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We define “A mod n” as the remainder when A is divided by n. What is the value of “A mod 6”?

1) A mod 4 = A mod 12 2) A mod 3 = 2

=> Condition 1) A = 0, 1, 2, 3. This is not sufficient.

Condition 2) A = 2, 5. This is not sufficient.

Condition 1) & 2) From the condition 2), we have four cases as follows. A = 12k + 2 A = 12k + 5 A = 12k + 8 A = 12k + 11 From the condition 1), A should be the case that A = 12k + 2 = 6*(2k) + 2. Thus both conditions together are sufficient.

The original condition x^3y^4z^5<0 is equivalent to xy < 0 after dividing both sides of the inequality by x^2y^4z^4. Then the question is equivalent to y>0, since xy < 0 from the equivalent condition.

Thus the condition 1) is sufficient, since the question is if y > 0.

The probability that event A occurs is 0.5 and the probability that event B occurs is 0.7. What is the probability that event A occurs but not event B?

1) Event A and event B are independent. 2) The probability that neither event A nor event B occurs is 0.15.

=>Condition 1) We call an event A and an event B independent if P(A∩B) = P(A)*P(B). P(A∩B) = P(A)*P(B) = 0.5 * 0.7 = 0.35. P(A∩B^C) = P(A) - P(A∩B) = 0.5 – 0.35 = 0.15. This is sufficient.

Condition 2) 1 – P(A∪B) = 0.15. It means P(A∪B) = 0.85. P(A∪B) = P(A) + P(B) - P(A∩B) = 0.5 + 0.7 - P(A∩B) = 0.85. Thus P(A∩B) = 0.15. P(A∩B^C) = P(A) - P(A∩B) = 0.5 – 0.35 = 0.15. This is also sufficient too.

Re: The Ultimate Q51 Guide [Expert Level] [#permalink]

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11 Sep 2017, 22:02

MathRevolution wrote:

Even for the mistake type 4 questions, they are becoming more complicated unlike simple questions in the past. For a question like the below, 1)=2), which is no and sufficient. Therefore, D is the answer.

(ex 3) If n is positive integer, is 3n+n2+1 not divisible by 3? 1) n is not multiple of 2 2) n is not multiple of 3

Answer: D

That is, questions are evolving like highly challenging questions, in which the mistake type 4 and 1 are combined. So, we should keep up with the evolving questions. In particular, integer questions are numerously given and you should carefully approach the questions, which will lead you to hit 51.

I agree with your answer but I think the answer can be obtained without any condition (ie) 3n + n2 + 1 is always not divisible by 3 and will leave remainder of 1 or 2. Is my solution right? And will GMAT give questions wherein the answer can be directly deduced?

Re: The Ultimate Q51 Guide [Expert Level] [#permalink]

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13 Sep 2017, 17:47

MathRevolution wrote:

If a, b, and c are positive integers, is (a+b)c divisible by 3?

1) 2-digit integer ab is divisible by 3. 2) When c is divided by 3, the remainder is 0.

==> If you modify the original condition and the question, in order to have (a+b)c to be divided by 3, a+b or c has to be divided by 3. However, if you look at con 2), c is divisible by 3, hence it is yes and sufficient. For con 1), ab is also divisible by 3, and thus a+b is also divisible by 3, hence it is yes and sufficient. Therefore, the answer is D. This type of question is a 5051-level question which applies CMT 4 (B: if you get A or B too easily, consider D).

Answer: D

Hi, I didn't get you when you said, "This type of question is a 5051-level question which applies CMT 4 (B: if you get A or B too easily, consider D)." Thanks.

I. 9/10, 10/11, 11/12, 12/13, 13/14 II. 10/9, 11/10, 12/11, 13/12, 14/13 III. 6/10, 7/10, 8/10, 9/10, 10/10

For which of the following options is the average (arithmetic mean) of the numbers less than the median of numbers?

A. I. only B. II. only C. III. only D. II and III E. I, II and III

=>

I. Differences between consecutive terms are 10/11 – 9/10 = 1/110, 11/12 – 10/11 = 1/132, 12/13 – 11/12 = 1/156, 13/14 – 12/13 = 1/182. The differences are getting smaller. The average is smaller than the median.

II. Difference between consecutive terms are 11/10 – 10/9 = -1/90, 12/11 – 11/10 = -1/110, 13/12 – 12/11 = -1/132, 14/13 – 13/12 = -1/156. The differences are getting bigger. The average is bigger than the media.

II Difference between consecutive terms are 7/10 – 6/10 = 1/10, 8/10 – 7/10 = 1/10, 9/10 – 8/10 = 1/10, 10/10 – 9/10 = 1/10. All differences are equal. The average is equal to the median.

3 girls and 4 boys line up in a row. If girls stand at the first and boys stand at the last, how many possible cases are there?

A. 96 B. 108 C. 144 D. 192 E. 256

=>

The number of ways that 3 girls permute is 3! = 6 and the number of ways that 4 boys permute is 4! = 24. Those cases happen together and so the number of cases that 3 girls stand at the first and 4 boys stand at the last is 3! * 4! = 6 * 24 = 144.

=> x^3y^4z^5 < 0 is equivalent to xz < 0. Then the question xyz > 0 is equivalent to y < 0. The condition 1) is sufficient. And the condition 2) is not sufficient.

Re: The Ultimate Q51 Guide [Expert Level] [#permalink]

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20 Sep 2017, 18:48

MathRevolution wrote:

1) As always, more developed questions combined with CMT 3 and 4 are being released. Look at the question below. This question, a 5051-level question including CMT 4(A), was released recently. To strengthen your skills for this type of questions, you need to know the relationship between variable approach and CMT.

If a and b are integers, is ab an odd?

1) a=0 2) b=1-a

==> In the original condition, there are 2 variables (a, b), and therefore C is most likely to be the answer. By solving con 1) and con 2), from a=0 and b=1, you get ab0*1=0=even, hence no, it is sufficient. Therefore, C is the answer. However, this is an integer question, one of the key questions. Thus, if you apply CMT 4 (A, B), if 1) a=0, you get ab=0 and it is always even, hence no, it is sufficient. Also, for con 2), from a+b=1=odd, and (a, b)=(odd, even), (even, odd), you get ab=even, hence yes, it is sufficient. Therefore, the answer is D. This question is related to CMT 4(B). In other words, con 1) is easy and con 2) is difficult, so you apply CMT 4(B) (If you get A and B easily, consider B). Answer: D

MathRevolution wrote:

It is well-known that the way to find out approximation value of a positive integer n’s square root is following; 1st approximation: select a positive integer "a" and n is divided by a. 2nd approximation: a positive integer n’s square root is the average (arithmetic mean) of a quotient and divisor. What is the approximation positive integer n’s square root, in terms of a and n? A. (a2+n)/2a B. (a2+n)/2 C. (a2-n)/2a D. (a2+n)/a E. (a2+2n)/a Answer: A

This is a thesis=like question. It is a very challenging 50-51 level question. N=aQ means √n=(a+Q)/2. Hence, the correct answer is A.

MathRevolution wrote:

Geometry questions are continuously increasing. Let's have a look at the recent question below.

Attachment:

CUBE.jpg

A cube has 4 as a side’s length. If A and B are midpoints of each side and C is a vertex of the cube, what is the length of AB? A. 2√3 B. 3√6 C. √6 D. 3√2 E. 2√6

Answer: E

In a case of this question, you can just use Pythagoras' theorem twice. This type of geometry questions are likely to have been increasing since 2014. If AC^2=2^2+4^2=20, AB=x, x^2=2^2+AC^2=24 is derived. Then, x= √24= 2√6. The answer is E.