Q11. A room is 9.68 m long and 6.2 m wide. Its floor is to be covered with rectangular tiles of size 22 cm by 10 cm. Find the total cost of the tiles at the rate of Rs. 2.50 per tile.
Sol. \begin{array}{l} \text {Area of the room }=\text { Length } \times \text { Breadth } =9.68 \times 6.2 =60.016 \text { m}^{2} \\\text {Area of 1 tile }=\text { Length } \times \text { Breadth } =22 \times 10 =220 \text { cm}^{2} \\=220 \times \frac {1}{10000} \text { m}^{2} \quad \quad [\because 1 \text { cm}^{2} = \frac {1}{10000} \text { m}^{2}] \\ =0.220 \text { m}^{2} \\\text {Total no. of tiles required to cover the room} = \frac {\text {Area of the room}}{\text {Area of 1 tile}} \\ =\frac {60.016}{0.220} =\frac {60.016 \times 1000}{220} = 2728 \text { tiles}\\\text {Given cost of one tile }=\text {Rs. } 2.50 \\ \therefore \text {Total cost to tile the wall } = 2728 \times 2.50 = \text {Rs.} 6820\end{array}Q12. One side of a square field is 179 m. Find the cost of raising a lawn on the field at the rate of Rs 1.50 per square metre.
Sol. \begin{array}{l} \text {Given length of side of the square field } =179 \text { m} \\ \text {Area of the field } =(\text { Side })^{2}=(179)^{2}=32041 \text { m}^{2} \\ \text {Rate of raising a lawn on the field per square metre }= \text { Rs. } 1.50 \\ \therefore \text {Total cost of raising a lawn on the field }=32041 \times 1.50 =\text {Rs. } 48061.50\end{array}Q13. A rectangular field is measured 290 m by 210 m. How long will it take for a girl to go two times round the field, if she walks at the rate of 1.5 m/sec ?
Sol.
Length of the rectangular field = 290 m
Breadth of the rectangular field = 210 m
Perimeter of the rectangular field = 2(Length + Breadth) = 2(290 + 210) = 1000 m
Total distance that girl need to cover}=2 \times \text {Perimeter of the rectangular field
=2 × 1000 = 2000 m
\text {Given that girl walks at the rate of 1.5 m/sec. \begin{array}{l}\therefore \text {Time required to cover a distance of 2000 m } = \frac {\text {Total distance}}{\text {Speed of girl}} \\ =\frac{2000}{1.5} \text { sec} \\ =\frac{2000}{1.5} \times \frac {1}{60} \text { min} \quad [\because 1 \text { sec} = \frac {1}{60} \text { min}] \\ =22 \frac{2}{9} \text { min} \\ \text {Hence, the girl will take } 22 \frac{2}{9} \text { min to complete two round of the field.}\end{array}Q14. A corridor of a school is 8 m long and 6 m wide. It is to be covered with convas sheets. If the available canvas sheets have the size 2 m × 1 m, find the cost of canvas sheets required to cover the corridor at the rate of Rs. 8 per sheet.
Sol. \begin{array}{l} \text {Area of the corridor of a school }=\text { Length } \times \text { Breadth } =8 \times 6 =48 \text { m}^{2} \\\text {Area of 1 canvas sheet }=\text { Length } \times \text { Breadth } =2 \times 1 =2 \text { m}^{2} \\\text {Total no. of canvas sheets required to cover the corridor} = \frac {\text {Area of the corridor of a school}}{\text {Area of 1 canvas sheet}} \\ =\frac {48}{2} = 24 \text { sheets}\\\text {Given cost of 1 canvas sheet }=\text {Rs. } 8 \\ \therefore \text {Total cost of the canvas sheets } = 24 \times 8 = \text {Rs.} 192\end{array}Q15. The length and breadth of a playground are 62 m 60 cm and 25 m 40 cm respectively. Find the cost of turfing it at Rs 2.50 per square metre. How long will a man take to go three times round the field, if he walks at the rate of 2 metres per second.
Sol.
Length of a playground = 62 m 60 cm.
\begin{array}{l}=62 + 60 \times \frac {1}{100} \text { m} \quad [\because \frac {1}{100} \text { m} =1 \text{ cm}] \\ =62.6 \text { m} \\\text {Breadth of a playground} =25 \text { m } 40 \text { cm} \\ =25 + 40 \times \frac {1}{100} \text { m} \quad [\because \frac {1}{100} \text { m} =1 \text{ cm}] \\ =25.4 \text { m} \\\text {Area of a playground }= \text {Length} \times \text {Breadth} =62.6 \times 25.4 =1590.04 \text { m}^{2} \\ \text {Rate of turfing per sq. metre }= \text {Rs. } 2.50 \\ \text {Total cost of turfing the playground } =\text {Rs. } 1590.04 \times 2.50=\text {Rs. } 3975.10 \\\text {Perimeter of the rectangular field } =2(\text { Length }+\text { Breadth })=2(62.6+25.4)=176 \text { m} \\ \text {Total distance that man need to cover}=3 \times \text {Perimeter of the rectangular field } \\ =3 \times 176=528 \text { m} \\\text {Given that man walks at the rate of 2 m/sec } \\ \therefore \text {Time required to cover a distance of 528 m } = \frac {\text {Total distance}}{\text {Speed of man}} \\ =\frac{528}{2} \text { sec} \\ =\frac{528}{2} \times \frac {1}{60} \text { min} \quad [\because 1 \text { sec} = \frac {1}{60} \text { min}] \\ =\frac {264}{60} \text { min} \\ =4 \frac {24}{60} \text { min} \\ = 4 \text { min} 24 \text { sec } \\ \text {Hence, the man will take 4 min 24 sec to complete 3 round of the playground.}\end{array}Q16. A lane 180 m long and 5 m wide is to be paved with bricks of length 20 cm and breadth 15 cm. Find the cost of bricks that are required, at the rate of Rs. 750 per thousand.
Sol.
Length of the lane = 180 m.
Width of the lane = 5 m.
\begin{array}{l} \text {Area of a lane }= \text {Length} \times \text {Breadth} =180 \times 5 =900 \text { m}^{2} \\\text {Length of the brick }=20 \text { cm} \\ \text {Breadth of the brick }=15 \text { cm} \\ \text {Area of 1 brick }= \text {Length} \times \text {Breadth} =20 \text { cm} \times 15 \text { cm} \\ =300 \text { cm}^{2} \\=300 \times \frac {1}{10000} \text { m}^{2} \quad [\because 1 \text{ cm}^{2} = \frac {1}{10000} \text { m}^{2}] \\=0.03 \text { m}^{2} \\\text {Total no. of bricks required } =\frac{900}{0.03}=30000 \text { bricks.}\text {Given that cost of 1000 bricks } = \text {Rs. } 750 \\ \Rightarrow \text {Cost of 1 brick} = \text {Rs. } \frac {750}{1000} \\ \Rightarrow \text {Cost of 30000 bricks} = \text {Rs. } \frac {750}{1000} \times 30000 \\ = \text {Rs. } 22500 \\ \end{array}Q17. How many envelopes can be made out of a sheet of paper 125 cm by 85 cm, supposing one envelope requires a piece of paper of size 17 cm by 5 cm ?
Sol.
Length of the sheet of paper = 125 cm.
Breadth of the sheet of paper = 85 cm.
\begin{array}{l} \text {Area of a sheet of paper }= \text {Length} \times \text {Breadth} =125 \times 85 =10625 \text { cm}^{2} \\ \text {Length of sheet required for an envelope } =17 \text { cm} \\ \text {Breadth of sheet required for an envelope } =5 \text { cm} \\ \text {Area of the sheet required for 1 envelope }= \text {Length} \times \text {Breadth} =17 \times 5 =85 \text { cm}^{2} \\ \text {Hence, total no. of envelopes that can be made }=\frac{10625}{85}=125 \text { envelopes.} \end{array}Q18. The width of a cloth is 170 cm. Calculate the length of the cloth required to make 25 diapers, if each diaper requires a piece of cloth of size 50 cm by 17 cm.
Sol.
Given that length of the diaper = 50 cm.
Breadth of the diaper = 17 cm.
\begin{array}{l}\text {Area of cloth to make 1 diaper} = \text { Length } \times \text { Breadth} =50 \times 17 =850 \text { cm}^{2} \\ \text {Hence, area of 25 such diapers }=25 \times 850 =21250 \text { cm}^{2} \\ \text {Area of cloth required for 25 diapers} =21250 \text { cm}^{2} \\ \text {Given that width of a cloth }=170 \text { cm} \\ \text {Lets assume that length of the cloth be x cm} \\ \therefore \text {Area of cloth for 25 diapers} = 170 \times x \\ \Rightarrow 21250 = 170 \times x \\ \Rightarrow x = \frac {21250}{170} \\ \Rightarrow x = 125 \text { cm} \\\end{array} Hence, length of the cloth will be 125 cm.
Q19. The carpet for a room 6.6 m by 5.6 m costs Rs. 3960 and it was made from a roll 70 cm wide. Find the cost of the carpet per metre.
Sol.
Length of a room = 6.6 m.
Breadth of a room = 5.6 m.
\begin{array}{l}\text {Area of a room }= \text {Length} \times \text {Breadth} =6.6 \times 5.6 =36.96 \text { m}^{2} \\\text {Given that Width of a carpet }=70 \text { cm}=0.7 \text { m} \quad [\because 1 \text { m}= 100 \text { cm}] \\\text {Length of a carpet} =\frac{\text {Area of the room}}{\text {Width of the carpet}}=\frac{36.96}{0.7}=52.8 \text { m} \\\text {Cost of 52.8 m long roll of carpet }=\text {Rs. } 3960 \\ \therefore \text {Cost of 1 m long roll of carpet }=\text {Rs. } \frac {3960}{52.8}=\text {Rs. } 75\end{array}Q20. A room is 9 m long, 8 m broad and 6.5 m high. It has one door of dimensions 2 m x 1.5 m and three windows each of dimensions 1.5 m x 1 m. Find the cost of white washing the walls at Rs. 3.80 per square metre.
Sol.
Given that length of the room = 9 m.
Breadth of the room = 8 m.
Height of the room = 6.5 m.
\begin{array}{l} \text {Area of 4 walls }= 2lh + 2bh = 2(l+b)h \\ = 2(9 +8 ) \times 6.5 =2 \times 17 \times 6.5 =221 \text { m}^{2} \\\text {Length of a door } =2 \text { m} \\ \text {Breadth of a door }=1.5 \text { m} \\ \text {Area of a door }= \text {Length} \times \text {Breadth} \\ =2 \times 1.5 =3 \text { m}^{2} \\\text {Length of a window }=1.5 \text { m} \\ \text {Breadth of a window }=1 \text { m} \\ \text {Area of one window }= \text {Length} \times \text {Breadth} =1.5 \times 1 =1.5 \text { m}^{2} \\\text {Thus, Area of 3 such windows} =3 \times 1.5 =4.5 \text { m}^{2} \\\text {Area to be white-washed } = \text {Area of 4 walls} – \text {Area of one door} – \text {Area of 3 windows} \\ =221-3+ 4.5 =213.5 \text { m}^{2} \\\end{array}Cost of white-washing for 1 sq. metre of area = Rs. 3.80
Cost of white-washing for 213.5 sq. metre area= Rs. 213.5 × 3.80 =Rs. 811.30
Q21. A hall 36 m long and 24 m broad allowing 80 sq metre for doors and windows, the cost of papering the walls at Rs. 8.40 per sq. metre is Rs. 9408. Find the height of the hall.
Sol.
Given that length of the hall = 36 m
Breadth of the hall = 24 m
Let’ assume that height of the hall be h m.
For papering the wall, we need to paper the four walls excluding the floor and roof of the hall along with doors and windows. \begin{array}{l}\Rightarrow \quad \text {The area of the wall which is to be papered }= {Area of 4 walls } – \text {Area of doors and windows} \\=2 h(1+b) – 80 \text { m}^{2} \\ =2 h(36+24) – 80 \text { m}^{2} \\ =120h – 80 \text { m}^{2} \\\text {Givne rate of papering the wall } = \text {Rs. 8.40 per sq. meter} \\\text {Total cost of papering the walls} = (120h – 80) \times 8.40 \\ \Rightarrow \quad 9408 = (120h – 80) \times 8.40 \\ \Rightarrow \quad 120h -80 = \frac {9408}{8.40} \\ \Rightarrow \quad 120h -80 = 1120 \\ \Rightarrow \quad 120h = 1120 +80 \\ \Rightarrow \quad h = \frac {1200}{120} \\ \Rightarrow \quad h = 10 \text { m} \\ \text {Hence, the height of the wall is 10 m.} \end{array}
Sol. \begin{array}{l} \text {Area of the room }=\text { Length } \times \text { Breadth } =9.68 \times 6.2 =60.016 \text { m}^{2} \\\text {Area of 1 tile }=\text { Length } \times \text { Breadth } =22 \times 10 =220 \text { cm}^{2} \\=220 \times \frac {1}{10000} \text { m}^{2} \quad \quad [\because 1 \text { cm}^{2} = \frac {1}{10000} \text { m}^{2}] \\ =0.220 \text { m}^{2} \\\text {Total no. of tiles required to cover the room} = \frac {\text {Area of the room}}{\text {Area of 1 tile}} \\ =\frac {60.016}{0.220} =\frac {60.016 \times 1000}{220} = 2728 \text { tiles}\\\text {Given cost of one tile }=\text {Rs. } 2.50 \\ \therefore \text {Total cost to tile the wall } = 2728 \times 2.50 = \text {Rs.} 6820\end{array}Q12. One side of a square field is 179 m. Find the cost of raising a lawn on the field at the rate of Rs 1.50 per square metre.
Sol. \begin{array}{l} \text {Given length of side of the square field } =179 \text { m} \\ \text {Area of the field } =(\text { Side })^{2}=(179)^{2}=32041 \text { m}^{2} \\ \text {Rate of raising a lawn on the field per square metre }= \text { Rs. } 1.50 \\ \therefore \text {Total cost of raising a lawn on the field }=32041 \times 1.50 =\text {Rs. } 48061.50\end{array}Q13. A rectangular field is measured 290 m by 210 m. How long will it take for a girl to go two times round the field, if she walks at the rate of 1.5 m/sec ?
Sol.
Length of the rectangular field = 290 m
Breadth of the rectangular field = 210 m
Perimeter of the rectangular field = 2(Length + Breadth) = 2(290 + 210) = 1000 m
Total distance that girl need to cover}=2 \times \text {Perimeter of the rectangular field
=2 × 1000 = 2000 m
\text {Given that girl walks at the rate of 1.5 m/sec. \begin{array}{l}\therefore \text {Time required to cover a distance of 2000 m } = \frac {\text {Total distance}}{\text {Speed of girl}} \\ =\frac{2000}{1.5} \text { sec} \\ =\frac{2000}{1.5} \times \frac {1}{60} \text { min} \quad [\because 1 \text { sec} = \frac {1}{60} \text { min}] \\ =22 \frac{2}{9} \text { min} \\ \text {Hence, the girl will take } 22 \frac{2}{9} \text { min to complete two round of the field.}\end{array}Q14. A corridor of a school is 8 m long and 6 m wide. It is to be covered with convas sheets. If the available canvas sheets have the size 2 m × 1 m, find the cost of canvas sheets required to cover the corridor at the rate of Rs. 8 per sheet.
Sol. \begin{array}{l} \text {Area of the corridor of a school }=\text { Length } \times \text { Breadth } =8 \times 6 =48 \text { m}^{2} \\\text {Area of 1 canvas sheet }=\text { Length } \times \text { Breadth } =2 \times 1 =2 \text { m}^{2} \\\text {Total no. of canvas sheets required to cover the corridor} = \frac {\text {Area of the corridor of a school}}{\text {Area of 1 canvas sheet}} \\ =\frac {48}{2} = 24 \text { sheets}\\\text {Given cost of 1 canvas sheet }=\text {Rs. } 8 \\ \therefore \text {Total cost of the canvas sheets } = 24 \times 8 = \text {Rs.} 192\end{array}Q15. The length and breadth of a playground are 62 m 60 cm and 25 m 40 cm respectively. Find the cost of turfing it at Rs 2.50 per square metre. How long will a man take to go three times round the field, if he walks at the rate of 2 metres per second.
Sol.
Length of a playground = 62 m 60 cm.
\begin{array}{l}=62 + 60 \times \frac {1}{100} \text { m} \quad [\because \frac {1}{100} \text { m} =1 \text{ cm}] \\ =62.6 \text { m} \\\text {Breadth of a playground} =25 \text { m } 40 \text { cm} \\ =25 + 40 \times \frac {1}{100} \text { m} \quad [\because \frac {1}{100} \text { m} =1 \text{ cm}] \\ =25.4 \text { m} \\\text {Area of a playground }= \text {Length} \times \text {Breadth} =62.6 \times 25.4 =1590.04 \text { m}^{2} \\ \text {Rate of turfing per sq. metre }= \text {Rs. } 2.50 \\ \text {Total cost of turfing the playground } =\text {Rs. } 1590.04 \times 2.50=\text {Rs. } 3975.10 \\\text {Perimeter of the rectangular field } =2(\text { Length }+\text { Breadth })=2(62.6+25.4)=176 \text { m} \\ \text {Total distance that man need to cover}=3 \times \text {Perimeter of the rectangular field } \\ =3 \times 176=528 \text { m} \\\text {Given that man walks at the rate of 2 m/sec } \\ \therefore \text {Time required to cover a distance of 528 m } = \frac {\text {Total distance}}{\text {Speed of man}} \\ =\frac{528}{2} \text { sec} \\ =\frac{528}{2} \times \frac {1}{60} \text { min} \quad [\because 1 \text { sec} = \frac {1}{60} \text { min}] \\ =\frac {264}{60} \text { min} \\ =4 \frac {24}{60} \text { min} \\ = 4 \text { min} 24 \text { sec } \\ \text {Hence, the man will take 4 min 24 sec to complete 3 round of the playground.}\end{array}Q16. A lane 180 m long and 5 m wide is to be paved with bricks of length 20 cm and breadth 15 cm. Find the cost of bricks that are required, at the rate of Rs. 750 per thousand.
Sol.
Length of the lane = 180 m.
Width of the lane = 5 m.
\begin{array}{l} \text {Area of a lane }= \text {Length} \times \text {Breadth} =180 \times 5 =900 \text { m}^{2} \\\text {Length of the brick }=20 \text { cm} \\ \text {Breadth of the brick }=15 \text { cm} \\ \text {Area of 1 brick }= \text {Length} \times \text {Breadth} =20 \text { cm} \times 15 \text { cm} \\ =300 \text { cm}^{2} \\=300 \times \frac {1}{10000} \text { m}^{2} \quad [\because 1 \text{ cm}^{2} = \frac {1}{10000} \text { m}^{2}] \\=0.03 \text { m}^{2} \\\text {Total no. of bricks required } =\frac{900}{0.03}=30000 \text { bricks.}\text {Given that cost of 1000 bricks } = \text {Rs. } 750 \\ \Rightarrow \text {Cost of 1 brick} = \text {Rs. } \frac {750}{1000} \\ \Rightarrow \text {Cost of 30000 bricks} = \text {Rs. } \frac {750}{1000} \times 30000 \\ = \text {Rs. } 22500 \\ \end{array}Q17. How many envelopes can be made out of a sheet of paper 125 cm by 85 cm, supposing one envelope requires a piece of paper of size 17 cm by 5 cm ?
Sol.
Length of the sheet of paper = 125 cm.
Breadth of the sheet of paper = 85 cm.
\begin{array}{l} \text {Area of a sheet of paper }= \text {Length} \times \text {Breadth} =125 \times 85 =10625 \text { cm}^{2} \\ \text {Length of sheet required for an envelope } =17 \text { cm} \\ \text {Breadth of sheet required for an envelope } =5 \text { cm} \\ \text {Area of the sheet required for 1 envelope }= \text {Length} \times \text {Breadth} =17 \times 5 =85 \text { cm}^{2} \\ \text {Hence, total no. of envelopes that can be made }=\frac{10625}{85}=125 \text { envelopes.} \end{array}Q18. The width of a cloth is 170 cm. Calculate the length of the cloth required to make 25 diapers, if each diaper requires a piece of cloth of size 50 cm by 17 cm.
Sol.
Given that length of the diaper = 50 cm.
Breadth of the diaper = 17 cm.
\begin{array}{l}\text {Area of cloth to make 1 diaper} = \text { Length } \times \text { Breadth} =50 \times 17 =850 \text { cm}^{2} \\ \text {Hence, area of 25 such diapers }=25 \times 850 =21250 \text { cm}^{2} \\ \text {Area of cloth required for 25 diapers} =21250 \text { cm}^{2} \\ \text {Given that width of a cloth }=170 \text { cm} \\ \text {Lets assume that length of the cloth be x cm} \\ \therefore \text {Area of cloth for 25 diapers} = 170 \times x \\ \Rightarrow 21250 = 170 \times x \\ \Rightarrow x = \frac {21250}{170} \\ \Rightarrow x = 125 \text { cm} \\\end{array} Hence, length of the cloth will be 125 cm.
Q19. The carpet for a room 6.6 m by 5.6 m costs Rs. 3960 and it was made from a roll 70 cm wide. Find the cost of the carpet per metre.
Sol.
Length of a room = 6.6 m.
Breadth of a room = 5.6 m.
\begin{array}{l}\text {Area of a room }= \text {Length} \times \text {Breadth} =6.6 \times 5.6 =36.96 \text { m}^{2} \\\text {Given that Width of a carpet }=70 \text { cm}=0.7 \text { m} \quad [\because 1 \text { m}= 100 \text { cm}] \\\text {Length of a carpet} =\frac{\text {Area of the room}}{\text {Width of the carpet}}=\frac{36.96}{0.7}=52.8 \text { m} \\\text {Cost of 52.8 m long roll of carpet }=\text {Rs. } 3960 \\ \therefore \text {Cost of 1 m long roll of carpet }=\text {Rs. } \frac {3960}{52.8}=\text {Rs. } 75\end{array}Q20. A room is 9 m long, 8 m broad and 6.5 m high. It has one door of dimensions 2 m x 1.5 m and three windows each of dimensions 1.5 m x 1 m. Find the cost of white washing the walls at Rs. 3.80 per square metre.
Sol.
Given that length of the room = 9 m.
Breadth of the room = 8 m.
Height of the room = 6.5 m.
\begin{array}{l} \text {Area of 4 walls }= 2lh + 2bh = 2(l+b)h \\ = 2(9 +8 ) \times 6.5 =2 \times 17 \times 6.5 =221 \text { m}^{2} \\\text {Length of a door } =2 \text { m} \\ \text {Breadth of a door }=1.5 \text { m} \\ \text {Area of a door }= \text {Length} \times \text {Breadth} \\ =2 \times 1.5 =3 \text { m}^{2} \\\text {Length of a window }=1.5 \text { m} \\ \text {Breadth of a window }=1 \text { m} \\ \text {Area of one window }= \text {Length} \times \text {Breadth} =1.5 \times 1 =1.5 \text { m}^{2} \\\text {Thus, Area of 3 such windows} =3 \times 1.5 =4.5 \text { m}^{2} \\\text {Area to be white-washed } = \text {Area of 4 walls} – \text {Area of one door} – \text {Area of 3 windows} \\ =221-3+ 4.5 =213.5 \text { m}^{2} \\\end{array}Cost of white-washing for 1 sq. metre of area = Rs. 3.80
Cost of white-washing for 213.5 sq. metre area= Rs. 213.5 × 3.80 =Rs. 811.30
Q21. A hall 36 m long and 24 m broad allowing 80 sq metre for doors and windows, the cost of papering the walls at Rs. 8.40 per sq. metre is Rs. 9408. Find the height of the hall.
Sol.
Given that length of the hall = 36 m
Breadth of the hall = 24 m
Let’ assume that height of the hall be h m.
For papering the wall, we need to paper the four walls excluding the floor and roof of the hall along with doors and windows. \begin{array}{l}\Rightarrow \quad \text {The area of the wall which is to be papered }= {Area of 4 walls } – \text {Area of doors and windows} \\=2 h(1+b) – 80 \text { m}^{2} \\ =2 h(36+24) – 80 \text { m}^{2} \\ =120h – 80 \text { m}^{2} \\\text {Givne rate of papering the wall } = \text {Rs. 8.40 per sq. meter} \\\text {Total cost of papering the walls} = (120h – 80) \times 8.40 \\ \Rightarrow \quad 9408 = (120h – 80) \times 8.40 \\ \Rightarrow \quad 120h -80 = \frac {9408}{8.40} \\ \Rightarrow \quad 120h -80 = 1120 \\ \Rightarrow \quad 120h = 1120 +80 \\ \Rightarrow \quad h = \frac {1200}{120} \\ \Rightarrow \quad h = 10 \text { m} \\ \text {Hence, the height of the wall is 10 m.} \end{array}