Answer: Option D. -> 1200
$$\eqalign{
& {\text{Number of tins}} \cr
& = \frac{{{\text{Voulme of the drum}}}}{{{\text{Volume of each tin}}}} \cr
& = \frac{{\left( {\frac{{22}}{7} \times \frac{{35}}{2} \times \frac{{35}}{2} \times 24} \right)}}{{\left( {\frac{{25}}{{10}} \times \frac{{22}}{{10}} \times \frac{{35}}{{10}}} \right)}} \cr
& = 1200 \cr} $$

Answer: Option B. -> decreases by 25%
Let the original radius and height of the cone be r and h respectively
Then, Original volume = $$\frac{1}{3}\pi {r^2}h$$
New radius = $$\frac{r}{2}$$ and new hight = 2h
New volume :
$$\eqalign{
& = \frac{1}{3} \times \pi \times {\left( {\frac{r}{2}} \right)^2} \times 3h \cr
& = \frac{3}{4} \times \frac{1}{3}\pi {r^2}h \cr} $$
∴ Decrease % :
$$\eqalign{
& = \left( {\frac{{\frac{1}{4} \times \frac{1}{3}\pi {r^2}h}}{{\frac{1}{3}\pi {r^2}h}} \times 100} \right)\% \cr
& = 25\% \cr} $$

Answer: Option A. -> 15 cm
Volume of bucket :
= 28.490 litres
= (28.490 ×1000) cm3
= 28490 cm3
Let the height of the bucket be h cm
We have : r = 21 cm. and R = 28 cm
$$\therefore \frac{\pi }{3}h\left[ {{{\left( {28} \right)}^2} + {{\left( {21} \right)}^2} + 28 \times 21} \right]$$     $$ = 28490$$
$$ \Rightarrow h\left( {784 + 441 + 588} \right) = $$     $$\frac{{28490 \times 21}}{{22}}$$
$$\eqalign{
& \Rightarrow 1813h = 27195 \cr
& \Rightarrow h = \frac{{27195}}{{1813}} \cr
& \Rightarrow h = 15\,cm \cr} $$

Answer: Option D. -> 80 min
Volume of cistern :
$$\eqalign{
& = \left( {\pi \times {{10}^2} \times 15} \right){m^3} \cr
& = 1500\pi {m^3} \cr} $$
Volume of water flowing through the pipe in 1 sec :
$$\eqalign{
& = \left( {\pi \times 0.25 \times 0.25 \times 5} \right){m^3} \cr
& = 0.3125\pi {m^3} \cr} $$
∴ Time taken to fill the cistern :
$$\eqalign{
& = \left( {\frac{{1500\pi }}{{0.3125\pi }}} \right) \cr
& = \left( {\frac{{1500 \times 10000}}{{3125}}} \right) \cr
& = 4800\,\sec \cr
& = \left( {\frac{{4800}}{{60}}} \right)\min \cr
& = 80\,\min \cr} $$

Answer: Option C. -> 44.0625 cm
External radius = 6 cm
Internal radius = (6 - 0.25) = 5.75 cm
Volume of material in hollow cylinder :
$$\eqalign{
& = \left[ {\frac{{22}}{7} \times \left\{ {{{\left( 6 \right)}^2} - {{\left( {5.75} \right)}^2}} \right\} \times 15} \right]{\text{c}}{{\text{m}}^3} \cr
& = \left( {\frac{{22}}{7} \times 11.75 \times 0.25 \times 15} \right){\text{c}}{{\text{m}}^3} \cr
& = \left( {\frac{{22}}{7} \times \frac{{1175}}{{100}} \times \frac{{25}}{{100}} \times 15} \right){\text{c}}{{\text{m}}^3} \cr
& = \left( {\frac{{11 \times 705}}{{56}}} \right){\text{c}}{{\text{m}}^3} \cr} $$
Let the length of solid cylinder be h
Then,
$$\eqalign{
& \frac{{22}}{7} \times 1 \times 1 \times h = \left( {\frac{{11 \times 705}}{{56}}} \right) \cr
& \Rightarrow h = \left( {\frac{{11 \times 705}}{{56}} \times \frac{7}{{22}}} \right) \cr
& \Rightarrow h = 44.0625\,{\text{cm}} \cr} $$

Answer: Option C. -> 3 units
$$\eqalign{
& \frac{4}{3}\pi {r^3} = 4\pi {r^2} \cr
& \Rightarrow r = 3 \text{ units} \cr} $$

Answer: Option B. -> 0.577 m3
Area of the base :
$$\eqalign{
& = \left( {\frac{{\sqrt 3 }}{4} \times {1^2}} \right){m^2} \cr
& = \frac{{\sqrt 3 }}{4}{m^2} \cr} $$
∴ Volume of pyramid :
$$\eqalign{
& = \left( {\frac{1}{3} \times \frac{{\sqrt 3 }}{4} \times 4} \right){m^3} \cr
& = \left( {\frac{{\sqrt 3 }}{3}} \right){m^3} \cr
& = \left( {\frac{{1.732}}{3}} \right){m^3} \cr
& = 0.577\,{m^3} \cr} $$

Answer: Option C. -> 64 : 27
Let the side of cube = 10 cm
∴ Original volume = 10 × 10 × 10 = 1000 cm3
Now, side of new cube :
= 10 - 25% of 10 = 7.5
∴ New volume = 7.5 × 7.5 × 7.5 = 421.875 cm3
∴ Required ratio :
$$\eqalign{
& = \frac{{1000}}{{421.875}} \cr
& = \frac{{1000000}}{{421875}} \cr
& = \frac{{64}}{{27}}\,Or\,64:27 \cr} $$

Answer: Option B. -> 3 cm
Let the radius of the sphere be r cm
Then,
$$\eqalign{
& \frac{4}{3}\pi {r^3} = \pi \times {\left( {0.1} \right)^2} \times 3600 \cr
& \Leftrightarrow {r^3} = 36 \times \frac{3}{4} \cr
& \Leftrightarrow {r^3} = 27 \cr
& \Leftrightarrow r = 3\,cm \cr} $$

Answer: Option C. -> 4 : 9
Let their radii be R and r
Then,
$$\eqalign{
& \frac{{\frac{2}{3}\pi {R^3}}}{{\frac{2}{3}\pi {r^3}}} = \frac{{6.4}}{{21.6}} \cr
& \Rightarrow {\left( {\frac{R}{r}} \right)^3} = \frac{8}{{27}} \cr
& \Rightarrow {\left( {\frac{R}{r}} \right)^3} = {\left( {\frac{2}{3}} \right)^3} \cr
& \Rightarrow \frac{R}{r} = \frac{2}{3} \cr} $$
∴ Ratio of curved surface area :
$$ = \frac{{2\pi {R^2}}}{{2\pi {r^2}}} = {\left( {\frac{R}{r}} \right)^2} = \frac{4}{9}\,or\,4:9$$