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[Math Revolution GMAT math practice question]

(sequence) The terms of \(a\) sequence are defined by an=an-2+3. Is 411 a term of the sequence?

1) a1=111
2) a2=112


From statement 1:

\(a_1\) = 111.

\(a_3 = a_1\) + 3 = 114

\(a_5 = a_3\) + 3 = 117.

The series becomes 111, 114, 117....

the terms are 1, 3, 5, 7....

The common difference between the terms 1,3,5... is 3.

Let's consider 411 as the last term. then 411 = 111 +(n-1)3.
Then 411 becomes 101th term.

A is sufficient.

From statement 2:

We get \(a_2\), \(a_4\) and \(a_6\) and so on..
But first term is unknown.
B is sinsufficient.

A is the answer.
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[Math Revolution GMAT math practice question]

(algebra) \((x^2-3x+2)(y^2-5y+6)=?\)

\(1) x=1\)
\(2) y=1\)


From statement 1:

If x = 1. Then the expression becomes zero.
Sufficient.

From statement 2:

If y = 1. then expression reduces to 2x^2-6x+4
or (x-2)(2x+2). Since the RHS side is unknown. It's insufficient.

A is the answer.
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[Math Revolution GMAT math practice question]

(algebra) \((x^2-3x+2)(y^2-5y+6)=?\)

\(1) x=1\)
\(2) y=1\)

=>

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

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

Factoring the question gives \((x^2-3x+2)(y^2-5y+6) = (x-1)(x-2)(y-2)(y-3).\)

Condition 1)
If \(x = 1\), then \((x-1)(x-2)(y-2)(y-3) = 0.\)
Condition 1) is sufficient.

Condition 2)
If \(y = 1\), then \((x-1)(x-2)(1-2)(1-3) = 2(x-1)(x-2).\)
Since we don’t know the value of \(x\), condition 2) is not sufficient.

Note: Tip 4) of VA (Variable Approach) method states that if both conditions are trivial, they are not usually sufficient. Thus, conditions 1) & 2) are not sufficient by Tip 4).

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

(number property) If \(m\) and \(n\) are integers, is \(m+m^2-n\) an even number?

1) \(m\) is an even number
2) \(n\) is an even number
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[Math Revolution GMAT math practice question]

(number property) If \(m\) and \(n\) are integers, is \(m+m^2-n\) an even number?

1) \(m\) is an even number
2) \(n\) is an even number


\(m+m^2\) = m(m+1) = always even.

So B is sufficient.

B is the answer.
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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
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MathRevolution
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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
Hi R2S

If we multiply \(1+x+x^2+x^3+x^4+x^5+x^6\) by 1-x, we get (\(1+x+x^2+x^3+x^4+x^5+x^6-x-x^2-x^3-x^4-x^5-x^6-x^7\))(1-x)

All the terms are canceled except \(1-x^7\)(1-x).

Hope it's clear.
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Afc0892
MathRevolution
[Math Revolution GMAT math practice question]

(sequence) The terms of \(a\) sequence are defined by an=an-2+3. Is 411 a term of the sequence?

1) a1=111
2) a2=112


From statement 1:

\(a_1\) = 111.

\(a_3 = a_1\) + 3 = 114

\(a_5 = a_3\) + 3 = 117.

The series becomes 111, 114, 117....

the terms are 1, 3, 5, 7....

The common difference between the terms 1,3,5... is 3.
Here the common difference is 2; 3-1=2; 5-3=2

Let's consider 411 as the last term. then 411 = 111 +(n-1)3.
Then 411 becomes 101th term.

Now the formula is, a = a1+(n-1)*d
411 = 111+(n-1)*2
411-111=(n-1)*2
300=(n-1)*2
150=n-1
151=n
This shows that 411 is the 151st term in the sequence. Is it correct?


Based on either, the answer choice will still remain the same i.e. A

A is sufficient.

From statement 2:

We get \(a_2\), \(a_4\) and \(a_6\) and so on..
But first term is unknown.
B is sinsufficient.

A is the answer.
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Afc0892
MathRevolution
[Math Revolution GMAT math practice question]

(sequence) The terms of \(a\) sequence are defined by an=an-2+3. Is 411 a term of the sequence?

1) a1=111
2) a2=112


From statement 1:

\(a_1\) = 111.

\(a_3 = a_1\) + 3 = 114

\(a_5 = a_3\) + 3 = 117.

The series becomes 111, 114, 117....

the terms are 1, 3, 5, 7....

The common difference between the terms 1,3,5... is 3.
Here the common difference is 2; 3-1=2; 5-3=2

Let's consider 411 as the last term. then 411 = 111 +(n-1)3.
Then 411 becomes 101th term.

Now the formula is, a = a1+(n-1)*d
411 = 111+(n-1)*2
411-111=(n-1)*2
300=(n-1)*2
150=n-1
151=n
This shows that 411 is the 151st term in the sequence. Is it correct?


Based on either, the answer choice will still remain the same i.e. A

A is sufficient.

From statement 2:

We get \(a_2\), \(a_4\) and \(a_6\) and so on..
But first term is unknown.
B is sinsufficient.

A is the answer.

Thanks R2S.

Missed it in urgency. :)
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Yes, it is pretty clear now :)
Thanx Afc0892
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[Math Revolution GMAT math practice question]

(sequence) The terms of \(a\) sequence are defined by an=an-2+3. Is 411 a term of the sequence?

1) a1=111
2) a2=112

Let me present the proof of the insufficiency of statement (2) alone, for the benefit of the most-rigorous students!

\(S\,\,{\rm{sequence:}}\,\,\left\{ \matrix{\\
{{\rm{a}}_{\rm{1}}},{a_2}, \ldots \hfill \cr \\
{a_n} = {a_{n - 2}} + 3\,\,,\,\,{\rm{for}}\,\,{\rm{all}}\,\,n \ge 3 \hfill \cr} \right.\,\,\,\,\,\left( * \right)\)

\(411\,\,\mathop \in \limits^? \,\,\,S\)


\(\left( 2 \right)\,\,{a_2} = 112\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,{a_4} = 112 + 3\,\,\,\, \Rightarrow \,\,\,{a_6} = 112 + 2 \cdot 3\,\,\, \Rightarrow \,\,\,\, \ldots\)

\(411 \ne 112 + k \cdot 3\,,\,\,{\rm{for}}\,\,{\rm{all}}\,\,k\,\, \ge 1\,\,{\mathop{\rm int}} \,\,\,\,\left\{ \matrix{\\
\,\left( {{\mathop{\rm Re}\nolimits} } \right){\rm{Take}}\,\,{{\rm{a}}_{\rm{1}}}{\rm{ = 111}}\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \hfill \cr \\
\,\rm{Take}\,\,{a_1} = 112\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \hfill \cr} \right.\)


Obs.: once a_1 and a_2 are given, the sequence S is uniquely defined.


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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[Math Revolution GMAT math practice question]

(sequence) The terms of \(a\) sequence are defined by an=an-2+3. Is 411 a term of the sequence?

1) a1=111
2) a2=112

=>

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 formula an=an-2+3 tells us that alternate terms have the same remainder when they are divided by 3.
Since a1 = 111 = 3*37 is a multiple of three, all multiples of three greater than 111 can be obtained as odd-numbered terms. Therefore, 411= 3*137 is one of the odd-numbered terms, and 411 is in the sequence.
Condition 1) is sufficient.

a2 = 112 = 3*37 + 1 and all even-numbered terms have a remainder of 1 when they are divided by 3. Since 411 = 3*137, it is not an even-numbered term. Since we don’t know any of the odd-numbered terms, condition 2) is not sufficient.

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

(number property) When a positive integer \(n\) is divided by \(19\), what is the remainder?

1) \(n-17\) is a multiple of \(19\)
2) \(n-19\) is a multiple of \(17\)
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[Math Revolution GMAT math practice question]

(number property) When a positive integer \(n\) is divided by \(19\), what is the remainder?

1) \(n-17\) is a multiple of \(19\)
2) \(n-19\) is a multiple of \(17\)

Target question: When n is divided by 19, what is the remainder?

Statement 1: n-17 is a multiple of 19
----ASIDE-------------------------------------------
If N is a multiple of d, then we can write N = dk (for some integer k)
For example, if N is a multiple of 5, then we can write N = 5k (for some integer k)
----BACK TO THE QUESTION-------------------

If n-17 is a multiple of 19, then we can write: n - 17 = 19k
Add 17 to both sides to get: n = 19k + 17
In other words, n is 17 greater than some multiple of 19
So, when we divide n by 19, the remainder will be 17
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: n-19 is a multiple of 17
We can write n-19 = 17k
Add 19 to both sides to get: n = 17k + 19
Hmmm, this information doesn't help us answer the target question
Consider these two contradictory cases:
Case a: if k = 1, then n = 17(1) + 19 = 36. When we divide 36 by 19, the answer to the target question is the remainder is 17
Case b: if k = 2, then n = 17(2) + 19 = 53. When we divide 53 by 19, the answer to the target question is the remainder is 15
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer: A

Cheers,
Brent
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Ask me Anything about GMAT Data Sufficiency


Hello, I am Max Lee, Founder and Lead Math Tutor at Math Revolution.

I have over 6,000 posts and almost 5,000 Kudos. This topic is a new feature on the DS Forum and a way for you to directly interact with me and ask anything about the DS, e.g. if you want a certain concept explained or have a particular you question you want me addressed, this is the place to post a link to it or your question. I intend to have this thread be as a "Everything You Need to Know about DS" type of thread. I will keep updating this post with links and resources that are helpful for the DS. Meanwhile, you can ask me anything ;-)


My other discussions you may be interested in:



Thank you all - good luck on the GMAT and look forward to seeing you in the DS forum!
- Max Lee.

Please review my logic to this question.
Is z even?
(1) 5z is even
(2) 3z is even

Solution
(1) 5z is even
z=2, 5z is even yes
z=2/5 , 5z is even no
Insufficient.
(2) 3z is even
z=2, 3z is even yes
z=2/3 , 3z is even no
Insufficient.

Combining (1) and (2)
since even + even =even
Adding both (1) and (2)
5z+3z= even
or 8z=even
hence z could be odd or even.

hence answer should be E
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[Math Revolution GMAT math practice question]

(number property) If \(m\) and \(n\) are integers, is \(m+m^2-n\) an even number?

1) \(m\) is an even number
2) \(n\) is an even number

=>
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.

\(m+m^2-n = m(m+1) – n\). Now, m(m+1) is an even number since \(m(m+1)\) is the product of two consecutive integers. Thus, the parity of \(m+m^2-n\) depends on the parity of \(n\) only.
Thus, condition 2) is sufficient.

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

(statistics) If the average (arithmetic mean) of \(p, q\), and \(r\) is \(6\), what is the value of \(r\)?

\(1) p=-r\)
\(2) p=-q\)
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