Latitudes and Longitudes

Latitude is a geographic coordinate that specifies the north-south position of a point on the earth's surface. Longitude refers to the imaginary line that bisects the globe through the North and the South poles. This note has explains about the longitudes and latitudes.

Summary

Latitude is a geographic coordinate that specifies the north-south position of a point on the earth's surface. Longitude refers to the imaginary line that bisects the globe through the North and the South poles. This note has explains about the longitudes and latitudes.

Things to Remember

  • Latitude is a geographic coordinate that specifies the north-south position of a point on the earth's surface. 
  • Longitude refers to the imaginary line that bisects the globe through the North and the South poles. 
  • The parts of the earth diametrically opposite is called antipode. 
  • As latitude increases, the temperature decreases.
  • Latitude is used to find out the location of any place.

MCQs

No MCQs found.

Subjective Questions

Q1:

Calculate the first quartile.
2, 18, 20, 14, 10, 4, 16 


Type: Very_short Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>Now,<br>Arranging in ascerding order<br>2, 4, 10, 14, 16, 18, 20<br>Total number(N) = 7<br>We know that,<br>or, Q<sub>1</sub> = \(\begin{pmatrix}\frac{n + 1}{4} \end{pmatrix}\)<sup>th</sup> term<br>or, Q<sub>1</sub> = \(\begin{pmatrix}\frac{7 + 1}{4} \end{pmatrix}\)<sup>th</sup> term<br>or, Q<sub>1</sub> = \(\begin{pmatrix}\frac{8}{4} \end{pmatrix}\)<sup>th</sup> term<br>or, Q<sub>1</sub> = 2<sup>th</sup> term<br>\(\therefore\) The value of Q<sub>1</sub>is 4.</p>

Q2:

Calculate the third quartile.
90, 42, 52, 82, 72, 62, 100


Type: Short Difficulty: Easy

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Answer: <p>Solution:</p> <p>Now,<br>Arranging in ascending order<br>42, 52, 62, 72, 82, 90, 100 <br>Total number (N) = 7<br>We know that,<br>or, Q<sub>3</sub> = 3\(\begin{pmatrix}\frac{n + 1}{4} \end{pmatrix}\)<sup>th</sup>item<br>or, Q<sub>3</sub> = 3\(\begin{pmatrix}\frac{7 + 1}{4} \end{pmatrix}\)<sup>th</sup>item<br>or, Q<sub>3</sub> = 3\(\begin{pmatrix}\frac{8}{4} \end{pmatrix}\)<sup>th</sup>item<br>or, Q<sub>3</sub> = 3 \(\times\) 2<sup>th</sup>item<br>or, Q<sub>3</sub> = 6<sup>th</sup>item<br>\(\therefore\) The value of Q<sub>3</sub>is 90.<br><br><br><br><br><br><br><br></p>

Q3:

Find the Q1, Q2 and Q3 from the following data.
72, 52, 50, 61, 56, 68, 64


Type: Long Difficulty: Easy

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Answer: <p>Solution:</p> <p>Now,<br>Arranging in ascending order<br>50, 52, 56, 61, 64, 68, 72<br>numbers = 7<br>We know that,<br>or, Q<sub>1</sub> = \(\begin{pmatrix}\frac{n + 1}{4} \end{pmatrix}\)<sup>th</sup>item<br>or, Q<sub>1</sub> = \(\begin{pmatrix}\frac{7 + 1}{4} \end{pmatrix}\)<sup>th</sup>item<br>or, Q<sub>1</sub> = \(\begin{pmatrix}\frac{8}{4} \end{pmatrix}\)<sup>th</sup>item<br>or, Q<sub>1</sub> = 2<sup>th</sup>item<br>or,Q<sub>1</sub> = 52<br>Again,<br>or, Q<sub>2</sub>= \(\begin{pmatrix}\frac{n + 1}{2} \end{pmatrix}\)<sup>th</sup>item<br>or, Q<sub>2</sub>= \(\begin{pmatrix}\frac{7 + 1}{2} \end{pmatrix}\)<sup>th</sup>item<br>or, Q<sub>2</sub>= \(\begin{pmatrix}\frac{8}{2} \end{pmatrix}\)<sup>th</sup>item<br>or, Q<sub>2</sub>= 4<sup>th</sup>item<br>or, Q<sub>2</sub>= 61<br>Similarly,<br>or, Q<sub>3</sub>= 3\(\begin{pmatrix}\frac{n + 1}{4} \end{pmatrix}\)<sup>th</sup>item<br>or, Q<sub>3</sub>= 3\(\begin{pmatrix}\frac{7 + 1}{4} \end{pmatrix}\)<sup>th</sup>item<br>or, Q<sub>3</sub>= 3\(\begin{pmatrix}\frac{8}{4} \end{pmatrix}\)<sup>th</sup>item<br>or, Q<sub>3</sub>= 3 \(\times\)2<sup>th</sup>item<br>or, Q<sub>3</sub>= 6<sup>th</sup>item<br>or, Q<sub>3 </sub>= 68</p> <p>\(\therefore\) The value ofQ<sub>1</sub>, Q<sub>2</sub> and Q<sub>3</sub>is 52, 61 and 68 respectively.</p>

Q4:

Find the lower quartile and upper quartile  of the following data set of scores:

18     20     23     20     23     27     24


Type: Short Difficulty: Easy

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Answer: <p>Solution:</p> <p>Arrange the values in ascending order of magnitude:</p> <p>18 20 20 23 23 23 24 27 29</p> <p>Here, n =7</p> <p>Lower quartile = value of\(\begin{pmatrix}\frac{n+1}{4} \end{pmatrix}\)<sup>th</sup>term</p> <p>= value of\(\begin{pmatrix}\frac{7+1}{4} \end{pmatrix}\)<sup>th</sup>term</p> <p>= value of\(\begin{pmatrix}\frac{8}{4} \end{pmatrix}\)<sup>th</sup>term</p> <p>= value of 2<sup><span style="font-size: xx-small;">nd</span></sup><span style="font-size: xx-small;"></span>item</p> <p>\(\therefore\) Lower Quartile = 20</p> <p>Upper Quartile = value of\(\begin{pmatrix}\frac{3(n+1)}{4} \end{pmatrix}\)<sup>th</sup>term</p> <p>= value of\(\begin{pmatrix}\frac{3(7+1)}{4} \end{pmatrix}\)<sup>th</sup>term</p> <p>= value of\(\begin{pmatrix}\frac{3 \times 8}{4} \end{pmatrix}\)<sup>th</sup>term</p> <p>= value of\(\begin{pmatrix}\frac{24}{4} \end{pmatrix}\)<sup>th</sup>term</p> <p>= value of 6<sup>th</sup>term</p> <p>\(\therefore\) Upper Quartile = 27</p>

Q5:

Compute Q1, Q2 and Q3 from the following data:

Marks 10 20 30 40 50 60 70
No. of Students 2 4 6 12 8 6 1

Type: Long Difficulty: Easy

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Answer: <p>Solution:</p> <table width="305"><tbody><tr><td>Marks</td> <td>No. of students(f)</td> <td>cumulative frequency (c.f.)</td> </tr><tr><td>10</td> <td>2</td> <td>2</td> </tr><tr><td>20</td> <td>4</td> <td>6</td> </tr><tr><td>30</td> <td>6</td> <td>12</td> </tr><tr><td>40</td> <td>12</td> <td>24</td> </tr><tr><td>50</td> <td>8</td> <td>32</td> </tr><tr><td>60</td> <td>6</td> <td>38</td> </tr><tr><td>70</td> <td>1</td> <td>39</td> </tr><tr><td></td> <td>N = 39</td> <td></td> </tr></tbody></table><p>We have,</p> <p>Q1= value of \(\begin{pmatrix}\frac{n + 1}{4} \end{pmatrix}\)<sup>th</sup> term</p> <p>= value of \(\begin{pmatrix}\frac{39 + 1}{4} \end{pmatrix}\)<sup>th</sup> term</p> <p>= value of 10<sup>th</sup> term</p> <p>In c.f. column just greater than 10 is 12 and its coressponding value is 30.</p> <p>\(\therefore\) Q<sub>1</sub> = 30</p> <p>Q<sub>2 </sub>=value of \(\begin{pmatrix}\frac{n + 1}{2} \end{pmatrix}\)<sup>th</sup> term</p> <p>= value of \(\begin{pmatrix}\frac{39 + 1}{2} \end{pmatrix}\)<sup>th</sup> term</p> <p>= value of 20<sup>th</sup> term</p> <p>In c.f. column just greater than 20 is 24and its coressponding value is 40.</p> <p>\(\therefore\) Q<sub>2</sub>=40</p> <p>Q<sub>3 </sub>=value of \(\begin{pmatrix}\frac{3(n + 1)}{4} \end{pmatrix}\)<sup>th</sup> term</p> <p>=value of \(\begin{pmatrix}\frac{3(39 + 1)}{4} \end{pmatrix}\)<sup>th</sup> term</p> <p>=value of \(\begin{pmatrix}\frac{3 * 40}{4} \end{pmatrix}\)<sup>th</sup> term</p> <p>=value of \(\begin{pmatrix}\frac{3(39 + 1)}{4} \end{pmatrix}\)<sup>th</sup> term</p> <p>=value of 30<sup>th</sup> term</p> <p>In c.f. column just greater than 30 is 32 and its coressponding value is 50.</p> <p>\(\therefore\) Q<sub>3</sub>= 50</p>

Q6:

The number 6, x+5, 12, 14, 17, 20, 21 are in ascending order. If Q1 of this data is 8, find the value of x.


Type: Short Difficulty: Easy

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Answer: <p>Solution:</p> <p>Here,<br>6, x+5, 12, 14, 17, 20, 21<br>Q<sub>1</sub>= 8<br>numbers = 7<br>We know that,<br>or, Q<sub>1</sub> = \(\begin{pmatrix}\frac{n + 1}{4} \end{pmatrix}\)<sup>th</sup> item<br>or, Q<sub>1</sub> = \(\begin{pmatrix}\frac{7 + 1}{4} \end{pmatrix}\)<sup>th</sup> item<br>or, Q<sub>1</sub> = \(\begin{pmatrix}\frac{8}{4} \end{pmatrix}\)<sup>th</sup> item<br>or, Q<sub>1</sub> = 2<sup>th</sup> item<br>or, Q<sub>1</sub> =x + 5<br>Now,<br>or,Q<sub>1</sub>= x + 5<br>or, 8= x + 5<br>or, x = 8 - 5<br>\(\therefore\) x = 5<br><br><br><br><br></p>

Q7:

Calculate Q1 and Q3 from the following data.

marks 15 25 35 45 55
No. of students 5 10 16 7 5

Type: Long Difficulty: Easy

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Answer: <p>Solution:</p> <table width="342"><tbody><tr><td>Marks</td> <td>Frequency (f)</td> <td>c.f</td> </tr><tr><td>15</td> <td>5</td> <td>5</td> </tr><tr><td>25</td> <td>10</td> <td>15</td> </tr><tr><td>35</td> <td>16</td> <td>31</td> </tr><tr><td>45</td> <td>7</td> <td>38</td> </tr><tr><td>55</td> <td>5</td> <td>43</td> </tr><tr><td></td> <td>N = 43</td> <td></td> </tr></tbody></table><p>Now,<br>or,Q<sub>1</sub>= \(\begin{pmatrix}\frac{N + 1}{4} \end{pmatrix}\)<sup>th</sup> item<br>or,Q<sub>1</sub>= \(\begin{pmatrix}\frac{43 + 1}{4} \end{pmatrix}\)<sup>th</sup> item<br>or,Q<sub>1</sub>= \(\begin{pmatrix}\frac{44}{4} \end{pmatrix}\)<sup>th</sup> item<br>or,Q<sub>1</sub>= 11<sup>th</sup> item</p> <p>In c.f column, c.f is just greater than 11 is 15 and its corresponding value is 25. So, Q<sub>1</sub> = 25<br>Again,<br>or, Q<sub>3</sub> = 3\(\begin{pmatrix}\frac{n + 1}{4} \end{pmatrix}\)<br>or, Q<sub>3</sub> = 3\(\begin{pmatrix}\frac{43 + 1}{4} \end{pmatrix}\)<br>or, Q<sub>3</sub> = 3\(\begin{pmatrix}\frac{44}{4} \end{pmatrix}\)<br>or, Q<sub>3</sub> = 3 \(\times\) 11<sup>th</sup>item<br>or, Q<sub>3</sub> = 33<sup>th</sup> item</p> <p>In c.f column, c.f is just greater than 33 is 38 and its corresponding value is 45.<br>So, Q<sub>3</sub> is 45.</p>

Q8:

Calculate the first quartile (lower quartile) from the given data.

Class  0 -10  10 -20  20 - 30  30 - 40  40 - 50
frequency  6 8 10 7 5




Type: Long Difficulty: Easy

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Answer: <p>Solution:</p> <table width="363"><tbody><tr><td>x</td> <td>f</td> <td>c.f</td> </tr><tr><td>0 -10</td> <td>6</td> <td>6</td> </tr><tr><td>10 - 20</td> <td>8</td> <td>14</td> </tr><tr><td>20 - 30</td> <td>10</td> <td>24</td> </tr><tr><td>30 - 40</td> <td>7</td> <td>31</td> </tr><tr><td>40 - 50</td> <td>5</td> <td>36</td> </tr><tr><td></td> <td>N = 36</td> <td></td> </tr></tbody></table><p>Now,<br>or, Q<sub>1</sub> = \(\begin{pmatrix}\frac{N}{4} \end{pmatrix}\)<sup>th</sup> item<br>or, Q<sub>1</sub> = \(\begin{pmatrix}\frac{36}{4} \end{pmatrix}\)<sup>th</sup> item<br>or, Q<sub>1</sub> = 9<sup>th</sup> item<br>In c.f column, c.f is just greater than 9 is 14 and its corresponding value is 10 - 20.<br>So, Q<sub>1</sub> lies in class 10 - 20<br>Then,<br>l = 10, c.f = 6, f = 8, and i = 10</p> <p>or, Q<sub>1</sub> =L + \(\frac{\frac{N}{4}-C.F}{f}\) &times; i<br>or,Q<sub>1</sub> = 10 + \(\frac{\frac{36}{4}-6}{8}\) &times; 10<br>or,Q<sub>1</sub> = 10 + \(\frac{9 - 6}{8}\) &times; 10<br>or,Q<sub>1</sub> = 10 + \(\frac{3}{8}\) &times; 10<br>or,Q<sub>1</sub> = 10 + \(\frac{3}{4}\) &times; 5<br>or,Q<sub>1</sub> = 10 + 0.75&times; 5<br>or,Q<sub>1</sub> = 10 +3.75<br>or,Q<sub>1</sub> = 13.75<br>\(\therefore\) The value of Q<sub>1</sub>is 13.75.<br><br></p>

Q9:

Calculate the first quartile and third quartile from the given data:

Class 0-5 5-10 10-15 15-20 20-25 25-30
Frequency 6 8 10 7 5 3

Type: Long Difficulty: Easy

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Answer: <p>Solution:</p> <table width="189"><tbody><tr><td>Class</td> <td>Frequency(f)</td> <td>c.f.</td> </tr><tr><td>0-5</td> <td>6</td> <td>6</td> </tr><tr><td>5-10</td> <td>8</td> <td>14</td> </tr><tr><td>10-15</td> <td>10</td> <td>24</td> </tr><tr><td>15-20</td> <td>7</td> <td>31</td> </tr><tr><td>20-25</td> <td>5</td> <td>36</td> </tr><tr><td>25-30</td> <td>4</td> <td>40</td> </tr><tr><td></td> <td>N =40</td> <td></td> </tr></tbody></table><p>Now,</p> <p>Q1 Class = value of (\(\frac{N}{4}\))<sup>th</sup> item</p> <p>= value of ( \(\frac{40}{4}\))<sup>th</sup> item</p> <p>= value of 10<sup>th</sup> item</p> <p>In c.f. column, c.f. just greater than 10 is 14 and its corresponding class is (5-10).</p> <p>\(\therefore\) Q<sub>1</sub> lies in the class 5-10</p> <p>Here,</p> <p>\(\frac{N}{4}\) = 10,l= 5, c.f. = 6, f = 8 and i = 5</p> <p>\(\therefore\)Q<sub>1</sub>= l + \(\frac{\frac{N}{4}&minus; c.f}{f}\) &times; i</p> <p>= 5 + \(\frac{10&minus; 6}{8}\) &times;5</p> <p>=5 + \(\frac{4}{8}\) &times;5</p> <p>= 5 + 2.5</p> <p>= 7.5</p> <p>Again,</p> <p>Q<sub>3</sub> Class = value of(\(\frac{3N}{4}\))<sup>th</sup> item</p> <p>= value of(\(\frac{3 * 40}{4}\))<sup>th</sup> item</p> <p>=value of30<sup>th</sup> item</p> <p>In c.f. column, c.f.just greater than 30 is 31 and its corresponding class is 15-20.</p> <p>Here,</p> <p>\(\frac{3N}{4}\) = 30,l= 5, c.f. = 24, f = 7and i = 5</p> <p>\(\therefore\)Q<sub>3</sub>= l + \(\frac{\frac{3N}{4}&minus; c.f}{f}\) &times; i</p> <p>= 5 + \(\frac{30&minus; 24}{7}\) &times;5</p> <p>= 5 + 4.28</p> <p>= 9.28</p> <p></p> <p></p>

Q10:

Calculate the third quartile (upper quartile) from the following data.

Class 5 -10 10 -15  15 - 20  20 - 25  25 - 30  30 - 35  35 - 40
Frequency 4 6 10 15 10 6 5

Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <table width="396"><tbody><tr><td>Class</td> <td>frequency</td> <td>c.f</td> </tr><tr><td>5 - 10</td> <td>4</td> <td>4</td> </tr><tr><td>10 - 15</td> <td>6</td> <td>10</td> </tr><tr><td>15 - 20</td> <td>10</td> <td>20</td> </tr><tr><td>20 - 25</td> <td>15</td> <td>35</td> </tr><tr><td>25 - 30</td> <td>10</td> <td>45</td> </tr><tr><td>30 - 35</td> <td>6</td> <td>51</td> </tr><tr><td>35 - 40</td> <td>5</td> <td>56</td> </tr><tr><td></td> <td>N = 56</td> <td></td> </tr></tbody></table><p>Now,<br>or, Q<sub>3</sub>= 3(\(\frac{N}{4}\))<sup>th</sup> item<br>or, Q<sub>3</sub>= 3(\(\frac{56}{4}\))<sup>th</sup> item<br>or, Q<sub>3</sub>= 3 \(\times\) 14<sup>th</sup> item<br>or, Q<sub>3</sub>= 42<sup>th</sup> item<br>In c.f column, c.f is just greater than 42 is 45 and its corresponding value is 25 - 30.<br>Now,<br>l = 25, c.f = 35, f = 10, and i = 5<br>Then,<br>or, Q<sub>3</sub>= L + \(\frac{\frac{3N}{4}-c.f}{f}\)&times; i<br>or, Q<sub>3</sub>= 25 + \(\frac{42 - 35}{10}\)&times; 5<br>or, Q<sub>3</sub>= 25 + \(\frac{7}{2}\)<br>or, Q<sub>3</sub>= 25 +3.5<br>or, Q<sub>3</sub>= 28.5<br>\(\therefore\) The value of Q<sub>3</sub> is 28.5.</p>

Q11:

The numbers 10, 20, 30, 40, 50, x + 25 and 70 are in ascerding order. If Q3 of this data is 60, find the value of x.


Type: Short Difficulty: Easy

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Answer: <p>Solution:</p> <p>Here,<br>numbers = 7<br>Q<sub>3</sub>= 60<br>Now,<br>or, Q<sub>3</sub>= 3(\(\frac{N + 1}{4}\))<sup>th</sup> item<br>or, Q<sub>3</sub>= 3(\(\frac{7 + 1}{4}\))<sup>th</sup> item<br>or, Q<sub>3</sub>= 3(\(\frac{8}{4}\))<sup>th</sup> item<br>or, Q<sub>3</sub>= 3 \(\times\) 2<sup>th</sup> item<br>or, Q<sub>3</sub>= 6<sup>th</sup> item<br>i.e.Q<sub>3</sub>= x + 25<br>Then,<br>or,Q<sub>3</sub>= x + 25<br>or, 60 = x + 25<br>or, x + 60 - 25<br>\(\therefore\) x = 35<br><br><br><br></p>

Q12:

Calculate the first quartile (lower quartile) from the give data.
42, 58, 36, 38, 56, 50, 48


Type: Very_short Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>Arranging in ascerding order<br>36, 38, 42, 48, 50, 56, 58<br>Total numbers (N) = 7<br>We know that,<br>or, Q<sub>1</sub>= (\(\frac{N+1}{4}\))<sup>th</sup> item<br>or, Q<sub>1</sub>= (\(\frac{7+1}{4}\))<sup>th</sup> item<br>or, Q<sub>1</sub>= (\(\frac{8}{4}\))<sup>th</sup> item<br>or, Q<sub>1</sub>= 2<sup>th</sup> item<br>i.e.Q<sub>1</sub>= 38</p>

Q13:

Calculate first, second and third quartile from the following data:

Marks 30-40 40-50 50-60 60-70 70-80 80-90
Students 5 12 25 15 8 3

Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <table width="189"><tbody><tr><td>Marks</td> <td>Students (f)</td> <td>c.f.</td> </tr><tr><td>30-40</td> <td>5</td> <td>5</td> </tr><tr><td>40-50</td> <td>12</td> <td>17</td> </tr><tr><td>50-60</td> <td>25</td> <td>42</td> </tr><tr><td>60-70</td> <td>15</td> <td>57</td> </tr><tr><td>70-80</td> <td>8</td> <td>65</td> </tr><tr><td>80-90</td> <td>3</td> <td>68</td> </tr><tr><td></td> <td>N =68</td> <td></td> </tr></tbody></table><p>Now,</p> <p>Q<sub>1</sub> Class = value of (\(\frac{N}{4}\))<sup>th</sup> item</p> <p>= value of ( \(\frac{68}{4}\))<sup>th</sup> item</p> <p>= value of 17<sup>th</sup> item</p> <p>In c.f. column, c.f. just greater than 17 is 42 and its corresponding class is (50-60).</p> <p>\(\therefore\) Q<sub>1</sub> lies in the class 50-60</p> <p>Here,</p> <p>\(\frac{N}{4}\) = 17, l= 50, c.f. = 17, f = 25 and i =10</p> <p>\(\therefore\)Q<sub>1</sub>= l + \(\frac{\frac{N}{4}&minus; c.f}{f}\) &times; i</p> <p>= 50 + \(\frac{17&minus; 17}{25}\) &times;10</p> <p>=50 + \(\frac{0}{25}\) &times;10</p> <p>= 50 +0</p> <p>= 50</p> <p>Again,</p> <p>Q<sub>2</sub>Class = value of (\(\frac{N}{2}\))<sup>th</sup> item</p> <p>= value of ( \(\frac{68}{2}\))<sup>th</sup> item</p> <p>= value of 34<sup>th</sup> item</p> <p>In c.f. column, c.f. just greater than 34 is 42 and its corresponding class is (50-60).</p> <p>\(\therefore\) Q<sub>2</sub>lies in the class 50-60</p> <p>Here,</p> <p>\(\frac{N}{2}\) = 34, l = 50, c.f. = 17, f = 25 and i =10</p> <p>\(\therefore\)Q<sub>1</sub>= l + \(\frac{\frac{N}{2}&minus; c.f}{f}\) &times; i</p> <p>= 50 + \(\frac{34&minus; 17}{25}\) &times;10</p> <p>=50 + \(\frac{17}{25}\) &times;10</p> <p>= 50 +6.8</p> <p>= 56.8</p> <p>Q<sub>3</sub> Class = value of(\(\frac{3N}{4}\))<sup>th</sup> item</p> <p>= value of(\(\frac{3 * 68}{4}\))<sup>th</sup> item</p> <p>=value of 51<sup>th</sup> item</p> <p>In c.f. column, c.f.just greater than 51 is 57 and its corresponding class is 60-70.</p> <p>Here,</p> <p>\(\frac{3N}{4}\) = 51, l= 60, c.f. = 42, f = 15 and i = 10</p> <p>\(\therefore\)Q<sub>3</sub>= l + \(\frac{\frac{3N}{4}&minus; c.f}{f}\) &times; i</p> <p>= 60 + \(\frac{51&minus; 42}{15}\) &times; 10</p> <p>= 60 + 6</p> <p>= 66</p> <p></p> <p></p>

Q14:

Find the third quartile (upper quartile) from the given data.
41, 82, 77, 52, 70, 62, 46


Type: Very_short Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>Arranging in ascerding order<br>41, 46, 52, 62, 70, 77, 82<br>Total numbers (N)= 7<br>We know that,<br>or, Q<sub>3</sub> = 3(\(\frac{N+1}{4}\))<sup>th</sup> item<br>or, Q<sub>3</sub> = 3(\(\frac{7+1}{4}\))<sup>th</sup> item<br>or, Q<sub>3</sub> = 3(\(\frac{8}{4}\))<sup>th</sup> item<br>or, Q<sub>3</sub> = 3 \(\times\) 2<sup>th</sup> item<br>or, Q<sub>3</sub> = 6<sup>th</sup> item<br>i.e.Q<sub>3</sub> = 68</p>

Q15:

Find the Q1 and Q3 of the given data.

Weight (in kg) 50 60 70 80 90
No. of teachers 8 16 12 4 5

Type: Long Difficulty: Easy

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Answer: <p>Solution:</p> <table width="425"><tbody><tr><td>x</td> <td>f</td> <td>c.f</td> </tr><tr><td>50</td> <td>8</td> <td>8</td> </tr><tr><td>60</td> <td>16</td> <td>24</td> </tr><tr><td>70</td> <td>12</td> <td>36</td> </tr><tr><td>80</td> <td>4</td> <td>40</td> </tr><tr><td>90</td> <td>3</td> <td>43</td> </tr><tr><td></td> <td>N = 43</td> <td></td> </tr></tbody></table><p>Now,<br>or, Q<sub>1</sub> =(\(\frac{N+1}{4}\))<sup>th</sup> item<br>or, Q<sub>1</sub> =(\(\frac{43+1}{4}\))<sup>th</sup> item<br>or, Q<sub>1</sub> =(\(\frac{44}{4}\))<sup>th</sup> item<br>or, Q<sub>1</sub> =11<sup>th</sup> item<br>In c.f column, c.f is just greater than 11 is 24 and its corresponding value is 60.<br>So,Q<sub>1</sub> = 60<br>Again,<br>or, Q<sub>3</sub> =3(\(\frac{N+1}{4}\))<sup>th</sup> item<br>or, Q<sub>3</sub> =3(\(\frac{43+1}{4}\))<sup>th</sup> item<br>or, Q<sub>3</sub> =3(\(\frac{44}{4}\))<sup>th</sup> item<br>or, Q<sub>3</sub> =3 \(\times\) 11<sup>th</sup> item<br>or, Q<sub>3</sub> =33<sup>th</sup> item<br>In c.f column, c.f is just greater than 33 is 36 and its corresponding value is 70.<br>i.e.Q<sub>3</sub> = 70<br><br><br><br><br><br><br><br></p>

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Latitudes and Longitudes

Latitudes and Longitudes

Latitude

The vertical and horizontal lines, drawn only on the globe or maps, are imaginary lines. The equator in the basis for all the horizontal lines are known as latitudes and along with each latitude makes a complete circle. The equator 0° latitude passes exactly through the halfway between the North Pole andSouth Pole, the two opposite ends of the earth. It thus divides the earth's surface into a northern and southern hemisphere. There are altogether 180 latitudes, 90 on each of the two hemispheres. They are marked N or S according to their hemispheres. The latitudes are of varied lengths; the equator is the longest and poles (90°) merely a point or no more a circle.

Latitudinal effects
  1. As latitude increases, the temperature decreases.
  2. As the latitude increases, there is different in the length of day and night.
  3. Latitude is used to find out the location of any place.

Longitude

While the latitudes run from the east to the west, the longitudes run north to south converging at the poles. There is no prime longitude at the equator, the longest and at the midway between the two poles could be. All longitudes are half circles and of equal lengths. They form a right angle at the equator. Of the total 360 longitudes, the one passing through Greenwich in London in arbitrarily taken as the prime Meridian, the basis for all longitudes.The prime Meridian and 180° longitude, just opposite, together divide the earth's surface into two hemispheres- the eastern and western hemisphere. Hence, longitude is the angular distance of a place east or west of the Greenwich meridian, or west of the standard meridian of a celestial object, usually expressed in degrees and minutes. It refers to the imaginary line that bisects the globe through the North and the South poles.The lines on the eastern hemisphere are marked E and the ones in the west carry W for their notification.

Longitudinal effects
  1. There is a difference of 1 hour in every 15-degree longitude. The time of some areas of the eastern side of some longitude may be fast whereas the time of western side may be late.
  2. There is a change in day and date after crossing International dateline. There is a difference of 24 hours in both left and right side of this line.

Describing position

First we say how far from the equator a place is. This is its latitude. Then we must say how far from the prime meridian to east or west, is the place situated. This is its longitude. We can see that the position of the southern tip of Africa in approximately 35°S, 24°E. Thus, latitudes and longitudes together make a geographical grid according to which a place on the earth make a geographical grid according to which a place on the earth can be designated precisely. This in an actually same system of coordinates in geometry. When we have to designate the location of a larger area, we take two farthest latitudes on its north and south and two longitudes passing along its western borders. We say Nepal lies between 26°N and 30° N latitudes and between 80 °E and 88°E longitudes. Each degree of latitude and longitude in divided further into 60 minutes and each minute into 60 seconds. This division allows designation the location of a place more precisely. It in more precise to say numerically that Nepal extends from 26°22'N to 30°27'N latitudes and from 80°4' E to 88°12'E longitudes.

Antipode

The parts of the earth diametrically opposite are called antipodes.

Antipode of Latitudes

If the antipode of northern latitude is southern latitude than the antipode southern latitude is northern latitude such as the antipode of 20-degree northern latitude is equal to the antipode of southern latitude.

Lesson

Our Earth

Subject

Social Studies and Population Education

Grade

Grade 8

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