Q: The barrel of a fountain-pen, cylindrical in shape, is 7 cm long and 5 mm in diameter. A full barrel of ink in the pen will be used up on writing 330 words on an average. How many words would use up a bottle of ink containing one fifth of a litre ?

Volume of a barrel = \$\dfrac{22}{7} \times 0.25 \times 0.25 \times 7 = 1.375\$ \$cm^{3}\$Volume of ink in the bottle = \$\dfrac{1}{5}\$ litre = \$\dfrac{1000}{5}=200\$ \$cm^{3}\$Therefore, total number of barrels that can be filled from the given volume of ink = \$\dfrac{200}{1.375}\$So, required number of words = \$\dfrac{200}{1.375}\times 330 = 48000\$

Q: A Rectangular piece of paper 11 cm \$\times\$ 4 cm is folded without overlapping to make a cylinder of height 4 cm. Find the volume of the cylinder.

Given a Rectangular piece of paper \$11 cm\times 4 cm\$ is folded without overlapping to make a cylinder of height 4 cmThe length of the paper becomes the circumference of the base of the cylinder and width becomes height.Let radius of the cylinder\$ = r\$ and height\$ = h\$Circumference of the base of cylinder \$=2 \pi r = 11 cm\$\$\Rightarrow 2 \times \dfrac{22}{7} \times r = 11\$\$\therefore r = \dfrac{7}{4} cm\$ Now \$h = 4 cm\$Volume of the cylinder \$(V) = \pi r^2 h\$\$= \dfrac{22}{7} \times \dfrac{7}{4} \times \dfrac{7}{4} \times 4 cm^3\$\$= 38.5 cm^3\$

Q: The internal measures of a cuboidal room are \$12m\times 8m\times 4m.\$ Find the total cost of whitewashing all four walls of a room, if the cost of white washing is 5 per \${ m }^{ 2 }.\$ What will be the cost of white washing if the ceiling of the room is also whitewashed.

The dimensions of room is \$12\times 8\times 4\$The Lateral surface area is \$2h(l+b)\$\$2(4)(12+8)\$\$8\times 20=160m^2\$Cost of white wash is \$5m^2\$Total cost is \$5\times 160=800Rs,\$

Q: A village, having a population of \$4000\$, requires \$150\$ litres of water per head per day. It has a tank measuring \$20\$ m \$\times 15\$ m \$\times 6\$ m. For how many days will the water of this tank last?

Given:\$\Rightarrow\$\$l\times b\times h=20\times 15\times 6\$\$\Rightarrow\$Capacity of tank \$=l\times b\times h=(20\times 15\times 6)\$m\$^{3}\$\$=1800\$m\$^{3}\$\$=1800000\$ litre\$\Rightarrow\$Consumption of water in \$1\$ day \$=4000\times 150\$ litres\$=600000\$ litres\$\Rightarrow\$Water of this tank will last for \$=\cfrac{1800000}{600000}\$\$=3\$ days

Q: The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.

Let \$d_{1}\$ and \$d_{2}\$ be the two diagonals of a rhombus.Then, \$d_{1}=7.5\$ cm and \$d_{2}=12\$ cmWe know that,Area of Rhombus \$=\dfrac{d_{1}d_{2}}{2}\$ \$=\dfrac{(7.5)(12)}{2}=45 cm^2\$Hence, the answer is \$45 cm^{2}\$.

Q: Find the area of four walls of a room having length, breadth and height as \$8\$ m, \$5\$ m and \$3\$ m respectively. Find the cost of white-washing the walls at the rate of Rs. \$15 / m^2\$.

It is given that length, breadth and height of a room are \$8\$ m, \$5\$ m and \$3\$ m respectively.Area of four walls of the room is: \$2h(l + b)\$\$A=2h(l + b)=2\times 3(8+5)=6\times 13=78\$ m\$^2\$Area of the ceiling is: \$A=l\times b=8\times 5=40\$ m\$^2\$Total area of the four walls and the ceiling of the room is \$78+40=118\$ m\$^2\$It is given that cost of white washing is Rs.\$15\$/m\$^2\$, therefore,Total cost of white washing is:\$118\times 15=1770\$Hence, the cost of white washing the walls is Rs.\$1770\$.

Q: A cylinder has radius \$7\$cm and height \$10\$cm.Find the total surface area of the cylinder.

Given: Radius\$=7\$cm,Height\$=10\$cmTotal surface area\$=2\pi r\left(h+r\right)\$sq.units\$=2\times\dfrac{22}{7}\times 7\left(10+7\right)\$\$=44\times 17=748\$sq.cm

Q: If two cubes of dimensions 2 cm by 2 cm by 2 cm are placed side by side, what would the dimensions of the resulting cuboid ?

The side of cube is \$2cm\$The resultant length of cuboid is \$2+2=4cm\$The breadth of cuboid is \$2cm\$The height of cuboid is \$2cm\$

Q: The area of trapezium shaped field is \$480\: m^2\$, the distance between two parallel sides is \$15\: cm\$ and one of the parallel sides is \$20\: m\$. Find the other parallel side.

Let \$ABCD\$ be a trapezium with parallel sides \$AB\$ and \$CD\$Let \$AB=20\$ ....... (Given) and \$CD=x\$\$A(\square ABCD)=480\ m^2, \text{height}(h)=15\ m\$ .......... (Given)We know that, \$A(\square ABCD)=\dfrac{(AB+CD)}{2}\times h \$\$\implies 480 =\dfrac{20+x}{2}\times 15\$\$\implies \dfrac{2\times 480}{15}=20+x\$\$\implies 64=20+x\$\$\implies x=44\$Hence, length of other parallel side \$CD\$ is \$44\ m\$.

Question 1

A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4 cm (see Fig. 13.11). Find its(i) inner curved surface area,(ii) outer curved surface area,(iii) total surface area.

Accepted Answer

(i) inner curved surface areaAccording to the question inner diameter is 4 cm,then,Inner radius of a cylindrical pipe $ = {r_1} = \dfrac{{innner\,diameter}}{2} = \left( {\dfrac{4}{2}} \right)cm = 2cm$Height (h) of cylindrical pipe=77 cm,Curved Surface Area of inner surface of pipe  $=2\pi {r_1}h$$=\left( {2 \times \dfrac{{22}}{7} \times 2 \times 77} \right)c{m^2}$$=968\,c{m^2}$(ii) Outer curved surface areaAccording to the question outer diameter is 4.4 cmOuter radius of cylindrical pipe $ = {r_2} = \dfrac{{outer\,diamterer}}{2} = \left( {\dfrac{{4.4}}{2}} \right)\,cm\, = \,2.2\,cm$Height of cylinder = h = 77 cm,Curved area of outer surface of pipe = $2\pi {r_2}h$$=\left( {2 \times \dfrac{{22}}{7} \times 2.2 \times 77} \right)c{m^2}$$=\left( {2 \times 22 \times 2.2 \times 11} \right)\,c{m^2}$$=1064.8\,c{m^2}$(iii) Total surface areaTotal surface area = curved surface area of inner cylinder + curved surface of outer cylinder + 2 $ \times $ Area of baseArea of base = area of circle with radius 2.2 cm - Area of circle with radius 2 cm$=\pi {r_2}^2 - \,\pi {r_1}^2$$=\dfrac{{22}}{7} \times ({(2.2)^2} - {(2)^2})$$=\dfrac{{22}}{7} \times (4.84 - 4)$$=\dfrac{{22}}{7} \times (0.84)$$=2.74\,c{m^2}$Total surface area = curved surface area of inner cylinder + curved surface area of outer cylinder + 2 $ \times $ Area of base$ = \,968\,+ \,1064.8\,+\,2\, \times \,2.74$$=$ 2032.8 + 5.76$ = 2039.08\,c{m^2}$Therefore, the total surface area of the cylindrical pipe is $ = 2039.08\,c{m^2}$

Question 2

The barrel of a fountain-pen, cylindrical in shape, is 7 cm long and 5 mm in diameter. A full barrel of ink in the pen will be used up on writing 330 words on an average. How many words would use up a bottle of ink containing one fifth of a litre ?

Accepted Answer

Volume of a barrel = $\dfrac{22}{7} \times 0.25 \times 0.25 \times 7 = 1.375$ $cm^{3}$Volume of ink in the bottle = $\dfrac{1}{5}$ litre = $\dfrac{1000}{5}=200$ $cm^{3}$Therefore, total number of barrels that can be filled from the given volume of ink = $\dfrac{200}{1.375}$So, required number of words = $\dfrac{200}{1.375}\times 330 = 48000$

Question 3

A Rectangular piece of paper 11 cm $\times$ 4 cm is folded without overlapping to make a cylinder of height 4 cm. Find the volume of the cylinder.

Accepted Answer

Given a Rectangular piece of paper $11 cm\times 4 cm$ is folded without overlapping to make a cylinder of height 4 cmThe length of the paper becomes the circumference of the base of the cylinder and width becomes height.Let radius of the cylinder$ = r$ and height$ = h$Circumference of the base of cylinder $=2 \pi r = 11 cm$$\Rightarrow 2 \times \dfrac{22}{7} \times r = 11$$\therefore r = \dfrac{7}{4} cm$ Now $h = 4 cm$Volume of the cylinder $(V) = \pi r^2 h$$= \dfrac{22}{7} \times \dfrac{7}{4} \times \dfrac{7}{4} \times 4 cm^3$$= 38.5 cm^3$

Question 4

The internal measures of a cuboidal room are $12m\times 8m\times 4m.$ Find the total cost of whitewashing all four walls of a room, if the cost of white washing is 5 per ${ m }^{ 2 }.$ What will be the cost of white washing if the ceiling of the room is also whitewashed.

Accepted Answer

The dimensions of room is $12\times 8\times 4$The Lateral surface area is $2h(l+b)$$2(4)(12+8)$$8\times 20=160m^2$Cost of white wash is $5m^2$Total cost is $5\times 160=800Rs,$

Question 5

A village, having a population of $4000$, requires $150$ litres of water per head per day. It has a tank measuring $20$ m $\times 15$ m $\times 6$ m. For how many days will the water of this tank last?

Accepted Answer

Given:$\Rightarrow$$l\times b\times h=20\times 15\times 6$$\Rightarrow$Capacity of tank $=l\times b\times h=(20\times 15\times 6)$m$^{3}$$=1800$m$^{3}$$=1800000$ litre$\Rightarrow$Consumption of water in $1$ day $=4000\times 150$ litres$=600000$ litres$\Rightarrow$Water of this tank will last for $=\cfrac{1800000}{600000}$$=3$ days

Question 6

The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.

Accepted Answer

Let $d_{1}$ and $d_{2}$ be the two diagonals of a rhombus.Then, $d_{1}=7.5$ cm and $d_{2}=12$ cmWe know that,Area of Rhombus $=\dfrac{d_{1}d_{2}}{2}$ $=\dfrac{(7.5)(12)}{2}=45 cm^2$Hence, the answer is $45 cm^{2}$.

Question 7

Find the area of four walls of a room having length, breadth and height as $8$ m, $5$ m and $3$ m respectively. Find the cost of white-washing the walls at the rate of Rs. $15 / m^2$.

Accepted Answer

It is given that length, breadth and height of a room are $8$ m, $5$ m and $3$ m respectively.Area of four walls of the room is: $2h(l + b)$$A=2h(l + b)=2\times 3(8+5)=6\times 13=78$ m$^2$Area of the ceiling is: $A=l\times b=8\times 5=40$ m$^2$Total area of the four walls and the ceiling of the room is $78+40=118$ m$^2$It is given that cost of white washing is Rs.$15$/m$^2$, therefore,Total cost of white washing is:$118\times 15=1770$Hence, the cost of white washing the walls is Rs.$1770$.

Question 8

A cylinder has radius $7$cm and height $10$cm.Find the total surface area of the cylinder.

Accepted Answer

Given: Radius$=7$cm,Height$=10$cmTotal surface area$=2\pi r\left(h+r\right)$sq.units$=2\times\dfrac{22}{7}\times 7\left(10+7\right)$$=44\times 17=748$sq.cm

Question 9

If two cubes of dimensions 2 cm by 2 cm by 2 cm are placed side by side, what would the dimensions of the resulting cuboid ?

Accepted Answer

The side of cube is $2cm$The resultant length of cuboid is $2+2=4cm$The breadth of cuboid is $2cm$The height of cuboid is $2cm$

Question 10

The area of trapezium shaped field is $480\: m^2$, the distance between two parallel sides is $15\: cm$ and one of the parallel sides is $20\: m$. Find the other parallel side.

Accepted Answer

Let $ABCD$ be a trapezium with parallel sides $AB$ and $CD$Let $AB=20$ ....... (Given) and $CD=x$$A(\square ABCD)=480\ m^2, \text{height}(h)=15\ m$ .......... (Given)We know that, $A(\square ABCD)=\dfrac{(AB+CD)}{2}\times h $$\implies 480 =\dfrac{20+x}{2}\times 15$$\implies \dfrac{2\times 480}{15}=20+x$$\implies 64=20+x$$\implies x=44$Hence, length of other parallel side $CD$ is $44\ m$.