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MathRevolution

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01 May 2020, 03:52

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[GMAT math practice question]

(Algebra) \(a, b\) and \(X\) are positive numbers where \(a > b\) and \(n\) is a positive integer. What is the value of \((\sqrt{X}-\sqrt{X-1})^{\frac{1}{n}}\)?

1) \(X=(\frac{a^n+b^n}{2})^2.\)

2) \(ab = 1.\)

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Updated on: 21 Aug 2021, 02:32

Originally posted by MathRevolution on 03 May 2020, 20:33.
Last edited by MathRevolution on 21 Aug 2021, 02:32, edited 1 time in total.

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[GMAT math practice question]

(Algebra) \(abc ≠ 0\). What is the value of \(a^2+b^2+c^2\)?

1) \(a+b+c=3.\)

2) \(a^3+b^3+c^3=27.\)

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Attachment:

5.1DS(A).png [ 24.73 KiB | Viewed 1708 times ]

Since there many possible values of b, both conditions together do not yield a unique solution, and they are not sufficient.

Therefore, E is the answer.
Answer: E

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]

(Algebra) What is the value of \(2x + y\)?

1) \(\frac{(x - y)(y + 3)}{4(x - y)^2 +(y + 3)^2}= - \frac{1}{4}\)

2) \(x – y = -1\)

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05 May 2020, 19:27

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[GMAT math practice question]

(Number Properties) \(a, b\) and \(c\) are positive integers. Is \(2(a^4+b^4+c^4)\) a perfect square?

1) \(a + b + c = 0\)

2) \(abc = 6\)

MathRevolution

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Updated on: 24 Mar 2021, 04:02

Originally posted by MathRevolution on 06 May 2020, 01:18.
Last edited by MathRevolution on 24 Mar 2021, 04:02, edited 1 time in total.

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[GMAT math practice question]

(Algebra) What is the value of \(2x + y\)?

1) \(\frac{(x - y)(y + 3)}{4(x - y)^2 +(y + 3)^2}= - \frac{1}{4}\)

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.
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 \(2\) 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.

Assume \(a = x – y\) and \(b = y + 3.\)

When we try to check both conditions together, we realize condition 1) alone is sufficient because condition 1) tells us \(2x – y = -3\) for the following reason.

\(\frac{(x-y)(y+3)}{4(x-y)^2+(y+3)^2}= - \frac{1}{4}\)

⇔ \(\frac{ab}{4a^2+b^2}=-\frac{1}{4}\) (\(a = (x – y)\) and \(b = (y + 3)\))

⇔ \(4a^2+b^2=-4ab\) (cross multiplying)

⇔ \(4a^2+4ab+b^2=0\) (adding \(4ab\) to both sides)

⇔\((2a+b)^2=0\) (factoring)

⇔\(2a+b=0\)

⇔ \(2(x - y) + (y + 3) = 0\) (substituting (\(x – y\)) and (\(y + 3\)) back in)

⇔ \(2x – 2y + y + 3 = 0\) (multiplying \(2\) through the first bracket)

⇔ \(2x – y + 3 = 0\) (adding like terms)

Thus, we have \(2x – y = -3\) from condition 1) alone.

Condition 2)
Condition 2) is not sufficient, obviously, since condition 2) does not yield a unique solution.

Therefore, A is the answer.
Answer: A

This question is an application of CMT 4(B): condition 2) is easy to work with, and condition 2) is difficult to work with. For CMT 4(B) questions, we may assume condition 1) is sufficient. Since we can figure out condition 2) is not sufficient, we should be able to choose A.

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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06 May 2020, 01:19

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[GMAT math practice question]

(Number Properties) Is \(\sqrt{A}\) an irrational number?

1) \(A = (n-1)^2+n^2+(n+1)^2\)

2) \(A\) is an integer, and the units digit of \(A\) is \(2\).

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Updated on: 24 Mar 2021, 04:02

Originally posted by MathRevolution on 07 May 2020, 00:56.
Last edited by MathRevolution on 24 Mar 2021, 04:02, edited 1 time in total.

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[GMAT math practice question]

(Number Properties) \(a, b\) and \(c\) are positive integers. Is \(2(a^4+b^4+c^4)\) a perfect square?

1) \(a + b + c = 0\)

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.

Condition 2)
Condition 2) tells us that \(abc = 6\). For three numbers to multiply to \(6\), the numbers must be \(1, 2,\) and \(3\) and the order does not matter. We can substitute these numbers into the given equation to get:
\(2(a^4 + b^4 + c^4) = 2(1^4 + 2^4 + 3^4) = 2(1 + 16 + 81) = 2*98 = 196 = 14^2,\) which is a perfect square.
Thus condition 2) is sufficient.

Condition 1)
\((a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)\)

Then, we have \(a^2 + b^2 + c^2 = -2(ab + bc + ca)\) since \(a + b + c = 0.\)

We have \((a^2 + b^2 + c^2)^2 = (-2(ab + bc + ca))^2\) when we square both sides.

Its left-hand side is
\((a^2 + b^2 + c^2)^2 \)

\(= (a^2 + b^2 + c^2) (a^2 + b^2 + c^2)\)

\(= a^4 + a^2b^2 +a^2c^2 + a^2b^2 + b^4 + b^2c^2 + a^2c^2 + b^2c^2 + c^4\)

\(= a^4 + b^4 + c^4 + 2a^2b^2 + 2b^2c^2 + 2c^2a^2.\)

Its right-hand side is
\((-2(ab + bc + -ca))^2 \)

\(= 4(ab + bc + ca)(ab + bc + ca)\)

\(= 4(a^2b^2+ ab^2c + a^2bc + ab^2c + b^2c^2 + abc^2 + a^2bc + abc^2 + c^2a^2 \)

\(= 4(a^2b^2 + b^2c^2 + c^2a^2 + 2a^2bc + 2ab^2c + 2abc^2)\)

\(= 4(a^2b^2 + b^2c^2 + c^2a^2) + 2abc(a + b + c))\)

\(= 4(a^2b^2 + b^2c^2 + c^2a^2),\) since \(a + b + c = 0\)

We have \(a^4 + b^4 + c^4 = 2(a^2b^2 + b^2c^2 + c^2a^2).\)

Then \(2(a^4 + b^4 + c^4) = 4(a^2b^2 + b^2c^2 + c^2a^2) = (-2(ab + bc + ca))^2 \)

Thus \(2(a^4 + b^4 + c^4)\) is a square of \(-2(ab + bc + ca).\)

We realize we have only used condition 1), but haven’t used condition 2).
It means condition 1) alone is sufficient.
Thus, condition 1) is sufficient.

Therefore, D is the answer.
Answer: D

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

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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07 May 2020, 00:57

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[GMAT math practice question]

(Geometry) We have a rectangle \(ABCD\) with \(AB = 6\) and \(BC = 10\) as below. What is the length of \(PQ\)?

Attachment:

5.7ds.png [ 10.58 KiB | Viewed 1607 times ]

1) \(AD=AP\)

2) \(∠DAQ = ∠PAQ\)

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Updated on: 18 Jan 2021, 03:41

Originally posted by MathRevolution on 08 May 2020, 01:21.
Last edited by MathRevolution on 18 Jan 2021, 03:41, edited 1 time in total.

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[GMAT math practice question]

(Number Properties) Is \(\sqrt{A}\) an irrational number?

1) \(A = (n-1)^2+n^2+(n+1)^2\)

2) \(A\) is an integer, and the units digit of \(A\) is \(2\).

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.

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)
\(A = (n-1)^2+n^2+(n+1)^2\)

\(= (n – 1)(n – 1) + n^2 + (n + 1)(n + 1)\)

\(= n^2 – n – n + 1 + n^2 + n^2 + n + n + 1\)

\(= 3n^2 + 2\)

Squares of integers do not have the remainder 2 when it is divided by 3, for the following reason:
Case 1) \(n = 3k\)

\(n^2 = (3k)^2 = 9k^2 = 3(3k^2) + 0\)

Case 2) \(n = 3k+1\)

\(n^2 = (3k+1)^2 = 9k^2 + 9k + 1 = 3(3k^2+3k) + 1\)

Case 3) \(n = 3k+2\)

\(n^2 = (3k+2)^2 = 9k^2 + 12k + 4 = 3(3k^2+4k+1) + 1\)

Thus, A is not a square, A is an irrational number, and the answer is ‘no’.

Since 'no' is also a unique answer, according to CMT (Common Mistake Type) 1, condition 1) is sufficient.

Condition 2)
\(0^2 = 0, 1^2 = 1, 2^2 = 4, 3^2 = 9, 4^2 = 16, 5^2=25, 6^2 = 36, 7^2 = 49, 8^2 = 64, 9^2 = 81.\)

Thus, squares do not have a units digit 2 and the answer is ‘no’.

Since 'no' is also a unique answer, according to CMT (Common Mistake Type) 1, condition 2) is sufficient.

Therefore, D is the answer.
Answer: D

This question is a CMT 4(B) question: condition 2) is easy to work with, and condition 1) 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.

MathRevolution

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08 May 2020, 01:21

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[GMAT math practice question]

(Number Properties) \(x\) is an integer. What is the value of \(x\)?

1) \(x^2+4x+9\) is a perfect square.

2) \(x\) is a non-zero integer.

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10 May 2020, 18:55

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[GMAT math practice question]

(Geometry) We have a rectangle \(ABCD\) with \(AB = 6\) and \(BC = 10\) as below. What is the length of \(PQ\)?

Attachment:

5.7ds.png

1) \(AD=AP\)

2) \(∠DAQ = ∠PAQ\)

MathRevolution

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Updated on: 18 Jan 2021, 03:43

Originally posted by MathRevolution on 10 May 2020, 18:58.
Last edited by MathRevolution on 18 Jan 2021, 03:43, edited 1 time in total.

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[GMAT math practice question]

(Number Properties) \(x\) is an integer. What is the value of \(x\)?

1) \(x^2+4x+9\) is a perfect square.

2) \(x\) is a non-zero 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 the answer. So, we should consider each condition on its own first.

Condition 1)
We have \(x^2 + 4x + 9 = k^2\) for some integer \(k.\)

Then we have
\(k^2 – (x^2 + 4x + 4) = 5\)

\(k^2 – (x + 2)^2 = 5\)

or \((k + x + 2)(k – x - 2) = 5.\)

We have four cases
Case 1) \(k + x + 2 = 1, k – x – 2 = 5\)

Adding both equations together gives us:
\(k + x + 2 + k – x – 2 = 1 + 5\)

\(2k = 6\)

\(k = 3\)

Then \(k + x + 2 = 1\) becomes

\(3 + x + 2 = 1\)

\(x = -4\)

Then we have \(k = 3, x = -4.\)

Case 2) \(k + x + 2 = 5, k – x – 2 = 1\)

Adding both equations together gives us:
\(k + x + 2 + k – x – 2 = 5 + 1\)

\(2k = 6\)

\(k = 3\)

Then \(k + x + 2 = 5 \)becomes

\(3 + x + 2 = 5\)

\(x = 0\)

Then we have \(k = 3, x = 0.\)

Case 3) \(k + x + 2 = -1, k – x – 2 = -5\)

Adding both equations together gives us:
\(k + x + 2 + k – x – 2 = -1 + -5\)

\(2k = -6\)

\(k = -3\)

Then \(k + x + 2 = -1\) becomes

\(-3 + x + 2 = -1\)

\(x = 0\)

Then we have \(k = -3, x = 0.\)

Case 4) \(k + x + 2 = -5, k – x – 2 = -1\)

Adding both equations together gives us:
\(k + x + 2 + k - x - 2 = -5 + -1\)

\(2k = -6\)

\(k = -3\)

Then \(k + x + 2 = -5\) becomes

\(-3 + x + 2 = -5\)

\(x = -4\)

Then we have \(k = -3, x = -4.\)

Thus, we have two solutions for \(x\), which are \(0\) and \(-4\).

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

Condition 2)

Since condition 2) does not provide enough information to yield a unique solution, it is not sufficient.

Conditions 1) & 2)
We have a unique solution \(-4.\)
Since both conditions together yield a unique solution, they are sufficient.

Therefore, C is the answer.
Answer: C

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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11 May 2020, 19:51

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[GMAT math practice question]

(Equation) What is the value of \(4a + 2b\)?

1) \((a - b)^2 - 3(a - b) – 18 = 0\)

2) \(a + b = 8\)

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12 May 2020, 04:35

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[GMAT math practice question]

(Geometry) The figure shows a semi-circle. \(AC\) and \(BD\) are perpendicular to \(AB\) and \(CD\) is a tangent line to the semi-circle. What is the radius of the semi-circle?

Attachment:

5.12DS.png [ 7.38 KiB | Viewed 1475 times ]

1) \(AC = 10\)

2) \(BD = 5\)

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Updated on: 20 Jan 2021, 04:30

Originally posted by MathRevolution on 13 May 2020, 03:28.
Last edited by MathRevolution on 20 Jan 2021, 04:30, edited 1 time in total.

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[GMAT math practice question]

(Equation) What is the value of \(4a + 2b\)?

1) \((a - b)^2 - 3(a - b) – 18 = 0\)

2) \(a + b = 8\)

=>

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)
From condition 1) we get:
\((a - b)^2 - 3(a - b) – 18 = 0\)

⇔ Let \(x = (a + b)\) giving us:

⇔ \(x^2 – 3x - 18 = 0\)

⇔ \((x + 3)(x – 6)\) (trinomial factoring)

⇔ \(((a-b) + 3)((a - b) - 6) = 0\) (substituting \((a + b)\) back in)

⇔ \(a - b = -3\) or \(a - b = 6\)

When we consider \(a - b = -3\) from condition 1) and \(a + b = 8\) from condition 2) and add them together we get \(2a = 5\), and \(a = \frac{5}{2}\). Substituting that back into the second equation gives us \(\frac{5}{2} + b = 8\), and \(b = \frac{11}{2}\). Then \(4a + 2b = 4(\frac{5}{2}) + 2(\frac{11}{20}) = 10 + 11 = 21.\)

When we consider \(a - b = 6\) from condition 1) and \(a + b = 8\) from condition 2) and add them together we get \(2a = 14\), and \(a = 7\). Substituting that back into the second equation gives us \(7 + b = 8\), and \(b = 1.\) Then \(4a + 2b = 4(7) + 2(1) = 28 + 2 = 30.\)

Since both conditions together do not yield a unique solution, they are not sufficient.

Therefore, E is the answer.
Answer: E

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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13 May 2020, 03:29

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[GMAT math practice question]

(Algebra) \(p, q, r,\) and \(s\) are integers. \(s\) and \(t\) are two roots of \(x^2 + px + q = 0\) where \(q\) is a prime number. What the value of \(p + q\)?

1) \(s\) and \(t\) are consecutive integers.

2) \(p\) is positive.

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MathRevolution

Attachment:

The attachment 5.12DS.png is no longer available

1) \(AC = 10\)

2) \(BD = 5\)

Attachment:

5.12ds(a).png [ 24.23 KiB | Viewed 1408 times ]

Conditions 1) & 2)

When we have \(DH\) perpendicular to \(AC\), we have a right triangle \(CDH\).

\(AB^2 = DH^2 = CD^2 – CH^2 = 15^2 – 5^2 = 200.\)

Thus, the radius = \(\frac{AB}{2} = \frac{√200}{2} = \frac{10√2}{2} = 5√2.\)

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]

(Function) \(f(x)\) is a function. What is the value of \(f(2020)\)?

1) \(f(10)=11\)

2) \(f(x+3)=\frac{f(x) - 1}{f(x) + 1}\)

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[1]

Updated on: 20 Jan 2021, 04:31

Originally posted by MathRevolution on 15 May 2020, 07:30.
Last edited by MathRevolution on 20 Jan 2021, 04:31, edited 1 time in total.

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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.
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 \(2\) 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)

Assume \(t = s + 1\) from condition 1).
Then we have \(x^2 + px + q = (x - s)(x - t) = (x - s)(x - (s + 1)) = x^2 – (2s + 1)x + s(s + 1) = 0\) and \(p = -(2s + 1), q = s(s + 1).\)

Since \(q = s(s + 1)\) is a product of two consecutive integers, \(q\) is an even number, and the unique even prime is \(2\). So, we have \(q = 2.\)

Then we have \(s(s + 1) = 2, s^2 + s = 2\) or \(s^2 + s - 2 = (s + 2)(s - 1) = 0\). We have \(s = -2\) or \(s = 1.\)

Then we have \(p = -(2s + 1) = -(-3) = 3\) for \(s = -2\) or \(p = -(2s + 1) = -3\) for \(s = 1.\)

Since \(p\) is positive, we have \(p = 3\) and \(p + q = 5.\)

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.