Difference between revisions of "2011 AMC 8 Problems/Problem 11"

Latest revision as of 15:00, 17 December 2023

Problem

The graph shows the number of minutes studied by both Asha (black bar) and Sasha (grey bar) in one week. On the average, how many more minutes per day did Sasha study than Asha?

$[asy] size(300); real i; defaultpen(linewidth(0.8)); draw((0,140)--origin--(220,0)); for(i=1;i<13;i=i+1) { draw((0,10*i)--(220,10*i)); } label("$0$",origin,W); label("$20$",(0,20),W); label("$40$",(0,40),W); label("$60$",(0,60),W); label("$80$",(0,80),W); label("$100$",(0,100),W); label("$120$",(0,120),W); path MonD=(20,0)--(20,60)--(30,60)--(30,0)--cycle,MonL=(30,0)--(30,70)--(40,70)--(40,0)--cycle,TuesD=(60,0)--(60,90)--(70,90)--(70,0)--cycle,TuesL=(70,0)--(70,80)--(80,80)--(80,0)--cycle,WedD=(100,0)--(100,100)--(110,100)--(110,0)--cycle,WedL=(110,0)--(110,120)--(120,120)--(120,0)--cycle,ThurD=(140,0)--(140,80)--(150,80)--(150,0)--cycle,ThurL=(150,0)--(150,110)--(160,110)--(160,0)--cycle,FriD=(180,0)--(180,70)--(190,70)--(190,0)--cycle,FriL=(190,0)--(190,50)--(200,50)--(200,0)--cycle; fill(MonD,grey); fill(MonL,lightgrey); fill(TuesD,grey); fill(TuesL,lightgrey); fill(WedD,grey); fill(WedL,lightgrey); fill(ThurD,grey); fill(ThurL,lightgrey); fill(FriD,grey); fill(FriL,lightgrey); draw(MonD^^MonL^^TuesD^^TuesL^^WedD^^WedL^^ThurD^^ThurL^^FriD^^FriL); label("M",(30,-5),S); label("Tu",(70,-5),S); label("W",(110,-5),S); label("Th",(150,-5),S); label("F",(190,-5),S); label("M",(-25,85),W); label("I",(-27,75),W); label("N",(-25,65),W); label("U",(-25,55),W); label("T",(-25,45),W); label("E",(-25,35),W); label("S",(-26,25),W);[/asy]$

$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 12$

Solution

Average the differences between each day. We get $10, -10,\text{ } 20,\text{ } 30,-20$ . We find the average of this list to get $\boxed{\textbf{(A)}\ 6}$ . ( In case you were wondering, the way to calculate the average is $\frac{(10+(-10)+20+30+(-20))}{5} = \frac{ 30}{5} = 6$ . So the answer is indeed $\boxed{\textbf{(A)}\ 6}$ )

Solution 2

This solution may take longer to do than the first solution. In total, Asha studied for 400 minutes a week (80 minutes per day) and Sasha studied for 430 minutes a week (86 minutes per day). 86 - 80 = 6. Therefore, the answer is $\boxed{\textbf{(A)}\ 6}$ .

-MiracleMaths

Video Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=mYn6tNxrWBU

@@ Line 1: / Line 1: @@
+==Problem==
 The graph shows the number of minutes studied by both Asha (black bar) and Sasha (grey bar) in one week. On the average, how many more minutes per day did Sasha study than Asha?
 <asy>
 size(300);
@@ Line 41: / Line 43: @@
 <math> \textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 12 </math>
+==Solution==
+Average the differences between each day. We get <math>10, -10,\text{ } 20,\text{ } 30,-20</math>. We find the average of this list to get <math>\boxed{\textbf{(A)}\ 6}</math>. ( In case you were wondering, the way to calculate the average is <math>\frac{(10+(-10)+20+30+(-20))}{5} = \frac{ 30}{5} = 6</math>. So the answer is indeed  <math>\boxed{\textbf{(A)}\ 6}</math>)
+==Solution 2==
+This solution may take longer to do than the first solution. In total, Asha studied for 400 minutes a week (80 minutes per day) and Sasha studied for 430 minutes a week (86 minutes per day).  86 - 80 = 6.  Therefore, the answer is <math>\boxed{\textbf{(A)}\ 6}</math>.
+-MiracleMaths
+==Video Solution 1 by SpreadTheMathLove==
+https://www.youtube.com/watch?v=mYn6tNxrWBU
 ==See Also==
-{{AMC8 box|year=2010|num-b=24|after=Last Problem}}
+{{AMC8 box|year=2011|num-b=10|num-a=12}}
+{{MAA Notice}}

2011 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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