Tag: problems on ratios

Questions Related to problems on ratios

The ratio of the number of boys to that of the girls in a school of $432$ students is $5\colon4$. How many new girls should join the school for the ratio to be $1\colon1$.

maths fundamental concept of ratio and proportion division problem dividing a quantity in a given ratio problems on ratios

  1. $\;16$

  2. $\;68$

  3. $\;116$

  4. $\;48$

Show answer
Correct Option: D
Explanation:

Number of boys $=\displaystyle\frac{5}{9}\times432=240$
$\;\;\;\;\;\;\;$ Number of girls $=432-240=192$.
$\;\;\;\;\;\;\;$ Let the number of new girls be $x$. Then,
$\;\;\;\;\;\;\displaystyle\frac{240}{192+x}=\displaystyle\frac{1}{1}\;\Rightarrow\,240=192+x\,\Rightarrow\,x=48$.

The sum of three numbers is $116.$ The ratio of the second to the third is $9 : 16$ and the first to third is $1 : 4.$ The second number is

maths fundamental concept of ratio and proportion division problem dividing a quantity in a given ratio problems on ratios

  1. $30$

  2. $32$

  3. $34$

  4. $36$

Show answer
Correct Option: D
Explanation:

Let the 1st number be $x$. Then, third number $= 4x$
Given, $\displaystyle \frac{\text{2nd number}}{\text{3rd number}} = \frac{9}{16} \Rightarrow \frac{\text{2nd number}}{4x} = \frac{9}{16}$
$\Rightarrow \displaystyle \text{2nd number} = \frac{9}{16} \times 4x = \frac{9}{4} x$
Given, $\displaystyle x + \frac{9x}{4} + 4x = 116$
$\Rightarrow 4x + 9x + 16x = 116 \times 4$
$\Rightarrow 29 x = 116 \times 4 \Rightarrow x = \displaystyle \frac{116 \times 4}{29} = 16$
$\therefore \text{2nd number}= \displaystyle \frac{9}{16} \times 4 \times 16 = 36$

A sum of money is divided among Peter, Anita and Janet in the ratio $13\colon12\colon7$. Calculate how much Anita gets, if the amount Peter gets more than Janet is $Rs.\,360$?

maths fundamental concept of ratio and proportion division problem dividing a quantity in a given ratio problems on ratios

  1. $\;Rs.\,180$

  2. $\;Rs.\,640$

  3. $\;Rs.\,720$

  4. $\;Rs.\,480$

Show answer
Correct Option: C
Explanation:

Ratio of the money divided among  Peter, Anita and Janet is $=13:12:7$
Difference of the ratio between Peter and Janet $=13-7=6$
If difference between Peter and Janet is $6$ then Anita gets $=6$
If the difference between Peter and Janet is $Rs.360$ then Anita gets $=\dfrac{12}{6}\times 360=720  Rs.$

Divide 170 into three parts such that the first part is 10 more than the second and its ratio with the third part is 2:5.

maths fundamental concept of ratio and proportion division problem dividing a quantity in a given ratio problems on ratios

  1. $40, 30, 100$

  2. $20, 30, 100$

  3. $40, 50, 100$

  4. $50, 30, 100$

Show answer
Correct Option: A
Explanation:

Let the first part =x. Then,
Second part = x-10
$\displaystyle \frac {First \ part}{Third \ part} = \frac {2}{5} \Rightarrow Third \ part = \frac {5x}{2} $
Given, $ \displaystyle x= x - 10 + \frac {5x}{2}= 170 \Rightarrow 2x+ 2x - 20 + 5x = 340 \Rightarrow 9x = 360 \Rightarrow x = 40$
$\displaystyle \therefore $ The  three  parts  are  40, 40- 10 , 5 $ \displaystyle \times \frac {40}{2}$, i.e. , 40, 30, 100

$A$ and $B$ have a monthly incomes in the ratio $5\,\colon\,6$ and monthly expenditures in the ratio $3\,\colon\,4$. If they save $Rs.\,1800$ and $Rs.\,1600$ respectively, find the monthly income of $B$.

maths fundamental concept of ratio and proportion division problem dividing a quantity in a given ratio problems on ratios

  1. $\;Rs.\,3400$

  2. $\;Rs.\,2700$

  3. $\;Rs.\,1720$

  4. $\;Rs.\,7200$

Show answer
Correct Option: D
Explanation:

Let the monthly incomes of $A$ and $B$ be $Rs.\,5x$ and $Rs.\,6x$ respectively.
$\;\;\;\;\;\;\;\;$ Then, $\displaystyle\frac{5x-1800}{6x-1600}=\displaystyle\frac{3}{4}$
$\;\;\;\;\;\;\;\;\;\Rightarrow\,20x-7200=18x-4800$
$\;\;\;\;\;\;\;\;\;\Rightarrow\,2x=2400\;\;\Rightarrow\;\;x=1200$
$\therefore$ Monthly income of B$=6\times\,Rs.\,1200=Rs.\,7200$.

A person divided Rs. $10,800$ among his three sons in the ratio $3\colon4\colon5$. Second son kept Rs. $1000$ for himself, gave Rs. $600$ to his wife and divided the remaining money among his two daughters in the ratio $11\colon9$. Then one of his daughter's received

maths fundamental concept of ratio and proportion division problem dividing a quantity in a given ratio problems on ratios

  1. Rs. $1000$

  2. Rs. $1050$

  3. Rs. $1100$

  4. Rs. $1150$

Show answer
Correct Option: C
Explanation:

Second son's share $=\displaystyle\frac{4}{(3+4+5)}\times $Rs. $10,800$
$=\dfrac{4}{12}\times $ Rs $,10,800=$ Rs. $3600$
Remaining money with him $=$ Rs. $3600-$ Rs. $(1000+600)$
$=$ Rs. $2000$.
Both the daughter's share are $\displaystyle\frac{11}{20}\times$ Rs. $2000 $ and $\displaystyle\frac{9}{20}\times $ Rs. $2000$

$=$ Rs. $1100$ and Rs. $900$.

The incomes of $A$ and $B$ are in the ratio $5 : 3.$ The expenses of $A, B$ and $C$ are in the ratio $8 : 5 : 2.$ If C spends $Rs. 2000$ and $B$ saves $Rs. 700,$ then $A$ saves

maths fundamental concept of ratio and proportion division problem dividing a quantity in a given ratio problems on ratios

  1. $Rs. 1500$

  2. $Rs. 1000$

  3. $Rs. 500$

  4. $Rs. 250$

Show answer
Correct Option: A
Explanation:

Let the expenses of A, B and C be Rs. 8x, Rs. 5x and Rs. 2x respectively. Given, 2x = 2000
$\Rightarrow x = Rs. 1000$
$\Rightarrow$ B's expenses $= 5 \times Rs. 1000 = Rs. 5000$,
A's expenses $=Rs. 8000$
Given, B's saving $= Rs. 700$
$\Rightarrow $ B's income $= Rs. 5000 + Rs. 700 = Rs. 5700$
Given, A's income : B's income = 5 : 3
$\Rightarrow \displaystyle \frac{\text{A's income}}{5700} = \frac{5}{3}$
$\Rightarrow $ A's income $= \displaystyle \frac{5}{3} \times Rs. 5700 = Rs. 9500$
$\therefore$ A's savings $= Rs 9500 - Rs. 8000 = Rs. 1500$

$Rs. 366$ is divided among A,B & C so that A get $\dfrac{1}{2}$ as much B & C together, B get $\dfrac{2}{3}$ as much A & C together , then the share of A is

maths fundamental concept of ratio and proportion division problem dividing a quantity in a given ratio problems on ratios

  1. Rs. $122$

  2. Rs. $129.60$

  3. Rs. $146.60$

  4. Rs. $183$

Show answer
Correct Option: A
Explanation:

$A:\begin{pmatrix}B+C\end{pmatrix}=1:2$
A"s share $=Rs.\dfrac{366\times1}{3}=Rs.122$

A sum of money is delivered among $A, B, C$ and $D$ in the ratio $3:4:9:10$. If the share of $C$ is Rs. $2580$ more than the share of $B$, What is the total amount of money of $A$ and $D$ together ?

maths fundamental concept of ratio and proportion division problem dividing a quantity in a given ratio problems on ratios

  1. 6400

  2. 6708

  3. 6700

  4. 6510

Show answer
Correct Option: B
Explanation:

Given ratio : $3 : 4 : 9 : 10$
Let $A$ share $= 3x$
$B$ share $= 4x$
$C$ share $= 9x$
$D$ share $ = 10x$
Given, $C$ share is $2580$ more than $B$ share i.e.
$9x - 2580 = 4x$
$5x = 2580$
$x = 516$
So, $A$ share $= 3x = 516\times 3 = 1548$
$D$ share  $= 10x = 516\times 10 = 5160$
$A + D = 1548 + 5160 =$ Rs. $6708$

$Rs. 2430$ are distributed among three persons so that their shares be diminished by $Rs. 5, Rs. 10$ and $Rs. 15$ respectively, the remainder shall be in the ratio $3 : 4 : 5.$ The share of $C$ is

maths fundamental concept of ratio and proportion division problem dividing a quantity in a given ratio problems on ratios

  1. $Rs. 1015$

  2. $Rs. 605$

  3. $Rs. 810$

  4. $Rs. 1415$

Show answer
Correct Option: A
Explanation:

Let $Rs. 2430$ be divided among three persons with their shares as $x, y$ and $z$ respectively so that $x + y + z = 2430$
Given, $x - 5 : y - 10 : z - 15 = 3 : 4 : 5$
$\Rightarrow x- 5 = 3k, y - 10 = 4k $ and $ z - 15 = 5 k$
$\Rightarrow x = 3k + 5, y = 4k + 10$ and $z = 5k + 15$
$\therefore 3k + 5 + 4k + 10 + 5k + 15 = 2430$
$\Rightarrow 12k + 30 = 2430    \Rightarrow 12 k = 2400 \Rightarrow k = 200$
$\therefore$ C's share $= z = 5k + 15 = 5 \times 200 + 15 = Rs. 1015$