Desired ratio $=\displaystyle\frac{1}{2}\colon\displaystyle\frac{1}{3}\colon\displaystyle\frac{1}{4}=\displaystyle\frac{1}{2}\times12\colon\displaystyle\frac{1}{3}\times12\colon\displaystyle\frac{1}{4}\times12$
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=6\colon4\colon3$
Ratio by mistake $=2\colon3\colon4=6\colon9\colon12$
Hence, it is clear that both $Q$ and $R$ gained in the transaction.

Let the cost of almond per kg be Rs. $x$ and cost of walnuts per kg be Rs. $y$. Then
$2x=8y\;\;\Rightarrow\;y=\displaystyle\frac{x}{4}$
Given, $5x+16y=1080$
$\Rightarrow\;5x+\displaystyle\frac{16\times\,x}{4}=1080$
$\Rightarrow\;9x=1080\,$

Let the number of 10 rupee notes = x. Then,
Number of 5 rupee notes = 72 - x
Given, $x \times 10 + 5 (72 - x) = 600$
$\Rightarrow 10x + 360 - 5x = 600 \Rightarrow 5x = 240 \Rightarrow x = 48$
$\therefore$ Required ratio $= \displaystyle \frac{48}{(72 - 48)} = \frac{48}{24} = \frac{2}{1} =2 : 1$

46.3 is the number we have to convert in a fraction. 

Let the numbers be $2x$ and $7x$.
$2x + 7x = 351$
$x = 39$
Therefore, product of the numbers is $2x \times 7x$ $=$ $14x^2$
$=$ $14\, \times\, (39)^2$
$= 21,294$

Height of A $\displaystyle =\frac{4}{3}\times$ height of B
                    

Let the three numbers be 2x, 3x and 4x.
Given, $(2x)^2 + (3x)^2 + (4x)^2 = 725$
$\Rightarrow 4x^2 + 9x^2 + 16 x^2 = 725 \Rightarrow 29 x^2 = 725$
$\Rightarrow x^2 = 25 \Rightarrow x = 5$
$\therefore$ The numbers are 10, 15 and 20.

Let the numbers be $2x$ and $6x$
We have,
$6x-2x=84$
 $4x=84$
 $x=21$
Largest number $=6x$
$=6\times 21=126$

Total age of $3$ boys $= (25 $ $\times$ $3)$ years $=75$ years. 

Let the shares of $P, Q$ and $R$ be Rs. $ 4x$, Rs. $7x$ and Rs. $ 9x$ respectively. 

The whole number must   be divided by sum of two ratio particularly

A.$5+7=12$
$72$ can be divided by $12$

B.$ 3+5=8$
$72$ can be divided by $8$

C.$ 4+3=7$
$72$ can not be divided by $7$

D.$4+5=9$
$72$ can be divided by $9$

Let number of Deer $= 5x$
And number of lions $= 2x$
Given number of deer $= 20$
$\Rightarrow 5x = 20$
$\Rightarrow x = \dfrac{20}{5} = 4$
Thus, number of lions $= 2x = 2\cdot 4 = 8$

For dividing $12$ into two whole numbers, the sum of the ratios must be a factor of $12$. So they cannot be in the ratio $3:2$

Given ratio = $\dfrac{1}{2} : \dfrac{2}{3} : \dfrac{3}{4} $     ...............   Multiplying by $12$ 

Given,
No of students who applied for scholarship $=126$
No of students who got scholarship $=9$
No of students who didn't get scholarship $=126-9=117$
Ratio of students who received a scholarship to those who didn't is $9:117=1:13$.

Let the three numbers b $3x, 2x, 5x$

According to the question;

Let C's speed = x km/h
B's speed = 3x km/h
A's speed=6x km/h
There fore Ratio of speed of ABC = 6x : x= 6 : 3 : 1
Ratio of times taken = $\dfrac{1}{6} : \dfrac{1}{3} : 1=1 : 2 : 6$
If C takes 42 min , A takes 1 min.
If C takes 42 min. A takes $(\dfrac{1}{6}\times{ 42})=7\, minutes$

Let the two numbers be $8x$ and $3x$ respectively

Given that the sum of the numbers is $143$

Portion of the book left unread after one day$=1-$ portion of the book read on one day
                                                                           $=1-\dfrac38$ 

The portion of the book left unread after one day$=1-$ portion of the book read in one day
                                                                           $=1-\dfrac38$

According to the question boy pays $6$ student bill.

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 $= x+y -30$

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

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

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 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 $

P share $= 3x$

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

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.

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

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

The ratio in which Rs $715$ is to be divided is $6:4:3$.

Let the shares of $A,B\,$and$\,C$ be $3x,x\,$and$\,5x$ respectively.
$\;\;\;\;\;\;\;$ Given, $5x-x=3600\;\;\;\;\;\Rightarrow\;4x=3600$
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\Rightarrow\,x=900$
$\;\;\;\;\;\;\;\;\therefore\;A's\;share+B's\;share=Rs.\,3\times900+Rs.\,900$
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=Rs.\,2700+Rs.\,900$
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\,Rs.\,3600$

The three numbers are in the ratio $\displaystyle\frac{1}{2}\colon\displaystyle\frac{2}{3}\colon\displaystyle\frac{3}{4}$,
$\;\;\;\;\;\;\;\;i.e.,\displaystyle\frac{1}{2}\times12\colon\displaystyle\frac{2}{3}\times12\colon\displaystyle\frac{3}{4}\times12,i.e.,\,6\,\colon8\,\colon\,9$.
$\;\;\;\;\;\;\;\;\;$ Let the number be $6x,8x$ and $9x$.
$\;\;\;\;\;\;\;\;\;$ Given, $9x-6x=36$
$\;\;\;\;\;\;\;\;\;\Rightarrow\,3x=36\;\Rightarrow\;x=12$
$\therefore$The numbers are $72,96$ and $108$.

Let Q's share be Q, and P's share be $(Q - 180)$

Let the monthly savings of the family be $Rs.\,x$.
Then,$\displaystyle\frac{9}{2}=\displaystyle\frac{2700}{x}\;\Rightarrow\;\;\;x=600$
$\therefore$ Monthly expenditure$=Rs.\,2700-Rs.\,600$
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=Rs.\,2100$.

Quantity of $A=\displaystyle\frac{2}{7}\times21:kg=6:kg$,
$\;\;\;\;\;\;\;$ Quantity of $B=\displaystyle\frac{5}{7}\times21:kg=15:kg$
$\;\;\;\;\;\;\;$ After adding $3:kg$ of $A$ in the given fertilizer,

Amount to be paid in taxes $=20\%$ of $Rs.\,9,00,000=Rs.\,1,80,000$
$\;\;\;\;\;\;\;$ Amount to be distributed among $P,Q\,and\,R=9,00,000-1,80,000=7,20,000$.
$\;\;\;\;\;\;\;\therefore\;Let\;the\;shares\;of\;P,Q\,and\,R\;be\,x,\displaystyle\frac{3x}{2}$ and $2x$ respectively. 

Let, Number of one-rupee coins = x,
Number of 50 paise coins = y,
Number of 25 paise coins = z,
Given, x + y + z = 378              ....... (1)
Also, $1 \times x : \displaystyle \dfrac{50}{100} \times y : \dfrac{25}{100} \times z = 13 : 11 : 7$
$\Rightarrow x : y/2 : z/4 = 13 : 11 : 7$
$\Rightarrow \displaystyle \dfrac{x}{13} = \dfrac{y/2}{11} = \dfrac{z/4}{7} =k $  (say)
$\Rightarrow x = 13k, y = 22 k, z = 28 k.$
Then, $13 k + 22k + 28 k = 378 $        (From (1))
$\Rightarrow 63 k = 378$  hence $ k = 6$
$\therefore $ Number of 50 paise coins $= 22 \times 6 = 132$

Let the number of ladies and gents in the party be 3x and 2x respectively. Then,
$\displaystyle \frac{3x}{2x + 20} = \frac{2}{3} \Rightarrow 9x = 4x + 40 \Rightarrow 5x = 40$
$\Rightarrow x = 8$
$\therefore$ Number of ladies in the party $= 3x = 3 \times 8 = 24$

Let the two numbers be $2x$ and $3x$. Then,
$\;\;\;\;\;\;\;\;\displaystyle\frac{2x-2}{3x+2}=\displaystyle\frac{1}{2}\;\Rightarrow\;4x-4=3x+2\;\Rightarrow\,x=6$
$\therefore$ The numbers are $12$ and $18$.
Required sum$=12+18=30$.

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

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, $\displaystyle x + \frac{9x}{4} + 4x = 116$
$\Rightarrow 4x + 9x + 16x = 116 \times 4$
$\Rightarrow 29 x = 116 \times 4 \Rightarrow x = \displaystyle \frac{116 \times 4}{29} = 16$
$\therefore \text{2nd number}= \displaystyle \frac{9}{16} \times 4 \times 16 = 36$

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 $=\dfrac{12}{6}\times 360=720  Rs.$

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

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$=6\times\,Rs.\,1200=Rs.\,7200$.

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. $2000$

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 $= Rs 9500 - Rs. 8000 = Rs. 1500$

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

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. $6708$

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 $= z = 5k + 15 = 5 \times 200 + 15 = Rs. 1015$

If the sum of whole no. is $24$ then the ratio cannot be $2:5$
Clearly, $\dfrac{24}{2+5}$ is not an integer value

Ratio of coins  = 1 : 2 : 3
Value of the total coin = Rs 154
Let number of one rupee coin is x, 50 paise coin is 2x and 25 paise coin is 3x.
Value of the one rupee coin = x
value of 50 paise coin  =  Rs. $\dfrac{2x}{2}$ = Rs.x
value of the 25 paise coin  = 3x/4
value of whole coins  = 154
x + x + $\dfrac{3x}{4}$ = 154
$8x + 3x = 154 *4$
$11x = 154 *4$
x = 14 *4 = 56. Then
Number of 25 paise coins  = 3x = 3 *56 = 168

As per the given data,

Initially, Milk = 81 litres and waterr = 0 litre Afte 1st operation,
Milk = $ \displaystyle[81 - \frac{1}{3} \times 81] litres = (81 - 27) litres = 54$ litres
Water =  $\displaystyle [ 0+ \frac{1}{3} \times 54] litres $ = $(54-18)$ litres $= 36$ litres
Water= $\displaystyle [27 - \frac{1}{3} \times 27] litres + [\frac{1}{3} \times 54 + \frac{1}{3} \times 27 ] litres$ = $(27-9)$ litres + $(18+9)$ litres $= 45 $litres 
$\therefore $ Required ratio of milk and water in the new mixture $= 36: 45 = 4:5 $

Let the on one part $=x$

Let C=x. Then $A=\begin{pmatrix}x-20\end{pmatrix}$ and $B=\begin{pmatrix}x-40\end{pmatrix}$.
$x+x-20+x-40=120$ Or $x=60$
$A:B:C=40:20:60=2:1:3$
B"s share $=Rs.120\times \dfrac{1}{6}=Rs.20$

$A:B:C=35000:45000:55000=7:9:11$
A"s share $=\dfrac{7}{27}\times40500=Rs.10,500$

Let $C=x$
Then $B=\dfrac{x}{2}$
And $A=\dfrac{x}{4}$
$A:B:C=1:2:4$
C"s share $Rs.\dfrac{4}{7} \times700=400$

Let the boys are $8x$ and Girls are $5x$
$\Rightarrow\;5x=200$
$\Rightarrow\;x=40$
Total students $=8x+5x=13x=13\begin{pmatrix}40\end{pmatrix}=520$

Ratio of their shares $=22500:35000$
$=9:14$
Deepak"s share $=Rs.\begin{pmatrix}13800\times \dfrac{14}{23}\end{pmatrix}=Rs.8400$

Now as per question, Kamal invested for $12$ months and Sameer invested for $7$ months.
So, Kamal:Sameer=$\begin{pmatrix}9000\times12\end{pmatrix}:\begin{pmatrix}8000\times7\end{pmatrix}$
$=108.56=27:14$
Sameer ratio in profit will be $=6970\times \dfrac{14}{41}=Rs.2380$

Quantity of milk $=60\times \dfrac{2}{3}=40$ litres
Quantity of water $=60-40=20$ litres
As per question we need to add water to get quantity $2:1$
$\Longrightarrow \dfrac{40}{20+x}=\dfrac{1}{2}$
$\Rightarrow\;20+x=80$
$\Rightarrow\;x=60$ litres

Since the ratio is $2:3$,

$ Let\quad number\quad of\quad 50\quad paise\quad coin=a\ \quad \quad \quad number\quad of\quad 1\quad Rs\quad coin=b\ \quad \quad \quad number\quad of\quad 2\quad Rs\quad coin=c\ \therefore \quad Given\quad a:b:c=8:17:15\ \therefore \quad b's\quad part\quad is\quad \frac { 17 }{ 8+17+15 } =\frac { 17 }{ 40 } \ \therefore \quad Actual\quad number\quad of\quad b\quad is\quad \frac { 17 }{ 40 } \times 240=102\ \therefore \quad Number\quad of\quad 1\ Rs\quad coin\quad is\quad 102.\quad \quad (Ans) $

According to the problem, there are $7$ men for every $5$ women in the office.
Let the number of men and women be $7x$ and $5x$ respectively.
This means that the total number of employees will always be a multiple of $(7x+5x)=12x$.
From the given options, $30$ is not a multiple of $12$.
So, the required answer is $30$.

$Let\quad the\quad numbers\quad be\quad 3x\quad and\quad 5x.\$ 

$\Rightarrow$  Consonants in the given word are = $M,T,H,M,T,C,S$

Given that,
Total money = Rs. 9000 
A: B: C = 4 :5:6 
As share= $\cfrac{4}{(4 + 5 + 6)} \times 9000 = 2400$
Bs share = $\cfrac{5}{(4 + 5 + 6)} \times 9000 = 3000$
Cs share = $\cfrac{6}{(4 + 5 + 6)}\times 9000=3600$

It is given that,
$\dfrac {l _{1}}{l _{2}} = \dfrac {5}{3}$ and $\dfrac {s _{1}}{s _{2}} = \dfrac {6}{5}\Rightarrow \dfrac {t _{1}}{t _{2}} = \dfrac {\dfrac {l _{1}}{s _{1}}}{\dfrac {l _{2}}{s _{2}}} = \dfrac {l _{1}}{l _{2}}\times \dfrac {s _{2}}{s _{1}} = \dfrac {5}{3}\times \dfrac {5}{6} = \dfrac {25}{18}$
Hence, $t _{1} : t _{2} = 25 : 18$.