Average Questions in English

(A) The initial sum of the ages of $4$ family members is $4 \times 32 = 128$ years.
The new average age after including the guest is $32 + (12.5 \% \text{ of } 32) = 32 + (0.125 \times 32) = 32 + 4 = 36$ years.
Let the age of the guest be $G$. The total number of people is now $5$.
The new average is given by $\frac{128 + G}{5} = 36$.
Multiplying both sides by $5$,we get $128 + G = 180$.
Solving for $G$,we get $G = 180 - 128 = 52$ years.

(D) The initial sum of the ages of $6$ members is $6 \times 20 = 120$ years.
The new average age after including the servant is $20 + 25\% \text{ of } 20 = 20 + 5 = 25$ years.
There are now $6 + 1 = 7$ people in the family including the servant.
The total sum of the ages of $7$ people is $7 \times 25 = 175$ years.
Let the age of the servant be $S$. Then,$120 + S = 175$.
Therefore,$S = 175 - 120 = 55$ years.

(B) Total revenue for $13$ years $= 13 \times 70 = 910$ lakhs.
Total revenue for the first $7$ years $= 7 \times 65 = 455$ lakhs.
Total revenue for the last $7$ years $= 7 \times 77 = 539$ lakhs.
The $7^{th}$ year is included in both the first $7$ years and the last $7$ years.
Therefore,revenue of the $7^{th}$ year $= (455 + 539) - 910 = 994 - 910 = 84$ lakhs.

(D) Let the total number of students be $N$. The total height of all students is $T = aN$.
The sum of heights of $10$ students is $10b$.
The number of remaining students is $(N-10)$,and their total height is $(N-10)c$.
The sum of the heights of the two groups equals the total height of the class:
$10b + (N-10)c = aN$
Expand the equation:
$10b + Nc - 10c = aN$
Rearrange to solve for $N$:
$10b - 10c = aN - Nc$
$10(b-c) = N(a-c)$
Therefore,the total number of students is:
$N = \frac{10(b-c)}{a-c}$

(A) Let the temperatures on Monday,Tuesday,Wednesday,Thursday,and Friday be $M, T, W, Th,$ and $F$ respectively.
The average temperature for Monday,Tuesday,Wednesday,and Thursday is $48^{\circ}$.
So,$(M + T + W + Th) / 4 = 48$,which implies $M + T + W + Th = 48 \times 4 = 192$.
The average temperature for Tuesday,Wednesday,Thursday,and Friday is $52^{\circ}$.
So,$(T + W + Th + F) / 4 = 52$,which implies $T + W + Th + F = 52 \times 4 = 208$.
Subtracting the first equation from the second:
$(T + W + Th + F) - (M + T + W + Th) = 208 - 192$
$F - M = 16$.
Given that the temperature on Monday $(M)$ is $42^{\circ}$,we substitute this value:
$F - 42 = 16$
$F = 16 + 42 = 58$.
Therefore,the temperature on Friday was $58^{\circ}$.

(B) Total marks of the class $= 60 \times 65 = 3900$.
Number of students in half of the class $= 60 / 2 = 30$.
Total marks of these $30$ students $= 30 \times 85 = 2550$.
Total marks of the remaining $30$ students $= 3900 - 2550 = 1350$.
Average marks of the remaining students $= 1350 / 30 = 45$.

(C) Let the number of passed candidates be $x$ and the number of failed candidates be $(100 - x)$.
According to the problem,the total marks obtained by all candidates is $100 \times 30 = 3000$.
The total marks obtained by passed candidates is $35x$.
The total marks obtained by failed candidates is $10(100 - x)$.
Therefore,$35x + 10(100 - x) = 3000$.
$35x + 1000 - 10x = 3000$.
$25x = 2000$.
$x = \frac{2000}{25} = 80$.
Thus,the number of candidates who passed the examination is $80$.

(C) The sum of $25$ results is $25 \times 20 = 500$.
The sum of the first $12$ results is $12 \times 15 = 180$.
The sum of the last $12$ results is $12 \times 18 = 216$.
Let the $13^{th}$ result be $x$.
Therefore,the sum of all $25$ results can be expressed as: (Sum of first $12$) + ($13^{th}$ result) + (Sum of last $12$) = $500$.
$180 + x + 216 = 500$.
$x + 396 = 500$.
$x = 500 - 396$.
$x = 104$.
Thus,the $13^{th}$ result is $104$.

(B) Given,the number of observations $n = 100$ and the initial average = $35$.
Sum of observations = $100 \times 35 = 3500$.
Since one observation was misread as $83$ instead of $53$,we need to subtract the incorrect value and add the correct value to the sum.
Correct sum = $3500 - 83 + 53 = 3500 - 30 = 3470$.
Correct average = $\frac{\text{Correct sum}}{n} = \frac{3470}{100} = 34.7$.

(A) The average of $x$ and $y$ is given by $\frac{x+y}{2}$.
The average of $y$ and $z$ is given by $\frac{y+z}{2}$.
According to the problem,the difference between these averages is $12$:
$\frac{x+y}{2} - \frac{y+z}{2} = 12$
Multiplying the entire equation by $2$ to clear the denominator:
$(x+y) - (y+z) = 12 \times 2$
$x + y - y - z = 24$
$x - z = 24$
Therefore,the difference between $x$ and $z$ is $24$.

(D) Let the $13$ consecutive odd integers be $x, x+2, x+4, \dots, x+24$.
The average of the first $7$ integers is given as $37$.
The first $7$ integers are $x, x+2, x+4, x+6, x+8, x+10, x+12$.
The average of these $7$ integers is the middle term, which is the $4^{\text{th}}$ term: $x+6 = 37$.
Thus, $x = 31$.
The entire series consists of $13$ consecutive odd integers starting from $31$: $31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55$.
The average of an arithmetic progression is the middle term. For $13$ terms, the middle term is the $\frac{13+1}{2} = 7^{\text{th}}$ term.
The $7^{\text{th}}$ term is $x + (7-1) \times 2 = 31 + 12 = 43$.
Alternatively, the $7^{\text{th}}$ term is $37 + (7-4) \times 2 = 37 + 6 = 43$.

(B) Total number of students = $35$.
Initial average marks = $35$.
Total marks calculated initially = $35 \times 35 = 1225$.
Since the marks of one student were incorrectly entered as $65$ instead of $35$,we need to subtract the incorrect value and add the correct value.
Correct total marks = $1225 - 65 + 35 = 1195$.
Correct average = $\frac{1195}{35} \approx 34.14$.

(B) Let the three consecutive even numbers be $x$,$x+2$,and $x+4$.
The sum of these numbers is $S = x + (x+2) + (x+4) = 3x + 6$.
The average of these numbers is $A = \frac{3x+6}{3} = x+2$.
According to the problem,the sum is $28$ more than the average:
$S = A + 28$
$(3x+6) = (x+2) + 28$
$3x + 6 = x + 30$
$2x = 24$
$x = 12$.
Thus,the smallest number is $12$.

(C) Let the runs scored in the $100^{\text{th}}$ innings be $x$.
Total runs scored in $99$ innings $= 99 \times 99 = 9801$.
Total runs required for an average of $100$ in $100$ innings $= 100 \times 100 = 10000$.
Runs to be scored in the $100^{\text{th}}$ innings $= 10000 - 9801 = 199$.

(C) Let the number of wickets taken by the cricketer before this match be $w$.
Total runs conceded before this match = $12.4w$.
In the current match,he takes $5$ wickets for $22$ runs.
Total wickets after the match = $w + 5$.
Total runs after the match = $12.4w + 22$.
The new average is $12.4 - 0.4 = 12.0$ runs/wicket.
According to the problem:
$\frac{12.4w + 22}{w + 5} = 12.0$
$12.4w + 22 = 12(w + 5)$
$12.4w + 22 = 12w + 60$
$12.4w - 12w = 60 - 22$
$0.4w = 38$
$w = \frac{38}{0.4} = 95$.
Thus,the number of wickets taken before this match was $95$.

(D) Total score of the player for $40$ innings $= 50 \times 40 = 2000$ runs.
Let the highest score be $H$ and the lowest score be $L$. Given,$H - L = 172$.
Total score of the remaining $38$ innings $= 38 \times 48 = 1824$ runs.
The sum of the highest and lowest scores is $H + L = 2000 - 1824 = 176$.
Now,we have a system of two linear equations:
$1) H + L = 176$
$2) H - L = 172$
Adding these two equations:
$(H + L) + (H - L) = 176 + 172$
$2H = 348$
$H = 174$.
Therefore,the highest score of the player is $174$.

(B) Let the present ages of the husband,wife,and child be $H$,$W$,and $C$ respectively.
The average age of the husband,wife,and child $3 \text{ years}$ ago was $27 \text{ years}$.
Sum of their ages $3 \text{ years}$ ago $= 27 \times 3 = 81$.
Sum of their present ages $(H + W + C) = 81 + (3 \times 3) = 81 + 9 = 90$.
The average age of the wife and child $5 \text{ years}$ ago was $20 \text{ years}$.
Sum of their ages $5 \text{ years}$ ago $= 20 \times 2 = 40$.
Sum of their present ages $(W + C) = 40 + (5 \times 2) = 40 + 10 = 50$.
Now,subtract the sum of the wife and child's present ages from the total sum:
$H = (H + W + C) - (W + C) = 90 - 50 = 40$.
Therefore,the present age of the husband is $40 \text{ years}$.

(C) Let the number of remaining workers be $x$.
The total salary of all workers is the sum of the salaries of the technicians and the remaining workers.
Using the formula for average: $\text{Average} = \frac{\text{Total Sum}}{\text{Number of items}}$.
$\frac{7 \times 15000 + x \times 9000}{7 + x} = 12000$
Multiply both sides by $(7 + x)$:
$105000 + 9000x = 12000(7 + x)$
$105000 + 9000x = 84000 + 12000x$
Rearrange the terms to solve for $x$:
$105000 - 84000 = 12000x - 9000x$
$21000 = 3000x$
$x = \frac{21000}{3000} = 7$
The total number of workers is $7 + x = 7 + 7 = 14$.

(A) Let the number of remaining associates be $x$.
The total salary of the team is the sum of the salaries of the senior associates and the remaining associates.
Total salary $= (7 \times 24000) + (x \times 12000)$.
The average salary of all associates is given by the formula: $\text{Average} = \frac{\text{Total Salary}}{\text{Total Number of Associates}}$.
Given,$16000 = \frac{(7 \times 24000) + (12000x)}{7 + x}$.
Multiplying both sides by $(7 + x)$:
$16000(7 + x) = 168000 + 12000x$.
$112000 + 16000x = 168000 + 12000x$.
Subtracting $12000x$ and $112000$ from both sides:
$4000x = 56000$.
$x = \frac{56000}{4000} = 14$.
Total number of associates $= 7 + x = 7 + 14 = 21$.

(A) Let the number of visitors on the three days be $2x, 5x,$ and $13x$ respectively.
Total number of visitors $= 2x + 5x + 13x = 20x$.
Total collection from the visitors $= (15 \times 2x) + (7.50 \times 5x) + (2.50 \times 13x)$.
$= 30x + 37.5x + 32.5x = 100x$.
Average charge per person $= \frac{\text{Total collection}}{\text{Total number of visitors}} = \frac{100x}{20x} = 5$.
Therefore,the average charge per person is $Rs. 5$.

(C) Let the temperatures of Monday,Tuesday,Wednesday,Thursday,and Friday be $M, T, W, Th,$ and $F$ respectively.
Given,the average of $M, T, W, Th$ is $60^{\circ}$.
Sum of temperatures $(M + T + W + Th) = 60 \times 4 = 240^{\circ}$.
Given,the average of $T, W, Th, F$ is $63^{\circ}$.
Sum of temperatures $(T + W + Th + F) = 63 \times 4 = 252^{\circ}$.
Subtracting the first equation from the second:
$(T + W + Th + F) - (M + T + W + Th) = 252 - 240$
$F - M = 12^{\circ}$.
Given the ratio $M : F = 21 : 25$. Let $M = 21x$ and $F = 25x$.
Substituting these into the difference equation:
$25x - 21x = 12$
$4x = 12$
$x = 3$.
Therefore,the temperature of Friday $F = 25 \times 3 = 75^{\circ}$.

(B) Total daily attendance for $15$ theatres $= 600 \times 15 = 9000$.
Since the total attendance remains the same after $6$ theatres close,the total attendance is still $9000$.
The number of remaining theatres $= 15 - 6 = 9$.
The new average daily attendance per theatre $= \frac{9000}{9} = 1000$.

(B) Total number of non-working days for $5$ companies last year is calculated as follows:
$3$ companies had $16$ non-working days each: $3 \times 16 = 48$.
$2$ companies had $5$ fewer non-working days than $16$,which is $16 - 5 = 11$ days each: $2 \times 11 = 22$.
Total non-working days $= 48 + 22 = 70$.
Average number of non-working days $= \frac{\text{Total days}}{\text{Total companies}} = \frac{70}{5} = 14$.

(D) Let the maximum temperature of Chennai be $x^{\circ}C$.
The average maximum temperature of the four cities is given by:
$\text{Average} = \frac{\text{Sum of temperatures}}{\text{Number of cities}}$
Given that the average is $35^{\circ}C$ for $4$ cities:
$35 = \frac{35 + 33 + 34 + x}{4}$
Multiply both sides by $4$:
$140 = 35 + 33 + 34 + x$
Sum the known temperatures:
$140 = 102 + x$
Solve for $x$:
$x = 140 - 102$
$x = 38^{\circ}C$
Therefore,the maximum temperature of Chennai was $38^{\circ}C$.

(A) To find the average number of candidates who qualified in $2004$ from states $Q, S, T,$ and $U$,we first identify the values from the provided table for the year $2004$:
State $Q = 240$
State $S = 620$
State $T = 840$
State $U = 420$
Now,calculate the sum of these candidates:
Sum $= 240 + 620 + 840 + 420 = 2120$
Since there are $4$ states,the average is:
Average $= \frac{\text{Sum}}{\text{Number of states}} = \frac{2120}{4} = 530$
Thus,the average number of candidates is $530$.

(B) Let the average weight of the $15$ oarsmen be $A\, kg$.
Total weight of the $15$ oarsmen $= 15A\, kg$.
When a man weighing $42\, kg$ is replaced by a new man of weight $x\, kg$,the new total weight becomes $(15A - 42 + x)\, kg$.
The new average weight is $(A + 1.6)\, kg$.
Therefore,the new total weight is $15(A + 1.6)\, kg$.
Equating the two expressions for the total weight:
$15A - 42 + x = 15(A + 1.6)$
$15A - 42 + x = 15A + 24$
$x = 42 + 24$
$x = 66\, kg$.
Thus,the weight of the new man is $66\, kg$.

(C) Let the five consecutive integers be $x, x+1, x+2, x+3, x+4$.
The average of these five integers is given by $\frac{x + (x+1) + (x+2) + (x+3) + (x+4)}{5} = \frac{5x + 10}{5} = x + 2$.
Given that the average is $n$,so $n = x + 2$.
Now,if the next two integers ($x+5$ and $x+6$) are included,the total number of integers becomes $7$.
The sum of these seven integers is $(5x + 10) + (x+5) + (x+6) = 7x + 21$.
The new average is $\frac{7x + 21}{7} = x + 3$.
Since the original average was $n = x + 2$,the new average $x + 3$ can be written as $(x + 2) + 1 = n + 1$.
Therefore,the average increases by $1$.

(D) Let the number of girls be $4x$ and the number of boys be $5x$.
The total marks obtained by girls $= 85 \times 4x = 340x$.
The total marks obtained by boys $= 87 \times 5x = 435x$.
The total number of students $= 4x + 5x = 9x$.
The average marks of the whole class $= \frac{\text{Total marks of girls} + \text{Total marks of boys}}{\text{Total number of students}}$.
Average $= \frac{340x + 435x}{9x} = \frac{775x}{9x} = \frac{775}{9}$.
Average $\approx 86.11$.
Therefore,the average marks of the whole class is closest to $86.1$.

(B) Let the $4$ numbers be $a, b, c, d$.
The average of the first three numbers $(a, b, c)$ is $16$,so:
$(a + b + c) / 3 = 16$
$a + b + c = 48$ --- (Equation $1$)
The average of the last three numbers $(b, c, d)$ is $15$,so:
$(b + c + d) / 3 = 15$
$b + c + d = 45$ --- (Equation $2$)
Subtracting Equation $2$ from Equation $1$:
$(a + b + c) - (b + c + d) = 48 - 45$
$a - d = 3$
Given that the last number $d = 20$,substitute this value into the equation:
$a - 20 = 3$
$a = 23$
Therefore,the first number is $23$.

(D) Let the initial average price of one book be $Rs. x$.
Total cost of $50$ books $= 50x$.
If he spends $Rs. 76$ more,the total cost becomes $(50x + 76)$ and the total number of books becomes $(50 + 14) = 64$.
The new average price is $(x - 1)$.
According to the problem: $\frac{50x + 76}{64} = x - 1$.
$50x + 76 = 64(x - 1)$.
$50x + 76 = 64x - 64$.
$64x - 50x = 76 + 64$.
$14x = 140$.
$x = 10$.
Therefore,the average price of each book he initially bought was $Rs. 10$.

(D) The sum of weights of $A, B,$ and $C$ is $3 \times 84 = 252 \, kg$.
When $D$ joins,the total weight of $A, B, C,$ and $D$ is $4 \times 80 = 320 \, kg$.
Therefore,the weight of $D = 320 - 252 = 68 \, kg$.
The weight of $E = D + 3 = 68 + 3 = 71 \, kg$.
When $E$ replaces $A$,the new group is $B, C, D,$ and $E$. The sum of their weights is $4 \times 79 = 316 \, kg$.
So,$B + C + D + E = 316$.
Substituting the values of $D$ and $E$: $B + C + 68 + 71 = 316$.
$B + C + 139 = 316 \implies B + C = 316 - 139 = 177 \, kg$.
Since $A + B + C = 252$,we have $A + 177 = 252$.
Therefore,$A = 252 - 177 = 75 \, kg$.

(A) The formula for the weighted average is given by:
$\text{Average} = \frac{\sum (x_i \cdot f_i)}{\sum f_i}$
Given the data:
$\text{Sum of products} = (8 \times 3) + (20 \times 2) + (26 \times m) + (29 \times 1) = 24 + 40 + 26m + 29 = 93 + 26m$
$\text{Sum of frequencies} = 3 + 2 + m + 1 = 6 + m$
Given that the average is $17$,we have:
$17 = \frac{93 + 26m}{6 + m}$
Multiplying both sides by $(6 + m)$:
$17(6 + m) = 93 + 26m$
$102 + 17m = 93 + 26m$
Rearranging the terms to solve for $m$:
$102 - 93 = 26m - 17m$
$9 = 9m$
$m = 1$

(B) Total annual expenditure of the family = $(4 \times 2570) + (3 \times 2490) + (5 \times 3030)$
= $10,280 + 7,470 + 15,150 = Rs.\, 32,900$
Total annual income = Total annual expenditure + Total annual savings
= $32,900 + 5,320 = Rs.\, 38,220$
Average monthly income = $\frac{\text{Total annual income}}{12}$
= $\frac{38,220}{12} = Rs.\, 3,185$

(C) Total expenditure of the man in a year is calculated as follows:
Expenditure for the first $4$ months $= 4 \times 1800 = Rs. 7200$.
Expenditure for the next $8$ months $= 8 \times 2000 = Rs. 16000$.
Total annual expenditure $= 7200 + 16000 = Rs. 23200$.
Total annual income is the sum of total expenditure and annual savings:
Total annual income $= 23200 + 5600 = Rs. 28800$.
Average monthly income $= \frac{\text{Total annual income}}{12} = \frac{28800}{12} = Rs. 2400$.

(B) The arithmetic mean is calculated by dividing the sum of all observations by the total number of observations.
Sum of observations $= (1 \times 1) + (2 \times 2) + (3 \times 3) + (4 \times 4) + (5 \times 5) + (6 \times 6) + (7 \times 7)$
Sum $= 1 + 4 + 9 + 16 + 25 + 36 + 49 = 140$
Total number of observations $= 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28$
Arithmetic Mean $= \frac{140}{28} = 5$.

(D) The sum of six numbers is $6 \times 20 = 120$.
After removing one number,there are five numbers remaining.
The sum of these five numbers is $5 \times 15 = 75$.
The removed number is the difference between the initial sum and the new sum.
Removed number $= 120 - 75 = 45$.

(D) Let the fixed expenditure be $x$ and the variable expenditure per student be $y$.
Total expenditure for $n$ students is given by $E = x + ny$.
The average expense per student is $A = \frac{E}{n} = \frac{x}{n} + y$.
For $24$ students,$A = 615$,so $\frac{x}{24} + y = 615 \Rightarrow x + 24y = 14760$ ...$(1)$
For $40$ students,$A = 465$,so $\frac{x}{40} + y = 465 \Rightarrow x + 40y = 18600$ ...$(2)$
Subtracting equation $(1)$ from $(2)$:
$(x + 40y) - (x + 24y) = 18600 - 14760$
$16y = 3840 \Rightarrow y = 240$.
Substituting $y = 240$ into equation $(1)$:
$x + 24(240) = 14760$
$x + 5760 = 14760 \Rightarrow x = 9000$.
For $60$ students,the average expense is:
$A = \frac{x}{60} + y = \frac{9000}{60} + 240 = 150 + 240 = 390$.
Thus,the average expense is $Rs.\,390$.

(C) Total marks of $40$ students $= 40 \times 86 = 3440$.
When the $5$ highest marks are removed,the number of students becomes $40 - 5 = 35$.
The new average is $86 - 1 = 85$.
Total marks of the remaining $35$ students $= 35 \times 85 = 2975$.
Total marks of the top $5$ students $= 3440 - 2975 = 465$.
Average marks of the top $5$ students $= \frac{465}{5} = 93$.

(A) Let the average of the $10$ numbers be $y$.
Then,the sum of the $10$ numbers is $10y$.
When the digits of one number are interchanged,the new average becomes $y + 3.6$.
Therefore,the new sum of the $10$ numbers is $10(y + 3.6) = 10y + 36$.
The increase in the sum is $(10y + 36) - 10y = 36$.
Let the two-digit number be $10a + b$,where $a$ is the tens digit and $b$ is the units digit.
After interchanging the digits,the new number is $10b + a$.
The difference between the new number and the original number is $(10b + a) - (10a + b) = 9b - 9a = 9(b - a)$.
Since the increase in the total sum is $36$,we have $9(b - a) = 36$.
Dividing both sides by $9$,we get $b - a = 4$.

(D) Let the initial sum of the ages of $30$ teachers be $S$.
The initial mean age is $\frac{S}{30}$.
When a teacher of age $60$ $years$ retires and a teacher of age $30$ $years$ is appointed,the new sum of ages becomes $S - 60 + 30 = S - 30$.
The number of teachers remains $30$.
The new mean age is $\frac{S - 30}{30} = \frac{S}{30} - 1$.
Therefore,the mean age decreases by $1$ $year$.

(A) The average age of $A, B$ and $C$ is $84 \text{ years}$,so $A + B + C = 84 \times 3 = 252$ ......$(1)$
When $D$ joins,the average age of $A, B, C$ and $D$ becomes $80 \text{ years}$,so $A + B + C + D = 80 \times 4 = 320$ ......$(2)$
Subtracting $(1)$ from $(2)$,we get $D = 320 - 252 = 68 \text{ years}$.
Given that $E$ is $4 \text{ years}$ older than $D$,so $E = 68 + 4 = 72 \text{ years}$.
When $E$ replaces $A$,the new average of $B, C, D$ and $E$ is $78 \text{ years}$,so $B + C + D + E = 78 \times 4 = 312$.
Substituting the values of $D$ and $E$,we get $B + C + 68 + 72 = 312$,which simplifies to $B + C + 140 = 312$,so $B + C = 172$ ......$(3)$
Finally,substituting $(3)$ into $(1)$,we get $A + 172 = 252$,so $A = 252 - 172 = 80 \text{ years}$.

(C) Average $= \frac{\text{Sum of elements}}{\text{Number of elements}}$
Given,initial number of elements $= 50$.
Initial average $= 38$.
Sum of $50$ numbers $= 50 \times 38 = 1900$.
When two numbers $45$ and $55$ are discarded,the new sum of the remaining numbers is:
Sum $= 1900 - (45 + 55) = 1900 - 100 = 1800$.
The number of remaining elements $= 50 - 2 = 48$.
New average $= \frac{1800}{48} = 37.5$.

(B) Let the total number of workers in the institution be $Z$.
The total salary of all workers is given by the product of the average salary and the total number of workers: $60Z$.
The total salary of $12$ officers is $12 \times 400 = 4800$.
The number of remaining workers is $(Z - 12)$,and their average salary is $56$. Thus,the total salary of the remaining workers is $56(Z - 12)$.
The sum of the total salary of the officers and the remaining workers must equal the total salary of all workers:
$4800 + 56(Z - 12) = 60Z$
Expanding the equation:
$4800 + 56Z - 672 = 60Z$
$4128 + 56Z = 60Z$
$60Z - 56Z = 4128$
$4Z = 4128$
$Z = \frac{4128}{4} = 1032$
Therefore,the total number of workers in the institution is $1032$.

(B) Let the four numbers be $a, b, c,$ and $d.$
According to the problem,the average of the first three numbers is double the fourth number:
$\frac{a+b+c}{3} = 2d$
$\Rightarrow a+b+c = 6d$ ....$(1)$
The average of all four numbers is $12$:
$\frac{a+b+c+d}{4} = 12$
$\Rightarrow a+b+c+d = 48$ ....$(2)$
Substitute the value of $(a+b+c)$ from equation $(1)$ into equation $(2)$:
$6d + d = 48$
$7d = 48$
$d = \frac{48}{7}$

(B) Let the $6$ consecutive even numbers be $x, x+2, x+4, x+6, x+8,$ and $x+10$.
The average of these numbers is given by the sum divided by the count:
$\text{Average} = \frac{x + (x+2) + (x+4) + (x+6) + (x+8) + (x+10)}{6} = 25$
Simplify the numerator:
$\frac{6x + 30}{6} = 25$
$x + 5 = 25$
$x = 20$
The smallest number is $x = 20$ and the largest number is $x + 10 = 30$.
The difference between the largest and the smallest number is $30 - 20 = 10$.

(B) Let the initial arithmetic mean be $\bar{x} = 24$.
When a constant value $k$ is added to each observation,the new mean becomes $\bar{x} + k$.
Here,$k = 6$,so the new mean after addition is $24 + 6 = 30$.
When each observation is multiplied by a constant factor $m$,the new mean becomes $\text{mean} \times m$.
Here,$m = 2.5$,so the final mean is $30 \times 2.5 = 75$.

(C) Let the new average after $12$ innings be $x$ runs.
Then,the average after $11$ innings was $(x - 5)$ runs.
Total runs scored in $11$ innings $= 11(x - 5)$.
In the $12^{th}$ innings,he scores $120$ runs,so the total runs after $12$ innings $= 11(x - 5) + 120$.
According to the problem,the average after $12$ innings is $x$,so the total runs $= 12x$.
Equating the two expressions for total runs:
$11(x - 5) + 120 = 12x$
$11x - 55 + 120 = 12x$
$11x + 65 = 12x$
$x = 65$.
Therefore,his new average is $65$ runs.

(D) The sum of $11$ results is $11 \times 50 = 550$.
The sum of the first $6$ results is $6 \times 49 = 294$.
The sum of the last $6$ results is $6 \times 52 = 312$.
The sixth result is counted twice when we add the sum of the first $6$ results and the last $6$ results.
Therefore,the sixth result $= (294 + 312) - 550 = 606 - 550 = 56$.

(B) The total weight of group $A$ is $42 \times 25 = 1050 \, kg$.
The total weight of group $B$ is $28 \times 40 = 1120 \, kg$.
The total number of students in the class is $42 + 28 = 70$.
The average weight of the whole class is calculated as the sum of total weights divided by the total number of students:
$\text{Average weight} = \frac{1050 + 1120}{70} = \frac{2170}{70} = 31 \, kg$.

(C) Let the number of male employees be $x$ and the number of female employees be $y$.
The total salary of all employees is the sum of the total salary of male employees and the total salary of female employees.
According to the problem:
$(x + y) \times 12,000 = (x \times 15,000) + (y \times 8,000)$
Dividing both sides by $1,000$:
$(x + y) \times 12 = 15x + 8y$
Expanding the equation:
$12x + 12y = 15x + 8y$
Rearranging the terms to solve for the ratio $x/y$:
$12y - 8y = 15x - 12x$
$4y = 3x$
Therefore,the ratio of male employees to female employees is:
$\frac{x}{y} = \frac{4}{3}$ or $4:3$.