Question 13·50 Super-Hard SAT Problem Solving and Data Analysis Questions·Problem Solving and Data Analysis

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The table summarizes the practice times, in minutes, for a data set of students. Each student's practice time is one of the values shown.

Practice time, in minutes	Number of students
$10$	$3$
$20$	$a$
$30$	$8$
$40$	$b$
$50$	$4$

The mean practice time is $29$ minutes, and the median practice time is $25$ minutes. The values of $a$ and $b$ are positive integers.

How many students had a practice time of $40$ minutes?

Your answer

A median between two practice times in a frequency table tells you where the ordered data split: the two middle students must fall in different rows. Use that split to relate the missing frequencies, then turn the weighted mean into a one-unknown equation for Desmos. The trap is to find the frequency of the wrong row and stop.

Hints

Hint 1
The median is the middle of the ordered data. Since no student practiced for 25 minutes, the two middle times must average $25$ . The 20- and 30-minute rows each have students, so which adjacent times can sit in the middle?
Hint 2
A frequency is a number of students, not a practice time. Count the students through the 20-minute row and those starting at the 30-minute row, then set those counts equal.
Hint 3
For a mean, add each practice time multiplied by its frequency, then divide by the total number of students. Once you relate the two missing frequencies, you can write this as an equation with one unknown.

Step-by-step

Use the median split, then solve the mean
Step 1
Locate the two middle students
Let $x=a$ , the number of 20-minute students. The median is $25$ , but no student practiced for $25$ minutes, so it must be the average of two middle times. The $20$ - and $30$ -minute rows both have students. So the two middle times are $20$ and $30$ ; a $10$ - and $40$ -minute pair couldn't be next to each other in the ordered data.
Step 2
Count students on each side of the split
The two middle students sit on opposite sides of the split between $20$ and $30$ , so each side has the same number of students. There are $3+x$ students through the 20-minute row and $8+b+4$ starting at the 30-minute row: $3+x=8+b+4$ . A median between two rows fixes where the middle positions split those rows.
Step 3
Combine the known frequencies
The $30$ - and $50$ -minute rows are both above the split. Add their frequencies: $3+x=12+b$ .
Step 4
Remove the 10-minute group
The $3$ on the left counts the 10-minute students. Subtract $3$ from both sides: $x=9+b$ .
Step 5
Write the 40-minute frequency using x
To leave $b$ by itself, subtract $9$ from both sides: $x-9=b$ .
Step 6
Translate the mean into one equation
A mean of $29$ means total practice minutes equal $29$ times the number of students. Multiply each time by its frequency, using $x-9$ for the 40-minute frequency: $10(3)+20x+30(8)+40(x-9)+50(4)=29\bigl(3+x+8+(x-9)+4\bigr)$ .
Step 7
Find the 20-minute frequency in Desmos
Type that equation with $x$ changed to $x_1$ and $=$ changed to $\sim$ . The $\sim$ tells Desmos to find the unknown value. Under PARAMETERS, Desmos shows $x_1=32$ , so $a=32$ .
Step 8
Find the requested frequency
The question asks for the 40-minute students, whose frequency is $b=x-9$ , not $a$ . Type $x_1-9$ below the regression. Desmos shows $23$ , so $23$ students practiced for $40$ minutes. Grid in 23.

Question 13·50 Super-Hard SAT Problem Solving and Data Analysis Questions·Problem Solving and Data Analysis

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