ORIGINAL WORKED EXAMPLE
Apply the idea
Five values have mean 18. Four are 12,16,21,24. Find the fifth.
- The total must be 5×18=90.
- The four known values sum to 73.
- The missing value is 90−73.
Answer 17.
GMAT · Counting, probability and statistical arithmetic · TOPIC LESSON
Averages, distributions and statistical comparisons
Original GMAT Problem Solving instruction: reason from constraints, calculate efficiently without a calculator, and check the requested quantity.
BY THE END, YOU CAN
- Explain and apply arithmetic mean.
- Explain and apply weighted mean.
- Explain and apply combined groups.
- Explain and apply median and mode.
- Explain and apply range.
- Explain and apply conceptual spread and standard deviation.
- Explain and apply effect of shifts and scaling.
- Explain and apply missing values and constraints.
- Explain and apply distinguish count from proportion.
BEFORE YOU START
- Arithmetic, ratios, signed values and square roots.
Original educational lesson. Work through the examples before revealing the practice answers. This original lesson supports current GMAT study through an editorial skill map. It does not reproduce official questions, predict exam content or guarantee a score.
LEARN THROUGH ORIGINAL PRACTICE
Original GMAT-style worked examples and practice
All passages, examples and exercises here are original. No official GMAC questions, choices or explanations are reproduced. This independent preparation resource is not affiliated with or endorsed by GMAC. GMAT is a trademark of the Graduate Management Admission Council.
01 / UNDERSTAND & APPLY
The mean is a balance point and a total divided by count
For n values, arithmetic mean=sum/n. Recovering a sum from a mean uses sum=n×mean. A mean need not appear in the data and need not be an integer even when all observations are integers.
Deviations from the mean sum to zero: values above it balance values below it. This can make no-calculator arithmetic efficient. The mean is sensitive to extreme values because every value contributes to the total.
ORIGINAL PRACTICE · 1
Find the mean of 47,49,50,52,57 without a long sum.
Reveal answer & reasoning
51.
Use reference 50: deviations are −3,−1,0,2,7, summing to 5. The average deviation is 1, so mean is 51. Choosing the middle listed value 50 would give the median, not the mean.
02 / UNDERSTAND & APPLY
Weights describe how much each value contributes
A weighted mean is sum(weight×value)/sum(weights). Frequency weights count repeated observations. For nonnegative weights with positive total, the weighted mean lies between the smallest and largest values carrying positive weight.
An ordinary mean of group means is valid only when groups have equal weights or when the question explicitly weights groups equally. A weight is not automatically a percentage; normalize by the total weight if weights do not sum to one.
ORIGINAL WORKED EXAMPLE
Apply the idea
A store sells 4 units at $15 and 6 units at $20. What is the average selling price per unit?
- Total revenue is 4×15+6×20=60+120=180.
- Total units sold are 10.
- Average price is 180/10.
Answer $18.
ORIGINAL PRACTICE · 1
A grade weights two tests in ratio 2:3. The scores are 70 and 90. What is the weighted mean?
Reveal answer & reasoning
82.
Use (2×70+3×90)/(2+3)=(140+270)/5=82. The simple mean 80 assumes equal weights, contrary to the stated ratio.
03 / UNDERSTAND & APPLY
Reconstruct group totals before combining
For groups of sizes n₁ and n₂ with means m₁ and m₂, the combined mean is (n₁m₁+n₂m₂)/(n₁+n₂). The group sizes provide the weights. If one group is removed, subtract its total and count rather than subtracting its mean.
A combined mean between two distinct group means is closer to the mean of the larger group. The distances to the combined mean are inversely proportional to group sizes. This can recover a group-size ratio without knowing either absolute size.
ORIGINAL WORKED EXAMPLE
Apply the idea
A group of 8 people averages 30 minutes on a task; another group of 12 averages 40 minutes. What is the combined mean?
- The group totals are 8×30=240 and 12×40=480 minutes.
- Combined total is 720 over 20 people.
- Divide 720 by 20.
Answer 36 minutes.
ORIGINAL PRACTICE · 1
Two groups have means 20 and 35, and their combined mean is 29. What is the ratio of the first group’s size to the second’s?
Reveal answer & reasoning
2:3.
If sizes are a,b, then 20a+35b=29(a+b), so 6b=9a and a:b=2:3. Distances from 29 are 9 and 6; the sizes are inversely proportional to those distances.
04 / UNDERSTAND & APPLY
Sort before finding the middle
The median is the middle value of a sorted odd-length list, or the average of the two middle values of a sorted even-length list. Sorting is essential. The mode is a most frequent value; there may be several tied modes or no unique mode.
Median depends on order position rather than the size of distant extremes. Changing one extreme value can alter the mean without changing the median. Do not assume mode, median and mean coincide unless the distribution supports it.
ORIGINAL WORKED EXAMPLE
Apply the idea
Find the median and mode of 8,3,5,8,10,2.
- Sort: 2,3,5,8,8,10.
- The middle positions are third and fourth, so median=(5+8)/2=6.5.
- The value 8 occurs twice; every other value occurs once.
Answer Median 6.5; mode 8.
ORIGINAL PRACTICE · 1
The sorted list is 2,4,4,7,13. If 13 is replaced by 100, which of mean, median and mode change?
Reveal answer & reasoning
Only the mean changes.
The median remains the third value 4 and the mode remains 4. The total rises by 87, so the mean rises by 87/5. An extreme change affects the mean without moving the middle position.
05 / UNDERSTAND & APPLY
Range uses only the extremes
Range=maximum−minimum. It is nonnegative and zero exactly when every value is equal. Knowing the range does not determine the mean, median or how the interior values are distributed.
Adding or changing an interior value may leave the range unchanged. Removing a current extreme can change the range, but only if no identical extreme remains. When a missing value is constrained by a range, consider whether it becomes a new minimum, a new maximum or stays inside.
ORIGINAL WORKED EXAMPLE
Apply the idea
The values are 5,8,12,x and their range is 10. Find every possible x.
- The known minimum and maximum are 5 and 12, with range 7.
- To increase range to 10, x must be outside that interval.
- If x is the new minimum, 12−x=10 gives 2; if the new maximum, x−5=10 gives 15.
Answer x=2 or x=15.
ORIGINAL PRACTICE · 1
A list has minimum −4 and maximum 9. What is its range?
Reveal answer & reasoning
13.
Range is 9−(−4)=13. Adding the signed endpoints gives 5, which is not the distance between them.
06 / UNDERSTAND & APPLY
Spread measures distances from the centre
Standard deviation summarizes distances from the mean by squaring deviations, averaging with a specified convention and taking a square root. Population variance divides the squared-deviation total by n; sample variance divides by n−1 when n>1. Use the convention supplied by the problem.
For conceptual comparisons, exact square roots are often unnecessary. The displayed sets A={4,5,6} and B={1,5,9} both have mean 5, but B has deviations −4,0,4 rather than −1,0,1. Its squared deviations are larger, so its standard deviation is larger under either consistent convention. Range alone cannot always rank standard deviations.
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ORIGINAL WORKED EXAMPLE
Apply the idea
Compare the population standard deviations of A={4,5,6} and B={1,5,9}. Give the ratio without evaluating irrational roots.
- Both means are 5.
- Squared-deviation totals are 1+0+1=2 for A and 16+0+16=32 for B.
- Population variances are 2/3 and 32/3, a ratio of 16. Taking square roots gives a standard-deviation ratio of 4.
Answer B’s standard deviation is 4 times A’s.
ORIGINAL PRACTICE · 1
Which has greater population standard deviation: C={0,0,10,10} or D={0,5,5,10}?
Reveal answer & reasoning
C.
Both means are 5 and both ranges are 10. Squared-deviation totals are 100 for C and 50 for D. Dividing by the same count preserves the comparison, so C has greater standard deviation. Equal ranges do not force equal spread.
07 / UNDERSTAND & APPLY
Uniform transformations have predictable effects
Adding a constant c to every observation adds c to the mean and median while leaving every deviation from the mean, and hence range and standard deviation, unchanged. Multiplying every observation by k multiplies the mean by k and scales range and standard deviation by |k|.
For a negative multiplier, the order reverses, but the median still scales by k. Variance scales by k² rather than |k|. These rules apply when every observation is transformed in the same way; changing just one value is a different operation.
ORIGINAL WORKED EXAMPLE
Apply the idea
A data set has mean 12, range 8 and standard deviation 3. Every value is transformed to 2x+5. Find the new mean, range and standard deviation.
- Mean becomes 2×12+5=29.
- Adding 5 does not affect spread; multiplication by 2 doubles it.
- Range becomes 16 and standard deviation becomes 6.
Answer Mean 29; range 16; standard deviation 6.
ORIGINAL PRACTICE · 1
A data set has mean 7 and standard deviation 4. Every value is multiplied by −3. What are the new mean and standard deviation?
Reveal answer & reasoning
Mean −21; standard deviation 12.
The mean scales by −3, but standard deviation is nonnegative and scales by |−3|=3. Reporting a negative standard deviation confuses signed location with spread.
08 / UNDERSTAND & APPLY
Use all statistical and contextual constraints
A mean condition gives a total; a median condition gives an order-position restriction; a range condition fixes extreme separation. Combine these with any integer or nonnegative requirements. One statistic alone may leave many possible data sets.
When optimizing a missing value, distribute the remaining total subject to their bounds. Do not assume the data are distinct or positive unless stated. A construction proves feasibility, while an inequality proves no larger or smaller value can occur.
ORIGINAL WORKED EXAMPLE
Apply the idea
Five nonnegative integers have mean 8. Four of them are at least 6. What is the greatest possible fifth integer?
- The total is 5×8=40.
- The other four consume at least 4×6=24.
- The fifth is at most 16, achieved by 6,6,6,6,16.
Answer 16.
ORIGINAL PRACTICE · 1
Three real numbers have mean 10, and two are 6 and 8. Find the third and the median.
Reveal answer & reasoning
Third value 16; median 8.
Total is 30, so the missing value is 30−14=16. Sorted values are 6,8,16, making median 8. A mean of 10 does not force the middle observation to equal 10.
09 / UNDERSTAND & APPLY
Counts and proportions can move differently
A proportion is a category count divided by a total. A category count can rise while its proportion falls if the total rises faster. Comparing counts across unequal populations does not directly compare rates.
To combine proportions, recover their category counts or weight them by group sizes. If only percentages are known and group sizes are unspecified, a combined percentage is generally not determined. Keep numerators and denominators paired to the same group and time.
ORIGINAL WORKED EXAMPLE
Apply the idea
A club grows from 80 to 120 members. The number of remote members rises from 24 to 30. How do the remote count and proportion change?
- The count rises by 6.
- The original proportion is 24/80=30%.
- The new proportion is 30/120=25%, a decrease of 5 percentage points.
Answer Count rises by 6; proportion falls from 30% to 25%.
ORIGINAL PRACTICE · 1
Group A has 20 people and a 50% participation rate. Group B has 80 people and a 25% rate. What is the combined participation rate?
Reveal answer & reasoning
30%.
Participants number 10+20=30 out of 100 people, giving 30%. Averaging 50% and 25% to get 37.5% weights the small and large groups equally rather than weighting people equally.
KEEP THE BIG IDEAS
Quick revision
- Recover totals from means before combining groups.
- Sort for median and distinguish central position from average value.
- Range and standard deviation measure different aspects of spread.
- Uniform shifts preserve spread; scaling multiplies spread by absolute scale.
- A changing count need not move in the same direction as its proportion.
READ WITH CONTEXT
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