Difference between revisions of "2020 AMC 8 Problems/em 1"
MRENTHUSIASM (talk | contribs) (Created page with "Since the amount of sugar is twice the amount of lemon juice, Luka uses <math>3\cdot2=6</math> cups of sugar. Since the amount of water is <math>4</math> times the amount of...") |
MRENTHUSIASM (talk | contribs) |
||
| Line 1: | Line 1: | ||
| + | ==Problem== | ||
| + | Four friends do yardwork for their neighbors over the weekend, earning <math>\$15, \$20, \$25,</math> and <math>\$40,</math> respectively. They decide to split their earnings equally among themselves. In total how much will the friend who earned <math>\$40</math> give to the others? | ||
| + | |||
| + | ==Solution== | ||
Since the amount of sugar is twice the amount of lemon juice, Luka uses <math>3\cdot2=6</math> cups of sugar. | Since the amount of sugar is twice the amount of lemon juice, Luka uses <math>3\cdot2=6</math> cups of sugar. | ||
Since the amount of water is <math>4</math> times the amount of sugar, he uses <math>6\cdot4=\boxed{\text{(E) }24}</math> cups of water. | Since the amount of water is <math>4</math> times the amount of sugar, he uses <math>6\cdot4=\boxed{\text{(E) }24}</math> cups of water. | ||
| + | |||
| + | ==See also== | ||
| + | {{AMC8 box|year=2020|before = First Problem|num-a=2}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 09:47, 12 February 2021
Problem
Four friends do yardwork for their neighbors over the weekend, earning 
and 
respectively. They decide to split their earnings equally among themselves. In total how much will the friend who earned 
give to the others?
Solution
Since the amount of sugar is twice the amount of lemon juice, Luka uses 
cups of sugar.
Since the amount of water is 
times the amount of sugar, he uses 
cups of water.
See also
| 2020 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by First Problem |
Followed by Problem 2 | |
| 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. 
