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Textbook | NCERT |

Class | Class 10 |

Subject | Maths |

Chapter | Chapter 9 |

Chapter Name | Mean, Median, Mode of Grouped Data |

Exercise | 9E |

Category | RS Aggarwal Solutions |

**RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9 E**

**Question 1:**

From the given table, we may prepare the cumulative frequency table as shown below:

Class interval | C.F |

Less than 220 | 7 |

Less than 240 | 10 |

Less than260 | 16 |

Less than 280 | 24 |

Less than 300 | 26 |

Less than 320 | 30 |

On a graph paper, we plot the point (220, 7), (240, 10), (260, 16), (280, 24), (300, 26), (320, 30)

We join these points successively with a freehand to get the cumulative frequency curve or an ogive

Here N = 30, N2= 15

Take point (0, 15) on the y-axis and draw AP || x-axis meeting the curve at P. Draw PM ⊥ x -axis intersecting the x-axis at M. Then OM = 258 and hence median = 258.

**Question 2:**

Cumulative frequency table is as follows:

Marks | C.F |

Marks less than 10 | 3 |

Marks less than 20 | 11 |

Marks less than 30 | 28 |

Marks less than 40 | 48 |

Marks less than 50 | 70 |

Scale :

Horizontal: 1 small div = 10 marks

Vertical : 1 small div = 10 students

The points (10, 3), (20, 11), (30, 28), (40, 48), (50, 70) are plotted and joined as shown above. This is required cumulative curve.

N = 70, N2= 35

Ob vertical line OY, take OA = 35

Through A, a horizontal line AP is drawn meeting the graph at P

Through P, a vertical line PM is drawn.

Now, OM = 34, therefore Median = 34

**Question 3:**

The following table gives height in metre of 360 tress:

Height in m | Number of trees |

Less than 7Less than 14Less than 21Less than 28Less than 35Less than 42Less than 49Less than 56 | 254595140235275320360 |

The points (7, 25), (14, 45), 21, 95), (28, 140), (35, 235), (42, 275), (49, 320), (56, 360) are plotted and joined one by one as shown in the figure.

This is the required cumulative curve

N = 360, N2= 180

At y = 180, affix A.

Through A, draw a horizontal line meeting the curve at P.

Through P, a vertical line is drawn which meets OX at M.

OM = 31.

Hence median = 31

**Question 4:**

Less than series

Less thanCapital(in lakh of Rs) | No. of Companies(C.F.) |

Less than 10Less than 20Less than 30Less than 40Less than 50Less than 60Less than 70Less than 80 | 25122338454750 |

Scale: Horizontal : 1 small div = capital of 1 lakh of Rs

Vertical: 1 small div. = 1 company.

Plot the points (10, 2), (20, 5), (30, 12), (40, 23), (50, 38), (60, 45), (70, 47) and (80, 50).

Through A(0, 25), AP is drawn parallel to OX and PM OX, OM = 41. Hence median = 41.

**Question 5:**

More than series

Score | No of Candidates |

More than 400 | 230 |

More than 450 | 210 |

More than 500 | 175 |

More than 550 | 135 |

More than 600 | 103 |

More than 600 | 79 |

More than 700 | 52 |

More than 750 | 34 |

We plot the points (400, 230), (450, 210), (500, 175), (550, 135), (600, 103), (650, 79), (700, 52), (750, 34)

Hence, N = 230, N2= 115

Take a point A(0, 115) on the y-axis and draw AP||x-axis meeting the curve at P, Draw PM x-axis intersecting x-axis at M

Then, OM = 590

Hence median = 590

**Question 6:**

(i) Less than series:

Marks | Number of Students |

Less than 5Less than 10Less than 15Less than 20Less than 25Less than 30Less than 35Less than 40Less than 45Less than 50 | 2713213156769498100 |

Plot the points (5, 2), (10, 7), (15, 13), (20, 21), (25, 31), (30, 56), (35, 76), (40, 94), (45, 98) and (50, 100)

Join these points free hand to get the curve representing “less than” cumulative curve.

(ii) From the given table we may prepare the ‘more than’ series as shown below

Marks | Number of Students |

More than 45More than 40More than 35More than 30More than 25More than 20More than 15More than 10More than 5More than 0 | 2624446979879398100 |

Now, on the same graph paper as above, we plot the point (0, 100), (5, 98), (10, 93), (15, 87), (20, 79), (25, 69), (30, 44), (35, 24) and (40, 6) and (45, 2)Plot the points (5, 2), (10, 7), (15, 13), (20, 21), (25, 31), (30, 56), (35, 76), (40, 94), (45, 98) and (50, 100)

Join these points free hand to get required curve

Here N = 100, N2= 50

Two curves intersect at point P(28, 50)

Hence, the median = 28

**Question 7:**

We may prepare less than series and more than series

(i) Less than series

Height in (cm) | Frequency |

Less than 140Less than 144Less than 148Less than 152Less than 156Less than 160Less than 164Less than 168Less than 172Less than 176Less than 180 | 03123667109173248330416450 |

Now on graph paper plot the points (140, 0), (144, 3), (148, 12), (152, 36), (156, 67), (160, 109), (164, 173), (168, 248), (172, 330), (176, 416), (180, 450)

(ii) More than series

Height in (cm) | C.F. |

More than 140More than 144More than 148More than 152More than 156More than 160More than 164More than 168More than 172More than 176More than 180 | 450447438414383341277202120340 |

Now on the same graph plot the points (140, 450), (144, 447), (148, 438), (152, 414), (156, 383), (160, 341), (164, 277), (168, 202), (172, 120), (176, 34), (180, 0)

The curves intersect at (167, 225).

Hence, 167 is the median.

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