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Difference between revisions of "1999 AMC 8 Problems/Problem 16"

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To get a <math>60\%</math> on her test overall, she needed to get <math>60\% \cdot 75 = 0.60 \cdot 75 = 45</math> questions right.
 
To get a <math>60\%</math> on her test overall, she needed to get <math>60\% \cdot 75 = 0.60 \cdot 75 = 45</math> questions right.
  
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Therefore, she needed to answer <math>45 - 40 = 5</math> more questions to pass, and the the correct answer is <math>\boxed{B}</math>
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Therefore, she needed to answer <math>45 - 40 = 5</math> more questions to pass, so the correct answer is <math>\boxed{(B) 5}</math>
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==Video Solution==
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https://youtu.be/3sBMI9cVhNs Soo, DRMS, NM
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==See Also==
 
==See Also==
 
{{AMC8 box|year=1999|num-b=15|num-a=17}}
 
{{AMC8 box|year=1999|num-b=15|num-a=17}}
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{{MAA Notice}}

Latest revision as of 19:30, 11 January 2024

Problem

Tori's mathematics test had 75 problems: 10 arithmetic, 30 algebra, and 35 geometry problems. Although she answered 70% of the arithmetic, 40% of the algebra, and 60% of the geometry problems correctly, she did not pass the test because she got less than 60% of the problems right. How many more problems would she have needed to answer correctly to earn a 60% passing grade?

$\text{(A)}\ 1 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 11$

Solution

First, calculate how many of each type of problem she got right:

Arithmetic: $70\% \cdot 10 = 0.70 \cdot 10 = 7$

Algebra: $40\% \cdot 30 = 0.40 \cdot 30 = 12$

Geometry: $60\% \cdot 35 = 0.60 \cdot 35 = 21$

Altogether, Tori answered $7 + 12 + 21 = 40$ questions correct. To get a $60\%$ on her test overall, she needed to get $60\% \cdot 75 = 0.60 \cdot 75 = 45$ questions right.

Therefore, she needed to answer $45 - 40 = 5$ more questions to pass, so the correct answer is $\boxed{(B) 5}$

Video Solution

https://youtu.be/3sBMI9cVhNs Soo, DRMS, NM

See Also

1999 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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