A pair of dice is tossed twice. What is the probability that the first toss gives a total of either 7 or 11 and the second toss gives a total of 7 ?
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Solution
There are a total of 62 = 36 possibilities for each toss.There are a total of 8 ways we can get a total of 7 or 11 on the first toss: 6 ways to get a total of 7—(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), or (6, 1)—plus 2 ways to get a total of 11—(5, 6) or (6, 5).Therefore, the probability of getting a total of either 7 or 11 on the first toss is \(\frac{8}{36}=\frac{2}{6}\).The probability of getting a total of 7 on the second toss is \(\frac{6}{36}=\frac{1}{6}\)so the probability that both of these independent events occur is the product \(\frac{6}{36}=\frac{1}{6}\), choice (A).
Water flows into a 25-liter bucket through a hose and out through a hole in the bottom of the bucket. The rate of flows through the hose is 1 liter per minute. If the bucket is filled to capacity in 40 minutes, at what rate, in liters per minute, was water flowing out of the bucket through the hole?
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Solution
Remember that amount = rate × time. So, 25 liters = rate × 40 minutes.The rate was \(\frac{25liters}{40minutes}=\frac{5}{8}\) liters/min.The net rate at which the bucket is filling is the difference between the hose’s rate and the leaking rate. So, (1 liter/min) – (leaking rate) = 5⁄8. Solve for the leaking rate to find the leaking rate is 5⁄8 liters/min; the answer is choice (A).
If the average (arithmetic mean) of 31, 41, and p is between 29 and 47, inclusive, what is the least possible value of (p – 7)2 =
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Solution
64
To find the least possible value of p, work with the lowest possible average, 29. Draw an Average Pie.You have 3 values with an average of 29, so your total is 3 × 29 = 87. Now you know that 31 + 41 + p = 87, so p = 15, and (p — 7)2 = 64.
A bag contains 12 marbles: 5 of the marbles are red, 3 are green, and the rest are blue.
| Quantity A | Quantity B |
| The probability of consecutively choosing two red marbles and a green marble without replacement | The probability of consecutively choosing a red and two blue marbles with replacement |
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Solution
Quantity A asks for the probability “without replacement,” so that means you have to take into account that there will be one marble less in the total after each draw.The probability of first choosing a red marble is \(\frac{5}{12}\), a second red marble is \(\frac{5}{12}\), and then a green marble is \(\frac{5}{12}\).This is an “and” probability problem, so you have to multiply the probability of each event together:\(\frac{5}{12}\times \frac{4}{11}\times \frac{3}{10}=\frac{60}{1320}=\frac{1}{22}\). For Quantity B, you do the same thing, but the total stays the same for each draw:\(\frac{5}{12}\times \frac{4}{12}\times \frac{4}{10}=\frac{80}{1720}=\frac{5}{108}\). Quantity B is greater.
Five numbers in a set are arranged from least to greatest. If the median of the first two numbers is 13 and the average (arithmetic mean) of the remaining numbers is 23, what is the average (arithmetic mean) of the entire set?
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Solution
19
Take the problem a piece at a time. If a set has only two numbers, the median is the average of those two numbers. Set up average wheels for each of the first two averages. For the first one, two numbers that average 13 total to 26. For the second average, three numbers that average 23, so the must have a total of 69. Set up one final average wheel for all five numbers: 5 numbers total 95. Solve for the average, and you should get 19 from 95 ÷ 5.
If the probability that the first event will occur is 1⁄4, and the probability that the second event will occur is 1⁄2, then what is the probability that both events will occur?
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Solution
Plug In to make this problem much simpler. If you plug in x = 2, then the probability for the second event is:\(\frac{1}{\sqrt{4}}=\frac{1}{2}\). Now, because this is an “and” probability problem, you multiply the two probabilities together to find the target answer:\(\frac{1}{\sqrt{4}}=\frac{1}{2}\).Choice (A) is the only one that works:\(\frac{1}{\sqrt{4}}=\frac{1}{2}\).
Trip A is 6x miles long and takes 5y hours.Trip B is 5x miles long and takes 4y hours.
| Quantity A | Quantity B |
| The rate of trip A in miles per hour | The rate of trip B in miles per hour |
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Solution
Rate is calculated by dividing the distance traveled by the time elapsed. Plug In to compare the two rates. If you plug in x = 4 and y = 2, the rate for trip A will be:\(\frac{6(4)}{5(2)}=\frac{24}{10}=2.4\) mph and the rate for trip B will be:\(\frac{5(4)}{4(2)}=\frac{20}{8}=2.5\) mph; eliminate choices (A) and (C). Any set of values will have a greater rate for trip B, so select choice (B).
x,x2,xy,xy-1,x4,x6
x > 1
| Quantity A | Quantity B |
| The mode of the numbers above when y = 4 | The median of the numbers above when y = 5 |
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Solution
In Quantity A, if y = 4, then the numbers (arranged in increasing order) become x, x2, x3, x4, x4, x6; the mode is x4. In Quantity B, if y = 5, then the numbers become x, x2, x4, x4, x5, x6. Usually, you’d need to take the average of the middle two numbers to find the median because there is an even number of values, but in this case they’re both x4.The median, then, is x4, so the quantities are equal. Because x > 1, you don’t have to worry about special cases such as 0, 1, negatives, or fractions, and the correct answer is choice (C).
If the average of 10, 12, n, and n is greater than 25, what is the least possible value of integer n ?
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Solution
To find the average, divide the total by the number of values. So,\(\frac{10+12+n+n}{4}\) > 25. Multiply both sides of the inequality by 4 and then subtract 22 (10 + 12) to find 2n > 78. Divide by 2 to find n > 39. So, n is NOT 39; it is the least integer greater than 39, which is 40. Alternatively, you could plug in the answers starting with choice (A) because the question asks for the least possible value. If n = 38 or 39, then the average is not greater than 25. If n = 40, the average is greater than 25.The answer is choice (C).
The average (arithmetic mean) number of passengers on a subway car is 60. If the number of passengers on a car has a normal distribution with a standard deviation of 20, approximately what percent of subway cars carry more than 80 passengers?
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Solution
Adding the standard deviation (20) to the mean (60) gives you the number of passengers in a car that carries exactly one standard deviation above the mean number of passengers (80).The first standard deviation above the mean represents 34% of the population in a normal distribution, and a further 50% falls below the mean, so 84% of the cars will carry 80 people or fewer. Subtracting this from the entire population (100%) gives you the percent of cars that carry greater than 80: 100% − 84% = 16%.The answer is choice (A).