← Arithmetical & Numerical Ability · DSSSB AE Civil

Chapter 5 of 19

Average, Mixture & Alligation

In the DSSSB AE Civil syllabus under Arithmetical & Numerical Ability · 2 parts

Averages · Mixtures & Alligation

CIVILGYAN TEST SERIES · civilgyantests.com — free study notes for civil engineering exams

📑 Contents (13 sections)

Part 1 of 2

Averages

Last reviewed 22 Sept 2026 · 6 min read

Meaning

The average (arithmetic mean) of $n$ values is

Average = \frac{Sum of the values}{Number of values}, Sum = Average \times n

Almost every average question is solved by moving between these two forms: turn each average into a sum, do the arithmetic on sums, then turn the answer back into an average.

How an average behaves

Change to every value	Effect on the average
Add $k$	Average increases by $k$
Subtract $k$	Average decreases by $k$
Multiply by $k$	Average is multiplied by $k$
Divide by $k$	Average is divided by $k$

The average always lies between the smallest and the largest value.

Averages of standard series

FormulaStandard results

First $n$ natural numbers: $\frac{n + 1}{2}$
First $n$ odd numbers ( $1, 3, 5, \dots$ ): $n$
First $n$ even numbers ( $2, 4, 6, \dots$ ): $n + 1$
Squares of the first $n$ naturals: $\frac{( n + 1 ) ( 2 n + 1 )}{6}$
Cubes of the first $n$ naturals: $\frac{n ( n + 1 ) ^{2}}{4}$
Any set of consecutive numbers (or numbers in AP): $\frac{first + last}{2}$ , which is also the middle term when the count is odd

Combined and weighted average

FormulaGroups put together

For two groups of sizes $n_{1}, n_{2}$ with averages $A_{1}, A_{2}$ :

A = \frac{n _{1} A _{1} + n _{2} A _{2}}{n _{1} + n _{2}}

The same formula with $n_{i}$ read as "weights" is the weighted average. Note that it is not $\frac{A _{1} + A _{2}}{2}$ unless the groups are equal in size.

Replacement, inclusion and correction

FormulaThe three standard moves

Someone joins: new average $= \frac{n A + x}{n + 1}$
Someone is replaced: if the average of $n$ values changes by $d$ when a value $x$ is replaced by $y$ , then $y = x + n d$
A value was read wrongly: correct average $=$ wrong average $+ \frac{correct value - wrong value}{n}$

Average speed

FormulaTwo legs of a journey

Equal distances at speeds $x$ and $y$ : $Average speed = \frac{2 x y}{x + y}$ (the harmonic mean)
Equal times at speeds $x$ and $y$ : $Average speed = \frac{x + y}{2}$
In general, average speed $= \frac{total distance}{total time}$ — never the plain average of the speeds when the distances differ

Worked examples

Worked ExampleExample 1 — sum from average

The average of 11 results is 50. The average of the first six is 49 and of the last six is 52. Find the sixth result.

Solution. The sixth is counted in both groups: $6 \times 49 + 6 \times 52 - 11 \times 50 = 294 + 312 - 550 =$ 56.

Worked ExampleExample 2 — one more person

The average age of 30 students is 14 years. Including the teacher, the average becomes 15. Find the teacher's age.

Solution. $30 \times 14 + T = 31 \times 15 \Rightarrow 420 + T = 465 \Rightarrow T =$ 45 years.

Worked ExampleExample 3 — replacement

The average weight of 8 people increases by 2.5 kg when a new person replaces one weighing 65 kg. Find the new person's weight.

Solution. $y = x + n d = 65 + 8 \times 2.5 =$ 85 kg.

Worked ExampleExample 4 — a wrong reading

The average of 10 numbers is 23. One number, 32, was read as 23. Find the correct average.

Solution. Correct average $= 23 + \frac{32 - 23}{10} = 23 + 0.9 =$ 23.9.

Worked ExampleExample 5 — two classes

Class A has 30 students averaging 60 marks and class B has 20 averaging 70. Find the combined average.

Solution. $\frac{30 \times 60 + 20 \times 70}{50} = \frac{1800 + 1400}{50} =$ 64 marks.

Worked ExampleExample 6 — a standard series

Find the average of the first 50 natural numbers.

Solution. $\frac{n + 1}{2} = \frac{51}{2} =$ 25.5.

Worked ExampleExample 7 — consecutive numbers

The average of five consecutive even numbers is 26. Find the largest.

Solution. The middle number is 26, so the numbers are 22, 24, 26, 28, 30 and the largest is 30.

Worked ExampleExample 8 — a batsman's average

A batsman averages 40 runs in 20 innings. He scores 82 in the 21st. Find the new average.

Solution. $\frac{20 \times 40 + 82}{21} = \frac{882}{21} =$ 42.

Worked ExampleExample 9 — average speed, equal distances

A man travels a distance at 40 km/h and returns at 60 km/h. Find the average speed.

Solution. $\frac{2 \times 40 \times 60}{100} =$ 48 km/h — not 50 km/h, because he spends longer on the slower leg.

Worked ExampleExample 10 — average speed in general

Someone covers 120 km at 60 km/h and the next 120 km at 40 km/h, then rests for 1 hour. Find the average speed for the whole outing.

Solution. Time $= 2 + 3 + 1 = 6$ h for 240 km, so $\frac{240}{6} =$ 40 km/h. Resting time counts.

Worked ExampleExample 11 — average of a group after a change

The average salary of 20 workers is ₹ 15,000. Two workers earning ₹ 12,000 and ₹ 18,000 leave. Find the new average.

Solution. Total $= 300, 000$ ; the two removed total 30,000, so 18 workers share ₹ 270,000 → ₹ 15,000. Removing values whose own average equals the group average leaves the average unchanged.

Frequently tested points

Sum $=$ average $\times$ count; convert every average into a sum first.
Adding $k$ to every value adds $k$ to the average; multiplying by $k$ multiplies it.
First $n$ naturals: $(n + 1) /2$ ; first $n$ odd: $n$ ; first $n$ even: $n + 1$ .
Consecutive numbers or an AP: average $=$ (first + last)/2 $=$ middle term.
Combined average is weighted by group size, never the plain mean of two averages.
Replacement rule: $y = x + n d$ .
Wrong reading: correct the average by (correct − wrong)/ $n$ .
Equal distances: average speed $= 2 x y / (x + y)$ ; equal times: $(x + y) /2$ .

Common MistakeCommon mistakes

Averaging two averages when the groups are of different sizes.
Using $(x + y) /2$ for a there-and-back journey; equal distance needs $2 x y / (x + y)$ .
Forgetting that the overlapping value is counted twice in "first six / last six" problems.
Ignoring halts when asked for average speed over a whole journey.

Revision SummaryChapter summary

An average is a sum divided by a count, and most problems are solved on the sums.
Shifting or scaling every value shifts or scales the average by the same amount.
Standard series have standard averages, and any set of consecutive numbers averages to its middle term.
Groups combine by a weighted average, and joining, replacing or correcting a value each has a one-line rule.
Average speed is total distance over total time — the harmonic mean for equal distances, the arithmetic mean for equal times.

Part 2 of 2

Mixtures & Alligation

Last reviewed 22 Sept 2026 · 6 min read

The rule of alligation

Alligation answers one question: in what ratio must two things be mixed so that the mixture has a required mean value? The "value" may be a price per kg, a strength (percent of alcohol or milk), a speed, or a marks average — anything that averages by weight.

FormulaRule of alligation

For a cheaper ingredient of value $c$ , a dearer one of value $d$ and a mean value $m$ (with $c < m < d$ ):

\frac{Quantity of cheaper}{Quantity of dearer} = \frac{d - m}{m - c}

The ratio is the opposite difference: the quantity of the cheaper is proportional to how far the dearer is above the mean.

The alligation cross

Cheaper $c$		Dearer $d$
	Mean $m$
$d - m$		$m - c$

Write the two values on top, the mean below, and cross-subtract; the bottom row is the ratio cheaper : dearer.

Mixing two mixtures

When two mixtures already contain the same two ingredients in known ratios, work with the fraction of one ingredient in each and then alligate. For milk : water of $a : b$ in the first and $p : q$ in the second, the milk fractions are $\frac{a}{a + b}$ and $\frac{p}{p + q}$ , and the required mean fraction sits between them.

Replacement — drawing off and topping up

FormulaRepeated replacement

A vessel holds $V$ units of pure liquid. If $x$ units are drawn off and replaced by water, and this is done $n$ times:

Liquid left = V (1 - \frac{x}{V})^{n}

and the liquid : water ratio at the end is $(1 - \frac{x}{V})^{n} : [1 - (1 - \frac{x}{V})^{n}]$ .

If the vessel starts as a mixture, apply the same factor to the quantity of each ingredient — replacement removes both in their current proportion.

Changing a ratio by adding one ingredient

Adding water does not change the quantity of milk. So fix the milk quantity, write the new total the required ratio demands, and the difference is the water to add. The same reasoning covers adding pure milk, or removing part of the mixture and topping up.

Worked examples

Worked ExampleExample 1 — mean price

Tea costing ₹ 180 per kg is mixed with tea costing ₹ 280 per kg so that the mixture is worth ₹ 220 per kg. Find the ratio.

Solution. $\frac{280 - 220}{220 - 180} = \frac{60}{40} =$ 3 : 2 (cheaper : dearer).

Worked ExampleExample 2 — strengths

In what ratio must a 20% alcohol solution be mixed with a 50% solution to get 60 litres of a 30% solution?

Solution. Ratio $= (50 - 30) : (30 - 20) = 20 : 10 = 2 : 1$ . Of 60 litres: 40 litres of the 20% solution and 20 litres of the 50%.

Worked ExampleExample 3 — mixing two mixtures

Vessel A has milk and water in the ratio 5 : 2, vessel B in the ratio 3 : 4. Equal quantities are taken from each and mixed. Find the milk : water ratio.

Solution. Milk fractions are $5/7$ and $3/7$ ; equal quantities give milk $= \frac{5/7 + 3/7}{2} = \frac{4}{7}$ . Ratio $=$ 4 : 3.

Worked ExampleExample 4 — unequal quantities of two mixtures

In what ratio must the two vessels of Example 3 be mixed to get milk : water $= 1 : 1$ ?

Solution. Milk fractions $5/7$ and $3/7$ , required mean $1/2$ . $\frac{from B}{from A} = \frac{5/7 - 1/2}{1/2 - 3/7} = \frac{3/14}{1/14} = 3$ , so A : B $=$ 1 : 3.

Worked ExampleExample 5 — adding water

A 60-litre mixture has milk and water in the ratio 2 : 1. How much water must be added to make it 1 : 1?

Solution. Milk $= 40$ L, water $= 20$ L. For 1 : 1 the water must also be 40 L, so add 20 litres.

Worked ExampleExample 6 — replacement, twice

A vessel holds 40 litres of milk. 8 litres are drawn off and replaced by water; this is done twice. How much milk is left?

Solution. $40 (1 - \frac{8}{40})^{2} = 40 \times 0. 8^{2} = 40 \times 0.64 =$ 25.6 litres.

Worked ExampleExample 7 — replacement, three times

From 16 litres of milk, 4 litres are drawn off and replaced by water three times over. Find the milk left and the final ratio.

Solution. $16 \times (0.75)^{3} = 16 \times 0.421875 =$ 6.75 litres of milk, so water $= 9.25$ L and the ratio is 27 : 37.

Worked ExampleExample 8 — finding the mean from a ratio

Rice at ₹ 42 per kg and ₹ 54 per kg are mixed in the ratio 3 : 1. Find the price of the mixture.

Solution. $\frac{3 \times 42 + 1 \times 54}{4} = \frac{126 + 54}{4} =$ ₹ 45 per kg.

Worked ExampleExample 9 — profit through mixing

A shopkeeper mixes rice at ₹ 42 with rice at ₹ 54 in the ratio 3 : 1 and sells the mixture at ₹ 54 per kg. Find the profit percent.

Solution. The mixture costs ₹ 45 per kg (Example 8), so the profit is $9/45 =$ 20%.

Worked ExampleExample 10 — replacement in a mixture

A 45-litre mixture has milk and water in the ratio 7 : 2. Nine litres are removed and replaced by water. Find the new ratio.

Solution. Removing 9 of 45 litres leaves $4/5$ of each: milk $= 35 \times 0.8 = 28$ L, water $= 10 \times 0.8 = 8$ L, plus 9 L of water → 17 L. Ratio $= 28 : 17$ .

Worked ExampleExample 11 — three ingredients

Three varieties costing ₹ 60, ₹ 75 and ₹ 100 per kg are to be mixed so that the mixture costs ₹ 80 per kg, using the first two in equal quantity.

Solution. Treat ₹ 60 and ₹ 75 mixed equally as one ingredient at ₹ 67.50. Then $\frac{cheaper pair}{dearest variety} = \frac{100 - 80}{80 - 67.5} = \frac{20}{12.5} = 8 : 5$ . So the varieties are in the ratio 4 : 4 : 5.

Frequently tested points

Alligation ratio is the opposite difference: cheaper : dearer $= (d - m) : (m - c)$ .
The mean must lie strictly between the two values, or no mixture is possible.
Alligation works for any quantity that averages by weight — price, strength, speed, marks.
Two mixtures combine through the fraction of one ingredient, not through their ratios directly.
Repeated replacement: liquid left $= V (1 - x / V)^{n}$ .
Replacement removes ingredients in their current proportion; adding water leaves the milk untouched.
To change a ratio by adding one ingredient, hold the other ingredient's quantity fixed.

Common MistakeCommon mistakes

Writing the alligation ratio the right way round but assigning it to the wrong ingredient.
Using $V (1 - x / V)^{n}$ when the vessel is topped up with the same liquid rather than water.
Mixing two mixtures by averaging their ratios (5:2 and 3:4 do not give 8:6).
Forgetting that after removing part of a mixture, both ingredients have shrunk in the same proportion.

Revision SummaryChapter summary

Alligation gives the mixing ratio from the two values and the required mean, as opposite differences.
The cross is a layout, not a separate method, and it extends to three ingredients by grouping two of them first.
Two mixtures are combined through the fraction of one ingredient in each.
Drawing off and topping up $n$ times leaves $V (1 - x / V)^{n}$ of the original liquid.
Adding one ingredient changes the total but not the other ingredient, which is the quickest way to set up ratio-change problems.

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