2018 AMC 8

Complete problem set with solutions and individual problem pages

Problem 13 Medium

Laila took five math tests, each worth a maximum of 100 points. Laila's score on each test was an integer between 0 and 100, inclusive. Laila received the same score on the first four tests, and she received a higher score on the last test. Her average score on the five tests was 82. How many values are possible for Laila's score on the last test?

  • A.

    4

  • B.

    5

  • C.

    9

  • D.

    10

  • E.

    18

Answer:A

Solution 1

Say Laila gets a value of x on her first 4 tests, and a value of y on her last test. Thus, 4x+y=82 \cdot 5=410.

Because x and y are different, x must be less than 82 and y must be greater than 82. When x decreases by 1, y must increase by 4 to keep the total constant.

The greatest value for y is 98 (as y=100 would make x non-integer). In the range 83\leqslant y\leqslant 98, only 4 values for y result in integer values for x: 86, 90, 94 and 98. Thus, the answer is \boxed{\textbf{(A) }4}.

 

Solution 2

The average score is 82 which leads us to suppose that Laila got all 82 points for the tests. We know that Laila got the same points in the first four tests and they are all lower than the last test. Let the first four tests be 81 points, then the last test must be 86 points to keep the average fixed. Continue to decrement the first four tests to identify other possible combinations. The possible points for the fifth test are 86, 90, 94, 98. The answer is \boxed{\textbf{(A) }4}.