Ratio and Proportion Questions in English

(A) Let the fourth proportional be $d$.
According to the definition of proportion,for four numbers $a, b, c, d$ to be in proportion,the ratio of the first two must equal the ratio of the last two: $a : b :: c : d$.
Substituting the given values: $24 : 120 :: 22 : d$.
This can be written as the equation: $\frac{24}{120} = \frac{22}{d}$.
Simplifying the fraction on the left side: $\frac{1}{5} = \frac{22}{d}$.
Cross-multiplying to solve for $d$: $d = 22 \times 5 = 110$.
Therefore,the fourth proportional is $110$.

(C) Given ratios are $B: A = 2: 3$ and $A: C = 5: 7$.
To find the combined ratio $B: A: C$,we equate the value of $A$ in both ratios.
The common term is $A$. In the first ratio,$A = 3$,and in the second ratio,$A = 5$. The least common multiple of $3$ and $5$ is $15$.
Multiply the first ratio by $5$: $B: A = (2 \times 5) : (3 \times 5) = 10: 15$.
Multiply the second ratio by $3$: $A: C = (5 \times 3) : (7 \times 3) = 15: 21$.
Thus,$B: A: C = 10: 15: 21$.
From this,we have $A = 15$,$B = 10$,and $C = 21$.
Now,calculate $(A + B) : (B + C) = (15 + 10) : (10 + 21) = 25 : 31$.
Therefore,the correct option is $C$.

(C) Given the ratio of marks of $X$ and $Y$ is $3: 11$.
Let the marks of $X$ be $3k$ and the marks of $Y$ be $11k$,where $k$ is a constant.
According to the problem,the marks of $X$ are $9$,so $3k = 9$.
Solving for $k$,we get $k = 9 / 3 = 3$.
Now,substitute the value of $k$ to find the marks of $Y$:
Marks of $Y = 11k = 11 \times 3 = 33$.
Therefore,the marks of $Y$ are $33$.

(D) Given ratios are $U: V = 6: 7$ and $V: W = 21: 6$.
To find $U: V: W$,we need to make the value of $V$ common in both ratios.
The value of $V$ in the first ratio is $7$ and in the second ratio is $21$.
To make them equal,multiply the first ratio by $3$:
$U: V = (6 \times 3) : (7 \times 3) = 18: 21$.
Now,we have $U: V = 18: 21$ and $V: W = 21: 6$.
Combining these,we get $U: V: W = 18: 21: 6$.
Simplifying the ratio by dividing by $3$,we get $6: 7: 2$.
Therefore,the correct option is $D$.

(B) Given the ratio $a-b : b-c : c-d = 1 : 2 : 3$.
Let $a-b = k$,$b-c = 2k$,and $c-d = 3k$ for some constant $k$.
Adding these equations:
$(a-b) + (b-c) + (c-d) = k + 2k + 3k$
$a - d = 6k$.
Now,we need to express $a$ and $d$ in terms of $c$ and $k$ to find $(a+d) : c$.
From $c-d = 3k$,we get $d = c - 3k$.
From $b-c = 2k$,we get $b = c + 2k$.
From $a-b = k$,we get $a = b + k = (c + 2k) + k = c + 3k$.
Now calculate $a+d$:
$a + d = (c + 3k) + (c - 3k) = 2c$.
Therefore,the ratio $(a+d) : c = 2c : c = 2 : 1$.

(C) Let the present age of the father be $F$ and the son be $S$.
Given that the sum of their present ages is $90$ years: $F + S = 90$ (Equation $1$).
$10$ years ago,the father's age was $(F - 10)$ and the son's age was $(S - 10)$.
The ratio of their ages $10$ years ago was $5:2$,so: $\frac{F - 10}{S - 10} = \frac{5}{2}$.
Cross-multiplying gives: $2(F - 10) = 5(S - 10) \Rightarrow 2F - 20 = 5S - 50 \Rightarrow 2F - 5S = -30$ (Equation $2$).
From Equation $1$,$S = 90 - F$. Substituting this into Equation $2$:
$2F - 5(90 - F) = -30$
$2F - 450 + 5F = -30$
$7F = 420$
$F = 60$.
Therefore,the present age of the father is $60$ years.

(B) Let the present ages of $R$ and $S$ be $11x$ and $17x$ respectively.
According to the problem,$11$ years ago,the ratio of their ages was $11:20$.
So,$\frac{11x - 11}{17x - 11} = \frac{11}{20}$.
Cross-multiplying,we get: $20(11x - 11) = 11(17x - 11)$.
$220x - 220 = 187x - 121$.
$220x - 187x = 220 - 121$.
$33x = 99$.
$x = 3$.
Therefore,the present age of $R$ is $11x = 11 \times 3 = 33$ years.

(C) Let the shares of Sunil,Anil,and Jamil be $S, A,$ and $J$ respectively.
Given that $S + A + J = 15525$.
After diminishing the shares by ₹ $22$,₹ $35$,and ₹ $48$,the remaining amounts are in the ratio $7: 10: 13$.
Let the remaining amounts be $7x, 10x,$ and $13x$.
Then,$S - 22 = 7x \implies S = 7x + 22$.
$A - 35 = 10x \implies A = 10x + 35$.
$J - 48 = 13x \implies J = 13x + 48$.
Summing these: $(7x + 22) + (10x + 35) + (13x + 48) = 15525$.
$30x + 105 = 15525 \implies 30x = 15420 \implies x = 514$.
Calculating original shares:
$S = 7(514) + 22 = 3598 + 22 = 3620$.
$A = 10(514) + 35 = 5140 + 35 = 5175$.
$J = 13(514) + 48 = 6682 + 48 = 6730$.
Now,add ₹ $16$,₹ $77$,and ₹ $37$ to their original shares:
New $S = 3620 + 16 = 3636$.
New $A = 5175 + 77 = 5252$.
New $J = 6730 + 37 = 6767$.
The ratio is $3636 : 5252 : 6767$.
Dividing by $101$: $3636/101 = 36$,$5252/101 = 52$,$6767/101 = 67$.
The ratio is $36: 52: 67$.

(A) Given that $\frac{1}{2} A = \frac{1}{3} B = \frac{1}{6} C = k$,where $k$ is a constant.
From this,we get $A = 2k$,$B = 3k$,and $C = 6k$.
The total sum is $A + B + C = 1980$.
Substituting the values: $2k + 3k + 6k = 1980$.
$11k = 1980$.
$k = \frac{1980}{11} = 180$.
$B$'s part is $3k = 3 \times 180 = ₹ 540$.

(A) Let the equal value be $x$.
Then,$\frac{2}{5} A + 40 = x \implies A = \frac{5}{2}(x - 40)$.
$\frac{2}{7} B + 20 = x \implies B = \frac{7}{2}(x - 20)$.
$\frac{9}{17} C + 10 = x \implies C = \frac{17}{9}(x - 10)$.
Since $A + B + C = 600$,we have:
$\frac{5}{2}(x - 40) + \frac{7}{2}(x - 20) + \frac{17}{9}(x - 10) = 600$.
$\frac{5x}{2} - 100 + \frac{7x}{2} - 70 + \frac{17x}{9} - \frac{170}{9} = 600$.
$6x - 170 + \frac{17x - 170}{9} = 600$.
$6x + \frac{17x}{9} = 770 + \frac{170}{9}$.
$\frac{54x + 17x}{9} = \frac{6930 + 170}{9}$.
$71x = 7100 \implies x = 100$.
Thus,$A$'s share = $\frac{5}{2}(100 - 40) = \frac{5}{2}(60) = 150$.

(A) Let the shares of $A, B,$ and $C$ be $a, b,$ and $c$ respectively.
Given that $a + b + c = 600$.
According to the problem:
$\frac{2}{5}a + 40 = \frac{2}{7}b + 20 = \frac{9}{17}c + 10 = k$ (let this be $k$).
Then,$\frac{2}{5}a = k - 40 \implies a = \frac{5}{2}(k - 40) = 2.5k - 100$.
$\frac{2}{7}b = k - 20 \implies b = \frac{7}{2}(k - 20) = 3.5k - 70$.
$\frac{9}{17}c = k - 10 \implies c = \frac{17}{9}(k - 10) = \frac{17}{9}k - \frac{170}{9}$.
Substituting these into $a + b + c = 600$:
$(2.5k - 100) + (3.5k - 70) + (\frac{17}{9}k - \frac{170}{9}) = 600$.
$6k - 170 + \frac{17}{9}k - \frac{170}{9} = 600$.
$\frac{54k + 17k}{9} = 770 + \frac{170}{9}$.
$\frac{71k}{9} = \frac{6930 + 170}{9} = \frac{7100}{9}$.
$71k = 7100 \implies k = 100$.
Now,$A$'s share $a = 2.5(100) - 100 = 250 - 100 = 150$.

(C) Let the shares of $A, B$ and $C$ be $6x, 8x$ and $7x$ respectively.
Total amount $= 6x + 8x + 7x = 21x$.
Given $21x = 8400$,so $x = 400$.
Shares are: $A = 6 \times 400 = 2400$,$B = 8 \times 400 = 3200$,$C = 7 \times 400 = 2800$.
Let the savings of $A, B$ and $C$ be $3y, 2y$ and $4y$ respectively.
Given $B$'s saving $= 2y = 400$,so $y = 200$.
Savings are: $A = 3 \times 200 = 600$,$B = 2 \times 200 = 400$,$C = 4 \times 200 = 800$.
Expenditure = Income - Saving.
Expenditure of $A = 2400 - 600 = 1800$.
Expenditure of $B = 3200 - 400 = 2800$.
Expenditure of $C = 2800 - 800 = 2000$.
Ratio of expenditures $= 1800 : 2800 : 2000 = 18 : 28 : 20 = 9 : 14 : 10$.

(C) Let $A$'s income $= 2x$ and $C$'s income $= 3x$.
Let $B$'s income $= y$ and $D$'s income $= 2y$.
Given that $A$'s income is ₹ $140$ more than $B$'s: $2x - y = 140$ (Equation $1$).
Given that $C$'s income is ₹ $80$ more than $D$'s: $3x - 2y = 80$ (Equation $2$).
Multiply Equation $1$ by $2$: $4x - 2y = 280$ (Equation $3$).
Subtract Equation $2$ from Equation $3$: $(4x - 2y) - (3x - 2y) = 280 - 80$,which gives $x = 200$.
Substitute $x = 200$ into Equation $1$: $2(200) - y = 140 \implies 400 - y = 140 \implies y = 260$.
Now,calculate the incomes:
$A = 2x = 2(200) = ₹ 400$.
$B = y = ₹ 260$.
$C = 3x = 3(200) = ₹ 600$.
$D = 2y = 2(260) = ₹ 520$.
Thus,the incomes are $₹ 400, ₹ 260, ₹ 600$ and $₹ 520$.

(C) The ratio of the original fare to the new fare is $9:11$.
Let the original fare be $9x$ and the new fare be $11x$.
Given that the original fare is ₹ $18,000$,we have:
$9x = 18,000$
$x = 18,000 / 9 = 2,000$
The increase in the fare is the difference between the new fare and the original fare:
Increase $= 11x - 9x = 2x$
Substituting the value of $x$:
Increase $= 2 \times 2,000 = 4,000$
Therefore,the increase in the fare is ₹ $4,000$.

(C) Let the weight of the original diamond be $15x$ and the price be $P = k(15x)^2 = 225kx^2$,where $k$ is a constant.
The diamond is broken into three pieces with weights $3x, 5x$,and $7x$.
The price of the broken pieces will be:
Price of $1^{st}$ piece $= k(3x)^2 = 9kx^2$
Price of $2^{nd}$ piece $= k(5x)^2 = 25kx^2$
Price of $3^{rd}$ piece $= k(7x)^2 = 49kx^2$
Total price of the broken pieces $= 9kx^2 + 25kx^2 + 49kx^2 = 83kx^2$.
The loss incurred is the difference between the original price and the total price of the broken pieces:
Loss $= 225kx^2 - 83kx^2 = 142kx^2$.
Given that the loss is ₹ $42600$,we have:
$142kx^2 = 42600$
$kx^2 = \frac{42600}{142} = 300$.
Therefore,the original price of the diamond is $225kx^2 = 225 \times 300 = ₹ 67500$.

(A) Let the total number of candidates who applied be $5x$.
Since the ratio of selected to unselected candidates is $4:1$,the number of selected candidates is $4x$ and the number of unselected candidates is $x$.
According to the problem,if $90$ fewer candidates had applied,the new total would be $5x - 90$.
If $20$ fewer candidates were selected,the new number of selected candidates would be $4x - 20$.
The new number of unselected candidates would be $(5x - 90) - (4x - 20) = x - 70$.
Given the new ratio is $5:1$,we have the equation:
$\frac{4x - 20}{x - 70} = \frac{5}{1}$
$4x - 20 = 5(x - 70)$
$4x - 20 = 5x - 350$
$x = 350 - 20 = 330$
Therefore,the total number of candidates who applied is $5x = 5 \times 330 = 1650$.

(B) Let the number of selected candidates be $5x$ and the number of unselected candidates be $x$.
Total candidates applied = $5x + x = 6x$.
According to the problem,if $100$ fewer candidates had applied,the total number of candidates would be $6x - 100$.
If $20$ fewer candidates were selected,the number of selected candidates would be $5x - 20$.
The number of unselected candidates would then be $(6x - 100) - (5x - 20) = x - 80$.
The new ratio of selected to unselected candidates is $6:1$,so:
$\frac{5x - 20}{x - 80} = \frac{6}{1}$
$5x - 20 = 6(x - 80)$
$5x - 20 = 6x - 480$
$6x - 5x = 480 - 20$
$x = 460$
The total number of candidates who applied is $6x = 6 \times 460 = 2760$.

(C) Speed and time are inversely proportional to each other when the distance is constant.
Given the ratio of speeds $v_1 : v_2 : v_3 = 1 : 3 : 5$.
The time taken $t$ is given by $t = \frac{d}{v}$.
Since the distance $d$ is the same for all three cars,the ratio of time taken is $t_1 : t_2 : t_3 = \frac{1}{v_1} : \frac{1}{v_2} : \frac{1}{v_3}$.
Substituting the values,we get $t_1 : t_2 : t_3 = \frac{1}{1} : \frac{1}{3} : \frac{1}{5}$.
To simplify this ratio,multiply each term by the least common multiple $(LCM)$ of $1, 3,$ and $5$,which is $15$.
Ratio $= (1 \times 15) : (\frac{1}{3} \times 15) : (\frac{1}{5} \times 15) = 15 : 5 : 3$.

(D) We know that for a constant distance,speed is inversely proportional to time,i.e.,$\text{Speed} \propto \frac{1}{\text{Time}}$.
Given the ratio of speeds of Aman,Kamal,and Manan is $4: 5: 6$.
Therefore,the ratio of time taken is the reciprocal of the speed ratio: $\frac{1}{4}: \frac{1}{5}: \frac{1}{6}$.
To simplify this ratio,we find the Least Common Multiple $(LCM)$ of the denominators $4, 5,$ and $6$,which is $60$.
Multiplying each term by $60$,we get: $(\frac{1}{4} \times 60) : (\frac{1}{5} \times 60) : (\frac{1}{6} \times 60)$.
This results in $15: 12: 10$.
Thus,the ratio of time taken is $15: 12: 10$.

(A) First,convert the mixed fractions into improper fractions:
$1 \frac{1}{2} = \frac{3}{2}$
$1 \frac{1}{3} = \frac{4}{3}$
Next,find the square of these numbers:
$(\frac{3}{2})^2 = \frac{9}{4}$
$(\frac{4}{3})^2 = \frac{16}{9}$
Now,find the reciprocals of these squares:
Reciprocal of $\frac{9}{4} = \frac{4}{9}$
Reciprocal of $\frac{16}{9} = \frac{9}{16}$
Finally,calculate the ratio of these reciprocals:
Ratio $= \frac{4}{9} : \frac{9}{16} = \frac{4 \times 16}{9 \times 9} = \frac{64}{81}$
Therefore,the ratio is $64: 81$.

(A) Let the amount with $A$ be $A$ and the amount with $B$ be $B$.
Given that $A + B = 6300$.
According to the problem,$\frac{5}{19} A = \frac{2}{5} B$.
Cross-multiplying,we get $25 A = 38 B$,which implies $\frac{A}{B} = \frac{38}{25}$.
Let $A = 38x$ and $B = 25x$.
Substituting these into the sum: $38x + 25x = 6300$.
$63x = 6300$,which gives $x = 100$.
Therefore,the amount of $B = 25x = 25 \times 100 = 2500$.
Wait,re-evaluating the ratio: $\frac{5}{19} A = \frac{2}{5} B \Rightarrow A/B = (2/5) \times (19/5) = 38/25$. The sum is $38x + 25x = 63x = 6300 \Rightarrow x = 100$. So $B = 25 \times 100 = 2500$.

(B) Let the required fraction be $x$.
According to the problem,the ratio of $x$ to $\frac{1}{27}$ is equal to the ratio of $\frac{3}{7}$ to $\frac{5}{9}$.
This can be written as: $x : \frac{1}{27} = \frac{3}{7} : \frac{5}{9}$.
Using the property of ratios,we have: $\frac{x}{1/27} = \frac{3/7}{5/9}$.
Simplifying the right side: $\frac{3}{7} \times \frac{9}{5} = \frac{27}{35}$.
So,$x \times 27 = \frac{27}{35}$.
$x = \frac{27}{35} \times \frac{1}{27}$.
$x = \frac{1}{35}$.
Thus,the required fraction is $\frac{1}{35}$.

(D) Let the number of successful students be $9x$ and unsuccessful students be $2x$.
Given that the total number of examinees is $132$,we have:
$9x + 2x = 132$
$11x = 132$
$x = 12$
Therefore,the number of successful students is $9 \times 12 = 108$ and the number of unsuccessful students is $2 \times 12 = 24$.
If $4$ more students had passed,the new number of successful students would be $108 + 4 = 112$.
The number of unsuccessful students would decrease by $4$,so it would be $24 - 4 = 20$.
The new ratio of successful to unsuccessful students is $112:20$.
Dividing both terms by $4$,we get $28:5$.

(B) Total number of students = $720$.
The ratio of boys to girls is $7: 5$.
Let the number of boys be $7x$ and the number of girls be $5x$.
Sum of parts = $7 + 5 = 12$.
$12x = 720 \implies x = 60$.
Number of boys = $7 \times 60 = 420$.
Number of girls = $5 \times 60 = 300$.
To make the ratio $1: 1$,the number of girls must equal the number of boys.
Required number of girls = $420$.
Additional girls to be admitted = $420 - 300 = 120$.

(B) Let the two numbers be $x$ and $y$,where $x > y$.
According to the problem,the sum of the numbers is three times their difference:
$x + y = 3(x - y)$
Expanding the right side:
$x + y = 3x - 3y$
Rearranging the terms to group $x$ and $y$:
$y + 3y = 3x - x$
$4y = 2x$
Dividing both sides by $2y$:
$\frac{x}{y} = \frac{4}{2} = \frac{2}{1}$
Therefore,the ratio of the two numbers is $2: 1$.

(D) Given the ratio: $\frac{x + \frac{1}{x}}{x - \frac{1}{x}} = \frac{5}{5} = 1$.
Cross-multiplying gives: $x + \frac{1}{x} = x - \frac{1}{x}$.
Subtracting $x$ from both sides: $\frac{1}{x} = -\frac{1}{x}$.
Adding $\frac{1}{x}$ to both sides: $\frac{2}{x} = 0$.
This implies $2 = 0$,which is a contradiction.
Therefore,there is no finite value of $x$ that satisfies the given equation. Thus,no solution exists.

(B) The ratio of sweets distributed between $A$ and $B$ is $3:4$.
Let the number of sweets received by $A$ be $3x$ and by $B$ be $4x$.
Given that $A$ received $36$ sweets,we have $3x = 36$.
Solving for $x$,we get $x = 36 / 3 = 12$.
Now,the number of sweets received by $B$ is $4x = 4 \times 12 = 48$.
The total number of sweets is the sum of sweets received by $A$ and $B$,which is $36 + 48 = 84$.

(B) Let the present ages of $P$ and $Q$ be $P$ and $Q$ respectively.
According to the problem, $4 \, \text{years}$ ago, their ages were $(P-4)$ and $(Q-4)$.
The ratio was given as $\frac{P-4}{Q-4} = \frac{5}{6}$.
Cross-multiplying gives $6(P-4) = 5(Q-4)$, which simplifies to $6P - 24 = 5Q - 20$, or $6P - 5Q = 4 \quad (1)$.
We are also given that the sum of their present ages is $52$, so $P + Q = 52$, which means $P = 52 - Q \quad (2)$.
Substituting $(2)$ into $(1)$: $6(52 - Q) - 5Q = 4$.
$312 - 6Q - 5Q = 4$.
$312 - 11Q = 4$.
$11Q = 308$.
$Q = 28$.
Now, find $P$: $P = 52 - 28 = 24$.
The ratio of their present ages is $P:Q = 24:28$.
Dividing both by $4$, we get $6:7$.

(A) Let the present ages of $A$ and $B$ be $5x$ and $6x$ years respectively.
According to the problem,after seven years,the ratio of their ages will be $6:7$.
So,the equation is: $\frac{5x + 7}{6x + 7} = \frac{6}{7}$.
Cross-multiplying the terms,we get: $7(5x + 7) = 6(6x + 7)$.
$35x + 49 = 36x + 42$.
Rearranging the terms to solve for $x$: $49 - 42 = 36x - 35x$.
$x = 7$.
The present age of $A$ is $5x = 5 \times 7 = 35$ years.

(B) Let the common ratio multiplier be $x$.
Then the shares of $A$,$B$,and $C$ are $5x$,$6x$,and $9x$ respectively.
Given that $A$ received ₹ $450$,we have $5x = 450$.
Solving for $x$,we get $x = 450 / 5 = 90$.
The total sum divided is the sum of the shares: $5x + 6x + 9x = 20x$.
Substituting the value of $x$,the total sum is $20 \times 90 = 1800$.
Therefore,the total sum divided was ₹ $1800$.

(B) Let the shares of the three brothers be $A$,$B$,and $C$ respectively.
According to the problem,the total amount is $A + B + C = 1620$.
We are given that the share of the second brother $B$ is $\frac{5}{13}$ of the sum of the shares of the other two,i.e.,$B = \frac{5}{13}(A + C)$.
Substituting $(A + C)$ from the total sum equation: $(A + C) = 1620 - B$.
Now,substitute this into the given condition:
$B = \frac{5}{13}(1620 - B)$
$13B = 5(1620 - B)$
$13B = 8100 - 5B$
$13B + 5B = 8100$
$18B = 8100$
$B = \frac{8100}{18} = 450$.
Therefore,the share of the second brother is ₹ $450$.

(A) Let the number of boys be $8x$ and the number of girls be $12x$.
Total number of students $= 8x + 12x = 20x$.
Number of boys getting scholarships $= 50 \% \text{ of } 8x = 0.5 \times 8x = 4x$.
Number of girls getting scholarships $= 25 \% \text{ of } 12x = 0.25 \times 12x = 3x$.
Total number of students getting scholarships $= 4x + 3x = 7x$.
Number of students not getting scholarships $= \text{Total students} - \text{Students getting scholarships} = 20x - 7x = 13x$.
Percentage of students not getting scholarships $= (13x / 20x) \times 100 = 0.65 \times 100 = 65 \%$.

(B) The volume of a cylinder is given by the formula $V = \pi r^2 h$.
Let the radii of the two cylinders be $r_1$ and $r_2$,and their heights be $h_1$ and $h_2$.
Given that $r_1:r_2 = 2:3$ and $h_1:h_2 = 5:3$.
The ratio of their volumes is $\frac{V_1}{V_2} = \frac{\pi r_1^2 h_1}{\pi r_2^2 h_2} = \left(\frac{r_1}{r_2}\right)^2 \times \left(\frac{h_1}{h_2}\right)$.
Substituting the given values: $\frac{V_1}{V_2} = \left(\frac{2}{3}\right)^2 \times \left(\frac{5}{3}\right) = \frac{4}{9} \times \frac{5}{3} = \frac{20}{27}$.
Therefore,the ratio of their volumes is $20:27$.

(B) The office opens at $10 \, AM$ and closes at $5 \, PM$.
Total duration $= 5 \, PM - 10 \, AM = 7 \, \text{hours}$.
Since $1 \, \text{hour} = 60 \, \text{minutes}$, total office hours $= 7 \times 60 = 420 \, \text{minutes}$.
The lunch interval is $30 \, \text{minutes}$.
The ratio of the lunch interval to the total office hours is $30 : 420$.
Dividing both sides by $30$, we get $1 : 14$.

(C) Let the fare for air-conditioned sleeper class be $4x$ and for ordinary sleeper class be $1x$.
Let the number of passengers in air-conditioned sleeper class be $3y$ and in ordinary sleeper class be $25y$.
The total collection is given by the sum of (fare $\times$ number of passengers) for both classes.
Total collection $= (4x \times 3y) + (1x \times 25y) = 12xy + 25xy = 37xy$.
Given that the total collection is ₹ $37,000$,we have $37xy = 37,000$,which implies $xy = 1,000$.
The amount paid by air-conditioned sleeper passengers is $4x \times 3y = 12xy$.
Substituting $xy = 1,000$,we get $12 \times 1,000 = ₹ 12,000$.

(D) The ratio of male to female qualified candidates is $11:7.$
Let the number of male candidates be $11x$ and the number of female candidates be $7x.$
Given that the number of female qualified candidates is $126,$
$7x = 126$
$x = \frac{126}{7} = 18$
Now,the total number of qualified candidates is the sum of male and female candidates:
Total candidates $= 11x + 7x = 18x$
Substituting the value of $x = 18$:
Total candidates $= 18 \times 18 = 324.$
Note: The original question asks for the total number of qualified candidates. Based on the provided options and the calculation,the total number of qualified candidates is $324.$ However,if the question implies a calculation related to the total appeared candidates based on a percentage (often found in such data interpretation sets),and assuming the qualified candidates represent $60\%$ of the appeared candidates,then $0.6 \times \text{Appeared} = 324$,which gives $540.$ Thus,option $D$ is the correct answer.

(D) Given ratios are $A: B = 2: 3$ and $B: C = 3: 7$.
Since the value of $B$ is common in both ratios and is equal to $3$,we can directly write the combined ratio as $A: B: C = 2: 3: 7$.
Let $A = 2k$,$B = 3k$,and $C = 7k$ for some constant $k$.
Now,calculate the required sums:
$A + B = 2k + 3k = 5k$
$B + C = 3k + 7k = 10k$
$C + A = 7k + 2k = 9k$
Therefore,the ratio $A + B : B + C : C + A = 5k : 10k : 9k = 5: 10: 9$.

(B) Let the monthly incomes of $A$ and $B$ be $8x$ and $5x$ respectively.
Let the monthly expenditures of $A$ and $B$ be $5y$ and $3y$ respectively.
We know that $\text{Income} - \text{Expenditure} = \text{Savings}$.
For $A$: $8x - 5y = 12,000$ --- (Equation $1$)
For $B$: $5x - 3y = 10,000$ --- (Equation $2$)
To solve for $x$,multiply Equation $1$ by $3$ and Equation $2$ by $5$:
$24x - 15y = 36,000$ --- (Equation $3$)
$25x - 15y = 50,000$ --- (Equation $4$)
Subtract Equation $3$ from Equation $4$:
$(25x - 24x) - (15y - 15y) = 50,000 - 36,000$
$x = 14,000$
Monthly income of $A = 8x = 8 \times 14,000 = 1,12,000$.
Monthly income of $B = 5x = 5 \times 14,000 = 70,000$.
Difference in monthly incomes $= 1,12,000 - 70,000 = 42,000$.

(D) Initially,the total number of students is $1554$ and the ratio of boys to girls is $4:3$.
Total parts $= 4 + 3 = 7$.
Number of boys $= (4/7) \times 1554 = 888$.
Number of girls $= (3/7) \times 1554 = 666$.
After $30$ girls joined,the new number of girls $= 666 + 30 = 696$.
Let the number of boys who left be $x$.
Then,the new number of boys $= 888 - x$.
According to the problem,the new ratio of boys to girls is $7:6$,so:
$(888 - x) / 696 = 7 / 6$.
Multiplying both sides by $696$:
$888 - x = (7 / 6) \times 696$.
$888 - x = 7 \times 116$.
$888 - x = 812$.
$x = 888 - 812 = 76$.
Therefore,$76$ boys left the school.

(B) Let the two numbers be $2x$ and $3x$.
According to the problem,when $8$ is added to both numbers,the ratio becomes $3:4$.
So,$\frac{2x + 8}{3x + 8} = \frac{3}{4}$.
Cross-multiplying,we get $4(2x + 8) = 3(3x + 8)$.
$8x + 32 = 9x + 24$.
$32 - 24 = 9x - 8x$.
$x = 8$.
The two numbers are $2x = 2(8) = 16$ and $3x = 3(8) = 24$.
The sum of the two numbers is $16 + 24 = 40$.

(D) Let the number to be added be $x$.
According to the problem,the ratio becomes $\frac{2+x}{5+x} = \frac{5}{6}$.
By cross-multiplying,we get:
$6(2+x) = 5(5+x)$
$12 + 6x = 25 + 5x$
Subtract $5x$ from both sides:
$12 + x = 25$
Subtract $12$ from both sides:
$x = 25 - 12$
$x = 13$.
Therefore,we must add $13$ to each term.

(D) Given the ratio of $A$ to $B$ is $4:5.$
Let $A = 4x$ and $B = 5x.$
The difference of their squares is given as $81.$
$(5x)^2 - (4x)^2 = 81$
$25x^2 - 16x^2 = 81$
$9x^2 = 81$
$x^2 = 9$
$x = 3$
Therefore,the value of $A = 4x = 4 \times 3 = 12.$

(A) Let the shares of the son,wife,and daughter be $S$,$W$,and $D$ respectively.
Given ratios are $S:W = 3:1$ and $W:D = 3:1$.
To combine these ratios,we make the wife's share $(W)$ common in both ratios.
$S:W = 3:1 = 9:3$ and $W:D = 3:1$.
Thus,the combined ratio is $S:W:D = 9:3:1$.
Let the shares be $9k$,$3k$,and $k$ respectively,where $k$ is a constant.
The total property value is $9k + 3k + k = 13k$.
According to the problem,the daughter gets ₹ $10,000$ less than the son:
$S - D = 10000$
$9k - k = 10000$
$8k = 10000$
$k = 1250$.
The total property value is $13k = 13 \times 1250 = 16250$.
Therefore,the total value of the property is ₹ $16,250$.

(A) proportion is a statement that two ratios are equal. To check if a proportion is correct,we simplify both sides to their lowest terms.
For option $A$: $12: 9 = (12/3) : (9/3) = 4: 3$ and $16: 12 = (16/4) : (12/4) = 4: 3$. Since $4: 3 = 4: 3$,this is a correct proportion.
For option $B$: $13: 11$ is already in simplest form,and $5: 4$ is also in simplest form. Since $13/11 \neq 5/4$,this is incorrect.
For option $C$: $30: 45 = (30/15) : (45/15) = 2: 3$,while $13: 24$ is in simplest form. Since $2/3 \neq 13/24$,this is incorrect.
For option $D$: $3: 5 \neq 2: 5$,so this is incorrect.
Therefore,the correct option is $A$.

(C) Let the two numbers be $3x$ and $4x$,where $x$ is a common multiplier.
The $LCM$ of $3x$ and $4x$ is calculated as $3 \times 4 \times x = 12x$.
Given that the $LCM$ is $180$,we set up the equation:
$12x = 180$
Solving for $x$:
$x = 180 / 12 = 15$
The second number is $4x = 4 \times 15 = 60$.

(C) Total number of students $= z$.
Number of boys $= x$.
Number of girls $= z - x$.
The part of the class composed of girls is the ratio of the number of girls to the total number of students.
Part of girls $= \frac{z - x}{z} = \frac{z}{z} - \frac{x}{z} = 1 - \frac{x}{z}$.

(C) Let the third proportional of $12$ and $18$ be $x$.
According to the definition of proportion,if $a$ and $b$ are in proportion,then $a:b = b:x$.
Substituting the given values,we get:
$12:18 = 18:x$
This can be written as:
$\frac{12}{18} = \frac{18}{x}$
Solving for $x$:
$12x = 18 \times 18$
$12x = 324$
$x = \frac{324}{12}$
$x = 27$
Therefore,the third proportional is $27$.

(A) Let the marks in Science be $x$.
Since Ram got twice as many marks in English as in Science,his marks in English are $2x$.
The ratio of his marks in English to Mathematics is $2:3$. Given English marks are $2x$,we can represent Mathematics marks as $3x$.
The total marks in English,Science,and Mathematics are $180$.
Therefore,$x + 2x + 3x = 180$.
$6x = 180$.
$x = 30$.
Thus,his marks in Science are $30$.

(B) Let the three numbers be $2x, 3x$,and $4x$.
According to the problem,the sum of their squares is $1856$.
Therefore,$(2x)^2 + (3x)^2 + (4x)^2 = 1856$.
Expanding the squares,we get $4x^2 + 9x^2 + 16x^2 = 1856$.
Combining the terms,$29x^2 = 1856$.
Dividing by $29$,$x^2 = 1856 / 29 = 64$.
Taking the square root,$x = \sqrt{64} = 8$.
Thus,the numbers are $2(8) = 16$,$3(8) = 24$,and $4(8) = 32$.

(B) Given the ratios $x: y = 3: 2$ and $y: z = 3: 2$.
To combine these,we make the value of $y$ common in both ratios.
Multiply the first ratio by $3$ and the second ratio by $2$:
$x: y = (3 \times 3): (2 \times 3) = 9: 6$
$y: z = (3 \times 2): (2 \times 2) = 6: 4$
Therefore,the combined ratio is $x: y: z = 9: 6: 4$.
Let the runs scored be $9a, 6a,$ and $4a$.
The total runs scored is $342$,so:
$9a + 6a + 4a = 342$
$19a = 342$
$a = 342 / 19 = 18$
Now,calculate the runs for each:
$A = 9 \times 18 = 162$
$B = 6 \times 18 = 108$
$C = 4 \times 18 = 72$
Thus,the runs scored by $A, B,$ and $C$ are $162, 108,$ and $72$ respectively.