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

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Bunuel
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During a particularly hectic week at Ballard High, Robert drank 5, 8, [#permalink] New post   31 Jan 2019, 01:32
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A
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E

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25% (medium)

Question Stats:

83% (01:46) correct 17% (02:18) wrong based on 29 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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E
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Re: During a particularly hectic week at Ballard High, Robert drank 5, 8, [#permalink] New post   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
answer
40.3
IMO E
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Re: During a particularly hectic week at Ballard High, Robert drank 5, 8, [#permalink] New post   31 Jan 2019, 08:00
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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

Answer: E

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Re: During a particularly hectic week at Ballard High, Robert drank 5, 8, [#permalink] New post   04 Feb 2019, 19:56
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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

Answer: E
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Re: During a particularly hectic week at Ballard High, Robert drank 5, 8, [#permalink] New post   04 Feb 2019, 20:30
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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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Re: During a particularly hectic week at Ballard High, Robert drank 5, 8, [#permalink] New post   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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