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Math Revolution GMAT Instructor
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Re: Math Revolution Approach (DS)
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20 Jul 2017, 00:01
When n is divided by 4, what is the remainder? 1) When n is divided by 3, the remainder is 1 2) When n+1 is divided by 4, the remainder is 2 ==> In the original condition, there is 1 variable (n) and in order to match the number of variables to the number of equations, there must be 1 equation. Since there is 1 for con 1) and 1 for con 2), D is most likely to be the answer. For remainder questions, you can directly substitute. Therefore, for con 1), from n=3p+1=1,4,…, the remainder when divided by 4 becomes 1=4(0)+1, which makes remainder=1, and from 4=4(1)+0, you get remainder=0, hence it is not unique and not sufficient. For con 2), from n+1=4q+2 and n=4q+1, the remainder when divided by 4 always becomes 1, hence it is unique and sufficient. Therefore, the answer is B. Answer: B
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Re: Math Revolution Approach (DS)
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21 Jul 2017, 00:03
Is 3x greater than x6? 1) x is greater than 4 2) x is greater than 0 ==> If you modify the original condition and the question, you get 3x>x6?, 2x>6?, x>3?. There is 1 variable (x) and in order to match the number of variables to the number of equations, there must be 1 equation. Since there is 1 for con 1) and 1 for con 2), D is most likely to be the answer. For con 1), from x>4, the range of the question doesn’t include the range of the condition, hence it is not sufficient. For con 2), from x>0, the range of the question includes the range of the condition, hence it is sufficient. Therefore, the answer is B. Answer: B
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Math Revolution Approach (DS)
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23 Jul 2017, 17:27
Is x^y=? 1) x=1 2) y=1 ➔ For this type of questions, the answer is A. ==> In the original condition, there is 2 variable (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), you get x^y=1^1=1, hence it is unique and sufficient. The answer is C. However, according to tip4, a trivial condition cannot be the answer, so if you solve con 1) and con 2) separately, for con 1), you substitute x=1, and get x^y=1^y=1, which is unique and sufficient. The answer is A. Answer: A
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Re: Math Revolution Approach (DS)
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26 Jul 2017, 00:09
Is u>w? 1) u=∣w∣ 2) w=∣u∣ ==> For con 2), w=u≥u for all u is established, so you get w≥ur, which is always no and sufficient. Therefore, the answer is B. Answer: B
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Re: Math Revolution Approach (DS)
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27 Jul 2017, 00:09
Mary buys several pencils that cost 30 cents and 24 cents. How many of 30 cent pencils did he buy? 1) Total cost is $1.02 2) He buys 4 pencils in total ==> In the original condition, it is a 2 by 2 question that appears most frequently on current gmat math. Attachment:
7.27.png [ 19.21 KiB  Viewed 338 times ]
According to the table above, 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), you get 30x+24y=102, 5x+4y=17 and x+y=4, so you get x=1 and y=3. The answer is C. However, this question is an integer question, one of the key questions, so you use CMT 4 (A: if you get C too easily, consider A and B). For con 1), from 5x+4y=17, 5 and 4 are mutually exclusive and x and y are positive integers, you only get x=1, y=3, hence it is unique and sufficient. Therefore, the answer is A, not C. Answer: A
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Re: Math Revolution Approach (DS)
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28 Jul 2017, 00:23
xy=? 1) y=x1 2) y^2=x1 ==> a^2+b^2=0 or a+b=0 is satisfied by a=b=0 only. Thus, according to the same logic, a^2+b=0 also must be a=b=0. Then, you get 2) y^2=x1, y^2+x1=0, which is y=0 and x=1. The answer is B. Answer: B
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Re: Math Revolution Approach (DS)
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30 Jul 2017, 17:56
If x and y are positive integers, is xy even? 1) x+y is odd 2) x^y is even ==> If you modify the original condition and the question, in order to get xy=even, it can either be x=even? or y=even?. For con 1), from (x,y)=(odd,even),(even,odd), you always get yes, hence it is sufficient. For con 2), you always get x=even, hence it is yes and sufficient. The answer is D. Answer: D
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Re: Math Revolution Approach (DS)
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30 Jul 2017, 17:57
What is the units digit of a positive integer n? 1) The units digit of n^2 is 4 2) The units digit of n^3 is 8 ==> In the original condition, there is 1 variable (n) and in order to match the number of variables to the number of equations, there must be 1 equation. Since there is 1 for con 1) and 1 for con 2), D is most likely to be the answer. For con 1), you get n=~2, ~8, hence it is not unique and not sufficient. For con 2), you only get n=~2, hence it is unique and sufficient. Therefore, the answer is B. Answer: B
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Re: Math Revolution Approach (DS)
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01 Aug 2017, 23:56
John travelled for 12hrs. What is the total average speed? 1) For the first 8hrs, he travelled 60miles per 1hr. 2) For the last 8hrs, he travelled 50miles per 1hr. ==> In the original condition, he is travelling 12 hours, and for con 1) and con 2), it is divided into 8hrs each. Thus, the distance of first 4hrs, the second 4hrs, and the last 4hrs are unknown, hence there are too many variables. The answer is E. Answer: E
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Re: Math Revolution Approach (DS)
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03 Aug 2017, 00:26
If m and n are nonnegative integers, mn=? 1) 9^n=3^m 2) 2^n=5^m ==> In the original condition, there are 2 variables (m,n) 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 con 1), you get 9^n=(3^2)^n=3^{2n}=3^m, which becomes 2n=m. In order for con 2) to satisfy as well, you only get m=n=0, hence it is unique and 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), the way to satisfy 9^n=(3^2)^n=3^{2n}=3^m to 2n=m is not unique and not sufficient. For con 2), from 2^n=5^m, you get 2^n=even and 5^m=odd, so even≠odd. Only m=n=0 satisfies this, hence it is unique and sufficient. Therefore, the answer is B, not C. Answer: B
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Re: Math Revolution Approach (DS)
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04 Aug 2017, 01:19
Is p/m>0? 1) p>m 2) pm>0. ==> If you modify the original condition and the question, from is p/m>0?, you multiply m^2 on both sides, you get (Squared number is always positive, so even if you multiply, the inequality sign doesn't change) p/m(m^2)>0(m^2)?, which becomes pm>0?. The answer is B. Answer: B
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Re: Math Revolution Approach (DS)
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06 Aug 2017, 17:15
If a, b, and c are positive integers, is (a+b)c divisible by 3? 1) 2digit 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 5051level question which applies CMT 4 (B: if you get A or B too easily, consider D). Answer: D
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Re: Math Revolution Approach (DS)
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06 Aug 2017, 17:16
There are 200 number of clothes. What is the number of clothes that are red but not made of cotton? 1) 10% of the total clothes are neither red nor made of cotton. 2) 50% of the total clothes are both red and made of cotton. ==> If you modify the original condition and the question, it is a 2 by 2 question which appears most frequently on current GMAT math. Attachment:
8.4.png [ 2.09 KiB  Viewed 284 times ]
As shown above, there are 4 variables (w,x,y,z) and 1 equation (W+X+Y+Z=200). In order to match the number of variables to the number of equations, there must be 3 more 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) & con 2), you get z=200(10%)=20 and w=200(505)=100, but the value of y is not unique and not sufficient. The answer is E. Answer: E
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Re: Math Revolution Approach (DS)
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09 Aug 2017, 17:08
Is x>(3)1/3? 1) x>(2)1/3 2) x>(4)1/3 Attachment:
8.9.png [ 15.33 KiB  Viewed 277 times ]
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Re: Math Revolution Approach (DS)
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10 Aug 2017, 00:12
The percent of employees participated in A program is 80% of the total employees. What is the total number of employees in the company? 1) 168 employees participated in this program. 2) 42 employees did not participate in this program. ==> If you modify the original condition and the question and set the number of people who participated in the program as a and number of people who did not participate in the program as b, from a=80%(a+b), there are 2 variables and 1 equation. Since there is 1 for con 1) and 1 for con 2), D is most likely to be the answer. For con 1), from 168=80%(a+b) and a+b=168/80%=210, it is unique and sufficient. For con 2), from 42=20%(a+b) and a+b=42/20%=210, it is unique and sufficient. Therefore, the answer is D. Answer: D
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Re: Math Revolution Approach (DS)
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11 Aug 2017, 00:01
If y≠0, is x/y=x/y? 1) xy=xy 2) x/y=x/y ==> If you modify the original condition and the question, in order to get x/y=x/y, you get x/y≥0?, and if you multiply y^2 on both sides, you get xy≥0?. Then, for con 1), to satisfy xy=xy, you get xy≥0, hence yes, it is sufficient. For con 2), if x=y=2, yes, but if x=2 and y=1, no, hence it is not sufficient. The answer is A. Answer: A
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Re: Math Revolution Approach (DS)
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13 Aug 2017, 17:17
If x, y are positive integers, is xy=1? 1) x^2y^2=xy 2) 1/y=x. ==> 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), for con 2), you get xy=1, hence it is unique and sufficient. For con 1), from (xy)^2xy=0, xy(xy1)=0, you get xy=0,1, but since x and y are positive, you get xy=1, and thus con 1) = con 2). The answer is D. Answer: D
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Re: Math Revolution Approach (DS)
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13 Aug 2017, 17:19
Is x=y? 1) x^4+y^4=2x^2y^2 2) x^4+y^4=0 ==> 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), for con 1), from x^4+y^42x^2y^2=0, (x^2y^2)^2=0, you get x^2=y^2, and from x=±y, yes and no coexists, hence it is not sufficient. For con 2), you only get x=y=0, hence yes, it is sufficient. The answer is B. Answer: B
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Re: Math Revolution Approach (DS)
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16 Aug 2017, 01:40
What is the remainder, when n(n+2) is divided by 24 for a positive integer x? 1) n is an even integer 2) n has remainder 0 or 1 when it is divided by 3. =>Condition 1) There are two kinds of even integers, which are n = 4k or n = 4k + 2 for some integer k. That means n could have 0 or 2 as a remainder when it is divided by n. If n = 4k, n(n+2) = 4k(4k+2) = 8k(2k+1) and n(n+1) is a multiple of 8. If n= 4k+2, n(n+2) = (4k+2)(4k+2+2) = 2(2k+1)*4(k+1) = 8(2k+1)(k+1) and n(n+1) is a multiple of 8. For both cases, n(n+1) is a multiple of 8. However, we can’t identify the remainder when it is divided by 3 from the condition 1). Condition 2) n = 3k or n = 3k +1. If n = 3k, n(n+2) = 3k(3k+2) is a multiple of 3. If n = 3k + 1, n(n+2) = (3k+1)(3k+3) = 3(3k+1)(k+1) is a multiple of 3. Thus, for both cases, n(n+1) is a multiple of 3. However, we can’t identify the remainder when it is divided by 8 from the condition 2). Condition 1) & 2) From the condition 1), n(n+1) is a multiple of 8. And n(n+1) is a multiple of 3 from the condition 2). Therefore, n(n+1) is a multiple of 24 from the both conditions 1) & 2) together. Ans: C
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Re: Math Revolution Approach (DS)
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16 Aug 2017, 23:52
If the average (arithmetic mean) of 7 numbers is 21, what is the standard deviation of the numbers? 1) The smallest number of them is 21 2) The greatest number of them is 21 =>Condition 1) Since the minimum number and the average are same, all numbers should be 21. Condition 2) Since the maximum number and the average are same, all numbers should be 21. Ans: D
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