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[GMAT math practice question]

(function) If operation \(#\) represents one of addition, subtraction, multiplication, and division, what is the value of \(0#1\)?

\(1) 2#1 = 2\)
\(2) 4#2 = 2\)

#1
2#1=2
# can be multiplication or division so sufficient
as doing both operations we would get answer as 0 for \(0#1\)

#2
4#2 =2
# can be subtraction of division; in sufficient


IMO A

#1
n can be 0,4,8,12... and always the sum of \(4^n+n^2+1\) would be odd which is not divisible by 2 ; sufficient

#2
n can be 0,6,12,18... and always the sum of \(4^n+n^2+1\) would be odd which is not divisible by 2 ; sufficient

IMO D

\(?\,\, = \,\,0\# 1\,\,\,\,\,\,\left( {\# \in \left\{ { + , - , \times , \div } \right\}} \right)\)


\(\left( 1 \right)\,\,2\# 1 = 2\,\,\,\, \Rightarrow \left\{ \matrix{\\
\,\,\# = \times \,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,? = 0 \hfill \cr \\
\,\,{\rm{OR}} \hfill \cr \\
\,\,\# = \div \,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,? = 0 \hfill \cr} \right.\,\,\,\, \Rightarrow \,\,\,\,\,\,{\rm{SUFF}}{\rm{.}}\)


\(\left( 2 \right)\,\,\,\,4\# 2 = 2\,\,\,\, \Rightarrow \left\{ \matrix{\\
\,\,\# = - \,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,? = - 1 \hfill \cr \\
\,\,{\rm{OR}} \hfill \cr \\
\,\,\# = \div \,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,? = 0 \hfill \cr} \right.\,\,\,\, \Rightarrow \,\,\,\,\,\,{\rm{INSUFF}}{\rm{.}}\)


The correct answer is therefore (A).


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.

=>

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.
Since \(4^n\) is a multiple of \(2\), we only need to look at \(n^2+1\).
If \(n\) is an odd number, \(4^n+n^2+1\) is divisible by \(2\).
If \(n\) is an even number, \(4^n+n^2+1\) is not divisible by \(2\).
The question asks if \(n\) is an odd number.
Thus, each of conditions is sufficient.

Therefore, D is the answer.
Answer: D

[GMAT math practice question]

Five data values are \(11, 14, 16, 18\) and \(x\). What is the value of \(x\)?

1) The mode of the \(5\) data values is \(11\)
2) The average (arithmetic mean) of the \(5\) data values is \(14\)

=>

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 operation is considered as a variable. Since we have \(1\) variable and \(0\) equations, D is most likely to be the answer. So, we should consider each condition on its own first.

Condition 1)
Since \(2#1 = 2\), \(#\) is one of the operations, multiplication and division.
If \(#\) is the multiplication operation, then \(0#1 = 0\).
If \(#\) is the division operation, then \(0#1 = 0.\)
Since condition 1) yields a unique solution, it is sufficient.

Condition 2)
Since \(4#2 = 2, #\) is one of the operations, subtraction and division.
If \(#\) is the subtraction operation, then \(0#1 = -1\).
If \(#\) is the division operation, then \(0#1 = 0\).
Since condition 2) doesn’t yield a unique solution, it is not sufficient.

Therefore, A is the answer.
Answer: A

[GMAT math practice question]

(number properties) Is \(3\) a factor of \(x\)?

1) \(x-3\) is divisible by \(6\)
2) \(x+3\) is divisible by \(6\)

#1
mode of set is 11
so x=11
mode most repeated no in a set
sufficient
#2
avg of set = 14
sum of digits = 59+x = 70
x=11
sufficient
IMO D

#1
sufficient
x= 9,15,21
x-3 divisible by 6 and x a factor of 3

#2
sufficient
x+3 divisible by 6
x=3,9,15,18
sufficient
IMO D

=>

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 condition on its own first.

Condition 1)
The data values include four different values and \(x\).
Since the mode of the five data values is \(11, x\) must equal \(11\).
Condition 1) is sufficient.

Condition 2)
Calculating the mean of the five data values yields
\(\frac{( 11 + 14 + 16 + 18 + x )}{5} = 14.\)
Solving for \(x\) gives
\(11 + 14 + 16 + 18 + x = 70\)
\(59 + x = 70\)
\(x = 11\)
Condition 2) is also sufficient.

Therefore, D is the answer.
Answer: D

[GMAT math practice question]

(exponent) If \(p, x,\) and \(y\) are integers, \(\frac{x^p}{x^q}\)=?

\(1) p=q+4\)
\(2) x^q=16\)

\(p,x,q\,\,{\rm{ints}}\)

\(?\,\, = \,\,{{{x^p}} \over {{x^q}}}\)

\(\left( {1 + 2} \right)\,\,\,\left\{ \matrix{\\
\,p = q + 4 \hfill \cr \\
\,{x^q} = 16 \hfill \cr} \right.\,\,\,\,\,\,::\,\,\,\,\,\,\left\{ \matrix{\\
\,{\rm{Take}}\,\,\left( {x,q,p} \right) = \left( {16,1,5} \right)\,\,\,\, \Rightarrow \,\,\,? = {16^4} \hfill \cr \\
\,{\rm{Take}}\,\,\left( {x,q,p} \right) = \left( {2,4,8} \right)\,\,\,\, \Rightarrow \,\,\,? = {2^4} \hfill \cr} \right.\)


The correct answer is therefore (E).


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.

=>

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 condition on its own first.

To solve remainder questions, plugging in numbers is recommended.

Condition 1)
If we plug in x = 9, then x – 3 = 6 is divisible by 6 and x is a multiple of 3. Condition 1) is sufficient.

Condition 2)
If we plug in x = 9, then x + 3 = 12 is divisible by 6 and x is a multiple of 3. Condition 2) is sufficient.

Therefore, the answer is D.
Answer: D

=>

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 3 variables (x, y and z) 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 \(x = 2, p = 8\) and \(q = 4\), then \(\frac{x^p}{x^q} = x^{p-q} = x^4 = 2^4 = 16.\)
If \(x = 16, p = 5\) and \(q = 1\), then \(\frac{x^p}{x^q} = x^{p-q} = x^4 = 16^4 = 2^{16} = 65536.\)
Since they do not yield a unique solution, both conditions are not sufficient, when considered together.

Therefore, the answer is E.
Answer: E

Note: This question is related to finding a hidden 1.
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.

[GMAT math practice question]

(number properties) If \(x, y\) are integers, is \((x-y)(x+y)(x^2+y^2)\) an odd number?

1) \(x\) is an odd number
2) \(x-y\) is an odd number

Statement 1 says X is odd number. X could be any odd number e.g. 1 and Y can be odd and even. If y is even then result will be odd but if Y is odd then result will be even. So Statement 1 is not sufficient.

Statement 2 says X-Y is odd then one number is odd and another is even, result will be odd.

statement 2 is sufficient.