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Math Reviewer for Civil Service Exam 1. Oil is pumped into a non-empty tank at a changing rate. The volume of oil in the tank doubles every minute and the tank is filled in 10 minutes. How many minutes did it take for the tank to be half full? A. 2 B. 5 C. 7 D. 8 E. 9 Answer: (e) Solution: At the 10 minute mark, the tank has twice as much oil in it as it did at the 9 minute mark. So the tank became half-full at the 9 minute mark. Area: Algebra 2. Alice and Bill are walking in opposite directions along the same route between A and B. Alice is going from A to B, and Bill from B to A. They start at the same time. They pass each other 3 hours later. Alice arrives at B 2.5 hours before Bill arrives at A. How many hours does it take for Bill to go from B to A? A. 6 B. 6.5 C. 7 d. 7.5 E. 8.5 Answer: (d) Area: Algebra Solution: Suppose that it takes Alice x hours to go from A to B. Then, it takes Bill x +2.5 hours to go from B to A. Assuming that Alice and Bill maintain constant speed, we get that Alice’s speed is d=x and Bill’s speed is d=(x +2.5), where d is the distance from A to B. Since Alice and Bill met 3 hours after they started walking, Dividing by d and solving for x we get x =5, so the answer is 5+2:5=7:5 hours. Area: Algebra 3. Two armies are advancing towards each other, each one at 1 mph. A messenger leaves the first army when the two armies are 10 miles apart and runs towards the second at 9 mph. Upon reaching the second army, he immediately turns around and runs towards the first army at 9 mph. How many miles apart are the two armies when the messenger gets back to the first army? A. 5.6 b. 5.8 C. 6 D. 6.2 E. 6.4 Answer: (e) Solution: The messenger reaches the second army in one hour (the messenger and the second army advance at each other at a combined speed of 10 mph). At that time, the two armies are 8 miles apart. It takes 0.8 hours for the messenger to get back to the first army. At that time the armies are 8 - 0:8 -0.8=6.4 miles apart. Area: Algebra
4. You play on a game show where a prize has been randomly put into one of five boxes, labeled A, B, C, D, and E, with each box equally likely to contain the prize. The boxes are closed when you first see them, and you guess that box A contains the prize. Two of the empty boxes from among the four other boxes are then opened. If these two boxes are C and E, what is the probability that box D contains the prize? A. 1/5 B. 1/3 C. 2/5 D. 3/5 E. 2/3 Answer: (c) Solution: The probability of the prize being in box A is equal to that of it being in any other box; hence this probability is 1=5. The remaining 4=5 is split between Box D and Box B; hence the probability that Box D contains the prize is 2=5. Area: Statistics 5. Six mountain climbers divide themselves into three teams for the final assault on a peak. One team has 3 climbers; the others have 1 and 2 climbers. All manners of different team deployments are considered, which team goes first, second, and third, and two deployments are considered different if the composition of any of the teams is different. (Disregard deployments within each team.) What is the total number of possible team deployments? A. 60 B. 180 c. 360 D. 720 E. none of the above Answer: (c) Solution: There are six ways to choose which team goes first, which goes second and which 6 goes third. Also, there are = 20 ways to select three climbers out of the six for the three man team, and 3 ways to pick a climber out of the remaining three for the one man team. So, the total number of ways is 6 .20 .3 = 360 ways. Area: Statistics 6. How many positive integers less than or equal to 2013 are divisible by at least one of 3, 11, and 61? A. 3 B. 813 C. 1006 D. 1198 E. 2013 Answer: (b) Solution: There are 2013/3 = 671 multiples of 3 from 1 to 2013, 2013/11 = 183 multiples of 11, and 2013/61 = 33 multiples of 61. To avoid double counting we subtract the 61 multiples of 3 .11, the 11 multiples of 3 .61, and the 3 multiples of 61 .11. We have to add 1 for 2013 which was subtracted one too many times. Thus, the answer is 671 + 183 + 33 – 61 – 11 – 3 + 1 = 813. Area: Algebra 7. How many two digit prime numbers are there in which both digits are prime numbers? (For example, 23 is one of these numbers but 31 is not, since 1 is not a prime number.) A. 3 B. 4 C. 5 D. 8 E. 15 Answer: (b)
Solution: The second digit can only be 3 or 7, so the choice quickly narrows down to 23, 27,33, 37, 53, 57, 73, and 77. Of these, 27, 33, and 57 are divisible by 3, and 77 by 7, leaving 23, 37, 53, and 73. It is easy to see that none of these is divisible by 2, 3, 5, or 7, and there is no need to look at greater prime divisors since √77 < 11. 8. You own thirteen pairs of socks, all different, and all of the socks are individually jumbled in a drawer. One morning you rummage through the drawer and continue to pull out socks until you have a matching pair. How many socks must you pull out to guarantee having a matching pair? A. 3 B. 12 C. 13 D. 14 E. 25 Answer: (d) Solution: You might be unlucky and have the first thirteen socks all different, but then the 14thhas to match one of them. 9. A jeweler has a 20 gram ring that is 60% gold and 40% silver. He wants to melt it down andadd enough gold to make it 80% gold. How many grams of gold should be added? A. 4 grams B. 8 grams C. 12 grams D. 16 grams E. 20 grams Answer: (e) Solution: 10. Consider the following game. A referee has cards labeled A, B, C, and D, and places them face down in some order. You point to each card in turn, and guess what letter is written on the bottom. You guess each of A, B, C, and D exactly once (otherwise there is no chance ofgetting them all right!). You play this game once, and then the referee tells you that you guessed exactly n of the letters correctly. Which value of n is not a possible value of n? A. 0 B. 1 C. 2 D. 3 E. 4 Answer: (d) Solution: If the first three are correct, then by process of elimination the fourth has to be correct also. The same reasoning holds no matter when the three correct answers occur. 11. Consider the following game. A referee has cards labeled A, B, C, and D, and places them face down in some order. You point to each card in turn, and guess what letter is written on the bottom. You guess each of A, B, C, and D exactly once (otherwise there is no chance of getting them all right!).You play this game once, and then the referee tells you that you guessed exactly n of the letters correctly. Which value of n is not a possible value of n? A. 0 B. 1 C. 2 D. 3 E. 4 Answer: (d) Solution: If the first three are correct, then by process of elimination the fourth has to be correct also. The same reasoning holds no matter when the three correct answers occur.
12. On a test the passing students had an average of 83, while the failing students had an average of 55. If the overall class average was 76, what percent of the class passed? A. 44% B. 66% C. 68% D. 72% E. 75% Answer: (e) Solution: Let p = proportion that passed. Then 83 p + 55(1 - p) = 76, so p = 21/ 28 = .75 13. Jack and Lee walk around a circular track. It takes Jack and Lee respectively 6 and 10 minutes to finish each lap. They start at the same time, at the same point on the track, and walk in the same direction around the track. After how many minutes will they be at the same spot again (not necessarily at the starting point) for the first time after they start walking? A. 15 B. 16 C. 30 D. 32 E. 60 Answer: (a) Solution 1: Experimenting with the numbers in turn, dividing 6 into 15 and 10 into 15, we get answers one whole number apart, so they are together again at 15 minutes. 14. A class has three girls and three boys. These students line up at random, one after another. What is the probability that no boy is right next to another boy, and no girl is right next to another girl? A. 1/20 B. 1/12 C. 1/10 D. 3/10 E. 1/2 Answer: (c) Solution: We pick the students in order. The first choice is always okay. There is a probability 3/5 that the second choice is okay; for example, if you picked a boy first then there are two boys and three girls left. Similarly the probabilities that the remaining choices are okay is 1/2, 2/3,1/2, 1. The answer is 1/10 the product of all these probabilities. 15. At a party, each person shakes hands with 5 other people. There are a total of 60 handshakes. How many people are at the party? A. 6 B. 12 C. 15 D. 24 E. 30 Answer: (d) Solution: Let x be the total number of people. Then the total number of handshakes is 5x/2where we have to divide by 2 since every handshake is counted twice. Thus 5x/2 = 60, which leads to x = 24. 16. Each of 10 students has a ticket to one of ten chairs in a row at a theater. How many ways are there to seat the students so that each student sits either in the chair specified on his/her ticket or in one right next to it? Each chair is to be occupied by exactly one student. A. 89 B. 144 C. 243 D. 512 E. 1024 Answer: (a) Solution: Let F(n) be the number of ways to seat n students in a row of n chairs as described in the problem. The student who has the ticket to seat number n has two options. He/she can either sit in his/her own chair, in which case the remaining n - 1 students can arrange themselves in F(n - 1) ways. Alternatively, if this student takes the (n - 1)st seat, the student with