SAT Algebra Word Problems

Correct Answer: B

Solution 1:

Let the number of marbles each girl owned at the beginning of the year be $100.$

By the end of the year:

Katie had $25 \%$ more marbles, so $100+\frac{25}{100}\cdot 100=100+25=125$ marbles.

Beth had $75 \%$ fewer marbles, so $100-\frac{75}{100}\cdot 100=100-75=25$ marbles.

On December 31st the number of marbles Beth owned was $\frac{25}{125}=20\%$ of the number of marbles Katie owned.

Solution 2:

Let $x$ be the number of marbles Katie owned on January 1st. By the end of the year she had $x+\frac{25}{100}x = x+0.25x=1.25x$ marbles.

Beth also had $x$ marbles on January 1st. By December 31st, she had $x-\frac{75}{100}x=x-0.75x = 0.25x$ marbles.

Let's assume that the number of marbles Beth owned at the end of the year was $y$ percent of the number of marbles Katie owned.

We set up a proportion:

$\begin{array}{l c l l} \frac{y}{100}&=&\frac{0.25x}{1.25x} &\quad \frac{\text{part}}{\text{whole}}\\ \frac{y}{100} &=& 0.20 &\quad \text{simplify right side}\\ y &=& 20\% & \quad \text{multiply both sides by}\ 100\\ \end{array}$

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Incorrect Choices:

If you got this problem wrong, you should review SAT Ratios, Proportions, and Percents .

(A)
You will get this wrong answer if you multiply $25\%$ and $75\%$ .

(C)
You will get this wrong answer if you divide 25 and 75 and call the result a percent.

(D)
Tip: Just because a number appears in the question doesn’t mean it is the answer.

(E)
If you just add $25\%$ and $75\%$ , you will get this wrong answer.

Since this is a percentage problem, let each girl start with 100 marbles. At the end of the year, Katie has 125 marbles and Beth has 25 marbles. 25=(x/100)(125) x=20