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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8141
GMAT 1: 760 Q51 V42 GPA: 3.82
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1
[GMAT math practice question]

If x<y, is x(1+x)<y(1+y)?

1) x>1/2
2) x+y>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.

Now,

x(1+x)<y(1+y)
=> x+x^2 - y - y^2 < 0
=> (x-y) + (x^2 - y^2) < 0
=> (x-y) + (x-y)(x+y) < 0
=> (x-y)(1+x+y) < 0
=> 1+x+y > 0, since x < y.

Condition 1)
Since y > x > 1/2, we have x + y + 1 > 0.
Thus, condition 1) is sufficient.

Condition 2)
Since x + y > 1, we have x + y > 0.
Thus, condition 2) is sufficient too.

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Joined: 16 Aug 2015
Posts: 8141
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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[GMAT math practice question]

Is |x+y|>|x+z|?

1) y>z
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.

Since we have 3 variables (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) together:
If x =1, y = 2, and z = 3, the answer is ‘yes’.
If x = 1, y = -2, and z = -3, the answer is ‘no’.

Since we don’t have a unique solution, both conditions together are 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 Instructor V
Joined: 16 Aug 2015
Posts: 8141
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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[GMAT math practice question]

If b≠0, and |ab|=ab, is b>0?

1) a<0
2) a/b>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.

|ab| = ab
=> ab ≥ 0

Condition 1)
Since a < 0, b≠0 and ab ≥ 0, we have b < 0.
Since ‘no’ is also a unique answer by CMT (Common Mistake Type) 1, condition 1) is sufficient.

Condition 2)
If a = 1 and b = 1, the answer is ‘yes’.
If a = -1 and b = -1, the answer is ‘no’.
Since we don’t have a unique solution, condition 2) is not sufficient.

_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8141
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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[GMAT math practice question]

In how many ways can two integers m and n, with m > n, be selected from the whole numbers from 12 to 32, inclusive?

A. 150
B. 180
C. 190
D. 210
E. 240

=>

Since the order of m and n is fixed, we only need to count the number of ways to choose 2 numbers from 12, 13, …, 32.

We have 21 numbers to choose from since 32 – 12 + 1 = 21.

The number of selections is
21C2 = (21*20) / (1*2) = 210.

_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8141
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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[GMAT math practice question]

Is x/yz>0？

1) yz>x^2
2) x<y+z

=>

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) together:
If x = 1, y = 2 and z = 3, then the answer is ‘yes’.
If x = -1, y = 2 and z = 3, then the answer is ‘no’.

Since we don’t have a unique answer, both conditions together are not sufficient.

_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8141
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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[GMAT math practice question]

Is 2x+y>0?

1) x+y>0
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.

Since we have 2 variables (x and y) and 0 equations, C 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) together:
If x = 2, and y = -1, then the answer is ‘yes’.
If x = -1, and y = 2, then the answer is ‘no’.

Since we don’t have a unique solution, both conditions together are not sufficient.

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.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8141
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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[GMAT math practice question]

Which of the following inequalities is equivalent to |2x-|x||<3?

A. 0<x<2
B. 0<x<3
C. -1<x<3
D. 0<x<1
E. -3<x<1

=>

|2x-|x||<3
=> -3 < 2x – |x| < 3

Case 1: If x ≥ 0, then |x| = x, and so
-3 < 2x – |x| < 3
=> -3 < x < 3
=> 0 ≤ x < 3, since x ≥ 0.

Case 2: If x < 0, then |x| = - x, and so
-3 < 2x – |x| < 3
=> -3 < 3x < 3
=> - 1< x < 1
=> - 1< x < 0, since x < 0.

Thus, -1 < x < 3.

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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8141
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[GMAT math practice question]

How many different three-digit numbers can be formed which contain two digits that are the same, and a third digit that is different from the other two?

A. 196
B. 216
C. 243
D. 256
E. 316

=>

These three-digit numbers can have one of the forms XXY, XYX and YXX.
Note that 0 cannot be the hundreds digit.

Case 1): XXY
There are 9 possibilities for X (X is not 0), and 9 possibilities for Y (Y≠X).
This gives a total of 9*9 = 81 possible three-digit numbers of this form.

Case 2): XYX
There are 9 possibilities for X (X is not 0), and 9 possibilities for Y (Y≠X).
This gives a total of 9*9 = 81 possible three-digit numbers of this form.

Case 3): YXX
There are 9 possibilities for Y (Y is not 0), and 9 possibilities for X (X≠Y).
This gives a total of 9*9 = 81 possible three-digit numbers of this form.

Thus, the total number of possible three-digit numbers is 81 + 81 + 81 = 243.

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Alternative Approach:

The required numbers can be counted as follows:

The 2 digits can be selected out of 0, 1, 2…9 in 10C2ways = 45.

Out of these 2 selected numbers, the number which occurs twice can be selected in 2C1 ways = 2.

Now we need to arrange the digits to form number which has 2 identical digits, 1 different digits. It can be arranged in 3!/(2!*1!)= 3 ways.

Total numbers = 45*2*3 = 270.

But this also includes the cases which has 0 at hundredth place, we need to eliminate these cases:

001, 002, 003, …, 009 = 9

010, 020, 030, …, 090 = 9

011, 022, 033,…, 099 = 9

the cases which has 0 at hundredth place = 9*3=27

Required numbers = 270 - 27 = 243. Answer C

MathRevolution wrote:
[GMAT math practice question]

How many different three-digit numbers can be formed which contain two digits that are the same, and a third digit that is different from the other two?

A. 196
B. 216
C. 243
D. 256
E. 316

=>

These three-digit numbers can have one of the forms XXY, XYX and YXX.
Note that 0 cannot be the hundreds digit.

Case 1): XXY
There are 9 possibilities for X (X is not 0), and 9 possibilities for Y (Y≠X).
This gives a total of 9*9 = 81 possible three-digit numbers of this form.

Case 2): XYX
There are 9 possibilities for X (X is not 0), and 9 possibilities for Y (Y≠X).
This gives a total of 9*9 = 81 possible three-digit numbers of this form.

Case 3): YXX
There are 9 possibilities for Y (Y is not 0), and 9 possibilities for X (X≠Y).
This gives a total of 9*9 = 81 possible three-digit numbers of this form.

Thus, the total number of possible three-digit numbers is 81 + 81 + 81 = 243.

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Joined: 16 Aug 2015
Posts: 8141
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The Ultimate Q51 Guide [Expert Level]  [#permalink]

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[GMAT math practice question]

Are there more girls than boys at a school?

1) 3/7 of the number of girls is more than 1/3 of the number of boys
2) 1/3 of the number of girls is more than 2/5 of the number of boys

=>

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 2 variables (b for boys and g for girls) and 0 equations, C 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) together:
From condition 1:
(3/7)g > (1/3)b
=> 9g > 7b

From condition 2:
(1/3)g > (2/5)b
=> 5g > 6b
=> 6g > 5g > 6b
=> g > b

Both conditions together are sufficient.

Since this question is an inequality 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):

(3/7)g > (1/3)b
=> 9g > 7b

If g = 10 and b = 8, then the answer is ‘yes’.
If g = 8 and b = 8, then the answer is ‘no’.

Thus, condition 1) is not sufficient on its own.

Condition 2):

(1/3)g > (2/5)b
=> 5g > 6b
=> 6g > 5g > 6b
=> g > b

Thus, condition 2) is sufficient on its own.

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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Joined: 16 Aug 2015
Posts: 8141
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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[GMAT math practice question]

Is x – y > 0?

1) |x| + y > 0
2) 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 2 variables (x and y) and 0 equations, C 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) together:
If x = 2, and y = 1, then the answer is ‘yes’.
If x = 1, and y = 2, then the answer is ‘no’.

Since we don’t have a unique solution, both conditions are not sufficient, when taken together.

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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[quote="MathRevolution"][GMAT math practice question]

If x + y > 0, is xy^2 + x^2y > 0?

1) x > y
2) xy > 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.

Question:
xy^2 + x^2y > 0
=> xy(x+y) > 0
=> xy > 0, since x + y > 0
So, the question asks if xy > 0.

Since xy > 1 > 0 can be derived from condition 2), it is sufficient.
Condition 1) gives us no information about the sign of xy.

Answer: I think ans should had been D, as using (1) alone also we can solve.

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Manager  D
Joined: 17 May 2015
Posts: 246
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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jayantbakshi wrote:
MathRevolution wrote:
[GMAT math practice question]

If x + y > 0, is xy^2 + x^2y > 0?

1) x > y
2) xy > 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.

Question:
xy^2 + x^2y > 0
=> xy(x+y) > 0
=> xy > 0, since x + y > 0
So, the question asks if xy > 0.

Since xy > 1 > 0 can be derived from condition 2), it is sufficient.
Condition 1) gives us no information about the sign of xy.

Answer: I think ans should had been D, as using (1) alone also we can solve.

Sent from my iPhone using GMAT Club Forum mobile app

jayantbakshi,

St.(1) is not sufficient. Given x+y >0 and we have to check whether xy > 0 or not.
St.(1) x > y

Case1: x = 3, y = -1, xy < 0

Case2: x = 3, y = 2, xy > 0

We don't have a unique answer.

Hope this helps.

Thanks.
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8141
GMAT 1: 760 Q51 V42 GPA: 3.82
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[GMAT math practice question]

If the average (arithmetic mean) of 100 numbers is 50, what is the standard deviation of the numbers?

1) The smallest number is 50
2) The largest number is 50

=>
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 VA(Variable Approach) method is modifying the original condition and the question, and rechecking the number of variables and the number of equations. We often encounter this type of question in the GMAT quant exam these days. If the mean is equal to either the maximum or the minimum, or the range ( = Max – Min ) is zero, then the standard deviation is zero.

Condition 1)
All data items are greater than or equal to 50, and their average is 50.
It means that all data items are equal to 50.
Since all of the data items are equal, their standard deviation is 0.
This is sufficient.

Condition 2)
All data items are less than or equal to 50, and their average is 50.
It means that all data items are equal to 50.
Since all of the data items are equal, their standard deviation is 0.
This is sufficient too.

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.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8141
GMAT 1: 760 Q51 V42 GPA: 3.82
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[GMAT math practice question]

Car A drives from P to Q at a constant rate of 100 km per hour. After car A has driven for 1 hour, train B begins traveling from Q to P at a constant rate of 150 km per hour. If the distance between P and Q is 600 km, then what distance has car A traveled when it meets train B?

A. 200 km
B. 220 km
C. 250 km
D. 270 km
E. 300 km

=>

After car A has driven for 1 hour, the distance between car A and train B is 500 km.

Car A and train B approach each other at a speed of 250 km/hr. This means that they will take 2 hours to meet each other.
When they meet, car A will have traveled for 3 hours, and have covered a distance of 3 * 100 = 300 km.

It is important that both vehicles will have traveled for the same amount of time after train B has started moving. This gives the equation 100 + 100t + 150t = 600, from which we may deduce that t = 2. Since 100 + 100 *2 = 300, E is the answer.
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Joined: 16 Aug 2015
Posts: 8141
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[GMAT math practice question]

If x > y >0, a < 0 and b > 0, then which of the following is (are) true?

Ⅰ. ax+by
Ⅱ.ax-by
Ⅲ. by-ax

A. Ⅰonly
B. Ⅱ only
C. Ⅲ only
D.Ⅰ& Ⅱ only
E. Ⅱ & Ⅲ only

=>

Statement I.
x = 2, y = 1, a = -1, b = 1: False

Statement II.
x = 2, y = 1, a = -1, b = 1: False

Statement III.
Since a < 0 and x > 0, -ax > 0. And by > 0 since b > 0 and y > 0.
by – ax = by + (-ax) > 0. True

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Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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[GMAT math practice question]

Is x^2-y^2>x+y?

1) x-y<0
2) 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 2 variables (x and y) and 0 equations, C 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) together:
x^2 – y^2 = (x+y)(x-y) > 0 since x+y < 0 and x-y < 0.
Therefore,
x^2 – y^2 > 0 > x + y
Thus, both conditions together are sufficient.

Since this question is an inequality 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 x = -2 and y = -1, then the answer is “yes”.
If x = 1 and y = 2, then the answer is “no”.

Thus, condition 1) is not sufficient.

Condition 2):
If x = -3 and y = -1, then the answer is “yes”.
If x = -1 and y = -3, then the answer is “no”.
Thus condition 2) is not sufficient.

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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Joined: 16 Aug 2015
Posts: 8141
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Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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[GMAT math practice question]

Is x>0?

1) (x+y)^2 > (x-y)^2
2) x+y > x-y

=>

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 2 variables (x and y) and 0 equations, C 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)
Condition 1) tells us that
(x+y)^2 > (x-y)^2
=> x^2 + 2xy + y^2 > x^2 - 2xy + y^2
=> 2xy > -2xy
=> 4xy > 0
=> xy > 0

Condition 2) tells us that
x + y > x – y
=> y > -y
=> 2y > 0
=> y > 0.

Since xy > 0 and y > 0, we have x > 0.

Thus, both conditions 1) & 2) together are sufficient.

In general, there are many questions involving integers and statistics to which we need to apply CMT(Common Mistake Type) 4.

Condition 1):
If x = -2 and y = -1, then the answer is “yes”.
If x = 1 and y = 2, then the answer is “no”.

Thus, condition 1) is not sufficient.

Condition 2):
If x = -3 and y = -1, then the answer is “yes”.
If x = -1 and y = -3, then the answer is “no”.
Thus condition 2) is not sufficient.

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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Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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[GMAT math practice question]

If the elements of set X are a, b, c and d, is the average (arithmetic mean) of a, b, c, and d contained in set X?

1) The average (arithmetic mean) of every pair of elements of set X is 10.
2) The average (arithmetic mean) of every three elements chosen from set X is 10.

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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, c and d) 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) and 2):
Condition 1) gives rise to the following equations:
( a + b ) / 2 = 10 => a + b = 20
( a + c ) / 2 = 10 => a + c = 20

( c + d ) / 2 = 10 => c + d = 20

Condition 2) gives rise to the following equations:
( a + b + c ) / 3 = 10 => a + b + c = 30
( a + b + d ) / 3 = 10 => a + b + d = 30
( a + c + d ) / 3 = 10 => a + c + d = 30
( b + c + d ) / 3 = 10 => b + c + d = 30

Combining these equations yields a = b = c = d = 10.
Therefore, the average is 10, which is an element of set X, and conditions 1) and 2) are sufficient, when taken together.

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)
Condition 1) gives rise to the following equations:
( a + b ) / 2 = 10 => a + b = 20
( a + c ) / 2 = 10 => a + c = 20

( c + d ) / 2 = 10 => c + d = 20

Combining these equations yields a = b = c = d = 10.

Condition 2)

Condition 2) gives rise to the following equations:
( a + b + c ) / 3 = 10 => a + b + c = 30
( a + b + d ) / 3 = 10 => a + b + d = 30
( a + c + d ) / 3 = 10 => a + c + d = 30
( b + c + d ) / 3 = 10 => b + c + d = 30

Combining these equations yields a = b = c = d = 10.

Since conditions 1) and 2) are equivalent, D is the answer by Tip 1) of the VA method.

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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Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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[GMAT math practice question]

When a positive integer n is divided by 5, the remainder is 2. What is the remainder when n is divided by 3?

1) n is divisible by 2
2) When n is divided by 15, the remainder is 2.

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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.

We have 1 variable (n) and 1 equation. So, D is most likely to be the answer, and we should consider each of the conditions on its own first.

Plugging-in numbers is the suggested approach to remainder questions.

Condition 1)
The possible values of n are
n = 2, 4, 6, 8, …
When these are divided by 3, the remainders are 0, 1 and 2.
Since the answer is not unique, condition 1) is not sufficient.

Condition 2)
The possible values of n are
n = 17, 32, 47, 62, …
When these are divided by 3, the remainder is always 2.
Since the answer is unique, condition 2) is sufficient.

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.
_________________ Re: The Ultimate Q51 Guide [Expert Level]   [#permalink] 17 Apr 2018, 18:16

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# The Ultimate Q51 Guide [Expert Level]

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