For smaller boxes: Length of box l = 25 cm, breadth b = 20 cm and height h = 5 cm Total surface area of 1 bigger box = 2(lb + bh + hl) = 2(25 × 20 + 20 × 5 + 5 × 25) cm² = 2(500 + 100 + 125) cm² = 2(725) cm² = 1450 cm² Area of cardboard for overlap = 5% of 1450 cm² = 1450 × (5/100 = 72.5 cm² Total aRead more
For smaller boxes:
Length of box l = 25 cm, breadth b = 20 cm and height h = 5 cm
Total surface area of 1 bigger box = 2(lb + bh + hl)
= 2(25 × 20 + 20 × 5 + 5 × 25) cm² = 2(500 + 100 + 125) cm² = 2(725) cm² = 1450 cm²
Area of cardboard for overlap = 5% of 1450 cm² = 1450 × (5/100 = 72.5 cm²
Total area of cardboard for 1 bigger box = 1450 + 72.5 = 1522.5 cm²
Therefore, the area of cardboard for 250 bigger boxes = 250 × 1522.5 cm² = 380625 cm²
For smaller boxes:
Length of box l = 15 cm, breadth b = 12 cm and height h = 5 cm
Total surface area of 1 smaller box = 2(lb + bh + hl)
= 2(15 × 12 + 12 × 5 + 5 × 15) cm² = 2(180 + 60 + 75) cm²
= 2(315) cm²
= 630 cm²
Area of cardboard for overlap = 5% of 630 cm² = 630 × (5/100) = 31.5 cm²
Total area of cardboard for 1 smaller box = 630 + 31.5 = 661.5 cm²
Therefore, the area of cardboard for 250 smaller boxes = 250 × 661.5 cm² = 165375 cm²
So, the area of cardboard for 500 boxes = 380625 + 165375 = 546000 cm²
Total cost of cardboard at the rate of Rs 4 per 1000 cm² = Rs 546000 × 4/1000 = Rs 2184
Hence, the total cost of cardboard for 500 boxes is Rs 2184.
Length of shelter l = 4 m, breadth b = 3 m and height h = 2.5 m Area of four walls and top of shelter = total area of shelter - area of floor = 2(lb + bh + hl) - lb = 2(4 × 3 + 3 × 2.5 × 4) - 4 × 3 m² = 2(12 + 7.5 + 10) - 12 m² = 2(29.5) - 12 m² = 59 - 12 = 47 m² Hence, 47 m² tarpaulin is required tRead more
Length of shelter l = 4 m, breadth b = 3 m and height h = 2.5 m
Area of four walls and top of shelter = total area of shelter – area of floor = 2(lb + bh + hl) – lb
= 2(4 × 3 + 3 × 2.5 × 4) – 4 × 3 m²
= 2(12 + 7.5 + 10) – 12 m²
= 2(29.5) – 12 m²
= 59 – 12 = 47 m²
Hence, 47 m² tarpaulin is required to make this shelter.
(i) Length of green house l = 30 cm, breadth b = 25 cm and height h = 25 cm Total surface area of green house = 2(lb + bh + hl) = 2(30 × 25 + 25 × 25 + 25 × 30) cm² = 2(750 + 625 + 750) cm² = 2(2125) cm² = 4250 cm² Hence, the area of glass for green house is 4250 cm² (ii) Length of green house l = 3Read more
(i) Length of green house l = 30 cm, breadth b = 25 cm and height h = 25 cm
Total surface area of green house = 2(lb + bh + hl)
= 2(30 × 25 + 25 × 25 + 25 × 30) cm² = 2(750 + 625 + 750) cm²
= 2(2125) cm²
= 4250 cm²
Hence, the area of glass for green house is 4250 cm²
(ii) Length of green house l = 30 cm, breadth b = 25 cm and height h = 25 cm
Length of tape for all 12 edges (4 lenght, 4 Breath and 4 Height) = 4(l + b + h)
= 4(30 + 25 + 25) cm = 4(80) cm
= 320 cm
Hence, for all 12 edges, 320 cm tape is required.
(i) Side of cubical box l = 10 cm Curved surface area of cubical box = 4l² = 4(10)² cm² = 4(100) cm² = 400 cm² Length of cuboidal box l = 12.5 cm, breadth b = 10 cm and height h = 8 cm Curved surface area of cuboidal box = 2(l + b)h = 2(12.5 + 10) × 8 cm² = 2(22.5) × 8 cm² =360 cm² Difference betweeRead more
(i) Side of cubical box l = 10 cm
Curved surface area of cubical box = 4l² = 4(10)² cm² = 4(100) cm² = 400 cm²
Length of cuboidal box l = 12.5 cm, breadth b = 10 cm and height h = 8 cm
Curved surface area of cuboidal box = 2(l + b)h = 2(12.5 + 10) × 8 cm² = 2(22.5) × 8 cm² =360 cm²
Difference between surface area of two boxes = 400 cm² – 360 cm² = 40cm²
Hence, the curved surface area of cubical box is more than cuboidal box by 40 cm².
(ii) Side of cubical box l = 10 cm
Total surface area of cubical box = 6l² = 6(10)² cm² = 6(100) cm² = 600 cm²
Length of cuboidal box l = 12.5 cm, breadth b = 10 cm and height h = 8 cm
Total surface area of cuboidal box = 2(lb + bh + hl)
= 2(12.5 × 10 + 10 × 8 + 8 × 12.5) cm² = 2(125 + 80 + 100) cm²
= 2(305) cm²
= 610 cm²
Difference between total surface areas of two boxes = 610 cm² – 600 cm² = 10 cm²
Hence, the total surface area of cubical box is more than cuboidal box by 10 cm².
Mean = Sum of all observation/Total number of observations = 2 + 3 + 4 + 5 + 0 + 1 + 3 + 3 + 4 + 3/10 = 28/10 = 2.8 goals Arranging the data in ascending order: O, 1, 2, 3, 3, 3, 3, 4, 4, 5 Number of observations = 10 10 is an even number. Therefore, the median = Mean of 5th[(n/2)th] term and 6th[(nRead more
Mean = Sum of all observation/Total number of observations
= 2 + 3 + 4 + 5 + 0 + 1 + 3 + 3 + 4 + 3/10 = 28/10 = 2.8 goals
Arranging the data in ascending order:
O, 1, 2, 3, 3, 3, 3, 4, 4, 5
Number of observations = 10
10 is an even number. Therefore, the median = Mean of 5th[(n/2)th] term and 6th[(n/2 + 1)th] term.
Hence,
Median = 1/2(5th observation + 6th observation) = 1/2(3 + 3) = 3
Mode is the observation having maximum frequency. Here, the number 3 occurs maximum (4) times.
Hence, the mode of the observations is 3.
Shanti Sweets Stall was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxes were required. The bigger of dimensions 25 cm × 20 cm × 5 cm and the smaller of dimensions 15 cm × 12 cm × 5 cm. For all the overlaps, 5% of the total surface area is required extra. If the cost of the cardboard is Rs 4 for 1000 cm², find the cost of cardboard required for supplying 250 boxes of each kind.
For smaller boxes: Length of box l = 25 cm, breadth b = 20 cm and height h = 5 cm Total surface area of 1 bigger box = 2(lb + bh + hl) = 2(25 × 20 + 20 × 5 + 5 × 25) cm² = 2(500 + 100 + 125) cm² = 2(725) cm² = 1450 cm² Area of cardboard for overlap = 5% of 1450 cm² = 1450 × (5/100 = 72.5 cm² Total aRead more
For smaller boxes:
See lessLength of box l = 25 cm, breadth b = 20 cm and height h = 5 cm
Total surface area of 1 bigger box = 2(lb + bh + hl)
= 2(25 × 20 + 20 × 5 + 5 × 25) cm² = 2(500 + 100 + 125) cm² = 2(725) cm² = 1450 cm²
Area of cardboard for overlap = 5% of 1450 cm² = 1450 × (5/100 = 72.5 cm²
Total area of cardboard for 1 bigger box = 1450 + 72.5 = 1522.5 cm²
Therefore, the area of cardboard for 250 bigger boxes = 250 × 1522.5 cm² = 380625 cm²
For smaller boxes:
Length of box l = 15 cm, breadth b = 12 cm and height h = 5 cm
Total surface area of 1 smaller box = 2(lb + bh + hl)
= 2(15 × 12 + 12 × 5 + 5 × 15) cm² = 2(180 + 60 + 75) cm²
= 2(315) cm²
= 630 cm²
Area of cardboard for overlap = 5% of 630 cm² = 630 × (5/100) = 31.5 cm²
Total area of cardboard for 1 smaller box = 630 + 31.5 = 661.5 cm²
Therefore, the area of cardboard for 250 smaller boxes = 250 × 661.5 cm² = 165375 cm²
So, the area of cardboard for 500 boxes = 380625 + 165375 = 546000 cm²
Total cost of cardboard at the rate of Rs 4 per 1000 cm² = Rs 546000 × 4/1000 = Rs 2184
Hence, the total cost of cardboard for 500 boxes is Rs 2184.
Parveen wanted to make a temporary shelter for her car, by making a box-like structure with tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small, and therefore negligible, how much tarpaulin would be required to make the shelter of height 2.5 m, with base dimensions 4 m × 3 m?
Length of shelter l = 4 m, breadth b = 3 m and height h = 2.5 m Area of four walls and top of shelter = total area of shelter - area of floor = 2(lb + bh + hl) - lb = 2(4 × 3 + 3 × 2.5 × 4) - 4 × 3 m² = 2(12 + 7.5 + 10) - 12 m² = 2(29.5) - 12 m² = 59 - 12 = 47 m² Hence, 47 m² tarpaulin is required tRead more
Length of shelter l = 4 m, breadth b = 3 m and height h = 2.5 m
See lessArea of four walls and top of shelter = total area of shelter – area of floor = 2(lb + bh + hl) – lb
= 2(4 × 3 + 3 × 2.5 × 4) – 4 × 3 m²
= 2(12 + 7.5 + 10) – 12 m²
= 2(29.5) – 12 m²
= 59 – 12 = 47 m²
Hence, 47 m² tarpaulin is required to make this shelter.
A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30 cm long, 25 cm wide and 25 cm high.
(i) Length of green house l = 30 cm, breadth b = 25 cm and height h = 25 cm Total surface area of green house = 2(lb + bh + hl) = 2(30 × 25 + 25 × 25 + 25 × 30) cm² = 2(750 + 625 + 750) cm² = 2(2125) cm² = 4250 cm² Hence, the area of glass for green house is 4250 cm² (ii) Length of green house l = 3Read more
(i) Length of green house l = 30 cm, breadth b = 25 cm and height h = 25 cm
See lessTotal surface area of green house = 2(lb + bh + hl)
= 2(30 × 25 + 25 × 25 + 25 × 30) cm² = 2(750 + 625 + 750) cm²
= 2(2125) cm²
= 4250 cm²
Hence, the area of glass for green house is 4250 cm²
(ii) Length of green house l = 30 cm, breadth b = 25 cm and height h = 25 cm
Length of tape for all 12 edges (4 lenght, 4 Breath and 4 Height) = 4(l + b + h)
= 4(30 + 25 + 25) cm = 4(80) cm
= 320 cm
Hence, for all 12 edges, 320 cm tape is required.
A cubical box has each edge 10 cm and another cuboidal box is 12.5 cm long, 10 cm wide and 8 cm high.
(i) Side of cubical box l = 10 cm Curved surface area of cubical box = 4l² = 4(10)² cm² = 4(100) cm² = 400 cm² Length of cuboidal box l = 12.5 cm, breadth b = 10 cm and height h = 8 cm Curved surface area of cuboidal box = 2(l + b)h = 2(12.5 + 10) × 8 cm² = 2(22.5) × 8 cm² =360 cm² Difference betweeRead more
(i) Side of cubical box l = 10 cm
Curved surface area of cubical box = 4l² = 4(10)² cm² = 4(100) cm² = 400 cm²
Length of cuboidal box l = 12.5 cm, breadth b = 10 cm and height h = 8 cm
Curved surface area of cuboidal box = 2(l + b)h = 2(12.5 + 10) × 8 cm² = 2(22.5) × 8 cm² =360 cm²
Difference between surface area of two boxes = 400 cm² – 360 cm² = 40cm²
Hence, the curved surface area of cubical box is more than cuboidal box by 40 cm².
(ii) Side of cubical box l = 10 cm
See lessTotal surface area of cubical box = 6l² = 6(10)² cm² = 6(100) cm² = 600 cm²
Length of cuboidal box l = 12.5 cm, breadth b = 10 cm and height h = 8 cm
Total surface area of cuboidal box = 2(lb + bh + hl)
= 2(12.5 × 10 + 10 × 8 + 8 × 12.5) cm² = 2(125 + 80 + 100) cm²
= 2(305) cm²
= 610 cm²
Difference between total surface areas of two boxes = 610 cm² – 600 cm² = 10 cm²
Hence, the total surface area of cubical box is more than cuboidal box by 10 cm².
The following number of goals were scored by a team in a series of 10 matches: 2, 3, 4, 5, 0, 1, 3, 3, 4, 3 Find the mean, median and mode of these scores.
Mean = Sum of all observation/Total number of observations = 2 + 3 + 4 + 5 + 0 + 1 + 3 + 3 + 4 + 3/10 = 28/10 = 2.8 goals Arranging the data in ascending order: O, 1, 2, 3, 3, 3, 3, 4, 4, 5 Number of observations = 10 10 is an even number. Therefore, the median = Mean of 5th[(n/2)th] term and 6th[(nRead more
Mean = Sum of all observation/Total number of observations
See less= 2 + 3 + 4 + 5 + 0 + 1 + 3 + 3 + 4 + 3/10 = 28/10 = 2.8 goals
Arranging the data in ascending order:
O, 1, 2, 3, 3, 3, 3, 4, 4, 5
Number of observations = 10
10 is an even number. Therefore, the median = Mean of 5th[(n/2)th] term and 6th[(n/2 + 1)th] term.
Hence,
Median = 1/2(5th observation + 6th observation) = 1/2(3 + 3) = 3
Mode is the observation having maximum frequency. Here, the number 3 occurs maximum (4) times.
Hence, the mode of the observations is 3.