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A tailor used 30 buttons that had an average (arithmetic mea

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A tailor used 30 buttons that had an average (arithmetic mea [#permalink] New post 17 Feb 2017, 06:53
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A tailor used 30 buttons that had an average (arithmetic mean) weight of x grams per button and 20 other buttons that had an average weight of 80 grams per button. Which of the following is the average weight per button, in grams, of the 50 buttons that the tailor used?


A) \(\frac{x + (20)(80)}{50}\)

B) \(\frac{x + 80}{50}\)

C) \(\frac{3}{5} x + \frac{8}{5}\)

D) \(\frac{3}{5}x + 32\)

E) \(\frac{3}{5} x\)
[Reveal] Spoiler: OA

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Re: A tailor used 30 buttons that had an average (arithmetic mea [#permalink] New post 28 Feb 2017, 15:57
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Explanation

Total number of buttons used = 50 (30 +20).

Number of buttons that weighs x grams per button = 30

The total mass of buttons that weighs x grams per button = 30x.

Number of buttons that weighs x grams per button = 20

Total mass of buttons that weighs 80 grams per button = \(80 \times 20\) = 1600.

Total mass of 50 buttons = 30x + 160.

Average mass of buttons = \(\frac{30x + 160}{50}= \frac{3}{5}x + 32\).

Hence option D is correct.
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Re: A tailor used 30 buttons that had an average (arithmetic mea [#permalink] New post 25 Oct 2017, 19:41
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@Sandy, really weird answer: B/C 80 x 20 would be 1600 not 160. But in my version of the question, the answer was 3/5x + 32. Hence 80x20 = 1600. Hence (30x + 1600) / 50. Which would equal out to 3/5x + 32.
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Re: A tailor used 30 buttons that had an average (arithmetic mea [#permalink] New post 30 Jul 2018, 10:45
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Carcass wrote:


A tailor used 30 buttons that had an average (arithmetic mean) weight of x grams per button and 20 other buttons that had an average weight of 80 grams per button. Which of the following is the average weight per button, in grams, of the 50 buttons that the tailor used?


A) \(\frac{x + (20)(80)}{50}\)

B) \(\frac{x + 80}{50}\)

C) \(\frac{3}{5} x + \frac{8}{5}\)

D) \(\frac{3}{5}x + 3.2\)

E) \(\frac{5}{8} x\)


Hi Sandy, Please check this answer or your question. 80 x 20 = 1600 not 160 (80 x 2). The answer should be 3/5(x) + 32.
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Re: A tailor used 30 buttons that had an average (arithmetic mea [#permalink] New post 18 Nov 2018, 18:25
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sandy wrote:
Explanation



Total mass of buttons that weighs 80 grams per button = \(80 \times 20\) = 160.

Total mass of 50 buttons = 30x + 160.

Average mass of buttons = \(\frac{30x + 160}{50}= \frac{3}{5}x + 3.2\).

Hence option D is correct.



Sandy, please fix the error 80*20 = 1600 not 160
so that instead of the part 3.2, becomes 32.
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Re: A tailor used 30 buttons that had an average (arithmetic mea [#permalink] New post 18 Nov 2018, 19:51
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Carcass wrote:


A tailor used 30 buttons that had an average (arithmetic mean) weight of x grams per button and 20 other buttons that had an average weight of 80 grams per button. Which of the following is the average weight per button, in grams, of the 50 buttons that the tailor used?


A) \(\frac{x + (20)(80)}{50}\)

B) \(\frac{x + 80}{50}\)

C) \(\frac{3}{5} x + \frac{8}{5}\)

D) \(\frac{3}{5}x + 32\)

E) \(\frac{5}{8} x\)



Another way is to take a value for x..
let x be 80, so the overall average will remain 80.
let us see which value gives us 80 as an answer..

A) \(\frac{x + (20)(80)}{50}...... => \frac{80 + (20)(80)}{50}=\frac{8+160}{5}\)...NO

B) \(\frac{x + 80}{50}.........=>160/50\)....NO

C) \(\frac{3}{5} x + \frac{8}{5}......=>\frac{3}{5} 80 + \frac{8}{5}=8(30+1)/5\)....NO

D) \(\frac{3}{5}x + 32........=>\frac{3}{5}80 + 32=3*16+32=48+32=80\)....YES

E) \(\frac{5}{8} x......=>80*5?8=50\)...NO

D
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Some useful Theory.
1. Arithmetic and Geometric progressions : https://greprepclub.com/forum/progressions-arithmetic-geometric-and-harmonic-11574.html#p27048
2. Effect of Arithmetic Operations on fraction : https://greprepclub.com/forum/effects-of-arithmetic-operations-on-fractions-11573.html?sid=d570445335a783891cd4d48a17db9825
3. Remainders : https://greprepclub.com/forum/remainders-what-you-should-know-11524.html
4. Number properties : https://greprepclub.com/forum/number-property-all-you-require-11518.html
5. Absolute Modulus and Inequalities : https://greprepclub.com/forum/absolute-modulus-a-better-understanding-11281.html

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Re: A tailor used 30 buttons that had an average (arithmetic mea [#permalink] New post 20 Nov 2018, 17:29
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Fixed above the explanation provided by sandy.

Thank you so much guys.

Regards
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Re: A tailor used 30 buttons that had an average (arithmetic mea [#permalink] New post 10 Jan 2019, 14:03
Carcass wrote:
Fixed above the explanation provided by sandy.

Thank you so much guys.

Regards

I think still typo in calculation part it is 1600 not 160
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Re: A tailor used 30 buttons that had an average (arithmetic mea [#permalink] New post 10 Jan 2019, 14:05
AE wrote:
Carcass wrote:
Fixed above the explanation provided by sandy.

Thank you so much guys.

Regards




sandy wrote:

Total mass of buttons that weighs 80 grams per button = \(80 \times 20\) = 1600.

Total mass of 50 buttons = 30x + 160.

Average mass of buttons = [m]\frac{30x + 160}{50}


I think still typo in calculation part it is 1600 not 160
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Re: A tailor used 30 buttons that had an average (arithmetic mea [#permalink] New post 25 Aug 2020, 04:29
Avg = sum / num

A tailor used 30 buttons that had an average (arithmetic mean) weight of x grams per button
Let b = buttons
x=b/30
b=30x

and 20 other buttons that had an average weight of 80 grams per button.
80=b/20
b=1600

Which of the following is the average weight per button, in grams, of the 50 buttons that the tailor used?
So, we got two equations, combine both
(30x + 1600) / 50

Simplify:
30x/50 + 1600/50
3x/5 + 32 - Option D
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Re: A tailor used 30 buttons that had an average (arithmetic mea   [#permalink] 25 Aug 2020, 04:29
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