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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8987
GMAT 1: 760 Q51 V42
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Re: Overview of GMAT Math Question Types and Patterns on the GMAT  [#permalink]

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[GMAT math practice question]

(Probability) Among integers from 1 to 50, inclusively, what is the number of the multiples of 4 or 5?

A. 14
B. 16
C. 18
D. 20
E. 22

=>

The number of multiples of 4 is 12 = (48 – 4)/4 + 1 = 11 + 1, since 4, 8, …, 48 are the multiples of 4 between 1 and 50, inclusive.
The number of multiples of 5 is 10 = (50 – 5)/5 + 1 = 9 + 1, since 5, 10, …, 50 are the multiples of 5 between 1 and 50, inclusive.
Multiples of 20 are double-counted, and the number of multiples of 20 is 2 since 20 and 40 are multiples of 20 between 1 and 50, inclusive.
Then we have 12 + 10 - 2 = 20.

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Joined: 16 Aug 2015
Posts: 8987
GMAT 1: 760 Q51 V42
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Re: Overview of GMAT Math Question Types and Patterns on the GMAT  [#permalink]

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[GMAT math practice question]

(Geometry) In the figure below, is triangle AEF an isosceles triangle?

1) AB = AC
2) DF is perpendicular to BC.

Attachment: 1.22ds.png [ 8.57 KiB | Viewed 557 times ]

=>

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.
Visit https://www.mathrevolution.com/gmat/lesson for details.

Since triangle AEF has three sides, we have 3 variables (AE, AF, and EF) and 0 equations, and E is most likely 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 have ∠B = ∠C since AB = AC, and the triangle is isosceles.
Assume ∠B = ∠C = x.
Then ∠DEC = ∠AEF = 90 – x since the triangle is a right triangle, and ∠DEC is congruent to ∠AEF.
Since triangle BDF is a right triangle, we have ∠AFE = 90 – x.
Thus we have ∠AEF = ∠AFE, which means the triangle is isosceles, and we have AE = AF.

Since both conditions together yield a unique solution, they are 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.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8987
GMAT 1: 760 Q51 V42
GPA: 3.82
Re: Overview of GMAT Math Question Types and Patterns on the GMAT  [#permalink]

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[GMAT math practice question]

(Geometry) What is the measure of ∠BIC in the figure?

1) Point I is the incenter (the point where the three angle bisectors meet) of triangle ABC.
2) ∠BAC = 50°

Attachment: 1.28ds.png [ 4.2 KiB | Viewed 534 times ]

=>

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.
Visit https://www.mathrevolution.com/gmat/lesson for details.

Since we have 4 variables (∠BIC, ∠ABC, ∠BCA, and ∠CAB) and 1 equation (∠ABC + ∠BCA + ∠CAB = 180°), C is most likely 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)
Since point O is the incenter of triangle ABC from condition 1), the segments IA, IB, and IC are bisectors of ∠BAC, ∠ABC and ∠BCA, respectively.

∠BIC = 180° – (∠IBC + ∠ICB), ∠BIC = 180° – (1/2)(180° - ∠BAC), ∠BIC = 180° – (1/2)(180° - 50°), ∠BIC = 180° – 65° = 115°.

Since both conditions together yield a unique solution, they are 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 in which the answer is A, B, C, or D.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8987
GMAT 1: 760 Q51 V42
GPA: 3.82
Re: Overview of GMAT Math Question Types and Patterns on the GMAT  [#permalink]

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[GMAT math practice question]

(Number Properties) If 100! = 100 x 99 x…x 2 x 1 can be written as 2^a3^b5^c7^d…., what is a?

A. 86
B. 97
C. 108
D. 119
E. 131

=>

We need to count the number of prime factors 2 in the prime factorization of 100!. We can do it by counting:

The number of multiples of 2 is [100/2] = 50.
The number of multiples of 4 is [100/4] = 25.
The number of multiples of 8 is [100/8] = 12.
The number of multiples of 16 is [100/16] = 6.
The number of multiples of 32 is [100/32] = 3.
The number of multiples of 64 is [100/64] = 1.

Thus the number of prime factors 2 is 50 + 25 + 12 + 6 + 3 + 1 = 97.

_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8987
GMAT 1: 760 Q51 V42
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Re: Overview of GMAT Math Question Types and Patterns on the GMAT  [#permalink]

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[GMAT math practice question]

(Geometry) The figure below shows the dimensions of triangle ABC. What is ∠BIC?

1) Point I is the incenter of △ABC.
2) Line DE is parallel to line BC.

Attachment: 2.4DS.png [ 8.92 KiB | Viewed 479 times ]

=>

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.
Visit https://www.mathrevolution.com/gmat/lesson for details.

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

Condition 1)

Since we can find an incenter of a triangle by the intersection of lines bisecting interior angles, ∠DBI is equal to ∠IBC, and ∠ECI is equal to ∠ICB.

Attachment: 2.4DS(A).png [ 10.96 KiB | Viewed 478 times ]

Then we have ∠IBC = 22° and ∠ICB = 30°.
Thus, we have ∠BIC = 180° - ∠IBC - ∠ICB = 180° – 22° – 30° = 128°.

Since condition 1) yields a unique solution, it is sufficient.

Condition 2)

Since we don’t know the position of I on segment DE, condition 2) does not yield a unique solution, and it is not sufficient.

_________________
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Joined: 16 Aug 2015
Posts: 8987
GMAT 1: 760 Q51 V42
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Re: Overview of GMAT Math Question Types and Patterns on the GMAT  [#permalink]

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[GMAT math practice question]

(Algebra) What is the value of [x] + [-x]? ([x] means the greatest integer less than or equal to x.)

1) 0 ≤ x.
2) x is not an integer.

=>

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.
Visit https://www.mathrevolution.com/gmat/lesson for details.

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

If x = n + h where n is an integer and 0 ≤ h < 1, then [x] = n. Here n is the integer part of n, and h is the decimal part of n.

If x is an integer, then we have x = n + h where h = 0, [x] = n, [-x] = -n and [x] + [-x] = n + (-n) = 0.

Assume x is not an integer.
Then we have x = n + h where 0 < h < 1, [x] = n.
We have -x = -n - h, -x = -n - 1 + 1 - h, -x = -(n + 1) + (1 - h) where 0 < 1 - h < 1.
Thus [-x] = -n - 1 and we have [x] + [-x] = n + (-n - 1) = -1.

Condition 2) tells us that x is not an integer. Therefore [x] + [-x] = -1 and condition 2) yields a unique solution.

Condition 2) is sufficient.

Condition 1)
If x = 0 which is an integer, then we have [x] + [-x] = 0.
If x = 1.5 which is not an integer, then we have [x] + [-x] = -1.

Since condition 1) does not yield a unique solution, it is not sufficient.

_________________
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Joined: 16 Aug 2015
Posts: 8987
GMAT 1: 760 Q51 V42
GPA: 3.82
Re: Overview of GMAT Math Question Types and Patterns on the GMAT  [#permalink]

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[GMAT math practice question]

(Geometry) In the figure below, point I is the incenter of △ABC and line AH is perpendicular to BC. If ∠ABC = 80 and ∠ACB = 50 what is ∠x + ∠y?

Attachment: 2.4PS.png [ 17.52 KiB | Viewed 445 times ]

A. 55
B. 60
C. 65
D. 75
E. 80

=>

∠BAC = 180 – 80 – 50 = 50.
An incenter of a triangle is the intersection of lines bisecting all interior angles.
∠x = ∠CAH - ∠CAI = 40 – 25 = 15.
∠y = ∠CAI + ∠ACI = 25 + 25 = 50.
Thus ∠x + ∠y = 15 + 50 = 65.

_________________
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Joined: 16 Aug 2015
Posts: 8987
GMAT 1: 760 Q51 V42
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Re: Overview of GMAT Math Question Types and Patterns on the GMAT  [#permalink]

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[GMAT math practice question]

(Geometry) The figure shows parallelogram ABCD and point E on line AD. What is the ratio of △ABE : △DCE?

1) The area of ABCD is 50.
2) AE : ED = 3 : 2.

Attachment: 2.11DS.png [ 15.24 KiB | Viewed 437 times ]

=>

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.
Visit https://www.mathrevolution.com/gmat/lesson for details.

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

Since triangles ABE and DCE have the same height, we have △ABE : △DCE = AE : ED.
Thus, condition 2) is sufficient.

Condition 1)
Since we don’t know AE : ED, condition 1) is not sufficient.

_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8987
GMAT 1: 760 Q51 V42
GPA: 3.82
Re: Overview of GMAT Math Question Types and Patterns on the GMAT  [#permalink]

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[GMAT math practice question]

(Number Properties) For positive integers A and B, G is the greatest common divisor of A and B, and L is the least common multiple of A and B. What is A + B?

1) L = 70
2) G/A + G/B = 7/10

=>

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.
Visit https://www.mathrevolution.com/gmat/lesson for details.

Since we have 2 variables (A and B) and 0 equations, C is most likely 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)

Assume A = aG and B = bG where a and b are relatively prime numbers.

Since G/A + G/B = 7/10 from condition 2), we have
G/A + G/B = 7/10,
G/aG + G/bG = 7/10 (substituting in A = aG and B = bG),
bG/abG + aG/abG = 7/10 (getting a common denominator),
(aG + bG)/abG = 7/10 (adding the fractions),
(A + B)/L = 7/10 (substituting A = Ag, B = bG, and L = abG).
Since L = 70 from condition 1), we have (A + B)/L = 7/10, (A + B)/70 = 7/10 or A + B = 49.

Since both conditions together yield a unique solution, they 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)
If A = 70 and B = 1, then we have A + B = 71.
If A = 70 and B = 2, then we have A + B = 72.

Since condition 1) does not yield a unique solution, it is not sufficient.

Condition 2)
Assume A = aG and B = bG where a and b are relatively prime numbers.
G/A + G/B = 1/a + 1/b = 7/10.
If a = 2, b = 5, and G = 1, then we have A = aG = 2(1) = 2, B = bG = 5(1) = 5 and A + B = 2 + 5 = 7.
If a = 2, b = 5, and G = 2, then we have A = aG = 2(2) = 4, B = bG = 5(2) = 10 and A + B = 4 + 10 = 14.

Since condition 2) does not yield a unique solution, it 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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Posts: 8987
GMAT 1: 760 Q51 V42
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Re: Overview of GMAT Math Question Types and Patterns on the GMAT  [#permalink]

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[GMAT math practice question]

(Geometry) The figure below shows parallelogram ABCD, where line DH is perpendicular to line BC. P is a point inside parallelogram ABCD. If AD = 7, DH = 4 and the area of triangle PBC is 5, what is the area of triangle PAD?

Attachment: 2.11PS.png [ 7.03 KiB | Viewed 394 times ]

A. 8
B. 9
C. 10
D. 11
E. 12
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8987
GMAT 1: 760 Q51 V42
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Re: Overview of GMAT Math Question Types and Patterns on the GMAT  [#permalink]

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[GMAT math practice question]

(Geometry) The figure below shows parallelogram ABCD and point E is on line BC. Line DE bisects ∠D. Moreover, BE = DE and ∠A = 120°. What is the measure of ∠BDE?

A. 20°
B. 30°
C. 33°
D. 40°
E. 43°

Attachment: 2.13PS.png [ 9.96 KiB | Viewed 394 times ]

=>

Since BE = DE, we have ∠DBE = ∠BDE and ∠DEC = 2*(∠BDE) = 2*(∠EDC).
Since ∠EDC + ∠DEC + ∠C = 180°, ∠EDC + 2*(∠EDC) + 120° = 180°, 3*(∠EDC) + 120° = 180°, 3*(∠EDC) = 60°, ∠EDC = 20°.
Then we have ∠BDE = ∠EDC = 20°.

_________________
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Joined: 16 Aug 2015
Posts: 8987
GMAT 1: 760 Q51 V42
GPA: 3.82
Re: Overview of GMAT Math Question Types and Patterns on the GMAT  [#permalink]

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[GMAT math practice question]

(Equation) What is the value of abc?

1) a + 4/b = 1
2) b + 1/c = 4

=>

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.
Visit https://www.mathrevolution.com/gmat/lesson for details.

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

Since we have 3 variables (a, b, and c) and 0 equations, E is most likely 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.
The question --------- is equivalent to ----------- for the following reason

Conditions 1) & 2)

When we multiply both sides of the equation a + 4/b = 1 by b, we have ab + 4 = b.
When we multiply both sides of the equation b + 1/c = 4 by c, we have bc + 1 = 4c or bc = 4c – 1.
When we multiply both sides of the equation ab + 4 = b by c, we have abc + 4c = bc.
When we replace bc of the equation abc + 4c = bc by 4c – 1, we have abc + 4c = 4c – 1 or abc = -1.

Since both conditions together yield a unique solution, they are 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 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.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8987
GMAT 1: 760 Q51 V42
GPA: 3.82
Re: Overview of GMAT Math Question Types and Patterns on the GMAT  [#permalink]

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[GMAT math practice question]

(Algebra) We have a polynomial x^2 + 4x - n for an integer n between 1 and 100, inclusive. The polynomial x^2 + 4x - n is factored into (x + a)(x + b) where a and b are integers. How many different values of n satisfy the polynomial x^2 + 4x – n?

A. 5
B. 6
C. 7
D. 8
E. 9

=>

We can assume a < b without loss of generality.
Since we have a + b = 4 and ab = -n, we have the following pairs of (a, b).
(-1, 5), (-2, 6), (-3, 7), (-4, 8), (-5, 9), (-6, 10), (-7, 11), (-8, 12)
Then, the possible values of n are 5, 12, 21, 32, 45, 60, 77 and 96.
We have 8 possible values of n.

_________________
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Joined: 16 Aug 2015
Posts: 8987
GMAT 1: 760 Q51 V42
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Re: Overview of GMAT Math Question Types and Patterns on the GMAT  [#permalink]

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[GMAT math practice question]

(Geometry) The figure below shows rectangle ABCD with AB = 6 and AO = 5. What is the circumference of ABCD?

A. 26
B. 27
C. 28
D. 29
E. 30

Attachment: 2.21ps.png [ 7.71 KiB | Viewed 297 times ]

=>

We have AC = 2AO = 2*5 = 10.
Since we have AC^2 = AB^2 + BC^2, we have BC^2 = AC^2 – AB^2 = 10^2 – 6^2 = 100 – 36 = 64 or BC = 8.

Thus, the perimeter of ABCD is 2(6+8) = 2*14 = 28.

_________________
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Joined: 16 Aug 2015
Posts: 8987
GMAT 1: 760 Q51 V42
GPA: 3.82
Re: Overview of GMAT Math Question Types and Patterns on the GMAT  [#permalink]

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1
[GMAT math practice question]

(Number Properties) A, B, and C are positive numbers. We have an equation A^2 + 2B^2 = C^2. What is the value of A + B + C?

1) A, B, and C are less than 11.
2) A, B, and C are integers.

=>

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.
Visit https://www.mathrevolution.com/gmat/lesson for details.

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

Since A^2 + 2B^2 = C^2 , we have 2B^2 = C^2 - A^2 or 2B^2 = (C + A)(C - A).
C^2 - A^2 = (C + A)(C - A) is an even number, both A and C must either odd numbers or even numbers.
Then 2B^2 is a product of even numbers, and 2B^2 is a multiple of 4.
B is an even number.

Since we have 3 variables (A, B, and C) and 0 equations, E is most likely 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)
A, B, and C are positive integers less than or equal to 10.

Since we have 0 < C^2 - A^2 = 2B^2 < 10^2 or 0 < C^2 - A^2 = 2B^2 < 100, we have 0 < B < √50 and B = 2, 4 or 6.
We have C + A > C – A.

Case 1: B = 2
Since we have (C + A)(C - A) = 8, we have C + A = 4 and C – A = 2.
Then we have C = 3, A = 1 and A + B + C = 1 + 2 + 3 = 6.

Case 2: B = 4
Since we have (C + A)(C - A) = 32, we have C + A = 16, C - A = 2 or C + A = 8, C – A = 4.
If C + A = 16, C – A = 2, then we have A = 7, C = 9 and A + B + C = 7 + 4 + 9 = 20.

Since both conditions together do not yield a unique solution, they 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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Joined: 16 Aug 2015
Posts: 8987
GMAT 1: 760 Q51 V42
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Re: Overview of GMAT Math Question Types and Patterns on the GMAT  [#permalink]

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1
[GMAT math practice question]

(Statistics) By adding n into {3, 6, 9, 10}, the average is equal to the median. What is the sum of possible values of n?

A. 24
B. 26
C. 28
D. 30
E. 32

=>

The possible medians are 6, n, and 9 only.

Case 1: Assume n ≤ 6.
The median of n, 3, 6, 9, 10 is 6 and we have (n + 28) / 5 = 6 or n + 28 = 30.
Thus, we have n = 2.

Case 2: Assume 6 < n ≤ 9.
The median of 3, 6, n, 9, 10 is n and we have (n + 28) / 5 = n or n + 28 = 5n.
Thus, we have 4n = 28 or n = 7.

Case 3: Assume 9 < n.
The median of 3, 6, 9, n, 10 is 9 and we have (n + 28) / 5 = 9 or n + 28 = 45.
Thus, we have n = 17.

Hence, the possible values of n are 2, 7, and 17.
Then we have 2 + 7 + 17 = 26.

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Re: Overview of GMAT Math Question Types and Patterns on the GMAT  [#permalink]

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1
[GMAT math practice question]

(Algebra) What is the value of x?

1) x^2 + 4x + 9 is a perfect square of an integer.
2) x is an integer.

=>

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.
Visit https://www.mathrevolution.com/gmat/lesson for details.

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)

x^2 + 4x + 9 = x^2 + 4x + 4 + 5 = (x + 2)^2 + 5 = k^2 for some integer k.

Then we have
5 = k^2 - (x+2)^2 (subtracting (x + 2)^2 from both sides)
5 = (k + x + 2)(k – x – 2) (factoring using difference of squares).

If k + x + 2 = 5 and k – x – 2 = 1, then we have 2x + 4 = ( k + x + 2 ) – ( k – x – 2 ) = 5 -1 = 4 and x = 0.

If k + x + 2 = 1 and k – x – 2 = 5, then we have 2x + 4 = ( k + x + 2 ) – ( k – x – 2 ) = 1 - 5 = -4 and x = -4.

Since condition 1) does not yield a unique solution, it is not sufficient.

Condition 2)

Since condition 2) does not yield a unique solution; obviously, it is not sufficient.

Conditions 1) & 2)

The reasoning in condition 1) can be applied to both conditions together too.

Since both conditions together do not yield a unique solution, they are not 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.
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Re: Overview of GMAT Math Question Types and Patterns on the GMAT  [#permalink]

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Quote:
What is the value of x?

1) x^2 + 4x + 9 is a perfect square of an integer.
2) x is an integer.

St1) x^2 + 4x + 9 is a perfect square of an integer.
or, (x+2)^2 +5 is a perfect square of an integer
x can be equal to 0, -4, \sqrt{21}-2
NOT Sufficient

St2) x is an integer, hence it can be equal to -1, 2, 3, 4, 7 any integer.
NOT Sufficient

Even after combining St 1 & 2, x can be 0 or -4
NOT Sufficient

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[GMAT math practice question]

(Algebra) What is (a-b)^2?

1) |a| = 4, |b| = 3
2) b/a < 0

=>

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Since we have 2 variables (p and q) and 0 equations, C is most likely 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)

Since |a| = 4, and |b| = 3, we have a = ±4, and b = ±3 from condition 1) and ab < 0 from condition 2).
Then we have a^2 = 16, b^2 = 9 and ab = -12.
Thus (a - b)^2 = a^2 - 2ab + b^2 = 16 – 2(-12) + 9 = 49.

Since both conditions together yield a unique solution, they are 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: Overview of GMAT Math Question Types and Patterns on the GMAT  [#permalink]

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[GMAT math practice question]

(algebra ) If x = 1/2+√3, what is x^2 - 4x?

A. -2
B. -1
C. 0
D. 1
E. 2

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We have x = 1/2+√3
x = 2-√3/(2+√3)(2-√3) (multiplying by the conjugate)
x = 2-√3/4-3 (foiling the denominator)
x = 2-√3 (simplifying)
x – 2 = -√3 (subtracting 2 from both sides)
(x - 2)^2 = (-√3)^2 (squaring both sides)
x^2 – 4x + 4 = 3 (simplifying)
x^2 – 4x = -1 (subtracting 4 from both sides)

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# Overview of GMAT Math Question Types and Patterns on the GMAT  