Percentage Practice Questions

This is a weighted average problem; we cannot simply average the percentage of silver cars for the two batches because each batch has a different number of cars.



The car dealership currently has \(40\) cars, \(30\%\) of which are silver. It receives \(80\) new cars, \(60\%\) of which are silver (the \(40\%\) figure given in the problem refers to cars which are not silver). Note that the first batch represents \(\frac{1}{3}\) of the total cars and the second batch represents \(\frac{2}{3}\) of the total cars. Put differently, in the new total group there is \(1\) first-batch car for every \(2\) second-batch cars.



We can calculate the weighted average, weighting each percent according to the ratio of the number of cars represented by that percent:



$$\text{Weighted average} = \frac{1(30\%) + 2(60\%)}{3} = 50\%$$



Alternatively, you can calculate the actual number of silver cars and divide by the total number of cars. \(40(0.3) + 80(0.6) = 12 + 48 = 60\). \(\frac{60}{120} = 50\%\).



The correct answer is E.

Notice that Paul’s income is expressed as a percentage of Rex’s and that the other
two incomes are expressed as a percent of Paul’s. Lets assign a value of $100 to
Rex’s income. Paul’s income is 40% less than Rex's income, so (0.6)($100) =
$60. Quentin’s income is 20% less than Paul's income, so (0.8)($60) = $48. Sam’s
income is 40% less than Paul's income, so (0.6)($60) = $36. If Rex gives 60% of
his income, or $60, to Sam, and 40% of his income, or $40, to Quentin, then: Sam
would have $36 + $60 = $96 and Quentin would have $48 + $40 = $88. Quentin’s
income would now be $88/$96 = 11/12 that of Sam's.

Let's denote the formula for the money spent on computers as:
$$p q = b$$
where:
$$p = \text{price of computers}$$
$$q = \text{quantity of computers}$$
$$b = \text{budget}$$

We can solve a percent question that doesn't involve actual values by using smart numbers. Let's assign:
$$b_{\text{last}} = 1000 \quad p_{\text{last}} = 100$$
(1000 and 100 are easy numbers to take a percent of.)

This year's budget:
$$1000 \times 1.6 = 1600$$

This year's computer price:
$$100 \times 1.2 = 120$$

Now calculate number of computers each year:
$$q = \frac{b}{p}$$

Last year:
$$q_{\text{last}} = \frac{1000}{100} = 10$$

This year:
$$q_{\text{this}} = \frac{1600}{120} = 13\frac{1}{3}$$
(while \(\frac{1}{3}\) of a computer doesn't make sense, it won't affect the calculation)

Table of values:
\[ \begin{array}{|c|c|c|c|} \hline \text{Year} & p & q & b \\ \hline \text{Last Year} & 100 & 10 & 1000 \\ \text{This Year} & 120 & 13\frac{1}{3} & 1600 \\ \hline \end{array} \]

The percent increase in quantity from last year to this year:
\[ \frac{\text{new} - \text{old}}{\text{old}} \times 100\% = \frac{\frac{40}{3} - 10}{10} \times 100\% = 33\frac{1}{3}\% \]

Algebraic method:
Last year:
$$p q = b \quad \Rightarrow \quad q = \frac{b}{p}$$

This year:
$$\left(\frac{6}{5}\right)p \cdot x = \left(\frac{8}{5}\right)b$$

Solving for \(x\):
\[ x = \frac{\frac{8}{5}}{\frac{6}{5}} \cdot \frac{b}{p} = \frac{4}{3} \cdot \frac{b}{p} \]

If this year's quantity is \(\frac{4}{3}\) of last year's quantity, that represents a \(33\frac{1}{3}\%\) increase.