The arithmetic mean of a collection of 5 positive integers, not necess

Sum of original 5 numbers = 45
One number is added so now N=6. The question is asking if the sum of 6 numbers is >=60.
That means the number has to be >=15.

Statement 1: Not Sufficient
Statement 2: Number can be 5,10,15...

Both Combined: 15 is first such possible number. So Sufficient.

Hence C

The solution below explores the homogeneity nature of the average.

\(\left( * \right)\,\,\,5\,\,{\text{ints}} \geqslant 1\)

\(\sum\nolimits_{\,5} {\, = } \,\,5 \cdot 9 = 45\)

\(\sum\nolimits_{\,5} {\, + \,\,x\,\,\left( {6{\text{th}}} \right)\,\,\,\,\mathop \geqslant \limits^? } \,\,\,\,6 \cdot 10\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x\,\,\,\mathop \geqslant \limits^? \,\,\,15\)

\(\left( 1 \right)\,\,x \geqslant 14\,\,\,\left\{ \begin{gathered}\\
\,x = 14\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle \hfill \\\\
\,x = 15\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \hfill \\ \\
\end{gathered} \right.\)

\(\left( 2 \right)\, + \,\left( * \right):\,\,\,\,\,x = 5,10,15, \ldots \,\,\,\,\,\,\left\{ \begin{gathered}\\
\,x = 5\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle \hfill \\\\
\,x = 15\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \hfill \\ \\
\end{gathered} \right.\)

\(\left( {1 + 2} \right) + \,\left( * \right):\,\,\,\,x = 15,20,25, \ldots \,\,\,\,\, \Rightarrow \,\,x \geqslant 15\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \,\,\)

The above follows the notations and rationale taught in the GMATH method.
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Initial Sum = 5 * 9 = 45
New Sum = 6 * 10 = 60

New number >= 15?

1) New number >= 14 => Not sufficient

2) New number = 5, 10, 15, ...=> Not sufficient

1+2)
New number = 15, 20, .... => Sufficient

ANSWER: C

Solution:

We need to determine whether the mean of 6 integers is at least 10. Notice that the sum of the original 5 integers is 9 x 5 = 45. If we let the additional integer be n, the question becomes

(45 + n) / 6 ≥ 10 ?

45 + n ≥ 60 ?

n ≥ 15 ?

Therefore, if we can determine whether the additional integer is at least 15, then we can determine whether the mean of the 6 integers is at least 10.

Statement One Alone:

Knowing that the additional integer is at least 14 does not mean it’s at least 15. If n = 15, then the mean of the 6 integers is 10; however, if n = 14, then the mean is less than 10. Statement one alone is not sufficient.

Statement Two Alone:

Knowing that the additional integer is a multiple of 5 does not mean it’s at least 15. If n = 15, then the mean of the 6 integers is 10; however, if n = 10, then the mean is less than 10. Statement two alone is not sufficient.

Statements One and Two Together:

Knowing that the additional integer is at least 14 and a multiple of 5 allows us to say it’s at least 15 since 15 is the smallest integer that satisfies both conditions. Therefore, we can say that indeed, the mean of the 6 integers is at least 10.

Answer: C

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You don't know what the 6th number is. You just know that it is at least 14. It could be 14/15/16/17... etc

If it is 14, the new avg will be less than 10. If it is 15, the new avg will be 10. If it is 16/17... etc, the new avg will be greater than 10.
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Answer should be "C" as with (1) and (2) the smallest of added no is 15 which makes the arith mean 10 (atleast).

Ans C:
From Question Stem, we can determine the required range of value of 6th Integer as >=15, using Formulae for Arithmetic Mean.
To conclude this, we need to use both Statements 1 and 2.
Hence, Answer C.

Hi, I got a doubt. Here in the question we were given avg of 5 positive integers as 9 and the question asked is avg of 6 positive integers >= 10. My doubt is why can't we take the average of 9 and the additional positive integer(sixth number) answer A then will be sufficient
chetan2u Bunuel VeritasKarishma

LOL'ing at the amount of explanations that are way too complicated for this problem.. There is a very simple way to solve this problem shown below:

using the average formula its safe to say that 9=#/5 so #=45

if GMAT wants to add another number, the formula will now be 45+x/6 which will be >= 10 to be sufficient.

Now let's look at answer choice A

>=14. Plug that into the new average formula that is highlighted and if it is >= 10 A will be sufficient. 14 plugged into the equation will result in an average of 9 which is not true, move to B.

B states that anything that is a multiple of 5 will have an average >=10 which is so false (14 doesnt even produce >= average of 10!)

Look at C-- 70 is the only multiple of 14 that is also a multiple of 5... this number will produce an average greater than 10.


Hope this helps all.

Thanks,
Nick

I understand why the answer is C), but I think there's a way to interpret the question such that A) makes sense.

If the additional integer is at least 14, then the arithmetic mean of the 6 integers is at least 59/6, which in turn means that the arithmetic mean is NOT at least 10, which also answers the question at hand.

I know this is a super gimmicky way of looking at this question, but why is this interpretation wrong?

We can't answer yes/no with certainty.

If 14, then the mean is not at least 10.

If 15, then the mean is at least 10.
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