Difference between revisions of "1982 AHSME Problems/Problem 15"
m (→Problem) |
Arcticturn (talk | contribs) (→Solution) |
||
(2 intermediate revisions by the same user not shown) | |||
Line 11: | Line 11: | ||
\text{(D) }\text{ between 15 and 16}\qquad | \text{(D) }\text{ between 15 and 16}\qquad | ||
\text{(E) } 16.5 </math> | \text{(E) } 16.5 </math> | ||
+ | |||
+ | ==Solution== | ||
+ | We simply ignore the floor of <math>x</math>. Then, we have <math>y</math> = <math>2x + 3</math> = <math>3(x-2)+5</math>. Solving for <math>3x - 1 = 2x + 3</math>, we get <math>x = 4</math>. For the floor of <math>x</math>, we have <math>x</math> is between <math>4</math> and <math>5</math>. Plugging in <math>8</math> + <math>3</math> = <math>11</math> for <math>y</math>, we have <math>y = 11</math>. We have <math>11 + 4.x</math> = <math>\boxed {(D)}</math> | ||
+ | |||
+ | ~Arcticturn |
Revision as of 19:48, 21 October 2021
Problem
Let denote the greatest integer not exceeding . Let and satisfy the simultaneous equations
If is not an integer, then is
Solution
We simply ignore the floor of . Then, we have = = . Solving for , we get . For the floor of , we have is between and . Plugging in + = for , we have . We have =
~Arcticturn