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78 divided into three parts

Source: 1959 AMC 12 Problem 4

August 13, 2013
AMCalgebraAMC 12

Problem Statement

If 7878 is divided into three parts which are proportional to 1,13,161, \frac13, \frac16, the middle part is: <spanclass=latexbold>(A)</span> 913<spanclass=latexbold>(B)</span> 13<spanclass=latexbold>(C)</span> 1713<spanclass=latexbold>(D)</span> 1813<spanclass=latexbold>(E)</span> 26 <span class='latex-bold'>(A)</span>\ 9\frac13 \qquad<span class='latex-bold'>(B)</span>\ 13\qquad<span class='latex-bold'>(C)</span>\ 17\frac13 \qquad<span class='latex-bold'>(D)</span>\ 18\frac13\qquad<span class='latex-bold'>(E)</span>\ 26