Averages — SSC CGL Quantitative Aptitude Formulas and Traps

Sum and Mean

Average, sum and count

\text{Average} = \frac{\text{Sum}}{\text{Count}} \qquad \text{Sum} = \text{Count} \times \text{Average}

Average from an assumed value

\bar{x} = A + \frac{\sum (x - A)}{n} \qquad \sum (x - \bar{x}) = 0

$A$ the assumed average, any convenient number
$n$ the number of values
$\bar{x}$ the true average

Pair averages: add all the statements

Total from three pair sums

a + b + c = \frac{(a + b) + (b + c) + (c + a)}{2}

Common traps

Averaging two averages of unequal groups

If 4 numbers average 10 and 6 other numbers average 20, the average of all 10 is not 15. Change both to sums first:

\frac{40 + 120}{10} = 16

.

Forgetting part of the count

In a mixed purchase the average price is total money over total items. Count every item, including the ones whose number you had to work out from the money left over.

Dropping the sign of a deviation

A value below the assumed average has a negative deviation. Adding every gap as positive pushes the average too high.

Adding the deviation total without dividing

The sum of deviations must be divided by the count before it is added to the assumed average. Adding the whole total overshoots by a factor of the count.

Forgetting to halve the total of the pair sums

When all three pair sums are added, each person appears in two of them, so the result is twice the total of the three people.

Reading an average as a sum

'The average of F and M is 5,000' means F and M together are 10,000. Using 5,000 as their sum halves every answer that follows.

Members Joining and Leaving

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A new member joins the group

A member joins

x = A + (n + 1)(A' - A) \qquad A' = A + \frac{x - A}{n + 1}

$n$ number of members before the newcomer joins
$A$ average before
$A'$ average after
$x$ the newcomer's value

A member leaves the group

A member leaves

x = A + (n - 1)(A - A') \qquad \text{Leavers' total} = nA - (n - k)A'

$n$ number of members before anyone leaves
$k$ number who leave
$A$ average before
$A'$ average of those who stay

Common traps

Multiplying the rise by the wrong count

Starting from the old average, the newcomer must lift all

n + 1

members, so the extra is

(n + 1) \times

rise. Starting from the new average, it is

n \times

rise. Using

n

with the old average is short by exactly one rise.

Taking the new average as the newcomer's value

If the average rises after someone joins, the newcomer is above the new average, not equal to it. Only when the average does not change is the newcomer equal to it.

Multiplying the fall by the old count

When someone leaves, only the members who stay carry the new average. From the old average, the leaver is old average plus

(n - 1) \times

fall, not

n \times

fall. The

n \times

fall form belongs with the new average.

Thinking a fall means the leaver was low

If the average falls after someone leaves, the leaver was above the old average: removing a high value pulls the average down.

Replacements and Corrections

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Replacing members, or changing every value

Replacement and shifts

\text{New} = \text{Old} + n \times \text{rise} \qquad \overline{x + c} = \bar{x} + c \qquad \overline{cx} = c\,\bar{x}

$n$ the size of the group, which does not change
$c$ a number added to, or multiplied into, every value

Correcting a wrong entry

Correcting an average

\text{Correct average} = \text{Wrong average} + \frac{\text{Correct value} - \text{Wrong value}}{n}

Common traps

Using n + 1 for a replacement

A replacement does not change the count, so the total changes by

n \times

rise. The

(n + 1)

form is only for a member who joins.

Multiplying the rise by the number replaced

If 2 members of a group of 9 are replaced and the average rises by 3, the total rises by

9 \times 3 = 27

, not

2 \times 3 = 6

: every member's share went up.

Subtracting the wrong way round

The error is correct minus wrong. If 91 was misread as 19, the total was 72 too low, so the correct average is higher than the one found.

Moving the average by the whole error

The difference between the correct and wrong value is shared by all n values, so the average moves by the difference divided by n, not by the difference itself.

Parts of a Total

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The members left after the known parts

Leftover total

\text{Leftover total} = nA - (n_1A_1 + n_2A_2 + \cdots)

Overlapping parts share a member

Shared member of two overlapping parts

\text{Shared} = n_1A_1 + n_2A_2 - nA

A member tied to the average of all

Average of all when one member is tied to it

a = \bar{x}_{\text{others}} + \frac{k}{n - 1}

$a$ average of all n members
$k$ how far the special member is above the average of all
$n$ number of members, the special one included

Common traps

Applying a percentage to the wrong quantity

When a part's average is '25% more' than another's, raise the average by 25% and then multiply by that part's count. Raising the other part's total only works when the two parts have the same count.

Missing the shared member

January has 31 days, so 'the first sixteen days' and 'the last sixteen days' overlap at day 16. Treating them as two separate halves leaves the totals impossible to match.

Adding k to the others' average

The special member is k above the average of all, not above the others' average. The average of all is the others' average plus k ÷ (n − 1), and the special member is k above that.

Weighted and Combined Averages

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Combined average of two groups

Combined average

\bar{x} = \frac{n_1A_1 + n_2A_2}{n_1 + n_2} \qquad \bar{x} = A_1 + \frac{n_2}{n_1 + n_2}(A_2 - A_1)

$n_1, n_2$ the sizes of the two groups
$A_1, A_2$ their averages

Weights given as percentages or a ratio

Weighted average

\bar{x} = \frac{w_1x_1 + w_2x_2 + w_3x_3}{w_1 + w_2 + w_3}

$w$ a weight: a count, a ratio term or a percentage
$x$ the value carrying that weight

Finding a missing group's average

Missing group's average

A_2 = \frac{nA - n_1A_1}{n - n_1} \qquad n_1(A_1 - A) = n_2(A - A_2)

$n, A$ size and average of everyone
$n_1, A_1$ the known group
$n_2, A_2$ the missing group, with $n_2 = n - n_1$

Alligation: group sizes from the averages

Rule of alligation

\frac{n_1}{n_2} = \frac{A_2 - A}{A - A_1}

$A$ the combined average, between $A_1$ and $A_2$

Common traps

Averaging the two averages

Groups of 3 and 7 averaging 40 and 80 do not average 60 together: weight by size,

\frac{120 + 560}{10} = 68

. The plain average is right only for equal groups.

Choosing an option outside the two averages

A combined average can never be below the smaller group average or above the larger one. An option outside that range is wrong at sight.

Dropping the sign of a loss

In a weighted average a 5% loss enters as −5. Adding it as +5 overstates the result by twice its weighted share.

Ignoring the weights

Marks of 75, 80 and 60 with weights 40%, 50% and 10% do not average

\frac{75 + 80 + 60}{3}

. Each mark counts by its weight, and the light 60 barely moves the result.

Dividing the missing total by the whole count

The total left after removing the known group belongs only to the remaining members, so divide by the number remaining, not by the size of the whole group.

Giving each group its own distance

In alligation each group's share is the OTHER group's distance from the combined average. A combined average close to group 1's average means group 1 is the bigger group.

Averages of Consecutive Numbers

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Average of numbers in equal steps

Average of an arithmetic progression

\text{Average} = \frac{\text{first} + \text{last}}{2} \qquad \text{Sum} = \text{count} \times \text{middle term}

Average of the multiples of a number in a range

Multiples of k in a range

\text{Average} = \frac{\text{first multiple} + \text{last multiple}}{2} \qquad \text{Count} = \frac{\text{last} - \text{first}}{k} + 1

Averages of the first n natural numbers and their squares

Sums and averages of the first n natural numbers

\frac{1 + 2 + \cdots + n}{n} = \frac{n + 1}{2} \qquad \frac{1^2 + 2^2 + \cdots + n^2}{n} = \frac{(n + 1)(2n + 1)}{6} \qquad \frac{1^3 + 2^3 + \cdots + n^3}{n} = \frac{n(n + 1)^2}{4}

Common traps

Expecting a whole-number average from an even count of consecutive integers

Six consecutive integers such as 4 to 9 average 6.5: an even count of consecutive integers always averages to a number ending in .5.

Taking a middle term that does not exist

With an even count there is no single middle term. Six consecutive even numbers averaging 25 have no term equal to 25; the two middle terms are 24 and 26.

Averaging the ends of the range instead of the multiples

Between 50 and 100 the multiples of 8 run from 56 to 96, so their average is 76, not 75. Use the first and last multiples, not the ends of the range.

Using n ÷ 2 for the average of 1 to n

The average of 1 to n is (n + 1) ÷ 2. The numbers 1 to 9 average 5, not 4.5.

Giving the sum of squares as the average

n(n + 1)(2n + 1) ÷ 6 is the SUM of the first n squares. Divide by n to get their average, which is (n + 1)(2n + 1) ÷ 6.

Averages formulas

Sum and Mean

Pair averages: add all the statements

Common traps

Members Joining and Leaving

A new member joins the group

A member leaves the group

Common traps

Replacements and Corrections

Replacing members, or changing every value

Correcting a wrong entry

Common traps

Parts of a Total

The members left after the known parts

Overlapping parts share a member

A member tied to the average of all

Common traps

Weighted and Combined Averages

Combined average of two groups

Weights given as percentages or a ratio

Finding a missing group's average

Alligation: group sizes from the averages

Common traps

Averages of Consecutive Numbers

Average of numbers in equal steps

Average of the multiples of a number in a range

Averages of the first n natural numbers and their squares

Common traps

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