Ratio and Proportion Questions in English

(C) Let the initial quantity of alcohol be $4x$ and water be $3x$.
When $5 \text{ liters}$ of water is added,the new quantity of water becomes $3x + 5$.
According to the problem,the new ratio of alcohol to water is $4:5$.
So,$\frac{4x}{3x + 5} = \frac{4}{5}$.
Cross-multiplying,we get $20x = 4(3x + 5)$.
$20x = 12x + 20$.
$8x = 20$.
$x = \frac{20}{8} = 2.5$.
The quantity of alcohol is $4x = 4 \times 2.5 = 10 \text{ liters}$.

(C) Let the number of coins be $x$,$2x$,and $3x$ for $25 \ p$,$10 \ p$,and $5 \ p$ respectively.
The total value of the coins is given by:
$(x \times 25) + (2x \times 10) + (3x \times 5) = 3000 \ p$ (since $Rs. 30 = 3000 \ p$)
$25x + 20x + 15x = 3000$
$60x = 3000$
$x = 3000 / 60 = 50$
The number of $5 \ p$ coins is $3x = 3 \times 50 = 150$.

(B) Let $C$'s share be $x$.
Then,$B$'s share $= \frac{1}{4}x$.
$A$'s share $= \frac{2}{3} \times (\frac{1}{4}x) = \frac{1}{6}x$.
The total amount is $Rs. 510$,so:
$\frac{1}{6}x + \frac{1}{4}x + x = 510$.
Taking the least common multiple $(LCM)$ of $6, 4, 1$ as $12$:
$\frac{2x + 3x + 12x}{12} = 510$.
$\frac{17x}{12} = 510$.
$17x = 510 \times 12$.
$x = \frac{510 \times 12}{17} = 30 \times 12 = 360$.
Thus,$C$'s share $= Rs. 360$.
$B$'s share $= \frac{1}{4} \times 360 = Rs. 90$.
$A$'s share $= \frac{2}{3} \times 90 = Rs. 60$.
Therefore,the shares are $Rs. 60, Rs. 90, Rs. 360$.

(A) Let the shares of $A, B,$ and $C$ be $A, B,$ and $C$ respectively.
Given that $A + B + C = 366$ (Equation $1$).
According to the problem,$A = \frac{1}{2}(B + C)$,which implies $2A = B + C$ (Equation $2$).
Substituting Equation $2$ into Equation $1$: $A + (B + C) = 366 \Rightarrow A + 2A = 366 \Rightarrow 3A = 366 \Rightarrow A = 122$.
Thus,the share of $A$ is $Rs. 122$.

(D) Let the initial earnings of $A$ and $B$ be $4x$ and $7x$ respectively.
After an increase of $50\%$,the new earnings of $A = 4x \times (1 + 0.50) = 4x \times 1.5 = 6x$.
After a decrease of $25\%$,the new earnings of $B = 7x \times (1 - 0.25) = 7x \times 0.75 = 5.25x$.
The new ratio is $6x : 5.25x = 6 : 5.25 = 600 : 525 = 8 : 7$.
Since the ratio $8:7$ is consistent with the given information for any value of $x$,the specific value of $A$'s earnings cannot be determined without an absolute value provided in the problem statement.
Therefore,the data is inadequate.

(D) Let the shares of $A, B$,and $C$ after reducing $Rs. 25$ each be $x, 3x$,and $2x$ respectively.
According to the problem,the total amount is $Rs. 735$.
If each had received $Rs. 25$ less,the total reduction would be $3 \times 25 = Rs. 75$.
So,the new total sum would be $735 - 75 = 660$.
Thus,$x + 3x + 2x = 660$.
$6x = 660 \Rightarrow x = 110$.
The share of $C$ is $2x + 25$.
Substituting $x = 110$,we get $C = 2(110) + 25 = 220 + 25 = Rs. 245$.

(D) Let the shares of $A$,$B$,and $C$ be $(3x + 5)$,$(4x + 10)$,and $(5x + 15)$ respectively.
According to the problem,the sum of these shares is $Rs. 2430$.
$(3x + 5) + (4x + 10) + (5x + 15) = 2430$
$12x + 30 = 2430$
$12x = 2430 - 30$
$12x = 2400$
$x = 200$
Now,calculate $B$'s share:
$B$'s share $= 4x + 10$
$B$'s share $= 4(200) + 10$
$B$'s share $= 800 + 10 = 810$
Thus,$B$'s share is $Rs. 810$.

(B) We can use the method of alligation to solve this problem.
Let the density of water be $1$.
Density of gold = $19$.
Density of copper = $9$.
Density of the required alloy = $15$.
Using the alligation rule:
(Density of Gold - Mean Density) : (Mean Density - Density of Copper)
$= (19 - 15) : (15 - 9)$
$= 4 : 6$
$= 2 : 3$
Therefore,the ratio in which they should be mixed is $2:3$.

(B) Initial volume of mixture = $15$ litres.
Quantity of alcohol = $20 \%$ of $15 = \frac{20}{100} \times 15 = 3$ litres.
Quantity of water = $15 - 3 = 12$ litres.
When $3$ litres of water is added,the new quantity of water = $12 + 3 = 15$ litres.
The quantity of alcohol remains the same,i.e.,$3$ litres.
Total volume of the new mixture = $3 + 15 = 18$ litres.
Percentage of alcohol in the new mixture = $(\frac{\text{Quantity of alcohol}}{\text{Total volume}}) \times 100 = (\frac{3}{18}) \times 100 = \frac{1}{6} \times 100 = \frac{50}{3} \% = 16 \frac{2}{3} \%$.

(A) Total volume of the mixture = $85 \, L$.
The ratio of milk to water is $27: 7$.
Sum of ratio terms = $27 + 7 = 34$.
Value of $1$ unit = $\frac{85}{34} \, L = 2.5 \, L$.
Initial amount of milk = $27 \times 2.5 = 67.5 \, L$.
Initial amount of water = $7 \times 2.5 = 17.5 \, L$.
Let $x$ liters of water be added to the mixture.
The new ratio of milk to water becomes $3: 1$.
So,$\frac{67.5}{17.5 + x} = \frac{3}{1}$.
$67.5 = 3 \times (17.5 + x)$.
$67.5 = 52.5 + 3x$.
$3x = 67.5 - 52.5 = 15$.
$x = \frac{15}{3} = 5 \, L$.
Therefore,$5 \, L$ of water must be added.

(B) The ratio of the sides is given as $\frac{1}{2}: \frac{1}{3}: \frac{1}{4}$.
To simplify this ratio,multiply each term by the least common multiple $(LCM)$ of the denominators $(2, 3, 4)$,which is $12$.
Ratio $= (\frac{1}{2} \times 12) : (\frac{1}{3} \times 12) : (\frac{1}{4} \times 12) = 6 : 4 : 3$.
Let the sides be $6x, 4x,$ and $3x$.
The perimeter of the triangle is the sum of its sides: $6x + 4x + 3x = 13x$.
Given that the perimeter is $104 \, cm$,we have $13x = 104$.
Solving for $x$: $x = \frac{104}{13} = 8$.
The longest side corresponds to the largest ratio value,which is $6x$.
Longest side $= 6 \times 8 = 48 \, cm$.

(C) Let the number of boys be $300$ and the number of girls be $200$. Total students $= 300 + 200 = 500$.
Number of boys who are scholarship holders $= 20\% \text{ of } 300 = \frac{20}{100} \times 300 = 60$.
Number of girls who are scholarship holders $= 25\% \text{ of } 200 = \frac{25}{100} \times 200 = 50$.
Total students who got the scholarship $= 60 + 50 = 110$.
Total students who did not get the scholarship $= 500 - 110 = 390$.
Percentage of students who did not get the scholarship $= \frac{390}{500} \times 100 = 78\%$.

(C) Let the volumes of the three containers be $3x, 4x,$ and $5x$ respectively.
In the first container,the amount of milk is $\frac{4}{5} \times 3x = \frac{12x}{5}$ and water is $\frac{1}{5} \times 3x = \frac{3x}{5}$.
In the second container,the amount of milk is $\frac{3}{4} \times 4x = 3x$ and water is $\frac{1}{4} \times 4x = x$.
In the third container,the amount of milk is $\frac{5}{7} \times 5x = \frac{25x}{7}$ and water is $\frac{2}{7} \times 5x = \frac{10x}{7}$.
Total milk in the fourth container = $\frac{12x}{5} + 3x + \frac{25x}{7} = \frac{84x + 105x + 125x}{35} = \frac{314x}{35}$.
Total water in the fourth container = $\frac{3x}{5} + x + \frac{10x}{7} = \frac{21x + 35x + 50x}{35} = \frac{106x}{35}$.
The ratio of milk to water is $\frac{314x}{35} : \frac{106x}{35} = 314 : 106 = 157 : 53$.

(D) Given that $x$ varies inversely as the square of $y$,we can write the relation as $x = \frac{K}{y^2}$,where $K$ is a constant of proportionality.
Substitute the given values $x = 1$ and $y = 2$ into the equation:
$1 = \frac{K}{2^2}$
$1 = \frac{K}{4}$
$K = 4$
Now,we have the specific equation $x = \frac{4}{y^2}$.
To find the value of $x$ when $y = 6$,substitute $y = 6$ into the equation:
$x = \frac{4}{6^2}$
$x = \frac{4}{36}$
$x = \frac{1}{9}$

(B) Let the fixed amount be $Rs. x$ and the cost per unit be $Rs. y$.
According to the problem:
$540y + x = 1800$ $(i)$
$620y + x = 2040$ (ii)
Subtracting equation $(i)$ from equation (ii):
$(620y - 540y) + (x - x) = 2040 - 1800$
$80y = 240$
$y = 3$
Substituting $y = 3$ into equation $(i)$:
$540(3) + x = 1800$
$1620 + x = 1800$
$x = 1800 - 1620 = 180$
So,the fixed charge is $Rs. 180$ and the charge per unit is $Rs. 3$.
For $500$ units,the total bill is:
$500 \times 3 + 180 = 1500 + 180 = 1680$
Thus,the bill for $500$ units is $Rs. 1680$.

(C) Let the incomes of $A$ and $B$ be $5x$ and $4x$ respectively.
Let the expenditures of $A$ and $B$ be $3y$ and $2y$ respectively.
We know that $\text{Income} - \text{Expenditure} = \text{Saving}$.
For $A$: $5x - 3y = 1600$ ---$(1)$
For $B$: $4x - 2y = 1600$ ---$(2)$
To solve these equations,multiply equation $(1)$ by $2$ and equation $(2)$ by $3$:
$10x - 6y = 3200$ ---$(3)$
$12x - 6y = 4800$ ---$(4)$
Subtracting equation $(3)$ from $(4)$:
$(12x - 10x) = 4800 - 3200$
$2x = 1600$
$x = 800$
Therefore,the income of $A = 5x = 5 \times 800 = 4000$ $Rs.$

(C) In alloy $A$,the ratio of gold to copper is $7:2$. The total parts are $7+2=9$.
In alloy $B$,the ratio of gold to copper is $7:11$. The total parts are $7+11=18$.
Since equal quantities of alloys $A$ and $B$ are mixed,we must make the total parts equal in both alloys.
To make the total parts equal to $18$,multiply the ratio of alloy $A$ by $2$:
Alloy $A = (7 \times 2) : (2 \times 2) = 14:4$.
Now,alloy $A$ has $14$ parts gold and $4$ parts copper (total $18$ parts).
Alloy $B$ has $7$ parts gold and $11$ parts copper (total $18$ parts).
When mixed to form alloy $C$,the total gold is $14+7=21$ and the total copper is $4+11=15$.
The ratio of gold to copper in alloy $C$ is $21:15$.
Simplifying the ratio by dividing by $3$,we get $7:5$.

(A) Let the present ages of $P$ and $Q$ be $5x$ and $7x$ respectively.
$P$'s age after $6$ years will be $(5x + 6)$.
According to the problem,the difference between $Q$'s present age and $P$'s age after $6$ years is $2$:
$7x - (5x + 6) = 2$
$7x - 5x - 6 = 2$
$2x - 6 = 2$
$2x = 8$
$x = 4$
Now,calculate the present ages:
$P$'s present age $= 5x = 5(4) = 20$ years.
$Q$'s present age $= 7x = 7(4) = 28$ years.
The total of their present ages is $20 + 28 = 48$ years.

(B) Let the present ages be $7x$ and $3x$ for the father and son,respectively.
Given that the product of their ages is $756$,we have:
$7x \times 3x = 756$
$21x^2 = 756$
$x^2 = \frac{756}{21} = 36$
$x = 6$
Therefore,the present ages are:
Father: $7 \times 6 = 42$ years
Son: $3 \times 6 = 18$ years
After $6$ years,their ages will be:
Father: $42 + 6 = 48$ years
Son: $18 + 6 = 24$ years
The ratio of their ages after $6$ years will be $48:24 = 2:1$.

(B) Let the present ages of the three persons be $4x$,$7x$,and $9x$ years respectively.
Eight years ago,each person's age was $8$ years less than their present age.
Therefore,the sum of their ages $8$ years ago was $(4x - 8) + (7x - 8) + (9x - 8) = 56$.
Simplifying the equation: $20x - 24 = 56$.
$20x = 56 + 24 = 80$.
$x = 80 / 20 = 4$.
Now,calculating their present ages:
First person: $4x = 4 \times 4 = 16$ years.
Second person: $7x = 7 \times 4 = 28$ years.
Third person: $9x = 9 \times 4 = 36$ years.
Thus,their present ages are $16, 28, 36$ years.

(C) Let the present ages of the man and his wife be $4x$ and $3x$ respectively.
After $4$ years,their ages will be $(4x + 4)$ and $(3x + 4)$.
According to the problem,the ratio after $4$ years is $9:7$:
$\frac{4x + 4}{3x + 4} = \frac{9}{7}$
$7(4x + 4) = 9(3x + 4)$
$28x + 28 = 27x + 36$
$x = 8$
So,the present ages are: Man $= 4(8) = 32$ years,Wife $= 3(8) = 24$ years.
Let them have been married $y$ years ago. At that time,their ages were $(32 - y)$ and $(24 - y)$.
The ratio at the time of marriage was $5:3$:
$\frac{32 - y}{24 - y} = \frac{5}{3}$
$3(32 - y) = 5(24 - y)$
$96 - 3y = 120 - 5y$
$2y = 24$
$y = 12$
Therefore,they were married $12$ years ago.

(C) Let the present ages of $A$ and $B$ be $5x$ and $3x$ respectively.
According to the problem:
$A$'s age $4$ years ago $= 5x - 4$
$B$'s age $4$ years hence $= 3x + 4$
Given that the ratio is $1:1$,we have:
$5x - 4 = 3x + 4$
$2x = 8$
$x = 4$
Now,calculate the required ages:
$A$'s age $4$ years hence $= 5x + 4 = 5(4) + 4 = 24$ years.
$B$'s age $4$ years ago $= 3x - 4 = 3(4) - 4 = 8$ years.
The ratio between $A$'s age $4$ years hence and $B$'s age $4$ years ago is:
$24 : 8 = 3 : 1$.

(C) Let the mother's present age be $M$ years.
The person's present age is $\frac{2}{5}M$.
After $8$ years,the mother's age will be $M + 8$ and the person's age will be $\frac{2}{5}M + 8$.
According to the problem,after $8$ years,the person's age will be one-half of the mother's age:
$\frac{2}{5}M + 8 = \frac{1}{2}(M + 8)$
Multiply both sides by $10$ to clear the fractions:
$4M + 80 = 5(M + 8)$
$4M + 80 = 5M + 40$
Rearranging the terms to solve for $M$:
$80 - 40 = 5M - 4M$
$M = 40$
Therefore,the mother's present age is $40$ years.

(D) Let the present ages of the father and the son be $f$ and $s$ respectively.
From the given information:
$1$. Four years ago: $(f - 4) = 3(s - 4) \Rightarrow f - 4 = 3s - 12 \Rightarrow f - 3s = -8$ ... $(i)$
$2$. After four years: $(f + 4) + (s + 4) = 64 \Rightarrow f + s + 8 = 64 \Rightarrow f + s = 56$ ... (ii)
From equation (ii),$s = 56 - f$. Substituting this into equation $(i)$:
$f - 3(56 - f) = -8$
$f - 168 + 3f = -8$
$4f = 160$
$f = 40$
Thus,the father's present age is $40$ years.

(A) The present age of Reena is $24$ years and the present age of Usha is $36$ years.
To find the ages $8$ years ago,we subtract $8$ from their present ages:
Age of Usha $8$ years ago $= 36 - 8 = 28$ years.
Age of Reena $8$ years ago $= 24 - 8 = 16$ years.
The ratio of Usha's age to Reena's age $8$ years ago is $28:16$.
Simplifying the ratio by dividing both terms by $4$,we get $7:4$.

(B) Let Purvi's present age be $x$ years.
Then,Anil's present age is $1.5x = \frac{3}{2}x$ years.
After $8$ years,Anil's age will be $(\frac{3}{2}x + 8)$ and Purvi's age will be $(x + 8)$.
According to the problem,the ratio of their ages after $8$ years is $25:18$:
$\frac{\frac{3}{2}x + 8}{x + 8} = \frac{25}{18}$
Multiply both sides by $18(x + 8)$:
$18(\frac{3}{2}x + 8) = 25(x + 8)$
$27x + 144 = 25x + 200$
$27x - 25x = 200 - 144$
$2x = 56$
$x = 28$
Therefore,Purvi's present age is $28$ years.

(C) Let the ages of Meena and Seema be $7x$ and $8x$ respectively.
According to the problem,the difference between their ages is $3 \ yr$.
So,$8x - 7x = 3$,which gives $x = 3$.
The sum of their ages is $7x + 8x = 15x$.
Substituting the value of $x$,we get $15 \times 3 = 45 \ yr$.

(A) Rehana's age after $7 \text{ yr}$ is $85 \text{ yr}$.
Therefore,Rehana's present age $= 85 - 7 = 78 \text{ yr}$.
Wasim is $12 \text{ yr}$ younger than Rehana,so Wasim's present age $= 78 - 12 = 66 \text{ yr}$.
The ratio of the present age of Manoj to Wasim is $3:11$.
Let Manoj's age be $3x$ and Wasim's age be $11x$.
Given $11x = 66$,so $x = 6$.
Manoj's present age $= 3 \times 6 = 18 \text{ yr}$.
Manoj's father is $25 \text{ yr}$ older than Manoj.
Therefore,the present age of Manoj's father $= 18 + 25 = 43 \text{ yr}$.

(B) Let the present ages of Indira and Lizzy be $3x$ and $8x$ years,respectively.
According to the problem,Indira's age after $8 \text{ yr}$ will be $20 \text{ yr}$.
So,$3x + 8 = 20$.
$3x = 20 - 8 = 12$.
$x = 4$.
Therefore,the present age of Lizzy is $8x = 8 \times 4 = 32 \text{ yr}$.
Lizzy's age $5 \text{ yr}$ ago was $32 - 5 = 27 \text{ yr}$.

(D) Let Sarita's present age be $s$ years.
Then,Kavita's present age is $2s$ years.
After $8$ years,Sarita's age will be $(s + 8)$ years and Kavita's age will be $(2s + 8)$ years.
According to the problem,the ratio of their ages after $8$ years will be $22:13$.
So,$\frac{2s + 8}{s + 8} = \frac{22}{13}$.
Cross-multiplying,we get: $13(2s + 8) = 22(s + 8)$.
$26s + 104 = 22s + 176$.
$26s - 22s = 176 - 104$.
$4s = 72$.
$s = 18$.
Therefore,Kavita's present age is $2s = 2 \times 18 = 36$ years.

(C) Let the ages of $A$ and $B$ $10 \ yr$ ago be $13x$ and $17x$ respectively.
Their present ages are $(13x + 10)$ and $(17x + 10)$.
After $17 \ yr$ from now,their ages will be $(13x + 10 + 17)$ and $(17x + 10 + 17)$,which simplifies to $(13x + 27)$ and $(17x + 27)$.
According to the problem,the ratio after $17 \ yr$ is $10: 11$:
$\frac{13x + 27}{17x + 27} = \frac{10}{11}$
$11(13x + 27) = 10(17x + 27)$
$143x + 297 = 170x + 270$
$297 - 270 = 170x - 143x$
$27 = 27x$
$x = 1$
Therefore,the present age of $B = 17x + 10 = 17(1) + 10 = 27 \ yr$.

(D) Let the number of $10$-paise coins be $17k$ and the number of $25$-paise coins be $6k$.
The total value of $10$-paise coins is $17k \times 0.10 = 1.7k$ rupees.
The total value of $25$-paise coins is $6k \times 0.25 = 1.5k$ rupees.
Given that the total money is $Rs. 112$,we have:
$1.7k + 1.5k = 112$
$3.2k = 112$
$k = \frac{112}{3.2} = 35$
Therefore,the number of $10$-paise coins is $17 \times 35 = 595$.

(C) Let the fraction of copper in the first alloy be $\frac{1}{3}$ and in the second alloy be $\frac{2}{5}$.
The fraction of copper in the final alloy is $\frac{3}{8}$.
Using the rule of alligation:
Difference for the first alloy: $\frac{2}{5} - \frac{3}{8} = \frac{16-15}{40} = \frac{1}{40}$.
Difference for the second alloy: $\frac{3}{8} - \frac{1}{3} = \frac{9-8}{24} = \frac{1}{24}$.
The required ratio of the two alloys is $\frac{1}{40} : \frac{1}{24} = 24 : 40 = 3 : 5$.
Thus,$3$ and $5$ parts of the two alloys must be taken.

(D) In the first alloy of $10 \text{ kg}$ with ratio $4:1$,the amount of gold is $\frac{4}{5} \times 10 = 8 \text{ kg}$ and silver is $\frac{1}{5} \times 10 = 2 \text{ kg}$.
In the second alloy of $16 \text{ kg}$ with ratio $1:3$,the amount of gold is $\frac{1}{4} \times 16 = 4 \text{ kg}$ and silver is $\frac{3}{4} \times 16 = 12 \text{ kg}$.
Let the amount of pure gold added be $x \text{ kg}$.
Total gold in the new alloy $= 8 + 4 + x = 12 + x \text{ kg}$.
Total silver in the new alloy $= 2 + 12 = 14 \text{ kg}$.
The ratio of gold to silver in the new alloy is $3:2$,so $\frac{12 + x}{14} = \frac{3}{2}$.
$2(12 + x) = 3 \times 14 \Rightarrow 24 + 2x = 42 \Rightarrow 2x = 18 \Rightarrow x = 9 \text{ kg}$.
The total weight of the new alloy $= 10 + 16 + 9 = 35 \text{ kg}$.

(B) Given: $A = \frac{1}{2}(B + C) \Rightarrow B + C = 2A$.
Since $A + B + C = 5625$,we substitute $(B + C) = 2A$: $A + 2A = 5625 \Rightarrow 3A = 5625 \Rightarrow A = 1875$.
Also given: $B = \frac{1}{4}(A + C) \Rightarrow A + C = 4B$.
Since $A + B + C = 5625$,we substitute $(A + C) = 4B$: $4B + B = 5625 \Rightarrow 5B = 5625 \Rightarrow B = 1125$.
The sum of $A$'s and $B$'s share is $A + B = 1875 + 1125 = 3000$.
Thus,the total share of $A$ and $B$ is $Rs. 3000$.

(D) Total expenditure for the first $8$ months $= 8 \times 2310 = 18480$.
Total expenditure for the next $4$ months $= 4 \times 1800 = 7200$.
Total expenditure for the year $= 18480 + 7200 = 25680$.
Average expenditure $= \frac{25680}{12} = 2140$.
Total income for the year $=$ Total expenditure $-$ Loan amount $= 25680 - 1680 = 24000$.
Average income $= \frac{24000}{12} = 2000$.
Ratio of average income to average expenditure $= \frac{2000}{2140} = \frac{100}{107} = 100:107$.

(A) Let the initial expense be $5x$ and initial savings be $3x$.
Initial income = Expense + Savings = $5x + 3x = 8x$.
New expense = $5x + 60\% \text{ of } 5x = 5x + 3x = 8x$.
New income = $8x + 25\% \text{ of } 8x = 8x + 2x = 10x$.
New savings = New income - New expense = $10x - 8x = 2x$.
Decrease in savings = Initial savings - New savings = $3x - 2x = x$.
Given that the decrease is $Rs. 3500$,so $x = 3500$.
The increased income is $10x = 10 \times 3500 = Rs. 35000$.

(A) Let the number of $Rs. 1$,$50 \text{ paise}$,and $25 \text{ paise}$ coins be $12x$,$10x$,and $7x$ respectively.
The value of $12x$ coins of $Rs. 1$ is $12x \times 1 = 12x$ rupees.
The value of $10x$ coins of $50 \text{ paise}$ is $10x \times 0.5 = 5x$ rupees.
The value of $7x$ coins of $25 \text{ paise}$ is $7x \times 0.25 = 1.75x$ rupees.
According to the question,the total value is $Rs. 75$:
$12x + 5x + 1.75x = 75$
$18.75x = 75$
$x = \frac{75}{18.75} = 4$
The number of $25 \text{ paise}$ coins is $7x = 7 \times 4 = 28$.

(D) Let $H$ be the cost of a horse,$D$ be the cost of a dog,$O$ be the cost of an ox,$S$ be the cost of a sheep,and $G$ be the cost of a goat.
Given equations:
$2H = 5D \implies D = \frac{2}{5}H$
$6D = 8O \implies O = \frac{6}{8}D = \frac{3}{4}D$
$10O = 50S \implies S = \frac{10}{50}O = \frac{1}{5}O$
$14S = 9G \implies G = \frac{14}{9}S$
Substituting the values back to find $H$ in terms of $G$:
$S = \frac{1}{5} \times (\frac{3}{4}D) = \frac{3}{20}D$
$S = \frac{3}{20} \times (\frac{2}{5}H) = \frac{6}{100}H = \frac{3}{50}H$
$G = \frac{14}{9} \times (\frac{3}{50}H) = \frac{14}{150}H = \frac{7}{75}H$
Given $G = 700$,so $700 = \frac{7}{75}H$
$H = 700 \times \frac{75}{7} = 100 \times 75 = 7500$.
Therefore,the cost of one horse is $Rs. 7500$.

(C) Let the shares of $A$,$B$,and $C$ after the decrease be $9x$,$13x$,and $8x$ respectively.
According to the problem,the total amount before the decrease was $Rs. 2186$.
Therefore,the equation is: $(9x + 26) + (13x + 28) + (8x + 32) = 2186$.
Combining the terms: $30x + 86 = 2186$.
$30x = 2186 - 86$.
$30x = 2100$.
$x = 2100 / 30 = 70$.
The amount given to $A$ is $9x + 26$.
Substituting $x = 70$: $9(70) + 26 = 630 + 26 = 656$.
Thus,the amount given to $A$ is $Rs. 656$.

(A) Let the cost of Maruti and Figo a year ago be $3x$ and $4x$ respectively.
Given the ratio of present to past year cost for Maruti is $5:4$,so present cost of Maruti $(PM)$ $= \frac{5}{4} \times 3x = 3.75x$.
Given the ratio of present to past year cost for Figo is $3:2$,so present cost of Figo $(PF)$ $= \frac{3}{2} \times 4x = 6x$.
The sum of their present costs is $7.8$ lacs:
$3.75x + 6x = 7.8$
$9.75x = 7.8$
$x = \frac{7.8}{9.75} = 0.8$
The cost of Figo a year ago was $4x = 4 \times 0.8 = 3.2$ lacs.

(B) Let the salaries of Ram and Shyam last year be $3x$ and $5x$ respectively.
Ram's salary ratio (last year : present year) $= 2:3$. Thus,Ram's present salary $= 3x \times (3/2) = 4.5x$.
Shyam's salary ratio (last year : present year) $= 4:5$. Thus,Shyam's present salary $= 5x \times (5/4) = 6.25x$.
The total present salary is $4.5x + 6.25x = 10.75x$.
Given $10.75x = 8600$,we find $x = 8600 / 10.75 = 800$.
Ram's present salary $= 4.5 \times 800 = Rs. 3600$.

(D) Let the initial height of each candle be $x$ and the time elapsed be $T$ hours.
The rate of burning for the first candle is $\frac{x}{8}$ per hour.
The rate of burning for the second candle is $\frac{x}{6}$ per hour.
After $T$ hours,the remaining height of the first candle is $x - \frac{Tx}{8} = x(1 - \frac{T}{8})$.
The remaining height of the second candle is $x - \frac{Tx}{6} = x(1 - \frac{T}{6})$.
According to the problem,the ratio of the heights is $2:1$:
$\frac{x(1 - T/8)}{x(1 - T/6)} = \frac{2}{1}$
$\frac{1 - T/8}{1 - T/6} = 2$
$1 - \frac{T}{8} = 2(1 - \frac{T}{6})$
$1 - \frac{T}{8} = 2 - \frac{T}{3}$
Rearranging the terms to solve for $T$:
$\frac{T}{3} - \frac{T}{8} = 2 - 1$
$\frac{8T - 3T}{24} = 1$
$\frac{5T}{24} = 1$
$T = \frac{24}{5} = 4.8 \text{ hours}$.
Since $0.8 \text{ hours} = 0.8 \times 60 \text{ minutes} = 48 \text{ minutes}$,the time is $4 \text{ hours } 48 \text{ minutes}$.

(A) Let the incomes of Ram,Shyam,and Mohan be $x$,$y$,and $z$ respectively.
Ram's saving $= (100 - 25) \% \text{ of } x = 75 \% \text{ of } x = \frac{3}{4}x$.
Shyam's saving $= (100 - 20) \% \text{ of } y = 80 \% \text{ of } y = \frac{4}{5}y$.
Mohan's saving $= (100 - 50) \% \text{ of } z = 50 \% \text{ of } z = \frac{1}{2}z$.
Given the ratio of savings is $9:8:4$,let their savings be $9k$,$8k$,and $4k$ respectively.
Equating the savings:
$\frac{3}{4}x = 9k \Rightarrow x = 12k$.
$\frac{4}{5}y = 8k \Rightarrow y = 10k$.
$\frac{1}{2}z = 4k \Rightarrow z = 8k$.
Given the total income is $Rs. 450$:
$x + y + z = 450 \Rightarrow 12k + 10k + 8k = 450$.
$30k = 450 \Rightarrow k = 15$.
Ram's income $= 12k = 12 \times 15 = Rs. 180$.

(A) Initial quantity of milk $= 1000 \, \text{litres}$.
Step $1$: After removing $100 \, \text{litres}$ of milk and adding $100 \, \text{litres}$ of water, the remaining milk is $900 \, \text{litres}$. The ratio of milk to water is $900:100 = 9:1$.
Step $2$: After removing $200 \, \text{litres}$ of the mixture, the amount of milk removed is $200 \times (9/10) = 180 \, \text{litres}$. Remaining milk $= 900 - 180 = 720 \, \text{litres}$. Total mixture volume remains $1000 \, \text{litres}$ (as $200 \, \text{litres}$ of water is added).
Step $3$: The new ratio of milk to water is $720:280 = 18:7$. The total parts $= 18 + 7 = 25$.
Step $4$: After removing $400 \, \text{litres}$ of the mixture, the amount of milk removed is $400 \times (18/25) = 16 \times 18 = 288 \, \text{litres}$.
Final quantity of milk $= 720 - 288 = 432 \, \text{litres}$.

(B) Let the total land area be $x$ and the total water area be $2x$. The total area of the Earth is $3x$.
The area of each hemisphere (Northern and Southern) is half of the total area,which is $\frac{3x}{2}$.
In the Northern Hemisphere,the ratio of land to water is $2:3$. The total parts are $2+3=5$.
Land in the Northern Hemisphere $= \frac{2}{5} \times \frac{3x}{2} = \frac{3x}{5}$.
Water in the Northern Hemisphere $= \frac{3}{5} \times \frac{3x}{2} = \frac{9x}{10}$.
Now,calculate the land and water in the Southern Hemisphere:
Land in the Southern Hemisphere $= \text{Total Land} - \text{Land in Northern Hemisphere} = x - \frac{3x}{5} = \frac{2x}{5}$.
Water in the Southern Hemisphere $= \text{Total Water} - \text{Water in Northern Hemisphere} = 2x - \frac{9x}{10} = \frac{11x}{10}$.
The required ratio of land to water in the Southern Hemisphere is $\frac{2x/5}{11x/10} = \frac{2}{5} \times \frac{10}{11} = \frac{4}{11}$,which is $4:11$.

(A) Given, the ratio of incomes of $A, B, C$ is $3: 7: 4$.
Income of $A = Rs. 2400$.
Since $3$ units $= 2400$, $1$ unit $= 800$.
Income of $B = 7 \times 800 = Rs. 5600$.
Income of $C = 4 \times 800 = Rs. 3200$.
Expenditure of $A = \text{Income} - \text{Saving} = 2400 - 300 = Rs. 2100$.
The ratio of expenditures is $4: 3: 5$.
Since $4$ units $= 2100$, $1$ unit $= 525$.
Expenditure of $B = 3 \times 525 = Rs. 1575$.
Expenditure of $C = 5 \times 525 = Rs. 2625$.
Saving of $B = \text{Income of } B - \text{Expenditure of } B = 5600 - 1575 = Rs. 4025$.
Saving of $C = \text{Income of } C - \text{Expenditure of } C = 3200 - 2625 = Rs. 575$.

(B) Let the initial quantity of wine be $3x$ and water be $x$. The total volume of the mixture is $4x$.
Let $y$ be the amount of mixture taken out. The amount of wine removed is $\frac{3}{4}y$ and the amount of water removed is $\frac{1}{4}y$.
After adding $y$ litres of water,the new amount of wine is $3x - \frac{3}{4}y$.
The new amount of water is $x - \frac{1}{4}y + y = x + \frac{3}{4}y$.
According to the problem,the new ratio is $1:1$,so:
$3x - \frac{3}{4}y = x + \frac{3}{4}y$
$2x = \frac{6}{4}y$
$2x = \frac{3}{2}y$
$y = \frac{4}{3}x$
Since the total initial volume is $4x$,the fraction of the mixture replaced is $\frac{y}{4x} = \frac{4x/3}{4x} = \frac{1}{3}$.

(A) Let the total volume of the actual mixture be $V$ litres.
Quantity of petrol $= 99 \text{ litres}$.
Concentration of petrol in the actual mixture $= \frac{99}{V} \times 100 \%$.
In the new mixture,the total volume is $(V - 198) \text{ litres}$.
Concentration of petrol in the new mixture $= \frac{99}{V - 198} \times 100 \%$.
According to the problem,the concentration in the actual mixture is $13.33 \%$ (which is $\frac{40}{3} \%$) less than the new mixture:
$\frac{99}{V - 198} \times 100 - \frac{99}{V} \times 100 = \frac{40}{3}$.
Dividing by $100$:
$99 \left( \frac{1}{V - 198} - \frac{1}{V} \right) = \frac{40}{300} = \frac{2}{15}$.
$99 \left( \frac{V - (V - 198)}{V(V - 198)} \right) = \frac{2}{15}$.
$99 \times \frac{198}{V(V - 198)} = \frac{2}{15}$.
$V(V - 198) = 99 \times 99 \times 15 = 147015$.
$V^2 - 198V - 147015 = 0$.
Solving the quadratic equation,we get $V = 495$.
Actual concentration $= \frac{99}{495} \times 100 = \frac{1}{5} \times 100 = 20 \%$.

(A) Given the ratio $A:B = 6:5$,we can write $B = \frac{5}{6}A$.
Given $B:E = 2:3$,we can write $E = \frac{3}{2}B = \frac{3}{2} \times \frac{5}{6}A = \frac{5}{4}A$.
We are given $E - A = 3$. Substituting $E = \frac{5}{4}A$,we get $\frac{5}{4}A - A = 3$,which simplifies to $\frac{1}{4}A = 3$,so $A = 12$.
Using $A = 12$,we find $B = \frac{5}{6} \times 12 = 10$ and $E = 15$.
We are given that the age of $F$ is less than $A$ $(12)$ but greater than $B$ $(10)$. Since all ages are integers,$F$ must be $11$.
Therefore,the ratio of the ages of $A$ and $F$ is $12:11$.