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 "2014 AMC 10A Problems/Problem 6"

(→‎Solution 4)
(→‎Video Solution (CREATIVE THINKING))
 
(12 intermediate revisions by 7 users not shown)
Line 9: Line 9:
 
==Solution 1==
 
==Solution 1==
  
−
We need to multiply <math>b</math> by <math>\frac{d}{a}</math> for the new cows and <math>\frac{e}{c}</math> for the new time, so the answer is <math>b\cdot \frac{d}{a}\cdot \frac{e}{c}=\frac{bde}{ac}</math>, or <math>\boxed{\textbf{(A)}}</math>.
+
We need to multiply <math>b</math> by <math>\frac{d}{a}</math> for the new cows and <math>\frac{e}{c}</math> for the new time, so the answer is <math>b\cdot \frac{d}{a}\cdot \frac{e}{c}=\frac{bde}{ac}</math>, or <math>\boxed{\textbf{(A) } \frac{bde}{ac}}</math>.
  
 
==Solution 2==
 
==Solution 2==
Line 19: Line 19:
  
 
==Solution 3==
 
==Solution 3==
−
We see that the the number of cows is inversely proportional to the number of days and directly proportional to the gallons of milk. So our constant is <math>\dfrac{ac}{b}</math>.
+
We see that the number of cows is inversely proportional to the number of days and directly proportional to the gallons of milk. So our constant is <math>\dfrac{ac}{b}</math>.
  
 
Let <math>g</math> be the answer to the question. We have <math>\dfrac{de}{g}=\dfrac{ac}{b}\implies gac=bde\implies g=\dfrac{bde}{ac}\implies\boxed{ \textbf{(A)}\ \frac{bde}{ac}}</math>
 
Let <math>g</math> be the answer to the question. We have <math>\dfrac{de}{g}=\dfrac{ac}{b}\implies gac=bde\implies g=\dfrac{bde}{ac}\implies\boxed{ \textbf{(A)}\ \frac{bde}{ac}}</math>
Line 26: Line 26:
 
The problem specifics "rate," so it would be wise to first find the rate at which cows produce milk. We can find the rate of work/production by finding the gallons produced by a single cow in a single day. To do this, we divide the amount produced by the number of cows and number of days <cmath>\implies\text{rate}=\dfrac{b}{ac}</cmath>
 
The problem specifics "rate," so it would be wise to first find the rate at which cows produce milk. We can find the rate of work/production by finding the gallons produced by a single cow in a single day. To do this, we divide the amount produced by the number of cows and number of days <cmath>\implies\text{rate}=\dfrac{b}{ac}</cmath>
  
−
Now that we have the gallons produced by a single cow in a single day, we simply multiply that rate by the number of cows and the number of days <cmath>\implies\dfrac{bde}{ac}\implies\boxed{\textbf{(A)}}</cmath>
+
Now that we have the gallons produced by a single cow in a single day, we simply multiply that rate by the number of cows and the number of days <cmath>\implies\boxed{\textbf{(A) } \dfrac{bde}{ac}}</cmath>
 +
==Solution 5==
 +
If <math>a</math> cows give <math>b</math> gallons of milk in <math>c</math> days, that means that one cow will give <math>\frac{b}{a}</math> gallons of milk in <math>c</math> days. Also, we want to find the number of gallons of milk will <math>d</math> cows give in <math>e</math> days, so in <math>\frac{e}{c}</math> days <math>d</math> cows give <math>\frac{bd}{a}</math> gallons of milk. Multiplying with the formula <math>d=rt</math>, we get <math>\boxed{\textbf{(A) } \dfrac{bde}{ac}}</math>
 +
 
 +
==Solution 6==
 +
 
 +
In the right formula, plugging in <math>d=a</math> and <math>e=c</math> should simplify to <math>b</math>, as if it doesn't, we'd essentially be saying "<math>a</math> cows give <math>b</math> gallons of milk in <math>c</math> days, but <math>a</math> cows don't give <math>b</math> gallons of milk in <math>c</math> days." The only one of the answer choices that simplifies like this is <math>\boxed{\textbf{(A)}}</math>
 +
 
 +
==Solution 7==
 +
 
 +
Call the rate at which a single cow produces milk <math>r</math>. Using the given information we get that <math>rac=b</math>, and we want to find <math>rde</math>. We solve for <math>r</math> in the first equation to get <math>r=b \cdot \tfrac{1}{ac}</math>. Plugging in this value for <math>r</math> into the expression we want, we get that <math>rde=\frac{bde}{ac}</math>, or <math>\boxed{\textbf{(A)}}</math>.
 +
 
 +
==Video Solution (CREATIVE THINKING)==
 +
https://youtu.be/4THtSkiSGMg
 +
 
 +
~Education, the Study of Everything
 +
 
 +
==Video Solution==
 +
https://youtu.be/OW4rHUTPgPA
 +
 
 +
~savannahsolver
  
 
==See Also==
 
==See Also==

Latest revision as of 23:09, 26 June 2023

The following problem is from both the 2014 AMC 12A #4 and 2014 AMC 10A #6, so both problems redirect to this page.

Problem

Suppose that $a$ cows give $b$ gallons of milk in $c$ days. At this rate, how many gallons of milk will $d$ cows give in $e$ days?

$\textbf{(A)}\ \frac{bde}{ac}\qquad\textbf{(B)}\ \frac{ac}{bde}\qquad\textbf{(C)}\ \frac{abde}{c}\qquad\textbf{(D)}\ \frac{bcde}{a}\qquad\textbf{(E)}\ \frac{abc}{de}$

Solution 1

We need to multiply $b$ by $\frac{d}{a}$ for the new cows and $\frac{e}{c}$ for the new time, so the answer is $b\cdot \frac{d}{a}\cdot \frac{e}{c}=\frac{bde}{ac}$, or $\boxed{\textbf{(A) } \frac{bde}{ac}}$.

Solution 2

We plug in $a=2$, $b=3$, $c=4$, $d=5$, and $e=6$. Hence the question becomes "2 cows give 3 gallons of milk in 4 days. How many gallons of milk do 5 cows give in 6 days?"

If 2 cows give 3 gallons of milk in 4 days, then 2 cows give $\frac{3}{4}$ gallons of milk in 1 day, so 1 cow gives $\frac{3}{4\cdot2}$ gallons in 1 day. This means that 5 cows give $\frac{5\cdot3}{4\cdot2}$ gallons of milk in 1 day. Finally, we see that 5 cows give $\frac{5\cdot3\cdot6}{4\cdot2}$ gallons of milk in 6 days. Substituting our values for the variables, this becomes $\frac{dbe}{ac}$, which is $\boxed{\textbf{(A)}\ \frac{bde}{ac}}$.

Solution 3

We see that the number of cows is inversely proportional to the number of days and directly proportional to the gallons of milk. So our constant is $\dfrac{ac}{b}$.

Let $g$ be the answer to the question. We have $\dfrac{de}{g}=\dfrac{ac}{b}\implies gac=bde\implies g=\dfrac{bde}{ac}\implies\boxed{ \textbf{(A)}\ \frac{bde}{ac}}$

Solution 4

The problem specifics "rate," so it would be wise to first find the rate at which cows produce milk. We can find the rate of work/production by finding the gallons produced by a single cow in a single day. To do this, we divide the amount produced by the number of cows and number of days \[\implies\text{rate}=\dfrac{b}{ac}\]

Now that we have the gallons produced by a single cow in a single day, we simply multiply that rate by the number of cows and the number of days \[\implies\boxed{\textbf{(A) } \dfrac{bde}{ac}}\]

Solution 5

If $a$ cows give $b$ gallons of milk in $c$ days, that means that one cow will give $\frac{b}{a}$ gallons of milk in $c$ days. Also, we want to find the number of gallons of milk will $d$ cows give in $e$ days, so in $\frac{e}{c}$ days $d$ cows give $\frac{bd}{a}$ gallons of milk. Multiplying with the formula $d=rt$, we get $\boxed{\textbf{(A) } \dfrac{bde}{ac}}$

Solution 6

In the right formula, plugging in $d=a$ and $e=c$ should simplify to $b$, as if it doesn't, we'd essentially be saying "$a$ cows give $b$ gallons of milk in $c$ days, but $a$ cows don't give $b$ gallons of milk in $c$ days." The only one of the answer choices that simplifies like this is $\boxed{\textbf{(A)}}$

Solution 7

Call the rate at which a single cow produces milk $r$. Using the given information we get that $rac=b$, and we want to find $rde$. We solve for $r$ in the first equation to get $r=b \cdot \tfrac{1}{ac}$. Plugging in this value for $r$ into the expression we want, we get that $rde=\frac{bde}{ac}$, or $\boxed{\textbf{(A)}}$.

Video Solution (CREATIVE THINKING)

https://youtu.be/4THtSkiSGMg

~Education, the Study of Everything

Video Solution

https://youtu.be/OW4rHUTPgPA

~savannahsolver

See Also

2014 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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
2014 AMC 12A (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

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=2014_AMC_10A_Problems/Problem_6&oldid=194836"