1999 CEMC Pascal Problems/Problem 5

Revision as of 18:43, 29 June 2025 by Anabel.disher (talk | contribs) (Created page with "==Problem== If <math>10\%</math> of <math>400</math> is decreased by <math>25</math>, the result is <math> \text{ (A) }\ 15 \qquad\text{ (B) }\ 37.5 \qquad\text{ (C) }\ 65 \q...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem

If $10\%$ of $400$ is decreased by $25$, the result is

$\text{ (A) }\ 15 \qquad\text{ (B) }\ 37.5 \qquad\text{ (C) }\ 65 \qquad\text{ (D) }\ 260 \qquad\text{ (E) }\ 3975$

Solution

Setting up an equation, we have:

$10\% \times 400 - 25$

$=0.1 \times 400 - 25$

$=40 - 25$

$=\boxed {\textbf {(A) } 15}$

~anabel.disher

Solution 2 (answer choices)

We can see that $10\%$ of $400$ and $25$ are integers, so $37.5$ doesn't make sense, and $10\%$ of $400$ is $40$, which is well below the other answer choices except for answer choice A.

Thus, the answer is $\boxed {\textbf {(A) } 15}$

~anabel.disher