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15 Mar 2018, 02:24
[GMAT math practice question] X is proportional to Y. If Y is increased by 30%, by approximately what percent is X^2 increased? A. 30% B. 40% C. 50% D. 60% E. 70% => X = kY for some k. Since (1.3kY)^2 = 1.69k^2Y^2 = 1.69(kY)^2 = 1.69X^2, we have (1.69X^2 – X^2) / X^2 = 0.69, which is approximately 70%. Therefore, the answer is E. Answer: E
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16 Mar 2018, 00:55
[GMAT math practice question] Out of 75 students, 17 students enrolled in a Physics class, 28 students enrolled in a Chemistry class, and 39 students enrolled in a Biology class. 5 students enrolled in both the Physics and Chemistry classes, 7 students enrolled in both the Chemistry and Biology classes, and 6 students enrolled in both the Biology and Physics classes. If 4 students enrolled in Physics, Chemistry, and Biology, how many students did not enroll in any of the three science classes? A. 2 B. 3 C. 4 D. 5 E. 6 => Attachment:
777.png [ 10.71 KiB  Viewed 400 times ]
The total number of students is given by a + b + c + d + e + f + g + h = 75. Of these, a + d + f + g = 17 are enrolled in Physics, b + d + e + g = 28 are enrolled in Chemistry, and c + e + f + g = 39 are enrolled in Biology. Adding these three equations gives ( a + b + c ) + 2( d + e + f ) + 3g = 84. We also know that d + g = 5 study both Physics and Chemistry, e + g = 7 study both Chemistry and Biology, and f + g = 6 study both Physics and Biology. Adding these three questions yields ( d + e + f ) + 3g = 18. Since 4 students are enrolled in all three classes, we have g = 4, Plugging this value for g into the above equation yields (d + e + f) + 3g = 18 (d + e + f) + 12 = 18 d + e + f = 6. So, ( a + b + c ) + 2( d + e + f ) + 3g = 84 yields (a + b + c) + 2(6) + 3(4) = 84 (a + b + c) + 12 + 12 = 84 a + b + c = 60. Finally, using the equation, a + b + c + d + e + f + g + h = 75, we see that (a + b + c) + (d + e + f) + g + h = 75 60 + 6 + 4 + h = 75 h = 5. Therefore, the answer is D.
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18 Mar 2018, 23:20
[GMAT math practice question] What is the minimum distance between the point (3,4) and points on the circle x2+y2=1? A. 1 B. 2 C. 3 D. 4 E. 5 Attachment:
19.png [ 6.84 KiB  Viewed 378 times ]
=> Since the distance between (0,0) the center of the circle x^2+y^2=1 and (3,4) is 5, and the radius of the circle is 1, the minimum distance between a point on the circle and (3,4) is 5 – 1 = 4. Therefore, D is the answer. Answer: D
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18 Mar 2018, 23:22
[GMAT math practice question] Suppose y=2^{x^2}+2^{x+3}. Which of these xvalues gives the smallest value of y? A. 3 B. 2 C. 1 D. 1 E. 2 => Since the exponent function y=2^x is an increasing function, the smallest value of x^2+2x+3 gives the smallest value of y=2^{x^2}+2^{x+3}. Since x^2+2x+3 = (x+1)^2 + 2, the exponent (x+1)^2 + 2 has a minimum value when x = 1. Therefore, the answer is C. Answer : C
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21 Mar 2018, 02:51
[GMAT math practice question] In how many ways can two integers m and n, with m > n, be selected from the whole numbers from 12 to 32, inclusive? A. 150 B. 180 C. 190 D. 210 E. 240 => Since the order of m and n is fixed, we only need to count the number of ways to choose 2 numbers from 12, 13, …, 32. We have 21 numbers to choose from since 32 – 12 + 1 = 21. The number of selections is 21C2 = (21*20) / (1*2) = 210. Therefore, D is the answer. Answer: D
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22 Mar 2018, 17:55
[GMAT math practice question] Machine A can produce balls at a constant rate of 2 balls per hour, and machine B can produce balls at a constant rate of 3 balls per hour. If at least one of machine A and machine B produces balls at any time, what is the smallest possible number of hours that machine A and machine B must work together at their constant rates to produce 70 balls in 20 hours? A. 5hrs B. 6hrs C. 7hrs D. 8hrs E. 9hrs => Since machine B has the faster working rate, it must work for the entire period to minimize the number of hours that machine A has to work. When machine B works for 20 hours, it produces 60 balls. This means that machine A must produce 70 – 60 = 10 balls, and it must work for 5 hours, since the work rate of machine A is 2 balls per hour. Therefore, the answer is A. Answer: A
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23 Mar 2018, 00:22
[GMAT math practice question] Alice and Bob traveled in the same direction along the same route at their respective constant speeds of 12 km per hour and 6 km per hour. They each started to travel from their own houses. Bob’s house is halfway between Alice’s house and the destination. After passing Bob, Alice took 10 minutes to reach the destination. How many minutes did it take Bob to reach the destination after Alice passed him? A. 5 min B. 6 min C. 8 min D. 10 min E. 20 min => After Alice passed Bob, she traveled for 10 more minutes. Since Bob’s speed is half of Alice’s speed, he took a further 20 minutes to reach the destination. Therefore, the answer is E. Answer: E
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25 Mar 2018, 18:21
[GMAT math practice question] When you buy a wrap from a certain store, you must select at least one of the three kinds of meat (beef, pork and fish), and one of lettuce, spinach and cucumber. You can also add one of oriental sauce, spicy mayo and sesame soy sauce. How many kinds of wraps can you buy? A. 6 B. 9 C. 12 D. 18 E. 27 => There are three ways to choose one meat (beef, pork or fish), three ways to choose one vegetable (lettuce, spinach or cucumber), and 3 ways to choose one sauce (oriental sauce, spicy mayo or sesame soy sauce). So, the total number of different wraps that can be bought is 3*3*3 = 27. Therefore, E is the answer. Answer: E
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25 Mar 2018, 18:22
[GMAT math practice question] 2018 has 365 days and the 294th day, Oct 31. 2018 is on a Wednesday. On which day will the last day of 2018 fall? A. Mon B. Tue C. Wed D. Thu E. Fri => Since we have 30 days in November and 31 days in December, we have 61 days remain in 2018 after Oct. 31, 2018. Now, 61 = 7*8 + 5. Since 61 has a remainder of 5 when it is divided by 7, the last day of the year 2018 must be a Monday (count 5 days on from Wednesday). Therefore, the answer is A. Answer: A
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28 Mar 2018, 02:02
[GMAT math practice question] Which of the following is the greatest? A. (1+2)^5 B. 10^3 C. (1^2+2^2)^4 D. (2^2+2^2+2^2)^3 E. (2^2+2^2+2^2+2^2)^2 => A. (1+2)^5 = 3^5 = 243 B. 10^3 = 1000 C. (1^2+2^2)^4 = 5^4 = 625 D. (2^2+2^2+2^2)^3 = 12^3 = 1728 E. (2^2+2^2+2^2+2^2)^2 = 16^2 = 256 Therefore, D is the answer. Answer: D
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30 Mar 2018, 02:07
[GMAT math practice question] Which of the following inequalities is equivalent to 2xx<3? A. 0<x<2 B. 0<x<3 C. 1<x<3 D. 0<x<1 E. 3<x<1 => 2xx<3 => 3 < 2x – x < 3 Case 1: If x ≥ 0, then x = x, and so 3 < 2x – x < 3 => 3 < x < 3 => 0 ≤ x < 3, since x ≥ 0. Case 2: If x < 0, then x =  x, and so 3 < 2x – x < 3 => 3 < 3x < 3 =>  1< x < 1 =>  1< x < 0, since x < 0. Thus, 1 < x < 3. Therefore, the answer is C. Answer: C
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01 Apr 2018, 18:46
[GMAT math practice question] How many different threedigit numbers can be formed which contain two digits that are the same, and a third digit that is different from the other two? A. 196 B. 216 C. 243 D. 256 E. 316 => These threedigit numbers can have one of the forms XXY, XYX and YXX. Note that 0 cannot be the hundreds digit. Case 1): XXY There are 9 possibilities for X (X is not 0), and 9 possibilities for Y (Y≠X). This gives a total of 9*9 = 81 possible threedigit numbers of this form. Case 2): XYX There are 9 possibilities for X (X is not 0), and 9 possibilities for Y (Y≠X). This gives a total of 9*9 = 81 possible threedigit numbers of this form. Case 3): YXX There are 9 possibilities for Y (Y is not 0), and 9 possibilities for X (X≠Y). This gives a total of 9*9 = 81 possible threedigit numbers of this form. Thus, the total number of possible threedigit numbers is 81 + 81 + 81 = 243. Therefore, C is the answer. Answer: C
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01 Apr 2018, 18:48
[GMAT math practice question] An organization constructs a committee of 3 people from A, B, C, D, E, F and G. A and B are relatives, so they cannot both be committee members at the same time. How many different committees can be formed? A. 25 B. 30 C. 35 D. 40 E. 45 => The total number of possible committees of three 3 people chosen from 7 people is 7C 3 = 35. However, we need to exclude the committees containing both A and B. The number of committees containing A, B and one other person is equal to the number of ways of choosing 1 person from the C, D, E, F, and G, which is 5C 1 = 5. Therefore, the total number of committees 35 – 5 = 30. Therefore, the answer is B. Answer: B
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04 Apr 2018, 02:52
[GMAT math practice question] The number 123,k50 is a 6digit integer, and k is a positive 1 digit integer. Which of the following cannot be a factor of 123,k50? A. 2 B. 3 C. 4 D. 5 E. 6 => The last two digits tell us whether the number is divisible by 4. Since 50 is not a multiple of 4, the number cannot be a multiple of 4. Therefore, the answer is C. Answer: C Let’s see why the number could be divisible by each of the other options: A: Since the units digit is an even number, the whole number is a multiple of 2. B: A number is divisible by 3 if the sum of its digits is divisible by 3. If k = 4, then the sum of the digits is 1 + 2 + 3 + 4 + 5 + 0 = 15, which is a multiple of 3, and so the number is a multiple of 3. D: Since the units digit is a multiple of 5, the number is a multiple of 5. E: If k = 4, the number is divisible by 3 as seen in part B. Since it is also divisible by 2 (see part A), the number is divisible by 6.
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06 Apr 2018, 00:52
[GMAT math practice question] Car A drives from P to Q at a constant rate of 100 km per hour. After car A has driven for 1 hour, train B begins traveling from Q to P at a constant rate of 150 km per hour. If the distance between P and Q is 600 km, then what distance has car A traveled when it meets train B? A. 200 km B. 220 km C. 250 km D. 270 km E. 300 km => After car A has driven for 1 hour, the distance between car A and train B is 500 km. Car A and train B approach each other at a speed of 250 km/hr. This means that they will take 2 hours to meet each other. When they meet, car A will have traveled for 3 hours, and have covered a distance of 3 * 100 = 300 km. Therefore, the answer is E. Answer: E It is important that both vehicles will have traveled for the same amount of time after train B has started moving. This gives the equation 100 + 100t + 150t = 600, from which we may deduce that t = 2. Since 100 + 100 *2 = 300, E is the answer.
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08 Apr 2018, 18:15
[GMAT math practice question] If x > y >0, a < 0 and b > 0, then which of the following is (are) true? Ⅰ. ax+by Ⅱ.axby Ⅲ. byax A. Ⅰonly B. Ⅱ only C. Ⅲ only D.Ⅰ& Ⅱ only E. Ⅱ & Ⅲ only => Statement I. x = 2, y = 1, a = 1, b = 1: False Statement II. x = 2, y = 1, a = 1, b = 1: False Statement III. Since a < 0 and x > 0, ax > 0. And by > 0 since b > 0 and y > 0. by – ax = by + (ax) > 0. True Therefore, C is the answer. Answer: C
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08 Apr 2018, 18:18
[GMAT math practice question] If R n+1R n=(1/2)^n for positive integers n, which of the following is true? A. R 1>R 3>R 2B. R 1>R 2>R 3C. R 3>R 1>R 2 D. R 2>R 3>R 1 E. R 3>R 2>R 1=> If n = 1, then R 2 – R 1 = (1/2) < 0, and we have R 1 > R 2. If n = 2, then R 3 – R 2 = 1/4 > 0, and we have R 3 > R 2. Furthermore, R 3 – R 1 = ( R 3 – R 2 ) + ( R 2 – R 1 ) = 1/4 + (1/2) = 1/4 < 0, and we have R 3 < R 1. Thus R 1 > R 3 > R 2. Therefore, C is the answer. Answer: C
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11 Apr 2018, 05:57
[GMAT math practice question] A jar contains 4 balls, labeled 1,2,3, and 4. A ball is selected from the jar, and its number is recorded before it is returned to the jar. If a second ball is then selected from the jar, what is the probability that the difference between the numbers on the two balls selected is 1? A. 1/3 B. 3/4 C. 1/2 D. 5/8 E. 3/8 => The total number of ways the two balls may be selected is 4C 2 = ( 4 * 3 ) / ( 1 * 2 ) = 6. There are three ways in which the numbers on the two balls selected can have a difference of 1: ( 1, 2 ), ( 2, 3 ), and ( 3, 4 ). Thus, the probability is 3/6 or 1/2 Therefore, C is the answer. Answer: C
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12 Apr 2018, 01:59
[GMAT math practice question] In the xy plane, what is the slope of the line segment joining the yintercept and the positive xintercept of the curve y=x^25x6? A. 1 B. 2 C. 3 D. 4 E. 5 => First, we find the xintercepts of the curve. These occur when y = 0. Now, y=x^25x6 => y = (x+1)(x6). Thus, the xintercepts are the points (1,0) and (6,0), and the positive xintercept is (6,0). As the yintercept occurs when x = 0, it is the point (0,6). Thus, the required slope is (6 – 0) / ( 0 – (6) ) = 1. Therefore, the answer is A. Answer: A
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Re: Math Revolution Approach (PS)
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13 Apr 2018, 02:44
[GMAT math practice question] Each letter of a 5letter password must be a different letter of the alphabet. Each password must contain 2 vowels and 3 consonants, and the vowels must not be adjacent to each other. How many different passwords of this form can be made? A. 95760 B. 128000 C. 159600 D. 256000 E. 720000 => The total number of passwords with 2 vowels and 3 consonants is determined as follows: 1) Choose the 2 different vowels in 5C 2 ways 2) Choose the 3 consonants in 21C 3 ways 3) Arrange the 5 letters in 5! ways So, the number of passwords is 21C3 * 5C2 * 5! = ( 21*20*19 ) / (1*2*3) * (5*4)/(1*2) * 120 = 133 * 10 * 120 = 159600. The number of passwords with 2 adjacent vowels and 3 consonants is determined as follows: 1) Choose the 2 different vowels in 5C 2 ways 2) Choose the 3 consonants in 21C 3 ways 3) Choose a position for the two adjacent vowels in 4! ways 4) Rearrange the adjacent vowels in 2! ways So, the number of passwords with two adjacent vowels is 21C3 * 5C2 * 4!*2! = ( 21*20*19 ) / (1*2*3) * (5*4)/(1*2) * 48 = 133 * 10 * 48 = 63840. Therefore, the number of passwords without two adjacent vowels is 159600 – 63840 = 95760. Therefore, the answer is A.
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