Average Questions in English

(C) Let the average speed of walking be $x \, km/hr$.
Total distance covered $= \text{Average speed} \times \text{Total time} = 32 \times 3 = 96 \, km$.
Distance walked $= 6 \, km$.
Distance covered by train $= 96 - 6 = 90 \, km$.
Time taken for walking $= \frac{6}{x} \, hours$.
Time taken by train $= \frac{90}{60} = 1.5 \, hours$.
Total time $= \frac{6}{x} + 1.5 = 3 \, hours$.
$\frac{6}{x} = 3 - 1.5 = 1.5$.
$\frac{6}{x} = \frac{3}{2}$.
$3x = 12$,so $x = 4 \, km/hr$.

(A) Let the average of prime numbers be $P$ and the average of composite numbers be $C$. Let the number of prime numbers be $x$,then the number of composite numbers is $2x$.
The average of all numbers is given by: $\frac{Px + C(2x)}{x + 2x} = 9 \Rightarrow \frac{Px + 2Cx}{3x} = 9 \Rightarrow P + 2C = 27$ (Equation $i$).
If the counts are exchanged,the number of prime numbers becomes $2x$ and the number of composite numbers becomes $x$. The new average is $9 + 2 = 11$:
$\frac{P(2x) + Cx}{2x + x} = 11 \Rightarrow \frac{2Px + Cx}{3x} = 11 \Rightarrow 2P + C = 33$ (Equation $ii$).
Solving the system of equations:
From $(i)$,$P = 27 - 2C$.
Substitute into (ii): $2(27 - 2C) + C = 33 \Rightarrow 54 - 4C + C = 33 \Rightarrow 3C = 21 \Rightarrow C = 7$.
Substitute $C = 7$ into $(i)$: $P + 2(7) = 27 \Rightarrow P + 14 = 27 \Rightarrow P = 13$.
The ratio of the average of composite numbers to the average of prime numbers is $\frac{C}{P} = \frac{7}{13}$.

(C) Let the initial number of subjects be $n$ and the initial average marks be $x.$ The total marks initially $= nx.$
When one subject with $40$ marks is replaced by two subjects with $23$ and $25$ marks,the new number of subjects is $(n+1)$ and the new average is $(x-1).$
So,$(n+1)(x-1) = nx - 40 + 23 + 25.$
$nx - n + x - 1 = nx + 8 \Rightarrow x - n = 9$ (Equation $I$).
Next,when $57$ marks of Computer Science are added,the total number of subjects becomes $(n+2)$ and the average becomes $(x-1) + 2 = x+1.$
So,$(n+2)(x+1) = (nx - 40 + 23 + 25) + 57.$
$nx + n + 2x + 2 = nx + 65 \Rightarrow n + 2x = 63$ (Equation $II$).
From Equation $I$,$x = n + 9.$ Substituting this into Equation $II$:
$n + 2(n + 9) = 63 \Rightarrow n + 2n + 18 = 63 \Rightarrow 3n = 45 \Rightarrow n = 15.$
Thus,there were $15$ subjects initially.

(A) Let $H$ be Donald's age,$W$ be his wife's age,$D$ be his daughter's age,and $S$ be his son's age.
The average age of the $4$ family members is $23 \text{ years}$,so the sum of their ages is $H + W + D + S = 23 \times 4 = 92 \text{ years}$.
Given: $H = W + 4$.
Let $x$ be the number of years passed since the son was born. At that time,$H = 32$,$W = 28$,$D = 4$,and $S = 0$. The sum of their ages was $32 + 28 + 4 + 0 = 64$.
Since $92 - 64 = 28$ years have passed for the whole family,and there are $4$ members,the time elapsed since the son was born is $28 / 4 = 7 \text{ years}$.
Present ages:
$H = 32 + 7 = 39 \text{ years}$.
$W = 28 + 7 = 35 \text{ years}$.
$D = 4 + 7 = 11 \text{ years}$.
$S = 0 + 7 = 7 \text{ years}$.
Check: $39 + 35 + 11 + 7 = 92$. This matches the total sum.
The average age of Donald $(39)$ and his daughter $(11)$ is $(39 + 11) / 2 = 50 / 2 = 25 \text{ years}$.

(B) Let the $6$ consecutive odd numbers be $(a-5), (a-3), (a-1), (a+1), (a+3), (a+5)$.
The average of all $6$ numbers is $\frac{(a-5)+(a-3)+(a-1)+(a+1)+(a+3)+(a+5)}{6} = a$.
Given that the sum of the squares of the first and last element is $178$:
$(a-5)^2 + (a+5)^2 = 178$
$(a^2 - 10a + 25) + (a^2 + 10a + 25) = 178$
$2a^2 + 50 = 178$
$2a^2 = 128$
$a^2 = 64$
$a = 8$ (since the numbers are positive).
Verification with the first condition:
Average of squares of first $4$ numbers: $\frac{(a-5)^2 + (a-3)^2 + (a-1)^2 + (a+1)^2}{4} = \frac{(3)^2 + (5)^2 + (7)^2 + (9)^2}{4} = \frac{9 + 25 + 49 + 81}{4} = \frac{164}{4} = 41$.
Average of squares of last $4$ numbers: $\frac{(a-1)^2 + (a+1)^2 + (a+3)^2 + (a+5)^2}{4} = \frac{(7)^2 + (9)^2 + (11)^2 + (13)^2}{4} = \frac{49 + 81 + 121 + 169}{4} = \frac{420}{4} = 105$.
Difference: $105 - 41 = 64$. This matches the given condition.
Therefore,the average of all $6$ numbers is $a = 8$.

(A) Let the total age of $10$ directors be $T_1 = 10 \times 48 = 480$ years.
Let the age of the director who died be $x$ years.
When one director aged $53$ resigned and one director aged $x$ died,and a new director aged $34$ joined,the number of directors became $9$.
The total age of these $9$ directors at that moment was $T_2 = 480 - 53 - x + 34 = 461 - x$.
One year later,each of the $9$ directors aged by $1$ year,so the total age increased by $9 \times 1 = 9$ years.
The new total age is $T_3 = (461 - x) + 9 = 470 - x$.
We are given that the average age of these $9$ directors after one year is $46$ years.
Therefore,$T_3 = 9 \times 46 = 414$.
Equating the two expressions for $T_3$: $470 - x = 414$.
Solving for $x$: $x = 470 - 414 = 56$.
Thus,the age of the deceased director was $56$ years.

(D) The sum of $6$ numbers $= 6 \times 20 = 120$.
After removing one number, the count of numbers becomes $5$ and the new average is $15$.
The sum of the remaining $5$ numbers $= 5 \times 15 = 75$.
The removed number $= (\text{Sum of } 6 \text{ numbers}) - (\text{Sum of } 5 \text{ numbers})$.
Removed number $= 120 - 75 = 45$.

(D) Let the sum of the ages of the $5$ members $3 \, \text{years}$ ago be $S$.
Then,the average age $3 \, \text{years}$ ago was $\frac{S}{5}$.
The current sum of the ages of the same $5$ members (without replacement) would be $S + (5 \times 3) = S + 15$.
Let the age of the old member be $X$ and the new member be $Y$.
After replacing the old member with the new member,the new sum of ages is $(S + 15) - X + Y$.
The new average age is $\frac{S + 15 - X + Y}{5}$.
According to the problem,this new average is equal to the average age $3 \, \text{years}$ ago,which is $\frac{S}{5}$.
So,$\frac{S + 15 - X + Y}{5} = \frac{S}{5}$.
$S + 15 - X + Y = S$.
$15 - X + Y = 0$.
$X - Y = 15$.
Therefore,the difference between the ages of the replaced member and the new member is $15 \, \text{years}$.

(D) The sum of the first $25$ numbers $= 25 \times 10 = 250$.
The sum of the next $25$ numbers $= 25 \times 12 = 300$.
The total sum of all $50$ numbers $= 250 + 300 = 550$.
The average of all $50$ numbers $= \frac{\text{Total Sum}}{\text{Total Count}} = \frac{550}{50} = 11$.

(B) The sum of $m$ numbers is given by $m \times n^{2} = mn^{2}$.
The sum of $n$ numbers is given by $n \times m^{2} = nm^{2}$.
The total sum of $(m + n)$ numbers is $mn^{2} + nm^{2}$.
The average of $(m + n)$ numbers is calculated as:
$\text{Average} = \frac{\text{Total Sum}}{\text{Total Count}} = \frac{mn^{2} + nm^{2}}{m + n}$.
Factoring out $mn$ from the numerator:
$\text{Average} = \frac{mn(n + m)}{m + n}$.
Since $(n + m) = (m + n)$,they cancel out:
$\text{Average} = mn$.

(B) The first $9$ integral multiples of $3$ form an arithmetic progression: $3, 6, 9, \dots, 27$.
Here, the first term $a = 3$, the common difference $d = 3$, and the number of terms $n = 9$.
The sum of the first $n$ terms is given by $S_n = \frac{n}{2}[2a + (n - 1)d]$.
$S_9 = \frac{9}{2}[2(3) + (9 - 1)3] = \frac{9}{2}[6 + 24] = \frac{9}{2} \times 30 = 9 \times 15 = 135$.
The average is given by $\frac{\text{Sum of terms}}{\text{Number of terms}} = \frac{135}{9} = 15$.
Alternatively, for an arithmetic progression, the average is the middle term: $\frac{a + l}{2} = \frac{3 + 27}{2} = \frac{30}{2} = 15$.

(B) Sum of all $11$ numbers $= 35 \times 11 = 385$.
Sum of the first $6$ numbers $= 32 \times 6 = 192$.
Sum of the last $6$ numbers $= 37 \times 6 = 222$.
The sixth number is counted in both the first $6$ and the last $6$ numbers.
Therefore, the sixth number $= (\text{Sum of first } 6 + \text{Sum of last } 6) - \text{Sum of all } 11$.
Sixth number $= 192 + 222 - 385 = 414 - 385 = 29$.

(A) Let the $5$ consecutive integers starting with $m$ be $m, m + 1, m + 2, m + 3, m + 4$.
The average is given by: $\frac{m + (m + 1) + (m + 2) + (m + 3) + (m + 4)}{5} = n$.
Simplifying the numerator: $\frac{5m + 10}{5} = n$,which gives $m + 2 = n$.
Now,we need the average of $6$ consecutive integers starting with $(m + 2)$. These integers are $(m + 2), (m + 3), (m + 4), (m + 5), (m + 6), (m + 7)$.
The average is: $\frac{(m + 2) + (m + 3) + (m + 4) + (m + 5) + (m + 6) + (m + 7)}{6}$.
Summing the terms: $\frac{6m + 27}{6} = m + \frac{27}{6} = m + 4.5$.
Since $m = n - 2$,substitute this into the expression: $(n - 2) + 4.5 = n + 2.5$.
Converting to fraction form: $n + \frac{5}{2} = \frac{2n + 5}{2}$.

(B) Initial sum of $100$ items $= 46 \times 100 = 4600$.
Correcting the misread items: The item $16$ was misread as $61$ (subtract $61$,add $16$) and $43$ was misread as $34$ (subtract $34$,add $43$).
Correct sum $= 4600 - 61 + 16 - 34 + 43 = 4564$.
Since the actual number of items is $90$,the correct mean $= \frac{4564}{90} = 50.711... \approx 50.7$.

(C) Given that the average of $x$ and $\frac{1}{x}$ is $M$.
$\frac{x + \frac{1}{x}}{2} = M$
Multiplying both sides by $2$,we get:
$x + \frac{1}{x} = 2M$
Now,we need to find the average of $x^{2}$ and $\frac{1}{x^{2}}$,which is given by:
$\text{Average} = \frac{x^{2} + \frac{1}{x^{2}}}{2}$
Using the algebraic identity $(a + b)^{2} = a^{2} + b^{2} + 2ab$,we can write:
$x^{2} + \frac{1}{x^{2}} = (x + \frac{1}{x})^{2} - 2(x)(\frac{1}{x}) = (x + \frac{1}{x})^{2} - 2$
Substituting the value of $(x + \frac{1}{x}) = 2M$:
$x^{2} + \frac{1}{x^{2}} = (2M)^{2} - 2 = 4M^{2} - 2$
Therefore,the required average is:
$\frac{4M^{2} - 2}{2} = 2M^{2} - 1$

(B) The average $\bar{x}$ of $n$ numbers $x_{1}, x_{2}, \ldots, x_{n}$ is given by the formula:
$\bar{x} = \frac{\sum_{i=1}^{n} x_{i}}{n}$
This implies that $\sum_{i=1}^{n} x_{i} = n\bar{x}$ .....$(i)$
Now,we need to find the value of $\sum_{i=1}^{n}(x_{i}-\bar{x})$:
$\sum_{i=1}^{n}(x_{i}-\bar{x}) = \sum_{i=1}^{n} x_{i} - \sum_{i=1}^{n} \bar{x}$
Since $\bar{x}$ is a constant,$\sum_{i=1}^{n} \bar{x} = n\bar{x}$.
Substituting the value from $(i)$:
$\sum_{i=1}^{n}(x_{i}-\bar{x}) = n\bar{x} - n\bar{x} = 0$
Therefore,the sum of deviations of observations from their mean is always $0$.

(C) The total sum of the six numbers is calculated as $32 \times 6 = 192$.
According to the problem,the first three numbers are increased by $2$ each,adding a total of $3 \times 2 = 6$ to the sum.
The remaining three numbers are decreased by $4$ each,subtracting a total of $3 \times 4 = 12$ from the sum.
The new sum of the numbers is $192 + 6 - 12 = 186$.
The new average is the new sum divided by the total count of numbers: $\frac{186}{6} = 31$.

(B) Let the three numbers be $A, B,$ and $C$,where $A$ is the largest number.
Given that the average of the three numbers is $135$,their sum is $135 \times 3 = 405$.
Since the largest number $A = 195$,the sum of the other two numbers is $B + C = 405 - 195 = 210$ $(i)$.
We are also given that the difference between the other two numbers is $20$,so $B - C = 20$ $(ii)$.
Adding equations $(i)$ and $(ii)$:
$(B + C) + (B - C) = 210 + 20$
$2B = 230$
$B = 115$.
Substituting $B = 115$ into equation $(i)$:
$115 + C = 210$
$C = 210 - 115 = 95$.
The three numbers are $195, 115,$ and $95$. Therefore,the smallest number is $95$.

(A) Let the ten numbers be $x_1, x_2, ..., x_{10}$.
The correct average is $A = \frac{\sum x_i}{10}$.
Let the number with interchanged digits be $N = 10a + b$. The original number was $M = 10b + a$.
The change in the sum of the numbers is $M - N = (10b + a) - (10a + b) = 9(b - a)$.
Given that the average decreases by $1.8$,the total sum decreases by $1.8 \times 10 = 18$.
Therefore,$9(b - a) = 18$.
Dividing by $9$,we get $b - a = 2$.
Thus,the difference between the digits of the number is $2$.

(D) The sum of the squares of the first $n$ natural numbers is given by the formula: $\frac{n(n+1)(2n+1)}{6}$.
The average of the squares of the first $n$ natural numbers is the sum divided by $n$:
$\text{Average} = \frac{n(n+1)(2n+1)}{6n} = \frac{(n+1)(2n+1)}{6}$.
For the first $10$ natural numbers,substitute $n = 10$ into the formula:
$\text{Average} = \frac{(10+1)(2 \times 10 + 1)}{6} = \frac{11 \times 21}{6}$.
$\text{Average} = \frac{231}{6} = 38.5$.

(D) Sum of all $30$ numbers $= 30 \times 12 = 360$.
Sum of the first $20$ numbers $= 20 \times 11 = 220$.
Sum of the next $9$ numbers $= 9 \times 10 = 90$.
Sum of the first $29$ numbers $= 220 + 90 = 310$.
The last number $= (\text{Sum of all } 30 \text{ numbers}) - (\text{Sum of the first } 29 \text{ numbers})$.
Last number $= 360 - 310 = 50$.

(B) Let the third number be $x$.
According to the problem, the second number is thrice the third, so the second number $= 3x$.
The second number is also twice the first, so $2 \times (\text{first number}) = 3x$, which means the first number $= \frac{3x}{2}$.
The average of the three numbers is given as $44$, so their sum is $44 \times 3 = 132$.
Therefore, $x + 3x + \frac{3x}{2} = 132$.
Multiplying by $2$ to clear the fraction: $2x + 6x + 3x = 264$.
$11x = 264$, which gives $x = 24$.
The three numbers are:
First number $= \frac{3(24)}{2} = 36$.
Second number $= 3(24) = 72$.
Third number $= 24$.
The largest number is $72$.

(D) The sum of $30$ numbers is calculated as: $30 \times 15 = 450$.
The sum of the first $18$ numbers is: $18 \times 10 = 180$.
The sum of the next $11$ numbers is: $11 \times 20 = 220$.
The sum of the first $29$ numbers is: $180 + 220 = 400$.
The last number is the difference between the total sum and the sum of the first $29$ numbers: $450 - 400 = 50$.

(D) The first $9$ prime numbers are $2, 3, 5, 7, 11, 13, 17, 19,$ and $23$.
To find the average,we calculate the sum of these numbers:
Sum $= 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 = 100$.
The number of terms is $9$.
Average $= \frac{\text{Sum of terms}}{\text{Number of terms}} = \frac{100}{9} = 11 \frac{1}{9}$.

(C) Total expenditure for the first four months $= 1800 \times 4 = Rs. 7200$.
Total expenditure for the next eight months $= 2000 \times 8 = Rs. 16000$.
Total savings for the year $= Rs. 5600$.
Total annual income $=$ Total expenditure $+$ Total savings.
Total annual income $= 7200 + 16000 + 5600 = Rs. 28800$.
Average monthly income $= \frac{\text{Total annual income}}{12} = \frac{28800}{12} = Rs. 2400$.

(B) Total number of children = $16$.
Sum of the marks of all $16$ children = $16 \times 76 = 1216$.
Sum of the marks obtained by group $A$ ($10$ children) = $10 \times 75 = 750$.
Sum of the marks obtained by group $B$ ($6$ children) = $1216 - 750 = 466$.
Average marks of group $B$ = $\frac{466}{6} = \frac{233}{3} = 77 \frac{2}{3}$.

(B) Total weight of the five persons $= 5 \times 38 = 190\, kg$.
Total weight of the boat and the five persons (total $6$ entities) $= 6 \times 52 = 312\, kg$.
Weight of the boat $= (\text{Total weight of boat and persons}) - (\text{Total weight of persons})$.
Weight of the boat $= 312 - 190 = 122\, kg$.

(A) Let the total pocket money of $A, B,$ and $C$ be $T_p$.
Given that the average pocket money is $Rs. 80,$ so $T_p = 80 \times 3 = 240$.
Let the total unspent money be $T_u$.
Given that the average unspent money is $Rs. 60,$ so $T_u = 60 \times 3 = 180$.
The total amount spent by the three friends is $T_p - T_u = 240 - 180 = 60$.
Let $A$ spend $Rs. x$. Then $B$ spends $Rs. 2x$ and $C$ spends $Rs. 3x$.
According to the problem,the sum of their expenditures is $x + 2x + 3x = 60$.
$6x = 60 \Rightarrow x = 10$.
Therefore,$A$ spends $Rs. 10$.

(C) Total runs required to win $= 20 \times 7.2 = 144$.
Runs scored by the end of the $15^{th}$ over $= 15 \times 6 = 90$.
Remaining runs required $= 144 - 90 = 54$.
Remaining overs $= 20 - 15 = 5$.
Required run rate for the remaining $5$ overs $= \frac{54}{5} = 10.8$.

(C) Let the average expenditure of all nine persons be $x$.
The total expenditure of eight persons is $8 \times 30 = 240$.
The expenditure of the ninth person is $x + 20$.
The total expenditure of all nine persons is $240 + (x + 20) = x + 260$.
Since the average is $x$,we have the equation:
$x = \frac{x + 260}{9}$
Multiplying both sides by $9$:
$9x = x + 260$
$8x = 260$
$x = \frac{260}{8} = 32.5$
The total expenditure is $9x = 9 \times 32.5 = 292.5$.

(D) Let $H$ be the highest score and $L$ be the lowest score.
Total runs for $64$ innings $= 64 \times 62 = 3968$.
After excluding the two innings,the number of remaining innings is $64 - 2 = 62$.
Total runs for the remaining $62$ innings $= 62 \times 60 = 3720$.
The sum of the highest and lowest scores is $H + L = 3968 - 3720 = 248$.
We are given that $H - L = 180$.
Adding the two equations: $(H + L) + (H - L) = 248 + 180$.
$2H = 428$.
$H = 214$.
Thus,the highest score is $214$ runs.

(D) Initial sum of the weights of $20$ boys $= 89.4 \times 20 = 1788 \, kg$.
The difference between the correct weight and the misread weight is $87 - 78 = 9 \, kg$.
Therefore,the correct sum of the weights $= 1788 + 9 = 1797 \, kg$.
The correct average weight $= \frac{1797}{20} = 89.85 \, kg$.

(D) Total price of $75$ pencils $= 75 \times 2 = Rs. \, 150$.
Total price of $30$ pens $= 510 - 150 = Rs. \, 360$.
Average price of $30$ pens $= \frac{360}{30} = Rs. \, 12$.

(C) Let the number of wickets taken by the cricketer before the last match be $x$.
The total runs conceded before the last match $= 12.4x$.
In the last match,he took $5$ wickets for $26$ runs.
New total wickets $= x + 5$.
New total runs $= 12.4x + 26$.
The new bowling average is $12.4 - 0.2 = 12.2$.
According to the problem:
$\frac{12.4x + 26}{x + 5} = 12.2$
$12.4x + 26 = 12.2(x + 5)$
$12.4x + 26 = 12.2x + 61$
$12.4x - 12.2x = 61 - 26$
$0.2x = 35$
$x = \frac{35}{0.2} = 175$.
Therefore,the number of wickets taken before the last match was $175$.

(B) Let the average runs of the cricketer for $64$ innings be $x$.
Total runs scored in $64$ innings $= 64x$.
In the $65^{th}$ innings,he scores $0$ runs.
Total runs after $65$ innings $= 64x + 0 = 64x$.
The new average after $65$ innings is given as $(x - 2)$.
Therefore,the equation is: $\frac{64x}{65} = x - 2$.
Multiplying both sides by $65$: $64x = 65(x - 2)$.
$64x = 65x - 130$.
$x = 130$.
The new average is $x - 2 = 130 - 2 = 128$.

(C) Sum of the ages of father and mother $= 38 \times 2 = 76 \text{ years}$.
Sum of the ages of father,mother,and daughter $= 28 \times 3 = 84 \text{ years}$.
Age of the daughter $= 84 - 76 = 8 \text{ years}$.

(B) Total marks of $28$ students $= 28 \times 50 = 1400$.
After $8$ students left,the number of remaining students $= 28 - 8 = 20$.
The new average of $20$ students $= 50 + 5 = 55$.
Total marks of $20$ students $= 20 \times 55 = 1100$.
Total marks of the $8$ students who left $= 1400 - 1100 = 300$.
Average marks of the $8$ students $= \frac{300}{8} = 37.5$.

(D) Let the initial average weight of $25$ persons be $A \, kg$.
Total initial weight $= 25 \times A = 25A \, kg$.
When a person of $60 \, kg$ is replaced by a new person of weight $W$,the new total weight becomes $25A - 60 + W$.
The new average weight is $A + 1 \, kg$.
Therefore,the new total weight is $25(A + 1) = 25A + 25$.
Equating the two expressions for the new total weight:
$25A - 60 + W = 25A + 25$.
$W - 60 = 25$.
$W = 60 + 25 = 85 \, kg$.
Thus,the weight of the new person is $85 \, kg$.

(C) Given,the number of measurements $n = 20$ and the initial average $= 56 \, cm$.
The initial sum of the measurements $= 56 \times 20 = 1120 \, cm$.
Since one measurement was recorded as $64 \, cm$ instead of $61 \, cm$,we need to adjust the sum.
Correct sum $= 1120 - 64 + 61 = 1117 \, cm$.
Correct average $= \frac{\text{Correct sum}}{n} = \frac{1117}{20} = 55.85 \, cm$.

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

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

(A) The sum of the ages of the four brothers $= 4 \times 12 = 48 \, \text{years}$.
When the mother's age is included, the total number of people becomes $5$ and the new average becomes $12 + 5 = 17 \, \text{years}$.
The sum of the ages of the five people (four brothers + mother) $= 5 \times 17 = 85 \, \text{years}$.
Therefore, the age of the mother $= 85 - 48 = 37 \, \text{years}$.

(D) Let the initial average expenditure per student be $x$.
Initial total expenditure = $35x$.
New number of students = $35 + 7 = 42$.
New average expenditure per student = $x - 1$.
New total expenditure = $42(x - 1)$.
According to the problem,the total expenditure increases by $Rs. \,42$:
$42(x - 1) = 35x + 42$.
$42x - 42 = 35x + 42$.
$42x - 35x = 42 + 42$.
$7x = 84$.
$x = 12$.
Initial total expenditure = $35 \times 12 = Rs. \,420$.

(D) Total weight of $34$ students $= 34 \times 42 = 1428 \, kg$.
After including the teacher,the total number of people becomes $34 + 1 = 35$.
The new mean weight $= 42 \, kg + 400 \, g = 42 \, kg + 0.4 \, kg = 42.4 \, kg$.
Total weight of $35$ people $= 35 \times 42.4 = 1484 \, kg$.
Weight of the teacher $=$ (Total weight of $35$ people) $-$ (Total weight of $34$ students).
Weight of the teacher $= 1484 - 1428 = 56 \, kg$.

(B) The odd numbers up to $100$ are $1, 3, 5, \dots, 99$.
This is an arithmetic progression where the first term $a = 1$, the last term $l = 99$, and the common difference $d = 2$.
The number of terms $n$ can be calculated as $l = a + (n - 1)d$, so $99 = 1 + (n - 1)2$, which gives $98 = 2(n - 1)$, so $n - 1 = 49$, and $n = 50$.
The average of an arithmetic progression is given by the formula $\text{Average} = \frac{\text{First term} + \text{Last term}}{2}$.
$\text{Average} = \frac{1 + 99}{2} = \frac{100}{2} = 50$.

(D) Let the seven numbers be $x_1, x_2, x_3, x_4, x_5, x_6, x_7$.
Given that the average of the first four numbers is $4$,so $(x_1 + x_2 + x_3 + x_4) / 4 = 4$,which implies $x_1 + x_2 + x_3 + x_4 = 16$.
Given that the average of the last four numbers is $4$,so $(x_4 + x_5 + x_6 + x_7) / 4 = 4$,which implies $x_4 + x_5 + x_6 + x_7 = 16$.
Given that the average of all seven numbers is $3$,so $(x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7) / 7 = 3$,which implies the sum of all seven numbers is $21$.
Adding the sums of the first four and last four numbers: $(x_1 + x_2 + x_3 + x_4) + (x_4 + x_5 + x_6 + x_7) = 16 + 16 = 32$.
This sum includes the fourth number $(x_4)$ twice. Therefore,the sum of all seven numbers plus the fourth number is $32$.
$x_4 = (x_1 + x_2 + x_3 + x_4 + x_4 + x_5 + x_6 + x_7) - (x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7) = 32 - 21 = 11$.
Thus,the fourth number is $11$.

(C) Let the total number of candidates be $n$.
The initial total marks = $60n$.
When the marks of $80$ candidates are changed from $95$ to $70$,the reduction in marks for each candidate is $(95 - 70) = 25$.
The total reduction in marks for $80$ candidates is $80 \times 25 = 2000$.
The new total marks = $60n - 2000$.
Given that the new average is $55$,the new total marks = $55n$.
Equating the two expressions for the new total marks:
$60n - 2000 = 55n$
$60n - 55n = 2000$
$5n = 2000$
$n = 400$.
Therefore,the total number of candidates is $400$.

(C) Initial number of students $= 15$.
Initial average age $= 10 \, \text{years}$.
Total age of $15$ students $= 15 \times 10 = 150 \, \text{years}$.
When $5$ more students join,the total number of students becomes $15 + 5 = 20$.
The new average age becomes $10 + 1 = 11 \, \text{years}$.
Total age of $20$ students $= 20 \times 11 = 220 \, \text{years}$.
Total age of the $5$ new students $= 220 - 150 = 70 \, \text{years}$.
Average age of the $5$ new students $(x)$ $= \frac{70}{5} = 14 \, \text{years}$.

(C) Let $S$ be the total distance of the journey.
Time taken for the first $\frac{1}{3}$ of the journey at $60\, km/h$ is $t_1 = \frac{S/3}{60} = \frac{S}{180}$ hours.
Time taken for the second $\frac{1}{3}$ of the journey at $30\, km/h$ is $t_2 = \frac{S/3}{30} = \frac{S}{90}$ hours.
Time taken for the remaining $\frac{1}{3}$ of the journey at $10\, km/h$ is $t_3 = \frac{S/3}{10} = \frac{S}{30}$ hours.
Total time taken $T = t_1 + t_2 + t_3 = \frac{S}{180} + \frac{S}{90} + \frac{S}{30}$.
Taking the least common multiple $(LCM)$ of $180, 90, 30$,which is $180$:
$T = \frac{S + 2S + 6S}{180} = \frac{9S}{180} = \frac{S}{20}$ hours.
Average speed = $\frac{\text{Total distance}}{\text{Total time}} = \frac{S}{S/20} = 20\, km/h$.

(C) Let the speed of the return journey be $v_r = x\, \text{km/h}$.
Then,the speed of the onward journey is $v_o = x - 0.25x = 0.75x = \frac{3}{4}x\, \text{km/h}$.
The total distance for the to and fro journey is $800 + 800 = 1600\, \text{km}$.
The total time spent is $17\, \text{hours}$,including a $1\, \text{hour}$ halt. Thus,the actual travel time is $17 - 1 = 16\, \text{hours}$.
Time taken for onward journey $t_o = \frac{800}{v_o} = \frac{800}{3x/4} = \frac{3200}{3x}$.
Time taken for return journey $t_r = \frac{800}{v_r} = \frac{800}{x}$.
Total travel time $t_o + t_r = 16\, \text{hours}$.
$\frac{3200}{3x} + \frac{800}{x} = 16$.
Multiply by $3x$: $3200 + 2400 = 48x$.
$5600 = 48x \implies x = \frac{5600}{48} = \frac{350}{3} \approx 116.67\, \text{km/h}$.
Speed of the onward journey $v_o = \frac{3}{4}x = \frac{3}{4} \times \frac{350}{3} = \frac{350}{4} = 87.5\, \text{km/h}$.
Wait,re-evaluating the problem statement: If the total distance covered is $800\, \text{km}$ for the whole trip (i.e.,$400\, \text{km}$ one way):
$t_o + t_r = 16 \implies \frac{400}{3x/4} + \frac{400}{x} = 16 \implies \frac{1600}{3x} + \frac{1200}{3x} = 16 \implies \frac{2800}{3x} = 16 \implies 48x = 2800 \implies x = 58.33$.
$v_o = \frac{3}{4} \times 58.33 = 43.75\, \text{km/h}$.
Thus,the speed of the onward journey is $43.75\, \text{km/h}$.