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

(Created page with "Average the differences between the two times. 10, -10, 20, 30, -20 The sum is 30. Divide by 5 = 6.")
 
 
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Average the differences between the two times.
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==Problem==
10, -10, 20, 30, -20
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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?
The sum is 30.  Divide by 5 = 6.
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<asy>
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size(300);
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real i;
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defaultpen(linewidth(0.8));
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draw((0,140)--origin--(220,0));
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for(i=1;i<13;i=i+1) {
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draw((0,10*i)--(220,10*i));
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}
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label("$0$",origin,W);
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label("$20$",(0,20),W);
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label("$40$",(0,40),W);
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label("$60$",(0,60),W);
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label("$80$",(0,80),W);
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label("$100$",(0,100),W);
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label("$120$",(0,120),W);
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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;
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fill(MonD,grey);
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fill(MonL,lightgrey);
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fill(TuesD,grey);
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fill(TuesL,lightgrey);
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fill(WedD,grey);
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fill(WedL,lightgrey);
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fill(ThurD,grey);
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fill(ThurL,lightgrey);
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fill(FriD,grey);
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fill(FriL,lightgrey);
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draw(MonD^^MonL^^TuesD^^TuesL^^WedD^^WedL^^ThurD^^ThurL^^FriD^^FriL);
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label("M",(30,-5),S);
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label("Tu",(70,-5),S);
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label("W",(110,-5),S);
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label("Th",(150,-5),S);
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label("F",(190,-5),S);
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label("M",(-25,85),W);
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label("I",(-27,75),W);
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label("N",(-25,65),W);
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label("U",(-25,55),W);
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label("T",(-25,45),W);
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label("E",(-25,35),W);
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label("S",(-26,25),W);</asy>
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<math> \textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 12 </math>
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==Solution==
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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>
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==See Also==
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{{AMC8 box|year=2011|num-b=10|num-a=12}}
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{{MAA Notice}}

Latest revision as of 12:42, 25 January 2021

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 $\boxed{\textbf{(A)}\ 6}$

See Also

2011 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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 AJHSME/AMC 8 Problems and Solutions

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