Difference between revisions of "1995 AHSME Problems/Problem 1"

Revision as of 22:12, 9 January 2008

Problem

Kim earned scores of 87,83, and 88 on her first three mathematics examinations. If Kim receives a score of 90 on the fourth exam, then her average will

$\mathrm{(A) \ \text{remain the same} } \qquad \mathrm{(B) \ \text{increase by 1} } \qquad \mathrm{(C) \ \text{increase by 2} } \qquad \mathrm{(D) \ \text{increase by 3} } \qquad \mathrm{(E) \ \text{increase by 4} }$

Solution

The average of the first three test scores is $\frac{88+83+87}{3}=86$ . The average of all four exams is $\frac{87+83+88+90}{4}=87$ . It increased by one point. $\mathrm{(B)}$

Page

Toolbox

Search

Difference between revisions of "1995 AHSME Problems/Problem 1"

Revision as of 22:12, 9 January 2008

Problem

Solution

See also

Revision as of 12:31, 8 January 2008 (view source) 1=2 (talk \| contribs) (→‎See Also) ← Older edit	Revision as of 22:12, 9 January 2008 (view source) Azjps (talk \| contribs) m (1995 AMC 12 Problems/Problem 1 moved to 1995 AHSME Problems/Problem 1) Newer edit →
(No difference)