Difference between revisions of "2012 CEMC Gauss (Grade 8) Problems/Problem 5"

Latest revision as of 20:55, 18 October 2025

How many more coins does it take to make one dollar (100) using only nickels(5 coins) than it takes to make one dollar using only dimes(10 coins)?

$\text{ (A) }\ 15\qquad\text{ (B) }\ 10\qquad\text{ (C) }\ 75\qquad\text{ (D) }\ 64\qquad\text{ (E) }\ 54$

Since a dollar is $100$ cents and a nickel is $5$ cents, there are $\frac{100}{5} = 20$ nickels in a dollar.

Using this logic for dimes and the fact that a dime is $10$ cents, there are $\frac{100}{10} = 10$ dimes in a dollar.

Thus, it takes $20 - 10 = \boxed{\textbf{(B) } 10}$ more coins to make a dollar using only nickels than it takes to make one dollar using only dimes.

~anabel.disher

2012 CEMC Gauss (Grade 8) (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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)

@@ Line 1: / Line 1: @@
 ==Problem==
-If <math>\frac{8}{12} = \frac{\framebox {}}{3}</math>, then the value represented by <math>\framebox {}</math> is
+How many more coins does it take to make one dollar (100) using only nickels(5 coins) than it takes to make one dollar using only dimes(10 coins)?<!--Note: this is how the question was written-->
-<math>  \text{ (A) }\  24\qquad\text{ (B) }\ 1\qquad\text{ (C) }\ 12\qquad\text{ (D) }\ 2\qquad\text{ (E) }\ 4 </math>
+<math>  \text{ (A) }\  15\qquad\text{ (B) }\ 10\qquad\text{ (C) }\ 75\qquad\text{ (D) }\ 64\qquad\text{ (E) }\ 54 </math>
 ==Solution==
-<math>\frac{12}{3} = 4</math>, meaning we can divide the numerator and denominator by <math>4</math> to arrive at our answer.
+Since a dollar is <math>100</math> cents and a nickel is <math>5</math> cents, there are <math>\frac{100}{5} = 20</math> nickels in a dollar.
-<math>\frac{8}{12} = \frac{8 \div 4}{12 \div 4} = \frac{2}{3}</math>
+Using this logic for dimes and the fact that a dime is <math>10</math> cents, there are <math>\frac{100}{10} = 10</math> dimes in a dollar.
-Since the numerator is <math>2</math>, the answer is <math>\boxed {\textbf {(D) }2}</math>.
+Thus, it takes <math>20 - 10 = \boxed{\textbf{(B) } 10}</math> more coins to make a dollar using only nickels than it takes to make one dollar using only dimes.
 ~anabel.disher
+{{CEMC box|year=2012|competition=Gauss (Grade 8)|num-b=4|num-a=6}}