Percent problems

Google Classroom

Problem

Ann increased the quantities of all the ingredients in a recipe by $60\%$60, percent. She used $80$80 grams $(\text{g})$left parenthesis, start text, g, end text, right parenthesis of cheese.

How much cheese did the recipe require?

 $\text{g}$start text, g, end text

Hint #11 / 6

Ann used $60\%$60, percent more cheese than the recipe required. So she used all $100\%$100, percent that the recipe asked for plus $60\%$60, percent more.

Ann used $100\%+60\%=160\%$100, percent, plus, 60, percent, equals, 160, percent of the cheese the recipe required.

Hint #22 / 6

Percent means per hundred, so $\blueD{160\%}$start color #11accd, 160, percent, end color #11accd is equivalent to $\blueD{\dfrac{160}{100}}$start color #11accd, start fraction, 160, divided by, 100, end fraction, end color #11accd which is also equal to $\blueD{160\div 100}$start color #11accd, 160, divided by, 100, end color #11accd.

$\blueD{160\div 100 = 1.6}$start color #11accd, 160, divided by, 100, equals, 1, point, 6, end color #11accd

Hint #33 / 6

To find the amount of cheese in the original recipe, we need to answer, $\greenD{80\,\text{g}}$start color #1fab54, 80, start text, g, end text, end color #1fab54 is $\blueD{160\%}$start color #11accd, 160, percent, end color #11accd of what amount?

Hint #44 / 6

We can rewrite that question as an equation.

$\begin{array}{ccccc} \greenD{80\,\text{g}}&\text{is}&\blueD{160\%}&\text{of}&\text{what amount}\\\\ \greenD{80}&=&\blueD{1.6}&\times&? \end{array}$

Hint #55 / 6

Let's solve for the unknown amount.

$\begin{aligned} \dfrac{\greenD{80}}{1.6}&=\dfrac{\blueD{1.6}\times?}{1.6}\\\\ 50&=? \end{aligned}$

Hint #66 / 6

The recipe required $50\,\text{g}$50, start text, g, end text of cheese.