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During a particularly hectic week at Ballard High, Robert drank 5, 8,

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Joined: 02 Sep 2009
Posts: 58335
During a particularly hectic week at Ballard High, Robert drank 5, 8,  [#permalink]

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31 Jan 2019, 01:32
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Difficulty:

15% (low)

Question Stats:

85% (01:51) correct 15% (02:28) wrong based on 20 sessions

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During a particularly hectic week at Ballard High, Robert drank 5, 8, 3, 6, 2, 9, and 14 cans of Slurp Soda, respectively, on each of the 7 days. What is the product of Robert's mean and median soda consumption for that week?

A. 1.12
B. 6
C. 6.71
D. 36
E. 40.3

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Re: During a particularly hectic week at Ballard High, Robert drank 5, 8,  [#permalink]

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31 Jan 2019, 03:24
Bunuel wrote:
During a particularly hectic week at Ballard High, Robert drank 5, 8, 3, 6, 2, 9, and 14 cans of Slurp Soda, respectively, on each of the 7 days. What is the product of Robert's mean and median soda consumption for that week?

A. 1.12
B. 6
C. 6.71
D. 36
E. 40.3

avg=6.71
and median = 6
40.3
IMO E
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Joined: 04 Jan 2015
Posts: 3074
Re: During a particularly hectic week at Ballard High, Robert drank 5, 8,  [#permalink]

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31 Jan 2019, 08:00

Solution

Given:
• Robert drank 5, 8, 3, 6, 2, 9 and 14 cans of Slurp Soda respectively on each of the the 7 days in a week

To find:
• The product of Robert’s mean and median soda consumption for that week

Approach and Working:
• Mean = $$\frac{(5 + 8 + 3 + 6 + 2 + 9 + 14)}{7} = \frac{47}{7} = 6$$
• Median = 6
• Therefore, mean * median = $$\frac{47}{7} * 6 = \frac{282}{7} = 40.3$$

Hence, the correct answer is Option E

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Re: During a particularly hectic week at Ballard High, Robert drank 5, 8,  [#permalink]

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04 Feb 2019, 19:56
Bunuel wrote:
During a particularly hectic week at Ballard High, Robert drank 5, 8, 3, 6, 2, 9, and 14 cans of Slurp Soda, respectively, on each of the 7 days. What is the product of Robert's mean and median soda consumption for that week?

A. 1.12
B. 6
C. 6.71
D. 36
E. 40.3

Let’s first determine the mean:

(5 + 8 + 3 + 6 + 2 + 9 + 14)/7 = 47/7

Ordering the numbers from least to greatest, we have:

2, 3, 5, 6, 8, 9, 14

The median is 6.

Thus, the product of the mean and median is 47/7 x 6 = 282/7 ≈ 40.3

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Re: During a particularly hectic week at Ballard High, Robert drank 5, 8,  [#permalink]

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04 Feb 2019, 20:30
Bunuel wrote:
During a particularly hectic week at Ballard High, Robert drank 5, 8, 3, 6, 2, 9, and 14 cans of Slurp Soda, respectively, on each of the 7 days. What is the product of Robert's mean and median soda consumption for that week?

A. 1.12
B. 6
C. 6.71
D. 36
E. 40.3

Alternative approach:

The mean is 5 + 8 + 3 + 6 + 2 + 9 + 14/7 = 47/7.

49/7 = 7 So, 47/7 is almost 7.

The median of 2. 3. 5. 6, 8, 9, 14 is 6.

6 x (almost 7) = almost 42.

The correct answer is E. 40.3.
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Marty Murray

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Re: During a particularly hectic week at Ballard High, Robert drank 5, 8,  [#permalink]

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04 Feb 2019, 20:39
Bunuel wrote:
During a particularly hectic week at Ballard High, Robert drank 5, 8, 3, 6, 2, 9, and 14 cans of Slurp Soda, respectively, on each of the 7 days. What is the product of Robert's mean and median soda consumption for that week?

A. 1.12
B. 6
C. 6.71
D. 36
E. 40.3

My first thought was, find the median and to get the answer just look for a number which is divisible by it, could have actually backfired.

Lucky for me, i didn't do that and you should not as well.( Why hurt your score in the actual exam ???)

Arrange the terms to get the median => 6

Add the terms and then divide by 7 to get the mean = 47/7 = 6.71

6*6.71
40.3

E
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Re: During a particularly hectic week at Ballard High, Robert drank 5, 8,   [#permalink] 04 Feb 2019, 20:39
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