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# Math Revolution DS Expert - Ask Me Anything about GMAT DS

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Math Revolution GMAT Instructor
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Math Revolution GMAT Instructor
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[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 $$14$$
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MathRevolution wrote:
[GMAT math practice question]

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

$$1) 2#1 = 2$$
$$2) 4#2 = 2$$

=>

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.

Originally posted by MathRevolution on 31 Jan 2019, 02:12.
Last edited by MathRevolution on 26 May 2021, 02:51, edited 2 times in total.
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[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 $$6$$
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MathRevolution wrote:
[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 $$14$$

#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
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MathRevolution wrote:
[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
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
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MathRevolution wrote:
[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.

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.

Originally posted by MathRevolution on 01 Feb 2019, 02:48.
Last edited by MathRevolution on 28 Jul 2021, 00:42, edited 1 time in total.
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[GMAT math practice question]

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

$$1) p=q+4$$
$$2) x^q=16$$
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MathRevolution wrote:
[GMAT math practice question]

(exponent) If $$p, x,$$ and $$q$$ 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.
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MathRevolution wrote:
[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 $$6$$

=>

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.

Originally posted by MathRevolution on 04 Feb 2019, 05:30.
Last edited by MathRevolution on 20 Jul 2021, 02:58, edited 1 time in total.
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MathRevolution wrote:
[GMAT math practice question]

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

$$1) p=q+4$$
$$2) x^q=16$$

=>

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.

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.

Originally posted by MathRevolution on 04 Feb 2019, 05:36.
Last edited by MathRevolution on 20 Jul 2021, 02:59, edited 1 time in total.
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[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
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MathRevolution wrote:
[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.
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[GMAT math practice question]

If $$a>b>c>d>0$$, is $$d<4$$?

$$1) \frac{1}{c} + \frac{1}{d} > \frac{1}{2}$$
$$2) (\frac{1}{a})+(\frac{1}{b})+(\frac{1}{c})+(\frac{1}{d})=1$$
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MathRevolution wrote:
[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

given
$$(x-y)(x+y)(x^2+y^2)$$
it would stand odd only when the resultant of each function given is odd

#1
x is an odd no
not sufficient ; since for the relation to stand true value of y should be know ; else we can have both odd & even as result

#2
x-y is odd
which means that either of x or y is an even and other is odd integer
which makes our relation
$$(x-y)(x+y)(x^2+y^2)$$
stand valid as odd integer

so sufficient
IMO B
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MathRevolution wrote:
[GMAT math practice question]

If $$a>b>c>d>0$$, is $$d<4$$?

$$1) \frac{1}{c} + \frac{1}{d} > \frac{1}{2}$$
$$2) (\frac{1}{a})+(\frac{1}{b})+(\frac{1}{c})+(\frac{1}{d})=1$$

#1
1/c+1/d = 1/2
c+d/cd = 1/2
for all values d<4 it would be sufficient
so #1 is sufficient
#2
fraction of 1/a+1/b+1/c+1/d = 1
possible when all a=b=c=d are equal
so sufficient to say that d<4
IMO D

Originally posted by Archit3110 on 05 Feb 2019, 02:58.
Last edited by Archit3110 on 06 Feb 2019, 09:03, edited 1 time in total.
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MathRevolution wrote:
[GMAT math practice question]

If $$a>b>c>d>0$$, is $$d<4$$?

$$1) \frac{1}{c} + \frac{1}{d} > \frac{1}{2}$$
$$2) (\frac{1}{a})+(\frac{1}{b})+(\frac{1}{c})+(\frac{1}{d})=1$$

$$a > b > c > d > 0\,\,\,\,\, \Rightarrow \,\,\,\,0 < {1 \over a} < {1 \over b} < {1 \over c} < {1 \over d}\,\,\,\,\left( * \right)$$

$$d\,\,\mathop < \limits^? \,\,4$$

$$\left( 1 \right)\,\,{1 \over c} + {1 \over d} > {1 \over 2}\,\,\,\,\mathop \Rightarrow \limits^{\left( {**} \right)} \,\,\,\left\langle {{\rm{YES}}} \right\rangle$$

$$\left( {**} \right)\,\,\,d \ge 4\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,c > 4\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{1 \over c} + {1 \over d} < {1 \over 4} + {1 \over 4} = {1 \over 2}\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\rm{impossible}}$$

$$\left( 2 \right)\,\,\,1 = {1 \over a} + {1 \over b} + {1 \over c} + {1 \over d}\,\,\mathop < \limits^{\left( * \right)} \,\,4\left( {{1 \over d}} \right)\,\,\,\,\,\mathop \Rightarrow \limits^{ \cdot \,d\, > \,0} \,\,\,\,\,1 \cdot d < 4\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{YES}}} \right\rangle$$

The correct answer is therefore (D).

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

Regards,
Fabio.
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Archit3110 wrote:
MathRevolution wrote:
[GMAT math practice question]

If $$a>b>c>d>0$$, is $$d<4$$?

$$1) \frac{1}{c} + \frac{1}{d} > \frac{1}{2}$$
$$2) (\frac{1}{a})+(\frac{1}{b})+(\frac{1}{c})+(\frac{1}{d})=1$$

#2
fraction of 1/a+1/b+1/c+1/d = 1
possible when all a=b=c=d are equal
so in sufficient to say that d<4
IMO A

Hi, Archit3110 !

Please check the parts in red.

Regards,
Fabio.