Bobs age is 4 times greater than Susanne's age. Dakota is three years younger than susanne. The sum of Bob's, Susanne's, and Dakota's ages is 93. What is than Susanne's age?

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Answer 1 · 2016-11-11T14:46:21

Susanne's age is 16

First let's call Susanne's age $S$ $S$ .

Because Bob is 4 times greater than Susanne's age his age must be $4 \times S$ $4 xx S$

And because Dakota is three years younger than Susanne her age must be $S - 3$ $S - 3$

The sum of all their ages is 93, so adding together the terms we have created gives us:

$S + 4 S + (S - 3) = 93$ $S + 4S + (S - 3) = 93$

Now solving for $S$ $S$ while keeping this equation balanced gives:

$S + 4 S + S - 3 = 93$ $S + 4S + S - 3 = 93$

$6 S - 3 = 93$ $6S - 3= 93$

$6 S - 3 + 3 = 93 + 3$ $6S - 3 + 3 = 93 + 3$

$6 S = 96$ $6S = 96$

$\frac{6 S}{6} = \frac{96}{6}$ $(6S)/6 = 96/6$

$S = 16$ $S = 16$