Ratio and Proportion Questions in English

(D) Let the present ages of Ramu and Somu be $R$ and $S$ respectively.
One year ago,the ratio of their ages was $(R-1) : (S-1) = 6 : 7$.
This implies $7(R-1) = 6(S-1)$,so $7R - 7 = 6S - 6$,which simplifies to $7R - 6S = 1$ (Equation $i$).
Four years hence,the ratio of their ages will be $(R+4) : (S+4) = 7 : 8$.
This implies $8(R+4) = 7(S+4)$,so $8R + 32 = 7S + 28$,which simplifies to $8R - 7S = -4$ (Equation $ii$).
To solve the system of equations:
Multiply Equation $(i)$ by $8$: $56R - 48S = 8$ (Equation $iii$).
Multiply Equation $(ii)$ by $7$: $56R - 49S = -28$ (Equation $iv$).
Subtract Equation $(iv)$ from Equation $(iii)$:
$(56R - 48S) - (56R - 49S) = 8 - (-28)$
$S = 36$.
Therefore,the present age of Somu is $36$ years.

(B) Given that $33 \%$ of $A = 55 \%$ of $B$.
This can be written as: $\frac{33}{100} \times A = \frac{55}{100} \times B$.
Multiplying both sides by $100$,we get: $33A = 55B$.
To find the ratio $A:B$,we rearrange the equation: $\frac{A}{B} = \frac{55}{33}$.
Simplifying the fraction by dividing both numerator and denominator by $11$,we get: $\frac{A}{B} = \frac{5}{3}$.
Therefore,the ratio of $A$ and $B$ is $5:3$.

(A) To find the value of $68 \%$ of two-fifths of $550$,we follow these steps:
$1$. First,calculate two-fifths of $550$: $\frac{2}{5} \times 550 = 2 \times 110 = 220$.
$2$. Next,calculate $68 \%$ of the result $(220)$:
$68 \% \text{ of } 220 = \frac{68}{100} \times 220$.
$3$. Simplify the expression: $\frac{68 \times 22}{10} = \frac{1496}{10} = 149.6$.

(D) Let the number be $n$.
According to the problem,$45 \%$ of $n$ minus $24$ equals $48$.
$\frac{45}{100} n - 24 = 48$
$\frac{9}{20} n = 48 + 24$
$\frac{9}{20} n = 72$
$n = 72 \times \frac{20}{9}$
$n = 8 \times 20 = 160$
Now,we need to find $\frac{3}{8}$ of the number $n$.
$\frac{3}{8} \times 160 = 3 \times 20 = 60$.

(A) Let the number be $x$.
According to the problem,$30 \%$ of $x = 190.8$.
This can be written as: $\frac{30}{100} \times x = 190.8$.
Solving for $x$: $x = \frac{190.8 \times 100}{30} = \frac{19080}{30} = 636$.
Now,we need to find $175 \%$ of this number $x$.
$175 \%$ of $636 = \frac{175}{100} \times 636$.
$= 1.75 \times 636 = 1113$.
Alternatively,$175 \%$ of the number $= \frac{190.8}{30} \times 175 = 6.36 \times 175 = 1113$.

(C) First,calculate the $(\frac{3}{8})^{th}$ part of $1000$:
$\frac{3}{8} \times 1000 = 375$.
Now,calculate $32 \%$ of $375$:
$32 \% \text{ of } 375 = \frac{32}{100} \times 375$.
$= 0.32 \times 375 = 120$.

(C) Let the third number be $x$.
The first number is $20 \%$ more than $x$,which is $x + 0.20x = 1.2x$.
The second number is $50 \%$ more than $x$,which is $x + 0.50x = 1.5x$.
The ratio of the two numbers is $\frac{1.2x}{1.5x} = \frac{1.2}{1.5} = \frac{12}{15} = \frac{4}{5}$.
Therefore,the ratio is $4:5$.

(C) The ratio of distribution among $A, B, C, D$ is $5: 2: 4: 3$.
Let the shares of $A, B, C, D$ be $5x, 2x, 4x, 3x$ respectively.
According to the problem,$C$ gets $Rs. 1000$ more than $D$.
So,$4x - 3x = 1000$.
$x = 1000$.
$B$'s share is $2x$.
Therefore,$B$'s share $= 2 \times 1000 = Rs. 2000$.

(B) Given the proportion $0.75 : x :: 5 : 8.$
This can be written as the equation $\frac{0.75}{x} = \frac{5}{8}.$
By cross-multiplying,we get $5x = 0.75 \times 8.$
Calculating the product: $5x = 6.00.$
Dividing both sides by $5$: $x = \frac{6}{5} = 1.2.$
Therefore,the value of $x$ is $1.2.$

(B) Let the three numbers be $I, II,$ and $III$.
Given $I + II + III = 98$.
The ratios are $I : II = 2 : 3$ and $II : III = 5 : 8$.
To combine these ratios,we make the value of $II$ common in both ratios.
Multiply the first ratio by $5$ and the second ratio by $3$:
$I : II = (2 \times 5) : (3 \times 5) = 10 : 15$
$II : III = (5 \times 3) : (8 \times 3) = 15 : 24$
Thus,the combined ratio is $I : II : III = 10 : 15 : 24$.
Let the numbers be $10x, 15x,$ and $24x$.
Sum $= 10x + 15x + 24x = 49x$.
Given $49x = 98$,so $x = 98 / 49 = 2$.
The second number is $15x = 15 \times 2 = 30$.

(D) The given ratio is $\frac{1}{2}: \frac{2}{3}: \frac{3}{4}$.
To simplify the ratio,find the Least Common Multiple ($L$.$C$.$M$.) of the denominators $2, 3,$ and $4$,which is $12$.
Multiply each term of the ratio by $12$:
$\frac{1}{2} \times 12 : \frac{2}{3} \times 12 : \frac{3}{4} \times 12 = 6 : 8 : 9$.
The sum of the ratio parts is $6 + 8 + 9 = 23$.
Given that the total amount is Rs. $872$,we have $23 \text{ units} = 872$.
Therefore,$1 \text{ unit} = \frac{872}{23} \approx 37.913$.
The first part is $6 \text{ units} = 6 \times \frac{872}{23} = \frac{5232}{23} \approx 227.48$.
Rounding to the nearest provided option,the first part is $227.46$.

(B) To find the fourth proportional to three numbers $a, b, c$,we use the relation $a : b :: c : x$,where $x$ is the fourth proportional.
This can be written as the equation $\frac{a}{b} = \frac{c}{x}$.
Given the numbers $5, 8, 15$,we set up the proportion as $5 : 8 :: 15 : x$.
This implies $\frac{5}{8} = \frac{15}{x}$.
By cross-multiplying,we get $5x = 8 \times 15$.
$5x = 120$.
$x = \frac{120}{5} = 24$.
Therefore,the fourth proportional is $24$.

(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 two numbers are $3x = 3 \times 11 = 33$ and $5x = 5 \times 11 = 55$.
The smaller number is $33$.

(D) Let the first number be $a$ and the second number be $b$.
According to the problem,reducing the first number by $40 \%$ gives $\frac{2}{3}$ of the second number.
$a - 0.40a = \frac{2}{3}b$
$0.60a = \frac{2}{3}b$
$\frac{60}{100}a = \frac{2}{3}b$
$\frac{3}{5}a = \frac{2}{3}b$
To find the ratio of the second number $(b)$ to the first number $(a)$,we rearrange the equation:
$\frac{b}{a} = \frac{3}{5} \times \frac{3}{2} = \frac{9}{10}$
Therefore,the ratio of the second number to the first number is $9:10$.

(D) Given equation: $\frac{5a+3b}{2a-3b} = \frac{23}{5}$
Divide the numerator and denominator of the left side by $b$:
$\frac{5(a/b) + 3}{2(a/b) - 3} = \frac{23}{5}$
Let $\frac{a}{b} = x$. Substituting this into the equation:
$\frac{5x + 3}{2x - 3} = \frac{23}{5}$
Cross-multiply to solve for $x$:
$5(5x + 3) = 23(2x - 3)$
$25x + 15 = 46x - 69$
Rearrange the terms:
$69 + 15 = 46x - 25x$
$84 = 21x$
$x = \frac{84}{21} = 4$
Since $x = \frac{a}{b} = 4$,the ratio $a:b$ is $4:1$.

(C) Given ratios are $P: Q = 8: 15$ and $Q: R = 3: 2$.
To find $P: Q: R$,we need to make the value of $Q$ common in both ratios.
The value of $Q$ in the first ratio is $15$ and in the second ratio is $3$.
To make them equal,multiply the second ratio $Q: R = 3: 2$ by $5$:
$Q: R = (3 \times 5) : (2 \times 5) = 15: 10$.
Now,we have $P: Q = 8: 15$ and $Q: R = 15: 10$.
Combining these,we get $P: Q: R = 8: 15: 10$.

(A) To find $P: S$,we can multiply the given ratios:
$P: S = (P/Q) \times (Q/R) \times (R/S)$
$P: S = (8/15) \times (5/8) \times (4/5)$
By canceling the common terms:
$P: S = (8 \times 5 \times 4) / (15 \times 8 \times 5)$
$P: S = 4 / 15$
Therefore,$P: S = 4: 15$.

(A) Let the fourth proportional to $4, 16, 7$ be $x$.
According to the definition of proportion,we have:
$4 : 16 :: 7 : x$
This can be written as:
$\frac{4}{16} = \frac{7}{x}$
Cross-multiplying gives:
$4x = 16 \times 7$
Dividing both sides by $4$:
$x = \frac{16 \times 7}{4}$
$x = 4 \times 7$
$x = 28$
Therefore,the $4^{th}$ proportional is $28$.

(B) Let the mean proportional be $r$.
By definition,if $r$ is the mean proportional between $a$ and $b$,then $a:r :: r:b$,which implies $\frac{a}{r} = \frac{r}{b}$ or $r^2 = a \times b$.
Here,$a = 9$ and $b = 64$.
Therefore,$r^2 = 9 \times 64$.
$r = \sqrt{9 \times 64} = \sqrt{9} \times \sqrt{64} = 3 \times 8 = 24$.
Thus,the mean proportional is $24$.

(A) The duplicate ratio of a given ratio $a:b$ is defined as $a^2:b^2$.
For the given ratio $2:7$,the duplicate ratio is calculated as:
$2^2 : 7^2 = 4 : 49$.
Therefore,the correct option is $A$.

(C) The sub-duplicate ratio of a given ratio $a:b$ is defined as the ratio of the square roots of its terms,which is $\sqrt{a}:\sqrt{b}$.
For the given ratio $81:64$,the sub-duplicate ratio is calculated as:
$\sqrt{81}:\sqrt{64} = 9:8$.
Therefore,the correct option is $C$.

(B) The triplicate ratio of a given ratio $a:b$ is defined as $a^3:b^3$.
For the given ratio $7:5$,the triplicate ratio is calculated as:
$7^3 : 5^3 = (7 \times 7 \times 7) : (5 \times 5 \times 5) = 343 : 125$.
Therefore,the correct option is $B$.

(A) The inverse ratio (or reciprocal ratio) of a given ratio $a:b$ is defined as $b:a$.
Given the ratio $17:19$,we identify $a = 17$ and $b = 19$.
Therefore,the inverse ratio is $19:17$.

(B) The compound ratio of multiple ratios is found by multiplying all the antecedent terms together and all the consequent terms together.
For the given ratios $2:7$,$5:3$,and $4:7$:
Antecedents are $2, 5, 4$.
Consequents are $7, 3, 7$.
Compound ratio $= (2 \times 5 \times 4) : (7 \times 3 \times 7)$.
$= 40 : 147$.

(D) Given ratios are $A: B = 3: 4$ and $B: C = 8: 9$.
To find $A: B: C$,we need to make the value of $B$ common in both ratios.
In the first ratio,$B = 4$. In the second ratio,$B = 8$.
To make $B$ equal,multiply the first ratio by $2$:
$A: B = (3 \times 2) : (4 \times 2) = 6: 8$.
Now,we have $A: B = 6: 8$ and $B: C = 8: 9$.
Since the value of $B$ is now $8$ in both,we can combine them:
$A: B: C = 6: 8: 9$.

(D) Given ratios are $a: b = 3: 5$ and $b: c = 4: 7.$
To find $a: c,$ we can multiply the two ratios:
$\frac{a}{c} = \frac{a}{b} \times \frac{b}{c}$
$\frac{a}{c} = \frac{3}{5} \times \frac{4}{7}$
$\frac{a}{c} = \frac{12}{35}$
Therefore,$a: c = 12: 35.$

(A) Given the ratio $P: Q: R = 2: 3: 4.$
Let $P = 2k, Q = 3k, R = 4k$ for some constant $k \neq 0.$
Now,calculate the individual ratios:
$\frac{P}{Q} = \frac{2k}{3k} = \frac{2}{3}$
$\frac{Q}{R} = \frac{3k}{4k} = \frac{3}{4}$
$\frac{R}{P} = \frac{4k}{2k} = \frac{4}{2} = 2$
Now,find the ratio $\frac{P}{Q}: \frac{Q}{R}: \frac{R}{P} = \frac{2}{3}: \frac{3}{4}: 2.$
To simplify this,multiply each term by the least common multiple $(LCM)$ of the denominators $3$ and $4$,which is $12$:
$\left(\frac{2}{3} \times 12\right): \left(\frac{3}{4} \times 12\right): (2 \times 12) = 8: 9: 24.$
Thus,the required ratio is $8: 9: 24.$

(A) Given that $\frac{a}{3} = \frac{b}{8}$.
Let $\frac{a}{3} = \frac{b}{8} = k$,where $k$ is a constant.
Then $a = 3k$ and $b = 8k$.
We need to find the ratio $(a + 3) : (b + 8)$.
Substituting the values of $a$ and $b$ in terms of $k$:
$(a + 3) : (b + 8) = (3k + 3) : (8k + 8)$.
Factor out the common terms:
$(3(k + 1)) : (8(k + 1))$.
Since $k + 1 \neq 0$,we can cancel $(k + 1)$ from both sides:
Ratio $= 3 : 8$.

(B) To find the ratio of $4^{3.5} : 2^{5}$,we first express both terms with the same base.
Since $4 = 2^{2}$,we can rewrite $4^{3.5}$ as $(2^{2})^{3.5}$.
Using the power rule $(a^{m})^{n} = a^{m \times n}$,we get $(2^{2})^{3.5} = 2^{2 \times 3.5} = 2^{7}$.
Now,the ratio becomes $2^{7} : 2^{5}$.
Using the division rule for exponents $\frac{a^{m}}{a^{n}} = a^{m-n}$,we have $\frac{2^{7}}{2^{5}} = 2^{7-5} = 2^{2} = 4$.
Thus,the ratio is $4 : 1$.

(B) Given the proportion $2: x :: 5: 7$.
This can be written as the equation $\frac{2}{x} = \frac{5}{7}$.
By cross-multiplying,we get $5 \times x = 2 \times 7$.
$5x = 14$.
Dividing both sides by $5$,we get $x = \frac{14}{5}$.
Therefore,$x = 2.80$.

(C) Let the initial salaries of $A, B,$ and $C$ be $2x, 3x,$ and $5x$ respectively.
New salary of $A = 2x + 15\% \text{ of } 2x = 2x + 0.3x = 2.3x$.
New salary of $B = 3x + 10\% \text{ of } 3x = 3x + 0.3x = 3.3x$.
New salary of $C = 5x + 20\% \text{ of } 5x = 5x + 1.0x = 6x$.
The new ratio of their salaries is $2.3x : 3.3x : 6x$.
To simplify,multiply each term by $10$:
$23 : 33 : 60$.

(D) First,find the $L.C.M.$ of the denominators $2, 3,$ and $4$,which is $12$.
Multiply each ratio term by $12$ to convert them into whole numbers:
$\left(\frac{1}{2} \times 12\right) : \left(\frac{2}{3} \times 12\right) : \left(\frac{3}{4} \times 12\right) = 6 : 8 : 9$.
The sum of the ratio parts is $6 + 8 + 9 = 23$.
Given that the total amount is $Rs. 782$,we have $23 \text{ units} = 782$.
Therefore,$1 \text{ unit} = \frac{782}{23} = 34$.
The first part corresponds to $6 \text{ units}$,so the first part $= 6 \times 34 = 204$.

(C) Let the two numbers be $x$ and $2x$ since they are in the ratio $1: 2$.
According to the problem,if $7$ is added to both,the new ratio becomes $3: 5$.
$\frac{x + 7}{2x + 7} = \frac{3}{5}$
Cross-multiplying gives:
$5(x + 7) = 3(2x + 7)$
$5x + 35 = 6x + 21$
Rearranging the terms to solve for $x$:
$35 - 21 = 6x - 5x$
$x = 14$
The two numbers are $x = 14$ and $2x = 2(14) = 28$.
The greatest number is $28$.

(C) Let the three numbers be $3x$,$4x$,and $5x$.
According to the problem,the sum of their squares is $1250$.
Therefore,$(3x)^2 + (4x)^2 + (5x)^2 = 1250$.
$9x^2 + 16x^2 + 25x^2 = 1250$.
$50x^2 = 1250$.
$x^2 = \frac{1250}{50} = 25$.
$x = 5$.
The sum of the numbers is $3x + 4x + 5x = 12x$.
Substituting the value of $x$,we get $12 \times 5 = 60$.

(C) Let the age of Sachin be $S$ and the age of Rahul be $R$.
Given that Sachin is $4$ years younger than Rahul:
$S = R - 4$ --- $(i)$
The ratio of their ages is $7:9$:
$\frac{S}{R} = \frac{7}{9} \Rightarrow 9S = 7R$ --- $(ii)$
Substitute $R = S + 4$ from $(i)$ into $(ii)$:
$9S = 7(S + 4)$
$9S = 7S + 28$
$2S = 28$
$S = 14$
Thus,Sachin is $14$ years old.
Shortcut method:
Ratio of Sachin to Rahul $= 7:9$.
The difference in ratio units is $9 - 7 = 2$ units.
Given difference $= 4$ years.
$2$ units $= 4$ years $\Rightarrow 1$ unit $= 2$ years.
Sachin's age $= 7$ units $= 7 \times 2 = 14$ years.

(B) Let the present ages of Arun and Deepak be $4x$ and $3x$ respectively,based on the given ratio $4:3$.
After $6$ years,Arun's age will be $4x + 6$.
According to the problem,Arun's age after $6$ years is $26$ years.
So,$4x + 6 = 26$.
Subtracting $6$ from both sides,we get $4x = 20$.
Dividing by $4$,we find $x = 5$.
Deepak's present age is $3x = 3 \times 5 = 15$ years.

(A) Let the present ages of $X$ and $Y$ be $5x$ and $6x$ respectively.
According to the problem,after $7$ years,the ratio of their ages will be $6: 7$.
So,the equation is: $\frac{5x + 7}{6x + 7} = \frac{6}{7}$.
By cross-multiplying,we get: $7(5x + 7) = 6(6x + 7)$.
$35x + 49 = 36x + 42$.
Rearranging the terms: $49 - 42 = 36x - 35x$.
$x = 7$.
Therefore,$X$'s present age is $5x = 5 \times 7 = 35$ years.

(A) Let the present ages of Sameer and Anand be $5x$ and $4x$ respectively.
After $3$ years,their ages will be $(5x + 3)$ and $(4x + 3)$ respectively.
According to the problem,the ratio of their ages after $3$ years will be $11 : 9$.
So,$\frac{5x + 3}{4x + 3} = \frac{11}{9}$.
Cross-multiplying,we get: $9(5x + 3) = 11(4x + 3)$.
$45x + 27 = 44x + 33$.
$45x - 44x = 33 - 27$.
$x = 6$.
Anand's present age is $4x = 4 \times 6 = 24$ years.

(D) Let the ages of Jayant,Prem,and Saransh ten years ago be $2x$,$3x$,and $4x$ respectively.
Therefore,their present ages are $(2x + 10)$,$(3x + 10)$,and $(4x + 10)$ years.
The sum of their present ages is given as $93$ years.
$(2x + 10) + (3x + 10) + (4x + 10) = 93$
$9x + 30 = 93$
$9x = 63$
$x = 7$
The present age of Saransh is $(4x + 10)$ years.
Substituting $x = 7$:
$4(7) + 10 = 28 + 10 = 38$ years.

(B) Let the number of years ago be $x$.
According to the problem,$(40 - x) / (60 - x) = 3 / 5$.
Cross-multiplying gives: $5(40 - x) = 3(60 - x)$.
$200 - 5x = 180 - 3x$.
$200 - 180 = 5x - 3x$.
$20 = 2x$.
$x = 10$.
Therefore,$10$ years ago,their ages were $30$ and $50$,which gives the ratio $30:50 = 3:5$.

(B) Let the tens digit be $x$ and the units digit be $y$.
The number is $10x + y$.
After interchanging the digits,the new number is $10y + x$.
Given the ratio of the digits is $x:y = 1:2$,so $y = 2x$.
The difference between the numbers is $|(10x + y) - (10y + x)| = 36$.
$|9x - 9y| = 36 \Rightarrow |x - y| = 4$.
Since $y = 2x$,we have $|x - 2x| = 4 \Rightarrow |-x| = 4 \Rightarrow x = 4$.
Then $y = 2(4) = 8$.
The number is $48$.
Sum of the digits $= x + y = 4 + 8 = 12$.
Difference of the digits $= y - x = 8 - 4 = 4$.
The difference between the sum and the difference of the digits is $12 - 4 = 8$.

(A) Let the initial number of seats for Mathematics,Physics,and Biology be $5x$,$7x$,and $8x$ respectively.
After the proposed increases of $40\%$,$50\%$,and $75\%$,the new number of seats will be:
Mathematics: $5x + (40\% \text{ of } 5x) = 5x + 2x = 7x$
Physics: $7x + (50\% \text{ of } 7x) = 7x + 3.5x = 10.5x$
Biology: $8x + (75\% \text{ of } 8x) = 8x + 6x = 14x$
The ratio of the increased seats is $7x : 10.5x : 14x$.
To simplify,multiply by $2$ to remove the decimal: $14x : 21x : 28x$.
Dividing by $7$,we get the ratio $2:3:4$.

(B) Total quantity of the mixture = $60 \ L$.
The ratio of milk to water is $2:1$.
Sum of ratio parts = $2 + 1 = 3$.
Quantity of milk = $(2/3) \times 60 = 40 \ L$.
Quantity of water = $(1/3) \times 60 = 20 \ L$.
Let the quantity of water to be added be $x \ L$.
After adding water,the new quantity of water = $(20 + x) \ L$.
The new ratio of milk to water is $1:3$.
So,$40 / (20 + x) = 1 / 3$.
$40 \times 3 = 20 + x$.
$120 = 20 + x$.
$x = 120 - 20 = 100 \ L$.
Therefore,$100 \ L$ of water should be added.

(C) Let the number of boys be $7x$ and the number of girls be $8x$.
After a $20\%$ increase in the number of boys,the new number of boys is $7x \times (1 + 0.20) = 7x \times 1.2 = 8.4x$.
After a $10\%$ increase in the number of girls,the new number of girls is $8x \times (1 + 0.10) = 8x \times 1.1 = 8.8x$.
The new ratio of boys to girls is $8.4x : 8.8x$.
Multiplying both sides by $10$,we get $84 : 88$.
Dividing by $4$,the simplified ratio is $21 : 22$.

(D) Let the initial salaries of Ravi and Sumit be $2x$ and $3x$ respectively.
According to the problem,the salary of each is increased by $Rs. 4000$.
New salary of Ravi $= 2x + 4000$
New salary of Sumit $= 3x + 4000$
The new ratio is given as $40:57$.
So,$\frac{2x + 4000}{3x + 4000} = \frac{40}{57}$
Cross-multiplying,we get:
$57(2x + 4000) = 40(3x + 4000)$
$114x + 228000 = 120x + 160000$
$120x - 114x = 228000 - 160000$
$6x = 68000$
$x = \frac{68000}{6} = \frac{34000}{3}$
Sumit's original salary $= 3x = 3 \times \left(\frac{34000}{3}\right) = Rs. 34000$.

(C) Let the initial salaries of $A, B$ and $C$ be $2x, 3x$ and $5x$ respectively.
After an increment of $15\%$ for $A$,the new salary is $2x \times (1 + 0.15) = 2x \times 1.15 = 2.3x$.
After an increment of $10\%$ for $B$,the new salary is $3x \times (1 + 0.10) = 3x \times 1.10 = 3.3x$.
After an increment of $20\%$ for $C$,the new salary is $5x \times (1 + 0.20) = 5x \times 1.20 = 6.0x$.
The new ratio is $2.3x : 3.3x : 6.0x$.
Multiplying by $10$ to remove decimals,we get $23 : 33 : 60$.

(D) Let the number of $25 p$,$10 p$,and $5 p$ coins be $2x$,$3x$,and $4x$ respectively.
The value of $25 p$ coins $= 25 \times 2x = 50x$ paise.
The value of $10 p$ coins $= 10 \times 3x = 30x$ paise.
The value of $5 p$ coins $= 5 \times 4x = 20x$ paise.
Total value $= 50x + 30x + 20x = 100x$ paise.
Since the total amount is $Rs. 50$,which is equal to $5000$ paise,we have:
$100x = 5000$
$x = 50$
The number of $5 p$ coins is $4x = 4 \times 50 = 200$.

(D) Let the first sum of money be divided among $C, A$ and $B$ in the ratio $4:5:6$. Thus,$C = 4x, A = 5x, B = 6x$ for some constant $x$.
Let the second sum of money be divided between $M$ and $N$ equally. Thus,$M = y$ and $N = y$ for some constant $y$.
According to the problem,$B$ got $2000$ more than $M$,which gives the equation: $6x - y = 2000$.
We are asked to find the value of $C$,which is $4x$.
From the equation $6x - y = 2000$,we have two variables $x$ and $y$ and only one equation. Therefore,it is impossible to determine the unique value of $x$ without further information about the relationship between the two sums of money or the value of $y$.
Thus,the value of $C$ cannot be determined.

(B) Given the ratio of numbers $A : B : C = 12 : 15 : 25$.
Let the numbers be $12x, 15x$ and $25x$.
The sum of the numbers is $12x + 15x + 25x = 52x$.
Given that the sum is $364$,we have $52x = 364$.
Solving for $x$,$x = \frac{364}{52} = 7$.
The numbers are:
$A = 12 \times 7 = 84$
$B = 15 \times 7 = 105$
$C = 25 \times 7 = 175$
The difference between $B$ and $A$ is $105 - 84 = 21$.
The difference between $C$ and $B$ is $175 - 105 = 70$.
The required ratio is $21 : 70$.
Dividing both by $7$,we get $3 : 10$.

(B) Given the equation: $\frac{625}{x} = \frac{x}{1156}$
By cross-multiplying,we get: $x^2 = 625 \times 1156$
We know that $625 = 25^2$ and $1156 = 34^2$.
Therefore,$x^2 = 25^2 \times 34^2 = (25 \times 34)^2$.
Taking the square root on both sides: $x = 25 \times 34$.
Calculating the product: $x = 850$.