0
$\begingroup$

I have the following exercise:

$$2\cdot \{100\cdot 3-3\cdot [100-3\cdot (100-3\cdot 33)]\}$$

I've looked into the text book and the correct answer is $18$ but I keep getting $-99$.

Here's how I solve it: $$2[100\cdot 3-3\cdot (100-3\cdot 100-99)] = 2\cdot (100\cdot 3-300-300-99) \\= 600-300-300-99=-99 $$

Could you please help me out? Can't find out what I am doing wrong.

$\endgroup$
  • $\begingroup$ Try finishing the calculation in the inner most parentheses before doing the multiply by $3$ outside of it. You don't appear to be distributing properly. $\endgroup$ – John Habert Mar 18 '14 at 14:52
2
$\begingroup$

Just take it in small steps. Do multiplication before addition: $$\begin{align} 2*\{100*3-3*[100-3*(100-\color{red}{3*33})]\} &= 2*\{100*3-3*[100-3*(\color{red}{100-99})]\} \\ &= 2*\{100*3-3*[100-\color{red}{3*(1)}]\} \\ &= 2*\{100*3-3*[\color{red}{100-3}]\} \\ &= 2*\{100*3-\color{red}{3*[97]}\} \\ &= 2*\{\color{red}{100*3}-291\} \\ &= 2*\{\color{red}{300-291}\} \\ &= \color{red}{2*\{9\}} \\ &= 18 \end{align} $$ So for the first equality you find $3*33 = 99$. For the second equality you find $100 - 99 = 1$ and so on.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.