Four quantities are such that their arithmetic mean (A.M.) is the same as the A.M. of the first three quantities. The fourth quantity is

Correct — B, (b) A.M. of the first three quantities.

THE STEP THAT DECIDES THIS QUESTION is a single fact about the arithmetic mean: adding a new observation leaves the mean unchanged if and only if that observation is EQUAL TO THE EXISTING MEAN. Nothing else in the question matters, and once that is seen the answer can be written down without algebra.

The algebra, for a reader who wants it, takes two lines. Let the first three quantities add up to S and let the fourth be d. The mean of the first three is S/3 and the mean of all four is (S + d)/4. The stem says these are equal:

(S + d) / 4 = S / 3 3(S + d) = 4S 3S + 3d = 4S 3d = S d = S / 3

And S/3 is precisely the arithmetic mean of the first three quantities. So the fourth quantity equals the average of the other three, which is option (b).

THE SAME RESULT SEEN THROUGH DEVIATIONS, which is the way worth remembering because it generalises. The defining property of the arithmetic mean is that the deviations from it sum to zero: if m is the mean of a set, then adding up (each value minus m) gives exactly nothing. Now suppose m is the mean of the first three and is also the mean of all four. Over the first three the deviations already cancel to zero. For them still to cancel over all four, the fourth deviation must itself be zero — that is, the fourth quantity must be m. The mean behaves like a balance point, and a weight placed exactly at the balance point does not tip the beam.

THE GENERAL FORMULA that contains this whole question. If n observations have mean m and one more observation x is added, the new mean is

(nm + x) / (n + 1) = m + (x − m) / (n + 1)

Read that second form carefully. The new mean is the old mean plus a correction, and the correction is the new value's distance from the old mean, shared out over the enlarged set. If x is above m the mean rises; if x is below m it falls; if x equals m nothing moves. With n = 3 this is exactly the situation the stem describes, and the condition for no movement is x = m.

A LAST NICETY that confirms the key rather than disturbing it. The identity d = S/3 holds for every set of values the stem permits, so option (b) is always true. The other three options are true only in the single degenerate case where the first three quantities happen to add up to nothing at all, since then every expression on the list collapses to zero together. An option that is right in general beats one that is right only in a special case, and here only one option is right in general.

The English stem ends on the words "The fourth quantity is" with no punctuation, and the divisors in two of the options are printed inline after a bracketed phrase rather than as stacked fractions.

Why the others are wrong

• (a)Sum of the first three quantities — This makes the fourth quantity three times too large. If the first three have mean m their sum is 3m, and appending 3m to them gives a total of 3m + 3m = 6m spread over four observations, so the new mean is 6m/4, that is 3m/2 — half as large again as the old mean, not equal to it. The option is the reflex of a candidate who reads "the fourth quantity is related to the first three" and reaches for the sum because the sum is the quantity that has just been computed on the way to the mean. It also fails a sanity check that costs no time at all: a sum of three numbers is normally much bigger than any one of them, so a fourth quantity equal to that sum would drag the average sharply upward, which is the opposite of what the stem describes. Try it with numbers — 2, 4 and 6 have mean 4 and sum 12, and the mean of 2, 4, 6 and 12 is 6, not 4.
• (c)(Sum of the first three quantities)/4 — This divides the sum of the first three by FOUR, and it is the most attractive of the wrong options because the number four is visible everywhere in the question. It comes from a half-remembered rule — there are four quantities, so divide by four — applied to the wrong numerator. If the first three have mean m, their sum is 3m and a quarter of it is 3m/4. Appending that gives a total of 3m + 3m/4 = 15m/4 over four observations, so the new mean is 15m/16, slightly BELOW the old mean rather than equal to it. That is what a value below the average always does. With the numbers 2, 4 and 6 the option proposes a fourth quantity of 3, and the mean of 2, 4, 6 and 3 is 3·75, not 4. The safeguard is to notice that dividing by four is the operation used to average FOUR numbers, and here the four numbers have not all been added up.
• (d)(Sum of the first three quantities)/2 — This makes the fourth quantity half the sum of the first three, which is one and a half times their mean. If the mean is m, the sum is 3m, and half of it is 3m/2. Appending that gives 3m + 3m/2 = 9m/2 over four observations, so the new mean is 9m/8 — an eighth larger than before. It is a near miss in the sense that it moves the mean by less than the other two wrong options do, which is exactly why it is dangerous: a candidate who tests it loosely with round numbers may not notice the drift. Test it precisely instead. With 2, 4 and 6 the option proposes a fourth quantity of 6, and the mean of 2, 4, 6 and 6 is 4·5, not 4. The general rule settles all three wrong options at once: any value above the existing mean raises it and any value below lowers it, so the only value that leaves it alone is the mean itself.

THE ARITHMETIC MEAN AS A BALANCE POINT is the idea the question rests on, and it is worth learning in that form rather than as a formula, because almost every mean-related question on a recruitment paper is an application of it.

The arithmetic mean of a set of values is their sum divided by their number. Its defining property is that the deviations from it cancel: add up how far each value lies above the mean and how far each lies below, and the two totals are equal. That is why the mean is called the centre of gravity of a distribution. Place a new observation exactly at that centre and the balance does not shift; place it to one side and the centre moves a little way towards it.

HOW FAR IT MOVES is given by the updating formula. For n observations with mean m and one new value x,

new mean = m + (x − m) / (n + 1)

The shift is the new value's distance from the mean, divided by the new count. Two consequences are worth carrying away. First, a single extreme value moves the mean less as the set gets larger, which is why the mean of a big sample is stable and the mean of a handful of readings is not. Second, the mean is nevertheless sensitive to every value in the set — unlike the median, which ignores how far away an outlier lies — and that sensitivity is the standard argument for using the median on skewed data such as incomes.

THE COMBINED MEAN generalises the same idea to two groups rather than one new point. If group one has n1 observations with mean m1 and group two has n2 with mean m2, the mean of the two together is

(n1·m1 + n2·m2) / (n1 + n2)

a weighted average in which each group's mean is weighted by its own size — not the simple average of m1 and m2, which is the commonest error in this area and is only correct when the groups happen to be equal in size. The question in front of us is this formula with n1 = 3 and n2 = 1.

FINALLY, THE FAMILY OF AVERAGES. The arithmetic mean is one of several: the median is the middle value in order, the mode the most frequent, the geometric mean the nth root of the product, used for ratios and growth rates, and the harmonic mean the reciprocal of the mean of reciprocals, used for rates such as speed over a fixed distance. A question that says simply "average" in an Indian competitive paper means the arithmetic mean unless it says otherwise.

Key facts

• Adding an observation to a set leaves the arithmetic mean unchanged if and only if that observation equals the existing mean — the mean is the balance point of the data.
• The algebra: with the first three summing to S and the fourth equal to d, (S + d)/4 = S/3 gives 3d = S, so d = S/3, which is the mean of the first three.
• The defining property of the arithmetic mean is that the deviations from it sum to zero, so a new observation with zero deviation cannot shift it.
• The updating formula for one new value x added to n observations with mean m is new mean = m + (x − m)/(n + 1); the shift is the distance from the mean shared over the enlarged set.
• That formula shows a value above the mean raises it, a value below lowers it, and a larger existing set is moved less by any single new observation.
• The combined mean of two groups is (n1·m1 + n2·m2)/(n1 + n2), a weighted average — the simple average of two group means is correct only when the groups are equal in size.
• Testing with numbers is the quickest elimination here: for 2, 4 and 6 the four options propose 12, 4, 3 and 6 as the fourth quantity, and only 4 keeps the mean at 4.
• The mean uses every value and is therefore sensitive to outliers, which is why the median is preferred for skewed data such as incomes.

Common traps

• Reaching for the sum of the first three because the sum is what was just computed. A value equal to the sum is three times the mean and pushes the average up by half.
• Dividing by four because there are four quantities. Four is the divisor for averaging all four values, not for producing the missing one.
• Trying to answer by formula alone when the options are verbal. Substituting three convenient numbers eliminates three of the four in seconds.
• Forgetting that the mean moves towards any new value that is not equal to it, so only one of the four proposals can leave it fixed.
• Averaging two group means without weighting them by the size of each group, which is the same error in a two-group setting.

• practice — not a real PYQ

  The arithmetic mean of nine observations is 40. A tenth observation is added to the set and the arithmetic mean of all ten observations is still 40. What is the value of the tenth observation ?

  • (a)0
  • (b)4
  • (c)40
  • (d)400

  Answer(c) 40 — a new observation leaves the mean unchanged only when it equals the existing mean, because the shift in the mean is the new value's distance from the old mean divided by the enlarged count, and that distance must therefore be zero.

• practice — not a real PYQ

  The arithmetic mean of five numbers is 18. A sixth number is added to the set and the arithmetic mean of the six numbers becomes 20. What is the value of the sixth number ?

  • (a)24
  • (b)28
  • (c)30
  • (d)32

  Answer(c) 30 — the five numbers total 90 and the six must total 120, so the sixth is 30. The updating formula gives the same result: 18 + (x − 18)/6 = 20, so x − 18 = 12 and x = 30.