Difference between revisions of "2020 AMC 8 Problems/Problem 5"
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==Solution 3== | ==Solution 3== | ||
− | + | Notice that each cup receives <math>\frac 34 \cdot \frac 15=\frac{3}{20}=\frac{15}{100}</math> | |
Notice that each cup receives <math>\frac 34 \cdot \frac 15=\frac{3}{20}=\frac{15}{100}</math> of the entire pitcher which is <math>\textbf{(C) }15</math> percent. | Notice that each cup receives <math>\frac 34 \cdot \frac 15=\frac{3}{20}=\frac{15}{100}</math> of the entire pitcher which is <math>\textbf{(C) }15</math> percent. | ||
Revision as of 18:59, 19 November 2020
Problem 5
Three fourths of a pitcher is filled with pineapple juice. The pitcher is emptied by pouring an equal amount of juice into each of cups. What percent of the total capacity of the pitcher did each cup receive?
Solution
To equally distribute to cups, we will simply divide
by
Simplifying, we get:
Converting that into a percent, we get an answer of
Solution 2
Assume that the pitcher has a total capacity of ounces. Since the pitcher is filled three fourths with pineapple juice, it follows that it contains
ounces of pineapple juice. The pineapple juice is then divided equally into 5 cups, which means that each cup will contain
ounces of pineapple juice. Since the total capacity of the pitcher was
ounces, it follows that each cup received
of the total capacity of the pitcher
.
~junaidmansuri
Solution 3
Notice that each cup receives
Notice that each cup receives
of the entire pitcher which is
percent.
-franzliszt
Solution 4
In the problem, it states that the pitcher is , or
full. So, we can just divide this by
to get
, which means that the answer is
~aaja3427
.
Video Solution
~savannahsolver
See also
2020 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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.