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Math Revolution GMAT Instructor
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]
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15 May 2017, 18:10
For a positive integer n, when 12n is divided by 15, which of the following cannot be the remainder? A. 0 B. 3 C. 5 D. 6 E. 9 ==>From 12n=15Q+r, 12 and 15 are the multiples of 3, so remainder=r must also be the multiple of 3. Therefore, 5 cannot be the remainder. The answer is C. Answer: C
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16 May 2017, 18:20
G(x) is the greatest integer less than or equal to x and L(x) is the least integer greater than or equal to x. When x is not an integer, which of the following is the value of L(x)G(x)? A. 2 B. 1 C. 0 D. 1 E. 2 ==> You get G(x)=round down and L(X)=round up. Then, x≠integer, so if you substitute x=1.2, you get L(1.2)G(1.2)=21=1. The answer is D. Answer: D
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17 May 2017, 17:55
If the average (arithmetic mean) of a, b, and c is (a+2b)/3, what is the value of b? 1) a=1 2) c=2 ==> If you modify the original condition and the question and check the question again, from is (a+2b)/3=(a+b+c)/3, you get a+2b=a+b+c, then b=c. In order to find b, you only need to know c. Therefore, the answer is B. Answer: B
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]
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21 May 2017, 18:02
If two integers x and y such x>y are selected at random between 8 and 11, inclusive, how many cases are possible? A. 150 B. 180 C. 190 D. 210 E. 240 ==> Since two integers from 8 to 11 are being randomly selected and x>y, you use combination. Thus, the number of integers from 8 to 11 becomes 11(8)+1=20, so 20, then 20C2=(20)(19)/2!=190. The answer is C. Answer: C
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22 May 2017, 18:13
Which of the following points reflect to y=x at (3,2)? A. (2, 3) B. (2,3) C. (3,2) D. (3,2) E. (2,3) ==> In order to become symmetrical to y=x, you need to substitute –y value on x coordinate, and –x value on y coordinate. Thus, you get (3,2)>(2,3). The answer is A. Answer: A
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]
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24 May 2017, 00:22
There are 6 red balls and 4 blue balls in a jar. If 2 balls are selected from the jar, what is the probability that 2 balls selected are red balls? A. 1/10 B. 1/9 C. 3/10 D. 3/15 E. 2/15 ==> You get 4C2/10C2==2/15. The answer is E. Answer: E
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The Ultimate Q51 Guide [Expert Level] [#permalink]
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25 May 2017, 01:09
5 people including A and B line up in a row. How many possible cases are there such that at least one person stands between A and B? A. 24 B. 36 C. 48 D. 60 E. 72 ==> Since it is the number of cases such that at least one person stands between A and B when 5 people are lined up in a row, you need to subtract the number of cases where A and B stands next to each other from the total number of cases. Then, you get ABCDE(AB)CDE, which becomes 5!4!(2)=72. You get 4!(2) because there are cases where A and B switch the order in the line. Therefore, the answer is E. Answer: E
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]
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25 May 2017, 17:28
If x and y are integers, is x+y an odd number? 1) y=3x+1 2) y=2x+3 ==> In the original condition, there are 2 variables (x,y) and in order to match the number of variables to the number of equations, there must be 2 equations. Since there is 1 for con 1) and 1 for con 2), C is most likely to be the answer. By solving con 1) and con 2), from 3x+1=2x+3, you get x=2 y=7. Then, you get x+y=2+7=9=odd, hence yes, it is sufficient. The answer is C. However, this is an integer question, one of the key questions, so you apply CMT 4 (A: if you get C too easily, consider A or B). For con 1), from y=3x+1, you get (x,y)=(even,odd) or (odd,even), which always becomes x+y=odd, hence yes, it is sufficient. For con 2), you get (x,y)=(1,5) no, (2,7) yes, it is not sufficient. Therefore, the answer is A. Answer: A
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]
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28 May 2017, 23:49
Is kr<0? 1) k^2r^3<0 2) k+r<k+r ==> For con 1), you ignore the square, so t<0, and for con 2), you get kr<0, hence yes, it is sufficient. The reason is that from (k+r)^2<(k+r)^2, you get k^2+r^2+2kr<k^2+r^2+2kr, and if you get rid of k^2+r^2 from both sides, you get 2kr<2kr, then kr<kr, which becomes kr<0. The answer is B. Answer: B
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]
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30 May 2017, 01:11
If a and b are 1digit positive integers, is 100a+10b+2 divisible by 4? 1) a=2. 2) b=3. ==> If you modify the original condition and the question, the remainder when an integer n is divided by 4 is equal to the remainder when only the units digit and the tens digit are divided by 4, and so you only need to find b. Thus, for con 2), you get b=3, which becomes 10(3)+2=32, and so it is always divisible by 4, hence yes, it is sufficient. The answer is B. Answer: B
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]
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31 May 2017, 00:46
[p] is the greatest integer less than or equal p. What is the value of [π1]? A. 2 B. 3 C. 4 D. 5 E. 6 ==>You get π1=3.141=2.14, and thus [π1]=[2.14]=2. The answer is A. Answer: A
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]
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01 Jun 2017, 01:21
If \(n=s^at^b\), where r, a, b and s are integers, is \(\sqrt{n}\) an integer? 1) a+b is an even number 2) a is an even number ==> In the original condition, there are 5 variables (n,s,t,a,b) and 1 equation (\(n=s^at^b\)). In order to match the number of variables to the number of equations, there must be 5 equations. Since there is 1 for con 1) and 1 for con 2), E is most likely to be the answer. By solving con 1) and con 2), s=t=2 and a=b=2 yes, but s=t=2 and a=b=2 no, hence it is not sufficient. Therefore, the answer is E. Answer: E
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]
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04 Jun 2017, 17:59
How many possible 6digit codes can be formed from three a’s, two b’s, and one c? A. 40 B. 50 C. 60 D. 70 E. 80 ==> Since it is the number of ways of listing a,a,a,b,b,c, you get \(\frac{6!}{(3!)(2!)(1)}=60\). The answer is C. Answer: C
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]
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06 Jun 2017, 02:34
If a and b are positive integers, what is the number of the different prime factors of a^b? 1) a has different 2 prime factors. 2) b has different 3 prime factors. ==> If you modify the original condition and the question, the number of prime factors doesn’t affect the exponent. In other words, the number of prime factors of a^b is equal to the number of prime factors of a. Therefore, the answer is A. Answer: A
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]
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07 Jun 2017, 18:05
Is xyz>0? 1) xy>0 2) yz>0 ==> In the original condition, there are 3 variables (x,y,z) and in order to match the number of variables to the number of equations, there must be 3 equations. Since there is 1 for con 1) and 1 for con 2), E is most likely to be the answer. By solving con 1) and con 2), if x=y=z=1, yes, but if x=y=x=1, no, hence it is not sufficient. Therefore, the answer is E. Answer: E
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]
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08 Jun 2017, 02:08
Thank you MathRevolution. I got 49 and the types of questions you mentioned are exactly what I am the weakest at.



Math Revolution GMAT Instructor
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GPA: 3.82

Re: The Ultimate Q51 Guide [Expert Level] [#permalink]
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11 Jun 2017, 18:30
For an investment, you start off with $1,000, the interest rate of 5%, and you deposited two years as compound interest. What is the total interest earned in 2 years? A. $96.5 B. $98.5 C. $100.5 D. $102.5 E. $105.2 ==> You get \(1,000(1+0.05)^21,000=102.5.\) The answer is D. Answer: D
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]
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12 Jun 2017, 17:47
What is the median of the consecutive multiples of 7 in the first 50 positive integers? A. 14 B. 21 C. 28 D. 35 E. 42 ==> Since it is the first 50 positive integers, you need to find the median of the consecutive multiples of 7, you get 7, 14, 21, 28, 35, 42, 49, and the median is 28. The answer is C. Answer: C
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]
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14 Jun 2017, 18:13
If [x] is the least integer greater than or equal to x, and [x/2]=3, what is the scope of x？ A. 3<x≤4 B. 8<x≤9 C. 8<x≤10 D. 9<x≤12 E. 4<x≤6 ==> For [x], you get [1.2]=2, which means rounded up. In order to get [x/2]=3, you get 2<x/2≤3 and then 4<x≤6. Therefore, the answer is E. Answer: E
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]
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15 Jun 2017, 18:13
What is the number of multiples of 6 from 35 to 69? A. 14 B. 16 C. 17 D. 20 E. 21 ==> 0 becomes the multiple of all integers. From 1~69, 11 of them are multiples of 6, from 1~35, 5 of them are multiples of 6, hence you get 11+5+1=17. The answer is C. Answer: C
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