Difference between revisions of "1950 AHSME Problems/Problem 1"

Revision as of 10:59, 31 May 2013

Problem

If $64$ is divided into three parts proportional to $2$ , $4$ , and $6$ , the smallest part is:

$\textbf{(A)}\ 5\frac{1}{3}\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 10\frac{2}{3}\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ \text{None of these answers}$

Solution

If the three numbers are in proportion to $2:4:6$ , then they should also be in proportion to $1:2:3$ . This implies that the three numbers can be expressed as $x$ , $2x$ , and $3x$ . Add these values together to get: $x+2x+3x=6x=64$ Divide each side by 6 and get that $x=\frac{64}{6}=\frac{32}{3}=10 \frac{2}{3}$ which is answer choice $\boxed{C}$ .

1950 AHSC (Problems • Answer Key • Resources)
Preceded by First Question	Followed by Problem 2
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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

@@ Line 7: / Line 7: @@
 ==Solution==
-If the three number are in proportion to <math>2:4:6</math>, then they should also be in proportion to <math>1:2:3</math>. This implies that the three numbers can be expressed as <math>x</math>, <math>2x</math>, and <math>3x</math>. Add these values together to get:
+If the three numbers are in proportion to <math>2:4:6</math>, then they should also be in proportion to <math>1:2:3</math>. This implies that the three numbers can be expressed as <math>x</math>, <math>2x</math>, and <math>3x</math>. Add these values together to get:
 <cmath>x+2x+3x=6x=64</cmath>
 Divide each side by 6 and get that

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Difference between revisions of "1950 AHSME Problems/Problem 1"

Revision as of 10:59, 31 May 2013

Problem

Solution

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