RRB NTPC Average Questions with Answers

RRB NTPC Average Questions with Answers: Quick Answer

This page solves 6 official RRB NTPC average questions with full step-by-step working. All 6 come from two official papers: the RRB NTPC Graduate 2026 CBT 1 Mathematics section papers held on 16 March 2026, Shift 2 and Shift 3. Four of the 6 are averages of numbers or groups, and 2 are average-speed questions, labelled as the average-speed sub-topic below. That count of 6 is the honest total for these two papers: all 60 questions across the two section papers were checked, and only these 6 are genuine average questions. Mode and median questions in the same papers are not averages and are not included here, and neither are the train and circular-track questions, which test relative speed and meeting times rather than averages.

The 6 answers, in the order solved below, are 79, 63, 159.25, 408, 34 m/min and 64 km/h. The method guide before the questions covers everything needed to solve them: the average formula, finding a missing number, combined averages, averages of number series, and the difference between average speed and the average of speeds. As of 2 November 2026, these 6 questions are the complete set of official average questions from these two papers on this site.

RRB NTPC Average Formula Box

Keep these six ideas in front of you. Every average question in the two papers below is one of them, and most average questions in any NTPC paper are one of them too.

• Average = sum of the numbers ÷ how many numbers there are. Rearranged: sum = average × count. If an average and a count are given, the total follows at once.
• Finding a missing number: add up everything known, subtract from the total implied by the average, and the remainder is the missing number. When two groups overlap, the overlapped item is counted twice, so subtract the grand total once to find it.
• Combined average: add the totals of all groups, add the counts of all groups, then divide total by count. Never average the group averages directly unless the groups are the same size.
• Average of an arithmetic series: (first term + last term) ÷ 2. The first n natural numbers average (n + 1) ÷ 2, and the first n even natural numbers average n + 1.
• Average speed = total distance ÷ total time. It is not the average of the speeds, except when the times spent at each speed are equal.
• Equal distances, different speeds: average speed = 2xy ÷ (x + y) for two parts, where x and y are the two speeds. More generally, work out each part's time, add the times, and divide total distance by total time.

Shortcut habit: whenever a question gives you an average, immediately write down the total it implies. "Average 61 over 11 results" becomes "total 671" before you read the rest of the question. That one step turns most average questions into simple addition and subtraction.

Average: The Core Method

The average of a set of numbers is their sum divided by how many of them there are. If five numbers add to 100, their average is 100 ÷ 5 = 20. The formula works in both directions. If the average of five numbers is 20, their sum must be 5 × 20 = 100, even if the individual numbers are unknown. That reverse direction is the more useful one in examinations, because questions usually give averages and ask for a number, a total or a new average, rather than asking for the average of a fully listed set.

Method illustration (not a previous-year question): The average of 4 numbers is 25, so their total is 4 × 25 = 100. Three of the numbers are 20, 30 and 28, which add to 78. The fourth number is 100 − 78 = 22. No other method is needed: total first, known part second, remainder last. Keep this three-step shape in mind, because the overlapping-groups question below uses exactly the same shape with one extra counting subtlety.

Averages hide the individual values, and that is both their use and their trap. Two sets can share an average while containing entirely different numbers, so an average alone never tells you any single value. When a question asks for a single value, it must give enough extra information to pin that value down, usually by giving the averages of overlapping groups or of all but one member. Read what is given as totals, and the missing piece usually appears by subtraction.

Finding a Missing Number and Overlapping Groups

When a value is removed from or added to a set, the total changes by exactly that value, and the average changes accordingly. If the average of 10 numbers is 40, their total is 400. If one number, 55, is removed, the remaining 9 numbers total 400 − 55 = 345, and their average is 345 ÷ 9 = 38.33. If a number is added instead, add it to the total and divide by the new count. The method never changes: track the total and track the count, and the average follows.

Overlapping groups need one extra care. Suppose the average of 11 results is known, and so are the averages of the first 6 and the last 6. The 6th result belongs to both groups, so adding the two group totals counts it twice. Subtracting the grand total of all 11 removes one copy of every result, leaving exactly the doubly-counted 6th result. That is the whole trick of the first solved question below: total of first 6, plus total of last 6, minus total of all 11, equals the 6th result. Whenever groups overlap by one member, the overlapped member is found by exactly this addition and subtraction.

Method illustration (not a previous-year question): The average of 9 scores is 50, so the total is 450. The average of the first 5 scores is 48, giving a total of 240, and the average of the last 5 scores is 54, giving a total of 270. The 5th score is counted in both groups, so it equals 240 + 270 − 450 = 60. Check: 240 includes the 5th score once and 270 includes it once, so their sum includes it twice, and removing the grand total of 450 removes one copy of every score, leaving the 5th score alone.

Combined Averages: Weight by Group Size

When groups of different sizes are combined, the larger group pulls the combined average further towards its own average. The combined average is the total of all values divided by the total count, which means each group's contribution is its average multiplied by its size. Averaging the group averages directly treats every group as if it had the same size, and that is wrong whenever the sizes differ.

Method illustration (not a previous-year question): A class of 10 students averages 70 marks and a class of 30 students averages 50 marks. The combined average is not (70 + 50) ÷ 2 = 60. The totals are 10 × 70 = 700 and 30 × 50 = 1,500, giving a combined total of 2,200 over 40 students, so the combined average is 2,200 ÷ 40 = 55. The larger class dominates, as it should, because it contributes three times as many students. The direct average of averages, 60, overstates the result by ignoring the sizes entirely.

The same weighting idea appears in everyday forms: an overall average spend per person across groups of different sizes, an overall average score across sections with different numbers of questions, or an overall average wage across departments with different headcounts. In each case, multiply each average by its count, add the products, add the counts, and divide. The solved combined-average question below follows this pattern with three groups and a total that divides exactly.

Averages of Number Series: First and Last Terms

When numbers form an evenly spaced series, their average is simply the average of the first and last terms. The numbers in the middle cancel out in pairs: the smallest pairs with the largest, the second smallest with the second largest, and every pair averages to the same value. This makes averages of the first n natural numbers, even numbers and odd numbers quick to find without adding the whole series.

The first n natural numbers run from 1 to n, so their average is (1 + n) ÷ 2. The first n even natural numbers run from 2 to 2n, so their average is (2 + 2n) ÷ 2 = n + 1. The first n odd natural numbers run from 1 to 2n − 1, so their average is (1 + 2n − 1) ÷ 2 = n. These three results are worth memorising, because they appear in NTPC papers in exactly this form, with the count n chosen so that the answer is a whole number.

Method illustration (not a previous-year question): The first 40 even natural numbers run from 2 to 80, so their average is (2 + 80) ÷ 2 = 41, which also equals n + 1 = 41. The first 40 natural numbers average (1 + 40) ÷ 2 = 20.5. The first 40 odd natural numbers average 40. In each case the answer comes from the two end terms alone, with no need to list or add 40 numbers. The second solved question below is this method applied to the first 62 even natural numbers.

Average Speed Versus the Average of Speeds

Average speed is total distance divided by total time. The average of the speeds is a different quantity, and it equals the average speed only in the special case where equal time is spent at each speed. In general, the part of the journey that takes longer weighs more heavily in the true average speed, because the vehicle spends more of its time travelling at that speed. A question that gives speeds for parts of a journey is testing whether you work out the times, not whether you can average two numbers.

When the parts cover equal distances, the average speed for two speeds x and y is 2xy ÷ (x + y). For speeds of 30 km/h and 60 km/h over equal distances, that gives 2 × 30 × 60 ÷ 90 = 40 km/h, not the plain average of 45 km/h. The general method is safer than memorising the formula: find each part's time from time = distance ÷ speed, add the distances, add the times, and divide. Both solved average-speed questions below use that general method, one with times given directly and one with distances given through a ratio.

Method illustration (not a previous-year question): A cyclist rides at 12 km/h for 1 hour and at 18 km/h for 2 hours. The distances are 12 km and 36 km, the total distance is 48 km and the total time is 3 hours, so the average speed is 48 ÷ 3 = 16 km/h. The plain average of the speeds would be 15 km/h, which understates the result because more time was spent at the higher speed. Here the times are given, so the distances follow; in the next section the distances are given, so the times follow. Either way, totals first, division last.

Solved Question 1: Overlapping Averages

Question: The average of 11 results is 61. If the average of the first 6 results is 64 and that of the last 6 is 61, find the 6th result.

(A) 81

(B) 80

(C) 78

(D) 79

Answer: (D) 79.

Previously asked in RRB NTPC Graduate 2026 CBT 1, 16 March 2026, Shift 2 (Mathematics section paper, Question 14).

Shortcut: 6th result = 6 × 64 + 6 × 61 − 11 × 61 = 384 + 366 − 671 = 79.

Full working:

Step 1 — Turn every average into a total. The 11 results average 61, so their total is 11 × 61 = 671. The first 6 results average 64, so their total is 6 × 64 = 384. The last 6 results average 61, so their total is 6 × 61 = 366. Writing the totals down first is the shortcut habit from the formula box, and it converts this question from a puzzle into arithmetic.

Step 2 — Notice the overlap. The first 6 results are results 1 to 6, and the last 6 results are results 6 to 11. Result 6 belongs to both groups. Adding the two group totals, 384 + 366 = 750, therefore counts result 6 twice and every other result once. The grand total of 671 counts every result once, including result 6.

Step 3 — Subtract to isolate the doubly-counted result. The difference between the sum of the group totals and the grand total is exactly the extra copy of result 6: 750 − 671 = 79. So the 6th result is 79, which is option (D).

Step 4 — Check by size. The first group's average, 64, is above the overall average of 61, which suggests the early results, including the 6th, run high. A 6th result of 79 is above both averages, which is consistent: it lifts the first 6 while the last 6, which also contain it, still average only 61 because the remaining five results in that group must run lower. The answer is plausible in size as well as in arithmetic.

Solved Question 2: Average of the First 62 Even Natural Numbers

Question: The average of the first 62 even natural numbers is:

(A) 61

(B) 63

(C) 62

(D) 64

Answer: (B) 63.

Previously asked in RRB NTPC Graduate 2026 CBT 1, 16 March 2026, Shift 2 (Mathematics section paper, Question 19).

Shortcut: The first 62 even numbers run from 2 to 124. Their average is (2 + 124) ÷ 2 = 63. Equivalently, for the first n even numbers the average is n + 1, and 62 + 1 = 63.

Full working:

Step 1 — Identify the series. The even natural numbers are 2, 4, 6, 8, and so on. The first 62 of them end at 2 × 62 = 124, because the nth even natural number is 2n. The series is evenly spaced with a gap of 2 between neighbours.

Step 2 — Use the first-and-last-term rule. In an evenly spaced series, pairs from the two ends average to the same value: 2 pairs with 124 to give 126, 4 pairs with 122 to give 126, and so on through all 31 pairs. The average of the whole series is therefore the average of the two end terms: (2 + 124) ÷ 2 = 126 ÷ 2 = 63.

Step 3 — Cross-check with the memorised result. The average of the first n even natural numbers is n + 1. With n = 62, that gives 63 directly, agreeing with the pairing method. Either route reaches option (B). The trap in this question is the option 62, which is simply the count n; the average of the first 62 even numbers is one more than the count, not the count itself.

Solved Question 3: Average of Four Numbers

Question: Find the average of 312, 126, 100 and 99.

(A) 157.25

(B) 159.25

(C) 156.25

(D) 158.25

Answer: (B) 159.25.

Previously asked in RRB NTPC Graduate 2026 CBT 1, 16 March 2026, Shift 3 (Mathematics section paper, Question 7).

Shortcut: (312 + 126 + 100 + 99) ÷ 4 = 637 ÷ 4 = 159.25.

Full working:

Step 1 — Add the numbers. Group them to keep the arithmetic easy: 312 + 126 = 438, and 100 + 99 = 199. The total is 438 + 199 = 637. Adding in pairs avoids carrying a four-number sum in one step, which is where small addition slips usually happen.

Step 2 — Divide by the count. There are 4 numbers, so the average is 637 ÷ 4 = 159.25. Dividing by 4 can be done by halving twice: 637 ÷ 2 = 318.5, and 318.5 ÷ 2 = 159.25. The answer is option (B).

Step 3 — Sanity-check the size. The largest number is 312 and the other three are all near 100, so the average must sit well below 312 and above 99, closer to the three smaller numbers than to 312. The value 159.25 fits that picture: it is roughly one quarter of the way from the small cluster towards 312, as it should be when one of four numbers is much larger than the rest. The options differ by exactly 1 or 2, so the check is about the addition, and 637 is the sum to verify if your answer disagrees.

Solved Question 4: Combined Average of Three Groups

Question: There are 72 members in group A, 12 members in group B and 87 members in group C. All the members of these groups went to a restaurant. The average amounts spent on each member of groups A, B and C are ₹404, ₹316 and ₹424, respectively. The overall average amount (in ₹) spent per member is:

(A) 404

(B) 412

(C) 408

(D) 409

Answer: (C) 408.

Previously asked in RRB NTPC Graduate 2026 CBT 1, 16 March 2026, Shift 3 (Mathematics section paper, Question 8).

Shortcut: Total spent = 72 × 404 + 12 × 316 + 87 × 424 = 69,768. Total members = 171. Overall average = 69,768 ÷ 171 = 408.

Full working:

Step 1 — Find each group's total spend. Group A: 72 × 404 = 29,088. Group B: 12 × 316 = 3,792. Group C: 87 × 424 = 36,888. Each product is an average multiplied by its group size, which is the weighting step that a direct average of the averages would skip.

Step 2 — Add the totals and the counts. The combined total spent is 29,088 + 3,792 + 36,888 = 69,768. The combined count is 72 + 12 + 87 = 171 members. Both totals are needed before any division happens.

Step 3 — Divide total by count. The overall average is 69,768 ÷ 171 = 408. The division is exact: 171 × 400 = 68,400, leaving 1,368, and 171 × 8 = 1,368, so the quotient is exactly 408. The answer is option (C).

Step 4 — Check against the trap. The plain average of the three group averages is (404 + 316 + 424) ÷ 3 = 1,144 ÷ 3 = 381.33, which is not among the options, and it would be wrong in any case because group B, with the lowest average, has only 12 members while groups A and C have 72 and 87. The true combined average of 408 sits between the two large groups' averages of 404 and 424, pulled slightly above 404 by group C, which is exactly where weighting by size puts it.

Solved Question 5: Average Speed (Time Given)

Question: A woman travelled at a speed of 20 m/min for 90 min, and at a speed of 55 m/min for 60 min. Her average speed (in m/min) for the entire journey is:

(A) 27

(B) 40

(C) 38

(D) 34

Answer: (D) 34 m/min.

Previously asked in RRB NTPC Graduate 2026 CBT 1, 16 March 2026, Shift 3 (Mathematics section paper, Question 25). Sub-topic: average speed.

Shortcut: Distance = 20 × 90 + 55 × 60 = 1,800 + 3,300 = 5,100 m. Time = 90 + 60 = 150 min. Average speed = 5,100 ÷ 150 = 34 m/min.

Full working:

Step 1 — Find the distance covered in each part. At 20 m/min for 90 minutes, the distance is 20 × 90 = 1,800 m. At 55 m/min for 60 minutes, the distance is 55 × 60 = 3,300 m. Distance = speed × time in each part, using the units exactly as given, so no conversion is needed.

Step 2 — Add distances and times. Total distance = 1,800 + 3,300 = 5,100 m. Total time = 90 + 60 = 150 minutes. These are the two totals that average speed is defined from.

Step 3 — Divide total distance by total time. Average speed = 5,100 ÷ 150 = 34 m/min. Dividing by 150 can be done by dividing by 15 and then by 10: 5,100 ÷ 15 = 340, and 340 ÷ 10 = 34. The answer is option (D).

Step 4 — See why the plain average of the speeds fails. The plain average of 20 and 55 is 37.5 m/min, which is higher than the true 34 m/min. The slower speed was held for 90 minutes against 60 minutes at the faster speed, so the slower part weighs more heavily, pulling the average down towards 20. Whenever the times differ, the longer-held speed dominates, and only total distance over total time captures that.

Solved Question 6: Average Speed (Distances in a Ratio)

Question: Three friends travel along three consecutive segments: from P to Q, Q to R, and R to S. Their speeds on these segments are 48 km/h, 72 km/h, and 96 km/h, respectively. The distances of the three segments are in the ratio 2 : 3 : 1, and the total distance from P to S is 480 km. Calculate the average speed for the entire journey.

(A) 58 km/h

(B) 64 km/h

(C) 65 km/h

(D) 68 km/h

Answer: (B) 64 km/h.

Previously asked in RRB NTPC Graduate 2026 CBT 1, 16 March 2026, Shift 3 (Mathematics section paper, Question 26). Sub-topic: average speed.

Shortcut: Distances are 160 km, 240 km and 80 km. Times are 160 ÷ 48 = 10/3 hours, 240 ÷ 72 = 10/3 hours, and 80 ÷ 96 = 5/6 hours, totalling 15/2 = 7.5 hours. Average speed = 480 ÷ 7.5 = 64 km/h.

Full working:

Step 1 — Split the total distance by the ratio. The ratio parts are 2 + 3 + 1 = 6 parts, and the total distance is 480 km, so one part is 480 ÷ 6 = 80 km. The three segment distances are therefore 2 × 80 = 160 km, 3 × 80 = 240 km, and 1 × 80 = 80 km. Check: 160 + 240 + 80 = 480 km, as given.

Step 2 — Find the time for each segment. Time = distance ÷ speed. The first segment takes 160 ÷ 48 = 10/3 hours. The second takes 240 ÷ 72 = 10/3 hours. The third takes 80 ÷ 96 = 5/6 hours. Keeping these as fractions avoids rounding: 10/3 is 3.33 hours, and rounding it early would carry a small error into the final division.

Step 3 — Add the times. The total time is 10/3 + 10/3 + 5/6 = 20/3 + 5/6 = 40/6 + 5/6 = 45/6 = 15/2 hours, which is 7.5 hours. The first two segments take equal times because their distance-to-speed ratios happen to match, which is a useful check: 160 ÷ 48 and 240 ÷ 72 both simplify to 10/3.

Step 4 — Divide total distance by total time. Average speed = 480 ÷ 7.5 = 64 km/h. In fractions: 480 ÷ (15/2) = 480 × 2/15 = 960 ÷ 15 = 64. The answer is option (B). The plain average of the three speeds would be (48 + 72 + 96) ÷ 3 = 72 km/h, well above the true value, because the slowest segment covers a large distance and the fastest segment covers only 80 km. Total distance over total time is the only route that respects how long each speed was actually held.

Practise the Full Papers and the Maths Mock

The 6 questions above come from two 30-question section papers. Attempting the full papers shows how average questions sit among the other Mathematics topics, and a timed mock tests whether the methods above can be recalled quickly rather than merely followed slowly.

• Read the hub note: RRB NTPC Previous Year Question Paper — the full list of playable NTPC papers on this site, including both 16 March 2026 shifts used here.
• Attempt a full paper: RRB NTPC Graduate 2026 Question Paper 16 March Shift 2 — the complete paper from Shift 2, with the official answer key as labelled on this site.
• Practise maths under exam conditions: RRB NTPC Maths Mock Test 02 — a timed maths mock for the same examination.

When you attempt either test, write the implied total beside every average you are given before you calculate anything else. That single habit turns overlapping-group and combined-average questions into the addition and subtraction they really are.

RRB NTPC Average Questions FAQs

How many official average questions are solved on this page?

Six. Two come from the 16 March 2026 Shift 2 Mathematics section paper and four from the Shift 3 paper, and all 60 questions across the two papers were checked to reach that count. Mode, median, train and circular-track questions from the same papers are not counted, because they test different topics.

What is the difference between average speed and the average of speeds?

Average speed is total distance divided by total time, while the average of speeds is just the speeds added and divided by how many there are. They agree only when equal time is spent at each speed. In the solved questions above, the plain average of the speeds gives 37.5 m/min against a true 34 m/min, and 72 km/h against a true 64 km/h, so the distinction is worth real marks.

How do I find a number that belongs to two overlapping groups?

Add the two group totals and subtract the grand total. The overlapped number is counted twice in the group totals and once in the grand total, so the difference leaves exactly that number. This is the method used for the 6th result in the first solved question, where 384 + 366 − 671 = 79.

What is the average of the first n even natural numbers?

It is n + 1. The first n even numbers run from 2 to 2n, and the average of the first and last terms is (2 + 2n) ÷ 2 = n + 1. For the first 62 even numbers, the average is 63. The matching results are (n + 1) ÷ 2 for the first n natural numbers and n for the first n odd natural numbers.

Why is the combined average not the average of the group averages?

Because groups have different sizes, and a larger group contributes more values to the combined total. The combined average must weight each group's average by its size: multiply each average by its count, add the products, add the counts, and divide. Averaging the averages directly treats a group of 12 members as equal to a group of 87, which misstates the combined result.

Sources

• This site's own test files: RRB NTPC Graduate 2026 Maths Question Paper 16 March Shift 2 and Shift 3, each marked official with the official answer key, 30 questions each, exam date 16 March 2026. All 6 solved questions, their options and their answers are taken from these two files, and the remaining 54 stems were read to confirm that only 6 are average questions.
• Railway Recruitment Board portal, rrbapply.gov.in and the regional RRB websites — examination notices and answer-key notices for the RRB NTPC Graduate 2026 CBT 1 examination.

Last checked: 2 November 2026. This page will be updated when further official RRB NTPC section papers are added to this site and their average questions can be verified from the site's own files.