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 | + | 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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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>. | 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>. | ||
| − | Therefore, the total number of feet Moe mows is <math>75 | + | 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 
-foot by 
-foot lawn. The swath he cuts is 
inches wide, but he overlaps each cut by 
inches to make sure that no grass is missed. He walks at the rate of 
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?
Solution
Since the swath Moe actually mows is 
inches, or 
feet wide, he mows 
square feet in one hour. His lawn has an area of 
, so it will take Moe 
hours to finish mowing the lawn. Thus the answer is 
.
Solution 2 (very easy to understand)
Let's assume that the swath moves back and forth; parallel to the feet side. Thus, the length of one strip is feet. Now we need to find out how many strips there are. In reality, the swath Moe mows is inches wide, which can be easily translated into feet. 
is the number of strips Moe needs to mow, which is equal to 
.
Therefore, the total number of feet Moe mows is 
. Since Moe's mowing rate is feet per hour, 
is the number of hours it takes him to do his job. Using basic calculations, we compute the answer. .
~sakshamsethi
See Also
| 2003 AMC 12B (Problems • Answer Key • Resources) | |
| 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 (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
