Difference between revisions of "2020 AMC 8 Problems/Problem 11"
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==Solution 1== | ==Solution 1== | ||
+ | Naomi travels <math>6</math> miles in a time of <math>10</math> minutes, which is equivalent to <math>\dfrac{1}{6}</math> of an hour. Since <math>\text{speed} = \frac{\text{distance}}{\text{time}}</math>, her speed is <math>\frac{6}{\left(\frac{1}{6}\right)} = 36</math> mph. By a similar calculation, Maya's speed is <math>12</math> mph, so the answer is <math>36-12 = \boxed{\textbf{(E) }24}</math>. | ||
− | + | ==Solution 2 (Variant of Solution 1)== | |
− | + | Naomi's speed of <math>6</math> miles in <math>10</math> minutes is equivalent to <math>6 \cdot 6 = 36</math> miles per hour, while Maya's speed of <math>6</math> miles in <math>30</math> minutes (i.e. half an hour) is equivalent to <math>6 \cdot 2 = 12</math> miles per hour. The difference is consequently <math>36-12=\boxed{\textbf{(E) }24}</math>. | |
− | |||
− | + | ==Video Solution by NiuniuMaths (Easy to understand!)== | |
+ | https://www.youtube.com/watch?v=bHNrBwwUCMI | ||
− | + | ~NiuniuMaths | |
− | + | ==Video Solution by Math-X (First understand the problem!!!)== | |
+ | https://youtu.be/UnVo6jZ3Wnk?si=MfI4I15mLBXNvGMm&t=1375 | ||
− | - | + | ~Math-X |
− | ==Solution | + | ==Video Solution (CLEVER MANIPULATIONS!!!)== |
+ | https://youtu.be/gQrRdpw_tu8 | ||
− | + | ~Education, the Study of Everything | |
− | ==Solution | + | ==Video Solution by North America Math Contest Go Go Go== |
− | |||
− | |||
− | ==See also== {{AMC8 box|year=2020|num-b=10|num-a=12}} {{MAA Notice}} | + | https://www.youtube.com/watch?v=ND0y051eYm0 |
+ | |||
+ | ~North America Math Contest Go Go Go | ||
+ | |||
+ | ==Video Solution by WhyMath== | ||
+ | https://youtu.be/y__IHWpXprY | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/xjwDsaRE_Wo | ||
+ | |||
+ | ==Video Solution by Interstigation== | ||
+ | https://youtu.be/YnwkBZTv5Fw?t=456 | ||
+ | |||
+ | ~Interstigation | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | ==See also== {{AMC8 box|year=2020|num-b=10|num-a=12}} | ||
+ | {{MAA Notice}} |
Latest revision as of 16:34, 26 January 2024
Contents
- 1 Problem 11
- 2 Solution 1
- 3 Solution 2 (Variant of Solution 1)
- 4 Video Solution by NiuniuMaths (Easy to understand!)
- 5 Video Solution by Math-X (First understand the problem!!!)
- 6 Video Solution (CLEVER MANIPULATIONS!!!)
- 7 Video Solution by North America Math Contest Go Go Go
- 8 Video Solution by WhyMath
- 9 Video Solution
- 10 Video Solution by Interstigation
- 11 See also
Problem 11
After school, Maya and Naomi headed to the beach, miles away. Maya decided to bike while Naomi took a bus. The graph below shows their journeys, indicating the time and distance traveled. What was the difference, in miles per hour, between Naomi's and Maya's average speeds?
Solution 1
Naomi travels miles in a time of
minutes, which is equivalent to
of an hour. Since
, her speed is
mph. By a similar calculation, Maya's speed is
mph, so the answer is
.
Solution 2 (Variant of Solution 1)
Naomi's speed of miles in
minutes is equivalent to
miles per hour, while Maya's speed of
miles in
minutes (i.e. half an hour) is equivalent to
miles per hour. The difference is consequently
.
Video Solution by NiuniuMaths (Easy to understand!)
https://www.youtube.com/watch?v=bHNrBwwUCMI
~NiuniuMaths
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/UnVo6jZ3Wnk?si=MfI4I15mLBXNvGMm&t=1375
~Math-X
Video Solution (CLEVER MANIPULATIONS!!!)
~Education, the Study of Everything
Video Solution by North America Math Contest Go Go Go
https://www.youtube.com/watch?v=ND0y051eYm0
~North America Math Contest Go Go Go
Video Solution by WhyMath
~savannahsolver
Video Solution
Video Solution by Interstigation
https://youtu.be/YnwkBZTv5Fw?t=456
~Interstigation
See also
2020 AMC 8 (Problems • Answer Key • Resources) | ||
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.