Difference between revisions of "2008 AMC 12A Problems/Problem 10"

(Duplicate, AMC 10 box, style, answer choices)
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==Solution==
 
==Solution==
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=== Solution 1 ===
 
Doug can paint <math>\frac{1}{5}</math> of a room per hour, Dave can paint <math>\frac{1}{7}</math> of a room in an hour, and the time they spend working together is <math>t-1</math>.
 
Doug can paint <math>\frac{1}{5}</math> of a room per hour, Dave can paint <math>\frac{1}{7}</math> of a room in an hour, and the time they spend working together is <math>t-1</math>.
  
 
Since rate times time gives output, <math>\left(\frac{1}{5}+\frac{1}{7}\right)\left(t-1\right)=1 \Rightarrow \mathrm{(D)}</math>
 
Since rate times time gives output, <math>\left(\frac{1}{5}+\frac{1}{7}\right)\left(t-1\right)=1 \Rightarrow \mathrm{(D)}</math>
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=== Solution 2 ===
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If one person does a job in <math>a</math> hours and another person does a job in <math>b</math> hours, the time it takes to do the job together is <math>\frac{ab}{a+b}</math> hours.
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Since Doug paints a room in 5 hours and Dave paints a room in 7 hours, they both paint in <math>\frac{5*7}{5+7} = \frac{35}{12}</math> hours. They also take 1 hour for lunch, so the total time <math>t = \frac{35}{12} + 1</math> hours.
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Looking at the answer choices, <math>(D)</math> is the only one satisfied by <math>t = \frac{35}{12} + 1</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2008|ab=A|num-b=9|num-a=11}}
 
{{AMC12 box|year=2008|ab=A|num-b=9|num-a=11}}
 
{{AMC10 box|year=2008|ab=A|num-b=12|num-a=14}}
 
{{AMC10 box|year=2008|ab=A|num-b=12|num-a=14}}

Revision as of 20:37, 5 February 2012

The following problem is from both the 2008 AMC 12A #10 and 2008 AMC 10A #13, so both problems redirect to this page.

Problem

Doug can paint a room in $5$ hours. Dave can paint the same room in $7$ hours. Doug and Dave paint the room together and take a one-hour break for lunch. Let $t$ be the total time, in hours, required for them to complete the job working together, including lunch. Which of the following equations is satisfied by $t$?

$\mathrm{(A)}\ \left(\frac{1}{5}+\frac{1}{7}\right)\left(t+1\right)=1\qquad\mathrm{(B)}\ \left(\frac{1}{5}+\frac{1}{7}\right)t+1=1\qquad\mathrm{(C)}\ \left(\frac{1}{5}+\frac{1}{7}\right)t=1\\\mathrm{(D)}\ \left(\frac{1}{5}+\frac{1}{7}\right)\left(t-1\right)=1\qquad\mathrm{(E)}\ \left(5+7\right)t=1$

Solution

Solution 1

Doug can paint $\frac{1}{5}$ of a room per hour, Dave can paint $\frac{1}{7}$ of a room in an hour, and the time they spend working together is $t-1$.

Since rate times time gives output, $\left(\frac{1}{5}+\frac{1}{7}\right)\left(t-1\right)=1 \Rightarrow \mathrm{(D)}$

Solution 2

If one person does a job in $a$ hours and another person does a job in $b$ hours, the time it takes to do the job together is $\frac{ab}{a+b}$ hours.

Since Doug paints a room in 5 hours and Dave paints a room in 7 hours, they both paint in $\frac{5*7}{5+7} = \frac{35}{12}$ hours. They also take 1 hour for lunch, so the total time $t = \frac{35}{12} + 1$ hours.

Looking at the answer choices, $(D)$ is the only one satisfied by $t = \frac{35}{12} + 1$.

See Also

2008 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions
2008 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions
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