Difference between revisions of "2003 AMC 10B Problems/Problem 5"

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==Problem==
 
==Problem==
  
Moe uses a mower to cut his rectangular <math>90</math>-foot by <math>150</math>-foot lawn. The swath he cuts is <math>28</math> inches wide, but he overlaps each cut by <math>4</math> inches to make sure that no grass is missed. He walks at the rate of <math>5000</math> feet per hour while pushing the mower. Which of the following is closest to the number of hours it will take Moe to mow the lawn.
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Moe uses a mower to cut his rectangular <math>90</math>-foot by <math>150</math>-foot lawn. The swath he cuts is <math>28</math> inches wide, but he overlaps each cut by <math>4</math> inches to make sure that no grass is missed. He walks at the rate of <math>5000</math> feet per hour while pushing the mower. Which of the following is closest to the number of hours it will take Moe to mow the lawn?
  
 
<math>\textbf{(A) } 0.75 \qquad\textbf{(B) } 0.8 \qquad\textbf{(C) } 1.35 \qquad\textbf{(D) } 1.5 \qquad\textbf{(E) } 3 </math>
 
<math>\textbf{(A) } 0.75 \qquad\textbf{(B) } 0.8 \qquad\textbf{(C) } 1.35 \qquad\textbf{(D) } 1.5 \qquad\textbf{(E) } 3 </math>
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==Solution 2==
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==Solution 2 (very easy to understand)==
  
Let's assume that the swath moves back and forth; parallel to the 150 feet side. The length we cover in one strip of grass is <math>90</math> feet or <math>150</math> feet, but it doesn't matter, because in later calculations, we get the same result. Now we need to find out how many strips there are. In reality, the swath Moe mows is <math>24</math> inches wide, which can be easily translated into <math>2</math> feet. \frac{150}{2} is the number of strips Moe needs to mow, which is equal to <math>75</math>. We now compute
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Let's assume that the swath moves back and forth; parallel to the <math>90</math> feet side. Thus, the length of one strip is <math>90</math> feet. Now we need to find out how many strips there are. In reality, the swath Moe mows is <math>24</math> inches wide, which can be easily translated into <math>2</math> feet. <math>\frac{150}{2}</math> is the number of strips Moe needs to mow, which is equal to <math>75</math>.  
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Therefore, the total number of feet Moe mows is <math>75\times90</math>. Since Moe's mowing rate is <math>5000</math> feet per hour, <math>\frac{75\times90}{5000}</math> is the number of hours it takes him to do his job. Using basic calculations, we compute the answer. <math>\boxed{\textbf{(C) } 1.35}</math>.
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~sakshamsethi
  
 
==See Also==
 
==See Also==

Latest revision as of 16:23, 18 October 2023

The following problem is from both the 2003 AMC 12B #4 and 2003 AMC 10B #5, so both problems redirect to this page.

Problem

Moe uses a mower to cut his rectangular $90$-foot by $150$-foot lawn. The swath he cuts is $28$ inches wide, but he overlaps each cut by $4$ inches to make sure that no grass is missed. He walks at the rate of $5000$ feet per hour while pushing the mower. Which of the following is closest to the number of hours it will take Moe to mow the lawn?

$\textbf{(A) } 0.75 \qquad\textbf{(B) } 0.8 \qquad\textbf{(C) } 1.35 \qquad\textbf{(D) } 1.5 \qquad\textbf{(E) } 3$

Solution

Since the swath Moe actually mows is $24$ inches, or $2$ feet wide, he mows $10000$ square feet in one hour. His lawn has an area of $13500$, so it will take Moe $1.35$ hours to finish mowing the lawn. Thus the answer is $\boxed{\textbf{(C) } 1.35}$.


Solution 2 (very easy to understand)

Let's assume that the swath moves back and forth; parallel to the $90$ feet side. Thus, the length of one strip is $90$ feet. Now we need to find out how many strips there are. In reality, the swath Moe mows is $24$ inches wide, which can be easily translated into $2$ feet. $\frac{150}{2}$ is the number of strips Moe needs to mow, which is equal to $75$. Therefore, the total number of feet Moe mows is $75\times90$. Since Moe's mowing rate is $5000$ feet per hour, $\frac{75\times90}{5000}$ is the number of hours it takes him to do his job. Using basic calculations, we compute the answer. $\boxed{\textbf{(C) } 1.35}$.

~sakshamsethi

See Also

2003 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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
2003 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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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