GRE_Math_Bible_eBook
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Graphs 371
Medium
18.Which one of the following exams has been marked as having the highest Stress Factor?
(A)CET 1990
(B)CET 1991
(C)CET 1992
(D)CET 1993
(E)CET 1994
The Difficulty Factor of the exam is the sum of the products of the number of questions of each type and the corresponding difficulty level.
Then
The Stress Factor = the Difficulty Factor divided by the average time per question.
Let’s calculate the Stress Factor for each answer-choice and choose the one that has the highest value:
Choice (A): In CET 1990,
The Difficulty Factor is (3 55 + 3 55 + 2 55) = 165 + 165 + 110 = 440
The Stress Factor is 440/44 = 10.
Choice (B): In CET 1991,
Difficulty Factor is (2 50 + 1 50 + 1 50) = 100 + 50 + 50 = 200.
The Stress Factor is 200/48 < Choice (A). Reject.
Choice (C): In CET 1992,
Difficulty Factor is (3 50 + 1 50 + 3 50) = 350.
The Stress Factor is 350/48 < Choice (A). Reject.
Choice (D): In CET 1993,
Difficulty Factor is (3 50 + 1 50 + 2 50) = 300.
The Stress Factor is 300/48 < Choice (A). Reject.
Choice (E): In CET 1994,
Difficulty Factor is (3 37 + 1 50 + 3 39) = 111 + 50 + 117 = 278.
The Stress Factor is 278/71 < Choice (A). Reject.
Hence, the answer is (A).
Hard
19.Which one of the following statements can be inferred from the table?
(I)As the Stress Factor increased, the cut off marks of the top five universities decreased
(II)As the Difficulty Factor of the exam decreased, the cut off marks of the top five universities increased
(III)As the Difficulty Factor increased, the Stress Factor increased
(A)(I) only
(B)(II) only
(C)(III) only
(D)(I), and (II)
(E)(I), and (III)
The increasing order of the Difficulty Factor is
CET 1991 (200) < CET 1994 (278) < CET 1993 (300) < CET 1992 (350) < CET 1990 (440).
The increasing order of the Stress Factor is
372GRE Math Bible
CET 1994 (278/71 = 3.92) < CET 1991 (200/48 = 4.16) < CET 1993(300/48 = 6.25) < CET 1992(350/48 = 7.29) < CET 1990 (440/44 = 10).
The decreasing order of the cut off marks is
CET 1994 (68) > CET 1991 (65) > CET 1993 (60) > CET 1992 (58) > CET 1990 (55). The decreasing order of the cut off marks matches the increasing order of the Stress Factor.
The decreasing order of the cut off marks does not match the increasing order of the Difficulty Factor.
As the Difficulty Factor increased, the Stress Factor did not increase. Hence, III is false. The answer is (A), only I is true.
374GRE Math Bible
Hard
22.Which one of the following industry groups employs more than 10 employees per establishment?
(A)Construction
(B)Manufacturing
(C)Wholesale Trade
(D)Retail Trade
(E)Transportation and Warehousing
The correct choice is the industry that employs more than 10 employees per establishment in an average. Hence, the industry with the criterion: The Number of Establishments 10 < the Number of Employees would be the correct choice.
Choice (A): Construction.
The number of establishments = 749.
The Number of Establishments 10 = 7490.
The number of Employees = 6,836.
Here, The Number of Establishments 10 is not less than The number of Employees.
Reject the choice.
Choice (B): Manufacturing.
The number of Establishments = 19,335.
The Number of Establishments 10 = 193,350.
The number of Employees = 188,420.
Here, The Number of Establishments 10 is not less than The number of Employees.
Reject the choice.
Choice (C): Wholesale Trade.
The number of Establishments = 2,320.
The Number of Establishments 10 = 23,200.
The number of Employees = 9,711.
Here, The Number of Establishments 10 is not less than The number of Employees.
Reject the choice.
Choice (D): Retail Trading.
The number of Establishments = 11,342.
The Number of Establishments 10 = 113,420.
The number of Employees = 94,997.
Here, The Number of Establishments 10 is not less than The number of Employees.
Reject the choice.
Choice (E): Transportation and Warehousing.
The number of Establishments = 18,593.
The Number of Establishments 10 =.185,930.
The number of Employees = 299,648.
Here, The Number of Establishments 10 is less than The number of Employees.
Accept.
The answer is (E).
Graphs 375
Hard
23.Which one of the following is a valid inference?
(I)The State Government can be inferred as employing the highest number of Employees per Establishment only because the Percentage Employment it provides is the highest. The number of Establishments is not important.
(II)The State Government can be inferred as employing the highest number of Employees per Establishment since it has the least number of organizations and offers the highest Employment.
(III)The State Government can be inferred as employing the highest number of Employees per Establishment since it has the least number of organizations and offers the highest Percentage of Employment.
(A)I only
(B)II only
(C)III only
(D)I and II
(E)II and III
The Employment per establishment is given as The Number of Employees/The Number of Establishments. The ratio is greatest when the numerator has the greatest positive value, and the denominator has the smallest positive value. Hence, II is true.
The highest employment can also be directly understood by the highest percentage employment. Hence, just as Statement II is true because of the highest employment, Statement III is also true because of the highest percentage employment.
Hence, II and III are correct and the answer is (E).
Word Problems 377
MOTION PROBLEMS
Virtually, all motion problems involve the formula Distance = Rate Time, or
D = R T
Overtake: In this type of problem, one person catches up with or overtakes another person. The key to
these problems is that at the moment one person overtakes the other they have traveled the same distance.
Example: Scott starts jogging from point X to point Y. A half-hour later his friend Garrett who jogs 1 mile per hour slower than twice Scott’s rate starts from the same point and follows the same path. If Garrett overtakes Scott in 2 hours, how many miles will Garrett have covered?
(A) 2 |
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(B) 3 |
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(C) 4 |
(D) 6 |
(E) 6 |
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Following Guideline 1, we let r = Scott's rate. Then 2r – 1 = Garrett's rate. Turning to Guideline 2, we look for two quantities that are equal to each other. When Garrett overtakes Scott, they will have traveled
the same distance. Now, from the formula D = R T , Scott’s distance is D = r 2 12
and Garrett’s distance is D = (2r – 1)2 = 4r – 2 |
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Opposite Directions: In this type of problem, two people start at the same point and travel in opposite
directions. The key to these problems is that the total distance traveled is the sum of the individual distances traveled.
Example: Two people start jogging at the same point and time but in opposite directions. If the rate of one jogger is 2 mph faster than the other and after 3 hours they are 30 miles apart, what is the rate of the faster jogger?
(A) 3 |
(B) 4 |
(C) 5 |
(D) 6 |
(E) 7 |
Let r be the rate of the slower jogger. Then the rate of the faster jogger isr + 2. Since they are jogging for 3 hours, the distance traveled by the slower jogger is D = rt = 3r, and the distance traveled by the faster jogger is 3(r + 2). Since they are 30 miles apart, adding the distances traveled gives
3r + 3(r + 2) = 30
3r + 3r + 6 = 30
6r + 6 = 30
6r = 24 r = 4
Hence, the rate of the faster jogger is r + 2 = 4 + 2 = 6. The answer is (D).
378GRE Math Bible
Round Trip: The key to these problems is that the distance going is the same as the distance returning.
Example: A cyclist travels 20 miles at a speed of 15 miles per hour. If he returns along the same path and the entire trip takes 2 hours, at what speed did he return?
(A) 15 mph |
(B) 20 mph |
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(C) 22 mph |
(D) 30 mph |
(E) 34 mph |
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Solving the formula D = R T for T yields T = |
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is (D).
Compass Headings: In this type of problem, typically two people are traveling in perpendicular
directions. The key to these problems is often the Pythagorean Theorem.
Example: At 1 PM, Ship A leaves port heading due west at x miles per hour. Two hours later, Ship B is 100 miles due south of the same port and heading due north at y miles per hour. At 5 PM, how far apart are the ships?
(A)
(4 x)2 + (100 + 2y)2
(B)x + y
(C)
x 2 + y 2
(D)
(4 x)2 + (2y)2
(E)
(4 x)2 + (100 2y)2
Since Ship A is traveling at x miles per hour, its distance traveled at 5 PM is D = rt = 4x. The distance traveled by Ship B is D = rt = 2y. This can be represented by the following diagram:
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Distance traveled by Ship B |
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Applying the Pythagorean Theorem yields s 2 = (4x)2 + (100 2y)2 . Taking the square root of this equation gives s = 
(4 x)2 + (100 2y)2 . The answer is (E).
380GRE Math Bible
Example: A tank is being drained at a constant rate. If it takes 3 hours to drain 67 of its capacity, how
much longer will it take to drain the tank completely?
(A) 1/2 hour (B) 3/4 hour (C) 1 hour (D) 3/2 hours (E) 2 hours
Since 6/7 of the tank’s capacity was drained in 3 hours, the formula W = R T becomes |
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Solving for R gives R = 2/7. Now, since 6/7 of the work has been completed, 1/7 of the work remains. Plugging this information into the formula W = R T gives 17 = 27 T . Solving for T gives T = 1/2. The answer is (A).
MIXTURE PROBLEMS
The key to these problems is that the combined total of the concentrations in the two parts must be the same as the whole mixture.
Example: How many ounces of a solution that is 30 percent salt must be added to a 50-ounce solution that is 10 percent salt so that the resulting solution is 20 percent salt?
(A) 20 |
(B) 30 |
(C) 40 |
(D) 50 |
(E) 60 |
Let x be the ounces of the 30 percent solution. Then 30%x is the amount of salt in that solution. The final solution will be 50 + x ounces, and its concentration of salt will be 20%(50 + x). The original amount of salt in the solution is 10% 50 . Now, the concentration of salt in the original solution plus the concentration of salt in the added solution must equal the concentration of salt in the resulting solution:
10% 50 + 30%x = 20%(50 + x)
Multiply this equation by 100 to clear the percent symbol and then solving for x yields x = 50. The answer is (D).
COIN PROBLEMS
The key to these problems is to keep the quantity of coins distinct from the value of the coins. An example will illustrate.
Example: |
Laura has 20 coins consisting of quarters and dimes. |
If she has a total of $3.05, how many |
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(A) 3 |
(B) 7 |
(C) 10 |
(D) 13 |
(E) 16 |
Let D stand for the number of dimes, and let Q stand for the number of quarters. Since the total number of coins in 20, we get D + Q = 20, or Q = 20 – D. Now, each dime is worth 10¢, so the value of the dimes is 10D. Similarly, the value of the quarters is 25Q = 25(20 – D). Summarizing this information in a table yields
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D |
20 – D |
20 |
Value |
10D |
25(20 – D) |
305 |
Notice that the total value entry in the table was converted from $3.05 to 305¢. Adding up the value of the dimes and the quarters yields the following equation:
10D + 25(20 – D) = 305
10D + 500 – 25D = 305
–15D = –195
D = 13
Hence, there are 13 dimes, and the answer is (D).