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17 May 2018, 01:43
[GMAT math practice question] In a certain theater, the first row has 12 seats, and each row has 1 more seat than the previous row. If the last row has 50 seats, what is the total number of seats in the theater? A. 1003 B. 1029 C. 1129 D. 1209 E. 1,339 => The question asks for the value of 12 + 13 + ... + 50. This is the sum of an arithmetic sequence with first term a = 12, and last term l = 50. The sum of n terms of an arithmetic sequence may be found using the formula n/2 (a + l). The number of rows is n = 50 – 12 + 1 = 39. So, the number of seats in the theater is 39 * ( 12 + 50 ) / 2 = 39 * 62 / 2 = 39 * 31 = 1209. Therefore, the answer is D. Answer: D
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18 May 2018, 01:48
[GMAT math practice question] If 10^n< 0.003456 <10^{n+1}, what is the value of the integer n? A. 4 B. 3 C. 2 D. 1 E. 0 => When we multiply all sides of the inequality by 10^6, we obtain 10^{n*}10^6 < 3,456 < 10^{n+1*}10^6 or 10^{n+6} < 3,456 < 10^{n+7}.. Since 1,000 < 3,456 < 10,000, 10^{n+6}=1,000=10^3 and n+6=3. So, n=3. Therefore, the answer is B. Answer: B
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20 May 2018, 18:14
[GMAT math practice question] How many 4digit numbers have only even digits? A. 500 B. 525 C. 600 D. 625 E. 800 => Suppoose abcd is 4digit number with only even digits. Then a is one of 2, 4, 6 and 8, and b, c and d may be selected from the digits 0, 2, 4, 6 and 8. So, there are 4 choices for a, and 5 choices for each of b, c and d. The total number of 4digit numbers with only even digits is thus 4*5*5*5 = 500. Therefore, A is the answer. Answer: A
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20 May 2018, 18:17
[GMAT math practice question] How many triples (a,b,c) of even positive integers satisfy a3 + b2 + c = 50? A. one B. two C. three D. four E. five => Consider the variable a first. Since 4^3 > 50, we can only have a = 2. If a = 2, then b^2 + c = 42 since a^3 = 2^3 = 8. Since 8^2 = 64 > 42, we can only have b = 2, 4 or 6. If b = 2, then b^2 + c = 2^2 + c = 4 + c = 42 and c = 38. If b = 4, then b^2 + c = 4^2 + c = 16 + c = 42 and c = 26. If b = 6, then b^2 + c = 6^2 + c = 36 + c = 42 and c = 6. Thus, there are three possible triples: ( 2, 2, 38 ), ( 2, 4, 26 ) and ( 2, 6, 6 ). Therefore, the answer is C. Answer: C
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23 May 2018, 18:12
[GMAT math practice question] If 123,456=123a+87 and 234,567=123b+6, how many multiples of 123 lie between 123,456 and 234,567? A. a B. b C. a+b D. ab E. ba => Since 123,456 = 123*1003 + 87, we must have a = 1003. Since 234,567 = 123*1907 + 6, we must have b = 1907. The multiples of 123 between 123,456 and 234,567 are 123*1004, 123*1005, …, 123*1907. Thus, 1907 – 1004 + 1 = 1907 – 1003 = b – a multiples of 123 lie between 123,456 and 234,567. Therefore, E is the answer. Answer E
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24 May 2018, 18:29
[GMAT math practice question] How many 4digit integers have the form abcd, where b is even and d >= 2b? A. 900 B. 1200 C. 1520 D. 2400 E. 2700 => Suppose abcd is a 4digit number. There are 9 possible values for a: a = 1, 2, …, 9. There are 10 possible values of c: c = 0,1,2,…,9. Since b is even, b can take on the values 0,2,4,6 and 8. However, the condition d >= 2b limits the possible values of b to 0, 2 and 4. Case 1 : b = 0 => d = 0, 1, … , 9 The number of possible values of d is 10. There are 10 * 9 * 10 = 900 4digit integers with b = 0. Case 2: b = 2 => d = 4, 5, …, 9 The number of possible values of d is 6. There are 6 * 9 * 10 = 540 4digit integers with b = 2. Case 3: b = 4 => d = 8, 9 The number of possible values of d is 2. There are 2 * 9 * 10 = 180 4digit integers with b = 4. In total, there are 900 + 540 + 180 = 1520 possible 4digit integers of this form. Therefore, the answer is C. Answer: C
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25 May 2018, 02:43
[GMAT math practice question] The median of 5 numbers is 50, and their range is 40. If the median of the 3 smallest numbers is 40, which of the following could not be the range of the 3 largest numbers? I. 0 II. 20 III. 40 A. I only B. II only C. I & II only D.I & III only E. I, II & III => Suppose a, b, c, and d satisfy a ≤ b ≤ 50 ≤ c ≤ d. Since the median of the 3 smallest numbers is 40, we must have b = 40. The range of the 3 largest numbers is d – 50. Since we are told that d – a = 40, the maximum range of the 3 largest numbers occurs when a = b = 40. Then d = 80, and the maximum range is d – 50 = 80 – 50 = 30. The minimum range of the 3 largest numbers occurs when a = 10 and b = 40. Then we have c = 50 and d = 50, and the range is d – 50 = 50 – 50 = 0. The range of the 3 largest numbers lies between 0 and 30, inclusive. Thus, 0 and 20 are the only possible values. Therefore, the answer is C. Answer: C
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27 May 2018, 18:22
[GMAT math practice question] What is the remainder when 7^{100} is divided by 50? A. 0 B. 1 C. 7 D. 21 E. 49 => The remainder when 7^{100} is divided by 50 depends only on the units and tens digits. The units digits of 7^n cycle through the four values 7, 9, 3, and 1. The tens digits of 7^n cycle through the four values 0, 4, 4, and 0. We have the following sequence of units and tens digits for 7^n: 7^1 = 07 ~ 07 7^2 = 49 ~ 49 7^3 = 343 ~ 43 7^4 = 2401 ~ 01 7^5 = 16807~ 07 … So, 7^{100} = (7^4)^}{25} has the same units and tens digits as 7^4, that is, 01. Thus, the remainder when 7^{100} is divided by 50 is 1. Therefore, B is the answer. Answer : B
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27 May 2018, 18:23
[GMAT math practice question] The alarm of a certain clock rings every 15 minutes. If the alarm first rings at 12:15, when will it ring for the 24th time? A. 17:30 B. 18:00 C. 18:30 D. 19:00 E. 19:30 => It takes 1 hour = 4 x 15 minutes for every fourth alarm to ring. The time period starts at fifteen minutes before 12:15, or 12:00. So: The 4th alarm rings at 13:00. The 8th alarm rings at 14:00. The 12th alarm rings at 15:00. The 16th alarm rings at 16:00. The 20th alarm rings at 17:00. The 24th alarm rings at 18:00. Therefore, the answer is B. Answer: B
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Re: Math Revolution Approach (PS)
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27 May 2018, 22:43
MathRevolution wrote: [GMAT math practice question]
If 3 persons are selected at random from 8 persons for 3 positions present, vicepresent, and secretary, how many such ways are possible?
A. 56 B. 70 C. 240 D. 336 E. 1680
=>
8P3 = 8 x 7 x 6 = 336
Ans: D Shouldn't the answer be A? Sent from my MotoG3TE using GMAT Club Forum mobile app



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30 May 2018, 01:27
[GMAT math practice question] John invested a total of $100,000 in two accounts. The first account had an annual interest rate of 4%, and the second account had an annual interest rate of 6%. If his investment earnings at the end of the first year were at least $5,000, what was the smallest possible amount that he could have invested at an annual interest rate of 6%? A. $40,000 B. $42,000 C. $45,000 D. $50,000 E. $54,000 => Let x be the principal invested at the 6% interest rate. Then we must have x*0.06 + ( 100,000 – x )*0.04 ≥ 5,000. Now, x*0.06 + ( 100,000 – x )*0.04 ≥ 5,000 => x*0.06 + ( 4,000 – x *0.04 ) ≥ 5,000 => x*0.02 + 4,000 ≥ 5,000 => x*0.02 ≥ 1,000 => x ≥ 50,000 Therefore, the smallest amount he could have invested at 6% was $50,000. Therefore, D is the answer. Answer: D
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31 May 2018, 04:39
[GMAT math practice question] 5.29 The operation # is defined to be multiplication. Which of the following must be true? Ⅰ. x#1=x Ⅱ. x#0=0 Ⅲ. x#y=y#x A. Ⅰ only B. Ⅲ only C. Ⅱ only D. Ⅰ and Ⅱ only E. Ⅰ,Ⅱ and Ⅲ => If # is the multiplication operation, then we must have x#1 = x, x#0 = 0 and x#y = y#x. Therefore, E is the answer. Answer: E
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31 May 2018, 23:16
karanoberoi93 wrote: MathRevolution wrote: [GMAT math practice question]
If 3 persons are selected at random from 8 persons for 3 positions present, vicepresent, and secretary, how many such ways are possible?
A. 56 B. 70 C. 240 D. 336 E. 1680
=>
8P3 = 8 x 7 x 6 = 336
Ans: D Shouldn't the answer be A? Sent from my MotoG3TE using GMAT Club Forum mobile appWe have 8 choices for president. After choosing president, 7 people are left. We have 7 choices for vicepresident. After choosing vicepresident, 6 people are left. We have 6 choices for secretary. Thus 8*7*6 is the number of the cases
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01 Jun 2018, 02:37
[GMAT math practice question] Attachment:
6.1.png [ 8.01 KiB  Viewed 146 times ]
If the distances between each pair of consecutive ticks are equal, what is the value of x? A. 2^12 B. 3(2^10) C. 5(2^10) D. (2^13) E. 3(2^12) => The distance between each pair of consecutive ticks is 2^11 – 2^10 = 2*2^10 – 2^10 = 2^10. Thus, x = 2^11 + 2^10 = 2*2^10 + 2^10 = 3(2^10). Therefore, the answer is B. Answer: B
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06 Jun 2018, 08:29
[GMAT math practice question] When x and n are positive integers, if x^{2n}>(3x)^n, which of the following must be true? A. x>3 B. n>1 C. x=3 D. n=3 E. x=n => x^{2n}>(3x)^n => x^{2n} > 3^nx^n => x^n > 3^n => x > 3 Therefore, A is the answer. Answer: A
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07 Jun 2018, 02:42
[GMAT math practice question] The test scores of students at a school have a distribution that is symmetric about the mean m. 68 percent of the test scores lie within one standard deviation d of the mean. If the average score of students at the school is 65 and the standard deviation is 5 points, what percent of the students at the school score more than 70 points? A. 10% B. 12% C. 14% D. 16% E. 18% Attachment:
6.5.png [ 27.47 KiB  Viewed 117 times ]
=> 50% – ( 68% / 2 ) = 16% Therefore, D is the answer. Answer: D
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08 Jun 2018, 02:43
[GMAT math practice question] If the sum of the minimum and the maximum of 7 consecutive odd integers is 114, what is the average (arithmetic mean) of the seven integers? A. 19 B. 38 C. 57 D. 76 E. 114 => Let the 7 consecutive odd integers be 2n5, 2n3, 2n1, 2n+1, 2n+3, 2n+5, and 2n+7. Then their minimum is 2n5 and their maximum is 2n+7. The sum of the maximum and minimum is ( 2n – 5 ) + ( 2n + 7 ) = 4n + 2 = 114 and so n = 28. Since the average and the median of consecutive odd integers are equal, the average is 2n + 1 = 57. Therefore, the answer is C. Answer: C Another way to solve this problem is to note that the average of the consecutive odd integers is equal to the average of the two end points. That is, 114 / 2 or 57.
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10 Jun 2018, 18:35
[GMAT math practice question] How many integers between 50 and 100, inclusive, are divisible by 2 or 3? A. 35 B. 37 C. 42 D. 47 E. 52 => Let A be the set of integers between 50 and 100 that are divisible by 2. Let B be the set of integers between 50 and 100 that are divisible by 3. Let C be the set of integers between 50 and 100 that are divisible by both 2 and 3. This is the same as the set of integers between 50 and 100 that are divisible by 6. Then A = { 50, 52, …, 100 } B = { 51, 54, …, 96, 99 } C = { 54, 60, …, 96 } The number of elements of the set A is A = ( 100 – 50 ) / 2 + 1 = 26. The number of elements of the set B is B = ( 99 – 51 ) / 3 + 1 = 17. The number of elements of the set C is C = ( 96 – 54 ) / 6 + 1 = 8. Using a Venn diagram, we can see that we need to find A + B  C as the integers in the intersection of sets A and B are counted twice. Attachment:
6.11.png [ 5.62 KiB  Viewed 102 times ]
A + B  C = 26 + 17 – 8 = 35. Therefore, the answer is A. Answer : A
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10 Jun 2018, 18:37
[GMAT math practice question] A college has a soccer club, a tennis club and a basketball club. Students can enroll in only one of these three clubs. The ratio of the number of soccer club members to the number of tennis club members is 2:3. The ratio of the number of tennis club members to the number of basketball club members is 4:5. A total of 350 students have joined one of these three clubs. How many students are enrolled in the soccer club? A. 80 B. 90 C. 100 D.120 E. 150 => Let S, T and B be numbers of members in the soccer club, the tennis club and the basketball club, respectively. Since S:T = 2:3 = 8:12 and T:B = 4:5 = 12:15, we have S:T:B = 8:12:15. Let S = 8k, T = 12k and B = 15k. Then S + T + B = 8k + 12k + 15k = 35k = 350, and k = 10. The number of members of the soccer club is S = 8k = 80. Therefore, the answer is A. Answer: A
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13 Jun 2018, 02:58
[GMAT math practice question] If n is the sum of the first 50 positive integers, what is the greatest prime factor of n? A. 3 B. 5 C. 7 D. 17 E. 51 => 1 + 2 + 3 + … + 50 = 50*(50 + 1)/2 = 25*51 = 52*3*17 17 is the greatest prime factor of n. Therefore, the answer is D. Answer: D
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