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Math Revolution GMAT Instructor
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Re: Math Revolution Approach (DS)
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30 Sep 2016, 06:48
When a positive integer n is divided by 7, what is the remainder? 1) When n294 is divided by 7, the remainder is 3 2) n3 is divisible by 7 ==> In the original condition, the answer is highly likely to be D since there is 1 variable(n), and it becomes 1)=2) so the remainder of both is 3, hence unique, and suffi. The answer is D. Answer: D
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Re: Math Revolution Approach (DS)
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01 Oct 2016, 01:10
Is xyz>0? 1) xy>0 2) yz>0 ==>In the original condition, the answer is highly likely to be E with 3 variables (x,y,z). If you do 1)&2), you get x=y=z=1 yes, and x=y=z=1, hence no, and not suffi. Thus, the answer is E. Answer: E  Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.
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Re: Math Revolution Approach (DS)
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02 Oct 2016, 03:03
If m is 2digit positive integer and n is a positive integer, what is the units digit of (2^n)(m)(5^n)? 1) m=17 2) n=3 ==> In the original condition, you get 2 variables(m,n) and the answer is C. However, since this is an integer problem, a key question, if you apply CMT 4(A), when you solve 2), the first digit of (2^3)(m)(5^3) is always 0, so suffi. The answer is B. Answer B
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Re: Math Revolution Approach (DS)
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03 Oct 2016, 00:01
If m and n are integers, is m  n= m  n? 1) m>n 2) m+2n>0 ==> If you modify the original condition and the problem, and square the bothe sides, you get (m  n)^2= (m  n)^2?. From m^22mn+n^2=m^2+n^22mn?, you get rid of m^2 and n^2, you 2mn=2mn?, mn=mn?. In other words, you get mn≥0?. However, there is no information regarding mn≥0 for 1) and 2), the answer is E. Answer: E
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Re: Math Revolution Approach (DS)
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04 Oct 2016, 06:24
Is x<y? 1) xy<4 2) 2x>2y ==> In the original condition, there are 2 variables, so C is likely to be the answer. As 1)=2), and in the case of 1), from xy<4<0, you get xy<0, x<y, hence yes. The answer is D. Answer: D
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Re: Math Revolution Approach (DS)
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06 Oct 2016, 04:54
For a positive integer n, what is the remainder when n(n+1) is divided by 12? 1) n is divisible by 3. 2) n is divisible by 4. ==> In the original condition, there is 1 variable (n), so D is highly likely to be the answer. In the case of 1), if =3, n(n+1)=12 with the remainder of 0,and if n=6, n(n+1)=42=12*3+6 with the remainder of 6, hence not sufficient. In the case of 2), if n=4, n(n+1)=20=12*1+8 with the remainder of 8, if n=8, n(n+1)=72=12*6 with the remainder of 0, hence not sufficient. Through 1) & 2), n=12, 24, 36… all have the remainder of 0, hence unique, and sufficient. C is the answer. Answer: C
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Re: Math Revolution Approach (DS)
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06 Oct 2016, 04:55
If r and s are positive integers, is r/s a terminating decimal? 1) r is a factor of 100 2) s is a factor of 500 ==> If you change the original condition and the problem, in order for r/s to be the terminating decimal, prime factors of s should be 2 or 5. In the case of 2), from 500=2^25^3, prime factors of 500 should always be either 2 or 5, hence yes, and sufficient. The answer is B. Answer: B
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Re: Math Revolution Approach (DS)
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10 Oct 2016, 04:43
2^m*2^n=? (1) mn=6 (2) m+n=5 ==> If you modify the original condition and the problem, 2^m2^n=2^m+n, , then you only need to know m+n, so B is the answer. Answer: B
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Re: Math Revolution Approach (DS)
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12 Oct 2016, 02:33
If x and y are positive integers, when x^4−y^4 is divided by 4, what is the remainder? 1) xy is divisible by 4 2) x^2+y^2 is divisible by 4 ==> If you modify the original condition and the problem, you get x^4y^4=(xy)(x+y)(x^2+y^2). Then, from 1)=2), the remainder of what are both divided by 4 is 0, hence unique, and sufficient. The answer is D. Answer: D
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Re: Math Revolution Approach (DS)
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13 Oct 2016, 06:29
If n and mare positive integers, is n^m+2n divisible by 3? 1)m=3 2)n=1 ==>In the original condition, there are 2 variables, and C is the answer. However, 1) and 2) each becomes yes, so D is the answer, a common CMT 4(B). Especially in the case of 1), if you substitute m=3, you get n^3+2n=n^3n+3n=(n1)n(n+1)+3n. Here, (n1)n(n+1) is the multiple of three consecutive integers, and you always get the multiples of 6, hence yes each time. Answer: D
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Re: Math Revolution Approach (DS)
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16 Oct 2016, 23:39
What is the median number of cloths sold in the last 15 days? 1) 21 cloths are sold every day in the last 8 days 2) The total number of clothes sold was 600 in the last 15 days. ANS: ==>If you change the condition and the problem, the median of the number of all the clothes sold for 15 days is the 8th number when all figures are arranged in an ascending order. In the case of 1), 21 clothes were sold each day for the last 8 days, hence median=21, and sufficient. The answer is A. This kind of questions(integer and statistics) is the current 5051question. Answer: A
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Re: Math Revolution Approach (DS)
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19 Oct 2016, 22:35
If a≠b, √[(a+2b)/(a+b)]=? 1) a=3 2) b/a=2/3 ==>In the original condition, there are 2 variables (a,b), so C is the answer. However, if you modify the problem, and divide the numerator and denominator of √[(a+2b)/(a+b)] by a, you get sqrt[1+2(b/a)][1+(b/a)], so you only need to know b/a. B is the answer. In general, if one condition is a number, and the other condition is a ratio, the answer is highly likely to be the ratio. Answer: B
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Re: Math Revolution Approach (DS)
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21 Oct 2016, 07:28
If a and b are integers, is a an odd number? 1) a+b is an even number 2) ab is an odd number ==>In the original condition, there are 2 variables, so C is the answer. Through 1) & 2), a=b=odd, so yes, hence sufficient. The answer is C. This problem, too, is an integer problem, which is a key question, so if you apply CMT 4(A), and tackle 2), a=b=odd, hence yes, and sufficient. B is the answer. Answer: B
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Re: Math Revolution Approach (DS)
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24 Oct 2016, 00:13
If 23,ab2 is 5digit positive integers, what is the remainder when 23,ab2 is divided by 25? 1) a=9 2) b=5 ==>If you modify the condition and the problem, in general, the remainder of a certain integer, n, divided by 25 is the same as the value when divided by 25 to the nearest hundreds digit. That is because 100=25*4. 2,3,a do not have any influence when you divide 23,ab2 by 25 like b2, so you only need to know what b is. The answer is B.
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Re: Math Revolution Approach (DS)
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26 Oct 2016, 02:01
If a>b, is a>b? 1) a>0 2) b<0 ==>If you look at the original condition, there are 2 variables (a,b) and 1 equation (a>b), so in order for them to match with the number of equations there has to be 1 more equation. Hence D is the answer. In the case of 1), you get a>b≥b, then a>b, hence always yes, and sufficient. In the case of 2), you get a>b=b>b, then a>b, hence always yes, and sufficient. The answer is D. Answer: D
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Re: Math Revolution Approach (DS)
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26 Oct 2016, 02:01
If r and s are positive integers is r/s a terminating decimal? 1) r is a factor of 100 2) s is a factor of 200 ==>If you modify the original condition and the problem, in order for r/s to become a terminating decimal, prime factors of s, the denominator, should be only 2 or 5. However, in the case of 2), in order for s to be a factor of 200, you always get 200=2^3*5^2, hence yes, and sufficient. The answer is B. Answer: B
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Re: Math Revolution Approach (DS)
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31 Oct 2016, 01:12
(ex 8) *(integer) If x and y are prime numbers, what is the number of the different factors of x^4*y^3? 1) xy=15 2) x and y are different ==>In the original condition, x and y are prime numbers, so if x and y are different, the number of different factors of x^4*y^3, is (4+1)(3+1)=20. However, in the case of 2), the condition is sufficient because of the word “different”. In the case of 1), too, (x,y)=(3,5),(5,3) then the number both two different factors is 20, hence sufficient. The answer is D. This is a common 5051level problem related with CMT 4(B) Answer: D
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Re: Math Revolution Approach (DS)
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02 Nov 2016, 00:21
If mn=3, is m>n? 1) mn<0 2) The distance between m and 0 is 2 ==>In the original condition, there are 2 variables(m,n) and 1 equation(mn=3), so D is highly likely to be the answer. In the case of 1), (m,n)=(2,1),(2,1), so yes and no coexist, hence not sufficient. In the case of 2), you get m0=m=2, m=±2, then (m,n)=(2,1),(2,1), hence yes/no, so not sufficient. Through 1) & 2), you get (m,n)=(2,1),(2,1), so yes/no, hence not sufficient. The answer is E. Answer: E
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Re: Math Revolution Approach (DS)
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03 Nov 2016, 00:26
If a, b, c are integers, is a+b+c an odd number? 1) a+b=odd 2) a+c=odd ==>In the original condition, there are 3 variables (a,b,c), so E is highly likely to be the answer. Through 1) & 2), if (a,b,c)=(1,2,2), it is yes, but if (a,b,c)=(2,1,1), it is no, hence not sufficient. The answer is E. Answer E
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Re: Math Revolution Approach (DS)
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04 Nov 2016, 20:46
If x and y are positive integers, is x divisible by y? 1) x is divisible by 5 2) y is divisible by 5 ==>In the original condition, there are 2 variables (x,y), so C is highly likely to be the answer. Through 1) & 2), x=y=5, it is yes, but if x=5, y=10, you get no, hence not sufficient. The answer is E. Answer: E
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