ORIGINAL PRACTICE · 1
Find the total and arithmetic mean of 4,6 and 8.
Reveal answer & reasoning
Total 18 and mean 6.
Add the three values, then divide 18 by the count 3. The denominator is the number of observations, not their largest value.
UPSC CSE · CSAT · Numeracy and data interpretation · TOPIC LESSON
Averages and mixtures
Recover totals before averaging, and conserve the named component when mixing or replacing quantities.
BY THE END, YOU CAN
- Calculate weighted means and membership updates.
- Distinguish mean, median and mode.
- Solve ideal mixture and average-rate problems with explicit assumptions.
BEFORE YOU START
- Use fractions, percentages and consistent units.
Original educational lesson. Work through the examples before revealing the practice answers. This original common-core lesson supports its declared paper scope. Optional subjects, qualifying language papers and the interview are outside this resource.
LEARN THROUGH ORIGINAL PRACTICE
Original UPSC-style worked applications and practice
All teaching examples, cases, prompts and model answers are original. Official exam questions and answer keys are not reproduced. This common-core lesson does not imply complete historical paper coverage or an official score.
01 / UNDERSTAND & APPLY
A mean is a total divided by a count
The arithmetic mean of n values is their sum divided by n, with n positive. Reverse the formula when a problem gives the mean: total=mean×count. This step makes adding, removing and combining groups much safer than manipulating the mean alone.
A mean need not be an observed value. The mean of 2 and 5 is 3.5 even though neither observation is 3.5. It summarizes a balance of numerical values, not necessarily a typical actual person or object. Keep the measured unit and the counted unit clear.
Changing the total and changing the count have different effects. A new observation alters both; replacing an old observation alters only the total. That distinction determines the denominator in the final calculation.
02 / UNDERSTAND & APPLY
Groups contribute according to their sizes
A combined mean equals the combined totals divided by the combined count. If one group has n observations with mean a and another m with mean b, the result is(na+mb)/(n+m). An unweighted average of a and b is valid only when the group sizes are equal, or in a special case where their means already coincide.
With positive weights, a weighted mean lies between the component means. If both means differ and both weights are positive, it lies strictly between them. A result outside that interval signals an arithmetic or denominator error.
Weights must correspond to the quantity being averaged. Student counts weight student scores; litres weight volume-based mixture concentrations under an additive-volume model; elapsed times weight speeds for a time-weighted mean.
ORIGINAL WORKED EXAMPLE
Unequal class sizes
Class A has 20 students with mean score 60. Class B has 30 students with mean score 80. Find the combined mean.
- Recover totals: A contributes 20×60=1200 score units; B contributes 30×80=2400.
- Combined total=3600 and count=50.
- Mean=3600/50=72.
- The result is closer to 80 because more students belong to that group; it lies between 60 and 80.
Answer 72. The unweighted value 70 would incorrectly give the two unequal groups equal influence.
ORIGINAL PRACTICE · 1
One group has 10 observations with mean 50 and another has 10 with mean 70. Find the combined mean.
Reveal answer & reasoning
60.
Equal counts make the weights equal: total 500+700=1200 divided by 20 gives 60. Here averaging the two means is valid because the group sizes match.
03 / UNDERSTAND & APPLY
Adding, removing and replacing
For addition, add the new value to the old total and increase the count by one. For removal, subtract the removed value and reduce the count. For replacement, subtract the old value, add the new one and retain the count. State which operation the question describes.
Adding a value above the old mean raises the mean; adding one below lowers it. Adding a value equal to the old mean leaves it unchanged. These direction checks help identify a wrong denominator before you finish.
If a removed or replaced observation is not specified, the new mean may remain undetermined. Do not assume it equals the old mean merely because that would simplify the calculation.
ORIGINAL WORKED EXAMPLE
Replacement is not addition
Five readings have mean 18. One reading of 10 is replaced by 25. Find the new mean, and compare it with adding 25 as a sixth reading instead.
- Original total=5×18=90.
- Replacement total=90−10+25=105, count remains 5, so mean=21.
- For addition instead, total=90+25=115 and count becomes 6, giving 115/6≈19.17.
- The two operations change different totals and denominators.
Answer Replacement gives 21. Adding a sixth reading of 25 would instead give 115/6, approximately 19.17.
ORIGINAL PRACTICE · 1
Four scores have mean 15. A fifth score is 25. What is the new mean?
Reveal answer & reasoning
17.
Old total=4×15=60; new total 85, new count 5. Thus 85/5=17. It rises because the added 25 exceeds the old mean 15.
04 / UNDERSTAND & APPLY
Median and mode answer different questions
The median is the middle observation after ordering. For an even count, this lesson uses the mean of the two middle values. Ordering is essential: the middle item in the original unsorted list need not be the median.
The mode is the most frequent value. Several values can tie for highest frequency. Under the convention used here, when every value occurs once, there is no mode rather than treating every value as informative. State the convention if a question uses a different one.
An extreme value can move the mean substantially while leaving the median unchanged. This does not make one measure always superior. The mean uses all magnitudes and is useful for totals; the median describes rank position and can be less sensitive to a single extreme observation.
ORIGINAL PRACTICE · 1
For the ordered values 2,3,9,10, find mean, median and mode under this lesson’s convention.
Reveal answer & reasoning
Mean 6, median 6 and no mode.
The total 24 divided by 4 gives 6. The middle pair 3 and 9 has mean 6. Every value occurs once, so no value is more frequent.
ORIGINAL PRACTICE · 2
For 1,2,2,100, find the mean, median and mode.
Reveal answer & reasoning
Mean 26.25, median 2 and mode 2.
Total 105 divided by 4 gives 26.25. The two middle values are 2 and 2, while 2 also occurs most often. The large 100 raises the mean without changing the middle pair.
05 / UNDERSTAND & APPLY
Conserve the named component
A concentration is the amount of a named component divided by total mixture, using the stated measurement basis. Mass fraction uses component mass and total mass. Volume fraction uses component volume and total volume under the model’s conventions. Kilograms and litres cannot be interchanged without density information.
For ideal mixtures, first calculate the amount of component in each input, add those amounts, then divide by the combined total. If volumes are assumed additive, say so: real mixtures may contract or expand, and a numerical exercise must not hide that physical assumption.
The component must be the same in both inputs. Two percentages describing different substances cannot be combined as though they measured one shared ingredient. Also distinguish concentration from the total amount of the component: a more dilute large mixture may contain more component overall.
ORIGINAL WORKED EXAMPLE
An ideal miscible-liquid mixture
Mix 20 litres of a hypothetical liquid mixture containing 30% by volume of component X with 30 litres containing 10% by volume of the same component. Assume the liquids are miscible and all mixture and component volumes add without contraction. Find the resulting percentage of X.
- First input contains 0.30×20=6 litres of X.
- Second contains 0.10×30=3 litres of X.
- Combined X=9 litres and combined mixture=50 litres under the explicit additive-volume assumption.
- Concentration=9/50=0.18=18%. It lies between 10% and 30% and closer to 10%, whose input volume is larger.
Answer 18% by volume in the stated ideal model. This is not a claim that dissolved salt can be measured or combined by the same simple volume convention.
ORIGINAL PRACTICE · 1
A 50 kg mixture contains 20% of component X by mass. How much X is present?
Reveal answer & reasoning
10 kg.
Use the stated mass basis:0.20×50 kg=10 kg. Calling the answer 10 litres would require a density conversion that is not supplied.
06 / UNDERSTAND & APPLY
Derive the mixing ratio
For two concentrations L and H bracketing a target T, let low and high quantities be x and y. Component balance gives Lx+Hy=T(x+y), using proportions consistently. Rearrangement gives(T−L)x=(H−T)y, so low:high=(H−T):(T−L).
This shortcut is a weighted-average equation, not a separate rule to memorize without conditions. Positive amounts of two components cannot produce a concentration outside their range. A target equal to an endpoint would require zero of the other component in this simple two-input model.
Use consistent units and a valid additive basis. If density or volume contraction makes the simple total invalid, the same shortcut cannot be applied without further information.
ORIGINAL WORKED EXAMPLE
A target inside and outside the range
Under a compatible additive-mixture model, combine 10% and 40% mixtures of the same component to obtain 25%. What low:high quantity ratio is required? Could those same two mixtures alone produce 50% using positive amounts?
- Let quantities be x and y.0.10 x+0.40 y=0.25(x+y).
- Rearrange 0.15 y=0.15 x, so x=y. Equivalently, low:high=(40−25):(25−10)=1:1.
- A positive weighted average of 10% and 40% must lie between them, so 50% is impossible from these inputs alone.
Answer Use equal quantities, ratio 1:1. A 50% target is impossible under the stated two-mixture model.
07 / UNDERSTAND & APPLY
Repeated replacement changes the base
Removing a fraction f of a uniformly mixed container removes that same fraction of every component. Refilling with a component-free liquid restores the total volume but not the removed component. After one cycle the original component retains fraction 1−f; after n identical cycles it retains(1−f)^n.
Uniform mixing before each removal is essential. Drawing only from a separated layer would not necessarily remove the same component fraction. The model also assumes the refill volume restores the same total and that volumes add as specified.
Do not subtract the first cycle’s component loss repeatedly. The second withdrawal is drawn from a more dilute mixture, so it contains less original component than the first.
ORIGINAL WORKED EXAMPLE
Two replacement cycles
A container holds 40 litres of a pure hypothetical coloured liquid. Remove 10 litres and refill with 10 litres of water. Mix uniformly, then repeat the 10-litre removal and water refill. Assume additive volumes and no other losses. How much original coloured liquid remains?
- Each withdrawal is 10/40=1/4 of the current well-mixed contents, retaining 3/4 of the original component present.
- After the first cycle, original component=40×3/4=30 litres.
- After the second, it is 30×3/4=22.5 litres. The second removal took 7.5 litres of original component, not 10.
- Final concentration=22.5/40=56.25%.
Answer 22.5 litres of original component remain, giving 56.25% concentration in the ideal mixture.
08 / UNDERSTAND & APPLY
Average speed uses total time
Average speed equals total distance divided by total elapsed time. It is not generally the simple mean of segment speeds. Equal-time segments give an arithmetic mean of their speeds because their time weights are equal. Equal-distance segments take different times and require calculating those times first.
For two equal distances at positive speeds u and v with no stops, average speed is 2 uv/(u+v), obtained by dividing total distance 2 d by d/u+d/v. The formula is optional; the total-distance/total-time method is more general and includes stops when needed.
Always check the denominator’s unit. Dividing kilometres by hours gives kilometres per hour. Averaging numbers that represent different kinds of rates without a common exposure can create a meaningless result.
ORIGINAL PRACTICE · 1
Travel 60 km at 30 km/h and another 60 km at 60 km/h, with no stop. Find average speed.
Reveal answer & reasoning
40 km/h.
Times are 2 hours and 1 hour. Total distance 120 km divided by total time 3 hours gives 40 km/h. The simple mean 45 ignores the longer time spent at the slower speed.
ORIGINAL PRACTICE · 2
Travel for one hour at 30 km/h and one hour at 60 km/h, with no stop. Find average speed.
Reveal answer & reasoning
45 km/h.
Distances are 30 km and 60 km, total 90 km in 2 hours. Equal times make the arithmetic mean of speeds valid in this case.
KEEP THE BIG IDEAS
Quick revision
- Recover total=mean×count before combining or updating observations.
- Weight group means by their group sizes.
- Mean, median and mode summarize different features.
- Conserve the named component with a stated mass or volume basis.
- Repeated replacement uses the current concentration; average rates use the correct exposure.
READ WITH CONTEXT
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