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Math Revolution Approach (DS)

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New post 30 Sep 2018, 21:52
[Math Revolution GMAT math practice question]

What is the standard deviation of a, b and c?

1) a^2+b^2+c^2 = 77
2) a+b+c =15

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

The mean of a, b and c is m = ( a + b + c ) / 3. So, a + b + c = 3m.
The variance of a, b and c, that is, the square of the standard deviation of a, b and c, is given by
VAR = SD^2 = { ( a – m )^2 + ( b – m )^2 + ( c – m )^2 } / 3
= { a^2 - 2am + m^2 + b^2 - 2bm + b^2 + c^2 - 2cm + m^2 } / 3
= { a^2 + b^2 + c^2 – 2m(a+b+c) + 3m^2 } / 3
= { a^2 + b^2 + c^2 – 2m*3m + 3m^2 } / 3
= { a^2 + b^2 + c^2 – 3m^2 } / 3
= ( a^2 + b^2 + c^2 ) / 3 – m^2
= ( a^2 + b^2 + c^2 ) / 3 – { ( a + b + c ) / 3 }^2

This value can be calculated using the information given in conditions 1) and 2). Thus, both conditions together are sufficient.

Therefore, C is the answer.
Answer: C

When we have a sum of data values and a sum of the squares of data values, we can always calculate the standard deviation.
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Re: Math Revolution Approach (DS)  [#permalink]

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New post 30 Sep 2018, 21:55
[Math Revolution GMAT math practice question]

Is |x+y|<|x|+|y|?

1) x<0
2) y>0

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

The question |x+y|<|x|+|y| is equivalent to xy<0 as shown below

|x+y|<|x|+|y|
=> |x+y|^2<(|x|+|y|)^2
=> |x+y|^2-(|x|+|y|)^2< 0
=> (x+y)^2-(|x|+|y|)^2< 0
=> x^2+2xy+y^2-(|x|^2 +2|x||y|+|y|^2) < 0
=> x^2+2xy+y^2-(x^2 +2|xy|+y^2) < 0
=> 2xy-2|xy| < 0
=> xy-|xy| < 0
=> xy < |xy|
=> xy < 0

This occurs if x < 0 and y > 0. Thus, both conditions together are sufficient.

Therefore, C is the answer.
Answer: C
_________________

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Re: Math Revolution Approach (DS)  [#permalink]

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New post 03 Oct 2018, 01:24
[Math Revolution GMAT math practice question]

If m and n are integers, is mn an odd integer?

1) m(n+1) is even
2) (m+1)n is even

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

Modifying the question:
mn is odd only when both m and n are odd. So, the question asks if both m and n are odd.

Conditions 1) and 2), when applied together, tell us that either
both m and n are odd numbers or both m and n are even numbers.

Since we don’t have a unique solution, both conditions, taken together, are not sufficient.

Therefore, E is the answer.
Answer: E
_________________

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Re: Math Revolution Approach (DS)  [#permalink]

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New post 03 Oct 2018, 23:54
[Math Revolution GMAT math practice question]

Is x^3-x>0?

1) x>1.
2) x>0

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

The question x^3-x>0 is equivalent to -1<x<0 or x > 1 as shown below:

x^3-x>0
=> x(x^2-1)>0
=> x(x+1)(x-1) > 0
=> -1<x<0 or x > 1

Attachment:
10.4.png
10.4.png [ 6.75 KiB | Viewed 184 times ]


Since we have 1 variable (x) and 0 equations, D is most likely to be the answer. So, we should consider each of the conditions on their own first.

In inequality questions, the law “Question is King” tells us that if the solution set of the question includes the solution set of the condition, then the condition is sufficient.

Condition 1)
Since the solution set of the question includes that of condition 1), condition 1) is sufficient.

Condition 2)
Since the solution set of the question does not include that of condition 2), condition 2) is not sufficient.


Therefore, A is the answer.
Answer: A

If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.
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Re: Math Revolution Approach (DS)  [#permalink]

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New post 05 Oct 2018, 00:35
[Math Revolution GMAT math practice question]

If the integers p, q, r, and s satisfy p<q<r<s, are they consecutive?

1) r-q=1
2) (p-q)(r-s)=1

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 4 variables (p, q, r and s) and 0 equations, E is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)
Since r – q = 1 or r = q + 1 from condition 1), q and r are consecutive integers.
Since we have (p-q)(r-s) = (q-p)(s-r) =1, where p < q, r < s and p-q and r-s are integers, we have q-p = 1 and s-r = 1, or q = p +1, and s = r +1.
Thus p, q, r and s are consecutive integers and both conditions are sufficient when applied together.

Since this question is an integer question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.

Condition 1)
p = 1, q = 2, r = 3 and s = 4 are consecutive integers satisfying r – q = 1, and the answer is ‘yes’.
P = 1, q = 2, r = 3 and s = 5 satisfy r – q = 1, but are not consecutive integers, and the answer is ‘no’.
Since we don’t have a unique solution, condition 1) is not sufficient.

Condition 2)
p = 1, q = 2, r = 3 and s = 4 are consecutive integers satisfying (p-q)(r-s)=1,
and the answer is ‘yes’.
p = 1, q = 2, r = 4 and s = 5 satisfy (p-q)(r-s)=1, but are not consecutive integers, and the answer is ‘no’.
Since we don’t have a unique solution, condition 2) is not sufficient.

Therefore, C is the answer.
Answer: C

In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions on which the answer is A, B, C or D.
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New post 07 Oct 2018, 17:33
[Math Revolution GMAT math practice question]

If p and q are prime numbers, is pq+1 an odd number?

1) p – q = 5
2) p = 7

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Modifying the question:
pq + 1 is odd only when pq is even. So, the question is asking whether either p or q is an even prime number. Since the only even prime number is 2, the question is asking whether p or q is equal to 2.


Condition 1):
For p – q = 5, either p or q must be even. Since the only even prime number is 2, we must have p = 7 and q = 2.
Thus, condition 1) is sufficient.

Condition 2)
Since it provides no information about q, condition 2) is not sufficient.

Therefore, A is the answer.
Answer: A
_________________

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Re: Math Revolution Approach (DS)  [#permalink]

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New post 07 Oct 2018, 17:36
[Math Revolution GMAT math practice question]

If xy=y, |x|+|y|=?

1) x=-1
2) y=0

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

Modifying the original condition:
The equality xy=y is equivalent to y = 0 or x = 1 as shown below:

xy=y
=> xy-y=0
=> y(x-1)=0
=> y = 0 or x = 1

Since we have 2 variables (x and y) and 1 equation (xy=y), D is most likely to be the answer.


Condition 1)
Since x = -1 from condition 1) and y = 0 or x = 1 from the original condition, y = 0.
Thus, |x| + |y| = |-1| + |0| = 1.
Condition 1) is sufficient.

Condition 2)
If x = 1 and y = 0, then |x|+|y| = 1.
If x = 2 and y = 0, then |x|+|y| = 2.
Since we don’t have a unique solution, condition 2) is not sufficient.

Therefore, A is the answer.
Answer: A

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
_________________

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Re: Math Revolution Approach (DS)  [#permalink]

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New post 09 Oct 2018, 23:58
[Math Revolution GMAT math practice question]

If x^2-3x=10, what is the value of x?

1) x^2-4 = 0
2) x<6

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

The original condition x^2-3x=10 is equivalent to x = -2 or x = 5 as shown below:

x^2-3x=10
=> x^2-3x-10= 0
=> (x+2)(x-5) = 0
=> x = -2 or x = 5

Condition 1)
x^2-4 = 0
=> (x-2)(x+2) = 0
=> x = -2 or x = 2
Only x = -2 also satisfies the original condition, so we have a unique solution.
Thus, condition 1) is sufficient.

Condition 2)
Since x < 6 from condition 2) and x = -2 or x = 5 from the original condition, x = -2 or x = 5.
Since we don’t have a unique solution, condition 2) is not sufficient.

Therefore, A is the answer.
Answer: A

If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.
_________________

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Re: Math Revolution Approach (DS)  [#permalink]

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New post 11 Oct 2018, 00:34
[Math Revolution GMAT math practice question]

The terminal zeros of a number are the zeros to the right of its last nonzero digit. For example, 30,500 has two terminal zeros because there are two zeros to the right of its last nonzero digit, 5. How many terminal zeros does n! have?

1) n^2 – 15n + 50 < 0
2) n > 5

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 1 variable (n) and 0 equations, D is most likely to be the answer. So, we should consider each of the conditions on their own first.

Consider 1)
n^2 – 15n + 60 < 0
=> (n-5)(n-10) < 0
=> 5 < n < 10

The number of terminal zeros of a number is determined by the number of 5s in its prime factorization.

The integers satisfying 5 < n < 10 are 6, 7, 8 and 9. We count the 5s in the prime factorizations of 6!, 7!, 8! and 9!:
6! has one 5 in its prime factorization.
7! has one 5 in its prime factorization.
8! has one 5 in its prime factorization.
9! has one 5 in its prime factorization.

Thus, for 5 < n < 10, n! has one terminal zero.
As it gives us a unique answer, condition 1) is sufficient.

Condition 2)

If n = 6, then 6 > 5 and 6! = 720 has one terminal 0.
If n = 10, then 10 > 5 and 10! = 3,628,800 has two terminal 0s.

Condition 2) is not sufficient since it does not give a unique solution.

Therefore, A is the answer.
Answer: A

If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.
_________________

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New post 12 Oct 2018, 07:33
[Math Revolution GMAT math practice question]

If x=0.abcabc........(a repeating infinite decimal), what is the value of a+b+c?

1) 1-x=0.123123123…….
2) 0.8<x<0.9

=>
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 4 variables (x, a, b and c) and 1 equation (x = 0.abcabc….), E is most likely to be the answer. However, condition 1) includes 3 hidden equations (matching the values of the decimal places) and allows us to determine the values of all four variables as follows:
x = 1 – 0.123123123… = 0.876876876… = 0.abcabcabcabc…
Thus a = 8, b = 7, c = 6 and a + b + c = 8 + 7 + 6 = 21.
Condition 1) is sufficient.

Condition 2)
If x = 0.811811811…, then a = 8, b = 1, c = 1 and a + b + c = 10.
If x = 0.812812812…, then a = 8, b = 1, c = 2 and a + b + c = 11.
Since we don’t have a unique solution, condition 2) is not sufficient.


Therefore, A is the answer.
Answer: A
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Re: Math Revolution Approach (DS)  [#permalink]

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New post 14 Oct 2018, 17:53
[Math Revolution GMAT math practice question]

When A and B are positive integers, is AB a multiple of 4?

1) The greatest common divisor of A and B is 6
2) The least common multiple of A and B is 30

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

Asking if AB is a multiple of 4 is equivalent to asking if AB = 4k for some integer k.

Condition 1)
Since A = 6a = 2*3*a and B = 6b = 2*3*b for some integers a and b, AB = 2^2*3^2*ab = 4*3^2ab.
Thus, AB is a multiple of 4 and condition 1) is sufficient.

Condition 2)
If A = 6 and B = 5, then lcm(A,B) = 30, but AB = 30 is not a multiple of 4, and the answer is ‘no’.
If A = 6 and B = 10, then lcm(A,B) = 30, and AB = 60 is a multiple of 4. The answer is ‘yes’.
Since we don’t have a unique solution, condition 2) is not sufficient.

Therefore, A is the answer.
Answer: A
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Re: Math Revolution Approach (DS)  [#permalink]

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New post 14 Oct 2018, 17:57
[Math Revolution GMAT math practice question]

Attachment:
11.11.png
11.11.png [ 6.48 KiB | Viewed 130 times ]


The above figure shows a sector of a circle. What is the area of the sector?

1) x = 120.
2) AB=6√3.

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Attachment:
10.15.png
10.15.png [ 9.94 KiB | Viewed 130 times ]


The area of a sector is (1/2) πr^2*(x/360), where r is the radius of the sector. Since we have two variables, r and x, C is most likely to be the answer and we need to check both conditions together first.

Conditions 1) and 2):
Quadrilateral OACB is a kite, so its diagonals bisect each other at right angles, and bisect the angles at the vertices.
Since AB = 6√3, AD = AB/2 = 3√3. Since x =120, angle AOD has measure 60, the triangle ODA is a right triangle and OD:OA:DA = 1:2:√3. This yields OA:DA = r: 3√3 = 2: √3, which implies that r = 6.
Thus, the area of the sector is (1/2) π6^2*(120/360) = (1/2) (1/3)36π = 6π.

Both conditions together are sufficient.

Therefore, C is the answer.
Answer: C

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
_________________

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Re: Math Revolution Approach (DS) &nbs [#permalink] 14 Oct 2018, 17:57

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Math Revolution Approach (DS)

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