Average Questions in English

(D) Let the three numbers be $x, y,$ and $z$.
According to the problem,the average of the first and second is greater than the average of the second and third by $15$.
Mathematically,this is expressed as: $\frac{x+y}{2} = \frac{y+z}{2} + 15$.
Subtracting $\frac{y}{2}$ from both sides,we get: $\frac{x}{2} = \frac{z}{2} + 15$.
Rearranging the terms: $\frac{x}{2} - \frac{z}{2} = 15$.
Multiplying both sides by $2$,we get: $x - z = 30$.
Thus,the difference between the first and the third number is $30$.

(D) Total score in $10$ innings $= 32 \times 10 = 320$.
To increase the average by $4$,the new average will be $32 + 4 = 36$.
Total score required after $11$ innings $= 36 \times 11 = 396$.
Runs required in the $11^{th}$ innings $= 396 - 320 = 76$.

(B) Let the total number of workers be $x$.
The number of workers excluding the technicians is $(x - 7)$.
The total salary of all workers is the sum of the total salary of technicians and the total salary of the remaining workers.
$8000x = (7 \times 12000) + ((x - 7) \times 6000)$
$8000x = 84000 + 6000x - 42000$
$8000x - 6000x = 42000$
$2000x = 42000$
$x = \frac{42000}{2000} = 21$
Therefore,the total number of workers in the workshop is $21$.

(C) The average age of $5$ members $3$ years ago was $17$ years.
Total age of $5$ members $3$ years ago $= 5 \times 17 = 85$ years.
Present total age of these $5$ members $= 85 + (5 \times 3) = 85 + 15 = 100$ years.
Let the present age of the baby be $x$ years.
Now,the family has $6$ members,and their average age is still $17$ years.
Total age of $6$ members $= 6 \times 17 = 102$ years.
Therefore,$100 + x = 102$.
$x = 102 - 100 = 2$ years.
The present age of the baby is $2$ years.

(D) Let the number of wickets taken by the cricketer till the last match be $x$.
The total runs conceded by him till the last match is $12.4x$.
In the current match,he takes $5$ wickets for $26$ runs.
So,the new total number of wickets is $x + 5$ and the new total runs conceded is $12.4x + 26$.
The new bowling average is $12.4 - 0.4 = 12$.
According to the problem,the new average is given by the ratio of total runs to total wickets:
$\frac{12.4x + 26}{x + 5} = 12$
Multiplying both sides by $(x + 5)$:
$12.4x + 26 = 12(x + 5)$
$12.4x + 26 = 12x + 60$
Subtracting $12x$ from both sides:
$0.4x + 26 = 60$
Subtracting $26$ from both sides:
$0.4x = 34$
Dividing by $0.4$:
$x = \frac{34}{0.4} = 85$
Thus,the number of wickets taken by him till the last match was $85$.

(C) Let the weight of the new person be $x\, kg$.
Weight of the person replaced $= 65\, kg$.
The total increase in the weight of the group $= \text{Number of persons} \times \text{Increase in average}$.
Total increase $= 8 \times 2.5 = 20\, kg$.
Weight of the new person $= \text{Weight of replaced person} + \text{Total increase}$.
$x = 65 + 20 = 85\, kg$.
Therefore,the weight of the new person is $85\, kg$.

(A) Total weight of $24$ students $= 24 \times 35 = 840 \, kg$.
Let the weight of the teacher be $x \, kg$.
When the teacher is included, the total number of people becomes $24 + 1 = 25$.
The new average weight is $35 \, kg + 400 \, g = 35 \, kg + 0.4 \, kg = 35.4 \, kg$.
The total weight of $25$ people $= 25 \times 35.4 = 885 \, kg$.
The weight of the teacher $x = (\text{Total weight of } 25 \text{ people}) - (\text{Total weight of } 24 \text{ students})$.
$x = 885 - 840 = 45 \, kg$.

(D) To find the average marks,we use the formula: $\text{Average} = \frac{\text{Sum of all observations}}{\text{Number of observations}}$.
Sum of marks $= 76 + 65 + 82 + 67 + 85 = 375$.
Number of subjects $= 5$.
$\text{Average marks} = \frac{375}{5} = 75$.

(D) Let the number of other workers be $n$.
Then,the number of agricultural workers is $11n$.
Total income of agricultural workers $= 11n \times S = 11nS$.
Total income of other workers $= n \times T = nT$.
Total number of workers $= 11n + n = 12n$.
Total income of all workers $= 11nS + nT = n(11S + T)$.
Average annual income of all workers $= \frac{\text{Total Income}}{\text{Total Number of Workers}} = \frac{n(11S + T)}{12n} = \frac{11S + T}{12}$.

(B) Let the number of boys be $m$ and the number of girls be $n$.
Given that the average age of boys is $16.4 \text{ years}$ and the average age of girls is $15.4 \text{ years}$.
The total average age of the class is $15.8 \text{ years}$.
Using the weighted average formula:
$\frac{m \times 16.4 + n \times 15.4}{m + n} = 15.8$
$16.4m + 15.4n = 15.8(m + n)$
$16.4m + 15.4n = 15.8m + 15.8n$
Rearranging the terms to group $m$ and $n$:
$16.4m - 15.8m = 15.8n - 15.4n$
$0.6m = 0.4n$
$\frac{m}{n} = \frac{0.4}{0.6} = \frac{2}{3}$
Thus,the ratio of the number of boys to the number of girls is $2: 3$.

(D) Let the ten numbers be $x_1, x_2, ..., x_{10}$.
The average is given by $A = \frac{x_1 + x_2 + ... + x_{10}}{10}$.
If each number is increased by $12 \%$,the new numbers become $x_i' = x_i + 0.12x_i = 1.12x_i$.
The new average $A'$ is given by $A' = \frac{1.12x_1 + 1.12x_2 + ... + 1.12x_{10}}{10}$.
$A' = 1.12 \times \frac{x_1 + x_2 + ... + x_{10}}{10} = 1.12A$.
Since $A' = A + 0.12A$,the average increases by $12 \%$.

(C) The total earnings for $7$ days are calculated by summing the daily amounts:
Total earning $= 60 + 65 + 70 + 52.50 + 63 + 73 + 68 = 451.50$
The average daily earning is calculated by dividing the total earnings by the number of days $(7)$:
Average daily earning $= \frac{451.50}{7} = 64.50$
Therefore,the average daily earning is $Rs. 64.50$.

(C) The average of $10$ numbers is $7$.
The sum of these $10$ numbers is $10 \times 7 = 70$.
If each number is multiplied by $8$,the new sum of the $10$ numbers becomes $70 \times 8 = 560$.
The new average is calculated by dividing the new sum by the total count of numbers:
New average $= \frac{560}{10} = 56$.
Alternatively,if each number in a set is multiplied by a constant $k$,the average of the set is also multiplied by $k$. Here,$7 \times 8 = 56$.

(B) Average weight of $5$ persons $= 38 \, kg$.
Total weight of these $5$ persons $= 38 \times 5 = 190 \, kg$.
Now,the total number of entities including the boat and the $5$ persons is $5 + 1 = 6$.
Average weight of (the boat $+$ $5$ persons) $= 52 \, kg$.
Total weight of (the boat $+$ $5$ persons) $= 52 \times 6 = 312 \, kg$.
Therefore,the weight of the boat $= 312 - 190 = 122 \, kg$.

(D) Let the original total expenditure be $Rs. x$.
Original average expenditure per student $= \frac{x}{35}$.
When the number of students increases by $7$,the new number of students $= 35 + 7 = 42$.
The new total expenditure $= Rs. (x + 42)$.
New average expenditure per student $= \frac{x + 42}{42}$.
According to the problem,the average expenditure decreased by $Rs. 1$,so:
$\frac{x}{35} - \frac{x + 42}{42} = 1$.
Taking the least common multiple $(LCM)$ of $35$ and $42$,which is $210$:
$\frac{6x - 5(x + 42)}{210} = 1$.
$6x - 5x - 210 = 210$.
$x = 210 + 210 = 420$.
Therefore,the actual original expenditure of the mess is $Rs. 420$.

(C) To find the average daily maximum temperature,we calculate the arithmetic mean of the given temperatures.
Average = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$
Sum = $42.7 + 44.6 + 42.0 + 39.1 + 43.0 + 42.5 + 38.5 = 292.4$
Number of days = $7$
Average = $\frac{292.4}{7} = 41.77^{\circ} C$
Thus,the average daily maximum temperature is $41.77^{\circ} C$.

(D) Let the total number of workers be $x$.
The total salary of all workers is $850x$.
The total salary of $7$ technicians is $7 \times 1000 = 7000$.
The number of remaining workers is $(x - 7)$,and their total salary is $(x - 7) \times 780$.
Equating the total salary:
$850x = 7000 + 780(x - 7)$
$850x = 7000 + 780x - 5460$
$850x - 780x = 1540$
$70x = 1540$
$x = \frac{1540}{70} = 22$.
Thus,the total number of workers in the workshop is $22$.

(A) The average speed is defined as the total distance traveled divided by the total time taken.
First,we calculate the time taken for each segment of the journey using the formula: $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$.

Distance	Speed	Time
$2500 \, km$	$500 \, km/h$	$5 \, hrs$
$1200 \, km$	$400 \, km/h$	$3 \, hrs$
$500 \, km$	$250 \, km/h$	$2 \, hrs$
Total: $4200 \, km$	-	Total: $10 \, hrs$

Now,calculate the average speed:
$\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{4200 \, km}{10 \, hrs} = 420 \, km/h$.

(A) Total number of students $= 20$.
Marks scored by $2$ students $= 2 \times 100 = 200$.
Marks scored by $3$ students $= 3 \times 0 = 0$.
Number of remaining students $= 20 - (2 + 3) = 15$.
Marks scored by $15$ students $= 15 \times 40 = 600$.
Total marks scored by all $20$ students $= 200 + 0 + 600 = 800$.
Average marks of the whole class $= \frac{\text{Total Marks}}{\text{Total Students}} = \frac{800}{20} = 40$.

(B) Average weight of $24$ students of section $A = 58 \,kg$.
Total weight of $24$ students of section $A = 58 \times 24 = 1392 \,kg$.
Average weight of $26$ students of section $B = 60.5 \,kg$.
Total weight of $26$ students of section $B = 60.5 \times 26 = 1573 \,kg$.
Total weight of $50$ students $= 1392 + 1573 = 2965 \,kg$.
Average weight of all $50$ students $= \frac{2965}{50} = 59.3 \,kg$.

(C) Total age of $5$ members $= 21 \times 5 = 105 \text{ years}$.
The youngest member is $5 \text{ years}$ old,which means $5 \text{ years}$ ago,the youngest member was just born.
At that time ($5 \text{ years}$ ago),each of the $5$ members was $5 \text{ years}$ younger. So,the total age of the family was $105 - (5 \times 5) = 105 - 25 = 80 \text{ years}$.
At the time of the birth of the youngest member,the family consisted of only $4$ members (excluding the youngest one).
Therefore,the average age of the $4$ members at that time $= \frac{80}{4} = 20 \text{ years}$.

(C) The sum of $7$ numbers is calculated as: $7 \times 5 = 35$.
The sum of the first $6$ numbers is calculated as: $6 \times 4 = 24$.
The seventh number is the difference between the sum of $7$ numbers and the sum of the first $6$ numbers.
Therefore,the seventh number $= 35 - 24 = 11$.

(B) The average age of $5$ members $3$ years ago was $27 \text{ years}$.
The total age of these $5$ members $3$ years ago was $5 \times 27 = 135 \text{ years}$.
The total age of these $5$ members at present is $135 + (5 \times 3) = 135 + 15 = 150 \text{ years}$.
Now,a child is added,so the total number of members becomes $5 + 1 = 6$.
The present average age of these $6$ members is $27 \text{ years}$.
The total age of these $6$ members at present is $6 \times 27 = 162 \text{ years}$.
The present age of the child is the difference between the total age of $6$ members and the total age of $5$ members: $162 - 150 = 12 \text{ years}$.

(A) Let the initial average weight of $10$ students be $A \, kg$.
Total initial weight $= 10A \, kg$.
When a student weighing $50 \, kg$ is replaced by a new student of weight $W$,the new total weight becomes $(10A - 50 + W) \, kg$.
The new average weight is $(A + 0.5) \, kg$.
Therefore,the new total weight is $10(A + 0.5) = 10A + 5$.
Equating the two expressions for the new total weight:
$10A - 50 + W = 10A + 5$.
$W - 50 = 5$.
$W = 55 \, kg$.
Alternatively,the increase in total weight is equal to the number of students multiplied by the increase in average weight: $10 \times 0.5 = 5 \, kg$.
Thus,the new student's weight $= 50 + 5 = 55 \, kg$.

(A) Average salary of $9$ persons $= Rs.\, 2450$.
Total salary of $9$ persons $= 2450 \times 9 = Rs.\, 22050$.
Salary of the transferred person $= Rs.\, 2650$.
Total salary of the remaining $8$ persons $= 22050 - 2650 = Rs.\, 19400$.
Average salary of the remaining $8$ persons $= \frac{19400}{8} = Rs.\, 2425$.

(A) The mean marks of $10$ boys $= 70 \%$.
Total marks of $10$ boys $= 70 \% \times 10 = 700 \%$.
The mean marks of $15$ girls $= 60 \%$.
Total marks of $15$ girls $= 60 \% \times 15 = 900 \%$.
Therefore,the sum of the total marks of $25$ students $= 700 + 900 = 1600 \%$.
Therefore,the mean marks of all the $25$ students $= \frac{1600}{25} = 64 \%$.

(D) Total income for $15$ days $= 15 \times 70 = 1050$.
Total income for the first $5$ days $= 5 \times 60 = 300$.
Total income for the last $9$ days $= 9 \times 80 = 720$.
Income for the $6^{th}$ day $=$ (Total income for $15$ days) $-$ (Total income for first $5$ days $+$ Total income for last $9$ days).
Income for the $6^{th}$ day $= 1050 - (300 + 720) = 1050 - 1020 = 30$.
Thus,$A$'s income for the $6^{th}$ day is $Rs. 30$.

(C) The five consecutive even numbers starting with $4$ are $4, 6, 8, 10,$ and $12$.
The average is calculated as the sum of the numbers divided by the count of numbers.
Average $= \frac{4 + 6 + 8 + 10 + 12}{5} = \frac{40}{5} = 8$.

(D) The sum of the ages of $5$ members $3$ $years$ ago was $5 \times 17 = 85$ $years$.
The sum of the present ages of these $5$ members is $85 + (5 \times 3) = 85 + 15 = 100$ $years$.
Currently,there are $6$ members in the family (including the baby),and the average age is still $17$ $years$.
The sum of the present ages of all $6$ members is $6 \times 17 = 102$ $years$.
The age of the baby is the difference between the sum of the ages of $6$ members and the sum of the ages of the $5$ original members:
Age of the baby $= 102 - 100 = 2$ $years$.

(A) Total sum of $17$ numbers $= 17 \times 10.9 = 185.3$.
Sum of the first $9$ numbers $= 9 \times 10.5 = 94.5$.
Sum of the last $9$ numbers $= 9 \times 11.4 = 102.6$.
The sum of the first $9$ and last $9$ numbers includes the middle number twice.
Sum of $18$ numbers $= 94.5 + 102.6 = 197.1$.
The middle number $= (\text{Sum of } 18 \text{ numbers}) - (\text{Sum of } 17 \text{ numbers})$.
Middle number $= 197.1 - 185.3 = 11.8$.

(C) Let the average of runs for $12$ innings be $x$.
Total runs for $12$ innings $= 12x$.
In the $13^{th}$ innings,he scores $96$ runs,so the total runs for $13$ innings $= 12x + 96$.
The new average after $13$ innings is $x + 5$.
Therefore,the equation is: $\frac{12x + 96}{13} = x + 5$.
$12x + 96 = 13(x + 5)$.
$12x + 96 = 13x + 65$.
$x = 96 - 65 = 31$.
The average after the $13^{th}$ innings is $x + 5 = 31 + 5 = 36$.

(B) Let the average score after $16$ innings be $x$.
Total runs scored in $16$ innings $= 16x$.
In the $17^{th}$ innings,he scores $85$ runs,so the total runs after $17$ innings $= 16x + 85$.
The new average after $17$ innings is $x + 3$.
Therefore,the equation is: $(16x + 85) / 17 = x + 3$.
$16x + 85 = 17(x + 3)$.
$16x + 85 = 17x + 51$.
$x = 85 - 51 = 34$.
The average after the $17^{th}$ innings is $x + 3 = 34 + 3 = 37$.

(A) Let the three numbers be $x, y,$ and $z$.
According to the problem,the sum of the numbers is $x + y + z = 98$.
The ratio between the first and second number is $x : y = 2 : 3$,which implies $x = \frac{2y}{3}$.
The ratio between the second and third number is $y : z = 5 : 8$,which implies $z = \frac{8y}{5}$.
Substituting these values into the sum equation: $\frac{2y}{3} + y + \frac{8y}{5} = 98$.
To solve for $y$,find a common denominator,which is $15$: $\frac{10y + 15y + 24y}{15} = 98$.
This simplifies to $\frac{49y}{15} = 98$.
Multiplying both sides by $15$ and dividing by $49$: $y = \frac{98 \times 15}{49} = 2 \times 15 = 30$.
Therefore,the second number is $30$.

(C) Let the initial average weight of the $8$ sailors be $A \, kg$.
Total weight of the $8$ sailors $= 8A \, kg$.
When a sailor weighing $56 \, kg$ is replaced by a new sailor of weight $W$,the new total weight becomes $(8A - 56 + W) \, kg$.
The new average weight is $(A + 1) \, kg$.
Therefore,the new total weight is $8(A + 1) \, kg$.
Equating the two expressions for the total weight:
$8A - 56 + W = 8(A + 1)$
$8A - 56 + W = 8A + 8$
$W = 8 + 56$
$W = 64 \, kg$.
Alternatively,the increase in total weight is equal to the number of sailors multiplied by the increase in average weight: $8 \times 1 = 8 \, kg$. Thus,the new sailor's weight is $56 + 8 = 64 \, kg$.

(C) The average of $5, 7, 14$ and $y$ is given by $\frac{5+7+14+y}{4} = \frac{26+y}{4}$.
Given that $x$ is $80 \%$ of this average,we have $x = \frac{80}{100} \times \frac{26+y}{4} = \frac{4}{5} \times \frac{26+y}{4} = \frac{26+y}{5}$.
So,$5x = 26+y$,which implies $y = 5x - 26$ ..... $(1)$.
Also,the average of $x$ and $y$ is $26$,so $\frac{x+y}{2} = 26$,which means $x+y = 52$ ..... $(2)$.
Substituting the value of $y$ from $(1)$ into $(2)$,we get $x + (5x - 26) = 52$.
$6x = 78$,so $x = 13$.
Substituting $x = 13$ into $(2)$,we get $13 + y = 52$,which gives $y = 52 - 13 = 39$.

(C) The average age of $A, B, C, D$ five years ago was $45$ years.
Therefore,the sum of their ages five years ago was $45 \times 4 = 180$ years.
The present sum of the ages of $A, B, C, D$ is $180 + (5 \times 4) = 180 + 20 = 200$ years.
When $x$ is included,the total number of people becomes $5$.
The present average age of all five is $49$ years.
Therefore,the present sum of the ages of all five is $49 \times 5 = 245$ years.
The present age of $x$ is the difference between the sum of the ages of all five and the sum of the ages of $A, B, C, D$.
Present age of $x = 245 - 200 = 45$ years.

(C) Let the rainfall on Wednesday be $x \, cm$.
Since it rained as much on Wednesday as on all the other $6$ days combined,the total rainfall for the other $6$ days is also $x \, cm$.
Total rainfall for the week $= x + x = 2x \, cm$.
The average rainfall for the week is given as $3 \, cm$.
Since there are $7$ days in a week,the total rainfall $= 3 \times 7 = 21 \, cm$.
Equating the two expressions for total rainfall: $2x = 21$.
Therefore,$x = 21 / 2 = 10.5 \, cm$.
Thus,it rained $10.5 \, cm$ on Wednesday.

(C) Total expenditure for the first $4$ months $= 4 \times 2750 = Rs. 11000$.
Total expenditure for the next $3$ months $= 3 \times 2940 = Rs. 8820$.
Total expenditure for the last $5$ months $= 5 \times 3130 = Rs. 15650$.
Total annual expenditure $= 11000 + 8820 + 15650 = Rs. 35470$.
Total annual income $=$ Total annual expenditure $+$ Annual savings.
Total annual income $= 35470 + 5330 = Rs. 40800$.
Average monthly income $= \frac{40800}{12} = Rs. 3400$.

(B) Let the initial average age of $8$ men be $x$ years.
Sum of the ages of $8$ men $= 8x$ years.
When $2$ men are replaced by $2$ women,the new average age becomes $(x + 2)$ years.
Sum of the ages of the new group $= 8(x + 2) = 8x + 16$ years.
The increase in the total sum of ages is $(8x + 16) - 8x = 16$ years.
This increase is due to the difference between the ages of the $2$ women and the $2$ men replaced.
Sum of the ages of $2$ women $=$ (Sum of the ages of $2$ replaced men) $+ 16$.
Sum of the ages of $2$ women $= (20 + 24) + 16 = 44 + 16 = 60$ years.
Average age of the $2$ women $= \frac{60}{2} = 30$ years.

(B) The sum of $50$ numbers is $50 \times 38 = 1900$.
After discarding two numbers $45$ and $55$,the new sum is $1900 - (45 + 55) = 1900 - 100 = 1800$.
The number of remaining values is $50 - 2 = 48$.
The new average is $\frac{1800}{48} = 37.5$.

(B) Time taken to cover the first $100 \, km = \frac{100}{30} = \frac{10}{3} \, h$.
Time taken to cover the second $100 \, km = \frac{100}{40} = \frac{5}{2} \, h$.
Time taken to cover the last $100 \, km = \frac{100}{50} = 2 \, h$.
Total time taken $= \frac{10}{3} + \frac{5}{2} + 2 = \frac{20 + 15 + 12}{6} = \frac{47}{6} \, h$.
Total distance covered $= 100 + 100 + 100 = 300 \, km$.
Average speed $= \frac{\text{Total distance}}{\text{Total time}} = \frac{300}{47/6} = \frac{300 \times 6}{47} = \frac{1800}{47} \approx 38.3 \, km/h$.

(C) The sum of the first $6$ observations is $6 \times 12 = 72$.
After including the seventh observation,the total number of observations becomes $7$.
The new average is $12 - 1 = 11$.
The sum of $7$ observations is $7 \times 11 = 77$.
The seventh observation is the difference between the sum of $7$ observations and the sum of $6$ observations.
Seventh observation $= 77 - 72 = 5$.

(B) Total age of $20$ boys $= 20 \times 15.6 = 312$ years.
Total number of boys after $5$ new boys join $= 20 + 5 = 25$.
Total age of $25$ boys $= 25 \times 15.56 = 389$ years.
Total age of the $5$ new boys $= 389 - 312 = 77$ years.
Average age of the $5$ new boys $= 77 \div 5 = 15.4$ years.

(C) $1$. The sum of weights of $A, B, C$ is $84 \times 3 = 252 \, kg$.
$2$. When $D$ joins,the total weight of $A, B, C, D$ is $80 \times 4 = 320 \, kg$.
$3$. Weight of $D = 320 - 252 = 68 \, kg$.
$4$. Weight of $E = 68 + 3 = 71 \, kg$.
$5$. The average weight of $B, C, D, E$ is $79 \, kg$,so the sum of their weights is $79 \times 4 = 316 \, kg$.
$6$. Sum of weights of $B, C = 316 - (D + E) = 316 - (68 + 71) = 316 - 139 = 177 \, kg$.
$7$. Since $A + B + C = 252 \, kg$,then $A = 252 - (B + C) = 252 - 177 = 75 \, kg$.

(C) Let the initial average expenditure per boarder be $Rs.\, x$.
Total initial expenditure for $30$ boarders $= 30x$.
When the number of boarders increases by $10$,the new number of boarders $= 30 + 10 = 40$.
The new total expenditure $= 30x + 40$.
The new average expenditure per head $= \frac{30x + 40}{40}$.
According to the problem,the average expenditure per head decreases by $Rs.\, 2$,so:
$\frac{30x + 40}{40} = x - 2$
$30x + 40 = 40(x - 2)$
$30x + 40 = 40x - 80$
$10x = 120$
$x = 12$.
The actual monthly expenses $= 30 \times x = 30 \times 12 = Rs.\, 360$.

(A) Let the third number be $x$.
Then,the second number is $3x$.
The first number is $2 \times (3x) = 6x$.
The average of the three numbers is given by $\frac{6x + 3x + x}{3} = 10$.
Simplifying the equation: $\frac{10x}{3} = 10$.
Multiplying both sides by $3$,we get $10x = 30$,so $x = 3$.
Therefore,the numbers are: First $= 6(3) = 18$,Second $= 3(3) = 9$,Third $= 3$.

(A) Total weight of $36$ students = $36 \times 50 = 1800 \,kg$.
Since $37 \,kg$ was misread as $73 \,kg$,the error in the total sum is $73 - 37 = 36 \,kg$.
Correct total weight = $1800 - 36 = 1764 \,kg$.
Correct average = $\frac{1764}{36} = 49 \,kg$.

(C) Let the earnings for the seven days be $d_1, d_2, d_3, d_4, d_5, d_6, d_7$.
Sum of the first four days: $d_1 + d_2 + d_3 + d_4 = 4 \times 18 = 72$.
Sum of the last four days: $d_4 + d_5 + d_6 + d_7 = 4 \times 22 = 88$.
Total sum for the week: $(d_1 + d_2 + d_3 + d_4) + (d_4 + d_5 + d_6 + d_7) - d_4 = 72 + 88 - 20 = 140$.
Average earning for the week: $\frac{140}{7} = 20$.

(A) Let the number of candidates who passed be $x.$
Then,the number of candidates who failed is $(120 - x).$
The total marks obtained by all candidates is $120 \times 35 = 4200.$
The sum of marks of passed candidates is $39x.$
The sum of marks of failed candidates is $15(120 - x).$
According to the problem,the sum of marks of passed and failed candidates equals the total marks:
$39x + 15(120 - x) = 4200$
$39x + 1800 - 15x = 4200$
$24x = 4200 - 1800$
$24x = 2400$
$x = \frac{2400}{24} = 100.$
Therefore,the number of candidates who passed the examination is $100.$

(A) The total number of boys is $20$.
The average age decreases by $2$ months when one boy is replaced.
Total decrease in age $= 20 \times 2 = 40$ months.
Convert $40$ months into years: $40 \div 12 = 3$ years and $4$ months.
Since the average age decreased,the new boy must be younger than the boy who left.
Age of the new boy $= 18$ years $- 3$ years $4$ months.
$= 14$ years and $8$ months.