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[GMAT math practice question]

(inequality) If \(a\) and \(b\) are integers, and \(x\) and \(y\) are positive integers, is \(a^x+b^y > 0?\)

\(1) a^{x+y}>0\)
\(2) b^{x+y}>0\)
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[GMAT math practice question]

(algebra) \(x^3-y^3=?\)

\(1) x-y=0\)
\(2) |x|=|y|\)


given
\(x^3-y^3=?\)

or say
x^3=y^3

#1
x-y=0
or x=y
so for each value of x =y ;x^3=y^3 ; sufficient

#2

\(2) |x|=|y|\)

will be true for x^3=y^3 when both x & y are -ve , but insufficient when either of x or y is -ve and other is +
so insufficient

IMOA
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[GMAT math practice question]

(absolute value) Is \(|a|<1\)?

\(1) a^2<1\)
\(2) \frac{1}{(1-a^2)}>0\)

=>

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.

\(|a|<1\)
\(=> |a|^2 < 1\)
\(=> a^2 < 1\)
\(=> a^2-1 < 0\)
\(=> (a+1)(a-1) < 0\)
\(=> -1 < a < 1\)

Condition 1) is sufficient, since it is equivalent to the question.

Condition 2)
\(\frac{1}{(1-a^2)} > 0\)
\(=> (1-a^2) > 0\)
\(=> a^2-1 < 0\)
\(=> (a+1)(a-1) < 0\)
\(=> -1 < a < 1\)
Thus, condition 2) is sufficient, since it is also equivalent to the question.

Therefore, the answer is D.
Answer: D

FYI: Tip 1) of the VA method states that D is most likely to be the answer if conditions 1) and 2) provide the same information.


MathRevolution
quoting your line:
\(\frac{1}{(1-a^2)} > 0\)
\(=> (1-a^2) > 0\)

how is the highlighted part above been derived?
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[GMAT math practice question]

(exponent) \(\frac{a^{n^2+n-1}}{a^{(n-1)(n+2)}}=?\)

\(1) n=5\)
\(2) a=5\)


given

[m]\frac{a^{n^2+n-1}}{a^{(n-1)(n+2)}}

which can be solved :
we get a^1
#1
value of a not known
in sufficeint

#2
a=5
means value of eqn is 5
sufficient
IMO B
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[GMAT math practice question]

(inequality) If \(a\) and \(b\) are integers, and \(x\) and \(y\) are positive integers, is \(a^x+b^y > 0?\)

\(1) a^{x+y}>0\)
\(2) b^{x+y}>0\)

given

\(a^x+b^y > 0\\
\\
and a,b are integers and x,y are +ve integers\\
so\\
#1\\
m]1) a^{x+y}>0\)

will be true
a = -1 and x,y are even +ve integers or both are odd
a=1 , for any +ve integer value of any x,y will make relation sufficient
but sine value of b is not known is insufficient

#2
\(2) b^{x+y}>0\)
again same reasoning as #1
value of a not known so insufficeint
from1 & 2
we cannot say whther a & b are +ve or -ve and x& y can be either odd or even

so in sufficient

IMO E
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[GMAT math practice question]

(statistics) If the median of \(5\) positive integers is \(10\), is their average (arithmetic mean) greater than \(10\)?

1) The largest number is \(40\)
2) The smallest number is \(1\)
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[GMAT math practice question]

(inequality) If \(a\) and \(b\) are integers, and \(x\) and \(y\) are positive integers, is \(a^x+b^y > 0?\)

\(1) a^{x+y}>0\)
\(2) b^{x+y}>0\)

=>

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.

Since we have \(4\) variables (\(a, b, x\) and \(y\)) 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):
If \(a = 1, b = 1, x = 1\) and \(y = 1\), then \(ax+by = 2\), and the answer is ‘yes’.
If \(a = -1, b = -1, x = 1\) and \(y = 1\), then \(ax+by = -2\), and the answer is ‘no’.

Thus, both conditions together are not sufficient since they do not yield a unique solution.

Therefore, E is the answer.
Answer: E

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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[GMAT math practice question]

(number properties) If \(a\) and \(b\) are positive integers, is \(a+b\) divisible by \(7\)?

1) \(a – b\) is divisible by \(7\)
2) \(a+b+7\) is divisible by \(91\)
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[GMAT math practice question]

(statistics) If the median of \(5\) positive integers is \(10\), is their average (arithmetic mean) greater than \(10\)?

1) The largest number is \(40\)
2) The smallest number is \(1\)

=>

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.

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

Suppose the numbers satisfy \(a ≤ b ≤ 10 ≤ c ≤ d\). The question asks if \(\frac{( a + b + 10 + c + d )}{5} > 10\) or \(a + b + c + d + 10 > 50.\)
This is equivalent to the inequality, \(a + b + c + d > 40.\)
If a question includes the words “greater than”, then it asks us to look for a minimum.
Since \(a, b c,\) and \(d\) are positive, and \(d = 40\) by condition 1), we must have \(a + b + c + d > 40.\)
Condition 1) is sufficient.

Condition 2)
If \(a = 1, b = 2, c = 11\), and \(d = 40\), then \(a + b + c + d > 40\), and the answer is ‘yes’.
If \(a = 1, b = 2, c = 11\), and \(d = 12\), then \(a + b + c + d < 40\), and the answer is ‘no’.
Thus, condition 2) is not sufficient since it does not yield a unique solution.

Therefore, A is the answer.
Answer: A
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[GMAT math practice question]

(exponent) \(f(x)=x^{2n}+x^n+1\), where \(n\) is an integer. Is \(f(x)=1\)?

\(1) x=-1\)
\(2) n\) is a multiple of \(5\).
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[GMAT math practice question]

(number properties) If \(a\) and \(b\) are positive integers, is \(a+b\) divisible by \(7\)?

1) \(a – b\) is divisible by \(7\)
2) \(a+b+7\) is divisible by \(91\)

=>

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.

Condition 2) tells us that \(a + b + 7 = 91k\) for some integer \(k\), and so \(a + b = 91k – 7 = 7(13k-1).\) Therefore, \(a + b\) is a multiple of \(7\).
Thus, condition 2) is sufficient.

Condition 1)
If \(a = 14\) and \(b = 7\), then \(a + b = 21\) is a multiple of \(7\), and the answer is ‘yes’.
If \(a = 8\) and \(b = 1\), then \(a + b = 9\) is not a multiple of \(7\), and the answer is ‘no’.
Thus, condition 1) is not sufficient since it does not yield a unique solution.

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

(number properties) \(a, b\), and \(c\) are positive integers. Is \(a+b+c\) an odd number?

1) \(ab\) is an odd number
2) \(c\) is an odd number
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[GMAT math practice question]

(number properties) \(a, b\), and \(c\) are positive integers. Is \(a+b+c\) an odd number?

1) \(ab\) is an odd number
2) \(c\) is an odd number


#1
ab is an odd no ; both a & b are odd integers
so a+b+c ; odd if c is odd and even if c is even
in sufficient

#2
c is odd ; value of a & b not know insufficient
from 1&2
a+b+c will be odd
sufficient
IMO C
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[GMAT math practice question]

(exponent) \(f(x)=x^{2n}+x^n+1\), where \(n\) is an integer. Is \(f(x)=1\)?

\(1) x=-1\)
\(2) n\) is a multiple of \(5\).

=>

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.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.
\(f(x)=1\)
\(=> x^{2n}+x^n+1=1\)
\(=> x^{2n}+x^n=0\)
\(=> x^n(x^n +1)=0\)
\(=> x^n= 0\) or \(x^n =-1\)
=> ( \(x = 0\)) or ( \(x = -1\) and \(n\) is odd )

Conditions 1) and 2)
If \(x = -1\) and \(n = 5\), then \(f(x) = (-1)^{10} + (-1)^5 + 1 = 1 + (-1) + 1 = 1\) and the answer is ‘yes’.
If \(x = -1\) and \(n = 10\), then \(f(x) = (-1)^{20} + (-1)^{10} + 1 = 1 + 1 + 1 = 1\) and the answer is ‘no’.

Thus, both conditions together are not sufficient, since they do not yield a unique solution.

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

(number properties) \(a, b\), and \(c\) are positive integers. Is \(a+b+c\) an odd number?

1) \(ab\) is an odd number
2) \(c\) is an odd number

=>

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 (\(a, b, x\) and \(y\)) 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 with the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)
Condition 1), \(ab\) is an odd integer, is equivalent to the statement that both \(a\) and \(b\) are odd numbers.
As \(c\) is also an odd number, \(a + b + c\) is an odd number since it is the sum of three odd numbers.

Both conditions together are sufficient.

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)
If \(a = 1, b = 1\), and \(c = 1\), then \(a + b + c = 3\) is an odd number, and the answer is ‘yes’.
If \(a = 1, b = 1\), and \(c = 2\), then \(a + b + c = 4\) is not an odd number, and the answer is ‘no’.
Thus, condition 1) is not sufficient.

Condition 2)
If \(a = 1, b = 1\), and \(c = 1\), then \(a + b + c = 3\) is an odd number, and the answer is ‘yes’.
If \(a = 2, b = 1\), and \(c = 1\), then we have \(a + b + c = 4\) is not an odd number, and the answer is ‘no’.
Thus, 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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If n is an integer, is (n+1)(n+2)(n+3) divisible by 12?

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


Answer: B
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n is a positive integer. What is the remainder when n is divided by 3?

1) n^2 has remainder 1 when it is divided by 3
2) n has remainder 7 when it is divided by 9
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