Average Questions and Answers

Q: The average attendance in a school for the first 4 days of the week is 30 and for the first 5 days of the week is 32. The attendance on the fifth day is

The attendance on the fifth day is 40.Convert each average into a total:\[ \begin{aligned} \text{first 4 days} &= 4 \times 30 = 120 \\ \text{first 5 days} &= 5 \times 32 = 160 \\ \text{fifth day} &= 160 - 120 = 40 \end{aligned} \]Shortcut: the fifth day must lift the average of five days by 2, so it is the old average plus \(5 \times 2\): \(30 + 10 = 40\). 32 is the new average itself, a common slip, and 36 and 38 do not give a five-day total of 160.

Q: The average of 5 consecutive even numbers is 42. Find the largest number is

The largest number is 46.For an odd count of consecutive even numbers, the average is the middle number. So the five numbers are centred on 42, going up and down in steps of 2:\[ 38,\ 40,\ 42,\ 44,\ 46 \]Check: \(38 + 40 + 42 + 44 + 46 = 210 = 5 \times 42\). 44 is the next number after the middle, not the largest. 43 and 45 are odd, so they cannot belong to a set of even numbers at all.

Q: The average of 5 consecutive integers starting with m is n. What is the average of 6 consecutive integers starting with (m+2) ?

The average is \(\frac{2n+5}{2}\).Five consecutive integers starting at m are m to m + 4, and their average is the middle one: \(n = m + 2\). Six consecutive integers starting at m + 2 run from m + 2 to m + 7, and their average is the midpoint of the ends:\[ \begin{aligned} \text{average} &= \frac{(m+2) + (m+7)}{2} = m + 4.5 \\ &= n + 2.5 = \frac{2n + 5}{2} \end{aligned} \]n + 2 and n + 3 are near misses on either side of n + 2.5, and \(\frac{n+5}{2}\) halves n by mistake.

Q: 45% of first number is equal to the 60% of the second number. If average of both the numbers is 10 less than the first number, then find 80% of second number.

80% of the second number is 48.Let the numbers be \(a\) and \(b\). From \(0.45a = 0.6b\), \(a = \frac{4}{3}b\). Their average is 10 less than \(a\).\[ \begin{aligned} \frac{a + b}{2} &= a - 10 \\ b &= a - 20 = \frac{4}{3}b - 20 \\ b &= 60,\ a = 80 \end{aligned} \]So 80% of 60 is 48. Check: 45% of 80 = 36 = 60% of 60, and the average 70 is 10 less than 80. 64 is 80% of the first number.

Q: The average age of 7 family members is 75 years, but the average age of 6 of them is 74 years. What is the age of the 7th family member ?

The seventh member is 81 years old.Work with totals rather than averages:\[ \begin{aligned} \text{total of 7} &= 7 \times 75 = 525 \\ \text{total of 6} &= 6 \times 74 = 444 \\ \text{7th member} &= 525 - 444 = 81 \end{aligned} \]75 is just the overall average, and 80 or 70 do not come from the totals.A shortcut: adding the seventh member raises the average of the other six from 74 to 75, so he must be \(74 + 7 \times 1 = 81\) years old.

Q: The average of three numbers is 77. The first number is twice the second and the second number is twice the third. Then the first number is

The first number is 132.Let the third number be \(t\). Then the second is \(2t\) and the first is \(4t\), and their total is three times the average:\[ \begin{aligned} t + 2t + 4t &= 3 \times 77 = 231 \\ 7t &= 231, \ t = 33 \end{aligned} \]So the numbers are 132, 66 and 33, and the first is 132. 33 and 66 are the third and second numbers, and 77 is the average.Check: \((132 + 66 + 33)/3 = 231/3 = 77\), the given average.

Q: The arithmatic mean of 12 observations is 7.5. If the arithmatic mean of 7 of these observations is 6.5, then the mean of the remaining observations is

The mean of the remaining observations is 8.9.\[ \begin{aligned} \text{total of 12} &= 12 \times 7.5 = 90 \\ \text{total of 7} &= 7 \times 6.5 = 45.5 \\ \text{mean of 5} &= \frac{90 - 45.5}{5} = 8.9 \end{aligned} \]The remaining observations number \(12 - 7 = 5\); dividing by 7 instead would give about 6.4, which is not offered. 5.5, 8.5 and 9.2 do not come from the totals.Working with totals rather than averages is the standard method for questions in which part of a group is removed.

Q: The average age of 20 number students in a class is 11 years. If the age of teacher is added, the average age is increased by 1 year. What is the age of the teacher ?

The teacher is 32 years old.Work with totals. The 20 students' ages add to \(20 \times 11 = 220\). With the teacher there are 21 people averaging 12 years.\[ \begin{aligned} \text{new total} &= 21 \times 12 = 252 \\ \text{teacher} &= 252 - 220 = 32 \end{aligned} \]Shortcut: the teacher brings his own share (12) plus 1 year for each of the 20 students, \(12 + 20 = 32\). 36 and 42 fit no step, so 'None of the above' is wrong too.

Q: The average age of 40 students is 16 years. If the teacher's age is also included, the average becomes 17 years. What is the age of the teacher ?

The teacher is 57 years old.Compare the total ages before and after the teacher is included:\[ \begin{aligned} \text{Students} &= 40 \times 16 = 640 \\ \text{With teacher} &= 41 \times 17 = 697 \\ \text{Teacher} &= 697 - 640 = 57 \end{aligned} \]Shortcut: the teacher must supply his own share of the new average (17) plus one extra year for each of the 40 students, which is 17 + 40 = 57.A teacher of 56 would make the new total 696, an average just under 17, and 48 or 60 miss it by more. Only 57 makes the average of all 41 people exactly 17.

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41.

The average attendance in a school for the first 4 days of the week is 30 and for the first 5 days of the week is 32. The attendance on the fifth day is

A.36
B.40
C.32
D.38

Exam Year(s)

Tripura Public Service Commission - 2026,

Answer: B

The attendance on the fifth day is 40.

Convert each average into a total:

\[ \begin{aligned} \text{first 4 days} &= 4 \times 30 = 120 \\ \text{first 5 days} &= 5 \times 32 = 160 \\ \text{fifth day} &= 160 - 120 = 40 \end{aligned} \]

Shortcut: the fifth day must lift the average of five days by 2, so it is the old average plus \(5 \times 2\): \(30 + 10 = 40\). 32 is the new average itself, a common slip, and 36 and 38 do not give a five-day total of 160.

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42.

The average of 5 consecutive even numbers is 42. Find the largest number is

A.46
B.43
C.44
D.45

Exam Year(s)

Tripura Public Service Commission - 2026,

Answer: A

The largest number is 46.

For an odd count of consecutive even numbers, the average is the middle number. So the five numbers are centred on 42, going up and down in steps of 2:

\[ 38,\ 40,\ 42,\ 44,\ 46 \]

Check: \(38 + 40 + 42 + 44 + 46 = 210 = 5 \times 42\). 44 is the next number after the middle, not the largest. 43 and 45 are odd, so they cannot belong to a set of even numbers at all.

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43.

The average of 5 consecutive integers starting with m is n. What is the average of 6 consecutive integers starting with (m+2) ?

A.\(\frac{2n+5}{2}\)
B.(n + 2)
C.(n+3)
D.\(\frac{n+5}{2}\)

Exam Year(s)

Tripura Public Service Commission - 2025,

Answer: A

The average is \(\frac{2n+5}{2}\).

Five consecutive integers starting at m are m to m + 4, and their average is the middle one: \(n = m + 2\). Six consecutive integers starting at m + 2 run from m + 2 to m + 7, and their average is the midpoint of the ends:

\[ \begin{aligned} \text{average} &= \frac{(m+2) + (m+7)}{2} = m + 4.5 \\ &= n + 2.5 = \frac{2n + 5}{2} \end{aligned} \]

n + 2 and n + 3 are near misses on either side of n + 2.5, and \(\frac{n+5}{2}\) halves n by mistake.

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44.

45% of first number is equal to the 60% of the second number. If average of both the numbers is 10 less than the first number, then find 80% of second number.

A.64
B.40
C.48
D.56

Exam Year(s)

Tripura Public Service Commission - 2026,

Answer: C

80% of the second number is 48.

Let the numbers be \(a\) and \(b\). From \(0.45a = 0.6b\), \(a = \frac{4}{3}b\). Their average is 10 less than \(a\).

\[ \begin{aligned} \frac{a + b}{2} &= a - 10 \\ b &= a - 20 = \frac{4}{3}b - 20 \\ b &= 60,\ a = 80 \end{aligned} \]

So 80% of 60 is 48. Check: 45% of 80 = 36 = 60% of 60, and the average 70 is 10 less than 80. 64 is 80% of the first number.

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45.

The average age of 7 family members is 75 years, but the average age of 6 of them is 74 years. What is the age of the 7th family member ?

A.80 years
B.81 years
C.75 years
D.70 years

Exam Year(s)

Tripura Public Service Commission - 2019,

Answer: B

The seventh member is 81 years old.

Work with totals rather than averages:

\[ \begin{aligned} \text{total of 7} &= 7 \times 75 = 525 \\ \text{total of 6} &= 6 \times 74 = 444 \\ \text{7th member} &= 525 - 444 = 81 \end{aligned} \]

75 is just the overall average, and 80 or 70 do not come from the totals.

A shortcut: adding the seventh member raises the average of the other six from 74 to 75, so he must be \(74 + 7 \times 1 = 81\) years old.

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46.

The average of three numbers is 77. The first number is twice the second and the second number is twice the third. Then the first number is

A.33
B.77
C.66
D.132

Exam Year(s)

Tripura Public Service Commission - 2019,

Answer: D

The first number is 132.

Let the third number be \(t\). Then the second is \(2t\) and the first is \(4t\), and their total is three times the average:

\[ \begin{aligned} t + 2t + 4t &= 3 \times 77 = 231 \\ 7t &= 231, \ t = 33 \end{aligned} \]

So the numbers are 132, 66 and 33, and the first is 132. 33 and 66 are the third and second numbers, and 77 is the average.

Check: \((132 + 66 + 33)/3 = 231/3 = 77\), the given average.

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47.

The arithmatic mean of 12 observations is 7.5. If the arithmatic mean of 7 of these observations is 6.5, then the mean of the remaining observations is

A.5.5
B.8.5
C.9.2
D.8.9.

Exam Year(s)

Tripura Public Service Commission - 2019,

Answer: D

The mean of the remaining observations is 8.9.

\[ \begin{aligned} \text{total of 12} &= 12 \times 7.5 = 90 \\ \text{total of 7} &= 7 \times 6.5 = 45.5 \\ \text{mean of 5} &= \frac{90 - 45.5}{5} = 8.9 \end{aligned} \]

The remaining observations number \(12 - 7 = 5\); dividing by 7 instead would give about 6.4, which is not offered. 5.5, 8.5 and 9.2 do not come from the totals.

Working with totals rather than averages is the standard method for questions in which part of a group is removed.

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48.

The average age of 20 number students in a class is 11 years. If the age of teacher is added, the average age is increased by 1 year. What is the age of the teacher ?

A.32 years
B.36 years
C.42 years
D.None of the above

Exam Year(s)

Tripura Public Service Commission - 2019,

Answer: A

The teacher is 32 years old.

Work with totals. The 20 students' ages add to \(20 \times 11 = 220\). With the teacher there are 21 people averaging 12 years.

\[ \begin{aligned} \text{new total} &= 21 \times 12 = 252 \\ \text{teacher} &= 252 - 220 = 32 \end{aligned} \]

Shortcut: the teacher brings his own share (12) plus 1 year for each of the 20 students, \(12 + 20 = 32\). 36 and 42 fit no step, so 'None of the above' is wrong too.

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49.

The average age of 40 students is 16 years. If the teacher's age is also included, the average becomes 17 years. What is the age of the teacher ?

A.48 years
B.56 years
C.60 years
D.57 years

Exam Year(s)

Tripura Public Service Commission - 2026,

Answer: D

The teacher is 57 years old.

Compare the total ages before and after the teacher is included:

\[ \begin{aligned} \text{Students} &= 40 \times 16 = 640 \\ \text{With teacher} &= 41 \times 17 = 697 \\ \text{Teacher} &= 697 - 640 = 57 \end{aligned} \]

Shortcut: the teacher must supply his own share of the new average (17) plus one extra year for each of the 40 students, which is 17 + 40 = 57.

A teacher of 56 would make the new total 696, an average just under 17, and 48 or 60 miss it by more. Only 57 makes the average of all 41 people exactly 17.

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50.

A grocer has a sale of Rs. 6435, Rs. 6927, Rs. 6855, Rs. 7230 and Rs. 6562 for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of Rs. 6500?

A.4800
B.4991
C.5004
D.5000

Answer: B

Let the sale in the sixth month \(=x\)
Then, \(\dfrac{6435+6927+6855+7230+6562+x}{6}=6500\)
\(\Rightarrow 6435+6927+6855+7230+6562+x=6 \times 6500\\ \Rightarrow 34009 + x = 39000\\ \Rightarrow x = 39000 - 34009 = 4991\)

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