Solution:
$$\eqalign{ {\text{Daily}}\,{\text{supply}}, \cr = \left( {100 - z} \right)\% \,{\text{of}}\,y \cr = \frac{{ {\left( {100 - z} \right)y} }}{{100}} \cr {\text{Thus,}}\,{\text{required}}\,{\text{number}}\,{\text{of}}\,{\text{days}} \cr = {\frac{{ {100x} }}{{\left( {100 - z} \right)y}}} \cr} $$

Solution:
Net growth on 1000
$$= (32 - 11)$$
$$= 21$$
Net growth on 100
$$\eqalign{ = \left( {\frac{{21}}{{1000}} \times 100} \right)\% \cr = 2.1\% \cr} $$

Solution:
Income of A is 80% of B
$$\eqalign{ A = B \times \frac{{80}}{{100}} \cr \frac{A}{B} = \frac{{4x}}{{5x}} \cr} $$
Expenditure of A is 60% of expenditure of B
$$\eqalign{ A = B \times \frac{{60}}{{100}} \cr \frac{A}{B} = \frac{{3y}}{{5y}} \cr} $$
Income of A is 90% of income of B
$$\eqalign{ 4x = \frac{{90}}{{100}} \times 5y \cr 8x = 9y \cr \frac{x}{y} = \frac{9}{8} \cr} $$
Saving ratio of A and B
= (4x - 3y) : (5x - 5y)   [∴ 8x = 9y]
= (4 × 9 - 3 × 8) : (5 × 9 - 5 × 8)
= (36 - 24) : (45 - 40)
= 12 : 5
More = 12 - 5 = 7
More% $$ = \frac{7}{5} \times 100 = 140\% $$

Alternate solution
\[\begin{array}{*{20}{c}} {}{{\text{A}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{B}}} \\ {{\text{Income}} \to }{4 \times 5 \times 9 = 180\,\,\,\,\,\,\,\,\,\,5 \times 5 \times 9 = 225} \\ {{\text{Expenditure}} \to }{3 \times 4 \times 10 = 120\,\,\,\,\,\,\,\,\,\,5 \times 4 \times 10 = 200} \\ {{\text{Saving}} \to }{\overline {\underline {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,60\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,:\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,25\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} } } \end{array}\]
Ratio of A saving to B saving = 60 : 25 = 12 : 5
More = 12 - 5 = 7
More% $$ = \frac{7}{5} \times 100 = 140\% $$

Solution:
\[\begin{array}{*{20}{c}} {\text{I}}{}{{\text{II}}}{{\text{III}}} \\ {125}:{160}{100} \\ {25}:{32}{} \end{array}\]

Solution:
Consumption of gas in the smaller burner in 1 hour :
$$\eqalign{ = \left( {\frac{{14.4}}{{104}}} \right){\text{kg}} \cr = \frac{9}{{65}}{\text{kg}} \cr} $$
Consumption of gas in the larger burner in 1 hour :
$$\eqalign{ = \left( {\frac{{14.4}}{{80}}} \right){\text{kg}} \cr = \frac{9}{{50}}{\text{kg}} \cr} $$
Difference in consumption :
$$\eqalign{ = \left( {\frac{9}{{50}} - \frac{9}{{65}}} \right){\text{kg}} \cr = \frac{{27}}{{650}}{\text{kg}} \cr} $$
Required percentage difference :
$$\eqalign{ {\text{ = }}\left( {\frac{{27}}{{650}} \times \frac{{50}}{9} \times 100} \right)\% \cr = \left( {\frac{{300}}{{13}}} \right)\% \cr = 23.07\% \cr} $$

Solution:
15% of 578 + 22.5% of 644
$$ = \left( {\frac{{15}}{{100}} \times 578} \right) + $$   $$\left( {\frac{{225}}{{10}} \times \frac{1}{{100}} \times 644} \right)$$
$$\eqalign{ = 86.7 + 144.9 \cr = 231.6 \cr} $$

Solution:
The production of cycles rose to 48400 from 40000 in 2 years
⇒ Present production = 40000
⇒ After two years = 48000
⇒ Time = 2 years
⇒ Rate of increment = ?
According to the question,
Production after 2 years = Present production $${\left( {1 + \frac{{\text{R}}}{{100}}} \right)^t}$$
$$\eqalign{ \Rightarrow 48400 = 40000{\left( {1 + \frac{{\text{R}}}{{100}}} \right)^2} \cr \Rightarrow \frac{{484}}{{400}} = {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^2} \cr \Rightarrow 1 + \frac{{\text{R}}}{{100}} = \frac{{22}}{{20}} \cr \Rightarrow \frac{{\text{R}}}{{100}} = \frac{1}{{10}} \cr \Rightarrow {\text{R}} = 10\% \cr} $$
⇔ Rate of increment = 10%