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A painter charges $12 an hour while his son charges $6 an hour. If fat

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Math Expert
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Joined: 02 Sep 2009
Posts: 55801
A painter charges $12 an hour while his son charges $6 an hour. If fat  [#permalink]

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New post 08 Jan 2019, 01:17
00:00
A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

84% (01:07) correct 16% (01:26) wrong based on 24 sessions

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VP
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Joined: 31 Oct 2013
Posts: 1376
Concentration: Accounting, Finance
GPA: 3.68
WE: Analyst (Accounting)
Re: A painter charges $12 an hour while his son charges $6 an hour. If fat  [#permalink]

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New post 08 Jan 2019, 01:21
Bunuel wrote:
A painter charges $12 an hour while his son charges $6 an hour. If father and son work the same amount of time together on a job, how many hours does each of them work if their combined charge for labor is $108?

A. 6
B. 8
C. 9
D. 12
E. 18



Both father and son spent equal amount of time.

A) 6

6*12 + 6*6 = 108.

A is the best answer.
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Re: A painter charges $12 an hour while his son charges $6 an hour. If fat  [#permalink]

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New post 08 Jan 2019, 04:19
Bunuel wrote:
A painter charges $12 an hour while his son charges $6 an hour. If father and son work the same amount of time together on a job, how many hours does each of them work if their combined charge for labor is $108?

A. 6
B. 8
C. 9
D. 12
E. 18


12x+6y=108
x=y
so
18x=108
x=6
IMO A
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Re: A painter charges $12 an hour while his son charges $6 an hour. If fat  [#permalink]

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New post 08 Jan 2019, 05:28

Solution


Given:
    • A painter charges $12 an hour
    • His son charges $6 an hour

To find:
    • If both work together on a job, what is the time taken, in hours, to earn $108

Approach and Working:
    • In one hour, both of them together can earn $12 + $6 = $18
    • $18 * number of hours = $108

Therefore, number of hours = \(\frac{108}{18} = 6\)

Hence, the correct answer is Option A

Answer: A

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Re: A painter charges $12 an hour while his son charges $6 an hour. If fat   [#permalink] 08 Jan 2019, 05:28
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