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Math Revolution GMAT Instructor V
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Hello, I am Max Lee, Founder and Lead Math Tutor at Math Revolution.

I have over 6,000 posts and almost 5,000 Kudos. This topic is a new feature on the DS Forum and a way for you to directly interact with me and ask anything about the DS, e.g. if you want a certain concept explained or have a particular you question you want me addressed, this is the place to post a link to it or your question. I intend to have this thread be as a "Everything You Need to Know about DS" type of thread. I will keep updating this post with links and resources that are helpful for the DS. Meanwhile, you can ask me anything My other discussions you may be interested in:

Thank you all - good luck on the GMAT and look forward to seeing you in the DS forum!
- Max Lee.

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Math Revolution GMAT Instructor V
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[Math Revolution GMAT math practice question]

(absolute value) If $$|x+1|=2|x-1|, x=?$$

$$1) x<1$$
$$2) x>0$$
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Originally posted by MathRevolution on 15 Oct 2018, 01:40.
Last edited by MathRevolution on 16 Oct 2018, 00:29, edited 1 time in total.
##### General Discussion
Math Revolution GMAT Instructor V
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MathRevolution wrote:
[Math Revolution GMAT math practice question]

(absolute value) If $$|x+1|=2|x-1|, x=?$$

$$1) x<1$$
$$2) x>0$$

=>

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 original condition $$|x+1|=2|x-1|$$ is equivalent to $$x=\frac{1}{3}$$ or $$x=3$$ as shown below:

$$|x+1|=2|x-1|$$
$$=> |x+1|^2=(2|x-1|)^2$$
$$=> (x+1)^2=4(x-1)^2$$
$$=> x^2+2x+1=4(x^2-2x+1)$$
$$=> x^2+2x+1=4x^2-8x+4$$
$$=> 3x^2-10x+3 = 0$$
$$=> (3x-1)(x-3) = 0$$
$$=> 3x-1=0$$ or $$x-3 = 0$$
$$=> x=\frac{1}{3}$$ or $$x=3$$

Thus, condition 1) is sufficient since it gives a unique solution.

Condition 2) is not sufficient, since it allows both possible solutions.

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[Math Revolution GMAT math practice question]

Is the sum of $$7$$ different positive integers greater than or equal to $$48$$?

1) Their median is $$9$$
2) The largest number is $$12$$
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Math Revolution GMAT Instructor V
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[Math Revolution GMAT math practice question]

(set) 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 $$|A∩B|$$?

$$1) |A∪B|＝50$$
$$2) |B|=40$$
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MathRevolution wrote:
[Math Revolution GMAT math practice question]

Is the sum of $$7$$ different positive integers greater than or equal to $$48$$?

1) Their median is $$9$$
2) The largest number is $$12$$

=>

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 many variables (x1, x2, …, x7) 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)
We write the numbers as x1≤x2≤x3≤x4≤x5≤x6≤x7. Then x4 is their median.
From condition 1), x4 = 9 and the smallest possible number is 1 + 2 + 3 + 9 + 10 + 11 + 12 = 48. Therefore, the answer is ‘yes’.
Both conditions 1) & 2) together are sufficient.

Since this question is a statistics 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)
Since we didn’t use condition 2) in the above argument, condition 1) is sufficient on its own.

By condition 1), x4 = 9 and the smallest possible number is 1 + 2 + 3 + 9 + 10 + 11 + 12 = 48. Therefore, the answer is ‘yes’.

Condition 2)
When the numbers are 6,7,8,9,10,11,12, their sum is 6 + 7 + 8 + 9 + 10 + 11 + 12 = 63 > 48, and the answer is ‘yes’.
When the numbers are 6,7,8,9,10,11,12, their sum is 1 + 2 + 3 + 4 + 5 + 6 + 12 = 33 < 48, and the answer is ‘no’.
Since we don’t obtain a unique answer, condition 2) is not sufficient.

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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[Math Revolution GMAT math practice question]

(inequality) Is $$1+x+x^2+x^3+x^4+x^5+x^6<\frac{1}{(1-x)}$$?

$$1) x>0$$
$$2) x<1$$
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Math Revolution GMAT Instructor V
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[Math Revolution GMAT math practice question]

(inequality) Is $$x^3-y^3>x^2+xy+y^2$$?

$$1) x > y + 1$$
$$2) 0 < y < x$$
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[Math Revolution GMAT math practice question]

(function) In the xy-plane, does the graph of $$y=ax^2+c$$ intersect the x-axis?

$$1) a>0$$
$$2) c>0$$
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Math Revolution GMAT Instructor V
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MathRevolution wrote:
[Math Revolution GMAT math practice question]

(inequality) Is $$x^3-y^>x^2+xy+y^2$$?

$$1) x > y + 1$$
$$2) 0 < y < x$$

=>

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 $$x^3-y^3>x^2+xy+y^2$$ is equivalent to $$x > y + 1$$ as shown below:
$$x^3-y^3 > x^2+xy+y^2$$
$$=> (x-y)(x^2+xy+y^2)>x^2+xy+y^2$$
$$=> x – y > 1$$ after dividing both sides by $$x^2+xy+y^2$$, since $$x^2+xy+y^2 > 0.$$

Since the final inequality is equivalent to $$x > y + 1$$, condition 1) is sufficient.

Condition 2)
If $$x = 3$$ and $$y = 1$$, then $$x – y = 2 > 1$$, and the answer is ‘yes’.
If $$x = 1$$ and $$y = \frac{1}{2},$$ then $$x – y = \frac{1}{2} < 1,$$ and the answer is ‘no’.
Since it doesn’t give a unique answer, condition 2) is not sufficient.

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MathRevolution wrote:
[Math Revolution GMAT math practice question]

(statistics) If the average (arithmetic mean) of $$5$$ numbers is $$20$$, what is their standard deviation?

1) Their minimum is $$20$$.
2) Their maximum is $$20$$.

=>

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.

Note that if the average and the maximum of a data set are the same, then all of the data values are the same and the standard deviation is 0. Similarly, if the average and the minimum of a data set are the same, all of the data values are the same and the standard deviation is 0.

Thus, each of conditions is sufficient on its own since the minimum and the maximum are the same as the average.

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R2S wrote:
MathRevolution wrote:
MathRevolution wrote:
[Math Revolution GMAT math practice question]

(inequality) Is $$1+x+x^2+x^3+x^4+x^5+x^6<\frac{1}{(1-x)}$$?

$$1) x>0$$
$$2) x<1$$

=>

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.

The question $$1+x+x^2+x^3+x^4+x^5+x^6<\frac{1}{(1-x)}$$ is equivalent to $$0 < x < 1$$ as shown below:

For $$x ≠1$$,
=>$$1+x+x^2+x^3+x^4+x^5+x^6<\frac{1}{(1-x)}$$
$$=> (1+x+x^2+x^3+x^4+x^5+x^6)(1-x)^2< (1-x)$$
$$=>(1 - x^7)(1 - x) < 1 – x$$
$$=> 1 - x^7 – x +x^8 < 1 - x$$
$$=> - x^7 + x^8 < 0$$
$$=> x^7( x – 1 ) < 0$$
$$=> x( x – 1 ) < 0$$
$$=> 0 < x < 1$$

Since both conditions must be applied together to obtain this inequality, both conditions 1) & 2) are sufficient, when applied together.

Can you please explain, how did you reach to (1-x^7)(1-x)<1-x step, I didn't get 1-x^7

Regards

Hi R2S

If we multiply $$1+x+x^2+x^3+x^4+x^5+x^6$$ by 1-x, we get ($$1+x+x^2+x^3+x^4+x^5+x^6-x-x^2-x^3-x^4-x^5-x^6-x^7$$)(1-x)

All the terms are canceled except $$1-x^7$$(1-x).

Hope it's clear.
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Afc0892 wrote:
MathRevolution wrote:
[Math Revolution GMAT math practice question]

(sequence) The terms of $$a$$ sequence are defined by an=an-2+3. Is 411 a term of the sequence?

1) a1=111
2) a2=112

From statement 1:

$$a_1$$ = 111.

$$a_3 = a_1$$ + 3 = 114

$$a_5 = a_3$$ + 3 = 117.

The series becomes 111, 114, 117....

the terms are 1, 3, 5, 7....

The common difference between the terms 1,3,5... is 3.
Here the common difference is 2; 3-1=2; 5-3=2

Let's consider 411 as the last term. then 411 = 111 +(n-1)3.
Then 411 becomes 101th term.

Now the formula is, a = a1+(n-1)*d
411 = 111+(n-1)*2
411-111=(n-1)*2
300=(n-1)*2
150=n-1
151=n
This shows that 411 is the 151st term in the sequence. Is it correct?

Based on either, the answer choice will still remain the same i.e. A

A is sufficient.

From statement 2:

We get $$a_2$$, $$a_4$$ and $$a_6$$ and so on..
But first term is unknown.
B is sinsufficient.

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MathRevolution wrote:
[Math Revolution GMAT math practice question]

(number property) When a positive integer $$n$$ is divided by $$19$$, what is the remainder?

1) $$n-17$$ is a multiple of $$19$$
2) $$n-19$$ is a multiple of $$17$$

Target question: When n is divided by 19, what is the remainder?

Statement 1: n-17 is a multiple of 19
----ASIDE-------------------------------------------
If N is a multiple of d, then we can write N = dk (for some integer k)
For example, if N is a multiple of 5, then we can write N = 5k (for some integer k)
----BACK TO THE QUESTION-------------------

If n-17 is a multiple of 19, then we can write: n - 17 = 19k
Add 17 to both sides to get: n = 19k + 17
In other words, n is 17 greater than some multiple of 19
So, when we divide n by 19, the remainder will be 17
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: n-19 is a multiple of 17
We can write n-19 = 17k
Add 19 to both sides to get: n = 17k + 19
Hmmm, this information doesn't help us answer the target question
Case a: if k = 1, then n = 17(1) + 19 = 36. When we divide 36 by 19, the answer to the target question is the remainder is 17
Case b: if k = 2, then n = 17(2) + 19 = 53. When we divide 53 by 19, the answer to the target question is the remainder is 15
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Cheers,
Brent
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[Math Revolution GMAT math practice question]

(inequality) Is $$x + \frac{1}{x} > 2$$?

$$1) x > 0$$
$$2) x ≠ 1$$
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[Math Revolution GMAT math practice question]

(number properties) If $$n$$ is a positive integer, is $$\sqrt{n+1}$$ an even integer?

1) $$n$$ is the product of $$2$$ consecutive odd numbers
2) $$n$$ is an odd number
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MathRevolution wrote:
[Math Revolution GMAT math practice question]

(number properties) If $$m$$ and $$n$$ are positive integers, is $$m + n$$ an odd number?

1) $$\frac{m}{n}$$ is an even number
2) $$m$$ or $$n$$ is an even number

#1:
m/n = even no
m=6, n= 3 m/n= 2 ; m+n= 6+3= 9 sufficient

m=4, n=2 ; m/n= 2 ; m+n=4+2 = 6 not sufficient

#1 not sufficient

#2 :
m or n is an even no

property of odd no in addition : even+ odd = odd no
so since m & n are + integers so either of m or n being even other has to be odd integer hence we would get m+n= odd only

sufficient

IMO B
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MathRevolution wrote:
[Math Revolution GMAT math practice question]

(number properties) What is the value of the integer $$n$$?

1) $$n$$ is a prime factor of $$21$$
2) $$n$$ is a factor of $$49$$

=>

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 we have $$1$$ variable ($$n$$) and $$0$$ equations, D is most likely to be the answer. So, we should consider each condition on its own first.

Condition 1)
$$n$$ is a prime factor of $$21 = 3*7$$ and $$n$$ is $$3$$ or $$7$$.
Since it does not give a unique answer, condition 1) is not sufficient.

Condition 2)
If $$n$$ is a factor of $$49 = 7^2$$, then $$n$$ is $$1, 7$$ or $$49$$.
Since it does not give a unique answer, condition 2) is not sufficient.

Conditions 1) & 2)
The unique integer satisfying both conditions is $$n = 7.$$
Both conditions are sufficient, when taken together.

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]

(number properties) If $$x$$ and $$y$$ are integers, is $$x^2-y^2$$ an even integer?

1) $$x^3-y^3$$ is an even integer
2) $$x+y$$ is an even integer
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[Math Revolution GMAT math practice question]

(absolute value) Is $$ab<0$$?

$$1) |a+b| = - ( a + b )$$
$$2) |a+b| + 1 = |a| + |b|$$
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