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The heights of the five starters on Redwood High's basketball team are

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The heights of the five starters on Redwood High's basketball team are  [#permalink]

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New post 01 Mar 2018, 00:51
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The heights of the five starters on Redwood High's basketball team are 5'11" (5 feet 11 inches), 6'3", 6', 6'6", and 6'2". The average height of these players is
(1 foot = 12 inches)

(A) 6'1"

(B) 6'2"

(C) 6' 3"

(D) 6'4"

(E) 6'5"

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Re: The heights of the five starters on Redwood High's basketball team are  [#permalink]

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New post 01 Mar 2018, 00:52
Bunuel wrote:
The heights of the five starters on Redwood High's basketball team are 5'11" (5 feet 11 inches), 6'3", 6', 6'6", and 6'2". The average height of these players is
(1 foot = 12 inches)

(A) 6'1"

(B) 6'2"

(C) 6' 3"

(D) 6'4"

(E) 6'5"


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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: The heights of the five starters on Redwood High's basketball team are  [#permalink]

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New post 01 Mar 2018, 01:18
Bunuel wrote:
The heights of the five starters on Redwood High's basketball team are 5'11" (5 feet 11 inches), 6'3", 6', 6'6", and 6'2". The average height of these players is
(1 foot = 12 inches)

(A) 6'1"

(B) 6'2"

(C) 6' 3"

(D) 6'4"

(E) 6'5"


The best way to do this problem is to convert the respective heights into inches
and finding the average, since the answers for the average height are very close!

When converted in inches, we will get 71,72,74,75, and 78 inches.

The average is \(\frac{70*5}{5} + \frac{1+2+4+5+8}{5} = 70 + \frac{20}{5} = 74\) inches, which is 6' 2"(Option B)
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Re: The heights of the five starters on Redwood High's basketball team are  [#permalink]

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New post 14 Mar 2018, 03:39
pushpitkc wrote:
Bunuel wrote:
The heights of the five starters on Redwood High's basketball team are 5'11" (5 feet 11 inches), 6'3", 6', 6'6", and 6'2". The average height of these players is
(1 foot = 12 inches)

(A) 6'1"

(B) 6'2"

(C) 6' 3"

(D) 6'4"

(E) 6'5"


The best way to do this problem is to convert the respective heights into inches
and finding the average, since the answers for the average height are very close!

When converted in inches, we will get 71,72,74,75, and 78 inches.

The average is \(\frac{70*5}{5} + \frac{1+2+4+5+8}{5} = 70 + \frac{20}{5} = 74\) inches, which is 6' 2"(Option B)


can't we just arrange the give heights in ascending order and take middle value as mean as there are odd no of elements ?
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Re: The heights of the five starters on Redwood High's basketball team are  [#permalink]

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New post 14 Mar 2018, 04:44
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utkarshrihand wrote:
pushpitkc wrote:
Bunuel wrote:
The heights of the five starters on Redwood High's basketball team are 5'11" (5 feet 11 inches), 6'3", 6', 6'6", and 6'2". The average height of these players is
(1 foot = 12 inches)

(A) 6'1"

(B) 6'2"

(C) 6' 3"

(D) 6'4"

(E) 6'5"


The best way to do this problem is to convert the respective heights into inches
and finding the average, since the answers for the average height are very close!

When converted in inches, we will get 71,72,74,75, and 78 inches.

The average is \(\frac{70*5}{5} + \frac{1+2+4+5+8}{5} = 70 + \frac{20}{5} = 74\) inches, which is 6' 2"(Option B)


can't we just arrange the give heights in ascending order and take middle value as mean as there are odd no of elements ?



Hi utkarshrihand

The middle value is the mean when the list is an arithmetic progression
(has a common difference). However, in this list, the elements do not
have a common difference and the middle value can't be the mean

Hope this helps you!
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Re: The heights of the five starters on Redwood High's basketball team are  [#permalink]

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New post 15 Mar 2018, 10:10
Bunuel wrote:
The heights of the five starters on Redwood High's basketball team are 5'11" (5 feet 11 inches), 6'3", 6', 6'6", and 6'2". The average height of these players is
(1 foot = 12 inches)

(A) 6'1"

(B) 6'2"

(C) 6' 3"

(D) 6'4"

(E) 6'5"


The sum of the heights is:

5 + 6 + 6 + 6 + 6 = 29 feet

plus

11 + 3 + 0 + 6 + 2 = 22 inches = 22/12 = 11/6 feet

So the total height is:

29 + 11/6 = 174/6 + 11/6 = 185/6 feet

Thus, the average is:

(185/6)/5 = 37/6 feet = 74/12 = 6 2/12 inches = 6’2”

Answer: B
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Re: The heights of the five starters on Redwood High's basketball team are &nbs [#permalink] 15 Mar 2018, 10:10
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