Difference between revisions of "2012 AMC 10A Problems/Problem 19"

(Solution 3 (GCD))
m (Solution 3 (GCD))
 
Line 35: Line 35:
  
 
==Solution 3 (GCD)==
 
==Solution 3 (GCD)==
We factor out the equations to be <math>(8-n)\left(\frac{1}{p}+\frac{1}{h}\right)=\frac{1}{2},\left(\frac{31}{5}-n\right)\left(\frac{1}{h}\right)=\frac{6}{25} \text{and } \left(\frac{56}{5}-n\right)\left(\frac{1}{p}=\frac{13}{50}\right)</math>, where n is the number of hours for the break, p is the time Paula requires, and h is the time her helpers require. We find that when we select <math>\mathbf{D}</math>, we have them being <math>10.4</math> and <math>5.4</math>, which correspond to being multiples of 13 and 6. Checking, we find that this satisfies the first equation, so multiplying <math>0.8\cdot 60= \boxed{48.}</math>
+
We factor out the equations to be <math>(8-n)\left(\frac{1}{p}+\frac{1}{h}\right)=\frac{1}{2},\left(\frac{31}{5}-n\right)\left(\frac{1}{h}\right)=\frac{6}{25} \text{ and } \left(\frac{56}{5}-n\right)\left(\frac{1}{p}=\frac{13}{50}\right)</math>, where n is the number of hours for the break, p is the time Paula requires, and h is the time her helpers require. We find that when we select <math>\mathbf{D}</math>, we have them being <math>10.4</math> and <math>5.4</math>, which correspond to being multiples of 13 and 6. Checking, we find that this satisfies the first equation, so multiplying <math>0.8\cdot 60= \boxed{48.}</math>
  
 
== See Also ==
 
== See Also ==

Latest revision as of 18:32, 8 November 2024

The following problem is from both the 2012 AMC 12A #13 and 2012 AMC 10A #19, so both problems redirect to this page.

Problem 19

Paula the painter and her two helpers each paint at constant, but different, rates. They always start at 8:00 AM, and all three always take the same amount of time to eat lunch. On Monday the three of them painted 50% of a house, quitting at 4:00 PM. On Tuesday, when Paula wasn't there, the two helpers painted only 24% of the house and quit at 2:12 PM. On Wednesday Paula worked by herself and finished the house by working until 7:12 P.M. How long, in minutes, was each day's lunch break?

$\textbf{(A)}\ 30\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 60$

Solution

Let Paula work at a rate of $p$, the two helpers work at a combined rate of $h$, and the time it takes to eat lunch be $L$, where $p$ and $h$ are in house/hours and L is in hours. Then the labor on Monday, Tuesday, and Wednesday can be represented by the three following equations:

\[(8-L)(p+h)=50\]

\[(6.2-L)h=24\]

\[(11.2-L)p=26\]

With three equations and three variables, we need to find the value of $L$. Adding the second and third equations together gives us $6.2h+11.2p-L(p+h)=50$. Subtracting the first equation from this new one gives us $-1.8h+3.2p=0$, so we get $h=\frac{16}{9}p$. Plugging into the second equation:

\[(6.2-L)\frac{16}{9}p=24\] \[(6.2-L)p=\frac{27}{2}\]

We can then subtract this from the third equation:

\[5p=26-\frac{27}{2}\] \[p=\frac{5}{2}\] Plugging $p$ into our third equation gives: \[L=\frac{4}{5}\]

Converting $L$ from hours to minutes gives us $L=48$ minutes, which is $\boxed{\textbf{(D)}\ 48}$.

Solution 2 (Modular Arithmetic)

Because Paula worked from \[8:00 \text{A.M.}\] to \[7:12 \text{P.M.}\], she worked for 11 hours and 12 minutes = 672 minutes. Since there is $100-50-24=26$% of the house left, we get the equation $26a=672$. Because $672$ is $22$ mod $26$, looking at our answer choices, the only answer that is $22$ $\text{mod}$ $26$ is $48$. So the answer is $\boxed{\textbf{(D)}\ 48}$.

Solution 3 (GCD)

We factor out the equations to be $(8-n)\left(\frac{1}{p}+\frac{1}{h}\right)=\frac{1}{2},\left(\frac{31}{5}-n\right)\left(\frac{1}{h}\right)=\frac{6}{25} \text{ and } \left(\frac{56}{5}-n\right)\left(\frac{1}{p}=\frac{13}{50}\right)$, where n is the number of hours for the break, p is the time Paula requires, and h is the time her helpers require. We find that when we select $\mathbf{D}$, we have them being $10.4$ and $5.4$, which correspond to being multiples of 13 and 6. Checking, we find that this satisfies the first equation, so multiplying $0.8\cdot 60= \boxed{48.}$

See Also

2012 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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
2012 AMC 12A (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 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png