Tag: problems on ratios

Questions Related to problems on ratios

In an examination, the ratio of passes to failures was 4 : 1. Had 30 less appeared and 20 less passed, the ratio of passes to failures would have been 5 : 1. Find the number of students who appeared for the examination.

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

  1. 150

  2. 140

  3. 130

  4. 120

Show answer
Correct Option: A
Explanation:

Suppose $x$ candidates passed and $y$ failed; therefore,
$\dfrac { x }{ y } =\dfrac { 4 }{ 1 } $
$x=4y$                                                                                    (1)
In the second case; no. of students appeared $= x+y -30$

and no. of those who passed  $ = x -20$
No of failed
$N=x+y-30-(x-20)$
$N=y-10$
from the question 
$\dfrac { x-20 }{ y-10 } =\dfrac{5}{1}$
$x-20=5y-50$
$x-5y=-30$
from ( 1)
$4y-5y=-30$
$y=30$
$x=120$
So, $x+y=150$
Total number of students who appeared for the examination $=150$

Geeta read $\dfrac{3}{8}$ of a book on one day and $\dfrac{4}{5}$ of the remaining on another day. Find the portion of the book read on another (second) day.

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

  1. $\dfrac{1}{3}$

  2. $\dfrac{5}{2}$

  3. $\dfrac{4}{3}$

  4. $\dfrac{1}{2}$

Show answer
Correct Option: D
Explanation:

Portion of the book left unread after one day$=1-$ Portion of the book read on one day

                                                                          $=1-\dfrac38$$=\dfrac58$

Portion of the book read on another day $= \dfrac{4}{5}$ of portion of the book left unread after one day
                                                                 
                                                                    $=\dfrac{4}{5}$$\times$ $\dfrac{5}{8}$

                                                                    $=\dfrac{1}{2}$


So the portion of the book read on second day= $\dfrac12$

On a particular day, $\dfrac{2}{15}$th of the total number of students in a school were absent. If $1950$ were present on that day, find the total strength of the school.

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

  1. $1570$

  2. $2250$

  3. $1925$

  4. $2750$

Show answer
Correct Option: B
Explanation:

Let total number of students in the school be $x$
$\therefore$ The number of students absent on particular day $=\dfrac2{15}x$


Number of students present on that day$=x-\dfrac2{15}x=\dfrac{13}{15}x$
According to question,
$\dfrac{13}{15}x=1950$
$\Rightarrow x=1950\times \dfrac{15}{13}=2250$
$\therefore $ Total strength of  school$=2250$

Geeta read $\dfrac{3}{8}$ of a book on one day and $\dfrac{4}{5}$ of the remaining on another day.  Find the total number of pages in the book, if $60$ pages are left unread after the second day.

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

  1. $480$

  2. $400$

  3. $430$

  4. $500$

Show answer
Correct Option: A
Explanation:

Portion of the book left unread after one day$=1-$ portion of the book read on one day
                                                                           $=1-\dfrac38$
                                                                           $=\dfrac{8-3}8$
                                                                           $=\dfrac58$
Portion of the book read on another da}$=\dfrac{4}{5}$ of portion of the book left unread after one day}
                                                                   $=\dfrac45\times \dfrac58$
                                                                   $=\dfrac12$

Portion of the book left unread after two days $=1 -$ (portion of book read on one day $+$ portion of book read on another day
                                                                              $=1-(\dfrac38+\dfrac12)$
                                                                              $=1-\dfrac78$
                                                                              $=\dfrac18$

Let there be '$x$' number of pages in the book
According to question,
$\dfrac18 \text{ of x}=60$
$\Longrightarrow  x=480$

Therefore, total number of page$=480$

The annual incomes of A and B are in the ratio 3 : 4 and their annual expenditures are in the ratio 5 : 7. If each saves Rs. 5, 000 then, find their annual incomes.

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

  1. A = Rs. 22,000 and B = Rs. 30,000

  2. A = Rs. 30,000 and B = Rs. 40,000

  3. A = Rs. 50,000 and B = Rs. 70,000

  4. A = Rs. 40,000 and B = Rs. 20,000

Show answer
Correct Option: B
Explanation:

Let the annual incomes of A and B be $ 3x $ and $ 4x $

And the annual expenditures of A and B be $ 5y $ and $ 7y $

Since each of them saves $ Rs  5000 $. 

$ 3x-5y = 5000 $ ----- (1)
$4x-7y = 5000 $ ---- (2)

Multiplying equation  $ (1) $ with $ 4 $ we get, $ 12x - 20y = 20000 $ ----- equation $ (3) $

Multiplying equation  $ (2) $ with $ 3 $ we get, $ 12x-21y=15000 $ ----- equation $ (4) $

Subtracting equation $ (4) $ from $ (3) $, we get $ y = 5000 $

Substituting $ y = 5000 $ in the equation $ (1) $, we get $ 3x-5(5000) = 5000 = x = 10000 $

Hence,, annual income of A $ = 3x =$ Rs $30000 $ and of B $ = 4x =$ Rs $ 40000 $

If Rs. $840$ is divided between $P$ and $Q$ in the ratio $3:4$, then P's share is

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

  1. Rs. $340$

  2. Rs. $480$

  3. Rs. $360$

  4. Rs. $400$

Show answer
Correct Option: C
Explanation:

P share $= 3x$


Q share $= 4x$


Total $= 7x$

$7x = 840$

$x = 120$

P share $= 3 (120)$

             $= 360$

A sum of money is divided into $2$ parts such that $6$ times of one part added to $15$ times the other gives $8$ times the whole. What is the ratio of one part to the other?

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

  1. $5 : 2$

  2. $4 : 9$

  3. $7 : 6$

  4. None of these

Show answer
Correct Option: D
Explanation:

Let two parts of money  be Rs. $ x$ and Rs. $ y$.
Then, according to the given condition,

$\Rightarrow 6x+15y=8(x+y)$
$\Rightarrow 6x+15y=8x+8y$
$\Rightarrow 2x=7y$
$\Rightarrow \displaystyle \frac{x}{y}=\frac{7}{2}$
$\Rightarrow  x:y=7:2$

Hence, option D.

Some one rupee, $50$ paise and $25$ paise coins make up $Rs.\,93.75$ and their numbers are in the ratio $3\,\colon\,4\,\colon\,5$. Find the number of each type of coins.

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

  1. $\;40,70,75$

  2. $\;46,58,75$

  3. $\;42,56,70$

  4. $\;45,60,75$

Show answer
Correct Option: D
Explanation:

Let the numbers of one rupee, $50:p$ and $25:p$ coins be $3x,4x$ and $5x$ respectively. Then,
$\;\;\;\;\;\;\;\;3x\times1+4x\times0.50+5x\times0.25=93.75$
$\;\;\;\;\;\;\;\;\Rightarrow\,3x+2x+1.25x=93.75$
$\;\;\;\;\;\;\;\;\Rightarrow\,6.25x=93.75$
$\;\;\;\;\;\;\;\;\Rightarrow\,x=\displaystyle\frac{93.75}{6.25}=\displaystyle\frac{9375}{625}=15$.
$\;\;\;\;\;\;\;\;\therefore\;$The number of onerupee, $50\ p$ and $25\ p$ coins are $45,60$ and $75$ respectively.

The product of two positive integers is $936$. Find the greater number, if the integers are in the ratio $13\,\colon\,18.$ 

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

  1. $27$

  2. $31$

  3. $36$

  4. $41$

Show answer
Correct Option: C
Explanation:

Let the two positive integers be $13x$ and $18x.$
Their product is $936.$

$\therefore 13x\times 18x=936$

$\Rightarrow x^{2}=\dfrac{936}{13\times 18}=4$

$\Rightarrow x=2$
Then two positive integers are $26$ and $36.$
$\therefore$ The greater number is $36.$

The ratio of the number of boys to girls of a school with $504$ students is $13\,\colon\,11$. What will be the new ratio if $12$ more girls are admitted?

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

  1. $\;91\,\colon\,81$

  2. $\;81\,\colon\,91$

  3. $\;9\,\colon\,10$

  4. $\;10\,\colon\,9$

Show answer
Correct Option: A
Explanation:

$Number\ of\ boys$ $=\displaystyle\frac{13}{(13+11)}\times504=\displaystyle\frac{13}{24}\times504$
   $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=13\times21=273$


$\;\;\;\;\;\;\;\therefore:Number:of:girls=504-273=231$

$\;\;\;\;\;\;\;\therefore:New:ratio=\displaystyle\frac{273}{(231+12)}=\displaystyle\frac{273}{243}=\displaystyle\frac{91}{81}$