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2000 CEMC Gauss (Grade 8) Problems/Problem 10: Difference between revisions

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{{Duplicate|[[2000 CEMC Gauss (Grade 8) Problems|2000 CEMC Gauss (Grade 8) #10]] and [[2000 CEMC Gauss (Grade 7) Problems|2000 CEMC Gauss (Grade 7) #13]]}}
==Problem==   
==Problem==   
Karl had his salary reduced by <math>10\%</math>. He was later promoted and his salary was increased by <math>10\%</math>. If his initial salary was <math>\$20~000</math>, what is his present salary?
Karl had his salary reduced by <math>10\%</math>. He was later promoted and his salary was increased by <math>10\%</math>. If his initial salary was <math>\$20~000</math>, what is his present salary?
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~anabel.disher
~anabel.disher
{{CEMC box|year=2000|competition=Gauss (Grade 8)|num-b=9|num-a=11}}
{{CEMC box|year=2000|competition=Gauss (Grade 7)|num-b=12|num-a=14}}

Latest revision as of 12:54, 20 October 2025

The following problem is from both the 2000 CEMC Gauss (Grade 8) #10 and 2000 CEMC Gauss (Grade 7) #13, so both problems redirect to this page.

Problem

Karl had his salary reduced by $10\%$. He was later promoted and his salary was increased by $10\%$. If his initial salary was $\$20~000$, what is his present salary?

$\text{ (A) }\ \$16~200 \qquad\text{ (B) }\ \$19~800 \qquad\text{ (C) }\ \$20~000 \qquad\text{ (D) }\ \$20~500 \qquad\text{ (E) }\ \$24~000$

Solution 1

If Karl's salary was reduced by $10\%$, that means that $90\%$ of the initial salary was remaining before the promotion. Thus, he had $90\% \times \$20~000 = \frac{9}{10} \times \$20~000 = \$18~000$ left.

After his promotion, this was then raised by $10\%$, so his salary is $110\%$ times what it was before he got promoted.

$\$18 000 \times 110\% = \$18~000 \times \frac{11}{10} = \boxed {\textbf {(B) } \$19~800}$

~anabel.disher

Solution 2

Let $x$ be the initial salary. After it was reduced, he had $\frac{9}{10}x$ for his salary.

After his promotion, he then had $\frac{9}{10}x \times \frac{11}{10} = \frac{99}{100}x$ for the salary.

Since we know the initial salary was $\$20~000$, we can plug-in $x = \$20~000$. We then have:

$\frac{99}{100} \times \$20~000 = \boxed {\textbf {(B) } \$19~800}$

~anabel.disher

2000 CEMC Gauss (Grade 8) (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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
CEMC Gauss (Grade 8)
2000 CEMC Gauss (Grade 7) (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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
CEMC Gauss (Grade 7)