples of these right triangles just listed are also pythagorean triples. Moreover, if you inscribe a circle in any triangle, then you can drop altitudes from the center of the circle to each side (each altitude having the length of the radius of the circle) and easily find that the area of the triangle is (1/2)*r*s, where r is the radius of the circle and s is the semiperimeter of the triangle (half the perimeter of the triangle). Thus, you also know that the sum of the three sides of the right triangle in question must be equal to the area, and then it's simple casework.

    Q4) The most common way to deal with sequences is to find a pattern. Work out the first few cases to see if you can discern a pattern.

    Q5)Since a^3 - b^3 = (a-b)(a^2 + ab + b^2), note that we can write each term listed as the reciprocal of a sum of three cube roots. For example, 7-6 = [7^(1/3) - 6^(1/3)]*[49^(1/3) + 42^(1/3) + 36^(1/3)] so 7^(1/3) - 6^(1/3) = 1/[49^(1/3) + 42^(1/3) + 36^(1/3)] . So essentially, each subsequent term in the list will have the same numerator but the denominator will increase.
     
  6. Aug 27, 2009 #5
    Re: Basic math problems(not homework & not urgent,but would like help anyway)

    Thanks for the quick response,Mark44 and Elucidus!

    Yeah,that makes more sense.thx.

    Now I'm having a bit of trouble with Q3) and Q5)

    I understand there are three triangles ,but I'm not sure how i could find the area for the 3rd triangle,"where no corners share an edge" ?

    For Q5)
    I'm a bit confused.Are you two suggesting the same thing?
    If so,then how do I find the minimum value for y from the graph & I'm sorry it's just not very clear to me how I could solve for Q5) even after knowing this.
     
    Last edited: Aug 28, 2009
  7. Aug 27, 2009 #6
    Re: Basic math problems(not homework & not urgent,but would like help anyway)

    thx for replying snipez90.
    I'm a bit too tired right now to really concentrate.
    I'll try to work this out tomorrow and if I still have a problem ,I'll post.

    Thank you all.
     
  8. Aug 27, 2009 #7
    Re: Basic math problems(not homework & not urgent,but would like help anyway)

    Heh, sorry, the explanation for Q2 is probably overly complicated. It seems a lot like an old SAT question. The best way to deal with these problems is not to think mathematically in a sense, but to play around with simple casework. For Q2 and Q5, you just have to know the facts and the formulas. On the other hand, for Q4, working with the small cases to discern a pattern is a basic problem solving technique that has far greater applicability to many problems in mathematics.
     
  9. Aug 27, 2009 #8
    Re: Basic math problems(not homework & not urgent,but would like help anyway)

    It is an equilateral triangle whose sides are all diagonals of a face with length [itex]\sqrt{2}[/itex].

    Somewhat. The sequence

    [tex]\sqrt[3]{n+1}-\sqrt[3]{n}[/tex]

    is decreasing as n grows. The least member from a list of this sequence is the one whose first radicand is largest.

    --Elucidus
     
  10. Aug 28, 2009 #9
    Re: Basic math problems(not homework & not urgent,but would like help anyway)

    OK.thx.

    Now for Q2)
    went through pages and pages of text and finally realised r= (a+b-c)/2 and r=radius of circle=2
    As suggested,tried using the pythagorean triples and I end up with an area of 24 ?
    I really hope this is correct.


    I'm still having problems with Q3)
    For the triangle where all corners are on one face,I get A= 1/2
    For the triangle where no corners share an edge,A= sqrt3/2
    But for the triangle where 2 corners are on the same edge,I only get A= sqrt2/2 ?
    Elucidus,
    sir I'm not sure if it's a typo ,or if my answer is wrong.
    If my answer is incorrect,pls do tell me where I've gone wrong.


    As for Q4),

    My method is a bit messy.I really don't know how to show it here,but it goes somewhat like this,
    for a3 , a4, a5 , a6 , a7 , a8 ,a9 ... ,I get values of
    -3 , 1/4 , 4/3 ,-3 , 1/4 , 4/3,...
    and after subtracting and dividing and all(I'm not sure if this is how you do it?) but I get 4/3 as the answer.
    Does this seem OK?
     
  11. Aug 28, 2009 #10
    Re: Basic math problems(not homework & not urgent,but would like help anyway)

    My apologies for the typo; the area of the second type of triangle is as you said sqrt(2)/2.

    The answer to Q2 is indeed 24.

    I think your answer to Q4 is off. The sequence in Q4 is cylcic of order 3 (not 4 as I mistakenly said earlier), so for all n > 0, an = a(n + 3).

    Using modular arithmetic (mod 3), the answer is very direct.

    --Elucidus
     
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