Question

Jane is playing a board game in which she must collect as many cards as possible. On her first turn, she loses $18$ percent of her cards. On the second turn, she increases her card count by $36$ percent. If her final card after these two turns is n, which of the following represents her starting card count in terms of n?

• A. $\frac{n}{\left(1.18\right)\left(0.64\right)}$
• B. $\left(1.18\right)\left(0.64\right)n$
• C. $\frac{n}{\left(1.36\right)\left(0.82\right)}$
• D. $\left(0.82\right)\left(1.36\right)n$

Answer 1

Answer: (C)

$\frac{n}{\left(1.36\right)\left(0.82\right)}$

Let her starting card be $x$

On her first turn she loses $18\%$ of card i.e. $0.18x$

Remaining cards $x-\left(0.18x\right)=0.82x$

On her second turn she gains $36\%$ of card ie

$=\left(0.82x\right)\left(0.36\right)=\left(0.36\right)\left(0.82x\right)$

now total cards $=\left(0.82\right)x+\left(0.36\right)\left(0.82x\right)$

$=\left(1.36\right)\left(0.82\right)x$

given after two cards no of cards aren so

$\left(1.36\right)\left(0.82\right)x=n$

$x=\frac{n}{\left(1.36\right)\left(0.82\right)}$