Math Hard Questions - Help please

Test Preparation SAT Preparation
1

<p>1) If P and Q are different points in a plane, the set if all points, the set of all points in this plane that are close to P than to Q is
a) The region of the plane on one side of a line</p>

<p>I guessed the answer, but why???</p>

<p>2) In a certain experiment, there is a 0.2 probability that any thermometer used in error by more than 1 deg C. If 4 thermometers are used, what is the probability that all of them are in error by more than 1 deg C.
a) .0016
b) .0081
c) .16
d) .25
e) .8</p>

<p>3) The magnitudes of vectors a and b are 5 and 12 respectively, then the magnitude of vector (b-a) could not be</p>

<p>a) 5
b) 7
c) 10
d) 12
e) 17</p>

<p>4) For all theta, sin (theta) + sin (- theta) + cos (theta) + cos (- theta) = </p>

<p>a) 0
b) 2
c) 2 sin theta
d) 2 cos theta
e) 2 (sin theta+ cos theta)</p>

<p>5) The radius of the base of a right circular cone is 6 and the radius of the parallel cross section is 4. If the distance between the base and the cross section is 8, what is the height of the cone</p>

<p>a) 11
b) 13 1/3
c) 16
d) 20
e) 24</p>

<p>6) An indirect proof of the statement “If x=2, then sq.rt of x is not a rational number could begin with the assumption that</p>

<p>a) X = Sq. rt 2
b) x.x = 2
c) sq. rt of x is rational
d) sq. rt of x is not rational
e) x is nonnegative</p>

<p>7) How many ways can 10 people be divided into two groups, one with 7 people and the other with 3 people</p>

<p>a) 120
b) 210
c) 240
d) 5040
e) 14400</p>

<p>8) Which of the following has an element that is less than any other element is that set?</p>

<p>I) The set of positive rational numbers
II) The set of positive rational numbers r such that r.r >= 2
III) The set of positive rational numbers r such that r.r >= 4</p>

<p>A) None
B) I only
C) II only
D) III only
E) I and III</p>

<p>Thanks so much.</p>

2

<p>They look like Math II questions.</p>

3

<p>For #4)
sin (theta) + sin (- theta) + cos (theta) + cos (- theta)</p>

<p>sin(-theta) = - sin(theta)
cos(theta) = cos(-theta)
so you have</p>

<p>sin(theta) - sin(theta) + cos(theta) + cos(theta)
= 2cos(theta)</p>

4

<h1>3:</h1>

<p>I would think that <b> - <a> would just be b - a, which would be 7. Still, I am assuming that these vectors are placed along the x or y axis (though placing them at any angle would probably yield the same result, if they were placed at the SAME angle).</a></b></p><b><a>
</a></b>

5

<p>For #3, this problem, though not a regular SAT problem, is a lot like the following (SAT-like) problem: “Two sides of a triangle are length 5 and 12, which of the following could not be the length of the third side?” For triangles, the third side needs to be bigger than the difference and smaller than the sum of the other two sides. For two vectors b and a, the length of b-a needs to be bigger than or equal to the difference in lengths (7) and smaller than or equal to the sum of the lengths (17). So, only 5 (A) doesn’t work.</p>

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<p>The only thing i’m confused on is if there’s a triangle, abc, doesn’t A+B <C? not less than or equal to, just less than. But 7+5=12 so it wouldn’t be able to make a triangle?</p>

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<h1>3 Is not a triangle question. It’s a question on vectors. You can get the magnitude of a vector if you do b-a except for a value less than b-a (7) or a value less than a+b (17). Both are when each vector is 180 degrees from each other.</h1>

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<h1>6 seems to be c. (sq.rt of 2 is rational). Indirect proof is synonimous with the proof by contradiction, which means you have to reverse the consequent in the original statement.</h1>

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<ol>
<li><p>Think about it.</p></li>
<li><p>is C. One way to prove that sqrt(2) is irrational is to assume it is rational and find a contradiction.</p></li>
<li><p>III only. The rationals are not a well-ordered set, so I is false. II is also false because there is no smallest rational r with r^2 >= 2. III is true since 2 is in that set and is the smallest element in the set.</p></li>
</ol>

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<p>Okay, I don’t really like the people who set those questions, their knowledge of mathematics is quite vague(?) judging by the nature of some of the questions. Anyway. </p>

<ol>
<li>I easiest determining thing I can tell you is the Formula for finding out the magnitude of the resultant of 2 vectors: - </li>
</ol>

<p>let the magnitude of the resultant vector be R. then – </p>

<pre><code> |R| = sqrt( a^2 + b^2 + abcos theta)
</code></pre>

<p>where theta is the angle between the two vectors. this formula is true for a+b as well as a-b or b-a (why? read the next answer).
we know,
-1<cos theta<1="" so–="" solving="" the="" formula="" with="" given="" data,="" we="" get="" that="" r="">=7 always. so (a). </cos></p>

<ol>
<li>I easiest the suggestion I can give you for such sums is the plug in method. So plug in theta = 60 (degree) and you’ll see that the answer is (d) 2cos(theta). </li>
</ol>

<p>If you require a further explanation - then I recommend searching for Plotting co-ordinates for Angles. In that, there is something called “all, sine, tan, cos” which shows the values of the trigonometric functions of the various ranges of theta angles.
If you wanna know further, because it may sound difficult but it’s quite easy, I am happy to help :slight_smile: . </p>

<ol>
<li>This is a “Similarity problem”. The answer is (d) or 24 (8+16). How do we get 16?
This requires a bit of drawing.
Since it’s a right circular cone, let’s take it in a 2D form.

  1. draw a right angled triangle. Then draw any line parallel to it’s base<br>
    within the triangle. You can see that the top vertex and the parallel
    line now forms a smaller triangle at the top.
  2. let the length of the base be 6. and the parallel line be 4.
  3. let the length of the distance from the base to the parallel line be 8.
  4. and from the parallel line to the top vertex of the triangle be a variable k. </li>
    </ol>

<p>Similarity states that, The entire right angled triangle and the triangle formed by the topmost vertex and the parallel line that too forms another rt. angled triangle (see it?) are similar and thus … </p>

<p>k/k+8 = 4/6 </p>

<p>Solve for k to get k=16 . so the height is 16+8 =24. </p>

<ol>
<li><p>The answer is C. We call it the elimination method :smiley: . It basically means that we assume that the square root of x=2 is rational (which of course it is not , sqrt(2) = 1.41…something something non-recurring) and then we prove that the assumption is wrong. </p></li>
<li><p>This is a permutation and combination sum. Wait I’ll get my calculator for this .Makes it easier. </p></li>
<li><p>This is a badly structured sum. I am guessing that it’s asking for the possible set with the least number of elements. </p></li>
</ol>

<p>So, the lowest member of I is between 0 and 1.
the lowest member of II is between 1 and 2.
the lowest member of III is 2.
So, obviously, the set III has the lowest members. so III only I think, if I’ve read it correctly.</p>

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<p>@rishavmadcapped, there is nothing wrong with #8. It’s not asking for the set with the least number of elements (in fact, I believe the three sets have the same number of elements). #8 is asking which set(s) has a smallest element.</p>

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<p>“Which of the following has an element that is less than any other element is that set?” </p>

<p>@MITer94 … does it make sense? really? :stuck_out_tongue:
Anyway, as far as the number of elements is considered, if we take the set s r = {rational no.} then yes, we can say that all the sets have a indeterminate Domain. Anyway, the question is overall a bit too ambiguous imo. </p>

<p>Btw. are you at MIT at the moment? what are you studying? how is it like?</p>

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<p>Yes, question makes perfect sense. For example, III has a smallest element (2), and no element in the set described by III is less than 2. However, I does not have a smallest element. If we let a/b be the smallest positive rational #, then a/2b is smaller, contradiction. II also doesn’t contain a smallest element, as sqrt(2) is irrational.</p>

<p>Sets that have a least element are said to be “well-ordered.”</p>

<p>Yes, I’m at MIT, majoring in math w/ computer science. Life can get busy at times, but the classes and atmosphere/student life are, overall, great.</p>

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<p>For #2. A Bernoulli trial (or binomial trial) is a random experiment with exactly two possible outcomes, “success” and “failure”, in which the probability of success is the same every time the experiment is conducted. Let p=0.2 be the probability of success in a Bernoulli trial (in this problem it is the probability that a thermometer is in error by more than 1 deg C). Then the probability of failure q is given by q=1-p=1-0.2=0.8. The probability of exactly k successes in n independent Bernoulli trials, each with a probability of success p, is given by
P<em>n (k)=n!/(k!(n-k)!)* p^k q^(n-k) ,
where n!/(k!(n-k)!) is a binomial coefficient. Since 4 thermometers are used, n is equal to 4, all four thermometres are in error by more than 1 deg C, therefore, k=4.
We substitute n=4 , k=4 , 4!/(4!0!)=1 into the formula of the probability and get
P</em>n (k)=4!/(4!0!)* p^4 q^(4-4)=p^4=(0.2)^4=0.0016.
So, the answer is a) .0016.</p>

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<p>Well it’s just (.2)^4 because the four thermometers are independent of each other…</p>