Average Questions in English

(C) The average age of $P$ and $Q$ five years ago was $15 \text{ years}$.
Therefore,their total age five years ago was $15 \times 2 = 30 \text{ years}$.
The present total age of $P$ and $Q$ is $30 + 5 + 5 = 40 \text{ years}$.
The present average age of $P$,$Q$,and $R$ is $20 \text{ years}$.
Therefore,their present total age is $20 \times 3 = 60 \text{ years}$.
Present age of $R = (P + Q + R) - (P + Q) = 60 - 40 = 20 \text{ years}$.
After $10 \text{ years}$,the age of $R$ will be $20 + 10 = 30 \text{ years}$.

(C) Let the number of non-officer staff be $x$.
The total salary of officers is $15 \times 750 = 11250$.
The total salary of non-officers is $x \times 250 = 250x$.
The total number of staff is $(15 + x)$.
The average salary of the entire staff is given by the formula: $\text{Average} = \frac{\text{Total Salary}}{\text{Total Number of Staff}}$.
Therefore,$500 = \frac{11250 + 250x}{15 + x}$.
Multiplying both sides by $(15 + x)$,we get: $500(15 + x) = 11250 + 250x$.
$7500 + 500x = 11250 + 250x$.
$500x - 250x = 11250 - 7500$.
$250x = 3750$.
$x = \frac{3750}{250} = 15$.
Hence,the number of non-officer staff is $15$.

(B) The sum of the ages of $5$ family members $3$ years ago was $27 \times 5 = 135 \, \text{years}$.
Since $3$ years have passed,each of the $5$ members has aged by $3$ years. Therefore,the total increase in age is $5 \times 3 = 15 \, \text{years}$.
The present total age of the $5$ members is $135 + 15 = 150 \, \text{years}$.
After the inclusion of a child,the family now has $6$ members,and the new average age is $27 \, \text{years}$.
The total present age of the family including the child is $27 \times 6 = 162 \, \text{years}$.
The age of the child is the difference between the total age with the child and the total age of the $5$ members: $162 - 150 = 12 \, \text{years}$.

(D) The total age of $5$ sisters is $5 \times 20 = 100 \, \text{years}$.
At the time of the birth of the youngest sister ($4 \, \text{years}$ ago),each of the $5$ sisters was $4 \, \text{years}$ younger.
Therefore,the total age of the group at that time was $100 - (5 \times 4) = 100 - 20 = 80 \, \text{years}$.
At the time of the youngest sister's birth,there were only $4$ sisters (excluding the youngest one who was $0 \, \text{years}$ old).
Thus,the average age of the $4$ sisters was $\frac{80}{4} = 20 \, \text{years}$.

(B) $1$. Initial sum of weights of $A, B, C, D = 4 \times 67 = 268 \, kg.$
$2$. After including $E,$ total men = $5,$ and new average = $67 - 2 = 65 \, kg.$
$3$. Total weight of $5$ men $(A+B+C+D+E) = 5 \times 65 = 325 \, kg.$
$4$. Weight of $E = 325 - 268 = 57 \, kg.$
$5$. Weight of $F = E + 4 = 57 + 4 = 61 \, kg.$
$6$. After replacing $A$ with $F,$ the new average weight is $64 \, kg.$
$7$. Total weight of $5$ men $(F+B+C+D+E) = 5 \times 64 = 320 \, kg.$
$8$. Since $(A+B+C+D+E) = 325,$ we have $(B+C+D+E) = 325 - A.$
$9$. Substituting this into the new total: $F + (325 - A) = 320.$
$10$. $61 + 325 - A = 320 \implies 386 - A = 320 \implies A = 66 \, kg.$

(A) Given that the average of $a, b, c$ is $11$,so the sum $a + b + c = 11 \times 3 = 33$.
Given that the average of $c, d, e$ is $17$,so the sum $c + d + e = 17 \times 3 = 51$.
Given that the average of $e, f$ is $22$,so the sum $e + f = 22 \times 2 = 44$.
Given that the average of $e, c$ is $17$,so the sum $e + c = 17 \times 2 = 34$.
We need to find the sum $a + b + c + d + e + f$.
From the sum $c + d + e = 51$ and $e + c = 34$,we can find $d = 51 - 34 = 17$.
Now,the total sum is $(a + b + c) + (d) + (e + f) = 33 + 17 + 44 = 94$.
The average of $a, b, c, d, e, f$ is $\frac{94}{6} = \frac{47}{3} = 15 \frac{2}{3}$.

(A) Let the age of the father be $F$ and the age of each twin son be $S$.
Since the sons are twins,their ages are equal.
The average age is given by $\frac{F + S + S}{3} = 30$.
This implies $F + 2S = 90$.
The ratio of the father's age to one son's age is $F:S = 5:2$,which means $F = \frac{5}{2}S$ or $S = \frac{2}{5}F$.
Substituting $S = \frac{2}{5}F$ into the equation $F + 2S = 90$:
$F + 2(\frac{2}{5}F) = 90$
$F + \frac{4}{5}F = 90$
$\frac{9}{5}F = 90$
$F = 90 \times \frac{5}{9} = 50$.
Therefore,the father's age is $50 \, \text{years}$.

(C) Let the ages of Rakesh,Mohan,and Ramesh be $R$,$M$,and $S$ respectively.
Given:
$(R + M) / 2 = 15 \implies R + M = 30$ (Equation $1$)
$(M + S) / 2 = 12 \implies M + S = 24$ (Equation $2$)
$(R + S) / 2 = 13 \implies R + S = 26$ (Equation $3$)
Adding all three equations:
$(R + M) + (M + S) + (R + S) = 30 + 24 + 26$
$2(R + M + S) = 80$
$R + M + S = 40$ (Equation $4$)
To find the age of Mohan $(M)$,subtract Equation $3$ from Equation $4$:
$M = (R + M + S) - (R + S)$
$M = 40 - 26 = 14$
Therefore,the age of Mohan is $14$ years.

(A) Let the second number be $x$.
Since the first number is half of the second,the first number $= \frac{x}{2}$.
Since the third number is twice the second,the third number $= 2x$.
The average of the three numbers is given as $28$,so their sum is $28 \times 3 = 84$.
Therefore,$\frac{x}{2} + x + 2x = 84$.
Multiplying by $2$ to clear the fraction: $x + 2x + 4x = 168$.
$7x = 168$.
$x = \frac{168}{7} = 24$.
The third number is $2x = 2 \times 24 = 48$.

(A) Let the temperatures on Monday,Tuesday,Wednesday,and Thursday be $M, T, W,$ and $Th$ respectively.
Given,the average temperature from Monday to Wednesday is $37^{\circ}C$:
$\frac{M + T + W}{3} = 37 \Rightarrow M + T + W = 111^{\circ}C$ (Equation $1$)
Given,the average temperature from Tuesday to Thursday is $34^{\circ}C$:
$\frac{T + W + Th}{3} = 34 \Rightarrow T + W + Th = 102^{\circ}C$ (Equation $2$)
Subtracting Equation $2$ from Equation $1$:
$(M + T + W) - (T + W + Th) = 111 - 102$
$M - Th = 9^{\circ}C \Rightarrow M = Th + 9^{\circ}C$
Given,$Th = \frac{4}{5}M$,which implies $M = \frac{5}{4}Th$.
Substituting $M$ in the equation $M - Th = 9$:
$\frac{5}{4}Th - Th = 9$
$\frac{1}{4}Th = 9 \Rightarrow Th = 36^{\circ}C$.
Thus,the temperature of Thursday is $36^{\circ}C$.

(B) Let the number of non-officer staff in the office be $x$.
The total salary of all staff is the sum of the total salary of officers and non-officers.
Total salary of officers $= 15 \times 460 = 6900$.
Total salary of non-officers $= x \times 110 = 110x$.
Total salary of all staff $= (15 + x) \times 120$.
Equating the two expressions:
$6900 + 110x = 120(15 + x)$
$6900 + 110x = 1800 + 120x$
$120x - 110x = 6900 - 1800$
$10x = 5100$
$x = 510$.
Therefore,the number of non-officer staff in the office is $510$.

(A) Let the number of candidates who passed be $x$.
Then,the number of candidates who failed is $(150 - x)$.
The total marks obtained by all candidates is $150 \times 50 = 7500$.
The sum of marks of passed candidates is $55x$.
The sum of marks of failed candidates is $25(150 - x)$.
According to the problem,the sum of marks of passed and failed candidates equals the total marks:
$55x + 25(150 - x) = 7500$
$55x + 3750 - 25x = 7500$
$30x = 7500 - 3750$
$30x = 3750$
$x = 3750 / 30 = 125$.
Therefore,the number of candidates who passed the examination is $125$.

(B) Total number of people = $24 \text{ boys} + 1 \text{ teacher} = 25 \text{ people}$.
Total age of $25$ people = $25 \times 15 = 375 \text{ years}$.
When the teacher is excluded,the number of people becomes $24$ and the average age becomes $15 - 1 = 14 \text{ years}$.
Total age of $24$ boys = $24 \times 14 = 336 \text{ years}$.
Age of the teacher = (Total age of $25$ people) - (Total age of $24$ boys).
Age of the teacher = $375 - 336 = 39 \text{ years}$.

(A) Initial total age of $8$ members $= 8 \times 40 = 320 \text{ years}$.
After the retirement of the $55 \text{ year}$ old member and the joining of a $39 \text{ year}$ old member,the new total age $= 320 - 55 + 39 = 304 \text{ years}$.
The new average age $= \frac{304}{8} = 38 \text{ years}$.
The change in average age $= 40 - 38 = 2 \text{ years}$.
Therefore,the average age decreased by $2 \text{ years}$.

(C) Let the initial average age of the $11$ players be $A$. The total age is $11A$.
When two players aged $18$ and $20$ are replaced by two new players with ages $x$ and $y$, the new total age becomes $11A - 18 - 20 + x + y$.
The new average age is $A + \frac{2}{12} = A + \frac{1}{6}$.
Thus, $\frac{11A - 38 + x + y}{11} = A + \frac{1}{6}$.
$11A - 38 + x + y = 11A + \frac{11}{6}$.
$x + y = 38 + \frac{11}{6} = 38 + 1 \text{ year } 10 \text{ months} = 39 \text{ years } 10 \text{ months}$.
The average age of the two new players is $\frac{x + y}{2} = \frac{39 \text{ years } 10 \text{ months}}{2} = 19 \text{ years } 11 \text{ months}$.

(A) Let the number of students be $x$.
The total age of $x$ students is $6x$.
The total age of $12$ teachers is $12 \times 40 = 480$.
The total number of people in the school is $(x + 12)$.
The average age of the combined group is given as $7 \, \text{years}$.
Therefore, the equation is:
$\frac{6x + 480}{x + 12} = 7$
Multiplying both sides by $(x + 12)$:
$6x + 480 = 7(x + 12)$
Expanding the right side:
$6x + 480 = 7x + 84$
Rearranging the terms to solve for $x$:
$480 - 84 = 7x - 6x$
$x = 396$
Thus, the number of students is $396$.

(A) Let the average of $11$ innings be $x$.
Total runs scored in $11$ innings $= 11x$.
In the $12^{\text{th}}$ innings,he scores $90$ runs,so the total runs after $12$ innings $= 11x + 90$.
The new average after $12$ innings is $(x - 5)$.
Therefore,the total runs after $12$ innings can also be expressed as $12(x - 5)$.
Equating the two expressions for total runs:
$11x + 90 = 12(x - 5)$
$11x + 90 = 12x - 60$
$12x - 11x = 90 + 60$
$x = 150$.
The average after $12$ innings $= x - 5 = 150 - 5 = 145$.

(A) Let the number of wickets taken by the cricketer before the last match be $x$.
The total runs conceded before the last match is $12.4x$.
In the last match,he gave $26$ runs and took $5$ wickets.
Total runs after the last match $= 12.4x + 26$.
Total wickets after the last match $= x + 5$.
According to the problem,the new average is $12$ runs per wicket:
$\frac{12.4x + 26}{x + 5} = 12$
$12.4x + 26 = 12(x + 5)$
$12.4x + 26 = 12x + 60$
$12.4x - 12x = 60 - 26$
$0.4x = 34$
$x = \frac{34}{0.4} = \frac{340}{4} = 85$
Therefore,the cricketer had taken $85$ wickets till the last match.

(C) Let the score of the batsman in the $26^{th}$ inning be $x$.
The total runs scored in $25$ innings $= 25 \times 56 = 1400$.
After the $26^{th}$ inning,the new average becomes $56 + 2 = 58$.
The total runs scored in $26$ innings $= 26 \times 58 = 1508$.
The score in the $26^{th}$ inning is the difference between the total runs after $26$ innings and the total runs after $25$ innings.
$x = 1508 - 1400 = 108$.
Therefore,the batsman scored $108$ runs in the $26^{th}$ inning.

(A) Let the number of students be $30$ and the teacher be $1$. Total number of people $= 30 + 1 = 31$.
Initial average age of $31$ people $= 14 \, \text{years}$.
Total age of $31$ people $= 31 \times 14 = 434 \, \text{years}$.
When the teacher is excluded,the number of students remains $30$.
The new average age reduces by $0.5 \, \text{years}$,so the new average $= 14 - 0.5 = 13.5 \, \text{years}$.
Total age of $30$ students $= 30 \times 13.5 = 405 \, \text{years}$.
Age of the teacher $= (\text{Total age of } 31 \text{ people}) - (\text{Total age of } 30 \text{ students})$.
Age of the teacher $= 434 - 405 = 29 \, \text{years}$.

(A) Let the marks of the new student be $x$.
The total number of students is $20$.
The change in the total marks of the group is given by the change in average multiplied by the number of students: $\text{Change} = -4 \times 20 = -80$.
This means the total marks of the group decreased by $80$ when the topper was replaced.
Let $T$ be the marks of the topper $(90)$ and $N$ be the marks of the new student $(x)$.
The relationship is: $N - T = \text{Change in total marks}$.
$x - 90 = -80$.
$x = 90 - 80 = 10$.
Therefore,the marks of the new student are $10$.

(A) Given that the average age of $x$ and $y$ is $20$ years,the sum of their ages is $x + y = 20 \times 2 = 40$ years.
If $z$ replaces $x$,the average age of $z$ and $y$ is $19$ years,so $z + y = 19 \times 2 = 38$ years.
If $z$ replaces $y$,the average age of $x$ and $z$ is $21$ years,so $x + z = 21 \times 2 = 42$ years.
Adding these three equations: $(x + y) + (z + y) + (x + z) = 40 + 38 + 42$,which simplifies to $2(x + y + z) = 120$,so $x + y + z = 60$.
Now,find the individual ages:
$x = (x + y + z) - (z + y) = 60 - 38 = 22$ years.
$y = (x + y + z) - (x + z) = 60 - 42 = 18$ years.
$z = (x + y + z) - (x + y) = 60 - 40 = 20$ years.
Thus,the ages are $x = 22, y = 18, z = 20$.

(A) Total age of a family of $6$ members $= 22 \times 6 = 132 \text{ years}$.
At the time of the birth of the youngest member (who is $7 \text{ years}$ old now),each of the $6$ members was $7 \text{ years}$ younger.
Total reduction in age $= 6 \times 7 = 42 \text{ years}$.
Total age of the family at the time of the birth of the youngest member $= 132 - 42 = 90 \text{ years}$.
At that time,the youngest member was not born,so there were only $5$ members in the family.
Average age at that time $= \frac{90}{5} = 18 \text{ years}$.

(B) Let the initial average age of $10$ men be $A$.
Total age of $10$ men $= 10A$.
When a man of $25 \, \text{years}$ is replaced by a new man of age $X$, the new total age becomes $10A - 25 + X$.
The new average age is $A + 2$.
Therefore, the new total age is $10(A + 2) = 10A + 20$.
Equating the two expressions for the new total age: $10A - 25 + X = 10A + 20$.
$X - 25 = 20$.
$X = 20 + 25 = 45$.
Thus, the age of the new man is $45 \, \text{years}$.

(A) The initial sum of marks for $100$ students is calculated as: $100 \times 60 = 6000$.
Since one student's marks were incorrectly recorded as $75$ instead of $65$,we must adjust the total sum.
The correct sum of marks is: $6000 - 75 + 65 = 5990$.
The correct mean is the correct sum divided by the number of students: $\frac{5990}{100} = 59.9$.

(C) Let the four consecutive even numbers be $x, x+2, x+4$ and $x+6$.
Given that their average is $65$,we have:
$\frac{x + (x+2) + (x+4) + (x+6)}{4} = 65$
$4x + 12 = 65 \times 4$
$4x + 12 = 260$
$4x = 248$
$x = 62$
Thus,the four numbers are $A = 62, B = 64, C = 66$ and $D = 68$.
The product of $A$ and $D$ is $62 \times 68 = 4216$.

(C) Initial total weight of $36$ students $= 36 \times 50 = 1800 \,kg$.
The difference between the misread weight and the actual weight is $73 - 37 = 36 \,kg$.
Since the misread weight was higher than the actual weight,we subtract the difference from the initial total weight.
Correct total weight $= 1800 - 36 = 1764 \,kg$.
Correct average $= \frac{1764}{36} = 49 \,kg$.

(C) Total runs scored in $25$ innings = $25 \times 46 = 1150$.
New average after $26$ innings = $46 + 2 = 48$ runs per innings.
Total runs scored in $26$ innings = $26 \times 48 = 1248$.
Score in the $26^{th}$ innings = (Total runs in $26$ innings) - (Total runs in $25$ innings).
Score in the $26^{th}$ innings = $1248 - 1150 = 98$ runs.

(A) The sum of the ages of $5$ members $3 \, \text{years}$ ago was $17 \times 5 = 85 \, \text{years}$.
The current sum of the ages of these $5$ members is $85 + (3 \times 5) = 85 + 15 = 100 \, \text{years}$.
The current average age of $6$ members (including the baby) is $17 \, \text{years}$.
Therefore,the current total sum of the ages of $6$ members is $17 \times 6 = 102 \, \text{years}$.
The age of the new baby is the difference between the total sum of $6$ members and the total sum of the original $5$ members: $102 - 100 = 2 \, \text{years}$.

(B) The sum of marks in the first $4$ subjects is calculated as: $4 \times 75 = 300$.
Adding the marks of the $5^{th}$ subject,the total sum of marks for $5$ subjects is: $300 + 80 = 380$.
The new average is calculated by dividing the total sum by the number of subjects: $\frac{380}{5} = 76$.

(B) The sum of the weights of $A, B$ and $C$ is $84 \times 3 = 252 \, kg$.
When $D$ joins,the sum of the weights of $A, B, C$ and $D$ is $80 \times 4 = 320 \, kg$.
Therefore,the weight of $D = 320 - 252 = 68 \, kg$.
Given that $E$ is $3 \, kg$ heavier than $D$,the weight of $E = 68 + 3 = 71 \, kg$.
When $E$ replaces $A$,the new group consists of $B, C, D$ and $E$. The sum of their weights is $79 \times 4 = 316 \, kg$.
Since $B + C + D + E = 316 \, kg$ and $E = 71 \, kg$,we have $B + C + D = 316 - 71 = 245 \, kg$.
From the equation $A + B + C + D = 320 \, kg$,we can find $A$ by subtracting the sum of $B, C$ and $D$:
$A = 320 - 245 = 75 \, kg$.

(A) Total runs of $42$ innings $= 42 \times 30 = 1260$.
Total runs excluding the maximum and minimum innings $= 40 \times 28 = 1120$.
Let $\text{max}$ be the maximum score and $\text{min}$ be the minimum score.
Then,$\text{max} + \text{min} = 1260 - 1120 = 140$ $.....(i)$.
We are given that $\text{max} - \text{min} = 100$ $.....(ii)$.
Adding equations $(i)$ and $(ii)$ gives $2 \times \text{max} = 240$,so $\text{max} = 120$.
Substituting $\text{max} = 120$ into equation $(i)$,we get $120 + \text{min} = 140$,which implies $\text{min} = 20$.
Hence,the minimum score is $20$.

(B) Let the present ages of the husband and wife be $H$ and $W$ respectively.
Five years ago,their average age was $23 \text{ years}$,so the sum of their ages five years ago was $23 \times 2 = 46 \text{ years}$.
The sum of their present ages is $(H + 5) + (W + 5) = 46 + 10 = 56 \text{ years}$.
Today,the average age of the husband,wife,and child is $20 \text{ years}$.
Thus,the sum of their present ages is $20 \times 3 = 60 \text{ years}$.
Let the age of the child be $C$. Then $H + W + C = 60$.
Substituting the sum of the husband and wife's ages: $56 + C = 60$.
Therefore,$C = 60 - 56 = 4 \text{ years}$.

(B) The sum of the ages of the husband and wife at the time of marriage was $25 \times 2 = 50 \, \text{years}$.
Since they were married $7 \, \text{years}$ ago,the current sum of their ages is $50 + (7 \times 2) = 50 + 14 = 64 \, \text{years}$.
The current average age of the family (husband,wife,and child) is $22 \, \text{years}$.
The total age of the family is $22 \times 3 = 66 \, \text{years}$.
The present age of the child is the difference between the total family age and the sum of the parents' ages: $66 - 64 = 2 \, \text{years}$.

(A) Let the ages of mother,father,and son at the time of marriage be $M, F,$ and $S$ respectively.
Given,$(M + F + S) / 3 = 42$,so $M + F + S = 126$.
Let the bride's age at the time of marriage be $B$.
After $6$ years,the ages of mother,father,son,and bride will be $(M+6), (F+6), (S+6),$ and $(B+6)$ respectively.
An infant was born $1$ year after the marriage,so the infant's age after $6$ years of marriage is $6 - 1 = 5$ years.
The total number of family members is $5$ (mother,father,son,bride,and infant).
The average age after $6$ years is $36$,so the total age is $36 \times 5 = 180$.
Equation: $(M+6) + (F+6) + (S+6) + (B+6) + 5 = 180$.
$(M + F + S) + 18 + B + 6 + 5 = 180$.
$126 + 29 + B = 180$.
$155 + B = 180$.
$B = 180 - 155 = 25$ years.

(A) Let the original average expenditure per student be $x$ $Rs$.
The original total expenditure for $35$ students is $35x$ $Rs$.
When the number of students increases by $7$,the new number of students becomes $35 + 7 = 42$.
The new total expenditure becomes $35x + 42$ $Rs$.
The new average expenditure per student is $(35x + 42) / 42$.
According to the problem,the new average expenditure is $(x - 1)$ $Rs$.
So,$(35x + 42) / 42 = x - 1$.
Multiplying both sides by $42$,we get $35x + 42 = 42(x - 1)$.
$35x + 42 = 42x - 42$.
$42x - 35x = 42 + 42$.
$7x = 84$.
$x = 12$ $Rs$.
The original expenditure of the mess is $35 \times x = 35 \times 12 = 420$ $Rs$.

(D) Ashish's total earnings for the first $11$ months $= 4200 \times 11 = 46,200$ dollars.
Total earnings required for $12$ months to maintain an average of $5000$ dollars per month $= 5000 \times 12 = 60,000$ dollars.
Earnings required in the last month $= 60,000 - 46,200 = 13,800$ dollars.

(A) Let the three numbers be $x, y,$ and $z$.
Given the ratios:
$x:y = 2:3$
$y:z = 5:8$
To combine these ratios,we make the value of $y$ common in both ratios. The least common multiple of $3$ and $5$ is $15$.
Multiplying the first ratio by $5$: $x:y = (2 \times 5) : (3 \times 5) = 10:15$.
Multiplying the second ratio by $3$: $y:z = (5 \times 3) : (8 \times 3) = 15:24$.
Thus,the combined ratio is $x:y:z = 10:15:24$.
The sum of the parts is $10 + 15 + 24 = 49$.
Given the sum of the numbers is $98$,the value of one part is $98 / 49 = 2$.
The second number $(y)$ corresponds to $15$ parts.
Therefore,the second number is $15 \times 2 = 30$.

(A) Let the total number of workers be $x$.
The total salary of all workers is $95x$.
The total salary of $15$ officers is $15 \times 525 = 7875$.
The number of remaining workers is $(x - 15)$,and their total salary is $(x - 15) \times 85$.
According to the problem:
$7875 + (x - 15) \times 85 = 95x$
Expanding the equation:
$7875 + 85x - 1275 = 95x$
$6600 + 85x = 95x$
$95x - 85x = 6600$
$10x = 6600$
$x = 660$
Therefore,the total number of workers is $660$.

(D) Let the average after $86$ innings be $A$. The total runs after $86$ innings is $86A$.
In the $87^{th}$ inning,he scores $270$ runs,so the new total is $86A + 270$.
The new average is $\frac{86A + 270}{87} = A + x$,where $x$ is a whole number.
$86A + 270 = 87A + 87x$
$270 - 87x = A$
Since $A$ must be positive,$270 - 87x > 0$,so $87x < 270$,which means $x$ can be $0, 1, 2, 3$.
If $x = 0$,$A = 270$,new average $= 270 + 0 = 270$.
If $x = 1$,$A = 270 - 87 = 183$,new average $= 183 + 1 = 184$.
If $x = 2$,$A = 270 - 174 = 96$,new average $= 96 + 2 = 98$.
If $x = 3$,$A = 270 - 261 = 9$,new average $= 9 + 3 = 12$.
Thus,the possible values for the new average are $12, 98, 184, 270$.

(D) Let the average expenditure of all $19$ persons be $x$.
Total expenditure of $19$ persons $= 19x$.
Expenditure of $13$ persons $= 13 \times 79 = 1027$.
Remaining persons $= 19 - 13 = 6$.
Expenditure of each of the remaining $6$ persons $= x + 4$.
Total expenditure of these $6$ persons $= 6(x + 4) = 6x + 24$.
According to the problem:
$1027 + 6x + 24 = 19x$
$1051 = 19x - 6x$
$1051 = 13x$
$x = \frac{1051}{13} \approx 80.846$.
Total expenditure $= 19x = 19 \times \frac{1051}{13} = \frac{19969}{13} \approx 1536.077$.
Rounding to two decimal places,the total money spent is $Rs. 1536.07$.

(B) Let the number of persons initially going for the picnic be $x$.
The total age of the initial group is $16.75x$.
The total age of the $20$ new persons is $20 \times 13.25 = 265$.
The total number of persons after joining is $(x + 20)$.
The new average age is $15 \text{ years}$,so the total age is $15(x + 20)$.
Equating the total ages: $16.75x + 265 = 15(x + 20)$.
$16.75x + 265 = 15x + 300$.
$16.75x - 15x = 300 - 265$.
$1.75x = 35$.
$x = \frac{35}{1.75} = \frac{3500}{175} = 20$.
Therefore,the number of persons initially going for the picnic is $20$.

(C) Let the number of faculty be $x$ and the number of management trainees be $y$.
Given that the average salary of faculty is $Rs. 4,000$ and management trainees is $Rs. 1,250$,and the average salary of all workers is $Rs. 2,000$.
The total salary equation is: $4000x + 1250y = 2000(x + y)$.
Expanding the equation: $4000x + 1250y = 2000x + 2000y$.
Rearranging the terms: $2000x = 750y$.
Simplifying the ratio: $x/y = 750/2000 = 3/8$.
This means the total number of workers $(x + y)$ must be a multiple of $(3 + 8) = 11$.
Among the given options,only $110$ is a multiple of $11$ $(110 = 11 \times 10)$.

(A) $1$. $10$ years ago,the average age of $25$ teachers was $45$ years. So,the total age was $25 \times 45 = 1125$ years.
$2$. $4$ years ago (i.e.,$6$ years after the initial point),the total age of these $25$ teachers was $1125 + (6 \times 25) = 1125 + 150 = 1275$ years.
$3$. When the principal retired at $60$ years,the total age of the remaining $24$ teachers was $1275 - 60 = 1215$ years.
$4$. One year later (i.e.,$3$ years ago from the present),the total age of the $24$ teachers was $1215 + (1 \times 24) = 1239$ years.
$5$. $A$ new principal aged $54$ years was recruited,so the total age of the $25$ staff members became $1239 + 54 = 1293$ years.
$6$. At the present time ($3$ years later),the total age of all $25$ staff members is $1293 + (3 \times 25) = 1293 + 75 = 1368$ years.
$7$. The present average age is $\frac{1368}{25} = 54 \frac{18}{25}$ years.

(A) Total weight of all $11$ players $= 11 \times 50 = 550 \, kg$.
Sum of the weights of the first six lightest players $= 6 \times 49 = 294 \, kg$.
Sum of the weights of the last six heaviest players $= 6 \times 52 = 312 \, kg$.
When we add the sum of the six lightest and the six heaviest players,the player at the $6^{th}$ position is counted twice.
Weight of the $6^{th}$ player $= (Sum \, of \, first \, six) + (Sum \, of \, last \, six) - (Total \, weight \, of \, 11 \, players)$.
Weight of the $6^{th}$ player $= 294 + 312 - 550 = 606 - 550 = 56 \, kg$.

(A) Let the ages be $S, G, K, D, I$.
Given: $(S + G) / 2 = 35 \Rightarrow S + G = 70$.
If $K$ replaces $S$: $(K + G) / 2 = 32 \Rightarrow K + G = 64$.
If $K$ replaces $G$: $(S + K) / 2 = 38 \Rightarrow S + K = 76$.
Adding these three equations: $2(S + G + K) = 70 + 64 + 76 = 210$.
Thus,$S + G + K = 105$.
The average age of $S, G, K$ is $105 / 3 = 35$.
The average age of $D$ and $I$ is half of this: $(D + I) / 2 = 35 / 2 = 17.5$.
Therefore,$D + I = 35$.
The sum of ages of all five is $(S + G + K) + (D + I) = 105 + 35 = 140$.
The average age of all five is $140 / 5 = 28 \text{ years}$.

(C) The total age of $6$ members $4$ years ago was $25 \times 6 = 150$ years.
The total present age of these $6$ members is $150 + (4 \times 6) = 150 + 24 = 174$ years.
Now,the family has $7$ members (including the child),and the average age is still $25$ years.
The total present age of the $7$ members is $25 \times 7 = 175$ years.
The present age of the child is the difference between the total age of $7$ members and the total age of $6$ members: $175 - 174 = 1$ year.

(A) The total expenditure from $January$ to $June$ is calculated as: $4200 \times 6 = 25200 \text{ Rs.}$
To find the total expenditure from $February$ to $July$,we subtract the expenditure of $January$ and add the expenditure of $July$ to the total of $January$ to $June$:
Total expenditure (Feb to July) $= 25200 - 1200 + 1500 = 25500 \text{ Rs.}$
The average expenditure for these $6$ months is:
Average $= \frac{25500}{6} = 4250 \text{ Rs.}$

(A) Total price of $80$ computers $= 80 \times 30,000 = 2,400,000$.
Total price of remaining $78$ computers $= 78 \times 29,500 = 2,301,000$.
Sum of the prices of the highest and lowest price computers $= 2,400,000 - 2,301,000 = 99,000$.
Given that the cost of the highest price computer is $80,000$.
Therefore,the cost of the lowest price computer $= 99,000 - 80,000 = 19,000$.

(A) $1$. Eleven years ago,the total age of $4$ members was $4 \times 28 = 112 \, \text{years}$.
$2$. Presently,these $4$ members have aged by $11$ years each,so their total age is $112 + (4 \times 11) = 112 + 44 = 156 \, \text{years}$.
$3$. Currently,the family has $6$ members,and their average age is $28 \, \text{years}$. Thus,the total age of the $6$ members is $6 \times 28 = 168 \, \text{years}$.
$4$. The sum of the ages of the $2$ children is $168 - 156 = 12 \, \text{years}$.
$5$. Let the present age of the youngest child be $x$. Since the children were born within the last $11$ years,the elder child's age is $x + k$,where $k$ is the age gap.
$6$. The problem states that at the time of the birth of the youngest child,the elder child's age was equal to the total number of family members $(6)$.
$7$. Let $t$ be the time elapsed since the birth of the youngest child. Then the youngest child's age is $t$,and the elder child's age is $6 + t$.
$8$. The sum of their ages is $t + (6 + t) = 12 \Rightarrow 2t + 6 = 12 \Rightarrow 2t = 6 \Rightarrow t = 3$.
$9$. Thus,the present age of the youngest member is $3 \, \text{years}$.