Average Questions in English

(D) Let the number of students be $n = 11$ (from statement $I$).
Let the average age of the students be $A_s$ and the age of the teacher be $T$.
From statement $II$,the average age of $11$ students and the teacher is $14$ years:
$\frac{11 \times A_s + T}{12} = 14 \implies 11A_s + T = 168$ (Equation $1$).
From statement $III$,the average age of the teacher and the students is $3$ years more than the average age of the students:
$\frac{11A_s + T}{12} = A_s + 3$.
Since $\frac{11A_s + T}{12} = 14$ from statement $II$,we have $14 = A_s + 3$,which gives $A_s = 11$ years.
Substituting $A_s = 11$ into Equation $1$:
$11(11) + T = 168$
$121 + T = 168$
$T = 168 - 121 = 47$ years.
Thus,all three statements $I, II,$ and $III$ are required to find the teacher's age.

(D) Let the contribution of the sixth man be $Rs.\,x.$
The sum of contributions of the first $5$ men is $5 \times 35 = 175.$
The new average of $6$ men is $\frac{175 + x}{6}.$
According to the problem,the sixth man pays $Rs.\,35$ more than the new average:
$x = \frac{175 + x}{6} + 35$
Multiply the entire equation by $6$:
$6x = 175 + x + 210$
$6x - x = 385$
$5x = 385$
$x = 77$
The total contribution of all $6$ men is $175 + 77 = 252.$
Thus,the total contribution is $Rs.\,252.$

(A) Let the five consecutive integers be $x, x+1, x+2, x+3, x+4$.
Their sum $a = 5x + 10$.
The next five consecutive integers are $x+5, x+6, x+7, x+8, x+9$.
Their sum $b = 5x + 35$.
Now,calculate $b - a = (5x + 35) - (5x + 10) = 25$.
Therefore,$\frac{b-a}{100} = \frac{25}{100} = \frac{1}{4}$.

(B) The average age of $P$ and $Q$ ten years ago was $20 \text{ years}$.
Therefore,the sum of their ages ten years ago was $20 \times 2 = 40 \text{ years}$.
Their present sum of ages is $(40 + 10 + 10) = 60 \text{ years}$.
The average age of $P, Q,$ and $R$ now is $30 \text{ years}$.
Therefore,the sum of their present ages is $30 \times 3 = 90 \text{ years}$.
The present age of $R$ is $(90 - 60) = 30 \text{ years}$.
After $10 \text{ years}$,the age of $R$ will be $(30 + 10) = 40 \text{ years}$.

(A) The sum of the given numbers is $15 + 21 + 32 + 35 + 46 + x + 59 + 65 + 72 = 345 + x$.
There are $9$ numbers in total.
The average is given by $\frac{345 + x}{9}$.
According to the problem,$43 \leq \frac{345 + x}{9} \leq 44$.
Multiplying the entire inequality by $9$,we get $387 \leq 345 + x \leq 396$.
Subtracting $345$ from all parts of the inequality,we get $387 - 345 \leq x \leq 396 - 345$.
Thus,$42 \leq x \leq 51$.

(C) Let the weight of the first member be $w_1$.
The average weight of the first member is $A_1 = w_1$.
Let the weight of the second member be $w_2$. The average weight of the first two members is $A_2 = \frac{w_1 + w_2}{2} = A_1 + 1 = w_1 + 1$.
Thus,$w_1 + w_2 = 2w_1 + 2$,which implies $w_2 = w_1 + 2$.
For the $n$-th member,the average weight $A_n = A_{n-1} + 1$.
The sum of weights of $n$ members is $S_n = n \times A_n$.
Since $A_n = A_1 + (n-1) = w_1 + n - 1$,we have $S_n = n(w_1 + n - 1)$.
The weight of the $n$-th member is $w_n = S_n - S_{n-1}$.
$w_n = n(w_1 + n - 1) - (n-1)(w_1 + n - 2) = nw_1 + n^2 - n - (nw_1 + n^2 - 2n - w_1 - n + 2) = w_1 + 2n - 2$.
For the first member $(n=1)$: $w_1 = w_1 + 2(1) - 2 = w_1$.
For the fifth member $(n=5)$: $w_5 = w_1 + 2(5) - 2 = w_1 + 8$.
The difference between the last and the first player is $w_5 - w_1 = (w_1 + 8) - w_1 = 8 \ kg$.

(C) Let the total expenditure of all $9$ persons be $S$.
Then the average expenditure is $\frac{S}{9}$.
Let the expenditure of the $9^{th}$ person be $x$.
We know that $8$ persons spent $Rs. 30$ each,so their total expenditure is $8 \times 30 = 240$.
Thus,$S = 240 + x$.
The average expenditure is $\frac{240 + x}{9}$.
According to the problem,the $9^{th}$ person spent $Rs. 20$ more than the average:
$x = \frac{240 + x}{9} + 20$
Multiply the entire equation by $9$:
$9x = 240 + x + 180$
$9x - x = 420$
$8x = 420$
$x = 52.5$
Now,the total expenditure $S = 240 + 52.5 = 292.5$.
Therefore,the total money spent by all of them is $Rs. 292.50$.

(B) Let the number of girls be $x$.
Then,the number of boys is $(600 - x)$.
The average age of the school is $11$ $years$ and $9$ $months$,which is equal to $11 + \frac{9}{12} = 11 + \frac{3}{4} = \frac{47}{4}$ $years$.
The total age of all students is the sum of the total age of boys and girls.
$(600 - x) \times 12 + x \times 11 = 600 \times \frac{47}{4}$
$7200 - 12x + 11x = 150 \times 47$
$7200 - x = 7050$
$x = 7200 - 7050 = 150$.
Therefore,the number of girls in the school is $150$.

(B) Step $1$: Calculate the initial sum of $100$ items.
Sum $= 100 \times 46 = 4600$.
Step $2$: Adjust the sum by removing the misread values ($61$ and $34$) and adding the correct values ($16$ and $43$).
Corrected Sum $= 4600 - 61 - 34 + 16 + 43 = 4600 - 95 + 59 = 4564$.
Step $3$: Since the actual number of items is $90$,divide the corrected sum by $90$.
Correct Mean $= \frac{4564}{90} = 50.711... \approx 50.7$.

(B) Let $M, T, W, Th, F$ represent the rainfall on Monday,Tuesday,Wednesday,Thursday,and Friday respectively.
From the given data:
$M + T + W + Th = 4 \times 420.5 = 1682\, cm$ ....$(1)$
$T + W + Th + F = 4 \times 440.5 = 1762\, cm$ ....$(2)$
Subtracting equation $(1)$ from equation $(2)$:
$(T + W + Th + F) - (M + T + W + Th) = 1762 - 1682$
$F - M = 80\, cm$
Given the ratio of rainfall on Monday and Friday is $20:21$,let $M = 20x$ and $F = 21x$.
Substituting these into the difference equation:
$21x - 20x = 80$
$x = 80$
Therefore,rainfall on Monday $= 20 \times 80 = 1600\, cm$.
Rainfall on Friday $= 21 \times 80 = 1680\, cm$.

(A) The $5$ consecutive integers starting with $m$ are $m, m+1, m+2, m+3, m+4$.
Their average is $\frac{m + (m+1) + (m+2) + (m+3) + (m+4)}{5} = n$.
$\frac{5m + 10}{5} = n \Rightarrow m + 2 = n \Rightarrow m = n - 2$.
Now,we need the average of $6$ consecutive integers starting with $(m+2)$.
These integers are $(m+2), (m+3), (m+4), (m+5), (m+6), (m+7)$.
Their average is $\frac{(m+2) + (m+3) + (m+4) + (m+5) + (m+6) + (m+7)}{6} = \frac{6m + 27}{6} = m + \frac{27}{6} = m + 4.5$.
Substituting $m = n - 2$ into the expression:
Average $= (n - 2) + 4.5 = n + 2.5 = \frac{2n + 5}{2}$.

(C) Let the three consecutive odd numbers be $x$,$x+2$,and $x+4$.
The average of these three numbers is $\frac{x + (x+2) + (x+4)}{3} = \frac{3x+6}{3} = x+2$.
According to the problem,the average is $12$ more than one-third of the first number $(x)$:
$x+2 = \frac{x}{3} + 12$
Multiply the entire equation by $3$ to clear the fraction:
$3(x+2) = x + 36$
$3x + 6 = x + 36$
Subtract $x$ from both sides:
$2x + 6 = 36$
Subtract $6$ from both sides:
$2x = 30$
$x = 15$
The three numbers are $15$,$17$,and $19$.
Therefore,the last of the three numbers is $19$.

(C) Let the four numbers be $a, b, c,$ and $d$.
Given that the average of the four numbers is $60$,so their sum is $a + b + c + d = 60 \times 4 = 240$.
According to the problem,the first number $a$ is one-fourth of the sum of the last three numbers $(b + c + d)$.
So,$a = \frac{1}{4}(b + c + d)$,which implies $b + c + d = 4a$.
Substitute this into the sum equation: $a + (b + c + d) = 240$.
$a + 4a = 240$.
$5a = 240$.
$a = \frac{240}{5} = 48$.
Therefore,the first number is $48$.

(D) Let the number of fruits in the second basket be $x$.
Therefore,the number of fruits in the first basket $= 2x$.
The number of fruits in the third basket $= 2x \times \frac{3}{4} = \frac{3}{2}x$.
The average of the fruits in all three baskets is given as $30$.
So,$\frac{2x + x + \frac{3}{2}x}{3} = 30$.
Multiplying by $3$,we get $2x + x + \frac{3}{2}x = 90$.
$\frac{4x + 2x + 3x}{2} = 90$.
$\frac{9x}{2} = 90$.
$9x = 180$.
$x = 20$.
The number of fruits in the first basket $= 2x = 2 \times 20 = 40$.

(D) Total marks of $24$ students $= 24 \times 56 = 1344$.
The sum of misread marks $= 44 + 45 + 61 = 150$.
The sum of actual marks $= 48 + 59 + 67 = 174$.
Difference in marks $= 174 - 150 = 24$.
Correct total marks $= 1344 + 24 = 1368$.
Correct average $= \frac{1368}{24} = 57$.

(B) The average of Set $A$ is calculated as $\frac{376}{8} = 47$.
The minimum number of the second set is $15$ more than the average of Set $A$,which is $47 + 15 = 62$.
The second set consists of $5$ consecutive numbers starting from $62$,which are $62, 63, 64, 65, 66$.
The sum of these $5$ numbers is $62 + 63 + 64 + 65 + 66 = 320$.

(D) Let the cricketer's highest score $= x$ runs.
Therefore,the minimum score $= (x - 172)$ runs.
Total runs scored in $40$ innings $= 40 \times 50 = 2000$ runs.
Total runs scored in $38$ innings $= 38 \times 48 = 1824$ runs.
The sum of the highest and lowest scores is the difference between the total runs of $40$ innings and $38$ innings:
$x + (x - 172) = 2000 - 1824$
$2x - 172 = 176$
$2x = 176 + 172$
$2x = 348$
$x = 174$
Thus,the highest score of the player is $174$ runs.

(A) Let the third number be $x$.
Then,the second number $= 2x$ and the first number $= 2(2x) = 4x$.
According to the problem,the average of these three numbers is $154$,so their sum is $154 \times 3 = 462$.
Therefore,$4x + 2x + x = 462$.
$7x = 462$.
$x = \frac{462}{7} = 66$.
The first number is $4x = 4 \times 66 = 264$.

(D) Let the number of officers be $x$. Then,the number of the rest of the staff is $(500 - x)$.
According to the problem,the total salary of all staff is the sum of the total salary of officers and the total salary of the rest of the staff.
Total salary of all staff $= 500 \times 5,000 = 2,500,000$.
Total salary of officers $= x \times 14,000$.
Total salary of the rest of the staff $= (500 - x) \times 4,000$.
Setting up the equation:
$14,000x + 4,000(500 - x) = 2,500,000$
Dividing the entire equation by $1,000$ to simplify:
$14x + 4(500 - x) = 2,500$
$14x + 2,000 - 4x = 2,500$
$10x = 2,500 - 2,000$
$10x = 500$
$x = 50$.
Therefore,the number of officers is $50$.

(D) Total sum of marks for $40$ students $= 40 \times 72 = 2880$.
The sum of the wrongly entered marks $= 68 + 65 + 73 = 206$.
The sum of the correct marks $= 64 + 62 + 84 = 210$.
The difference in the sum $= 210 - 206 = 4$.
Since the correct sum is greater than the wrong sum,we add the difference to the total sum: Correct total sum $= 2880 + 4 = 2884$.
Correct average $= \frac{2884}{40} = 72.1$.

(B) Let the first number be $x$.
Then,the second number $= 3x$ and the third number $= \frac{3x}{4}$.
The average of the three numbers is given as $114$,so their sum is $3 \times 114 = 342$.
According to the problem:
$x + 3x + \frac{3x}{4} = 342$
Multiply the entire equation by $4$ to clear the fraction:
$4x + 12x + 3x = 342 \times 4$
$19x = 1368$
Solving for $x$:
$x = \frac{1368}{19} = 72$
The three numbers are:
First number $= 72$
Second number $= 3 \times 72 = 216$
Third number $= \frac{3 \times 72}{4} = 3 \times 18 = 54$
Comparing the three numbers $(72, 216, 54)$,the largest number is $216$.

(D) The initial total marks of $24$ students $= 24 \times 56 = 1344$.
The sum of the misread marks $= 44 + 45 + 61 = 150$.
The sum of the actual marks $= 48 + 59 + 67 = 174$.
The difference between the actual sum and the misread sum $= 174 - 150 = 24$.
The correct total marks $= 1344 + 24 = 1368$.
The correct average $= \frac{1368}{24} = 57$.

(D) Ritu's marks $= 875 \times \frac{56}{100} = 490$
Smita's marks $= 875 \times \frac{92}{100} = 805$
Rina's marks $= 634$
Total marks $= 490 + 805 + 634 = 1929$
Average marks $= \frac{1929}{3} = 643$

(B) Let the five numbers be $x_1, x_2, x_3, x_4, x_5$.
Given that the sum of the five numbers is $x_1 + x_2 + x_3 + x_4 + x_5 = 260$.
The average of the first two numbers is $30$,so $x_1 + x_2 = 30 \times 2 = 60$.
The average of the last two numbers is $70$,so $x_4 + x_5 = 70 \times 2 = 140$.
Substituting these values into the sum equation:
$60 + x_3 + 140 = 260$.
$x_3 + 200 = 260$.
$x_3 = 260 - 200 = 60$.
Therefore,the third number is $60$.

(C) Let the four consecutive odd numbers be $x, x+2, x+4,$ and $x+6$.
Their average is given by $\frac{x + (x+2) + (x+4) + (x+6)}{4} = 42$.
$\frac{4x + 12}{4} = 42$.
$x + 3 = 42$,which gives $x = 39$.
Thus,the numbers are $A = 39, B = 41, C = 43,$ and $D = 45$.
The product of $B$ and $D$ is $41 \times 45 = 1845$.

(B) Let the scores of Rahul,Manish,and Suresh be $R, M,$ and $S$ respectively.
Given,the average score of Rahul,Manish,and Suresh is $63$,so $R + M + S = 63 \times 3 = 189$.
Ajay's score $(A)$ is $30$ more than the average score,so $A = 63 + 30 = 93$.
Rahul's score is $15$ less than Ajay,so $R = A - 15 = 93 - 15 = 78$.
Substituting the value of $R$ in the sum equation: $78 + M + S = 189$.
Therefore,the sum of Manish's and Suresh's scores is $M + S = 189 - 78 = 111$.

(A) Initial sum of weights $= 45 \times 36 = 1620 \, kg$.
Correction for the first student: Actual weight is $32 \, kg$,calculated weight is $34 \, kg$. Difference $= 32 - 34 = -2 \, kg$.
Correction for the second student: Actual weight is $45 \, kg$,calculated weight is $40 \, kg$. Difference $= 45 - 40 = +5 \, kg$.
Total change in sum $= -2 + 5 = +3 \, kg$.
Actual sum of weights $= 1620 + 3 = 1623 \, kg$.
Actual average weight $= \frac{1623}{45} = 36.0666... \, kg$.
Rounding to two decimal places,we get $36.07 \, kg$.

(B) First,calculate the unit price for each item:
Cost of $1 \,kg$ of apples $= \frac{450}{5} = Rs. \, 90$.
Cost of $1$ dozen mangoes $= \frac{4320}{12} = Rs. \, 360$.
Cost of $1 \,kg$ of oranges $= \frac{240}{4} = Rs. \, 60$.
Now,calculate the total cost for the required quantities:
Cost of $8 \,kg$ of apples $= 8 \times 90 = Rs. \, 720$.
Cost of $8$ dozen mangoes $= 8 \times 360 = Rs. \, 2880$.
Cost of $8 \,kg$ of oranges $= 8 \times 60 = Rs. \, 480$.
Total cost $= 720 + 2880 + 480 = Rs. \, 4080$.

(D) Suresh's annual salary $= Rs.\, 1,08,000$.
Suresh's monthly salary $= \frac{1,08,000}{12} = Rs.\, 9,000$.
Since Suresh's monthly salary is half of Nandini's,Nandini's monthly salary $= 9,000 \times 2 = Rs.\, 18,000$.
Let Kaushal's monthly salary be $K$.
According to the problem,$12 \% \text{ of } K = 16 \% \text{ of } 18,000$.
$K \times \frac{12}{100} = 18,000 \times \frac{16}{100}$.
$K \times 12 = 18,000 \times 16$.
$K = \frac{18,000 \times 16}{12}$.
$K = 1,500 \times 16 = 24,000$.
Thus,Kaushal's monthly salary is $Rs.\, 24,000$.

(B) The sum of the ages of the father and mother is $2 \times 35 = 70 \text{ years}$.
The sum of the ages of the father,mother,and son is $3 \times 27 = 81 \text{ years}$.
The age of the son is the difference between the total age of the three family members and the total age of the parents.
$\text{Age of the son} = 81 - 70 = 11 \text{ years}$.

(C) The sum of the first $n$ positive integers is given by the formula $S_n = \frac{n(n+1)}{2}$.
For the first $100$ positive integers,$n = 100$.
Sum $= \frac{100(100+1)}{2} = \frac{100 \times 101}{2} = 5050$.
The average is calculated as $\frac{\text{Sum}}{\text{Number of terms}}$.
Average $= \frac{5050}{100} = 50.5$.

(D) The initial total height of $35$ girls is calculated as: $35 \times 160 = 5600 \,cm$.
To find the actual total height,subtract the incorrect value and add the correct value: $5600 - 144 + 104 = 5560 \,cm$.
Now,calculate the actual average height by dividing the actual total height by the number of girls: $\frac{5560}{35} \approx 158.8571... \,cm$.
Rounding off to two decimal places,we get $158.86 \,cm$. (Note: Based on the provided options,the closest value is $158.85$ or $158.86$. Re-calculating $5560/35 = 158.857$,which rounds to $158.86$. Given the options,$158.86$ is the mathematically correct rounded value,but $158.85$ is provided as option $D$).

(A) Let the average of runs be $x$ till $11$ innings.
$\therefore$ Total runs after $11$ innings $= 11x$.
In the $12^{th}$ innings,he scores $63$ runs.
$\therefore$ Total runs after $12$ innings $= 11x + 63$.
According to the problem,the average increases by $2$ after the $12^{th}$ innings.
New average $= x + 2$.
Total runs after $12$ innings $= 12(x + 2)$.
Equating the two expressions for total runs:
$12(x + 2) = 11x + 63$
$12x + 24 = 11x + 63$
$12x - 11x = 63 - 24$
$x = 39$.
The average after $12^{th}$ innings is $x + 2 = 39 + 2 = 41$.

(B) According to the given information:
$\frac{A+B}{2} = 20 \Rightarrow A+B = 40$ .....$(1)$
$\frac{B+C}{2} = 19 \Rightarrow B+C = 38$ .....$(2)$
$\frac{C+A}{2} = 21 \Rightarrow C+A = 42$ .....$(3)$
Adding equations $(1), (2),$ and $(3)$:
$(A+B) + (B+C) + (C+A) = 40 + 38 + 42$
$2(A+B+C) = 120$
$A+B+C = 60$ .....$(4)$
To find the value of $A,$ subtract equation $(2)$ from equation $(4)$:
$A = (A+B+C) - (B+C)$
$A = 60 - 38 = 22$

(A) Total age of $80$ boys $= 80 \times 15 = 1200$ years.
Total age of the first group of $15$ boys $= 15 \times 16 = 240$ years.
Total age of the second group of $25$ boys $= 25 \times 14 = 350$ years.
Number of remaining boys $= 80 - (15 + 25) = 80 - 40 = 40$ boys.
Total age of the remaining $40$ boys $= 1200 - (240 + 350) = 1200 - 590 = 610$ years.
Average age of the remaining boys $= \frac{610}{40} = 15.25$ years.

(D) Let the marks obtained in Physics,Chemistry,and Mathematics be $Ph$,$Ch$,and $Ma$ respectively.
According to the problem,the total marks in all three subjects is $120$ more than the marks in Chemistry:
$Ph + Ch + Ma = Ch + 120$
Subtracting $Ch$ from both sides,we get:
$Ph + Ma = 120$
The average marks obtained in Physics and Mathematics together is given by the sum of marks in these two subjects divided by $2$:
$\text{Average} = \frac{Ph + Ma}{2} = \frac{120}{2} = 60$
Therefore,the average marks obtained by him in Physics and Mathematics together is $60$.