Average Questions in English

(B) Given that the average of $X_{1}, X_{2}$ and $X_{3}$ is $14$.
Therefore,the sum $X_{1} + X_{2} + X_{3} = 14 \times 3 = 42$ ......$(1)$
It is given that twice the sum of $X_{2}$ and $X_{3}$ is $30$,which means $2(X_{2} + X_{3}) = 30$.
Dividing by $2$,we get $X_{2} + X_{3} = 15$ ......$(2)$
Subtracting equation $(2)$ from equation $(1)$,we get $X_{1} = 42 - 15 = 27$.
Thus,the value of $X_{1}$ is $27$.

(C) When a person travels a distance $d$ at a speed $v_1$ and returns the same distance $d$ at a speed $v_2$,the average speed for the entire journey is given by the harmonic mean formula: $\text{Average Speed} = \frac{2v_1v_2}{v_1 + v_2}$.
Given $v_1 = 40 \, km/h$ and $v_2 = 50 \, km/h$.
Substituting these values into the formula:
$\text{Average Speed} = \frac{2 \times 40 \times 50}{40 + 50} \, km/h$.
$\text{Average Speed} = \frac{4000}{90} \, km/h$.
$\text{Average Speed} = \frac{400}{9} \, km/h$.

(A) cricket team consists of $11$ players.
Let the initial average age of the team be $x$.
The total age of the team is $11x$.
When two players aged $32$ and $30$ are replaced by two youngsters of age $y$ each,the new total age becomes $11x - 32 - 30 + 2y$.
The new average age is $(x - 2)$.
Thus,the new total age is $11(x - 2)$.
Equating the two expressions for the new total age:
$11x - 62 + 2y = 11x - 22$
Subtracting $11x$ from both sides:
$-62 + 2y = -22$
$2y = 62 - 22$
$2y = 40$
$y = 20$
Therefore,the age of each youngster is $20$ years.

(C) The total height of the initial $20$ students $= 20 \times 105 = 2100 \, cm$.
The total height of the $10$ new students $= 10 \times 120 = 1200 \, cm$.
The total number of students in the class now $= 20 + 10 = 30$.
The total height of all $30$ students $= 2100 + 1200 = 3300 \, cm$.
The new average height of the class $= \frac{3300}{30} = 110 \, cm$.

(B) The formula for the sum of the first $n$ natural numbers is $S_n = \frac{n(n+1)}{2}$.
To find the average,we divide the sum by $n$: $\text{Average} = \frac{S_n}{n} = \frac{n(n+1)}{2n} = \frac{n+1}{2}$.
Given $n = 46$,the average is $\frac{46+1}{2} = \frac{47}{2} = 23.5$.

(D) The sum of $13$ results $= 13 \times 50 = 650$.
The sum of the first $7$ results $= 7 \times 52 = 364$.
The sum of the last $7$ results $= 7 \times 49 = 343$.
The $7^{th}$ result is counted twice when we add the sum of the first $7$ results and the last $7$ results.
Therefore,the $7^{th}$ result $= (364 + 343) - 650$.
$= 707 - 650 = 57$.

(B) The average age of $18$ persons is $32.5$ years.
Total age of $18$ persons $= 18 \times 32.5 = 585$ years.
Total age of the three persons who left $= (65 \times 2) + 50 = 130 + 50 = 180$ years.
Total age of the remaining $15$ persons $= 585 - 180 = 405$ years.
Average age of the remaining $15$ persons $= \frac{405}{15} = 27$ years.

(D) The average marks of $3$ subjects (Physics,Chemistry,Biology) is $70.$
Sum of marks in $3$ subjects $= 3 \times 70 = 210.$
Marks obtained in Mathematics $= 90.$
Total marks for $4$ subjects $= 210 + 90 = 300.$
New average for $4$ subjects $= \frac{300}{4} = 75.$

(D) Average weight of $3$ persons $= 65 \, kg$.
Total weight of $3$ persons $= 65 \times 3 = 195 \, kg$.
Weight of the $4^{th}$ person $= 45 \, kg$.
Total weight of $4$ persons $= 195 + 45 = 240 \, kg$.
Average weight of the new group $= \frac{\text{Total weight}}{\text{Number of persons}} = \frac{240}{4} = 60 \, kg$.

(C) Let the total distance of the journey be $3d$.
Each part of the journey is $d$.
Time taken for the first part $t_1 = \frac{d}{80}$.
Time taken for the second part $t_2 = \frac{d}{60}$.
Time taken for the third part $t_3 = \frac{d}{30}$.
Total time $T = t_1 + t_2 + t_3 = d(\frac{1}{80} + \frac{1}{60} + \frac{1}{30})$.
$T = d(\frac{3 + 4 + 8}{240}) = d(\frac{15}{240}) = \frac{d}{16}$.
Average speed $= \frac{\text{Total distance}}{\text{Total time}} = \frac{3d}{d/16} = 3 \times 16 = 48\, km/h$.

(D) Let the initial average weight of $6$ persons be $A \, kg$.
Total weight of $6$ persons = $6A \, kg$.
When a person weighing $40 \, kg$ is replaced by a new man of weight $W$,the new total weight becomes $(6A - 40 + W) \, kg$.
The new average weight is $(A + 5) \, kg$.
Therefore,the new total weight is $6(A + 5) \, kg$.
Equating the two expressions for the total weight:
$6A - 40 + W = 6(A + 5)$
$6A - 40 + W = 6A + 30$
$W = 30 + 40 = 70 \, kg$.
Alternatively,using the shortcut formula:
Weight of new man = Weight of replaced man + (Increase in average $\times$ Number of persons)
$= 40 + (5 \times 6) = 40 + 30 = 70 \, kg$.

(D) Initial number of students $= 9$.
Initial average age $= 20\, \text{years}$.
Total age of $9$ students $= 9 \times 20 = 180\, \text{years}$.
New number of students $= 9 + 6 = 15$.
New average age $= 20 - 2 = 18\, \text{years}$.
Total age of $15$ students $= 15 \times 18 = 270\, \text{years}$.
Total age of $6$ new students $= 270 - 180 = 90\, \text{years}$.
Average age of $6$ new students $= \frac{90}{6} = 15\, \text{years}$.

(B) The overall average score is calculated by finding the total marks of all students and dividing it by the total number of students.
Total marks of the first batch $= 60 \times 50 = 3000$.
Total marks of the second batch $= 40 \times 45 = 1800$.
Total marks of both batches $= 3000 + 1800 = 4800$.
Total number of students $= 60 + 40 = 100$.
Overall average score $= \frac{4800}{100} = 48$.

(C) The formula for average speed when the distance traveled is the same for both trips is given by $\text{Average Speed} = \frac{2xy}{x+y}$,where $x$ and $y$ are the speeds of the two trips.
Here,$x = 60 \, km/h$ and $y = 20 \, km/h$.
Substituting the values into the formula:
$\text{Average Speed} = \frac{2 \times 60 \times 20}{60 + 20}$
$\text{Average Speed} = \frac{2400}{80}$
$\text{Average Speed} = 30 \, km/h$.

(C) Let the total number of students be $100$.
Then,the total score of all students $= 52 \times 100 = 5200$.
The total score of the brightest $20\%$ students $= 20 \times 80 = 1600$.
The total score of the dullest $25\%$ students $= 25 \times 31 = 775$.
The total score of the remaining $55\%$ students $= 5200 - 1600 - 775 = 2825$.
Therefore,the mean score of the remaining $55\%$ students $= \frac{2825}{55} \approx 51.36$,which is approximately $51.4$.

(C) The average age of $4$ members $10$ years ago was $24$ years.
Therefore,the sum of their ages $10$ years ago was $24 \times 4 = 96$ years.
At present,each of the $4$ members has aged by $10$ years,so the sum of their present ages is $96 + (4 \times 10) = 96 + 40 = 136$ years.
Let the ages of the two children be $x$ and $x+2$ years,where $x$ is the age of the youngest child.
The total number of family members is now $4 + 2 = 6$.
The present average age of the family is given as $24$ years.
Thus,the sum of the ages of all $6$ members is $24 \times 6 = 144$ years.
We can write the equation: $136 + x + (x + 2) = 144$.
$138 + 2x = 144$.
$2x = 144 - 138$.
$2x = 6$.
$x = 3$.
Therefore,the present age of the youngest child is $3$ years.

(C) Average weight of $A, B$ and $C = 84 \, kg$.
Sum of weights of $A, B, C = 84 \times 3 = 252 \, kg$ $(1)$.
When $D$ joins,the average weight of $4$ men becomes $80 \, kg$.
Sum of weights of $A, B, C, D = 80 \times 4 = 320 \, kg$ $(2)$.
Weight of $D = (2) - (1) = 320 - 252 = 68 \, kg$.
Weight of $E = D + 3 = 68 + 3 = 71 \, kg$.
When $E$ replaces $A$,the new group is $B, C, D, E$ with an average of $79 \, kg$.
Sum of weights of $B, C, D, E = 79 \times 4 = 316 \, kg$.
$B + C + D + E = 316 \implies B + C + 68 + 71 = 316$.
$B + C + 139 = 316 \implies B + C = 316 - 139 = 177 \, kg$.
From $(1)$,$A + B + C = 252$.
$A + 177 = 252$.
$A = 252 - 177 = 75 \, kg$.

(A) Let the average of the cricketer for $10$ innings be $x$.
Total runs scored in $10$ innings $= 10x$.
In the $11^{th}$ inning,he scored $108$ runs.
Total runs after $11$ innings $= 10x + 108$.
New average after $11$ innings $= x + 6$.
According to the problem,the average after $11$ innings is $\frac{10x + 108}{11} = x + 6$.
Solving for $x$:
$10x + 108 = 11(x + 6)$
$10x + 108 = 11x + 66$
$11x - 10x = 108 - 66$
$x = 42$.
The new average is $x + 6 = 42 + 6 = 48$ runs.

(D) Let the initial average age of $8$ men be $x$.
Total age of $8$ men $= 8x$.
Let the ages of the two new men be $a$ and $b$. The sum of their ages is $(a + b)$.
When two men of ages $21$ and $23$ are replaced by two new men, the new total age becomes $(8x - 21 - 23 + a + b)$.
The new average is given as $(x + 2)$.
So, $\frac{8x - 44 + (a + b)}{8} = x + 2$.
$8x - 44 + (a + b) = 8(x + 2)$.
$8x - 44 + (a + b) = 8x + 16$.
$(a + b) = 16 + 44 = 60$.
The average age of the two new men is $\frac{a + b}{2} = \frac{60}{2} = 30\, \text{years}$.

(C) Initial average weight of $45$ students $= 52 \, kg.$
Total weight of $45$ students $= 45 \times 52 = 2340 \, kg.$
Weight of $5$ students who left $= 5 \times 48 = 240 \, kg.$
Weight of $5$ students who joined $= 5 \times 54 = 270 \, kg.$
New total weight $= 2340 - 240 + 270 = 2370 \, kg.$
Since the number of students remains $45$,the new average weight $= \frac{2370}{45} = \frac{474}{9} = \frac{158}{3} = 52 \frac{2}{3} \, kg.$

(C) The total age of $36$ students $= 36 \times 14 = 504$ years.
When the teacher's age is included, the total number of people becomes $36 + 1 = 37$.
The new average age becomes $14 + 1 = 15$ years.
The total age of $37$ people $= 37 \times 15 = 555$ years.
Teacher's age $= (\text{Total age of } 37 \text{ people}) - (\text{Total age of } 36 \text{ students})$.
Teacher's age $= 555 - 504 = 51$ years.

(D) The sum of weights of $A, B$ and $C$ is $3 \times 54 \frac{1}{3} = 3 \times \frac{163}{3} = 163 \, kg.$
The sum of weights of $B, D$ and $E$ is $3 \times 53 = 159 \, kg.$
Adding these two sums gives $(A + B + C) + (B + D + E) = 163 + 159 = 322 \, kg.$
This simplifies to $A + 2B + C + D + E = 322 \, kg.$
To find the average of $A, B, C, D$ and $E$,we need the sum $(A + B + C + D + E)$.
Since the value of $B$ is unknown,we cannot determine the total sum $(A + B + C + D + E) = 322 - B$.
Therefore,the average weight cannot be calculated with the given information.

(C) Given that the mean of $50$ observations is $36$.
Sum of $50$ observations $= 50 \times 36 = 1800$.
It was found that an observation $48$ was wrongly taken as $23$.
Corrected sum $= 1800 - 23 + 48 = 1825$.
Corrected new mean $= \frac{1825}{50} = 36.5$.

(C) Let the third number be $x$.
Then,the second number $= 2x$.
The first number $= 2 \times (2x) = 4x$.
The reciprocals of the numbers are $\frac{1}{4x}, \frac{1}{2x}, \text{ and } \frac{1}{x}$.
The average of these reciprocals is given by $\frac{1}{3} \left( \frac{1}{4x} + \frac{1}{2x} + \frac{1}{x} \right) = \frac{7}{72}$.
Simplifying the expression inside the parentheses: $\frac{1+2+4}{4x} = \frac{7}{4x}$.
So,$\frac{1}{3} \times \frac{7}{4x} = \frac{7}{12x} = \frac{7}{72}$.
By cross-multiplying,$12x = 72$,which gives $x = 6$.
Therefore,the numbers are $4x = 24$,$2x = 12$,and $x = 6$.

(A) Score in the first $10$ overs $= 3.2 \times 10 = 32$ runs.
To achieve a target of $282$ runs,the remaining runs required $= 282 - 32 = 250$ runs.
The remaining overs available $= 40$ overs.
Therefore,the required run rate for the remaining $40$ overs $= \frac{250}{40} = 6.25$ runs per over.

(B) The sum of $6$ numbers $= 3.95 \times 6 = 23.70$.
The sum of the first two numbers $= 3.4 \times 2 = 6.8$.
The sum of the next two numbers $= 3.85 \times 2 = 7.7$.
The sum of the remaining two numbers $= 23.70 - (6.8 + 7.7) = 23.70 - 14.50 = 9.20$.
The average of the remaining two numbers $= \frac{9.20}{2} = 4.60$.

(C) Let the average marks of group $B$ be $x$.
Total marks of all $16$ children $= 16 \times 76 = 1216$.
Total marks of group $A$ $= 10 \times 75 = 750$.
Total marks of group $B$ $= 6 \times x = 6x$.
Since the total marks of group $A$ and group $B$ equal the total marks of all $16$ children:
$750 + 6x = 1216$.
$6x = 1216 - 750 = 466$.
$x = \frac{466}{6} = \frac{233}{3} = 77 \frac{2}{3} \%$.

(D) The sum of $5$ numbers is calculated as $27 \times 5 = 135$.
When one number is excluded,the count of numbers becomes $4$ and the new average is $25$.
The sum of the remaining $4$ numbers is $25 \times 4 = 100$.
The excluded number is the difference between the original sum and the new sum:
Excluded number $= 135 - 100 = 35$.

(C) Let the total money spent by all $9$ persons be $x$.
The average expenditure of all $9$ persons is $\frac{x}{9}$.
The expenditure of the first $8$ persons is $8 \times 30 = 240$.
The expenditure of the ninth person is $\frac{x}{9} + 20$.
According to the problem,the sum of the expenditures of all $9$ persons equals the total expenditure $x$:
$240 + (\frac{x}{9} + 20) = x$
$260 + \frac{x}{9} = x$
$260 = x - \frac{x}{9}$
$260 = \frac{8x}{9}$
$x = \frac{260 \times 9}{8}$
$x = 32.5 \times 9 = 292.5$
Thus,the total money spent by all of them is $Rs. 292.50$.

(B) The family consists of $2$ grandparents,$2$ parents,and $3$ grandchildren,making a total of $2 + 2 + 3 = 7$ members.
Total age of the grandparents $= 2 \times 67 = 134 \text{ years}$.
Total age of the parents $= 2 \times 35 = 70 \text{ years}$.
Total age of the grandchildren $= 3 \times 6 = 18 \text{ years}$.
Total age of the family $= 134 + 70 + 18 = 222 \text{ years}$.
Average age of the family $= \frac{\text{Total age}}{\text{Total number of members}} = \frac{222}{7} = 31 \frac{5}{7} \text{ years}$.

(B) Let the three consecutive odd numbers be $(x-2)$,$x$,and $(x+2)$.
The sum of these numbers is $(x-2) + x + (x+2) = 3x$.
The average of these numbers is $\frac{3x}{3} = x$.
According to the problem,the sum is $38$ more than the average:
$3x = x + 38$
Subtract $x$ from both sides:
$2x = 38$
Divide by $2$:
$x = 19$.
The first number is $(x-2) = 19 - 2 = 17$.

(C) Let the $7$ consecutive numbers be $x, (x+1), (x+2), (x+3), (x+4), (x+5),$ and $(x+6)$.
The average of these numbers is given by the sum divided by the count:
$\text{Average} = \frac{x + (x+1) + (x+2) + (x+3) + (x+4) + (x+5) + (x+6)}{7} = \frac{7x + 21}{7} = x + 3$.
Given that the average is $20$,we have:
$x + 3 = 20$
$x = 20 - 3 = 17$.
The largest number is $(x + 6)$.
Substituting $x = 17$,we get:
$\text{Largest number} = 17 + 6 = 23$.

(C) two-digit number $xy$ (where $x$ is the tens digit and $y$ is the units digit) remains the same when its digits are interchanged if $x = y$.
These numbers are $11, 22, 33, 44, 55, 66, 77, 88, 99$.
There are $9$ such numbers.
The sum of these numbers is $11 + 22 + 33 + 44 + 55 + 66 + 77 + 88 + 99 = 11(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)$.
Using the sum formula for the first $n$ natural numbers,$\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$,we get $11 \times \frac{9(10)}{2} = 11 \times 45 = 495$.
The average is $\frac{\text{Sum}}{\text{Count}} = \frac{495}{9} = 55$.

(B) Given that the mean of $a, b, c$ is $M$,we have $\frac{a+b+c}{3} = M$,which implies $a+b+c = 3M$.
Squaring both sides,we get $(a+b+c)^2 = (3M)^2 = 9M^2$.
Expanding the left side,we have $a^2 + b^2 + c^2 + 2(ab + bc + ca) = 9M^2$.
Given that $ab + bc + ca = 0$,the equation simplifies to $a^2 + b^2 + c^2 = 9M^2$.
The mean of $a^2, b^2, c^2$ is $\frac{a^2 + b^2 + c^2}{3}$.
Substituting the value,we get $\frac{9M^2}{3} = 3M^2$.

(C) The first nine prime numbers are $2, 3, 5, 7, 11, 13, 17, 19$,and $23$.
The sum of these numbers is $2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 = 100$.
The average is calculated as the sum of the numbers divided by the count of numbers.
Average $= \frac{100}{9} = 11 \frac{1}{9}$.

(B) The numbers between $6$ and $34$ that are divisible by $5$ are $10, 15, 20, 25,$ and $30$.
To find the average,we sum these numbers and divide by the count of numbers:
Sum $= 10 + 15 + 20 + 25 + 30 = 100$.
Number of terms $= 5$.
Average $= \frac{\text{Sum}}{\text{Number of terms}} = \frac{100}{5} = 20$.

(B) The formula for the arithmetic mean is $\text{Mean} = \frac{\text{Sum of observations}}{\text{Number of observations}}$.
Given that there are $12$ numbers in total and their mean is $12$.
Therefore,the sum of these $12$ numbers is $12 \times 12 = 144$.
The sum of the given $11$ numbers is $3 + 11 + 7 + 9 + 15 + 13 + 8 + 19 + 17 + 21 + 14 = 137$.
Let the unknown number be $x$.
Then,$137 + x = 144$.
Solving for $x$,we get $x = 144 - 137 = 7$.

(B) The mean of $5$ observations is given by the sum of observations divided by the number of observations.
$\frac{x + (x+2) + (x+4) + (x+6) + (x+8)}{5} = 11$
$\frac{5x + 20}{5} = 11$
$x + 4 = 11$
$x = 11 - 4 = 7$
The last three observations are $(x+4), (x+6),$ and $(x+8).$
Mean of the last three observations $= \frac{(x+4) + (x+6) + (x+8)}{3} = \frac{3x + 18}{3} = x + 6.$
Substituting $x = 7,$ we get $7 + 6 = 13.$

(A) Let the non-zero number be $x$.
The average of the number $x$ and its square $x^2$ is given by $\frac{x + x^2}{2}$.
According to the problem,this average is equal to $5$ times the number:
$\frac{x + x^2}{2} = 5x$
Multiply both sides by $2$:
$x + x^2 = 10x$
Rearrange the equation:
$x^2 - 9x = 0$
Factor out $x$:
$x(x - 9) = 0$
Since the number is non-zero $(x \neq 0)$,we have:
$x - 9 = 0$
$x = 9$
Therefore,the number is $9$.

(D) The average age of the whole class is calculated using the formula: $\text{Average} = \frac{\text{Total sum of ages}}{\text{Total number of students}}$.
Let the number of boys be $n_b$ and the number of girls be $n_g$.
The total age of boys is $16 \times n_b$ and the total age of girls is $15 \times n_g$.
The average age of the whole class is $\frac{16n_b + 15n_g}{n_b + n_g}$.
Since the ratio or the exact number of boys and girls is not provided,the weighted average cannot be determined.
Therefore,the answer is that it cannot be computed with the given information.

(D) In a month of $30$ days starting with a Sunday,the Sundays will fall on the $1^{st}, 8^{th}, 15^{th}, 22^{nd},$ and $29^{th}$ days. Thus,there are $5$ Sundays and $30 - 5 = 25$ other days.
The total number of visitors in the month is $(5 \times 510) + (25 \times 240)$.
Total visitors $= 2550 + 6000 = 8550$.
The average number of visitors per day $= \frac{8550}{30} = 285$.

(D) The total number of students is $55 + 60 + 45 = 160$.
The total marks obtained by all students is calculated as:
Total Marks $= (55 \times 50) + (60 \times 55) + (45 \times 60)$
$= 2750 + 3300 + 2700 = 8750$.
The average marks of all students is given by:
Average $= \frac{\text{Total Marks}}{\text{Total Students}} = \frac{8750}{160} = \frac{875}{16} = 54.6875$.

(A) In the first year,petrol consumed $= \frac{4000}{7.5}$ litres.
In the second year,petrol consumed $= \frac{4000}{8}$ litres.
In the third year,petrol consumed $= \frac{4000}{8.5}$ litres.
Total expenditure $= 4000 + 4000 + 4000 = 12000$ Rs.
Total petrol consumed $= 4000 \times (\frac{1}{7.5} + \frac{1}{8} + \frac{1}{8.5})$ litres.
Average cost per litre $= \frac{\text{Total Expenditure}}{\text{Total Petrol Consumed}} = \frac{12000}{4000 \times (\frac{1}{7.5} + \frac{1}{8} + \frac{1}{8.5})} = \frac{3}{\frac{1}{7.5} + \frac{1}{8} + \frac{1}{8.5}}$.
Calculating the denominator: $\frac{1}{7.5} + \frac{1}{8} + \frac{1}{8.5} = \frac{1}{7.5} + 0.125 + \frac{1}{8.5} \approx 0.1333 + 0.125 + 0.1176 = 0.3759$.
Average cost $= \frac{3}{0.3759} \approx 7.98$ Rs./litre.

(D) The sum of $50$ numbers is calculated as $50 \times 30 = 1500$.
When two numbers,$35$ and $40$,are discarded,the new sum becomes $1500 - (35 + 40) = 1500 - 75 = 1425$.
The number of remaining values is $50 - 2 = 48$.
The average of the remaining $48$ numbers is $\frac{1425}{48} = 29.6875$.
Rounding this to two decimal places,we get approximately $29.69$,which is closest to $29.68$.

(D) Total age of $35$ students $= 35 \times 16 = 560$ years.
Total age of $21$ students $= 21 \times 14 = 294$ years.
Total age of the remaining $14$ students $= 560 - 294 = 266$ years.
Average age of the remaining $14$ students $= \frac{266}{14} = 19$ years.

(C) The average monthly salary of $20$ employees is $Rs. 1500$.
Total salary of $20$ employees $= 20 \times 1500 = Rs. 30,000$.
Let the manager's salary be $x$.
After adding the manager, the total number of employees becomes $20 + 1 = 21$.
The new average salary increases by $Rs. 100$, so the new average $= 1500 + 100 = Rs. 1600$.
Total salary of $21$ employees $= 21 \times 1600 = Rs. 33,600$.
Manager's salary $x = (\text{Total salary of } 21 \text{ employees}) - (\text{Total salary of } 20 \text{ employees})$.
$x = 33,600 - 30,000 = Rs. 3,600$.

(B) Let $P$,$Q$,and $R$ represent the monthly incomes of the three individuals.
Given:
$P + Q = 2 \times 5050 = 10100$ ....$(1)$
$Q + R = 2 \times 6250 = 12500$ ....$(2)$
$P + R = 2 \times 5200 = 10400$ ....$(3)$
Adding equations $(1)$,$(2)$,and $(3)$:
$2(P + Q + R) = 10100 + 12500 + 10400 = 33000$
$P + Q + R = 16500$ ....$(4)$
To find the monthly income of $P$,subtract equation $(2)$ from equation $(4)$:
$P = (P + Q + R) - (Q + R)$
$P = 16500 - 12500 = 4000$
Thus,the monthly income of $P$ is $Rs. 4000$.

(A) Total age of $15$ students $= 15 \times 15 = 225 \text{ years}$.
Sum of ages of $5$ students $= 5 \times 14 = 70 \text{ years}$.
Sum of ages of $9$ students $= 9 \times 16 = 144 \text{ years}$.
Age of the $15^{th}$ student $= \text{Total age} - (\text{Sum of ages of } 5 \text{ students} + \text{Sum of ages of } 9 \text{ students})$.
Age of the $15^{th}$ student $= 225 - (70 + 144) = 225 - 214 = 11 \text{ years}$.

(D) Given that the average weight of $A, B$ and $C$ is $45\, kg$.
Total weight of $A + B + C = 45 \times 3 = 135\, kg$ .....$(1)$
The average weight of $A$ and $B$ is $40\, kg$.
Total weight of $A + B = 40 \times 2 = 80\, kg$ .....$(2)$
The average weight of $B$ and $C$ is $43\, kg$.
Total weight of $B + C = 43 \times 2 = 86\, kg$ .....$(3)$
Adding equations $(2)$ and $(3)$:
$(A + B) + (B + C) = 80 + 86 = 166\, kg$
$A + 2B + C = 166\, kg$ .....$(4)$
Subtracting equation $(1)$ from equation $(4)$:
$(A + 2B + C) - (A + B + C) = 166 - 135$
$B = 31\, kg$
Therefore,the weight of $B$ is $31\, kg$.

(C) Sum of $8$ numbers $= 8 \times 20 = 160$ ....$(1)$
Sum of first $2$ numbers $= 2 \times 15 \frac{1}{2} = 31$ ....$(2)$
Sum of next $3$ numbers $= 3 \times 21 \frac{1}{3} = 64$ ....$(3)$
Let the sixth number be $x$.
Then,the seventh number $= x + 4$ and the eighth number $= x + 7$.
Sum of the last $3$ numbers (sixth,seventh,and eighth) $= 160 - (31 + 64) = 160 - 95 = 65$.
Therefore,$x + (x + 4) + (x + 7) = 65$.
$3x + 11 = 65$.
$3x = 54$.
$x = 18$.
The eighth number $= x + 7 = 18 + 7 = 25$.