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2009 CEMC Gauss (Grade 8) Problems/Problem 17: Difference between revisions

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{{Duplicate|[[2009 CEMC Gauss (Grade 8) Problems|2009 CEMC Gauss (Grade 8) #17]] and [[2009 CEMC Gauss (Grade 7) Problems|2009 CEMC Gauss (Grade 7) #19]]}}
==Problem==
==Problem==
A jar contains quarters (worth <math>\$0.25</math> each), nickels (worth <math>\$0.05</math> each), and pennies (worth <math>\$0.01</math> each).  The value of the quarters is <math>\$10.00</math>. The value of the nickels is <math>\$10.00</math>. The value of the pennies is <math>\$10.00</math>. If Judith randomly chooses one coin from the jar, what is the probability that it is a quarter?
A jar contains quarters (worth <math>\$0.25</math> each), nickels (worth <math>\$0.05</math> each), and pennies (worth <math>\$0.01</math> each).  The value of the quarters is <math>\$10.00</math>. The value of the nickels is <math>\$10.00</math>. The value of the pennies is <math>\$10.00</math>. If Judith randomly chooses one coin from the jar, what is the probability that it is a quarter?
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~anabel.disher
~anabel.disher
{{CEMC box|year=2009|competition=Gauss (Grade 8)|num-b=16|num-a=18}}
{{CEMC box|year=2009|competition=Gauss (Grade 7)|num-b=18|num-a=20}}

Latest revision as of 20:27, 19 October 2025

The following problem is from both the 2009 CEMC Gauss (Grade 8) #17 and 2009 CEMC Gauss (Grade 7) #19, so both problems redirect to this page.

Problem

A jar contains quarters (worth $\$0.25$ each), nickels (worth $\$0.05$ each), and pennies (worth $\$0.01$ each). The value of the quarters is $\$10.00$. The value of the nickels is $\$10.00$. The value of the pennies is $\$10.00$. If Judith randomly chooses one coin from the jar, what is the probability that it is a quarter?

$\text{ (A) }\ \frac{25}{31} \qquad\text{ (B) }\ \frac{1}{31} \qquad\text{ (C) }\ \frac{1}{3} \qquad\text{ (D) }\ \frac{5}{248} \qquad\text{ (E) }\ \frac{1}{30}$

Solution

Using the total value of the coins divided by how much each coin is worth, we can find out how many quarters, nickels, and pennies there are. This would allow us to find the probability of picking a quarter by dividing the number of quarters by the total number of coins.

There are $\frac{10.00}{0.25} = \frac{10.00 \times 4}{0.25 \times 4} = \frac{40.00}{1} = 40$ quarters.

There are $\frac{10.00}{0.05} = \frac{10.00 \times 20}{0.05 \times 20} = \frac{200.00}{1} = 200$ nickels.

There are $\frac{10.00}{0.01} = \frac{10.00 \times 100}{0.01 \times 100} = \frac{1000.00}{1} = 1000$ pennies.

This means there are $40 + 200 + 1000 = 1240$ coins altogether.

Dividing the number of quarters by the total number of coins, we get:

$\frac{40}{1240} = \frac{1}{31}$

~anabel.disher

2009 CEMC Gauss (Grade 8) (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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)
2009 CEMC Gauss (Grade 7) (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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)
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