2004 CEMC Gauss (Grade 7) Problems/Problem 1

Problem

The value of $\frac{10 + 20 + 30 + 40}{10}$ is

$\textbf{(A)}\ 90 \qquad\textbf{(B)}\ 91 \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ 64 \qquad\textbf{(E)}\ 9$

$\frac{10 + 20 + 30 + 40}{10} = \frac{30 + 30 + 40}{10} = \frac{60 + 40}{10}$

$=\frac{100}{10} = \boxed {\textbf {(C) } 10}$

~anabel.disher

We can see that all of the numbers in the numerator are divisible by $10$ , meaning we can factor out $10$ from the start:

$\frac{10 + 20 + 30 + 40}{10} = \frac{10(1 + 2 + 3 + 4)}{10}$

$=1 + 2 + 3 + 4 = 3 + 3 + 4 = 6 + 4 = \boxed {\textbf {(C) } 10}$

~anabel.disher

We can use the fact that $1 + 2 + 3 + ... + n = \frac{n(n + 1)}{2}$ , and do the same thing as solution 2 until we get $1 + 2 + 3 + 4$

$1 + 2 + 3 + 4 = \frac{4 \times 5}{2} = \frac{20}{2} = \boxed {\textbf {(C) } 10}$

~anabel.disher

2004 CEMC Gauss (Grade 7) (Problems • Answer Key • Resources)
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CEMC Gauss (Grade 7)