Library / /45 of 356

, 1999

Algebra Difficulty 1.0 Junior Find the answer Canada

$1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8}$ is equal to

(A) $\frac{15}{8}$ (B) $1\frac{3}{14}$ (C) $\frac{11}{8}$ (D) $1\frac{3}{4}$ (E) $\frac{7}{8}$

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Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project. Solutions are the publisher's, linked not copied.