2010 AMC 10B Problem 2

P1-10~1 minPrint

Ratios, Percents & Averages Set up the equation/formula and compute; no special trick needed

Makayla attended two meetings during her $9$ -hour work day. The first meeting took $45$ minutes and the second meeting took twice as long. What percent of her work day was spent attending meetings?

Show solution without answering

Key insight

Put everything in minutes. The meetings take 45 + 90 = 135 minutes, and the work day is 9 times 60 = 540 minutes, exactly four times as long.

Solution

A percent compares a part with the whole, and both must be in the same unit. Minutes keep every number whole, so use minutes.

1. The second meeting took twice as long as the first: $2 \cdot 45 = 90$ minutes.
2. Both meetings together: $45 + 90 = 135$ minutes.
3. The work day: $9$ hours is $9 \cdot 60 = 540$ minutes.
4. The fraction of the day in meetings is
$\frac{135}{540}.$
Since $4 \cdot 135 = 540$ , this is $\frac14$ , which is $25\%$ .

The answer is $\boxed{\textbf{(C)}\ 25}$ .

Why this works

A percent is just a fraction of the whole, written out of $100$ . Before dividing, make sure the part and the whole use the same unit; choosing the smaller unit (minutes) avoids decimals.

Alternative approach

Work in hours instead. $135$ minutes is $2$ hours and $15$ minutes, or $2.25$ hours. Then $2.25 \div 9 = 0.25 = 25\%$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

The second meeting lasted $2 \cdot 45 = 90$ minutes, so the meetings took $45 + 90 = 135$ minutes in total.

Her work day is $9 \cdot 60 = 540$ minutes. The fraction spent in meetings is
$\frac{135}{540} = \frac{1}{4} = 25\%.$

The answer is $\boxed{\textbf{(C)}\ 25}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 6: Ratios and Percents
Introduction to Algebra, Chapter 7: Proportion (§7.1, §7.2, §7.3)
Prealgebra, Chapter 13: Data and Statistics (§13.1)