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Math Revolution GMAT Instructor
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10 Aug 2018, 02:35
[ Math Revolution GMAT math practice question] If x, y, z are positive integers and x<y<z, which of the following must be greater than (z^x)(z^y)? A. y^{2z} B. y^{2y} C. x^{2y} D. x^{zx} E. z^{2y} => Since z^x < z^y, we have (z^x)(z^y)< (z^y)(z^y) = z^{2y}. Therefore, the answer is E. Answer: E
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13 Aug 2018, 06:27
[ Math Revolution GMAT math practice question] he lengths of two sides of a certain obtuse triangle are 9 and 12. Which of the following could be the length of the third side of the triangle? A. 12 B. 13 C. 14 D. 15 E. 16 => If a, b and c are sides of a right triangle and c is the longest side, then c^2 = a^2 + b^2. If a, b and c are sides of an acute triangle and c is the longest side, then c^2 < a^2 + b^2. If a, b and c are sides of an obtuse triangle and c is the longest side, then c^2 > a^2 + b^2. Since 15^2 = 9^2 + 12^2 and 16 > 15, 16 could be the third side of the obtuse triangle. Therefore, the answer is E. Answer : E
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Re: Math Revolution Approach (PS)
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13 Aug 2018, 06:29
[ Math Revolution GMAT math practice question] In the xy coordinate plane, the distance between (p,q) and (1,1) is 5. If p and q are integers, how many possibilities are there for the point (p,q)? A. 2 B. 4 C. 8 D. 12 E. 16 => (p1)^2 + (q1)^2 = 5^2 If p – 1 = ±3, and q  1 = ±4, then p = 1 ± 3, and q = 1 ± 4. There are four possible points: ( p, q ) = ( 4, 5 ), ( 4, 3 ), ( 2, 5 ), ( 2, 3 ). If p – 1 = ±4, and q  1 = ±3, then p = 1 ±4, and q = 1 ± 3. There are four possible points: ( p, q ) = ( 5, 4 ), ( 5, 2 ), ( 3, 4 ), ( 3, 2 ). If p – 1 = 0, and q  1 = ±5, then p = 1, and q = 1 ±5. There are two possible points: ( p, q ) = ( 1, 6 ), ( 1, 4 ). If p – 1 = ±5, and q  1 = 0, then p – 1 = ±5, and q = 1. There are two possible points: ( p, q ) = ( 6, 1 ), ( 4, 1 ). There are a total of 4 + 4 + 2 + 2 = 12 possibilities for the point (p,q). Therefore, the answer is D. Answer: D
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15 Aug 2018, 07:50
[ Math Revolution GMAT math practice question] If the greatest common divisor of (n1)!, n!, and (n+1)! is 5040, what is the value of n? A. 4 B. 5 C. 6 D. 7 E. 8 => Since n! and (n+1)! are multiples of (n1)!, (n1)! is their gcd. It follows that n – 1 = 7 or n = 8, since 7! = 5040. Therefore, E is the answer. Answer: E
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16 Aug 2018, 08:56
[ Math Revolution GMAT math practice question] If n is a positive integer, which of the following can’t be the value of (n+1)^4n^4? A. 2465 B. 4641 C. 6096 D. 7825 E. 9855 => If n + 1 is an even number, then n is an odd number and (n+1)^4n^4 must be an odd number. If n + 1 is an odd number, then n is an even number and (n+1)^4n^4 must be an odd number. All answer choices except for C) are odd numbers. Therefore, the answer is C. Answer: C
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17 Aug 2018, 02:29
[ Math Revolution GMAT math practice question] If (2/3)^n=(9/4)^2, what is the value of n? A. 4 B. 2 C. 0 D. 2 E. 4 => (9/4)^2 = ((3/2)^2)^2 = (3/2)^4 = (2/3)^{4}. Thus, n = 4. Therefore, the answer is A. Answer: A
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19 Aug 2018, 18:45
[ Math Revolution GMAT math practice question] The median of a data set is x and its maximum is 40. The range of the set is 10 greater than the median. What is the minimum of the set in terms of x? A. x  20 B. 2x  40 C. 2x D. 20 – x E. 40 – 2x => Attachment:
8.20.png [ 1.12 KiB  Viewed 580 times ]
The range of the set is 2 times 40 – x, since 40 – x is the distance between the median and the maximum. The minimum is the median minus the distance between the median and the maximum, which is x – ( 40 – x ) = 2x – 40. Therefore, the answer is B. Answer : B
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19 Aug 2018, 18:47
[ Math Revolution GMAT math practice question] What is the 777th decimal digit of 0.345634563456……..? A. 3 B. 4 C. 5 D. 6 E. 7 => The digits of 0.345634563456…. repeat with period 4. Since the 1st, 5th, 9th, … decimal digits are 3, the nth decimal digit is 3 if n has a remainder of 1 when it is divided by 4. Since the 2nd, 6th, 10th, … decimal digits are 4, the nth decimal digit is 4 if n has a remainder of 2 when it is divided by 4. Since the 3rd, 7th, 11th, … decimal digits are 5, the nth decimal digit is 5 if n has a remainder of 3 when it is divided by 4. Since the 4th, 8th, 12th, … decimal digits are 6, the nth decimal digit is 6 if n is a multiple of 4. Since 777 has a remainder of 1 when it is divided by 4, the 777th decimal digit is 3. Therefore, the answer is A. Answer: A
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Re: Math Revolution Approach (PS)
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22 Aug 2018, 01:45
[ Math Revolution GMAT math practice question] Let z be the harmonic mean of x and y. If 1/z=(1/2)((1/x)+(1/y)), which of the following is an expression for z, in terms of x and y? A. 2xy / ( x + y ) B. 2( x + y ) / ( x – y ) C. 2( x – y ) / ( x + y ) D. 2( x + y ) / xy E. xy / ( x + y ) => 1/z = (1/2)(1/x + 1/y) = (1/2)( (x+y)/xy ) = (x+y)/(2xy) Thus, z = 2xy / ( x + y ). Therefore, A is the answer. Answer: A
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23 Aug 2018, 02:13
[ Math Revolution GMAT math practice question] Which of the following is the best approximation for 11^{11}9^9? A. 11^8 B. 11^9 C. 11^{10} D. 11^{11} E. 11^{12} => 9^9 is much smaller than 11^{11}. Thus, 11^{11}9^9 is closest to 11^{11}. Therefore, the answer is D. Answer: D
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24 Aug 2018, 00:53
[ Math Revolution GMAT math practice question] Alice, Bob, Cindy, Dave and Eddie joined a threepersonaside basketball tournament. In how many ways can be the three starters be chosen? A. 5 B. 6 C. 8 D. 9 E. 10 => The number of ways of choosing the three starters is the number of ways of choosing three people from a set of five people, where order does not matter and repetition is not allowed. The number of possible selections of three starters is: 5C 3 = 5C 2 = (5*4)/(1*2) = 10. Therefore, the answer is E. Answer: E
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26 Aug 2018, 18:23
[ Math Revolution GMAT math practice question] A whole number greater than 1 has remainders of 1 when it is divided by each of the numbers of 2, 3, 4 and 5. What is the smallest such number? A. 31 B. 51 C. 61 D. 91 E. 121 => Let x be the smallest number satisfying the original condition. Then x – 1 is the least common multiple of 2, 3, 4 and 5. So, x – 1 = 60. Thus, x = 61. Therefore, the answer is C. Answer: C
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26 Aug 2018, 18:29
[ Math Revolution GMAT math practice question] The average (arithmetic mean) of the five numbers in a data set is 72. The average (arithmetic mean) of the first three numbers is 98. What is the average (arithmetic mean) of the last two numbers? A. 31 B. 32 C. 33 D. 34 E. 35 => Let a, b, c, d and e be the five numbers. Now, the average of these numbers is ( a + b + c + d + e ) / 5 = 72, so a + b + c + d + e = 72 * 5 = 360. The average of the first three numbers is ( a + b + c ) / 3 = 98, so a + b + c = 98 * 3 = 294. Thus, the sum of the last two numbers is d + e = ( a + b + c + d + e ) – ( a + b + c ) = 360  294 = 66. Their average is ( d + e ) / 2 = 66 / 2 = 33. Therefore, the answer is C. Answer: C
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29 Aug 2018, 01:55
[ Math Revolution GMAT math practice question] If x and y are positive integers and xy=36, what is the smallest possible value of x+y? A. 6 B. 8 C. 10 D. 12 E. 18 => When we know the product of two numbers, the sum of those numbers is a minimum when the two numbers are equal. Since xy = 36, the minimum value of x + y occurs when x = y = 6. It is 6 + 6 = 12. Therefore, D is the answer. Answer: D
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30 Aug 2018, 00:52
[ Math Revolution GMAT math practice question] What is the sum of all multiples of 3 between 1 and 200? A. 3300 B. 6600 C. 6633 D. 10100 E. 20100 => We need to find the sum of 3, 6, …, 198. The number of data values is (198 – 3)/3 + 1 = 195/3 + 1 = 65 + 1 = 66. Thus the sum of 3, 6, …, 198 is 3 + 6 + … + 198 = 3+6+…+198=3(1+2+…+66)=3(66)(66+1)/2=3(33)(67)=6,633. Therefore, the answer is C. Answer: C
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31 Aug 2018, 01:16
[ Math Revolution GMAT math practice question] What is the units digit of (3^{101})(7^{103})? A. 1 B. 3 C. 5 D. 7 E. 9 => The units digit is the remainder when (3^{101})(7^{103}) is divided by 10. The remainders when powers of 3 are divided by 10 are 3^1: 3, 3^2: 9, 3^3: 7, 3^4: 1, 3^5: 3, … So, the units digits of 3n have period 4: They form the cycle 3 > 9 > 7 > 1. Thus, 3^n has the units digit of 3 if n has a remainder of 1 when it is divided by 4. The remainder when 101 is divided by 4 is 1, so the units digit of 3^{101} is 3. The remainders when powers of 7 are divided by 10 are 7^1: 7, 7^2: 9, 7^3: 3, 7^4: 1, 7^5: 7, … So, the units digits of 7^n have period 4: They form the cycle 7 > 9 > 3 > 1. Thus, 7^n has the units digit of 3 if n has a remainder of 3 when it is divided by 4. The remainder when 103 is divided by 4 is 3, so the units digit of 7^{103} is 3. Thus, the units digit of (3^{101})(7^{103}) is 3*3 = 9. Therefore, the answer is E. Answer: E
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02 Sep 2018, 18:51
[ Math Revolution GMAT math practice question] A jar contains 16 red balls and 8 white balls. If 3 balls are selected at random from the jar, what is the approximate probability that all balls selected are white? A. 0.02 B. 0.03 C. 0.04 D. 0.05 E. 0.06 => There are 24C 3 ways of choosing 3 balls from the 24 in the jar, and 8C 3 ways of choosing 3 balls from the 8 white balls. Therefore, the probability of choosing 3 white balls from the jar is: 8C 3 / 24C 3 = { (8*7*6) / (1*2*3) } / { (24*23*22) / (1*2*3) } = (8*7*6) / (24*23*22) = 7 / ( 23* 11 ) ≒ 1/ (3*11) ≒ 0.03 Therefore, the answer is B. Answer: B
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02 Sep 2018, 18:57
[ Math Revolution GMAT math practice question] When n is an odd integer, f(n)=3^n and when n is an even integer, f(n)=4^{n/2}. What is the value of f(2^5)=? A. 2^5 B. 3^{11} C. 3^{12} D. 3^{13} E. 2^{32} => Since n = 2^5 is an even integer, f(2^5) = 4^2^5/2=4^2^4=(2^2)^16=2^32. Therefore, the answer is E. Answer: E
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Re: Math Revolution Approach (PS)
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05 Sep 2018, 02:20
[ Math Revolution GMAT math practice question] Which of the following lines is perpendicular to y=(1/2)x? A. y=2x B. y=x/2 C. y=2x D. y=x/2 E. y=x => Two lines are perpendicular if and only if the product of their slopes is 1. Since (1/2)*(2) = 1, the slope of the line perpendicular to y = (1/2)x is 2. The only answer choice with a slope of 2 is A. Therefore, A is the answer. Answer: A
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Math Revolution GMAT Instructor
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Re: Math Revolution Approach (PS)
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06 Sep 2018, 01:35
[ Math Revolution GMAT math practice question] If x and y are integers, what is the least possible positive value of 21x+35y? A. 1 B. 3 C. 5 D. 7 E. 9 => 21x + 35y = 7*3x + 7*5y = 7(3x+5y) is a multiple of 7. The smallest positive multiple of 7 is 7. So, we look for x and y such that 21x + 35y = 7. This will occur when 3x + 5y = 1. When x = 2 and y = 1, 3x + 5y = 3*2 + 5*(1) = 6 – 5 = 1, and 21x + 35y = 7(3x+5y) = 7(1) = 7. Thus, 7 is the answer. Therefore, the answer is D. Answer: D
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