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In a weight-lifting competition, the average (arithmetic mean) of Jena

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In a weight-lifting competition, the average (arithmetic mean) of Jena  [#permalink]

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New post 28 Dec 2017, 21:49
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Question Stats:

73% (02:04) correct 27% (02:35) wrong based on 22 sessions

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In a weight-lifting competition, the average (arithmetic mean) of Jenai's three lifts was 500 pounds. If twice the weight of her first lift was 200 more pounds than the weight of her second lift, and her third lift was 40 more pounds than her second lift, what is the weight, in pounds of her first lift?

A. 1360
B. 744
C. 544
D. 372
E. 228

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In a weight-lifting competition, the average (arithmetic mean) of Jena  [#permalink]

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New post Updated on: 31 Dec 2017, 22:07
E.g
#1 list= a
#2 lift= b
#3 lift= c

(a+b+c)/3= 500........... (1)
2a=200+b........…(2)
c=40+b…………(3)

(1)(2)(3)
(a+b+c)/3= 500
(100+1/2b)+b+(40+b)=500.3
140+5/2=1500
b=544

(2)

2a=200+b
2a=200+544
a= 372

Answer: D


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Originally posted by tuvois10 on 28 Dec 2017, 22:57.
Last edited by tuvois10 on 31 Dec 2017, 22:07, edited 1 time in total.
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Re: In a weight-lifting competition, the average (arithmetic mean) of Jena  [#permalink]

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New post 28 Dec 2017, 23:41
Total weight=1500 let 1, 2 ,3 weights be F,S and T
2F=200+S
T=40+S
1500=100+ 0.5s+40+s+s
1360=5/2S
272*2=s
445
Ist = 645/2=372.5


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Re: In a weight-lifting competition, the average (arithmetic mean) of Jena  [#permalink]

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New post 29 Dec 2017, 05:00
Bunuel wrote:
In a weight-lifting competition, the average (arithmetic mean) of Jenai's three lifts was 500 pounds. If twice the weight of her first lift was 200 more pounds than the weight of her second lift, and her third lift was 40 more pounds than her second lift, what is the weight, in pounds of her first lift?

A. 1360
B. 744
C. 544
D. 372
E. 228


In addition to the direct calculation approaches seen above, we'll show a different approach using the answers.
This is an Alternative method.

Let's start with the median - (C).
If Jenai's first lift was 544 then her second lift was 544*2 - 200 = 888 and her third was 888 + 40 = 928.
Then the average of Jenai's lifts is (544+888+928)/3.
Since all of these are above 500, then the average is also above 500 which is too large!
(C) is eliminated as are the larger (A) and (B).
Let's try (D).
If Jenai's first lift was 372 then her second lift was 372*2 - 200 = 544 and her third was 544 + 40 = 584.
Then the average of Jenai's lifts is (372+544+584)/3 = 500.
Yes!
(B) is our answer.
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Re: In a weight-lifting competition, the average (arithmetic mean) of Jena &nbs [#permalink] 29 Dec 2017, 05:00
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