6 dozen eggs = 72 eggs
Let the required amount be Rs. x.

Let food stock is sufficient for $x$ days.

Time taken to cover $600km$ is `12 Hrs

From the given statements we can make the following equations.
(I) $\Rightarrow y=x+6$
(II) $\Rightarrow 0.4x=0.2y\Rightarrow \frac {x}{y}=\frac {x}{y}=\frac {3}{4}$
(III) $\Rightarrow \frac {y/2}{x/3}=\frac {2}{1}\Rightarrow \frac {y}{x}=\frac {4}{3}\Rightarrow \frac {x}{y}=\frac {3}{4}$
Obviously, question can be solved by using (I) and either (II) or (III) because equations (II) and (III) are same.

Suppose $sum=P$
Given $CI-SI=Rs. 122$
$P\left (1+\frac {5}{100}\right )^3-P-\left (\frac {P\times 5\times 3}{100}\right )=122$
$P\left (\frac {105^3}{100^3}-1-\frac {15}{100}\right )=122$
$P\left (\frac {7,625}{100^3}\right )=122$
$P=\frac {122\times 100^3}{7,625}=Rs. 16,000$

Ratio of catches by Billy and Kitty is 3 : 2. Therefore,
$3x+2x=60$
$5x=60$ or $x=\frac {60}{5}=12$
Number of mice caught by Kitty $=2x=2\times 12=24$

In questions of this kind, it is essential to have all quantities expressed in the same denomination; in the present instance, it will be convenient to express the money in rupees.
Let x be the number of geese. Then 108 -x is the number of ducks.
Since each goose costs 7 rupees, x geese cost 7x rupees.
And since each duck costs 3 rupees, 108 -x ducks cost 3(108 -x) rupees.
Therefore, the amount spent is 7x + 3(108 -x) rupees; but the question states that the amount is Rs 564. Hence,
$7x + 3(108 -x) = 564$
or $7x + 324 -3x = 564$
or $4x = 240$
Therefore, the number of geese, $x = 60$, and the number of ducks, $108 -x = 48$.

$P _1=1,600, t _1=3$
$P _2=1,1000, t _2=4$
$SI _1+SI _2=506$
or $\frac {1,600\times 3\times r}{100}+\frac {1,100\times 4\times r}{100}=506$
or $r(4,800+4,400)=506\times 100$
or $r=\frac {506\times 100}{92,000}=5.5$%

Let the salary be Rs 100
Savings $=$ Rs 20 ; Expenditure $=$ Rs 80
New expenditure $=$ 120% of Rs 80 $=$ Rs 96.
New savings $=Rs:100-Rs:96=Rs:4$
$\implies4\%x=Rs:800\implies x=Rs:20,000$

Cost Price $(C.P.) = Rs. (4700 + 800) = Rs. 5500$.
Selling Price $(S.P.) = Rs. 5800$.
$Gain = (S.P.) - (C.P.) = Rs.(5800 - 5500) = Rs. 300$.
Gain % $= \left (\dfrac {300}{5500}\times 100\right )$% $= 5\dfrac {5}{11}$%

Let the required weight be x kg.
Less weight, Less cost (Direct Proportion)
$\therefore 250 : 200 :: 60 : x \Leftrightarrow 250 \times x = (200 \times 60)$
$\Rightarrow x = \dfrac{(200 \times 60)}{250}$
$\Rightarrow x = 48$

Let the required number of revolutions made by larger wheel be x.
Then, More cogs, Less revolutions (Indirect Proportion)
$\therefore 14 : 6 :: 21 : x \Leftrightarrow 14 \times x = 6 \times 21$
$\Rightarrow x = \dfrac{6 \times 21}{14}$
$\Rightarrow x = 9$

Let the height of the building x metres.
Less lengthy shadow, Less in the height (Direct Proportion)
$\therefore 40.25 : 28.75 :: 17.5 : x \Leftrightarrow 40.25 \times x = 28.75 \times 17.5$
$x = \dfrac{28.75 \times 17.5}{40.25}$
$\Rightarrow x = 12.5$

Let the required number of days be x.
Less men, More days (Indirect Proportion)
$\therefore 27 : 36 :: 18 : x \Leftrightarrow 27 \times x = 36 \times 18$
$\Rightarrow x = \dfrac{36 \times 18}{27}$
$\Rightarrow x = 24$

Let the required number days be x.
Less spiders, More days (Indirect Proportion)
Less webs, Less days (Direct Proportion)
$\left.\begin{matrix}Spiders & 1 : 7 \ Webs  &  7 : 1 \end{matrix}\right} :: 7 : x$
$\therefore 1 \times 7 \times x = 7 \times 1 \times 7$
$\Rightarrow x = 7$

Number of pages typed Rave in $1$ hour $=\cfrac{32}{6}=\cfrac{16}{3}$
Number of pages typed by Kumar in $1$ hour $=\cfrac{40}{5}=8$.
Number of pages typed by both in $1$ hour $=\left( \cfrac { 16 }{ 3 } +8 \right) =\cfrac { 40 }{ 3 } $.
$\therefore$ Time taken by both to type $110$ pages $=\left( 110\times \cfrac { 3 }{ 40 }  \right) $ hours
$=8\cfrac{1}{4}$ hours (or) $8$ hours $15$ minutes.

Ratio of times taken by Sakshi and Tany $=125:100=5:4$.
Suppose Tanya takes $x$ days to do the work.
$5:4::20:x$ $\Rightarrow \left( \cfrac { 5\times 20 }{ 5 }  \right) $
$\Rightarrow$ $x=16$ days.
Hence, Tanya takes $16$ days to complete the work.

Let the cost price of the T.V.$ = x$

$A = \pi r^2$
$\displaystyle \frac{\Delta A}{A} = 2 \frac{\Delta A}{r}$
$\displaystyle \frac{\Delta A}{A}$% $= 2 \displaystyle \left ( \frac{\Delta A}{r} \right ) \times 100$
$\displaystyle \frac{\Delta A}{A}$% $= 2 \times 0.15 = 0.30$%

$\displaystyle \frac{V'}{V} = \frac{172.8}{100} = \frac{\displaystyle \frac{4}{3} \pi R^{.3}}{\displaystyle \frac{4}{3} \pi R^3}$
$\displaystyle \frac{R'}{R} = 1.2$
Now, ratio of surface area $= \displaystyle \frac{S'}{S} = \frac{4 \pi R^{.2}}{4 \pi R^3}$
$= \displaystyle \frac{S'}{S} = 1.44$
Hence surface area increased by 44%

Let the total number of eggs=x.

$(C.P.\ of\ 17\ balls) - (S.P.\ of\ 17\ balls) = (C.P.\ of\ 5\ balls)$
$\Rightarrow$ C.P. of $12\ balls = S.P.\ of\ 17\ balls = Rs. 720$.
$\Rightarrow C.P.\ of\ 1\ ball = Rs. \left (\dfrac {720}{12}\right )= Rs. 60$.

Let C.P. be $Rs. x$ and S.P. be $Rs. y$.
Then, $3(y - x) = (2y - x) \Rightarrow y = 2x$.
$Profit = Rs. (y - x) = Rs. (2x - x) = Rs. x$.
$\therefore Profit$ % $= \left (\dfrac {x}{x}\times 100\right )$% $= 100$%

Part filled in $4$ minutes $=4\left(\displaystyle\frac{1}{15}+\frac{1}{20}\right)=\displaystyle \frac{7}{15}$.
Remaining part$=\left(1-\displaystyle\frac{7}{15}\right)=\displaystyle\frac{8}{15}$.
Part filled by B in $1$ minute $=\displaystyle\frac{1}{20}$
$\therefore \displaystyle\frac{1}{20}:\frac{8}{15}::1:x$
$x=\left(\displaystyle\frac{8}{15}\times 1\times 20\right)=10\displaystyle\frac{2}{3}$min$=10$ min. $40$ sec.
$\therefore$ The tank will be full in $(4$ min. $+10$ min. $+40$ sec.)$=14$min. $40$sec.

I. Let C.P. be $Rs. x$.
Then, $M.P. = 130$% of $x = Rs. \left (\dfrac {13x}{10}\right )$
III. $S.P. = 90$% of M.P.
Thus, I and III give, $S.P. = Rs. \left (\dfrac {90}{100}\times \dfrac {3x}{10}\right ) = Rs. \left (\dfrac {117x}{100}\right )$
$Gain = Rs. \left (\dfrac {117x}{100} - x\right ) = Rs. \dfrac {17x}{100}$
Thus, from I and III, gain % can be obtained.
Clearly, II is redundant.

Curved surface area of sphere$=4\pi r^2$
diametre after decreases by $25\%$
$A _D=\pi d^2$
$d _1=\displaystyle\frac{75}{100}d=\frac{3d}{4}$
$A _1=\pi \left(\displaystyle\frac{3d}{4}\right)^2=\frac{9}{16}\pi d^2$
$\%$ decrease=$\displaystyle\frac{\displaystyle\frac{9}{16}\pi d^2-\pi d^2}{\pi d^2}\times 100=-43.75\%$