1989 AJHSME Problems/Problem 8

Revision as of 10:10, 28 August 2021 by Fn106068 (talk | contribs) (Solution)

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 $24(\frac12+\frac13+\frac14)= 12 + 8 + 6 = 26 \rightarrow \boxed{\text{E}}$

See Also

1989 AJHSME (ProblemsAnswer KeyResources)
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

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