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Difference between revisions of "2008 AMC 12A Problems/Problem 1"

(New page: ==Problem == What is the reciprocal of <math>\frac{1}{2}+\frac{2}{3}</math>? <math>\textbf{(A)} \frac{6}{7} \qquad \textbf{(B)} \frac{7}{6} \qquad \textbf{(C)} \frac{5}{3} \qquad \textb...)
 
(→‎Solution)
 
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{{duplicate|[[2008 AMC 12A Problems|2008 AMC 12A #1]] and [[2008 AMC 10A Problems/Problem 1|2008 AMC 10A #1]]}}
 
==Problem ==
 
==Problem ==
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What is the reciprocal of <math>\frac{1}{2}+\frac{2}{3}</math>?
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A bakery owner turns on his doughnut machine at <math>\text{8:30}\ {\small\text{AM}}</math>. At <math>\text{11:10}\ {\small\text{AM}}</math> the machine has completed one third of the day's job. At what time will the doughnut machine complete the job?
  
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<math>\textbf{(A)} \frac{6}{7} \qquad \textbf{(B)} \frac{7}{6} \qquad \textbf{(C)} \frac{5}{3} \qquad \textbf{(D)} 3  \qquad \textbf{(E)} \frac{7}{2} </math>
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<math>\mathrm{(A)}\ \text{1:50}\ {\small\text{PM}}\qquad\mathrm{(B)}\ \text{3:00}\ {\small\text{PM}}\qquad\mathrm{(C)}\ \text{3:30}\ {\small\text{PM}}\qquad\mathrm{(D)}\ \text{4:30}\ {\small\text{PM}}\qquad\mathrm{(E)}\ \text{5:50}\ {\small\text{PM}}</math>
  
 
==Solution==
 
==Solution==
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<math>\left(\frac{1}{2}+\frac{2}{3}\right)^{-1}=\left(\frac{3}{6}+\frac{4}{6}\right)^{-1}=\left(\frac{7}{6}\right)^{-1}=\frac{6}{7}\Rightarrow A</math>.
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The machine completes one-third of the job in <math>\text{11:10}-\text{8:30}=\text{2:40}</math> hours. Thus, the entire job is completed in <math>3\cdot(\text{2:40})=\text{8:00}</math> hours.
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Since the machine was started at <math>\text{8:30 AM}</math>, the job will be finished <math>8</math> hours later, at <math>\text{4:30 PM}</math>. The answer is <math>\mathrm{(D)}</math>.
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Note: <math>\text{2:40}</math> means <math>2</math> hours and <math>40</math> minutes. <math>3</math> multiplied by this time interval is <math>8</math> hours.
  
 
==See Also==
 
==See Also==
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{{AMC12 box|year=2008|ab=A|before=First Question|num-a=2}}
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{{AMC10 box|year=2008|ab=A|before=First Question|num-a=2}}
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{{MAA Notice}}

Latest revision as of 16:18, 4 June 2021

The following problem is from both the 2008 AMC 12A #1 and 2008 AMC 10A #1, so both problems redirect to this page.

Problem

A bakery owner turns on his doughnut machine at $\text{8:30}\ {\small\text{AM}}$. At $\text{11:10}\ {\small\text{AM}}$ the machine has completed one third of the day's job. At what time will the doughnut machine complete the job?

$\mathrm{(A)}\ \text{1:50}\ {\small\text{PM}}\qquad\mathrm{(B)}\ \text{3:00}\ {\small\text{PM}}\qquad\mathrm{(C)}\ \text{3:30}\ {\small\text{PM}}\qquad\mathrm{(D)}\ \text{4:30}\ {\small\text{PM}}\qquad\mathrm{(E)}\ \text{5:50}\ {\small\text{PM}}$

Solution

The machine completes one-third of the job in $\text{11:10}-\text{8:30}=\text{2:40}$ hours. Thus, the entire job is completed in $3\cdot(\text{2:40})=\text{8:00}$ hours.

Since the machine was started at $\text{8:30 AM}$, the job will be finished $8$ hours later, at $\text{4:30 PM}$. The answer is $\mathrm{(D)}$.

Note: $\text{2:40}$ means $2$ hours and $40$ minutes. $3$ multiplied by this time interval is $8$ hours.

See Also

2008 AMC 12A (Problems • Answer Key • Resources)
Preceded by
First Question 		Followed by
Problem 2
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 (Problems • Answer Key • Resources)
Preceded by
First Question 		Followed by
Problem 2
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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