<p>^Thank you! :)</p>
<p>Hi, I have a few questions on counting. </p>
<p>1) Six balls, each labeled with a different integer from 1 to 6, are placed in a container. Two balls are drawn simultaneously, at random, and the sum of their numbers is recorded. How many different sums are possible? </p>
<p>2) In a bus with a total of n people, 2 people must stand because all of the 15 equal rows of seats are filled. When 14 people leave the bus and no others board, 12 rows of seats are filled, 3 rows are empty, and no one is standing. What is the value of n? </p>
<p>3) Set X = {1,3, 5}
Set Y = {2,4,6}
If x and y are to be selected at random from sets X and Y, respectively, what is the probability that (2x)/y will be a member of set X? </p>
<p>4) There are 5 members in the MacDonald family. On their vacation, they plan to rent a canoe and a Jon boat. If each boat safely seats 3 people, how many ways can the MacDonalds group themselves in the canoe and the Jon boat?
a) 3 d) 12
b) 5 e) 20
c) 10</p>
<p>Thanks! :)</p>
<p>1) The smallest possible sum is 3 (1+2), and the largest sum is 11 (6+5). The sums 3,4,…,11 can be made using two different numbers 1 through 6. The answer is 9.</p>
<p>2) Let k be the number of seats in a row. We know n = 15k + 2, n - 14 = 12k → n = 12k + 14. Hence 15k + 2 = 12k + 14 → k = 4. n = 15(4) + 2 = 62.</p>
<p>3) 2x = 2, 6, or 10 (with equal probability) and y = 2, 4, 6. It’s pretty easy to brute force, the only possible cases are (2x,y) = (2,2), (6,2), (10,2), (6,6). There are nine possible cases, so the probability is 4/9.</p>
<p>4) The canoe can hold either 2 or 3 people (so that the Jon boat holds at most 3). Choose two people for the canoe, choose three people for the canoe. 5C2 + 5C3 = 20, E.</p>
<p>Thank you :)</p>
<p>I came across a practice question that went like ‘What is the product of the first five even integers?’ The answer was: 2 x 4 x 6 x 8 x 10 = 3840. As long as I’ve known, 0 is an even number. So shouldn’t it be counted as the first even number? If that’s the case, the product should be 0. Please correct me.</p>
<p>2 is the first POSITIVE even integer.</p>
<p>0 is the first NONNEGATIVE even integer.</p>
<p>There is NO first even integer.</p>
<p>Here is a simple proof: Let n be an even integer. Then n - 2 is an even integer less than n.</p>
<p>Note that the set of even integers is {…,-6, -4, -2, 0, 2, 4, 6, …}</p>
<p>So, in the practice question you’re referring to there is most likely a typo - the word “positive” was probably ommitted.</p>
<p>Hi, I have a few questions: </p>
<p>1) If the lengths of the sides of a triangle are x, x+1, and x+2, which of the following is a possible measure for the angle opposite the side with length x? </p>
<p>A) 58 degrees
B) 60 degrees
C) 61 degrees
D) 62 degrees
E) 89 degrees</p>
<p>2) A square and a triangle each have area 1. If the length of a diagonal of the square is equal to the length of one side of the triangle, what is the length of the altitude to that side of the triangle? </p>
<p>3) In rectangle ABCD, point E is the mid point of side BC. If the area of quadrilateral ABED is 2/3, what is the area of rectangle ABCD? </p>
<p>4) A square, 5 cm on a side, is cut out of each corner of a rectangular piece of sheet metal 20 cm by 15 cm. The sides of the pieces are then bent up to form an open box 5 cm high. What is the volume of the box, in cubic centimeters?
A) 1500
B) 1000
C) 750
D) 500
E) 250</p>
<p>Thanks!</p>
<p>1) The side with length x is the smallest side of the triangle. Therefore the opposite angle is the smallest angle of the triangle. Since a triangle has 180 degrees, this angle must have measure less than 180/3 = 60 degrees. Only choice (A) satisfies this condition.</p>
<p>2) Use the nifty little area formula A = d^2/2 where d is the length of the diagonal of the square. We have d^2/2 = 1, so d^2 = 2, and d=sqrt(2). </p>
<p>Now, the area of a triangle is A = 1/2 bh. So we have 1/2 sqrt(2) h = 1. So h = 2/sqrt(2). Put this in your calculator and grid in .707.</p>
<p>3) After drawing the picture, split the rectangle into four congruent traingles. Each triangle has area (2/3)/3 = 2/9. So the area of the rectangle is 4(2/9) = 8/9.</p>
<p>4) The height of the box is equal to the length of the side of each little square. So the length of the box is 20 - 2(5) = 10, and the width of the box is 15 - 2(5) = 5. So the volume is (10)(5)(5) = 250, choice (E).</p>
<p>I am not sure if this is how to create a link to the image. If anyone knows how to create a link to an image could you tell me.</p>
<p><a href=“https://sendnow.acrobat.com/?i=COiJWBf6FBIfu9JaKhoH9A[/url]”>https://sendnow.acrobat.com/?i=COiJWBf6FBIfu9JaKhoH9A</a></p>
<p>Thanks in advance.</p>
<ol>
<li><p>A craftsman creates necklaces out of beads. He uses colored beads in a repeating pattern of gray, lime, opal, ruby, clear, white, black; gray, lime, opal, ruby, clear, white, black and so on. If the first bead on a necklace is gray, what is the color of the 86th bead?
Answer is Lime. Please provide explanation</p></li>
<li><p>What is the remainder when 2^400 is divided by 10?
A. 0
B. 2
C. 4
D. 6
E. 8</p>
<pre><code> Don’t know what the answer is. Please provide explanation again.
</code></pre></li>
</ol>
<p>Thanks</p>
<ol>
<li>Sequence of colors alternates every 7. 86 ≡ 2 (mod 7) so the 86th bead will be the same color as the 2nd bead, lime.</li>
</ol>
<p>18.</p>
<p>2^1 ≡ 2 (mod 10)
2^2 ≡ 4
2^3 ≡ 8
2^4 ≡ 6
2^5 ≡ 2</p>
<p>This repeats every 4. 2^400 ≡ 2^4 ≡ 6 (mod 10). Remainder is 6.</p>
<ol>
<li>When 15 is divided by the positive integer k, the remainder is 3. For how many different values of k is this true?</li>
</ol>
<p>A) One
B) Two
C) Three
D) Four
E) Five</p>
<p>Why is this C?</p>
<p>4, 6, and 12 work.</p>
<p>You can just check each positive integer less than 15 - do the division by hand. The following observations can save you some time as you do this.</p>
<p>(1) a remainder is always less than the divisor, so you can start with 4.
(2) once you get to 8 the remainders decrease by one as you increase the divisor. So you just need to find the value of k so that 15 - k = 3. This value is k = 12.</p>
<p>Note that if the divisor is greater than 15, the remainder is always 15 (in particular it’s not 3). That is, if k>15, then k goes into 15 zero times with 15 left over.</p>
<p>Thanks rspence! Could you explain what mod means? 86 ≡ 2 (mod 7)</p>
<ol>
<li>If (x+y)(x^2 - y^2) = 0, which of the following must be true?
A. x=y
B. x= -y
C. x^2 = y^2
D. x^2 = -y^2
E. x^3 = y^3</li>
</ol>
<p>The answer is C but I don’t understand why A and B are wrong?</p>
<ol>
<li>A two digit number XY, where X and Y are digits is 3 times the sum of its digits. Which of the following equations could be used to represent the statement above?
A. 3(X) + Y= 10(X) +Y
B. 3(X) + Y= X + Y
C. 3(X + Y)= 10(X + Y)
D. 3(X + Y)= 10(X) + Y
E. 3(X + Y)= X + Y</li>
</ol>
<p>The answer is D. Please provide explanation.</p>
<ol>
<li>In a certain parking lot that contains 200 cars, 50 percent of the cars are red, 60 percent are four door cars, and 70 percent have alloy rims. What is the greatest number of cars in the parking lot that could be green two door cars with alloy rims?</li>
</ol>
<p>The answer is 80. Please explain.</p>
<ol>
<li>How many 3 digit positive integers have only odd integers as digits.</li>
</ol>
<p>The answer is 125. Please explain</p>
<p>Thanks for helping.</p>
<p>@tranman26: basically, 86 divided by 7 leaves a remainder of 2. We use the word “mod” (modulo) for remainders, and ≡ means “congruent to.” So 86 ≡ 2 (mod 7).</p>
<ol>
<li><p>Either x = -y or x^2 = y^2. Both A and B <em>could</em> be true, but C includes both cases.</p></li>
<li><p>Just turn it into an algebra question. “A two digit number XY” is equal to 10X + Y, “3 times the sum of its digits” is 3(X+Y). 10X + Y = 3(X+Y).</p></li>
</ol>
<p>37.
50% are red so at most 50% (100 cars) are green.
60% have four doors so at most 40% (80 cars) have two doors.
70% (140 cars) have alloy rims.</p>
<p>Here, the answer is at most 80 cars because at most 80 cars have two doors. It is possible for those 80 cars to satisfy the first and third conditions, so the answer is 80.</p>
<p>37 (second). There are five odd digits 1,3,5,7,9. Five choices for the first digit, five for the second, five for the third. 5<em>5</em>5 = 125.</p>