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[Math Revolution GMAT math practice question]

Is n an odd number?

1) n is the sum of 2 prime numbers
2) n is a multiple of 11

=>
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.

Condition 1)
If n = 2 + 3, then n = 5 is odd and the answer is ‘yes’.
If n = 3 + 5, then n = 8 is even and the answer is ‘no’.
Since we don’t have a unique solution, condition 1) is not sufficient.

Condition 2)
If n = 11, then n is odd and the answer is ‘yes’.
If n = 22, then n is even and the answer is ‘no’.
Since we don’t have a unique solution, condition 2) is not sufficient.

Conditions 1) & 2)
If n = 2 + 31, then n = 33 is odd and the answer is ‘yes’.
If n = 13 + 31, then n = 44 is even and the answer is ‘no’
Since we don’t have a unique solution, both conditions 1) and 2) are not sufficient, when taken together.

Therefore, E is the answer.

Answer: E
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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[Math Revolution GMAT math practice question]

If r is a positive integer, n=^r3 and 4,14, and 27 are factors of n, which of the following must be a factor of n?

A. 16
B. 32
C. 36
D. 48
E. 64

=>

Since 4 = 2^2,14 = 2*7, 27 = 3^3 are factors of n and n is a perfect cube, the smallest possible value of n is 2^3*3^3*7^3.

When n = 2^3*3^3*7^3, 16 = 2^4, 32 = 2^5, 48=2^4*3 and 64 = 2^6 can’t be factors of n. The only answer choice that is a factor of n is 36 = 2^2*3^2.


Therefore, the answer is C.
Answer: C
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[Math Revolution GMAT math practice question]

x=?

1) x^3+x^2+x=0
2) x=-2x

=>

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

Condition 1)
x^3+x^2+x=0
=> x(x^2+x+1)=0
=> x = 0 since x^2+x+1 ≠ 0
Condition 1) is sufficient.

Condition 2)
x = -2x
=> 3x = 0
=> x = 0
Condition 2) is sufficient.

Therefore, D is the answer.
Answer: D

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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[Math Revolution GMAT math practice question]

What is the value of 22C19?

A. 770
B. 1540
C. 3080
D. 4620
E. 6160

=>

Since nCn-r = nCr,
22C19 = 22C3 = (22*21*20)/(1*2*3) = 11*7*20 = 1540.

Therefore, the answer is B.
Answer: B
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[Math Revolution GMAT math practice question]

If n is an integer, is n(n+2) divisible by 8?

1) n is an even number.
2) n is a multiple of 4.

=>
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.

Condition 1)
If n is an even integer, then n and n + 2 are two consecutive even integers.
Products of two consecutive even integers are multiples of 8 since one of them must be a multiple of 4, and the other a multiple of 2.
Condition 1) is sufficient.

Condition 2)
Since n is a multiple of 4, n + 2 is an even integer.
Thus, n(n+2) is a multiple of 8.
Condition 2) is sufficient.

Therefore, D is the answer.

Answer: D

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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[Math Revolution GMAT math practice question]

x^4+12x^3+49x^2+78x+40 can be factored as a product of four linear polynomials with integer coefficients. Which of following cannot be a factor of x^4+12x^3+49x^2+78x+40?

A. x+1
B. x+2
C. x+3
D. x+4
E. x+5

=>

Assume x^4+px^3+qx^2+rx+s = (x+a)(x+b)(x+c)(x+d).
Then, s = abcd.
The constant terms a, b, c and d of the linear factors are factors of the constant term s of the original polynomial.

Since 3 is not a factor of 40, x + 3 cannot be a factor of the original polynomial x^4+12x^3+49x^2+78x+40.

Therefore, C is the answer.
Answer: C
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Are you sure that statistics is up there with the most important concepts? I have seen various books and online courses which stress upon Number Properties, Algebra and geometry as the three main areas to concentrate from.
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[Math Revolution GMAT math practice question]

Is x>x/y?

1) y>1
2) xy>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>x/y is equivalent to xy(y-1) > 0 as shown below:
x>x/y
=> xy^2 > xy since y^2 > 0
=> xy^2 – xy > 0
=> xy(y – 1) > 0

Condition 1) tells us that y – 1 > 0, and condition 2) tells us that xy > 0. Thus, both conditions together are sufficient.

Therefore, C is the answer.
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[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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[Math Revolution GMAT math practice question]

The 2 lines x+2y=3, 2x+py=q have infinitely many points of intersection in the xy-plane. Which of the following could be the value of p?

A. 0
B. 1
C. 2
D. 3
E. 4

=>

If the 2 lines have infinitely many points of intersection, their equations must specify the same straight line.
The equation x+2y=3 is equivalent to 2x+4y=6.
So, p = 4 and q = 6.

Therefore, the answer is E.
Answer: E
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[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
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[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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[Math Revolution GMAT math practice question]

Tom’s Mom is five times as old as him. In 6 years, Tom’s Mom will be three times as old as him. How old is Tom now?

A. 3
B. 4
C. 5
D. 6
E. 7

=>

We can set up this question using the IVY approach. “Is”
can be converted to “=” and “times” can be converted to the multiplication sign “x”.

Let m and t be Tom’s Mom’s age and Tom’s age now, respectively.
Then m = 5t since Tom’s mom is five times as old as Tom.
We also have
m + 6 = 3(t + 6) since Tom’s mom will be three times as old as Tom in 6 years.

Plugging m = 5t into the second equation yields
5t + 6 = 3t + 18 or 2t = 12.
So, t = 6.

Therefore, the answer is D.
Answer: D
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I will impart the post to my companions. I'm extremely awed with your article, such incredible and useful learning you specified here
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[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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[Math Revolution GMAT math practice question]

The points (1,6) and (7,-2) are two vertices of one diagonal of a square in the xy-plane. What is the area of the square?

A. 16
B. 25
C. 32
D. 36
E. 50

=>

The length of the diagonal of the square is √{ (7-1)^2 + (6-(-2))^2 } = √100 = 10.
The side-length of the square is 10/ √2 = 5 √2 since the side-length of a square is equal to the length of its diagonal divided by √2. Thus, the area of the square is (5 √2)^2 = 50.

Therefore, E is the answer.
Answer: E
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[Math Revolution GMAT math practice question]

If |X| is the number of elements in set X, and “∪” is the union and “∩” is the intersection of 2 sets, what is the value of |A∩B|?

1) |A∪B|=50
2) |B|=40

=>

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.
Note that
|A∪B| = |A| + |B| - |A∩B| and |A∩B| = |A| + |B| - |A∪B|.

Since we have 4 variables (|A∩B|, |A|, |B|, |A∪B|) and 1 equation (|A∩B| = |A| + |B| - |A∪B|), 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)
Suppose A and B are disjoint sets, |A∪B| = 50, |A| = 10, and |B| = 40. Then |A∩B| = |A| + |B| - |A∪B| = 0.
Suppose A contains B, |A∪B| = 50, |A| = 50, and |B| = 40. Then |A∩B| = |A| + |B| - |A∪B| = 40.
Since we don’t have a unique solution, both conditions together are not sufficient.


Therefore, E is the answer.
Answer: E
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