Average (Arithmetic Mean) for UP Police Constable Exam

Concept

Average (औसत) is one of the most reliably tested topics in UP Police Constable Quant. Every exam cycle brings at least two or three questions from this chapter — and the good news is they are all built on the same two or three ideas.

Here is the core idea. If you have a group of numbers and you want to replace every single one of them with one representative number, such that the total stays the same, that representative number is the average.

Formally:

$\text{Average} = \frac{\text{Sum of all observations}}{\text{Number of observations}}$

Or flipped around — the equation you will actually use more often in exams:

$\text{Sum} = \text{Average} \times \text{Number of observations}$

Analogy that sticks. Imagine a class of 8 students. Some are tall, some short. You want to cut all of them to the same height without changing the total "height material". The height each student ends up with after you do that redistribution — that is the average. No extra material, no loss of material.

Now, every UP Police Constable average question is essentially one of these four types:

1. Find the excluded/included number — someone leaves the group, average changes, find who left.
2. Find the required score — you know the target average over N items, you know N-1 of them, find the last.
3. Weighted average — two groups with different averages combine into one overall average.
4. Average of consecutive numbers — integers, even numbers, odd numbers in sequence.
5. Average speed — a special case where most students use the wrong formula.

Each type has a pattern. Learn the pattern, not just the formula.

Deep Dive

Type 1 — Excluded / Included Number

This is the single most common average question in UP Police Constable papers. The setup is always: average was X over N numbers, one number is removed (or added), average becomes Y. Find that number.

The logic is clean:

$\text{Excluded number} = \text{Old Sum} - \text{New Sum}$ $= N \times X - (N-1) \times Y$

You do not need to reconstruct the other numbers. Just compute two sums and subtract. That is it.

Key observation: If average increases after removing a number, the removed number was below the original average. If average decreases, the removed number was above the original average. Use this as a sanity check.

Type 2 — Required Value to Hit a Target Average

Target average over N items is given. Sum of (N-1) items is known. Required Nth item:

$\text{Required value} = (N \times \text{Target}) - \text{Sum of known items}$

This is pure arithmetic. No formula to memorize — just one multiplication and one subtraction.

Type 3 — Weighted Average

Two groups have averages $A_1$ and $A_2$ with sizes $n_1$ and $n_2$. Combined average:

$\bar{A} = \frac{n_1 \cdot A_1 + n_2 \cdot A_2}{n_1 + n_2}$

In the exam, they often give you the combined average and one group's data and ask for the total count. Set up the equation and solve for the unknown.

Shortcut via alligation. Draw the alligation cross:

A1          A2
   \      /
    \    /
     Ā (combined)
   /    \
(Ā-A2) (A1-Ā)

The ratio $n_1 : n_2 = (\bar{A} - A_2) : (A_1 - \bar{A})$. This saves significant algebra time when you only need the ratio of group sizes.

Type 4 — Consecutive Numbers

For n consecutive integers starting from $a$: average = $a + \frac{n-1}{2}$

Look — the average of any consecutive sequence is always the middle term (for odd count) or the average of the two middle terms (for even count). Memorize this instead of the formula.

• Odd count (e.g., 7 consecutive integers): average = middle (4th) term directly.
• Even count (e.g., 6 consecutive even numbers): average = average of 3rd and 4th terms = smallest + $\frac{n}{2} \times \text{step}$ ... but honestly just use: if smallest = $n$, average = $n + (n-1) \times \frac{\text{step}}{2}$ ...

Let's keep it concrete. For 6 consecutive even numbers with smallest $= n$: the numbers are $n, n+2, n+4, n+6, n+8, n+10$. Average $= n + 5$. So if average is given, subtract 5 to get smallest. Done.

For 6 consecutive odd numbers with smallest $= n$: same structure, average $= n + 5$.

For 7 consecutive integers: average is the 4th one, so smallest = average $- 3$.

Type 5 — Average Speed

Here's the trap that catches 40% of students. If you travel the same distance at speed $v_1$ and return at speed $v_2$, the average speed is NOT $(v_1 + v_2)/2$.

The correct formula for equal distances:

$\text{Average Speed} = \frac{2 v_1 v_2}{v_1 + v_2}$

This is the harmonic mean of the two speeds, not the arithmetic mean. The arithmetic mean always overestimates average speed when distances are equal.

Why? You spend more time at the slower speed, so that speed weighs more heavily. The correct formula accounts for this.

If distances are unequal, you must go back to the fundamental formula: Total Distance / Total Time.

Memory Tricks & Shortcuts

Fast-Solving Framework

In the exam hall, classify the question in the first 5 seconds. Here is your decision tree:

Step 1 — What is being asked?

• "One number excluded/included, average changed" → Type 1. Use: Old Sum − New Sum.
• "Need Nth score to achieve target average" → Type 2. Use: Target × N − Sum of known.
• "Two groups combine" → Type 3. Use alligation cross for ratio, then set up one equation.
• "Consecutive integers/even/odd" → Type 4. Use anchor: smallest = average ± offset.
• "Speed there and back" → Type 5. Use 2v1v2/(v1+v2), never (v1+v2)/2.

Step 2 — Sanity check before marking.

• Excluded number below original average? Then average should have gone up after removal. Check direction.
• Average speed below both given speeds? Then you made an error. Average speed must lie between the two individual speeds.
• Required score within logical bounds (not negative, not absurdly large)? If it is, re-check your target sum.

Step 3 — If stuck, try substituting answer options back into the condition. For exclusion problems, this takes under 20 seconds.

Solved PYQs

Why this question: This is the purest form of Type 1 exclusion — the entry-level question for this topic. Every UP Police Constable aspirant must solve this in under 20 seconds.

Solving path: Total of 7 numbers = $7 \times 30 = 210$. Total of remaining 6 = $6 \times 31 = 186$. Excluded = $210 - 186 = 24$. Note: average increased from 30 to 31 after removal, so the excluded number must be below 30. Indeed 24 < 30. Sanity check passes.

Why this question: Age-based exclusion — the same Type 1 logic applied to a group-age setting. Very common in UP Police papers.

Solving path: Total age of 8 = $8 \times 20 = 160$. Total age of remaining 7 = $7 \times 18 = 126$. Age of person who left = $160 - 126 = 34$ years. Average dropped from 20 to 18, so the person who left must have been older than 20. Indeed 34 > 20.

Why this question: Type 2 — required score to hit a target. Cricket match context makes this feel different but the math is identical.

Solving path: Target total = $200 \times 6 = 1200$. Use deviation method on known scores: reference = 200. Deviations: $-35, +10, -15, +40, -25$. Sum of deviations = $-25$. Actual sum = $1000 + (-25) = 975$. Required 6th score = $1200 - 975 = 225$.

Why this question: Weighted average — find total count. This tests whether you know the alligation shortcut.

Solving path: Alligation cross: Reference = ₹8,500. Group 1 avg = ₹10,000 (excess = 1,500). Group 2 avg = ₹7,800 (deficit = 700). Ratio of group sizes = 700 : 1500 = 7 : 15. Group 1 has 7 workers, so Group 2 has $7 \times \frac{15}{7} = 15$ workers. Total = $7 + 15 = 22$. Verify: $7 \times 10000 + 15 \times 7800 = 70000 + 117000 = 187000$. Average $= 187000/22 = 8500$. Confirmed.

Why this question: Average speed trap — the most frequently missed question type. Forces you to use harmonic mean.

Solving path: Equal distance in both directions. Do NOT use $(50 + 30)/2 = 40$ — that is the trap. Use: $\frac{2 \times 50 \times 30}{50 + 30} = \frac{3000}{80} = 37.5$ km/hr. Estimation check: 37.5 < 40 (arithmetic mean). Correct.

Common Mistakes

• Using (v1 + v2)/2 for average speed. This is wrong whenever equal distances are covered at different speeds. Always use the harmonic mean formula $2v_1v_2/(v_1+v_2)$ for same-distance return trips.

• Mixing up "old count" and "new count" in exclusion problems. When one number is removed, the new sum uses (N−1) as the count, not N. Writing $N \times$ new average instead of $(N-1) \times$ new average is the single most common arithmetic error in this chapter.

• Assuming the average of consecutive numbers is the first or last term. The average of consecutive integers is the middle term (for odd count) or the mean of the two middle terms (for even count). It is never the first term.

• In weighted average problems, solving for ratio but stopping there. The question asks for total count. Once you have the ratio $n_1 : n_2$ and you know one group's actual size, multiply to find the other. Students who only write "7:15" without computing the total miss the mark.

• Sign error in deviation method. If you pick a reference that is too high, deviations will mostly be negative. Negative sum of deviations means actual average is below your reference. Double-check the sign before adding to the reference.

• In complex multi-group problems (like the 12-number question), forgetting that the "middle chunk" sum = Total − (first group sum) − (last group sum). Students subtract only one group and get a wrong base, leading to cascading errors. Always verify: first group + middle + last group = total count.

Related Topics

• ratio and proportion
• percentage
• time speed distance
• age problems
• number series
• mixture and alligation

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