Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2008 AMC 10B Problems/Problem 18"

(→‎Solution 1)
m (→‎Solution)
 
(22 intermediate revisions by 8 users not shown)
Line 1: Line 1:
 
==Problem==
 
==Problem==
−
Bricklayer Brenda would take nine hours to build a chimney alone, and bricklayer Brandon would take <math>10</math> hours to build it alone. When they work together, they talk a lot, and their combined output decreases by <math>10</math> bricks per hour. Working together, they build the chimney in <math>5</math> hours. How many bricks are in the chimney?
+
Bricklayer Brenda takes <math>9</math> hours to build a chimney alone, and bricklayer Brandon takes <math>10</math> hours to build it alone. When they work together, they talk a lot, and their combined output decreases by <math>10</math> bricks per hour. Working together, they build the chimney in <math>5</math> hours. How many bricks are in the chimney?
  
 
<math>\mathrm{(A)}\ 500\qquad\mathrm{(B)}\ 900\qquad\mathrm{(C)}\ 950\qquad\mathrm{(D)}\ 1000\qquad\mathrm{(E)}\ 1900</math>
 
<math>\mathrm{(A)}\ 500\qquad\mathrm{(B)}\ 900\qquad\mathrm{(C)}\ 950\qquad\mathrm{(D)}\ 1000\qquad\mathrm{(E)}\ 1900</math>
  
 
==Solution==
 
==Solution==
−
Let <math>x</math> be the number of bricks in the chimney. Using <math>distance = rate \cdot time</math>, we get
+
Let <math>x</math> be the number of bricks in the chimney. The work done is the rate multiplied by the time.
−
<math>x = (x/9 + x/10 - 10)\cdot(5)</math>. Solving for <math>x</math>, we get <math>\boxed{900}</math>.
+
 
 +
Using <math>w = rt</math>, we get <math>x = \left(\frac{x}{9} + \frac{x}{10} - 10\right)\cdot(5)</math>. Solving for <math>x</math>, we get <math>900</math>, or <math>\boxed{B}</math>.
  
 
==See also==
 
==See also==
 
{{AMC10 box|year=2008|ab=B|num-b=17|num-a=19}}
 
{{AMC10 box|year=2008|ab=B|num-b=17|num-a=19}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 18:39, 3 November 2024

Problem

Bricklayer Brenda takes $9$ hours to build a chimney alone, and bricklayer Brandon takes $10$ hours to build it alone. When they work together, they talk a lot, and their combined output decreases by $10$ bricks per hour. Working together, they build the chimney in $5$ hours. How many bricks are in the chimney?

$\mathrm{(A)}\ 500\qquad\mathrm{(B)}\ 900\qquad\mathrm{(C)}\ 950\qquad\mathrm{(D)}\ 1000\qquad\mathrm{(E)}\ 1900$

Solution

Let $x$ be the number of bricks in the chimney. The work done is the rate multiplied by the time.

Using $w = rt$, we get $x = \left(\frac{x}{9} + \frac{x}{10} - 10\right)\cdot(5)$. Solving for $x$, we get $900$, or $\boxed{B}$.

See also

2008 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_10B_Problems/Problem_18&oldid=231419"