Why multiply the mean by the count?

A mean is the total divided by the count, so the total is the mean times the count. Totals can be added, subtracted and compared. Means cannot, unless the groups are the same size. Getting back to the total is the first step of almost every mean question.

Do I need a calculator for mean problems on the Digital SAT?

Rarely. The multiplications are small, and the method is the same on paper: rebuild the total, add or subtract, then divide if asked. A calculator helps when a list is long or has decimals, because total(L) adds it without a slip, but it cannot decide which totals to compare.

Which count do I use when a value joins or leaves?

The count of the group you are totalling. A value that joins raises the count by one, and a value that leaves lowers it by one. Dividing a new total by the old count is the commonest slip on this skill.

Can I just average the two means when two groups are combined?

Only when the groups are the same size. Otherwise the larger group counts for more, so multiply each mean by its group size, add the totals and divide by the combined size. The combined mean always lies between the two group means, nearer the larger group's.

How do I find the size of a group from a combined mean?

Call the size n and write the combined total two ways: each mean times its count, added, equals the combined mean times the combined count. For a group of 6 with mean 10 joined by n with mean 16, giving a mean of 13, 60 + 16n = 13(6 + n), so n = 6.

Solve mean problems by turning the mean back into a total

A mean times its count is the total. Work in totals: add or subtract them, and divide by a count only when the question asks for a mean.

1.Write $\text{total} = \text{mean} \times \text{count}$ for every group or stage the question mentions, including the one with the missing value. A mean on its own cannot be added to anything; a total can.
2.Add and subtract totals. A missing value is the target total minus the values you can see. A value that joins or leaves is the gap between the total before and the total after.
3.Count again. A value that joins raises the count by one, a value that leaves lowers it by one, and two groups add their counts.
4.Divide only if the question asks for a mean. If it asks for a value or the size of a group, the subtraction or the equation is the answer. Then check the direction: a new value above the mean pulls the mean up.

Try it on a question

Work it below before you open the working. The steps say what to do, and you can do each one yourself.

A planetarium counted the audience at its evening show on five nights. On the first four nights there were 78, 85, 91 and 80 people. The mean audience over all five nights was 84 people. How many people came on the fifth night?

Choice A: $2$
Choice B: $82$
Choice C: $84$
Choice D: $86$

1
Type the four audiences you are given as a list called $L$ . These are the nights you can see.
Type: L=[78,85,91,80]
2
total(L) adds them: 334. That is what four nights have already brought in.
Type: total(L)
3
Five nights at a mean of 84 must total $5 \times 84 = 420$ . This row subtracts the 334 you already have. You wrote the line; the calculator only does the sum.
Type: 5*84-total(L)
4
Find the answer in the rows. The number a row prints is beside it.

Open in the full calculator

Loading the calculator…

The full working, and why each wrong choice is there

Five nights with a mean of 84 add up to $5 \times 84 = 420$ people. The four nights you can see add up to $78 + 85 + 91 + 80 = 334$ . The fifth night is the gap: $420 - 334 = 86$ . Check it: $334 + 86 = 420$ , and $420 \div 5 = 84$ .

In the calculator: store the four counts as a list, then type $5 \times 84$ minus total(L). It is the same arithmetic you have just done, typed. By hand it takes about as long, and the calculator is no part of the method.

Choice A: $2$
Multiplying the mean by 4, the number of nights you were shown, gives a total of 336 and a fifth night of $336 - 334 = 2$ . The mean of 84 belongs to all five nights, so the count is 5.
Choice B: $82$
The first four audiences add up to 334, which is 2 short of $4 \times 84$ . The fifth night has to make that 2 up, so it lands 2 above 84, not 2 below it.
Choice C: $84$
This assumes the fifth night matches the mean. That works only if the first four nights already average 84, and they average 83.5.
Choice D: $86$ the answer
Five nights at a mean of 84 total $5 \times 84 = 420$ . The first four total 334, so the fifth night is $420 - 334 = 86$ .

Use it when

A question gives a mean and asks for a missing value, a target score or the size of a group.
A value is added, removed, replaced or corrected, and the question asks for the new mean or for that value.
Two groups with their own means and sizes are merged, and the question asks for the combined mean.
The data is a long list or has decimals, and typing total(L) is safer than adding it in your head.

Do not use it when

The numbers are small and whole. Here $5 \times 84 = 420$ and four additions take less time by hand than typing a list. This Move needs no calculator, and the calculator adds no idea of its own: it only does the sums the totals tell it to.
Every value is given and the question asks for the mean. Add and divide. There is no mean to turn back into a total, and one row, mean(L), does it.
The question is about the median, the range, or which statistic changes. Totals say nothing about order or spread, so reason from the values themselves.

Five questions on this move

Untimed, and nothing is saved. A wrong first answer gets one more try before the working is shown.

Question 1 of 5

A skateboarder practised for 38, 45, 41, 40, 44 and 44 minutes on six days in a row. How many minutes must she practise on the seventh day for her mean practice time over all seven days to be 44 minutes?

Five is a start

In your account there is a focused set of ten on One-variable data: distributions and measures of center and spread from the question bank, with the same one more try. And after each practice test, the drills your plan sets for the skills you missed name the moves for them.

Practise One-variable data: distributions and measures of center and spread in your account

How this page was checked

Every maths answer, by our own calculator
The answer to each of the 6 questions on this page is worked out again by our calculator engine in an automated test, which fails when a key disagrees with the engine.
In the calculator the test uses
Claude (an AI assistant) typed this page's calculator steps into the College Board version of the Desmos calculator on 14 October 2026 and read the same answers. Both list routes on this page were typed in exactly as printed and printed the keyed answer.
The whole page, by a person
Not yet. Every question and every step of its walk-through was worked through in a browser by Claude, an AI assistant, on 14 October 2026. The page was published on 14 October 2026 before a person had read it. If something here is wrong, tell us and it is put right.

The questions were drafted with AI and are original: none is a College Board question, and none comes from the question bank behind our practice tests.

Where this comes from

“Turn the mean back into a total” is one card from the Preptics lesson The mean: turn it back into a total, which sits under the Digital SAT skill One-variable data: distributions and measures of center and spread. The lesson has the other cards around it, and its questions come back for review on a schedule. The lessons are part of the Digital SAT track.

Common questions

Why multiply the mean by the count?: A mean is the total divided by the count, so the total is the mean times the count. Totals can be added, subtracted and compared. Means cannot, unless the groups are the same size. Getting back to the total is the first step of almost every mean question.
Do I need a calculator for mean problems on the Digital SAT?: Rarely. The multiplications are small, and the method is the same on paper: rebuild the total, add or subtract, then divide if asked. A calculator helps when a list is long or has decimals, because total(L) adds it without a slip, but it cannot decide which totals to compare.
Which count do I use when a value joins or leaves?: The count of the group you are totalling. A value that joins raises the count by one, and a value that leaves lowers it by one. Dividing a new total by the old count is the commonest slip on this skill.
Can I just average the two means when two groups are combined?: Only when the groups are the same size. Otherwise the larger group counts for more, so multiply each mean by its group size, add the totals and divide by the combined size. The combined mean always lies between the two group means, nearer the larger group's.
How do I find the size of a group from a combined mean?: Call the size n and write the combined total two ways: each mean times its count, added, equals the combined mean times the combined count. For a group of 6 with mean 10 joined by n with mean 16, giving a mean of 13, 60 + 16n = 13(6 + n), so n = 6.

Keep going

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