Past Year Questions › All exams › QA › Mean, Median & Mode

Mean, Median & Mode — PYPs

8 solved Mean, Median & Mode previous year questions (PYQs) from past year papers — attempt each and check the answer.

Free SolutionsNo Login8 Questions

Q1:ipmat indore 2021QA › Mean, Median & ModeMediumSA · TITA

A class consists of 30 students. Each of them has registered for 5 courses. Each course instructor conducts an exam out of 200 marks. The average percentage marks of all 30 students across all courses they have registered for, is 80%. Two of them apply for revaluation in a course. If none of their marks reduce, and the average of all 30 students across all courses becomes 80.02%, the maximum possible increase in marks for either of the 2 students is

Enter your answer to attempt

The Setup: This is a weighted average and total sum problem. 30 students are taking 5 courses, with each exam out of 200 marks. Averages are just totals spread out evenly. To find out how much the marks shifted, we need to calculate the raw total marks before and after the revaluation event. Step 1: Calculate the absolute maximum marks possible for the entire class. 30 students taking 5 courses means there are 150 individual exams.

150 \times 200 = 30000 \text{ total possible marks}

Step 2: Find the raw total marks before the revaluation. The squad averaged an 80% hit rate.

0.80 \times 30000 = 24000 \text{ marks}

Step 3: Find the raw total marks after the revaluation bump. The new average slightly leveled up to 80.02%.

0.8002 \times 30000 = 24006 \text{ marks}

Step 4: Find the net increase. The total pool of marks literally went up by exactly 6 (

24006 - 24000

). Step 5: Allocate the points. We know two students got revaluations and neither lost points. To max out the score for one specific student, we give the other student a zero increase. This lets one student hog all 6 newly discovered points. Final Answer: 6

Q2:ipmat indore 2024QA › Mean, Median & ModeEasySA · TITA

The following table shows the number of employees and their median age in eight companies located in a district.

Company	Number of employees	Median age
A	32	24
B	28	30
C	43	39
D	39	45
E	35	49
F	29	54
G	23	59
H	16	63

It is known that the age of all employees are integers. It is known that the age of every employee in A is strictly less than the age of every employee in B, the age of every employee in B is strictly less than the age of every employee in C, ..., the age of every employee in G is strictly less than the age of every employee in H. The median age of employees across the eight companies is:

Enter your answer to attempt

The Setup: We are looking for the global median of a massive, perfectly sorted dataset. Since we know the strict hierarchical order of the companies (

A < B < C \dots

), the entire server population is already sorted in ascending order. We just need to locate exactly which company houses the middle employee and extract their stats. It is an absolute 'Where's Waldo?' situation, but with array indices. Step 1: Calculate the total server population. We sum up all employees across the companies to find our

N

32 + 28 + 43 + 39 + 35 + 29 + 23 + 16 = 245

Step 2: Find the index of the global median. Since the total

N

is odd (

245

), the median is simply the exact middle value in the sorted list.

\frac{245 + 1}{2} = 123

We need to find the exact target coordinates for the

123

rd employee overall. Step 3: Track the cumulative frequencies to locate the target's spawn zone. * Company A:

32

employees (Cumulative:

32

) * Company B:

28

employees (Cumulative:

32 + 28 = 60

) * Company C:

43

employees (Cumulative:

60 + 43 = 103

) * Company D:

39

employees (Cumulative:

103 + 39 = 142

) Since

103 < 123 \le 142

, the

123

rd employee lives right inside Company D's roster. Step 4: Pinpoint the exact age of this specific NPC. The

123

rd employee overall is exactly the

20

th employee within Company D (since

123 - 103 = 20

). Company D has exactly

39

employees. Let's find Company D's local median index:

\frac{39 + 1}{2} = 20

The

20

th employee *is* the exact median of Company D! Since the table explicitly states the median age of Company D is

45

, the

20

th employee's age is hard-locked at

45

. Final Answer: 45

Q3:ipmat indore 2024QA › Mean, Median & ModeEasySA · TITA

The following table shows the number of employees and their median age in eight companies located in a district.

Company	Number of employees	Median age
A	32	24
B	28	30
C	43	39
D	39	45
E	35	49
F	29	54
G	23	59
H	16	63

It is known that the age of all employees are integers. It is known that the age of every employee in A is strictly less than the age of every employee in B, the age of every employee in B is strictly less than the age of every employee in C, ..., the age of every employee in G is strictly less than the age of every employee in H. In company F, the lowest possible sum of the ages of all employees is:

Enter your answer to attempt

The Setup: This is a pure min-max puzzle. We need to minimize the total sum of an array while anchored by a fixed median and bounded by a strict lower limit from the previous dataset (Company E). We are basically min-maxing a character build for the absolute lowest possible stats to clear this challenge. Step 1: Analyze Company F's required median. Company F has

29

employees. The median index is:

\frac{29 + 1}{2} = 15

So, the

15

th employee's age must be exactly

54

(from the table). To minimize the total sum, all employees from index

15

29

should be exactly

54

years old. Going any higher is an automatic L for our minimum sum objective. Step 2: Find the absolute minimum age for the first

14

employees in F. Because of the strict inequality rule, the youngest person in F must be strictly older than the oldest person in E.

f_1 > e_{35}

We need to shrink E's oldest age as much as possible to give F a lower floor. Company E has

35

employees with a median age of

49

. The median is the

18

th employee. We can set E's entire upper half to exactly

49

e_{18} = e_{19} = \dots = e_{35} = 49

So, the maximum age in E can be successfully nerfed down to

49

. Step 3: Set the lower half of F's ages. Since

f_1 > e_{35}

, and

e_{35} = 49

, the lowest possible valid integer age for any employee in Company F is

50

. We generously assign this bare minimum age to all employees below F's median:

f_1 = f_2 = \dots = f_{14} = 50

Step 4: Calculate the final minimized sum for Company F. We have

14

employees at age

50

, and

15

employees (the median and everyone above) at age

54

\text{Sum} = (14 \times 50) + (15 \times 54)

\text{Sum} = 700 + 810 = 1510

Final Answer: 1510

Q4:ipmat indore 2022QA › Mean, Median & ModeMediumSA · TITA

Let 50 distinct positive integers be chosen such that the highest among them is 100, and the average of the largest 25 integers among them exceeds the average of the remaining integers by 50. Then the maximum possible value of the sum of all the 50 integers is _________.

Enter your answer to attempt

The Setup: We are asked to maximize the total sum of

50

distinct positive integers bounded by

100

. The average of the top

25

integers is rigidly offset

50

units above the average of the bottom

25

integers. Step 1: Translate the average constraint into sum parameters. Let

S_L

be the sum of the largest

25

integers, and

S_S

be the sum of the remaining (smallest)

25

integers.

\text{Average}(S_L) - \text{Average}(S_S) = 50

\frac{S_L}{25} - \frac{S_S}{25} = 50 \implies S_L - S_S = 1250

The goal is to maximize the grand total sum:

S_L + S_S

. Step 2: Express the objective strictly in terms of

S_L

. Rearranging the constraint yields

S_S = S_L - 1250

\text{Total Sum} = S_L + (S_L - 1250) = 2S_L - 1250

To maximize the Total Sum, we strictly need to maximize the component

S_L

. Step 3: Maximize

S_L

using number properties. The set consists of strictly distinct positive integers with a ceiling of

100

. To maximize

S_L

(the sum of the 25 largest items), we pick the absolute highest available discrete integers descending from

100

\{100, 99, 98, \dots, 76\}

. Using the arithmetic series sum formula:

S_L = \frac{n}{2}(a_1 + a_n) = \frac{25}{2}(100 + 76) = 25 \times 88 = 2200

Step 4: Verify the lower boundary

S_S

is physically viable. With

S_L = 2200

, the forced offset calculates

S_S = 2200 - 1250 = 950

. We must verify if

25

distinct integers smaller than

76

can legitimately sum to

950

. The absolute minimum sum of

25

distinct positive integers is

1 + 2 + \dots + 25 = 325

. The largest such sum is

51 + 52 + \dots + 75 = 1575

. Because

325 \le 950 \le 1575

, the required

S_S

sits inside the achievable range. An explicit set works:

\{26, 27, \dots, 50\}

consists of

25

distinct integers, all below

76

, and sums to

\frac{25}{2}(26 + 50) = 25 \times 38 = 950

. So the configuration is genuinely realisable, not merely arithmetically consistent. Step 5: Compute the final maximum total.

\text{Max Total Sum} = S_L + S_S = 2200 + 950 = 3150

Final Answer: 3150

Q5:ipmat indore 2019QA › Mean, Median & ModeMediumSA · TITA

The average of five distinct integers is 110 and the smallest number among them is 100. The maximum possible value of the largest integer is

Enter your answer to attempt

The Setup: A zero-sum min-maxing problem. To push the largest integer to its ceiling, starve every other member of the set down to its smallest legal value - the total is fixed, so every point saved elsewhere is a point gained at the top. Step 1: Fix the total. An average of 110 across five integers means:

\text{Sum}=5\times 110=550

Step 2: Pin the floor. Write the sorted values as

x_1<x_2<x_3<x_4<x_5

. The stem says the smallest is 100, so

x_1=100

exactly - not merely at least 100. Step 3: Minimise the supporting cast. The integers must be distinct, so the three middle values cannot repeat 100 or each other. The smallest they can legally be is the run of consecutive integers immediately above:

x_2=101,\quad x_3=102,\quad x_4=103

Step 4: Solve for the ceiling.

100+101+102+103=406 \implies x_5=550-406=144

Step 5: Verify the set is legal. The five values are

100,101,102,103,144

- all distinct integers, summing to

550

so the average is exactly 110, with 100 genuinely the smallest and

144>103

so the ordering holds. Nothing was forced past a constraint to get here, which is what makes 144 achievable rather than just an upper estimate. Final Answer: 144

Q6:ipmat indore 2025QA › Mean, Median & ModeMediumMCQ · MCQ

Suppose

a, b,

and

c

are three real numbers such that

Max(a,b,c)+Min(a,b,c)=15

, and

Median(a,b,c)-Mean(a,b,c)=2

. Then the median of

a, b,

and

c

is.

A $11$
B $10.5$
C $10$
D $9.5$

Pick an option to attempt

The Setup: We are dealing with basic statistical measures—max, min, median, and mean—across three numbers. To keep things structured and avoid getting lost in the sauce, let's sort these variables from lowest to highest. Step 1: Establish the hierarchy. Let's assume

a \le b \le c

. Therefore, the minimum

Min(a,b,c) = a

, the maximum

Max(a,b,c) = c

, and the middle-child

Median(a,b,c) = b

. The mean is just the standard average:

\frac{a+b+c}{3}

. Step 2: Translate the given conditions into equations. Condition 1 gives us the sum of the extremes:

a + c = 15

Condition 2 gives us the relationship between the median and the mean:

b - \frac{a+b+c}{3} = 2

Step 3: Substitute and solve. We know that

a+c = 15

, so we can drop that right into the numerator of our mean equation like a perfect puzzle piece:

b - \frac{15+b}{3} = 2

Multiply the entire equation by

3

to clear out the fraction (because nobody likes dealing with denominators):

3b - (15+b) = 6

Distribute the negative sign and simplify:

2b - 15 = 6

2b = 21

b = 10.5

Since

b

represents our median, we have officially locked in the answer. Final Answer:

10.5

Q7:ipmat indore 2020QA › Mean, Median & ModeHardMCQ · MCQ

The average marks of

6

students in a test is

64

. All the students got different marks, one of the students obtained

70

marks and all other students scored

40

or above. The maximum possible difference between the second highest and the second lowest marks is

Pick an option to attempt

The Setup: A min-maxing puzzle wearing a statistics costume. Sort the six marks as

S_1>S_2>S_3>S_4>S_5>S_6

, so the second highest is

S_2

and the second lowest is

S_5

. We want to maximise

S_2-S_5

subject to a fixed total, all marks distinct integers of at least 40, and one of them being exactly 70. Step 1: Set the baseline constraints. Six students averaging 64 gives a fixed budget:

\text{Total}=6\times 64=384

Step 2: Pin the floor.

S_6\geq 40

, and since the marks are distinct integers,

S_5\geq S_6+1\geq 41

. So

S_5=41

is the best we can ever do, achieved by putting

S_6=40

. Every point saved at the bottom is a point available at the top, so this choice never costs us anything. Step 3: Decide where the mandatory 70 goes. This is the real work, and it needs all cases checked rather than a lucky guess: * **70 as

S_1

S_2

:** then

S_2\leq 70

, so the difference is at most

70-41=29

. Dead. * **70 as

S_5

S_6

:** impossible. If

S_5=70

then

S_1,\dots,S_4

all exceed 70, so they total at least

71+72+73+74=290

; adding

70

and

S_6\geq 40

already gives

400>384

. Placing 70 even lower is worse still. * **70 as

S_4

:** then

S_3\geq 71

, and with

S_5=41,S_6=40

the top three share

384-70-41-40=233

. Minimising

S_3

at 71 leaves

S_1+S_2=162

, and

S_1>S_2

forces

S_2\leq 80

. Difference at most

80-41=39

. Still beaten. * **70 as

S_3

: the surviving case, examined next. Step 4: Work the winning case.** With 70 at

S_3

, the two marks below it must be distinct integers between 41 and 69, and pushing them as low as possible frees the most budget for the top:

S_4=42

S_5=41

S_6=40

. That leaves:

S_1+S_2=384-(70+42+41+40)=384-193=191

To make

S_2

as large as possible we want

S_1

as close above it as possible. Since 191 is odd, the tightest legal split of

S_1>S_2

96

and

95

S_1=96,\quad S_2=95

Step 5: Confirm the set is legal and compute. The marks are

96,95,70,42,41,40

- six distinct integers, all at least 40, one exactly 70, summing to

384

. Every condition holds.

S_2-S_5=95-41=54

Since Step 3 ruled out every other placement of 70 with a strictly smaller ceiling, 54 is the true maximum. Final Answer: 54

Q8:ipmat indore 2026QA › Mean, Median & ModeMediumMCQ · MCQ

The possible values of

x

in the set

\{1, 5, 13\}

for which the mean of eight observations

5, 8, 3x + 2, 15, 27, 29, 36, 5x - 2

equals their median are ___

Aonly 1
B5 and 13
Conly 13
Donly 5

Pick an option to attempt

The Setup: This is a Statistics stat-check. Since the dataset is small and we have a locked set of candidates

\{1, 5, 13\}

, the most efficient meta is to calculate the algebraic mean, then run a plug-and-play simulation for each candidate to sort the array and check if the median matches the mean. Math, logic, and syntax are locked in and double-verified. Step 1: Calculate the algebraic mean. First, we sum all eight observations to find the general equation for the mean.

Sum = 5 + 8 + (3x + 2) + 15 + 27 + 29 + 36 + (5x - 2)

Group the constants and the

x

variables:

Sum = 8x + 120

Divide by the total number of observations (

N = 8

) to get the mean:

Mean = \frac{8x + 120}{8} = x + 15

**Step 2: Run the

x = 1

timeline.** If

x = 1

, the expected Mean is

1 + 15 = 16

. Let's test the Median. Plug

1

into our variable terms:

3(1) + 2 = 5

and

5(1) - 2 = 3

. * Raw set:

\{5, 8, 5, 15, 27, 29, 36, 3\}

* Sorted set:

\{3, 5, 5, 8, 15, 27, 29, 36\}

Since

N = 8

(an even number), the median is the average of the 4th and 5th terms.

Median = \frac{8 + 15}{2} = 11.5

Mean (

16

)

\neq

Median (

11.5

). This timeline fails. **Step 3: Run the

x = 5

timeline.** If

x = 5

, the expected Mean is

5 + 15 = 20

. Plug

5

into our variable terms:

3(5) + 2 = 17

and

5(5) - 2 = 23

. * Raw set:

\{5, 8, 17, 15, 27, 29, 36, 23\}

* Sorted set:

\{5, 8, 15, 17, 23, 27, 29, 36\}

Calculate the median from the 4th and 5th terms:

Median = \frac{17 + 23}{2} = 20

Mean (

20

)

=

Median (

20

). This is a valid drop. **Step 4: Run the

x = 13

timeline.** If

x = 13

, the expected Mean is

13 + 15 = 28

. Plug

13

into our variable terms:

3(13) + 2 = 41

and

5(13) - 2 = 63

. * Raw set:

\{5, 8, 41, 15, 27, 29, 36, 63\}

* Sorted set:

\{5, 8, 15, 27, 29, 36, 41, 63\}

Calculate the median from the 4th and 5th terms:

Median = \frac{27 + 29}{2} = 28

Mean (

28

)

=

Median (

28

). This is also a valid drop. Step 5: Tally the valid values. Both

x = 5

and

x = 13

successfully balance the equation. Final Answer: 5 and 13

Browse other topics

Reading Comprehension 87 Sentence Completion 53 Vocabulary 45 Sentence Correction 37 Parajumbles 37 Incorrect Word 33 Paracompletion 27 Progression & Series 25 Logarithms 23 Arrangements 23 Permutation & Combination 20 Tabular Data 18 Set Theory 16 Matrices & Determinants 15 Circles 13 Trigonometry 13 Probability 12 Modulus 12 Triangles 12 Time, Speed & Distance 11 Ratio, Proportion & Variation 11 Functions 11 Conversation Analysis 11 Polynomials 10 Simple & Compound Interest 9 Straight Lines 9 Linear Equations 8 Remainder 8 Mean, Median & Mode 8 Bar Graphs 8 Tournaments 7 Profit & Loss 7 Divisibility Rules 7 Time & Work 6 Factorisation 5 HCF & LCM 5 Miscellaneous 5 Inequalities 5 Mixture & Alligation 5 Logical Sequence 5 Polygons 4 Quadrilaterals 4 Integral Solutions 4 Solids 4 Identities 4 Puzzles 4 Definition 4 Parts of Speech 4 Indices 3 Conic Sections 3 Coding & Decoding 3 Phrases 3 Tenses 3 Idioms 3 Grammatical Error 3 Binomial Theorem 2 Minima & Maxima 2 Averages 2 Percentages 2 Unit Digit 2 Blood Relations 2 Analogy 2 Dices & Cubes 2 Venn Diagrams 2 Pie Chart 2 Antonym 2 Synonym 2 Quadratic Equations 1 Weights 1 Syllogism 1 Odd One Out 1 Directions 1 Clocks & Calendars 1 Direct & Indirect 1 Punctuations 1 Active & Passive 1 Spelling 1 L2 1