1989 AJHSME Problems/Problem 8: Difference between revisions

Latest revision as of 10:15, 28 August 2021

Problem

$(2\times 3\times 4)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right) =$

$\text{(A)}\ 1 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 24 \qquad \text{(E)}\ 26$

Solution 1

We use the distributive property to get $3\times 4+2\times 4+2\times 3 = 26 \rightarrow \boxed{\text{E}}$

Solution 2

Since $\frac12+\frac13+\frac14 > \frac12+\frac14+\frac14 = 1$ , we have $(2\times 3\times 4)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right) > 2\times 3\times 4 \times 1 = 24$ The only answer choice greater than $24$ is $\boxed{\text{E}}$ .

Solution 3(Bash)

We can just bash it out, getting $24(\frac12+\frac13+\frac14)= 12 + 8 + 6 = 26 \Longrightarrow \boxed{\text{E}}$ -fn106068

1989 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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
All AJHSME/AMC 8 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination

@@ Line 12: / Line 12: @@
 ==Solution 3(Bash)==
-We can just bash it out, getting <math>24(\frac12+\frac13+\frac14)= 12 + 8 + 6 = 26 \longrightarrow \boxed{\text{E}}</math>
+We can just bash it out, getting <math>24(\frac12+\frac13+\frac14)= 12 + 8 + 6 = 26 \Longrightarrow \boxed{\text{E}}</math>
 -fn106068

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