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1989 AJHSME Problems/Problem 8

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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

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}}.

See Also

1989 AJHSME (Problems • 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
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