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# The average of 6 numbers is 8. One new number is now added

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Joined: 04 Jan 2015
Posts: 2306
The average of 6 numbers is 8. One new number is now added  [#permalink]

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Updated on: 12 Aug 2018, 22:02
00:00

Difficulty:

15% (low)

Question Stats:

81% (01:25) correct 19% (01:30) wrong based on 107 sessions

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Solve any Mean and Standard deviation question under a minute- Exercise Question #2

2- The average of 6 numbers is 8. One new number is now added to the set of 6 numbers and the arithmetic mean of the 7 numbers is now calculated.
Is the arithmetic mean of 7 numbers greater than 10?

1. The additional number is at least 23.
2. The additional number is a multiple of 4.

Options

A. Statement (1) ALONE is sufficient, but statement (2) ALONE is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) ALONE is not sufficient.
C. Both statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statement (1) and (2) TOGETHER are NOT sufficient.

To solve question 3: Question 3

To read the article: Solve any Mean and Standard deviation question under a minute

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Originally posted by EgmatQuantExpert on 04 Jul 2018, 05:49.
Last edited by EgmatQuantExpert on 12 Aug 2018, 22:02, edited 2 times in total.
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Joined: 04 Jan 2015
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Re: The average of 6 numbers is 8. One new number is now added  [#permalink]

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Updated on: 07 Jul 2018, 06:13
1

Solution

Given:

• The mean of 6 positive integers = 8

To find:

• We need to find if the mean of 6 integers is less than 10 or not.
o 8= $$\frac{Sum\ of\ 6\ integers}{6}$$
o Sum of 6 integers= 48

•Now, for average of 7 integers to be at least 10, the sum of 7 integers must be 70.
o Or, new integer must be 22.
o Thus, we need to find if the new integer is ≥ 22 or not.

Statement-1: “The additional integer is at least 23. “

Since new integer is greater than 22, statement 1 alone is sufficient to answer the question.

Statement-2: “The additional integer is multiple of 4 “

The additional integer can be 4 or 8 or 16 or 20 or 24 and so on.

Since new integer can be both, less than 22 and greater than 22, Statement 2 alone is not sufficient to answer the question.

Hence, the correct answer is option A.

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Originally posted by EgmatQuantExpert on 04 Jul 2018, 05:50.
Last edited by EgmatQuantExpert on 07 Jul 2018, 06:13, edited 3 times in total.
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Joined: 27 Feb 2018
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Re: The average of 6 numbers is 8. One new number is now added  [#permalink]

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04 Jul 2018, 05:55
EgmatQuantExpert wrote:
Solve any Mean and Standard deviation question under a minute- Exercise Question #2

2- The average of 6 numbers is 8. One new number is now added to the set of 6 numbers and the arithmetic mean of the 7 numbers is now calculated.
Is the arithmetic mean of 7 numbers greater than 10?

1. The additional number is at least 23.
2. The additional number is a multiple of 4.

Options

A. Statement (1) ALONE is sufficient, but statement (2) ALONE is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) ALONE is not sufficient.
C. Both statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statement (1) and (2) TOGETHER are NOT sufficient.

Given Avg. of Six numbers is 8 i.e sum of 6 numbers is 48. Now, the avg of 7 numbers to be greater than 10, their sum must be at least 70.

Option A: new number is at least 23 i.e sum=48+23=71 Sufficient

Option B: new number is multiple of 4, it can be 4,8,12,16,20,24 etc Not sufficient

IMO Option A is correct
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Re: The average of 6 numbers is 8. One new number is now added  [#permalink]

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06 Jul 2018, 09:42
Average of 6 numbers = 8
Total = 8x6 = 48

Is AM>10?

I. Atleast 23. so the number can be either 23 or greater.

Case 1: 48+23/7 = 71/7 = 10.14>10 - Clearly sufficient.

II. Multiple of 4: 4,8,12,16,20,24

Let's take 2 cases here. 8 & 24

Case 1: 48+8/7 = 56/7 = 8<10
case 2: 48+24/7 = 72/7 = 10.28>10

Clearly insufficient as we got multiple answers for option II.

Therefore A alone is sufficient.

Thank you
Arjun
Re: The average of 6 numbers is 8. One new number is now added &nbs [#permalink] 06 Jul 2018, 09:42
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