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[GMAT math practice question]

(inequality) Which one of (a + 2b)^2 and 9ab is greater?

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

(a + 2b)^2 - 9ab = a^2 + 4ab + 4b^2 - 9ab = a^2 – 5ab + 4b^2 = (a - b)(a - 4b)
If (a - b)(a - 4b) > 0, then (a + 2b)^2 is greater than 9ab.
If (a - b)(a - 4b) < 0, then 9ab is greater than (a + 2b)^2.

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)
Since b < 1 < a, we have b < a or a – b > 0.
Since a < 2 < 4b, we have a < 4b or a – 4b < 0.
Then we have (a - b)(a - 4b) < 0 and (a + 2b)2 is greater than 9ab.

Since both conditions together yield a unique solution, they are sufficient.

Therefore, C is the answer.
Answer: C


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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[GMAT math practice question]

(Geometry) In rectangle ABCD, as shown in the figure, the length of line AB was increased by 10% to make line AB’ and line AD was decreased by 10% to make line AD’. What percentage is the reduced area of rectangle ABCD of the original area?

A. 1%
B. 5%
C. 10%
D. 15%
E. 17%

Attachment:
12.11PS.png
12.11PS.png [ 19.41 KiB | Viewed 2493 times ]

=>


Assume the width and the length of the rectangle are a and b.
Since a is increased by 10%, a becomes 1.1a.
Since b is decreased by 10%, b becomes 0.9b.

The area of the rectangle ab becomes (1.1)(0.9)ab = 0.99ab.
So, we have (0.99ab – ab) / ab = -0.01, meaning the area decreased by 1%.

Therefore, A is the answer.
Answer: A
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[GMAT math practice question]

(algebra) What is the value of xy? (max(x, y) denotes the maximum between x and y, and min(x, y) denotes the minimum between x and y)

1) x + y = 7 and x - y = 1.
2) max(x, y) = 2x + 3y - 13 and min(x, y) = 3x - y - 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.
Visit https://www.mathrevolution.com/gmat/lesson for details.

Since we have 2 variables (x and y) and each condition has 2 equations, D is most likely the answer. So, we should consider each condition on its own first.

Condition 1)
We have x + y = 7 and x – y = 1.
Adding the 2 equations together gives us (x + y) + (x – y) = 7 + 1, 2x = 8, or x = 4.
When we substitute x with 4 in the first equation, we have 4 + y = 7 or y = 3.
Therefore, we have xy = 4*3 = 12.

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

Condition 2)
Case 1: x ≥ y
Then we have 2x + 3y - 13 = x and 3x - y - 6 = y.
Then we have x + 3y - 13 = 0 and 3x - 2y - 6 = 0.
When we subtract three times the first equation from the second equation, we have (3x - 2y - 6) - 3(x + 3y - 13) = 3x - 2y - 6 - 3x - 9y + 39 = -11y + 33 = 0 or y = 3.
When we substitute y with 3 in the first equation, we have x + 3*3 - 13 = 0 or x = 4.
Then, we have xy = 4*3=12.


Case 2: x < y

Then we have 2x + 3y - 13 = y and 3x - y - 6 = x.
Then we have 2x + 2y - 13 = 0 and 2x - y - 6 = 0.
When we subtract the second equation from the first equation, we have (2x - y - 6) - (2x + 2y - 13) = 2x - y - 6 - 2x - 2y + 13 = -3y + 7 = 0 or y = 7/3.
When we substitute y with 7/3 in the second equation, we have 2x - 7/3 - 6 = 2x - 7/3 - 18/3 = 2x – 25/3 = 0, 2x = 25/3, or x = 25/6.
However, we have x > y in this case so we don’t have a solution in this case.

Therefore, we have a unique solution of x = 4 and y = 3.

Since condition 2) yields a unique solution, it is sufficient.

Therefore, D is the answer.
Answer: D

This question is a CMT 4(B) question: condition 1) is easy to work with, and condition 2) is difficult to work with. For CMT 4(B) questions, D is most likely the answer.

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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[GMAT math practice question]

(algebra) A sea food shop sells mackerel. A pack of fifty mackerel, a pack of thirty mackerel and a pack of ten mackerel is sold for $97, $67 and $25, respectively. The total number of mackerel in those packs is 1000. The total price of those packs of mackerel is $2000. What is the number of packs of ten mackerel?

A. 4
B. 6
C. 8
D. 10
E. 12

=>

Assume x, y and z are the numbers of packs with fifty mackerel, thirty mackerel and ten mackerel, respectively.
We have 50x + 30y + 10z = 1,000 and 97x + 67y + 25z = 2,000.
Then we have 5x + 3y + z = 100 (dividing everything by 10) and 97x + 67y + 25z = 2,000.

When we subtract the second equation from 25 times the first equation, we have 25(5x + 3y + z) - (97x + 67y + 25z) = 25(100) - 2000, 125x + 75y + 25z – 97x - 67y - 25z) = 2500 - 2000, 28x + 8y = 500 or 7x + 2y = 125.
Since 2y is an even number and 125 is an odd number, x must be an odd number.

Then we can put x = 2n + 1 for a positive integer n.
Rearranging the equation 7x + 2y = 125, gives us y = (125 – 7x) / 2, y = (125 – 7(2n + 1)), y = (125 - 14n - 7) / 2 = (-14n + 118) / 2 = 59 – 7n.
When we substitute x and y with 2n – 1 and 59 – 7n in the first equation, we have 5(2n + 1) + 3(59 – 7n) + z = 100, 10n + 5 + 177 - 21n + z = 100, 182 – 11n + z = 100 or z = 11n – 82.
Since x ≥ 0 and n is an integer, we have 2n – 1 ≥ 0 or n ≥ 1.
Since y = 59 – 7n ≥ 0, we have -7n ≥ -59, n ≤ 8 (the sign changes direction because we divide by a negative number).
Since z = 11n - 82 ≥ 0, we have 11n ≥ 82, n ≥ 8.
Then we have n = 8 and x = 17, y = 3 and z = 6.

Therefore B is the answer.
Answer: B
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[GMAT math practice question]

(Inequalities) What is the minimum value of a?

1) The solution of 3 - 3x ≥ 2x - 7 and x + 3 > a is ø.
2) a 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.

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

Condition 1)
3 - 3x ≥ 2x - 7
=> 10 ≥ 5x
=> x ≤ 2

X + 3 > a
=> x > a – 3

Looking at the intersection of x ≤ 2 and x > a - 3, we have a - 3 ≥ 2 or a ≥ 5.
Thus, the minimum value of a is 5.

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

Condition 2)
Condition 2) tells us that a is an integer, which gives us an infinite number of possibilities.

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

Therefore, A is the answer.
Answer: A

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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[GMAT math practice question]

(Number Properties) a, b and c are positive integers with a ≤ b ≤ c. What is the value of a + b + c?

1) abc = a + b + c.
2) abc = 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.
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.

Conditions 1) & 2)
Since a, b, and c are positive integers and we have abc = 6 with a ≤ b ≤ c, we have a unique solution of a = 1, b = 2, c = 3.
Therefore, we have a + b + c = 6.

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 2)
If a = 1, b = 2 and c = 3, then we have a + b + c = 1 + 2 + 3 = 6.
If a = 1, b = 1 and c = 6, then we have a + b + c = 1 + 1 + 6 = 8.

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

Condition 1)
Since abc = a + b + c and 1 ≤ a ≤ b ≤ c, we have c ≤ abc = a + b + c ≤ 3c or c ≤ abc ≤ 3c.
Therefore, we have 1 ≤ ab ≤ 3.
If ab = 1, then we have a = b = 1 and c = 2 + c, which doesn’t make sense.
If ab = 2, then we have a = 1, b = 2 and 2c = c + 3 or c = 3, which tells us that a + b + c = 6.
If ab = 3, then we have a = 1, b = 3 and 3c = 4 + c or c = 2, which doesn’t make sense since c < b.
Therefore, a = 1, b = 2 and c = 3 is the unique case and we have a + b + c = 6.

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

Therefore, A is the answer.
Answer: A

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.

This question is a CMT 4(B) question: We easily figured out condition 2) is not sufficient, and condition 1) is difficult to work with. For CMT 4(B) questions, we assume condition 1) is sufficient. Then A is most likely an answer.
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[GMAT math practice question]

(Number Properties) A teacher distributes n apples to some students. If she gives 4 apples to each student, 7 apples remain. If she tried to give 5 apples to each student, 3 students would not get anything. What is the range of possible values of n?

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

=>

Assume s is the number of students.
We have n = 4s + 7 and 5(s - 4) < n ≤ 5(s - 3).
Then we have 5s – 20 < 4s + 7 ≤ 5s – 15.
5s – 20 < 4s + 7 is equivalent to s < 27 and 4s + 7 ≤ 5s – 15 is equivalent to 22 ≤ s.
Therefore we have 22 ≤ s < 27 or 22 ≤ s ≤ 26.
We have 88 ≤ 4s ≤ 104 and 95 ≤ 4s + 7 ≤ 111.
The range is 111 – 95 = 16.

Therefore, B is the answer.
Answer: B
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[GMAT math practice question]

(Geometry) There is a point P(a, b). What is the value of a+b?

1) P is on the line x/3 + y/4 = 1.
2) The line x/3 + y/4 = 1 is parallel to the line (a/3)x + (b/4)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.
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)

We have a/3 + b/4 = 1 since the point P(a, b) is on the line x/3 + y/4 = 1.
We have a = b since x/3 + y/4 = 1 is parallel to the line (a/3)x + (b/4)y = 1 or 1/3 : a/3 = 1/4 : b/4.
Then, we have a/3 + a/4 = 1, 4a/12 + 3a/12 = 1, (7/12)a = 1 or a = 12/7.
Thus we have a + b = 12/7 + 12/7 = 24/7.

Since both conditions together yield a unique solution, they are sufficient.

Therefore, C is the answer.
Answer: C

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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[GMAT math practice question]

(Geometry) As the figure below shows, line l is perpendicular to BD and CE. What is the length of DE?

1) △ABC is a right isosceles triangle with ∠BAC = 90
2) BD = 3, and CE = 4

Attachment:
1.2ds.png
1.2ds.png [ 8.15 KiB | Viewed 2341 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 9 variables from 3 triangles and 2 equations from two right triangles, 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)

Since we have AB = AC from condition 1) and ∠DBA = ∠EAC, the triangles ADB and CEA are congruent.
Thus, DE = DA + AE = CE + BD = 3 + 4 = 7.

Since both conditions together yield a unique solution, they are sufficient.

Therefore, C is the answer.
Answer: C

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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[GMAT math practice question]

(Number Properties) Annie multiplied the date of her birthday by 5, then subtracted 4 from that. She multiplied it by 2 after that and added the month of her birthday again. She got the result of 232. When is her birthday?

A. Aug. 20th
B. Aug. 23rd
C. Sep. 21st
D. Sep. 22nd
E. Oct. 23rd

=>

Assume m and d are the month and date of Annie’s birthday.
Then we have 2(5d - 4) + m = 232 or m = 240 – 10d = 10(24-d).
Since m is a multiple of 10, we have m = 10. (Note that m cannot be 20, 30, etc. because there are only 12 months in a year.)
Then we have 10d = 240 – m, 10d = 240 – 10, 10d = 230 or d = 23.

Thus, Annie’s birthday is Oct. 23rd.

Therefore, E is the answer.
Answer: E
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[GMAT math practice question]

(Graphs) A line passes through two points (a, -2), and (2a-6, 2). This line is parallel to the y-axis. What is a?

A. 2
B. 4
C. 6
D. 8
E. 10

=>

The x-coordinate of all points on a vertical line must be equal.
Then we have a = 2a – 6 or a = 6.

Therefore, C is the answer.
Answer: C
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[GMAT math practice question]

(Geometry) The figure shows a parallelogram. What is the measure of ∠APC?

1) ∠B : ∠C = 2 : 3
2) ∠BAP = ∠DAP

Attachment:
1.7ds.png
1.7ds.png [ 11.78 KiB | Viewed 2302 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 a parallelogram, we have 3 variables, 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)

∠B + ∠C = 180° and ∠C = ∠BAD = ∠BAP + ∠DAP.
Since ∠B : ∠C = 2 : 3 and ∠B + ∠C = 180° from condition 1), we have ∠B = 72° and ∠C = 108°, because:
∠B + ∠C = 180° can be rewritten as ∠B = 180° - ∠C
Substitute into ∠B : ∠C = 2 : 3
180 - ∠C : ∠C = 2 : 3
3(180° - ∠C) = 2C (by cross multiplying)
540° - 3∠C = 2∠C
540° = 5∠C (adding 3∠C to both sides)
∠C = 108° (dividing both sides by 5)
∠B = 180° - ∠C, B = 180° - 108°, or ∠B = 72°

Since ∠C = 108° = ∠BAP + ∠DAP and ∠BAP = ∠DAP, we have ∠BAP = 54°.
Thus the exterior angle ∠APC of the triangle ABP is the sum of ∠ABP and ∠BAP and we have ∠APC = ∠ABP + ∠BAP = 72° + 54° = 126°.

The answer is unique, and the conditions combined are sufficient.

Therefore, C is the answer.
Answer: C
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[GMAT math practice question]

(Inequalities) What is the value of y? ([x] denotes the greatest integer less than or equal to x.)

1) y = 2[x] + 3
2) y = 3[x - 2] + 5

=>

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 (x and y) 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 we have n = [x].
Since we have y = 2n + 3 and y = 3(n - 2) + 5 since we have [x - 2] = [x] - 2.
Then, substituting the first equation into the second equation we have
2n + 3 = 3(n – 2) + 5
2n + 3 = 3n - 6 + 5
2n +3 = 3n - 1
-n = -4
n = 4

Then:
y = 2n + 3
y = 2(4) + 3
y = 11.

Since both conditions together yield a unique solution, they are sufficient.

Since this question is an integer 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)
x = 1, y = 5 and x = 2, y = 7 are solutions.
Since condition 1) does not yield a unique solution, it is not sufficient.


Condition 2)
x = 1, y = 2 and x = 2, y = 5 are solutions.
Since condition 2) does not yield a unique solution, it is not sufficient.

Therefore, C is the answer.
Answer: C

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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[GMAT math practice question]

(Function) The function is defined as follows: f(x+1) = 3x + 2. What is f(1) + f(2) + f(3) +…+ f(9) + f(10)?

A. 149
B. 153
C. 155
D. 157
E. 159

=>

Since f(x+1) = 3x + 2, we have f(x) = f((x-1) + 1) = 3(x - 1) + 2 = 3x - 3 + 2 = 3x - 1.
f(1) + f(2) + f(3) + … + f(9) + f(10)
= ( 3*1 – 1 ) + ( 3*2 – 1 ) + … + ( 3*10 – 1 )
= 3(1 + 2 + … + 10) – (1 + 1 +…+ 1)
= 3*55 – 10 = 155.

Therefore, C is the answer.
Answer: C
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[GMAT math practice question]

(Coordinate Geometry) A quadrilateral P has (1, 1), (3, 1), (3, 5) and (1, 5) as 4 vertices and another quadrilateral Q has (-1, -1), (-5, -1), (-5, -5) and (-1, -5) as 4 vertices. A line divides these two quadrilaterals evenly at the same time. What is this line?

A. y = - 1/6x + 5/6
B. y = - 5/6x + 1/6
C. y = 6/5x + 3/5
D. y = 1/3x + 5/6
E. y = - 1/6x + 7/5

=>

Attachment:
1.9ps(a).png
1.9ps(a).png [ 29.51 KiB | Viewed 2255 times ]

The quadrilaterals are rectangles, and bisecting lines of rectangles pass through the center of the rectangles.
Thus we have to find the line passing through the centers of those two rectangles.
The centers of the rectangles are (2, 3) and (-3, -3).
The slope of the line passing through (2, 3) and (-3, -3) is (3- (-3)) / (2 - (-3)) = 6/5.
The line passing through them is y – 3 = (6/5)(x - 2) or y = (6/5)x + (3/5).
Therefore, C is the answer.
Answer: C
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[GMAT math practice question]

(Inequality) Which is greater between (a + 2b)2 and 9ab?

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

The question is equivalent to the statement (a - b)(a - 4b) is greater than or less than 0 for the following reason:

(a + 2b)^2 - 9ab > 0
=> a^2 + 4ab + 4b^2 – 9ab > 0
=> a^2 - 5ab + 4b^2 > 0
=> (a - b)(a - 4b) > 0

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)
Since 1 < a < 2 and 1/2 < b < 1, we have 1/2 < b < 1 < a < 2 or b < a.
Since 1 < a < 2 and 2 < 4b < 4 (by multiplying the equation given in condition 2) by 4), we have 1 < a < 2 < 4b < 4 or a < 4b.

Then we have a – b > 0, and a – 4b < 0 or (a - b)(a - 4b) < 0.
Thus, we have (a + 2b)^2 - 9ab > 0 and (a + 2b)^2 is greater than 9ab.

Since both conditions together yield a unique solution, they are sufficient.

Therefore, C is the answer.
Answer: C

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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[GMAT math practice question]

(Algebra) Thomas went to a grocery store and bought some fruit with $25. The prices of the fruit are $5, $1, and $0.50 for each watermelon, pear, and apple, respectively. How many apples did he buy?

1) He bought 10 pieces of fruit, and he didn’t get any change back.
2) He had at least one of each fruit.

=>

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.

Assume x, y and z are the number of watermelons, pears, and apples, respectively.
Then we have 5x + y + 0.5z = 25 or 10x + 2y + z = 50.

Since we have 3 variables (x, y, and z) and 1 equation, 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)
We have x + y + z = 10 and 5x + y + 0.5z = 25 or 10x + 2y + z = 50. Subtracting the first equation from the second equation gives us (10x + 2y + z) – (x + y + z) = 50 - 10, or 9x + y = 40.
Then y = 40 – 9x.
Then the possible values of (x, y) are (1, 31), (2, 22), (3, 13), and (4, 4).
If x = 1, and y = 31, then z = 10 – x – y = -22 doesn’t make sense, since z must be positive.
If x = 2, and y = 22, then z = 10 – x – y = -14 doesn’t make sense, since z must be positive.
If x = 3, and y = 13, then z = 10 – x – y = -6 doesn’t make sense, since z must be positive.
If x = 4, and y = 4, then we have z = 10 – x – y = 2.

Since both conditions together yield a unique solution, they are sufficient.

Therefore, C is the answer.
Answer: C

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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