Difference between revisions of "2003 AMC 8 Problems/Problem 7"
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===Solution 2=== | ===Solution 2=== | ||
The total point difference between Blake's abd Jenny's scores is <math> 10-10+20+20=40 </math>. The average of it is <math> \frac{40}{4}=\boxed{\mathrm{(A)}\ 10} </math>. | The total point difference between Blake's abd Jenny's scores is <math> 10-10+20+20=40 </math>. The average of it is <math> \frac{40}{4}=\boxed{\mathrm{(A)}\ 10} </math>. | ||
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+ | {{AMC8 box|year=2003|num-b=6|num-a=8}} |
Revision as of 08:50, 25 November 2011
Contents
[hide]Problem
Blake and Jenny each took four -point tests. Blake averaged on the four tests. Jenny scored points higher than Blake on the first test, points lower than him on the second test, and points higher on both the third and fourth tests. What is the difference between Jenny's average and Blake's average on these four tests?
Solution
Solution 1
Blake scored a total of points. Jenny scored points higher than Blake, so her average is . the difference is .
Solution 2
The total point difference between Blake's abd Jenny's scores is . The average of it is .
2003 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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 |