Quantity A | Quantity B |
The number of hours in d days | The number of minutes in d/2.5 hours |
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Solution
Quantity A | Quantity B |
x2 + 3 | 3x − 2 |
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Solution
Account M earns 4.3% interest, compounded annually.
Account N earns 2.9% interest, compounded annually.
Joe opens one account of each type on January 1, 2000.
Quantity A | Quantity B |
The amount of interest that Joe earns from leaving $X in account M for 6 years. | The amount of interest that Joe earns from leaving $Y in account N for 9 years. |
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Solution
(D) The amounts of money, X and Y, that are invested in the two accounts are unknown, so we cannot determine the dollar amount of interest each account earns, (D).
The number of pull-ups performed
in 1 minute by 10 kids in a class is
given by set S = {0, 0, 0, 2, 9, 11, 11, 14, 16, 17}
Quantity A | Quantity B |
The median number of pull-ups performed by the 10 kids | The mean (average) number of pull-ups performed by the 10 kids |
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Solution
(x 2 + 9)(x + 9)(x – 5) = 0
Quantity A | Quantity B |
x | 7 |
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Solution
(B) The product of the three quantities is zero, so one of the individual factors (x2 + 9), (x + 9), or (x – 5) must equal zero. x2 + 9 = 0 has no real solution, because x2 can’t equal –9, or any negative. If x + 9 = 0 then x = –9, and if x – 5 = 0, then x = 5. So the two possible values of x are –9 and 5, both less than 7. Then Quantity (B), 7, is greater.
Quantity A | Quantity B |
(-2)16 | 85 |
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Solution
Quantity A | Quantity B |
The greatest common factor of 84 and 700 | The least common multiple of 6 and 15 |
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Solution
Team A has a 50% chance of winning any game that it plays.
Team B has a 60% chance of winning any game that it plays.
Quantity A | Quantity B |
The probability that Team A will win its next three games in a row | The probability that Team B will win its next four games in a row |
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Solution
(B) To find the probability of several independent events all occurring, multiply the probabilities of each event. For Team A to win all three of its next three games, the probability is .5 × .5 × .5 = .125. For Team B to win four games in a row, the probability is .6 × .6 × .6 × .6 = .1296, so (B) is greater. (Note: 50% = .5 and 60% = .6.)
The measure of one angle in an isosceles triangle is 120°. If the length of the side opposite this angle equals 12√3 , then what is the perimeter of the triangle?
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Solution
The diagram above represents a rectangular picture (inner region) surrounded by a picture frame of uniform width (darker region). The picture measures 8 inches by 12 inches. The area of the picture frame is equal to the area of the picture itself. What is the width of the frame?
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Solution
How many whole numbers between 100 and 200 are not equal to the square of another integer?
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Solution
In the figure above, the shaded region is a square with area equal to 256. If AB = 8, then what is the length of diagonal AC?
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Solution
A business is currently selling 30 chairs per day for a price of $25 per chair. A worker predicts that if the business chooses to lower the cost of the chairs, then for every $1 the price is lowered, one more chair will be sold. If the prediction is accurate, what is the maximum revenue the business can earn in a day from selling chairs? (Revenue is the amount of money that the business takes in, without consideration of expenses or other costs.)
$
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Solution
In the figure above, what is the value of y?
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Solution
(126) The sum of the measures of the three angles in a triangle is 180°, so 2x + 3x + 5x = 180. Then 10x = 180, and x = 18. Then the 3x angle measures 3(18) = 54°. Since the sum of angles that comprise a straight line is 180°, 54 + y = 180, and y = 126.
The ratio of boys to girls in a room is 7 to 5. The room contains 312 children. How many more boys than girls are in the room?
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Solution
In a certain game, a player flips a coin to determine movement. Every toss of the coin lands with either a head facing up or a tail facing up. If a head is facing up, the player moves forward 10 squares, and if a tail is facing up, he moves backward 13 squares. After flipping the coin 120 times, the player is four squares forward of his starting position. How many of the 120 flips showed a head facing up?
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Solution
A nurse has 80 ml of a solution that has a concentration of 5% Drug X in water. How many milliliters of pure water containing none of the drug must the nurse add to dilute the concentration to 2% Drug X in water?
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Solution
(120 ml) Drug X is 5% of the 80 ml solution, so there are .05 × 80 = 4 ml of Drug X. We want to dilute the solution to a 2% concentration, so this 4 ml of the drug will be 2% of some total volume, say T. That is, 4 = .02T. Then T is 4 ÷ .02 = 200. If the total volume after adding water will be 200 ml, then the amount added must be 200 minus the initial volume: 200 – 80 = 120 ml are added.
From the menu above, a restaurant offers customers two choices of appetizer and one choice each of entrée and dessert. If the appetizers must be different from each other, how many different meal combinations are available to customers?
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Solution
For the animal whose final weight differed most from its starting weight, what was the approximate change in weight from the beginning of the study to the end?
For the animal whose final weight differed most from its starting weight, what was the approximate change in weight from the beginning of the study to the end?
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Solution
(C) Examine only the beginning and ending weights for all three animals. For the cat and hamster the difference between the original and final weights was less than 2 kg. For the dog, the initial weight was a little more than 10 kg, while the final weight was a little more than 8 kg. The dog’s weight changed by 10 – 8 = 2 kg, which is the greatest change for the three animals. (C).
For the animal whose final weight differed most from its starting weight, what was the approximate change in weight from the beginning of the study to the end?
The data supports which of the following claims?
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Solution
(D) The hamster’s weight changed from nearly 6 kg in week 1 to just above 2 kg in week 2, a change of over 50% in 1 week, so (A) is not correct. The dog’s weight parallels the cat’s fairly closely throughout the study and is usually about 2 kg greater than the cat’s, except in week 4, when their weight graphs move closer together, showing that (B) is not correct. Each animal had a lower final weight than starting weight, so (C) is not correct. An examination of the week 3 weights for the animals shows that the dog did reach its maximum weight in week 3, so (D) is correct. The dog’s and cat’s weights did follow the “decrease-increase-decrease” pattern throughout the study, but the hamster’s weight simply decreased then increased, so (E) is not correct.