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Re: What is the average (arithmetic mean) of j and k ? [#permalink]
Walkabout wrote:
What is the average (arithmetic mean) of j and k ?

(1) The average (arithmetic mean) of j + 2 and k + 4 is 11.
(2) The average (arithmetic mean) of j, k, and 14 is 10.



To find: (j+k)/2
(1) (j+2+k+4)/2 = 11
j+k+6=22
j+k=22-6

Sufficient

AD

(2) (j+k+14)/3=10
Again we get the value for j+k.

Sufficient

D
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Re: What is the average (arithmetic mean) of j and k ? [#permalink]
Expert Reply
Walkabout wrote:
What is the average (arithmetic mean) of j and k ?

(1) The average (arithmetic mean) of j + 2 and k + 4 is 11.
(2) The average (arithmetic mean) of j, k, and 14 is 10.


Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

There are 2 variables \(i\), \(j\) and 0 equation are given from the conditions, so there is high chance (C) will be our answer.

For cases where we need 2 more equation, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority.

However, this question asks a relation between 2 variables, that is, \(\frac{j + k}{2}\). It might need less than 2 equations to solve.

Condition (1)
\(\frac{j + 2 + k + 4}{2} = 11\)
\(j + k + 6 = 22\)
\(j + k = 16\)
Hence, the average of \(j\) and \(k\) is \(\frac{j + k}{2} = 8\).

Condition (2)
\(\frac{j + k + 14}{3} = 10\)
\(j + k + 14 = 30\)
\(j + k = 16\)
Hence, the average of \(j\) and \(k\) is \(\frac{j + k}{2} = 8\).

Therefore, the correct answer is D.
Re: What is the average (arithmetic mean) of j and k ? [#permalink]
Walkabout wrote:
What is the average (arithmetic mean) of j and k ?

(1) The average (arithmetic mean) of j + 2 and k + 4 is 11.
(2) The average (arithmetic mean) of j, k, and 14 is 10.

Bunuel
Can I solve this question by the following way?

Question:
j+k=multiple of 2?
1) j+k=16
Yes--> 16 is a multiple of 2. Sufficient

2) j+k=16
Yes--> 16 is a multiple of 2. Sufficient
Thanks__
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Re: What is the average (arithmetic mean) of j and k ? [#permalink]
Expert Reply
Asad wrote:
Walkabout wrote:
What is the average (arithmetic mean) of j and k ?

(1) The average (arithmetic mean) of j + 2 and k + 4 is 11.
(2) The average (arithmetic mean) of j, k, and 14 is 10.

Bunuel
Can I solve this question by the following way?

Question:
j+k=multiple of 2?
1) j+k=16
Yes--> 16 is a multiple of 2. Sufficient

2) j+k=16
Yes--> 16 is a multiple of 2. Sufficient
Thanks__


No, that's not correct. This is not an YES/NO question, this is a value questions. "What is the value of (j + k)/2" is not the same as "is j + k a multiple of 2".
What is the average (arithmetic mean) of j and k ? [#permalink]
Bunuel wrote:
Asad wrote:
Walkabout wrote:
What is the average (arithmetic mean) of j and k ?

(1) The average (arithmetic mean) of j + 2 and k + 4 is 11.
(2) The average (arithmetic mean) of j, k, and 14 is 10.

Bunuel
Can I solve this question by the following way?

Question:
j+k=multiple of 2?
1) j+k=16
Yes--> 16 is a multiple of 2. Sufficient

2) j+k=16
Yes--> 16 is a multiple of 2. Sufficient
Thanks__


No, that's not correct. This is not an YES/NO question, this is a value questions. "What is the value of (j + k)/2" is not the same as "is j + k a multiple of 2".

Thanks for the response..
Question:
j+k=multiple of 2 of any digit?
1) j+k=16 (8X2)
Yes--> multiple of 2 of any digit . Sufficient

2) j+k=16 (8X2)
Yes--> multiple of 2 of any digit . Sufficient
Edited.....
Does it make sense now Bunuel?

Originally posted by TheUltimateWinner on 02 Mar 2020, 00:49.
Last edited by TheUltimateWinner on 02 Mar 2020, 00:54, edited 1 time in total.
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Re: What is the average (arithmetic mean) of j and k ? [#permalink]
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Expert Reply
Asad wrote:
Thanks for the response..
Question:
j+k=multiple of 2 of any digit?
1) j+k=16 (8X2)
Yes--> multiple of 2 of any digit . Sufficient

2) j+k=16 (8X2)
Yes--> multiple of 2 of any digit . Sufficient. Sufficient
Edited.....
Does it make sense now Bunuel?


No. This is not an YES/NO question, this is a value questions.
Re: What is the average (arithmetic mean) of j and k ? [#permalink]
Bunuel wrote:
Asad wrote:
Thanks for the response..
Question:
j+k=multiple of 2 of any digit?
1) j+k=16 (8X2)
Yes--> multiple of 2 of any digit . Sufficient

2) j+k=16 (8X2)
Yes--> multiple of 2 of any digit . Sufficient. Sufficient
Edited.....
Does it make sense now Bunuel?


No. This is not an YES/NO question, this is a value questions.

Honestly speaking, I did not find any mistake in converting value question to YES/NO question!
If a question prompt ask: What is the value of \(x\) if any digit multiplied by 2?
This one is value question..
Converted version:
Is \(x\) even?

It seems that both version are SAME.
Could you help me to understand whole the scenario, please-Bunuel?
Thanks__
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Re: What is the average (arithmetic mean) of j and k ? [#permalink]
Expert Reply
Asad wrote:
Bunuel wrote:
Asad wrote:
Thanks for the response..
Question:
j+k=multiple of 2 of any digit?
1) j+k=16 (8X2)
Yes--> multiple of 2 of any digit . Sufficient

2) j+k=16 (8X2)
Yes--> multiple of 2 of any digit . Sufficient. Sufficient
Edited.....
Does it make sense now Bunuel?


No. This is not an YES/NO question, this is a value questions.

Honestly speaking, I did not find any mistake in converting value question to YES/NO question!
If a question prompt ask: What is the value of \(x\) if any digit multiplied by 2?
This one is value question..
Converted version:
Is \(x\) even?

It seems that both version are SAME.
Could you help me to understand whole the scenario, please-Bunuel?
Thanks__


Sorry, but I cannot understand what does "What is the value of \(x\) if any digit multiplied by 2? " mean. It does not make sense.
Re: What is the average (arithmetic mean) of j and k ? [#permalink]
Bunuel wrote:
Asad wrote:
Thanks for the response..
Question:
j+k=multiple of 2 of any digit?
1) j+k=16 (8X2)
Yes--> multiple of 2 of any digit . Sufficient

2) j+k=16 (8X2)
Yes--> multiple of 2 of any digit . Sufficient. Sufficient
Edited.....
Does it make sense now Bunuel?


No. This is not an YES/NO question, this is a value questions.

Yes, i got my answer. It seems that there is no way to convert YES/NO question to VALUE question, right?
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Re: What is the average (arithmetic mean) of j and k ? [#permalink]
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