Mode and Range

Mode is the statistical term that refers to the most frequently occurring number found in a set of numbers and the "range" is just the difference between the largest and smallest values.

Summary

Mode is the statistical term that refers to the most frequently occurring number found in a set of numbers and the "range" is just the difference between the largest and smallest values.

Things to Remember

  • The mode is the number repeated most often.
  • This list has two values that are repeated three times.
  • The largest value is 13 and the smallest is 8, so the range is 13 - 8 = 5.

MCQs

No MCQs found.

Subjective Questions

Q1:

Find the length of a carpet of width 50cm to carpet a hall 8m long and 6m wide. Also, find the cost of carpeting at the rate of Rs 450/metre.


Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>Length of carpet(l<sub>1</sub>) = ?</p> <p>Breadth of carpet (b<sub>1</sub>) = 50cm = 0.5m</p> <p>Length of room (l) = 8m</p> <p>Breadth of room (b) = 6m</p> <p>While carpeting a room,</p> <p>Area of carpet (A<sub>1</sub>) = Area of room (A)</p> <p>i.e. l<sub>1</sub> * b<sub>1</sub> = l*b</p> <p>or, l<sub>1</sub> = \(\frac{l*b}{b<sub>1</sub>}\)</p> <p>= \(\frac{8m*6m}{0.5m}\)</p> <p>= \(\frac{48m^2}{0.5m}\)</p> <p>= 96m</p> <p>&there4; 96m long carpet is required.</p> <p>Again,</p> <p>Length of carpet (l<sub>1</sub>) = 96m</p> <p>Cost of carpet/metre (C) = Rs 450/m</p> <p>Toal cast (T) = l<sub>1</sub> * C</p> <p>= 96 * Rs 450</p> <p>Rs 43,200</p> <p>Hence, it costs Rs 43,200 in order to carpet the room.</p>

Q2:

Find the perimeter of the given window. Also, find its area, (\(\pi\) = \(\frac{22}{7}\))


Type: Long Difficulty: Easy

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Answer: <p>Solution:</p> <p>The given window frame consists of a rectangle and a semi-circle.</p> <p>For semi-circle,</p> <p>diameter (d) = 280cm</p> <p>radius (r) = \(\frac{1}{2}\)d</p> <p>=\(\frac{1}{2}\) * 280cm</p> <p>= 140cm</p> <p>Semicircle (c) =\(\frac{1}{2}\)\(\pi\)d</p> <p>=\(\frac{1}{2}\)\(\frac{22}{7}\)280cm</p> <p>=\(\frac{22 * 280}{14}\)cm</p> <p>= 440cm</p> <p>Now, perimeter of the window frame = l + b + b + semicircle</p> <p>= 280cm + 150cm + 150cm + 440cm</p> <p>= 1,020cm</p> <p>Now, Area of rectangular part (A<sub>1</sub>) = l * b</p> <p>= 280cm * 150cm</p> <p>= 42,000cm<sup>2</sup></p> <p>Area of semi circle (A<sub>2</sub>) =\(\frac{1}{2}\)\(\pi\)r<sup>2</sup></p> <p>=\(\frac{1}{2}\)*\(\frac{22}{7}\)*(140cm)<sup>2</sup></p> <p>= 30,800cm<sup>2</sup></p> <p>&there4; Area of window frame (A) = A<sub>1</sub> + A<sub>2</sub></p> <p>= 42,000cm<sup>2</sup> + 30,800cm<sup>2</sup></p> <p>= 72,800cm<sup>2</sup></p> <p>Hence,</p> <p>Perimeter of window frame (P) = 1020cm</p> <p>Area ofwindow frame (A) = 72,800cm<sup>2</sup> = 7.28m<sup>2</sup></p> <p></p>

Q3:

Find the perimeter of the given plane figure.



Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>Length (l) = 7.5cm</p> <p>Breadth (b) = 5cm</p> <p>Now,</p> <p>Area of given rectangle = l * b</p> <p>= 7.5cm * 5cm</p> <p>= 37.5cm<sup>2</sup></p> <p>Also,</p> <p>Perimeter = 2(l + b)</p> <p>= 2 (7.5 + 5)cm</p> <p>= 2 * 12.5cm</p> <p>= 25cm</p>

Q4:

A rectangular room is 8m long and 6m wide. What length of carpet of 3.5 m wide is needed to carpet room?


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>Length of carpet (l<sub>1</sub>) = ?</p> <p>Breadth of carpet (b<sub>1</sub>) = 3.5m</p> <p>Length of room (l) = 8m</p> <p>Breadth of room (b) = 6m</p> <p>Area of carpet (A<sub>1</sub>) = Area of room (A)</p> <p>i.e. l<sub>1</sub> * b<sub>1</sub> = l * b</p> <p>or, l<sub>1</sub> = \(\frac{l*b}{b<sub>1</sub>}\)</p> <p>= \(\frac{8m*6m}{3.5}\)</p> <p>= \(\frac{48m^2}{3.5m}\)</p> <p>= 13.71m</p> <p>&there4; 13.71m long carpet is required.</p>

Q5:

What length of wall paper of breadth  50cm is reqiured to cover a wall of 5m long and 3m high?


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>Length of wall paper (l<sub>1</sub>) = ?</p> <p>Breadth of wall paper (b<sub>1</sub>) = 50cm = 0.5m</p> <p>Length of wall (l) =5m</p> <p>Breadth of wall (b) =3m</p> <p>While carpeting a room,</p> <p>Area of wall paper (A<sub>1</sub>) = Area of wall (A)</p> <p>i.e. l<sub>1</sub> * b<sub>1</sub> = l*b</p> <p>or, l<sub>1</sub> = \(\frac{l*b}{b<sub>1</sub>}\)</p> <p>= \(\frac{5m*3m}{0.5m}\)</p> <p>= \(\frac{15m^2}{0.5m}\)</p> <p>= 30m</p> <p>&there4; 30m long wall paper is required.</p>

Q6:

Find the number of stone to pave 12m long and 8m wide courtyard if the length  of the stone is 2m and width is 1.6m?


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>Length of courtyard (l) = 12m</p> <p>Breadth of courtyard(b) = 8m</p> <p>Area of courtyard (A) = l * b</p> <p>= 12m * 8m</p> <p>= 96m<sup>2</sup></p> <p>Again,</p> <p>Length of stone (l<sub>1</sub>) = 2m</p> <p>Breadth of stone (b<sub>1</sub>) = 1.6m</p> <p>Area of stone (a) = l<sub>1</sub> * b<sub>1</sub></p> <p>= 2m * 1.6m</p> <p>= 3.2m<sup>2</sup></p> <p>Now,</p> <p>Required number of stone(N) = \(\frac{Area of courtyard}{Area of stone}\)</p> <p>= \(\frac{96m^2}{3.2m^2}\)</p> <p>= 30</p> <p>Hence, 30 stones are needed to pave the courtyard.</p>

Q7:

Find the area of the shaded region. 


Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <figure class="inline-right" style="width: 224px;"><img src="/uploads/Untitled58.png" alt=" " width="224" height="133"><figcaption></figcaption></figure><p>From the figure,</p> <p>Length of rectangle (l) = 50m</p> <p>Breadth (b) = 30m</p> <p>Width of the path (d) = 5m</p> <p>Now,</p> <p>Area of EFGH = l * d</p> <p>= 50m * 5m</p> <p>= 250m<sup>2</sup></p> <p>Again,</p> <p>Area of WXYZ = b * d</p> <p>= 30m * 5m</p> <p>= 150m<sup>2</sup></p> <p>Also,</p> <p>Area of crossing path ABCD = d<sup>2</sup></p> <p>= (5m)<sup>2</sup></p> <p>= 25m<sup>2</sup></p> <p>Hence,</p> <p>Area of shaded region = Area of EFGH + Area of WXYZ- Area of crossing path ABCD</p> <p>= 250m<sup>2</sup> + 150m<sup>2</sup>- 25m<sup>2</sup></p> <p>= 375m<sup>2</sup></p> <p>Alternatively,</p> <p>Area of crossing paths/shaded region = d(l + b - d)</p> <p>= 5m (50m - 30 - 5m)</p> <p>= 5m * 15m</p> <p>= 375m<sup>2</sup></p> <figure class="" style="width: 224px;"><figcaption></figcaption></figure>

Q8:

The path having area 75m2 and width 2.5m is surrounded inside a garden. Find the length of the path.


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>Area of square garden (A) = 75m<sup>2</sup></p> <p>Breadth (b) = 2.5m</p> <p>Length (l) = ?</p> <p>Now,</p> <p>Area of square = l * b</p> <p>or, 75m<sup>2</sup> = l * 2.5m</p> <p>or, l = \(\frac{75m^2}{2.5m}\)</p> <p>&there4; l = 30m</p> <p>Hence, the length of the path is 30m.</p>

Q9:

A room is 8m long, 6m wide and 4m height. It has door of 2m × 1.5m and two windows of size 2m × 3m. Find the

  1. Area of 4 walls
  2. Area of 4 walls and ceiling
  3. Area of 4 walls excluding door and windows
  4. Cost of painting 4 walls and ceiling at Rs 80/m2

Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>Given,</p> <p>Length of wall (<em>l</em>) = 8m</p> <p>Breadth of wall (b) = 6m</p> <p>Height of wall (h) = 4m</p> <p>Length of door (<em>l</em><sub>1</sub>) = 2m</p> <p>Height of door (h<sub>1</sub>) = 1.5m</p> <p>Length of window (<em>l</em><sub>2</sub>) = 2m</p> <p>Height of window (h<sub>2</sub>) = 3m</p> <p>Cost of painting walls (C) = Rs 80/m<sup>2</sup></p> <p>Now,</p> <p>Area of 4 walls = 2h(<em>l</em> +b)</p> <p>= 2 &times; 4m (8m + 6m)</p> <p>= 8m &times; 14m</p> <p>= 112m<sup>2</sup></p> <p>Area of 4 walls and ceiling =2h(<em>l</em> +b) +<em>l</em>b</p> <p>= 2 &times; 4m (8m + 6m) + 8m &times; 6m</p> <p>= 8m &times; 14m + 48m<sup>2</sup></p> <p>= 112m<sup>2</sup> + 48m<sup>2</sup></p> <p>= 160m<sup>2</sup></p> <p>Area of 4 walls excluding door and windows (A) =2h(<em>l</em> +b) -<em>l</em><sub>1</sub><span style="font-size: 11px;">h</span><sub>1</sub>- 2( l<sub>2</sub>h<sub>2</sub>)</p> <p>= 2 &times; 4m (8m + 6m) -2m &times; 1.5m - 2(2m &times; 3m)</p> <p>=8m &times; 14m - 3m<sup>2</sup> - 2 &times; 6m<sup>2</sup></p> <p>= 112m<sup>2</sup> - 3m<sup>2</sup> - 12m<sup>2</sup></p> <p>= 97m<sup>2</sup></p> <p>Cost of painting 4 walls and ceiling = C * A</p> <p>= Rs 80/m<sup>2</sup> &times; 97m<sup>2</sup></p> <p>= Rs 7760</p> <p>Hence,</p> <p>Area of 4 walls = 112m<sup>2</sup></p> <p>Area of 4 walls and ceiling = 160m<sup>2</sup></p> <p>Area of 4 walls excluding door and windows = 97m<sup>2</sup></p> <p>Cost of painting 4 walls and ceiling = Rs 7760</p> <p></p> <p></p> <p></p> <p></p>

Q10:

Calculate the volume and total surface area of the cylinder shown.


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <table><tbody><tr><td>Volume</td> <td>=</td> <td><em>&pi;&thinsp;r</em>&sup2;<em>h</em> = <em>&pi;</em> &times; 4&sup2; &times; 6 = 96 <em>&pi;</em></td> </tr><tr><td></td> <td>=</td> <td>301.5928947 cm&sup3;</td> </tr><tr><td></td> <td>=</td> <td>302 cm&sup3; (to 3 significant figures)</td> </tr></tbody></table><table><tbody><tr><td>Area of curved surface</td> <td>=</td> <td>2<em>&pi;&thinsp;rh</em> = 2 &times; <em>&pi;</em> &times; 4 &times; 6</td> </tr><tr><td></td> <td>=</td> <td>48<em>&pi;</em></td> </tr><tr><td></td> <td>=</td> <td>150.7964474 cm&sup2;</td> </tr></tbody></table><table><tbody><tr><td>Area of each end</td> <td>=</td> <td><em>&pi;&thinsp;r</em>&sup2; = <em>&pi;</em> &times; 4&sup2;</td> </tr><tr><td></td> <td>=</td> <td>16<em>&pi;</em></td> </tr><tr><td></td> <td>=</td> <td>50.26548246 cm&sup2;</td> </tr></tbody></table><table><tbody><tr><td>Total surface area</td> <td>=</td> <td>150.7964474 + (2 &times; 50.26548246)</td> </tr><tr><td></td> <td>=</td> <td>251.3274123 cm&sup2;</td> </tr><tr><td></td> <td>=</td> <td>251 cm&sup2; (to 3 significant figures)</td> </tr></tbody></table><p></p>

Q11:

The cost of carpeting a room of 7m long, 6m wide at the rate of Rs.200/m is Rs.2400. Find the breadth of the carpet.


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Here,</p> <p>Length of the room (l) = 7m</p> <p>Breadth of room (b0 = 6m</p> <p>Cost of carpeting the room (C) = Rs.200/m</p> <p>Total cost (T= Rs.2600</p> <p>&there4; Length of carpet (l<sub>1</sub>) = T/C</p> <p> = Rs.2400/Rs.200 = 12m</p> <p>Breadth of carpet (b<sub>1</sub>) = ?</p> <p>When we carpet the room,</p> <p>Area of carpet = Area of room</p> <p>i.e. l<sub>1</sub>*b<sub>1</sub> = l*b</p> <p>or, 12m*b<sub>1</sub> = 7m*6m</p> <p>or, b<sub>1</sub> = 42m<sup>2</sup>/12m</p> <p>or, b<sub>1</sub> = 3.5 m</p> <p>&there4; The breadth of the carpet is 3.5m.</p>

Q12:

Find the  number of stones of size 40cm long and 30 cm wide to cover a courtyard of length 30 m and breadth is 15 m?


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Here,</p> <p>Length of stone (l<sub>1</sub>) = 40 cm = 0.4m</p> <p>Breadth of stone (b<sub>1</sub>) = 30 cm = 0.3m</p> <p>&there4; Area of each stone (a) = l<sub>1</sub>*b<sub>1</sub></p> <p>= 0.4m * 0.3m = 0.12m<sup>2</sup></p> <p>And, length of courtyard (l) = 30m</p> <p>breadth of courtayard (b) = 15m</p> <p>&there4; Area of courtyard (A) = l*b</p> <p>= 30 m * 15 m</p> <p>= 450 m<sup>2</sup></p> <p>Now,</p> <p>Required number of stones (N) = Area of courtyard(A) / Area of stone(a)</p> <p>= 450m<sup>2</sup> / 0.12m<sup>2</sup></p> <p>= 3750</p> <p>Hence, 3750 stones of given dimension are required to cover the courtyard.</p>

Q13:

Find the length of carpet of width 50cm to carpet a hall 8m long and 6 m wide. Also find the cost of carpeting at the rate of Rs.400/m.


Type: Long Difficulty: Easy

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Answer: <p>Here,</p> <p>Length of carpet (l<sub>1</sub>) = ?</p> <p>Breadth of carpet (b<sub>1</sub>) = 50cm = 0.5m</p> <p>Length of room (l) = 8m</p> <p>Breadth of room (b) = 6m</p> <p>While carpeting a room,</p> <p>Area of carpet (A<sub>1</sub>) = Area of room (A)</p> <p>i.e. l<sub>1</sub>*b<sub>1</sub>= l*b</p> <p>or, l<sub>1</sub> = l*b / b<sub>1</sub></p> <p><sub></sub> = 8m * 6m / 0.5 m</p> <p> = 96m</p> <p>&there4; 96m long carpet is required.</p> <p>Again,</p> <p>Length of carpet (l<sub>1</sub>) = 96m</p> <p>cost of carpet/ metre (C) = Rs.400/m</p> <p>Total cost (T) = l<sub>1</sub>*C</p> <p> = 96 * Rs.400</p> <p> = Rs.38400</p> <p>Hence, it costs Rs.38400 in order to carpet the room.</p>

Q14:

A hall is 30 m long and 20 wide. The area of four wall is 400 m2. Find the height of the wall. Also, find

a) The area of four wall and ceiling.

b) The area of the four wall, ceiling and floor (T.S.A).

c) The area of floor and ceiling.

d) Find the cost of painting four wall and ceiling at the rate of Rs.200/m2


Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p>Hence,</p> <p>length (l) = 40m</p> <p>breadth (b) = 30 m</p> <p>area of 4 walls (A) = 600m<sup>2</sup></p> <p>height (h) =?</p> <p>Area of 4 walls (A) = 400m<sup>2</sup></p> <p>i.e. 2h(l + b) = 400 m<sup>2</sup></p> <p>or, 2h( 30m + 20m) = 400m<sup>2</sup></p> <p>or, 2h * 50m = 400m<sup>2</sup></p> <p>or, h = 4m</p> <p>&there4; height of the wall (h) = 4m</p> <p>a) Now, Area of 4 walls and ceiling</p> <p>= 2h(l + b) + lb</p> <p>= 2*4m( 30m + 20m) + 30m*20m</p> <p>= 400m<sup>2</sup> + 600m<sup>2</sup></p> <p>= 1000m<sup>2</sup></p> <p>b) Area of 4 walls, ceiling and floor (T.S.A)</p> <p>= 2h (l + b) + 2lb</p> <p>= 2*4m (30m + 20m ) + 2*30m*20m</p> <p>= 400m<sup>2</sup> + 1200m<sup>2</sup></p> <p>= 1600m<sup>2</sup></p> <p>c) Area of floor and ceiling</p> <p>= 2 lb</p> <p>= 2*30m*20m</p> <p>=1200m<sup>2</sup></p> <p>d) Area of 4 walls and ceiling (A) = 1000m<sup>2</sup></p> <p>cost of painting the wall (C) = Rs.200/m<sup>2</sup></p> <p>Total cost (T) = C*A</p> <p> = Rs.200/m<sup>2</sup>*1000m<sup>2</sup></p> <p> = Rs.200000</p> <p>Total cost of painting four wall and ceiling is Rs.200000.</p> <p></p>

Q15:

A room is 5m long, 4m wide and 3m high. Find the following:

a) Area of floor of a room

b) Area of 4 walls of a room

c) Area of ceiling

d) Area of 4 wall and ceiling

e) Area of 4 wall, ceiling and floor (TSA of a room)


Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p>Here,</p> <p>length (l) = 5m</p> <p>breadth (b) = 4 m</p> <p>height (h) = 3m</p> <p>Now,</p> <p>a) Area of floor (A)</p> <p>= l*b</p> <p>= 5m * 4m</p> <p>= 20m<sup>2</sup></p> <p>b) Area of 4 walls (A)</p> <p>= 2h (l + b)</p> <p>= 2*3m (5m + 4m)</p> <p>= 54m<sup>2</sup></p> <p>c) Area of ceiling (A)</p> <p>= l * b</p> <p>= 5m * 4m</p> <p>= 20m<sup>2</sup></p> <p>d) Area of 4 wall and ceiling (A)</p> <p>= 2h (l + b) + lb</p> <p>= 2*3m (5m + 4m) + 5m * 4m</p> <p>= 54 m<sup>2</sup>+ 20m<sup>2</sup></p> <p>= 74 m<sup>2</sup></p> <p>e) Area of 4 wall, ceiling and floor (TSA)</p> <p>= 2h (l + b) + lb + lb</p> <p>= 2*3m(5m + 4m) + 5m * 4m + 5m * 4m</p> <p>=54m<sup>2</sup> + 20m<sup>2</sup>+ 20m<sup>2</sup></p> <p>=94m<sup>2</sup></p>

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Mode and Range

Mode and Range

Mode

.

The value which appears most often is called mode. If there is no repeated numbers then there is no mode. The mode f the of data is denoted by Mo.

Examples

  • In the data, 4, 5, 6, 6, 6,7, 8. The mode is 6.
  • In the data, 5, 6, 7, 8, 9, 10,11. There is no mode.
  • In the data 5, 8, 8, 15, 18, 18. There are two modes 8 and 18.

Range

.

The range is the difference between the smallest and highest values in a set.To find the range, subtract the smallest values in the data set for the largest.

Examples

  1. Five boys have weights 45 kg, 55 kg, 50kg, 60 kg and 65 kg. What is the range of their heights?
    Solution:
    Arranging the data in order,
    45 kg, 50 kg, 55 kg, 60 kg and 65 kg
    Largest value = 65 kg
    Smallest value = 45 kg
    Range = largest value - smallest value
    = 65 kg - 45 kg
    = 20 kg

  2. What is the range of this group of numbers: 70, 20, 77, 50, 80, 60, 75, 40?
    Solution:
    Arranging the data in order,
    20, 40, 50, 60, 70, 75, 77, 80
    Largest value = 80
    Smallest value = 20
    Range = largest value - smallest value
    = 80 - 20
    = 20

Lesson

Statistics

Subject

Compulsory Maths

Grade

Grade 8

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