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)

Revision as of 15:33, 20 April 2025 (view source) Anabel.disher (talk \| contribs) (Undo revision 247023 by Anabel.disher (talk)) (Tag: Undo) ← Older edit		Latest revision as of 20:55, 18 October 2025 (view source) Anabel.disher (talk \| contribs)
Line 11:		Line 11:

	~anabel.disher		~anabel.disher
		+	{{CEMC box\|year=2012\|competition=Gauss (Grade 8)\|num-b=4\|num-a=6}}