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.

Mean = Sum of all observation/Total number of observations = 41 + 39 + 48 + 52 + 46 + 62 + 54 + 40 + 96 + 52 + 98 + 40 + 42 + 52 + 60/15 = 822/10 = 54.8 Arranging the data in ascending order: 39, 40, 40, 41, 42, 46, 48, 52, 52, 54, 60, 62, 96, 98 Number of observations = 15 15 is an odd number. TherRead more

Mean = Sum of all observation/Total number of observations
= 41 + 39 + 48 + 52 + 46 + 62 + 54 + 40 + 96 + 52 + 98 + 40 + 42 + 52 + 60/15
= 822/10 = 54.8
Arranging the data in ascending order:
39, 40, 40, 41, 42, 46, 48, 52, 52, 54, 60, 62, 96, 98
Number of observations = 15
15 is an odd number. Therefore, the mode = 8th[(n+1/2)th] term of the observation.
Hence,
Median = 8th term = 52
Mode is the observation having maximum frequency. Here, the number 52 occurs maximum (3) times.
Hence, the mode of the observations is 52.

Number of observations = 10 10 is an even number. Therefore, the modian = Mean of 5th[(n/2)th] term and 6th[(n/2 + 1)th] term. Hence, Median = 1/2(5th observation + 6th observation) ⇒ 63 = 1/2 [(x) + (x + 2)] ⇒ 63 = 1/2 [2x + 2] ⇒ 63 = x + 1 ⇒ x = 62 Hence, the valueof x is 62.

Number of observations = 10
10 is an even number. Therefore, the modian = Mean of 5th[(n/2)th] term and 6th[(n/2 + 1)th] term.
Hence,
Median = 1/2(5th observation + 6th observation)
⇒ 63 = 1/2 [(x) + (x + 2)]
⇒ 63 = 1/2 [2x + 2]
⇒ 63 = x + 1
⇒ x = 62
Hence, the valueof x is 62.

Arranging the data in ascending order: 14, 14, 14, 14, 17, 18, 18, 18, 22, 23, 25, 28 Mode is the observation having maximum frequency. Here, the number 14 occurs maximum (4) times. Hence, the mode of the observations is 14.

Arranging the data in ascending order:
14, 14, 14, 14, 17, 18, 18, 18, 22, 23, 25, 28
Mode is the observation having maximum frequency. Here, the number 14 occurs maximum (4) times.
Hence, the mode of the observations is 14.

(i) Lenth of plastic box l = 1.5 m, breadth b = 1.25 m height h = 65 cm = 0.65 m Area of sheet required for making a plastic box = total surface area of box - Area of top of box = 2(lb + bh + hl) - lb = 2(1.5 × 1.25 + 1.25 × 0.65 + 0.65 × 1.5) - 1.5 × 1.25m² = 2(1.875 + 0.8125 + 0.975) - 1.875 m² =Read more

(i) Lenth of plastic box l = 1.5 m, breadth b = 1.25 m height h = 65 cm = 0.65 m
Area of sheet required for making a plastic box = total surface area of box – Area of top of box
= 2(lb + bh + hl) – lb = 2(1.5 × 1.25 + 1.25 × 0.65 + 0.65 × 1.5) – 1.5 × 1.25m²
= 2(1.875 + 0.8125 + 0.975) – 1.875 m²
= 2(3.6625) – 1.875 m²
= 7.325 – 1.875 = 5.45 m²
Hence, the area of the sheet required for making the box is 5.45 m².
(ii) Total cost of cheet = Rs 20 × 5.45 = Rs 109.

Lenght of room l = 5 m, breadth b = 4 m and height h = 3 m Area of four walls and ceilling = Total surface area of room - area of floor = 2(lb + bh + hl) - lb = 2(5 × 4 + 4 × 3 + 3 × 5) - 5 × 4 m² = 2(20 + 12 + 15) - 20 m² = 2(47) - 20 m² = 94 - 20 = 74 m² Therefore, the area of four walls and ceillRead more

Lenght of room l = 5 m, breadth b = 4 m and height h = 3 m
Area of four walls and ceilling = Total surface area of room – area of floor
= 2(lb + bh + hl) – lb
= 2(5 × 4 + 4 × 3 + 3 × 5) – 5 × 4 m²
= 2(20 + 12 + 15) – 20 m²
= 2(47) – 20 m²
= 94 – 20 = 74 m²
Therefore, the area of four walls and ceilling = 74 m²
Hence, the cost of white washing the walls of walls of the room and the ceilling
= Rs 7.50 × 74 = Rs. 555.00

## 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.

## In a mathematics test given to 15 students, the following marks (out of 100) are recorded: 41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 40, 42, 52, 60 Find the mean, median and mode of this data.

Mean = Sum of all observation/Total number of observations = 41 + 39 + 48 + 52 + 46 + 62 + 54 + 40 + 96 + 52 + 98 + 40 + 42 + 52 + 60/15 = 822/10 = 54.8 Arranging the data in ascending order: 39, 40, 40, 41, 42, 46, 48, 52, 52, 54, 60, 62, 96, 98 Number of observations = 15 15 is an odd number. TherRead more

Mean = Sum of all observation/Total number of observations

See less= 41 + 39 + 48 + 52 + 46 + 62 + 54 + 40 + 96 + 52 + 98 + 40 + 42 + 52 + 60/15

= 822/10 = 54.8

Arranging the data in ascending order:

39, 40, 40, 41, 42, 46, 48, 52, 52, 54, 60, 62, 96, 98

Number of observations = 15

15 is an odd number. Therefore, the mode = 8th[(n+1/2)th] term of the observation.

Hence,

Median = 8th term = 52

Mode is the observation having maximum frequency. Here, the number 52 occurs maximum (3) times.

Hence, the mode of the observations is 52.

## The following observations have been arranged in ascending order. If the median of the data is 63, find the value of x. 29, 32, 48, 50, x, x + 2, 72, 78, 84, 95.

Number of observations = 10 10 is an even number. Therefore, the modian = Mean of 5th[(n/2)th] term and 6th[(n/2 + 1)th] term. Hence, Median = 1/2(5th observation + 6th observation) ⇒ 63 = 1/2 [(x) + (x + 2)] ⇒ 63 = 1/2 [2x + 2] ⇒ 63 = x + 1 ⇒ x = 62 Hence, the valueof x is 62.

Number of observations = 10

See less10 is an even number. Therefore, the modian = Mean of 5th[(n/2)th] term and 6th[(n/2 + 1)th] term.

Hence,

Median = 1/2(5th observation + 6th observation)

⇒ 63 = 1/2 [(x) + (x + 2)]

⇒ 63 = 1/2 [2x + 2]

⇒ 63 = x + 1

⇒ x = 62

Hence, the valueof x is 62.

## Find the mode of 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18.

Arranging the data in ascending order: 14, 14, 14, 14, 17, 18, 18, 18, 22, 23, 25, 28 Mode is the observation having maximum frequency. Here, the number 14 occurs maximum (4) times. Hence, the mode of the observations is 14.

Arranging the data in ascending order:

See less14, 14, 14, 14, 17, 18, 18, 18, 22, 23, 25, 28

Mode is the observation having maximum frequency. Here, the number 14 occurs maximum (4) times.

Hence, the mode of the observations is 14.

## A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is to be open at the top. Ignoring the thickness of the plastic sheet, determine:

(i) Lenth of plastic box l = 1.5 m, breadth b = 1.25 m height h = 65 cm = 0.65 m Area of sheet required for making a plastic box = total surface area of box - Area of top of box = 2(lb + bh + hl) - lb = 2(1.5 × 1.25 + 1.25 × 0.65 + 0.65 × 1.5) - 1.5 × 1.25m² = 2(1.875 + 0.8125 + 0.975) - 1.875 m² =Read more

(i) Lenth of plastic box l = 1.5 m, breadth b = 1.25 m height h = 65 cm = 0.65 m

See lessArea of sheet required for making a plastic box = total surface area of box – Area of top of box

= 2(lb + bh + hl) – lb = 2(1.5 × 1.25 + 1.25 × 0.65 + 0.65 × 1.5) – 1.5 × 1.25m²

= 2(1.875 + 0.8125 + 0.975) – 1.875 m²

= 2(3.6625) – 1.875 m²

= 7.325 – 1.875 = 5.45 m²

Hence, the area of the sheet required for making the box is 5.45 m².

(ii) Total cost of cheet = Rs 20 × 5.45 = Rs 109.

## The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of Rs 7.50 per m².

Lenght of room l = 5 m, breadth b = 4 m and height h = 3 m Area of four walls and ceilling = Total surface area of room - area of floor = 2(lb + bh + hl) - lb = 2(5 × 4 + 4 × 3 + 3 × 5) - 5 × 4 m² = 2(20 + 12 + 15) - 20 m² = 2(47) - 20 m² = 94 - 20 = 74 m² Therefore, the area of four walls and ceillRead more

Lenght of room l = 5 m, breadth b = 4 m and height h = 3 m

See lessArea of four walls and ceilling = Total surface area of room – area of floor

= 2(lb + bh + hl) – lb

= 2(5 × 4 + 4 × 3 + 3 × 5) – 5 × 4 m²

= 2(20 + 12 + 15) – 20 m²

= 2(47) – 20 m²

= 94 – 20 = 74 m²

Therefore, the area of four walls and ceilling = 74 m²

Hence, the cost of white washing the walls of walls of the room and the ceilling

= Rs 7.50 × 74 = Rs. 555.00