Ratio and Proportion Questions in English

(C) Let the amounts received by $A, B,$ and $C$ be $A, B,$ and $C$ respectively.
Given that $\frac{1}{2} A = \frac{1}{3} B = \frac{1}{4} C = k$ (where $k$ is a constant).
Then $A = 2k, B = 3k, C = 4k$.
The total amount is $A + B + C = 900$.
Substituting the values: $2k + 3k + 4k = 900$.
$9k = 900 \Rightarrow k = 100$.
Therefore,the amounts are:
$A = 2 \times 100 = ₹ 200$
$B = 3 \times 100 = ₹ 300$
$C = 4 \times 100 = ₹ 400$
Thus,the amounts received by $A, B,$ and $C$ are $₹ 200, ₹ 300,$ and $₹ 400$ respectively.

(B) Given ratio $A : B : C = 2 : 5 : 4$.
Sum of the ratio terms $= 2 + 5 + 4 = 11$.
Total amount $= ₹ 126.50$.
Share of $A = \frac{2}{11} \times 126.50 = 2 \times 11.50 = ₹ 23.00$.
Share of $B = \frac{5}{11} \times 126.50 = 5 \times 11.50 = ₹ 57.50$.
The difference between the share of $B$ and $A = 57.50 - 23.00 = ₹ 34.50$.
Alternatively,the difference in ratio $= 5 - 2 = 3$.
Required difference $= \frac{3}{11} \times 126.50 = 3 \times 11.50 = ₹ 34.50$.

(B) Let the number of one-rupee coins be $x$.
According to the problem,the number of $50-paise$ coins is $4x$.
Since the number of $50-paise$ coins is double the number of $25-paise$ coins,the number of $25-paise$ coins is $\frac{4x}{2} = 2x$.
The total value of the coins is ₹ $56$.
Value of one-rupee coins $= x \times 1 = x$.
Value of $50-paise$ coins $= 4x \times 0.50 = 2x$.
Value of $25-paise$ coins $= 2x \times 0.25 = 0.5x$.
Total value $= x + 2x + 0.5x = 3.5x$.
Given $3.5x = 56$,so $x = \frac{56}{3.5} = 16$.
The number of $50-paise$ coins is $4x = 4 \times 16 = 64$.

(A) Total number of students = $150$.
Ratio of boys to girls = $7:8$.
Total parts = $7 + 8 = 15$.
Number of boys = $(7/15) \times 150 = 70$.
Number of girls = $(8/15) \times 150 = 80$.
Students enrolled in all three disciplines = $2$ boys + $3$ girls = $5$ students.
Percentage of students enrolled in all three disciplines = $(\text{Total students in all three} / \text{Total students}) \times 100$.
Percentage = $(5 / 150) \times 100 = (1 / 30) \times 100 = 100 / 30 = 3.33\%$.

(B) Total students $= 150$. Ratio of boys to girls $= 7:8$. Total boys $= (7/15) \times 150 = 70$. Total girls $= (8/15) \times 150 = 80$.
Marketing: Boys $= 40\% \text{ of } 70 = 28$,Girls $= 50\% \text{ of } 80 = 40$.
$HR$: Boys $= 30\% \text{ of } 70 = 21$,Girls $= 30\% \text{ of } 80 = 24$.
Finance: Boys $= 30\% \text{ of } 70 = 21$,Girls $= 20\% \text{ of } 80 = 16$.
Let $M, H, F$ be the sets of boys in Marketing,$HR$,and Finance. Using the Principle of Inclusion-Exclusion for boys: Only Marketing $= M - (M \cap H) - (M \cap F) + (M \cap H \cap F) = 28 - 7 - 5 + 2 = 18$.
For girls: Only Marketing $= 40 - 9 - 8 + 3 = 26$.
Ratio of boys to girls only in Marketing $= 18:26 = 9:13$.

(C) Total students $= 150$. Ratio of boys to girls $= 7:8$. Total boys $= (7/15) \times 150 = 70$. Total girls $= (8/15) \times 150 = 80$.
Let $M, H, F$ represent Marketing, $HR$, and Finance. Boys in $M \cap F = 5$. Girls in $F = 80 \times 20\% = 16$. Girls in $(M \cap F) = 8$. Girls in $(H \cap F) = 7$. Girls in $(M \cap H \cap F) = 3$. Girls only in $F = (\text{Total girls in } F) - (\text{Girls in } M \cap F) - (\text{Girls in } H \cap F) + (\text{Girls in } M \cap H \cap F) = 16 - 8 - 7 + 3 = 4$.
The number of boys in both Marketing and Finance is given as $5$.
The required ratio $= 5:4$.

(D) Total students $= 150$. Ratio of boys to girls $= 7:8$.
Number of boys $= \frac{7}{15} \times 150 = 70$.
Number of girls $= \frac{8}{15} \times 150 = 80$.
Let $M, H, F$ be the sets of students in Marketing,$HR$,and Finance respectively.
For boys: Total $= 70$. $40\%$ of $70 = 28$ boys in Marketing. $30\%$ of $70 = 21$ boys in $HR$.
For girls: Total $= 80$. $50\%$ of $80 = 40$ girls in Marketing. $30\%$ of $80 = 24$ girls in $HR$.
Number of boys in Marketing $= 28$.
Number of girls in $HR = 24$.
Required percentage $= \frac{28 - 24}{24} \times 100 = \frac{4}{24} \times 100 = \frac{1}{6} \times 100 = 16 \frac{2}{3} \%$.

(A) Total students $= 150$. Ratio of boys to girls $= 7:8$.
Number of boys $= (7/15) \times 150 = 70$.
Number of girls $= (8/15) \times 150 = 80$.
Let $M, H, F$ denote Marketing,$HR$,and Finance.
For boys: $n(M) = 0.4 \times 70 = 28$,$n(H) = 0.3 \times 70 = 21$,$n(F) = 0.3 \times 70 = 21$.
Intersections: $n(H \cap M) = 7$,$n(H \cap F) = 6$,$n(M \cap F) = 5$,$n(M \cap H \cap F) = 2$.
Boys only in $HR = n(H) - [n(H \cap M) + n(H \cap F) - n(M \cap H \cap F)] - n(M \cap H \cap F) = 21 - [7 + 6 - 2] - 2 = 21 - 11 - 2 = 8$. Wait,recalculating: Boys only in $HR = 21 - (7-2) - (6-2) - 2 = 21 - 5 - 4 - 2 = 10$.
For girls: $n(M) = 0.5 \times 80 = 40$,$n(H) = 0.3 \times 80 = 24$,$n(F) = 0.2 \times 80 = 16$.
Intersections: $n(H \cap M) = 9$,$n(H \cap F) = 7$,$n(M \cap F) = 8$,$n(M \cap H \cap F) = 3$.
Girls only in $HR = 24 - (9-3) - (7-3) - 3 = 24 - 6 - 4 - 3 = 11$.
Ratio of boys to girls only in $HR = 10:11$.

(D) Let $X$ be subtracted from the numbers $9, 15$ and $27$ such that the resulting numbers are in continued proportion.
This means $(9-X), (15-X)$ and $(27-X)$ are in continued proportion.
For three numbers $a, b, c$ to be in continued proportion,they must satisfy the condition $b^2 = ac$.
Here,$a = (9-X)$,$b = (15-X)$,and $c = (27-X)$.
Substituting these values into the condition:
$(15-X)^2 = (9-X)(27-X)$
Expanding both sides:
$225 - 30X + X^2 = 243 - 9X - 27X + X^2$
$225 - 30X + X^2 = 243 - 36X + X^2$
Subtracting $X^2$ from both sides:
$225 - 30X = 243 - 36X$
Rearranging the terms to solve for $X$:
$36X - 30X = 243 - 225$
$6X = 18$
$X = 3$
Thus,the value of $X$ is $3$.

(C) Let the amounts received by $P$,$Q$,and $R$ be $₹ 3x$,$₹ 5x$,and $₹ 7x$,respectively.
According to the problem,the amount received by $R$ is ₹ $4,000$ more than the amount received by $Q$.
So,$7x - 5x = 4000$.
$2x = 4000$.
$x = 2000$.
The total amount received by $P$ and $Q$ together is $3x + 5x = 8x$.
Substituting the value of $x$,we get $8 \times 2000 = ₹ 16,000$.

(B) Let the original number of students in the three classes be $4x$,$6x$,and $9x$ respectively.
According to the problem,when $12$ students are added to each class,the new ratio becomes $7:9:12$.
We can set up the equation using the first two classes: $\frac{4x + 12}{6x + 12} = \frac{7}{9}$.
Cross-multiplying gives: $9(4x + 12) = 7(6x + 12)$.
$36x + 108 = 42x + 84$.
Rearranging the terms: $108 - 84 = 42x - 36x$.
$24 = 6x$,which gives $x = 4$.
The total number of students before the increase is $4x + 6x + 9x = 19x$.
Substituting $x = 4$,we get $19 \times 4 = 76$.

(A) Let the two numbers be $5x$ and $4x$ respectively.
According to the problem,$40\%$ of the first number is $12$.
So,$5x \times \frac{40}{100} = 12$.
$5x \times 0.4 = 12$.
$2x = 12$.
$x = 6$.
Now,the second number is $4x = 4 \times 6 = 24$.
We need to find $50\%$ of the second number.
$50\% \text{ of } 24 = 24 \times \frac{50}{100} = 24 \times 0.5 = 12$.
Therefore,the correct answer is $12$.

(D) Let the annual income of Amit and Veer be $3x$ and $2x$ respectively.
Let the annual expenditure of Amit and Veer be $5y$ and $3y$ respectively.
We know that $\text{Income} - \text{Expenditure} = \text{Savings}$.
For Amit: $3x - 5y = 1000$ --- (Equation $1$)
For Veer: $2x - 3y = 1000$ --- (Equation $2$)
From Equation $1$ and $2$,since both savings are equal:
$3x - 5y = 2x - 3y$
$3x - 2x = 5y - 3y$
$x = 2y$
Substitute $x = 2y$ into Equation $1$:
$3(2y) - 5y = 1000$
$6y - 5y = 1000$
$y = 1000$
Now,find $x$:
$x = 2(1000) = 2000$
Amit's annual income is $3x = 3 \times 2000 = ₹ 6000$.

(B) Given that $P$ varies inversely with the product of $Q$ and $R$,we can write:
$P \propto \frac{1}{Q \times R}$
This implies $P \times Q \times R = k$,where $k$ is a constant.
Using the given values $P = 75$,$Q = 6$,and $R = 12$:
$k = 75 \times 6 \times 12 = 5400$.
Now,we need to find $P$ when $Q = 5$ and $R = 10$:
$P \times 5 \times 10 = 5400$
$P \times 50 = 5400$
$P = \frac{5400}{50} = 108$.
Therefore,the value of $P$ is $108$.

(B) Let the shares of $A, B$,and $C$ be $A, B$,and $C$ respectively.
Given that $8A = 12B = 6C$.
To find the ratio $A:B:C$,divide the entire equation by the least common multiple of $8, 12$,and $6$,which is $24$.
$\frac{8A}{24} = \frac{12B}{24} = \frac{6C}{24}$
$\frac{A}{3} = \frac{B}{2} = \frac{C}{4}$
Therefore,the ratio $A:B:C = 3:2:4$.
The total sum is $₹ 864$.
The sum of the ratio parts is $3 + 2 + 4 = 9$.
$B$'s share $= \frac{2}{9} \times 864 = 2 \times 96 = ₹ 192$.

(B) Total population $= 3,11,250$.
Ratio of women to men $= 43:40$.
Sum of ratio terms $= 43 + 40 = 83$.
Number of women $= \frac{43}{83} \times 3,11,250 = 43 \times 3,750 = 1,61,250$.
Number of men $= \frac{40}{83} \times 3,11,250 = 40 \times 3,750 = 1,50,000$.
Literate women $= 8\% \text{ of } 1,61,250 = \frac{8}{100} \times 1,61,250 = 12,900$.
Literate men $= 24\% \text{ of } 1,50,000 = \frac{24}{100} \times 1,50,000 = 36,000$.
Total literate persons $= 12,900 + 36,000 = 48,900$.

(A) Let the total amount be $x$.
In the first case,the ratio is $2:3:4$. The sum of the ratio parts is $2 + 3 + 4 = 9$. Thus,$B$'s share is $\frac{3}{9}x = \frac{1}{3}x$.
In the second case,the ratio is $7:2:5$. The sum of the ratio parts is $7 + 2 + 5 = 14$. Thus,$B$'s share is $\frac{2}{14}x = \frac{1}{7}x$.
According to the problem,$B$ received ₹ $40$ less in the second case,so:
$\frac{1}{3}x - \frac{1}{7}x = 40$
$\frac{7x - 3x}{21} = 40$
$\frac{4x}{21} = 40$
$x = \frac{40 \times 21}{4} = 10 \times 21 = 210$.
Therefore,the actual amount is ₹ $210$.

(D) Let $A$'s share be $4x$ and $B$'s share be $7x$.
Given that the total amount is ₹ $73,689$.
Therefore,$4x + 7x = 73689$.
$11x = 73689$.
$x = 73689 / 11 = 6699$.
$A$'s share $= 4 \times 6699 = 26796$.
$B$'s share $= 7 \times 6699 = 46893$.
Thrice the share of $A = 3 \times 26796 = 80388$.
Twice the share of $B = 2 \times 46893 = 93786$.
Difference $= 93786 - 80388 = 13398$.

(D) Let the incomes of $A, B$ and $C$ be $₹ 7x, ₹ 9x$ and $₹ 12x$ respectively,and their expenditures be $₹ 8y, ₹ 9y$ and $₹ 15y$ respectively.
Given that $A$ saves one-fourth of his income,so $A$'s saving $= \frac{1}{4} \times 7x = \frac{7x}{4}$.
Since $\text{Saving} = \text{Income} - \text{Expenditure}$,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 $= 9x - 9y = 9x - 9(\frac{21x}{32}) = 9x(1 - \frac{21}{32}) = 9x(\frac{11}{32}) = \frac{99x}{32}$.
$C$'s saving $= 12x - 15y = 12x - 15(\frac{21x}{32}) = \frac{384x - 315x}{32} = \frac{69x}{32}$.
The ratio of savings of $A, B$ and $C$ is $\frac{7x}{4} : \frac{99x}{32} : \frac{69x}{32}$.
Multiplying by $32$,we get $56x : 99x : 69x$,which simplifies to $56: 99: 69$.

(A) Let the incomes of $A$ and $B$ be $2x$ and $x$ respectively.
Let the expenditures of $A$ and $B$ be $5y$ and $3y$ respectively.
The ratio of savings is given as $4:1$. Total savings $= ₹ 5,000$.
$A$'s savings $= \frac{4}{4+1} \times 5,000 = ₹ 4,000$.
$B$'s savings $= \frac{1}{4+1} \times 5,000 = ₹ 1,000$.
We have the equations:
$2x - 5y = 4,000$ --- $(1)$
$x - 3y = 1,000$ --- $(2)$
Multiply equation $(2)$ by $2$:
$2x - 6y = 2,000$ --- $(3)$
Subtract equation $(3)$ from equation $(1)$:
$(2x - 5y) - (2x - 6y) = 4,000 - 2,000$
$y = 2,000$.
Substitute $y = 2,000$ into equation $(2)$:
$x - 3(2,000) = 1,000$
$x - 6,000 = 1,000$
$x = 7,000$.
The monthly income of $B$ is $x = ₹ 7,000$.

(B) Let the two numbers be $x$ and $y$,where $x > y$.
According to the problem,the ratio of their sum to their difference is given by:
$\frac{x+y}{x-y} = \frac{5}{1}$
Applying the property of componendo and dividendo:
$\frac{(x+y) + (x-y)}{(x+y) - (x-y)} = \frac{5+1}{5-1}$
Simplifying the expression:
$\frac{2x}{2y} = \frac{6}{4}$
$\frac{x}{y} = \frac{3}{2}$
Thus,the ratio of the greater number to the smaller number is $3:2$.

(A) Let the initial number of employees be $9x$ and the initial wage per employee be $14y$.
The original total wage bill is given by the product of the number of employees and the wage per employee:
Original Bill $= 9x \times 14y = 126xy$.
According to the problem,the number of employees is reduced to $8x$ and the wage per employee is increased to $15y$.
The new total wage bill is:
New Bill $= 8x \times 15y = 120xy$.
The ratio of the new wage bill to the original wage bill is:
Ratio $= 120xy : 126xy = 120 : 126$.
Dividing both terms by their greatest common divisor,which is $6$:
$120 / 6 = 20$ and $126 / 6 = 21$.
Thus,the ratio in which the wage bill is changed is $20 : 21$.

(D) Let the number be $x$.
Given that $53 \%$ of $x$ is $358$ less than the square of $26$.
The square of $26$ is $26^2 = 676$.
So,the equation is: $0.53x = 676 - 358$.
$0.53x = 318$.
$x = \frac{318}{0.53} = 600$.
Now,we need to find $\frac{3}{4}$ of $23 \%$ of $x$.
$23 \%$ of $600 = 0.23 \times 600 = 138$.
$\frac{3}{4}$ of $138 = 138 \times 0.75 = 103.5$.

(B) Let the initial number of students in schools $A, B,$ and $C$ be $5x, 4x,$ and $7x$ respectively.
After an increase of $20\%$ in school $A$,the new number of students is $5x \times (1 + 0.20) = 5x \times 1.2 = 6x$.
After an increase of $25\%$ in school $B$,the new number of students is $4x \times (1 + 0.25) = 4x \times 1.25 = 5x$.
After an increase of $20\%$ in school $C$,the new number of students is $7x \times (1 + 0.20) = 7x \times 1.2 = 8.4x$.
The new ratio is $6x : 5x : 8.4x$.
To simplify,multiply by $5$ to remove the decimal: $6 \times 5 : 5 \times 5 : 8.4 \times 5 = 30 : 25 : 42$.

(C) Let the total number of sweets be $x$.
Initially,each child was supposed to receive $\frac{x}{450}$ sweets.
Due to the absence of $150$ children,the number of children present was $450 - 150 = 300$.
Now,each child receives $\frac{x}{300}$ sweets.
According to the problem,each child received $3$ extra sweets:
$\frac{x}{300} - \frac{x}{450} = 3$
Taking the least common multiple $(LCM)$ of $300$ and $450$,which is $900$:
$\frac{3x - 2x}{900} = 3$
$\frac{x}{900} = 3$
$x = 2700$
The total number of sweets is $2700$.
The number of sweets each child actually received is $\frac{2700}{300} = 9$.

(A) $1$. Calculate the speed of the tractor: $\text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{360 \text{ km}}{12 \text{ h}} = 30 \text{ km/h}$.
$2$. Given the ratio of speeds (Car : Jeep : Tractor) is $3:5:2$. Let the speeds be $3x, 5x, 2x$ respectively.
$3$. From the problem,the speed of the tractor is $2x = 30 \text{ km/h}$,so $x = 15 \text{ km/h}$.
$4$. Speed of the car = $3x = 3 \times 15 = 45 \text{ km/h}$.
$5$. Speed of the jeep = $5x = 5 \times 15 = 75 \text{ km/h}$.
$6$. The average speed of the car and the jeep = $\frac{\text{Speed of car} + \text{Speed of jeep}}{2} = \frac{45 + 75}{2} = \frac{120}{2} = 60 \text{ km/h}$.

(A) Let the number of coins remaining with Lalita,Amita,and Neela be $20x$,$73x$,and $83x$ respectively.
Before the distribution/usage,the number of coins they had were:
Lalita: $20x + 25$
Amita: $73x + 15$
Neela: $83x + 30$
The total number of coins is $950$.
Therefore,$(20x + 25) + (73x + 15) + (83x + 30) = 950$.
$176x + 70 = 950$.
$176x = 880$.
$x = 880 / 176 = 5$.
Amita received from Mr. Pandit = $73x + 15$.
Amita's coins = $73(5) + 15 = 365 + 15 = 380$.

(D) Let the first number be $x$ and the second number be $y$.
According to the problem,when $30$ percent of the first number is subtracted from the second number,the result is four-fifth of the second number.
This can be written as: $y - 0.3x = \frac{4}{5}y$.
Rearranging the terms to group $y$ on one side: $y - \frac{4}{5}y = 0.3x$.
Simplifying the left side: $\frac{1}{5}y = 0.3x$.
Converting $0.3$ to a fraction: $\frac{1}{5}y = \frac{3}{10}x$.
Multiplying both sides by $10$: $2y = 3x$.
Therefore,the ratio of the first number $(x)$ to the second number $(y)$ is $\frac{x}{y} = \frac{2}{3}$,which is $2:3$.

(B) Given that the total amount is $A + B + C = 1050$.
From this,we can express the combined share of $B$ and $C$ as $(B + C) = 1050 - A$.
According to the problem,the share of $A$ is $\frac{2}{5}$ of the combined share of $B$ and $C$,so $A = \frac{2}{5}(B + C)$.
Substituting the value of $(B + C)$,we get $A = \frac{2}{5}(1050 - A)$.
Multiplying both sides by $5$,we get $5A = 2(1050 - A)$.
Expanding the equation,$5A = 2100 - 2A$.
Adding $2A$ to both sides,$7A = 2100$.
Dividing by $7$,$A = 300$.
Therefore,the share of $A$ is ₹ $300$.

(B) Given ratios are $\frac{A}{B} = \frac{2}{3}$,$\frac{B}{C} = \frac{4}{5}$,and $\frac{C}{D} = \frac{5}{9}$.
To find $A : D$,we multiply the given ratios:
$\frac{A}{D} = \frac{A}{B} \times \frac{B}{C} \times \frac{C}{D}$
Substituting the values:
$\frac{A}{D} = \frac{2}{3} \times \frac{4}{5} \times \frac{5}{9}$
Canceling the common factor $5$ from the numerator and denominator:
$\frac{A}{D} = \frac{2 \times 4}{3 \times 9} = \frac{8}{27}$
Therefore,$A : D = 8 : 27$.