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[Math Revolution GMAT math practice question]

(inequality) Is \(x^2>xy?\)

\(1) x>y\)
\(2) y>0\)
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[Math Revolution GMAT math practice question]

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

\(1) x>0\)
\(2) x<1\)

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

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

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

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

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

Therefore, C is the answer.
Answer: C

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

Regards

\((1+x+x^2+x^3+x^4+x^5+x^6)(1-x)^2< (1-x)\)
\(⇔(1+x+x^2+x^3+x^4+x^5+x^6)(1-x)(1-x)< (1-x)\)
\(⇔(1+x+x^2+x^3+x^4+x^5+x^6-x-x^2-x^3-x^4-x^5-x^6-x^7)(1-x)< (1-x)\)
\(⇔(1-x^7)(1-x)< (1-x)\)
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[Math Revolution GMAT math practice question]

(number property) What is the value of \(A\)?

1) The four-digit number \(A77A\) is a multiple of \(4\).
2) The four-digit number \(A77A\) is a multiple of \(9\)
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[Math Revolution GMAT math practice question]

(inequality) Is \(x^2>xy?\)

\(1) x>y\)
\(2) y>0\)

=>

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.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.
\(x^2>xy\)
\(=> x^2-xy > 0\)
\(=> x(x-y) > 0\)
\(=> x>0, x>y\) or \(x<0, x<y\)

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)
Since we have \(x > y\) and \(x > 0\) from \(x > y > 0\) which are a combined inequality of both conditions, 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) \(x > y\)
If \(x = 2, y = 1\), then we have \(x^2>xy\) and the answer is ‘yes’.
If \(x = -1, y = -2,\) then we have \(x^2<xy\) and the answer is ‘no’.

Condition 2) \(y > 0\)
If \(x = 2, y = 1\), then we have \(x^2>xy\) and the answer is ‘yes’.
If \(x = 1, y = 2,\) then we have \(x^2<xy\) and the answer is ‘no’.


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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[Math Revolution GMAT math practice question]

(work rate) Machine A takes 20 minutes and B takes 15 minutes to warm up. The machine A produces 300 pins per 1 minute and the machine B produces 200 parts per 1 minute. Including the time for warming up, is the number of pins produced by machine A smaller than the number of pins from the machine B?

1) The number of pins is greater than 3000
2) The number of pins is less than 4000
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[Math Revolution GMAT math practice question]

(number property) What is the value of \(A\)?

1) The four-digit number \(A77A\) is a multiple of \(4\).
2) The four-digit number \(A77A\) is a multiple of \(9\)

=>

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 1 variable (A) and 0 equations, D is most likely to be the answer. So, we should consider each of the conditions on their own first.

Condition 1)
If the last two digit is a multiple of \(4\), the original number is a multiple of \(4\).
Thus, \(A\) is \(2\) or \(6\) since \(72\) and \(76\) are multiples of \(4\).
Condition 1) is not sufficient, because we don’t have a unique answer

Condition 2)
If the sum of all digits is a multiple of \(9\), the original number is a multiple of \(9\).
\(A + 7 + 7 + A = 2A + 14\)
If \(2A + 14 = 18\), we have \(2A = 4\) or \(A = 2.\)
We don’t have any digit integer \(A\) such that \(2A = 9, 27\) or \(36.\)
Thus \(A = 2\) is the unique integer.
Condition 2) is sufficient.

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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[Math Revolution GMAT math practice question]

(number property) What is the units digit of a positive integer \(n\)?

1) \(n\) is a common multiple of \(13\) and \(14.\)
2) \(n\) is a common multiple of \(13\) and \(15\).
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[Math Revolution GMAT math practice question]

(work rate) Machine A takes 20 minutes and B takes 15 minutes to warm up. The machine A produces 300 pins per 1 minute and the machine B produces 200 parts per 1 minute. Including the time for warming up, is the number of pins produced by machine A smaller than the number of pins from the machine B?

1) The number of pins is greater than 3000
2) The number of pins is less than 4000

=>

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.

Let \(w\) be the number of pins.
\(20 + \frac{w}{300}\) is the time the machine A takes to produce w pins and \(15 + \frac{w}{200}.\) The question asks if \(20 + \frac{w}{300}\) is less than \(15 + \frac{w}{200}\).
\(20 + \frac{w}{300} < 15 + \frac{w}{200}\)
\(=> 5 < \frac{w}{200} – \frac{w}{300}\)
\(=> 5 < \frac{3w}{600} – \frac{2w}{600}\)
\(=> 5 < \frac{w}{600}\)
\(=> 3000 < w\)

Since condition 1) is same as the question, condition 1) is sufficient.

Condition 2)
In inequality questions, the law “Question is King” tells us that if the solution set of the question includes the solution set of the condition, then the condition is sufficient
The solution set \(3000 < w\) of the question does not include the solution set of condition 2) \(w < 400\). Thus condition 2) is not sufficient.

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

(set) If \(n\) is contained in the set \(A\), \(n+5\) is contained in set \(A\). Is \(50\) contained in set \(A\)?

1) \(30\) is contained in the set \(A\).
2) \(100\) is contained in the set \(S\)
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[Math Revolution GMAT math practice question]

(number property) What is the units digit of a positive integer \(n\)?

1) \(n\) is a common multiple of \(13\) and \(14.\)
2) \(n\) is a common multiple of \(13\) and \(15\).

=>

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 1 variable (n) and 0 equations, D is most likely to be the answer. So, we should consider each of the conditions on their own first.

Condition 1)
Since n is a common multiple of \(13\) and \(14, n\) is a multiple of \(lcm(13,14) = 182.\) Thus units digits of multiples of \(n\) are \(0, 2, 4, 6, 8\) since the units digit of \(182\) is \(2\).
Condition 1) is not sufficient.

Condition 2)
Since n is a common multiple of \(13\) and \(15, n\) is a multiple of \(lcm(13,15) = 195\). Thus units digits of multiples of \(n\) are \(0, 5\) since the units digit of \(182\) is \(2\).
Condition 2) is not sufficient.

Conditions 1) & 2)
The common units digit of multiples of \(182\) and \(195\) is \(0\) only.
Both conditions together are sufficient.

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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[Math Revolution GMAT math practice question]

(set) If \(n\) is contained in the set \(A\), \(n+5\) is contained in set \(A\). Is \(50\) contained in set \(A\)?

1) \(30\) is contained in the set \(A\).
2) \(100\) is contained in the set \(S\)

=>

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.

Condition 1)
Since \(30\) is in the set \(A\), \(35, 40, 45, 50\) are in the set \(A\).
Condition 1) is sufficient.

Condition 2)
\(A = { 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100, … }\)
The set \(A\) has \(40\).
But the set \(A = { 100, 105, 110, … }\) does not have \(40\).
Condition 2) is not sufficient.

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

(function) What is the value of \(f(10)\)?

\(1) f(x) = 2\) for \(0 ≤ x < 3\)
\(2) f(x+3) = 2f(x)\)
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[Math Revolution GMAT math practice question]

(number property) When \(m\) and \(n\) are positive integers, is \(m!*n!\) an integer squared?

\(1) m = n + 1\)
\(2) m\) is an integer squared
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Hello,
If the average (AM) of 4 numbers is 10, how many of the numbers are greater than 10?

1) Precisely 2 of the numbers are greater than 10.
2) The largest of 4 numbers is 10 greatest than the smallest of the 4 numbers.
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[Math Revolution GMAT math practice question]

(function) What is the value of \(f(10)\)?

\(1) f(x) = 2\) for \(0 ≤ x < 3\)
\(2) f(x+3) = 2f(x)\)

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

When asked about a function, we assume there are many variables so that E is most likely to be the answer.

Conditions 1) & 2):
Repeatedly applying condition 2) gives \(f(10) = 2f(7), f(7) = 2f(4)\) and \(f(4) = 2f(1)\), so that \(f(10) = 2f(7) = 4f(4) = 8f(1).\)
Since \(f(1) = 2\) by condition 1), \(f(10) = 8*2 = 16.\)
Both conditions together are sufficient.

It is easy to see that neither condition is sufficient on its own.

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

(absolute value) Is \(\frac{|x|}{x}\) equal to \(-1\)?

\(1) x>0\)
\(2) x<1\)
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[Math Revolution GMAT math practice question]

(number property) When \(m\) and \(n\) are positive integers, is \(m!*n!\) an integer squared?

\(1) m = n + 1\)
\(2) m\) is an integer squared

=>

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 (\(m\) and \(n\)) 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)
By condition 1),
\(m!*n! = (n+1)!*n! = (n+1)(n!)(n!) = (n+1)(n!)^2 = m(n!)^2.\)
Since \((n!)^2\) is an integer squared, and \(m\) is an integer squared by condition 2), \(m!*n!\) is an integer squared.
Thus, both conditions together are sufficient.

Since this question is a statistics question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.

Condition 1)
If \(m = 4\) and \(n = 3\), then \(4!*3! = 4*3!*3! = (2(3!))^2\) and the answer is ‘yes’.
If \(m = 3\) and \(n = 2\), then \(3!*2! = 6*2 = 12\), and the answer is ‘no’.
Since it does not give a unique answer, condition 1) is not sufficient on its own.

Condition 2)
If \(m = 4\) and \(n = 3\), then \(4!*3! = 4*3!*3! = (2(3!))^2\) and the answer is ‘yes’.
If \(m = 4\) and \(n = 2,\) then \(4!*2! = 24*2 = 48\) and the answer is ‘no’.
Since it does not give a unique answer, condition 2) is not sufficient on its own.

Therefore, the answer is C.
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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