Ratio and Proportion Questions in English

(D) Let the volumes of the three vessels be $2x, 3x$,and $5x$ respectively.
In the first vessel (volume $2x$),the ratio of water to milk is $1: 3$.
Water $= 2x \times \frac{1}{4} = 0.5x$,Milk $= 2x \times \frac{3}{4} = 1.5x$.
In the second vessel (volume $3x$),the ratio of water to milk is $2: 3$.
Water $= 3x \times \frac{2}{5} = 1.2x$,Milk $= 3x \times \frac{3}{5} = 1.8x$.
In the third vessel (volume $5x$),the ratio of water to milk is $2: 5$.
Water $= 5x \times \frac{2}{7} = \frac{10}{7}x \approx 1.428x$,Milk $= 5x \times \frac{5}{7} = \frac{25}{7}x \approx 3.571x$.
Total water $= 0.5x + 1.2x + \frac{10}{7}x = 1.7x + \frac{10}{7}x = \frac{11.9x + 10x}{7} = \frac{21.9x}{7} = \frac{219x}{70}$.
Total milk $= 1.5x + 1.8x + \frac{25}{7}x = 3.3x + \frac{25}{7}x = \frac{23.1x + 25x}{7} = \frac{48.1x}{7} = \frac{481x}{70}$.
Ratio of milk to water $= \frac{481x/70}{219x/70} = 481: 219$.

(C) Let the price of the refrigerator be $5x$ and the price of the television set be $3x$.
According to the problem,the refrigerator costs $Rs. 5500$ more than the television set.
So,$5x - 3x = 5500$.
$2x = 5500$.
$x = 2750$.
Therefore,the price of the refrigerator is $5x = 5 \times 2750 = 13750$.

(D) Let the shares of $P, Q$ and $R$ be $2x, 7x$ and $9x$ respectively.
According to the problem,the sum of $P$'s and $Q$'s share is equal to $R$'s share,which gives $2x + 7x = 9x$,i.e.,$9x = 9x$.
This condition is satisfied for any value of $x$.
Since the total amount of money is not specified,we cannot determine the exact numerical value of the shares or their difference.
Therefore,the information provided is inadequate to find the difference between the shares of $P$ and $Q$.

(B) Given that $\frac{2}{3} A = \frac{4}{5} B$.
To find the ratio $A : B$,we rearrange the equation:
$\frac{A}{B} = \frac{4}{5} \times \frac{3}{2}$.
$\frac{A}{B} = \frac{12}{10} = \frac{6}{5}$.
Therefore,$A : B = 6 : 5$.

(C) Let the three numbers be $3x$,$4x$,and $5x$.
According to the problem,the sum of the largest $(5x)$ and the smallest $(3x)$ is equal to the sum of the second number $(4x)$ and $52$.
So,$5x + 3x = 4x + 52$.
$8x = 4x + 52$.
$8x - 4x = 52$.
$4x = 52$.
$x = 13$.
The smallest number is $3x = 3 \times 13 = 39$.

(A) Given the ratio $x : y = 2 : 1$,we can write this as $\frac{x}{y} = \frac{2}{1}$.
To find the value of $\frac{x^{2} - y^{2}}{x^{2} + y^{2}}$,we can divide both the numerator and the denominator by $y^{2}$:
$\frac{x^{2} - y^{2}}{x^{2} + y^{2}} = \frac{(\frac{x}{y})^{2} - 1}{(\frac{x}{y})^{2} + 1}$.
Substituting the value $\frac{x}{y} = 2$:
$\frac{2^{2} - 1}{2^{2} + 1} = \frac{4 - 1}{4 + 1} = \frac{3}{5}$.
Therefore,the ratio is $3 : 5$.

(A) Given the ratio $a: b: c = 3: 4: 7.$
Let $a = 3k, b = 4k,$ and $c = 7k$ for some constant $k \neq 0.$
We need to find the ratio $(a+b+c): c.$
Substitute the values of $a, b,$ and $c$ into the expression:
$\frac{a+b+c}{c} = \frac{3k + 4k + 7k}{7k}$
$= \frac{14k}{7k}$
$= \frac{2}{1}$
Therefore,the ratio is $2: 1.$

(D) The given ratio of the three numbers is $\frac{1}{2}: \frac{2}{3}: \frac{3}{4}$.
To simplify the ratio,multiply each term by the least common multiple $(LCM)$ of the denominators $(2, 3, 4)$,which is $12$.
Ratio $= (\frac{1}{2} \times 12) : (\frac{2}{3} \times 12) : (\frac{3}{4} \times 12) = 6: 8: 9$.
Let the numbers be $6x, 8x,$ and $9x$.
The greatest number is $9x$ and the smallest number is $6x$.
According to the problem,the difference between the greatest and smallest numbers is $36$:
$9x - 6x = 36$
$3x = 36$
$x = 12$.
Substituting the value of $x$ back into the expressions:
First number $= 6 \times 12 = 72$.
Second number $= 8 \times 12 = 96$.
Third number $= 9 \times 12 = 108$.
Therefore,the numbers are $72, 96, 108$.

(C) Given that $\frac{m}{n} = \frac{3}{2}$.
We need to find the value of $\frac{4m + 5n}{4m - 5n}$.
Divide both the numerator and the denominator by $n$:
$\frac{4(\frac{m}{n}) + 5}{4(\frac{m}{n}) - 5}$.
Substitute $\frac{m}{n} = \frac{3}{2}$ into the expression:
$= \frac{4(\frac{3}{2}) + 5}{4(\frac{3}{2}) - 5}$
$= \frac{6 + 5}{6 - 5}$
$= \frac{11}{1} = 11 : 1$.

(A) Let the quantity of milk be $M$ litres and the quantity of water be $W$ litres. The cost price of $1 \text{ L}$ of pure milk is $Rs. 10$,so the cost price of $M$ litres of milk is $10M$.
Since water is assumed to be free,the total cost price of the mixture is $10M$.
The total volume of the mixture is $(M + W)$ litres.
The selling price of the mixture is $9$ per litre,so the total selling price is $9(M + W)$.
The profit percentage is given as $20\%$.
Using the formula: $\text{Selling Price} = \text{Cost Price} \times (1 + \frac{\text{Profit}\%}{100})$
$9(M + W) = 10M \times (1 + \frac{20}{100})$
$9(M + W) = 10M \times 1.2$
$9M + 9W = 12M$
$9W = 3M$
$\frac{M}{W} = \frac{9}{3} = \frac{3}{1}$
Therefore,the ratio of milk to water is $3:1$.

(C) The given ratio is $1: \frac{1}{3}: \frac{1}{6}$.
To simplify this ratio,multiply each term by the least common multiple $(LCM)$ of the denominators ($3$ and $6$),which is $6$.
Ratio $= (1 \times 6) : (\frac{1}{3} \times 6) : (\frac{1}{6} \times 6) = 6: 2: 1$.
The sum of the parts is $6 + 2 + 1 = 9$.
The middle part corresponds to the value $2$.
Therefore,the middle part $= \frac{2}{9} \times 78 = \frac{2 \times 26}{3} = \frac{52}{3} = 17 \frac{1}{3}$.

(A) Let the two numbers be $x$ and $y$ respectively.
According to the problem:
$x + y = 25$ (Equation $1$)
$x - y = 20$ (Equation $2$)
Adding Equation $1$ and Equation $2$:
$(x + y) + (x - y) = 25 + 20$
$2x = 45$
$x = 22.5$ or $\frac{45}{2}$
Substituting $x$ in Equation $1$:
$22.5 + y = 25$
$y = 25 - 22.5 = 2.5$ or $\frac{5}{2}$
The ratio of the two numbers is $\frac{x}{y} = \frac{45/2}{5/2} = \frac{45}{5} = 9:1$.

(A) Let the first part be $x$. Then the second part is $(94 - x)$.
According to the problem,the ratio of the fifth part of the first to the eighth part of the second is $3:4$.
$\frac{x/5}{(94 - x)/8} = \frac{3}{4}$
$\frac{x}{5} \times \frac{8}{94 - x} = \frac{3}{4}$
$\frac{8x}{5(94 - x)} = \frac{3}{4}$
$32x = 15(94 - x)$
$32x = 1410 - 15x$
$47x = 1410$
$x = \frac{1410}{47} = 30$
Therefore,the first part is $30$.

(C) Let the two numbers be $4x$ and $7x$.
According to the problem,if each number is increased by $4$,the ratio becomes $3:5$.
So,the equation is: $\frac{4x + 4}{7x + 4} = \frac{3}{5}$.
By cross-multiplying,we get: $5(4x + 4) = 3(7x + 4)$.
$20x + 20 = 21x + 12$.
Rearranging the terms to solve for $x$: $20 - 12 = 21x - 20x$.
$x = 8$.
The two numbers are $4 \times 8 = 32$ and $7 \times 8 = 56$.
The larger number is $56$.

(C) Let the three numbers be $5x$,$6x$,and $7x$ respectively.
According to the problem,the product of these numbers is $5670$.
So,$(5x) \times (6x) \times (7x) = 5670$.
$210x^3 = 5670$.
$x^3 = \frac{5670}{210} = 27$.
$x = \sqrt[3]{27} = 3$.
The greatest number is $7x = 7 \times 3 = 21$.

(C) Let the three numbers be $3x$,$2x$,and $5x$.
According to the problem,the sum of their squares is $1862$.
$(3x)^2 + (2x)^2 + (5x)^2 = 1862$
$9x^2 + 4x^2 + 25x^2 = 1862$
$38x^2 = 1862$
$x^2 = \frac{1862}{38} = 49$
$x = 7$
Since the numbers are $3x$,$2x$,and $5x$,the smallest number is $2x$.
Smallest number $= 2 \times 7 = 14$.

(A) Let the two numbers be $10x$ and $7x$ respectively.
Given that the difference between the numbers is $105$.
So,$10x - 7x = 105$.
$3x = 105$.
$x = 105 / 3 = 35$.
The sum of the two numbers is $10x + 7x = 17x$.
Substituting the value of $x$,we get $17 \times 35 = 595$.
Therefore,the sum of the numbers is $595$.

(D) Let the total amount be $A + B + C + D = 60$.
Given:
$A = \frac{1}{2}(B + C + D) \Rightarrow 2A = B + C + D$. Adding $A$ to both sides,$3A = A + B + C + D = 60 \Rightarrow A = 20$.
$B = \frac{1}{3}(A + C + D) \Rightarrow 3B = A + C + D$. Adding $B$ to both sides,$4B = A + B + C + D = 60 \Rightarrow B = 15$.
$C = \frac{1}{4}(A + B + D) \Rightarrow 4C = A + B + D$. Adding $C$ to both sides,$5C = A + B + C + D = 60 \Rightarrow C = 12$.
Now,$D = 60 - (A + B + C) = 60 - (20 + 15 + 12) = 60 - 47 = 13$.
Thus,the amount paid by $D$ is $Rs. 13$.

(D) Let the annual incomes of Amit and Varun be $3x$ and $2x$ respectively.
Let the annual expenditures of Amit and Varun be $5y$ and $3y$ respectively.
We know that $\text{Income} - \text{Expenditure} = \text{Savings}.$
For Amit: $3x - 5y = 1000$ (Equation $1$)
For Varun: $2x - 3y = 1000$ (Equation $2$)
Since both save the same amount,we can equate the two expressions:
$3x - 5y = 2x - 3y$
$3x - 2x = 5y - 3y$
$x = 2y$
Substitute $x = 2y$ into Equation $1$:
$3(2y) - 5y = 1000$
$6y - 5y = 1000$
$y = 1000$
Now,find $x$:
$x = 2(1000) = 2000$
The annual income of Amit is $3x = 3 \times 2000 = Rs. 6000.$

(B) Let the annual incomes of $A$,$B$,and $C$ be $x$,$3x$,and $7x$ respectively.
According to the problem,the total annual income of $A$ and $C$ is $Rs. 800,000$.
So,$x + 7x = 800,000$.
$8x = 800,000$.
$x = 100,000$.
Therefore,the annual income of $B$ is $3x = 3 \times 100,000 = Rs. 300,000$.
The monthly salary of $B$ is the annual income divided by $12$.
Monthly salary of $B = \frac{300,000}{12} = Rs. 25,000$.

(B) Let the monthly salaries of $A, B$,and $C$ be $2x, 3x$,and $5x$ respectively.
According to the problem,$C$'s monthly salary is $Rs. 12000$ more than $A$'s monthly salary.
So,$5x - 2x = 12000$.
$3x = 12000$.
$x = 4000$.
Now,$B$'s monthly salary is $3x = 3 \times 4000 = 12000$.
Since there are $12$ months in a year,$B$'s annual salary is $12 \times 12000 = 144000$.

(D) Let the incomes of $A, B$ and $C$ be $7x, 9x$ and $12x$ respectively.
Let the spendings of $A, B$ and $C$ be $8y, 9y$ and $15y$ respectively.
Given that $A$ saves $\frac{1}{4}$th of his income,so his saving is $\frac{7x}{4}$.
Since $\text{Saving} = \text{Income} - \text{Spending}$,for $A$ we have: $7x - 8y = \frac{7x}{4}$.
$28x - 32y = 7x \Rightarrow 21x = 32y \Rightarrow y = \frac{21x}{32}$.
Now,calculate the savings for $B$ and $C$:
$B$'s saving $= 9x - 9y = 9x - 9(\frac{21x}{32}) = 9x - \frac{189x}{32} = \frac{288x - 189x}{32} = \frac{99x}{32}$.
$C$'s saving $= 12x - 15y = 12x - 15(\frac{21x}{32}) = 12x - \frac{315x}{32} = \frac{384x - 315x}{32} = \frac{69x}{32}$.
Ratio of savings of $A, B$ and $C = \frac{7x}{4} : \frac{99x}{32} : \frac{69x}{32}$.
Multiply by $32$ to simplify: $(\frac{7x}{4} \times 32) : (\frac{99x}{32} \times 32) : (\frac{69x}{32} \times 32) = 56x : 99x : 69x$.
Thus,the ratio is $56: 99: 69$.

(A) Let the income be $10x$ and the expenditure be $7x$.
Given that the expenditure is $Rs. 10500$.
So,$7x = 10500$.
$x = 10500 / 7 = 1500$.
Savings are calculated as Income - Expenditure.
Savings $= 10x - 7x = 3x$.
Substituting the value of $x$,Savings $= 3 \times 1500 = 4500$.
Therefore,the savings of the family is $Rs. 4500$.

(C) Let the income of $P$ and $Q$ be $3x$ and $4x$ respectively.
Let the expenditure of $P$ and $Q$ be $2y$ and $3y$ respectively.
We know that $\text{Income} - \text{Expenditure} = \text{Savings}$.
For $P$: $3x - 2y = 6000$ (Equation $1$)
For $Q$: $4x - 3y = 6000$ (Equation $2$)
From Equation $1$,$2y = 3x - 6000$,so $y = 1.5x - 3000$.
Substitute $y$ into Equation $2$: $4x - 3(1.5x - 3000) = 6000$.
$4x - 4.5x + 9000 = 6000$.
$-0.5x = -3000$.
$x = 6000$.
The income of $P$ is $3x = 3 \times 6000 = 18000$ $Rs.$

(C) Let the number of students in the three classes be $2x$,$3x$,and $5x$ respectively.
According to the problem,when $20$ students are added to each class,the new ratio becomes $4:5:7$.
Taking the first two classes:
$\frac{2x + 20}{3x + 20} = \frac{4}{5}$
Cross-multiplying gives:
$5(2x + 20) = 4(3x + 20)$
$10x + 100 = 12x + 80$
$100 - 80 = 12x - 10x$
$20 = 2x$
$x = 10$
The total number of students before the increase is the sum of the initial ratios:
Total $= 2x + 3x + 5x = 10x$
Substituting $x = 10$:
Total $= 10 \times 10 = 100$.

(D) Let the number to be added or subtracted be $x$.
According to the problem,the ratio becomes $1:2$ after the operation:
$\frac{17 + x}{24 + x} = \frac{1}{2}$
By cross-multiplying,we get:
$2(17 + x) = 1(24 + x)$
$34 + 2x = 24 + x$
Rearranging the terms to solve for $x$:
$2x - x = 24 - 34$
$x = -10$
Since the result is $-10$,it means that $10$ should be subtracted from each term.

(A) Let the initial number of students in the three classes be $2x$,$3x$,and $4x$ respectively.
According to the problem,when $12$ students are added to each class,the new ratio becomes $8:11:14$.
We can set up the equation using the first two classes: $\frac{2x + 12}{3x + 12} = \frac{8}{11}$.
Cross-multiplying gives: $11(2x + 12) = 8(3x + 12)$.
$22x + 132 = 24x + 96$.
$2x = 36$,which means $x = 18$.
The total number of students in the beginning was $2x + 3x + 4x = 9x$.
Substituting $x = 18$,we get $9 \times 18 = 162$.

(B) Let the two numbers be $3x$ and $5x$.
According to the problem,if $9$ is subtracted from each,the ratio becomes $12:23$.
So,$\frac{3x - 9}{5x - 9} = \frac{12}{23}$.
Cross-multiplying gives: $23(3x - 9) = 12(5x - 9)$.
$69x - 207 = 60x - 108$.
$69x - 60x = 207 - 108$.
$9x = 99$.
$x = 11$.
The smaller number is $3x = 3 \times 11 = 33$.

(A) Total number of students $= 504$.
Ratio of boys to girls $= 13: 11$.
Sum of ratio terms $= 13 + 11 = 24$.
Number of boys $= \frac{13}{24} \times 504 = 13 \times 21 = 273$.
Number of girls $= \frac{11}{24} \times 504 = 11 \times 21 = 231$.
If $12$ more girls are admitted,the new number of girls $= 231 + 12 = 243$.
The number of boys remains $273$.
New ratio of boys to girls $= 273: 243$.
Dividing both by $3$,we get $\frac{273}{3} : \frac{243}{3} = 91: 81$.

(B) Given that for every $100$ paise ($1$ rupee) that $A$ gets,$B$ gets $90$ paise.
So,the ratio $A:B = 100:90 = 10:9$.
Also,for every $100$ paise ($1$ rupee) that $B$ gets,$C$ gets $110$ paise.
So,the ratio $B:C = 100:110 = 10:11$.
To find the combined ratio $A:B:C$,we equate the $B$ term in both ratios.
Multiply $A:B$ by $10$ and $B:C$ by $9$:
$A:B = 100:90$
$B:C = 90:99$
Thus,$A:B:C = 100:90:99$.
The sum of the ratio parts is $100 + 90 + 99 = 289$.
$B$'s share $= \frac{90}{289} \times 86700$.
Since $86700 / 289 = 300$,
$B$'s share $= 90 \times 300 = 27000$.

(A) Let the amount received by $N$ be $x$.
Given that $L$ gets $Rs. 5.72$ more than $N$,so $L = x + 5.72$.
Given that $M$ gets $Rs. 2.24$ more than $L$,so $M = (x + 5.72) + 2.24 = x + 7.96$.
The total sum is $Rs. 340.68$,so $L + M + N = 340.68$.
Substituting the expressions in terms of $x$: $(x + 5.72) + (x + 7.96) + x = 340.68$.
$3x + 13.68 = 340.68$.
$3x = 340.68 - 13.68 = 327$.
$x = 327 / 3 = 109$.
Therefore,$N$ gets $Rs. 109$.

(D) Let the value of $Rs. 1$ coins be $13x$ and the value of $50$ paise coins be $11x$.
Number of $Rs. 1$ coins $= \frac{13x}{1} = 13x$.
Number of $50$ paise coins $= \frac{11x}{0.5} = 22x$.
Total number of coins $= 13x + 22x = 35x$.
Given that the total number of coins is $210$,so $35x = 210$,which gives $x = 6$.
Number of $Rs. 1$ coins $= 13x = 13 \times 6 = 78$.

(A) Let the values of $Rs. 1$,$50$ paise,and $25$ paise coins be $13x$,$11x$,and $7x$ respectively.
The number of $Rs. 1$ coins $= 13x / 1 = 13x$.
The number of $50$ paise coins $= 11x / 0.5 = 22x$.
The number of $25$ paise coins $= 7x / 0.25 = 28x$.
Total number of coins $= 13x + 22x + 28x = 63x$.
Given that the total number of coins is $378$,we have $63x = 378$,which gives $x = 378 / 63 = 6$.
The number of $50$ paise coins $= 22x = 22 \times 6 = 132$.

(D) Let the present age of Sumit be $2x$ and the present age of Prakash be $3x$.
According to the problem,Sumit is $6 \text{ yr}$ younger than Prakash,so $3x - 2x = 6$,which gives $x = 6$.
Therefore,the present age of Sumit is $2 \times 6 = 12 \text{ yr}$ and the present age of Prakash is $3 \times 6 = 18 \text{ yr}$.
After $6 \text{ yr}$,Sumit's age will be $12 + 6 = 18 \text{ yr}$ and Prakash's age will be $18 + 6 = 24 \text{ yr}$.
The required ratio of their ages after $6 \text{ yr}$ is $18:24$,which simplifies to $3:4$.

(B) Let the ages of Ram and Rahim $10 \text{ years}$ ago be $x$ and $3x$ years respectively.
Their present ages are $(x+10)$ and $(3x+10)$ years.
After $5 \text{ years}$ from now,their ages will be $(x+10+5)$ and $(3x+10+5)$,which is $(x+15)$ and $(3x+15)$.
According to the problem,the ratio after $5 \text{ years}$ is $2:3$:
$\frac{x+15}{3x+15} = \frac{2}{3}$
$3(x+15) = 2(3x+15)$
$3x + 45 = 6x + 30$
$45 - 30 = 6x - 3x$
$3x = 15 \Rightarrow x = 5$
Now,calculate the present ages:
Ram's present age $= x + 10 = 5 + 10 = 15 \text{ years}$.
Rahim's present age $= 3x + 10 = 3(5) + 10 = 25 \text{ years}$.
The ratio of their present ages $= 15:25 = 3:5$.

(C) The ratio of pens is given as $\frac{1}{3}: \frac{1}{4}: \frac{1}{5}: \frac{1}{6}$.
To simplify this ratio,find the least common multiple $(LCM)$ of the denominators $3, 4, 5,$ and $6$.
The $LCM$ of $3, 4, 5,$ and $6$ is $60$.
Multiply each term by $60$:
$\frac{60}{3}: \frac{60}{4}: \frac{60}{5}: \frac{60}{6} = 20: 15: 12: 10$.
Since the number of pens must be an integer,the minimum number of pens is the sum of the parts of the ratio:
$20 + 15 + 12 + 10 = 57$.

(B) Given that $A = \frac{4}{5} B$,which implies $\frac{A}{B} = \frac{4}{5}$.
Also,$B = \frac{5}{2} C$,which implies $\frac{B}{C} = \frac{5}{2}$.
To find the ratio $A : C$,we multiply the two ratios:
$\frac{A}{C} = \frac{A}{B} \times \frac{B}{C} = \frac{4}{5} \times \frac{5}{2}$.
$\frac{A}{C} = \frac{4}{2} = \frac{2}{1}$.
Therefore,the ratio $A : C$ is $2 : 1$.

(B) Given ratios are $\frac{A}{B} = \frac{3}{4}$ and $\frac{B}{C} = \frac{12}{13}$.
To find the ratio $A:C$,we multiply the two ratios:
$\frac{A}{C} = \frac{A}{B} \times \frac{B}{C}$
$\frac{A}{C} = \frac{3}{4} \times \frac{12}{13}$
$\frac{A}{C} = \frac{3 \times 3}{1 \times 13} = \frac{9}{13}$
Therefore,the ratio $A:C$ is $9: 13$.

(A) Let the two numbers be $2x$ and $3x$.
According to the problem,if $2$ is subtracted from the first and $2$ is added to the second,the ratio becomes $1:2$.
So,$\frac{2x - 2}{3x + 2} = \frac{1}{2}$.
Cross-multiplying gives: $2(2x - 2) = 1(3x + 2)$.
$4x - 4 = 3x + 2$.
$4x - 3x = 2 + 4$.
$x = 6$.
The numbers are $2(6) = 12$ and $3(6) = 18$.
The sum of the numbers is $12 + 18 = 30$.

(B) First,simplify the given ratios:
$a:b = \frac{2}{9} : \frac{1}{3} = \frac{2}{9} : \frac{3}{9} = 2:3$
$b:c = \frac{2}{7} : \frac{5}{14} = \frac{4}{14} : \frac{5}{14} = 4:5$
$d:c = \frac{7}{10} : \frac{3}{5} = \frac{7}{10} : \frac{6}{10} = 7:6$,which implies $c:d = 6:7$
Now,to find $a:b:c:d$,we equate the common terms:
$a:b = 2:3 = (2 \times 4) : (3 \times 4) = 8:12$
$b:c = 4:5 = (4 \times 3) : (5 \times 3) = 12:15$
So,$a:b:c = 8:12:15$
Now,combine with $c:d = 6:7$:
$a:b:c = 8:12:15 = (8 \times 2) : (12 \times 2) : (15 \times 2) = 16:24:30$
$c:d = 6:7 = (6 \times 5) : (7 \times 5) = 30:35$
Therefore,$a:b:c:d = 16:24:30:35$.

(A) Let the son's share be $S$,wife's share be $W$,and daughter's share be $D$.
Given ratios: $S:W = 3:1$ and $W:D = 3:1$.
To find the combined ratio $S:W:D$,we multiply the terms to equate $W$:
$S:W = 3:1 = 9:3$
$W:D = 3:1$
Thus,$S:W:D = 9:3:1$.
Let the shares be $9x, 3x,$ and $x$ respectively.
The difference between the son's share and the daughter's share is $9x - x = 8x$.
Given $8x = 10000$,we find $x = 10000 / 8 = 1250$.
The total property value is $9x + 3x + x = 13x$.
Total value $= 13 \times 1250 = 16250$.

(B) Given $A : B = 7 : 9$ and $B : C = 8 : 11$.
To find $A : B : C$,we need to make the value of $B$ common in both ratios.
The value of $B$ in the first ratio is $9$ and in the second ratio is $8$.
The least common multiple $(LCM)$ of $9$ and $8$ is $72$.
Multiply the first ratio by $8$:
$A : B = (7 \times 8) : (9 \times 8) = 56 : 72$
Multiply the second ratio by $9$:
$B : C = (8 \times 9) : (11 \times 9) = 72 : 99$
Now,combining both ratios,we get:
$A : B : C = 56 : 72 : 99$.

(C) Let the two numbers be $a$ and $b$.
Given that $a + b = 25$ and $a - b = 7 \frac{1}{2} = \frac{15}{2} = 7.5$.
We need to find the ratio $a:b$.
Using the componendo and dividendo rule:
$\frac{a+b}{a-b} = \frac{25}{7.5} = \frac{250}{75} = \frac{10}{3}$.
Applying the rule $\frac{(a+b)+(a-b)}{(a+b)-(a-b)} = \frac{10+3}{10-3}$.
This simplifies to $\frac{2a}{2b} = \frac{13}{7}$.
Therefore,the ratio $\frac{a}{b} = \frac{13}{7}$.

(D) Let the share of a boy be $x$.
According to the problem:
Share of $1$ man $= 2x$.
Share of $2$ women $= 3x$, so share of $1$ woman $= 1.5x$.
Total amount $= 6(\text{men}) + 8(\text{women}) + 6(\text{boys}) = 120$.
Substituting the values in terms of $x$:
$6(2x) + 8(1.5x) + 6(x) = 120$.
$12x + 12x + 6x = 120$.
$30x = 120$.
$x = 120 / 30 = 4$.
Therefore, the share of a boy is ₹ $4$.

(A) Let the number to be subtracted from each term be $x$.
According to the problem,we have the equation:
$\frac{19 - x}{23 - x} = \frac{3}{4}$
Cross-multiplying the terms:
$4(19 - x) = 3(23 - x)$
$76 - 4x = 69 - 3x$
Rearranging the terms to solve for $x$:
$76 - 69 = 4x - 3x$
$x = 7$
Therefore,the number to be subtracted is $7$.

(B) Let the initial number of employees be $9x$ and the initial wage per employee be $14y$.
Initial total wage bill $= 9x \times 14y = 126xy$.
New number of employees $= 7x$.
New wage per employee $= 15y$.
New total wage bill $= 7x \times 15y = 105xy$.
The ratio of the new wage bill to the initial wage bill is $105xy : 126xy$.
Dividing both by $21xy$,we get $5:6$.
Thus,the total bill of wages decreases in the ratio $6:5$ to $5:6$,which is a decrease.

(C) Let the total cost be $C$. The cost of wheat is proportional to the quantity consumed.
In the first case,the cost for $75$ men for $15$ days is $C = 75 \times 15 \times 13$.
In the second case,let $x$ be the number of men. The cost for $x$ men for $45$ days at ₹ $1$ per kg is $C = x \times 45 \times 1$.
Since the cost is the same,we equate the two expressions:
$75 \times 15 \times 13 = x \times 45 \times 1$
$x = \frac{75 \times 15 \times 13}{45}$
$x = \frac{1125 \times 13}{45}$
$x = 25 \times 13 = 325$
Therefore,$325$ men can be fed.

(B) Let $x$ be the number of reams of paper required.
The amount of paper used is directly proportional to the number of copies and the number of sheets per book.
We can set up the proportion as:
$\frac{\text{Copies}_1 \times \text{Sheets}_1}{\text{Reams}_1} = \frac{\text{Copies}_2 \times \text{Sheets}_2}{\text{Reams}_2}$
Substituting the given values:
$\frac{1000 \times 25}{50} = \frac{5000 \times 32}{x}$
Solving for $x$:
$x = \frac{5000 \times 32 \times 50}{1000 \times 25}$
$x = 5 \times 32 \times 2$
$x = 320$
Therefore,$320$ reams of paper are required.

(C) Let the share of $Q$ be $x$.
According to the problem,$P$'s share is $\frac{2}{3}x$ and $R$'s share is $\frac{5}{3}x$.
The total amount is $₹ 1,000$,so:
$\frac{2}{3}x + x + \frac{5}{3}x = 1000$
Combining the terms:
$\frac{2x + 3x + 5x}{3} = 1000$
$\frac{10x}{3} = 1000$
Solving for $x$:
$10x = 3000$
$x = 300$
Therefore,the share of $Q$ is $₹ 300$.

(A) Let the two numbers be $x$ and $y$.
According to the problem,$0.6x = 0.025y$.
To find the ratio $x/y$,we rearrange the equation:
$\frac{x}{y} = \frac{0.025}{0.6}$
Multiply both numerator and denominator by $1000$ to remove the decimals:
$\frac{x}{y} = \frac{25}{600}$
Simplifying the fraction by dividing both by $25$:
$\frac{x}{y} = \frac{1}{24}$
Therefore,the ratio of the two numbers is $1:24$.