RD Sharma Solutions Chapter 7 Statistics Exercise 7.6 Class 10 Maths
Chapter Name  RD Sharma Chapter 7 Statistics 
Book Name  RD Sharma Mathematics for Class 10 
Other Exercises 

Related Study  NCERT Solutions for Class 10 Maths 
Exercise 7.6 Solutions
1. Draw an given by less than method for the following data:
No. of rooms: 
1 
2 
3 
4 
5 
6 
7 
8 
9 
10 
No. of houses: 
4 
9 
22 
28 
24 
12 
8 
6 
5 
2 
Solution
We first prepare the cumulative frequency distribution table by less than method as given be :
Now, we mark the upper class limits along x  axis and cumulative frequency along y  axis.
Thus we plot the point (1,4), (2, 3), (3, 35), (4, 63), (5, 87), (6, 99), (7,107), (8, 113), (9, 118), (10, 120).
2. The marks scored by 750 students in an examination are given in the form of a frequency distribution table:
Marks 
No. of students 
Marks 
No. of students 
600 – 640 
16 
760 – 800 
172 
640 – 680 
45 
800 – 840 
59 
680 – 720 
156 
840 – 800 
18 
720 – 760 
284 
Prepare a cumulative frequency table by less than method and draw an ogive.
Solution
We first prepare the cumulative frequency table by less than method as given below :
Now, we mark the upper class limits along xaxis and cumulative frequency along yaxis on a suitable gear.
Thus, we plot the points (640, 16) (680, 61), (720, 217), (760, 501), (600, 673), (840, 732) and (880, 750).
3. Draw an ogive to represent the following frequency distribution:
Class – interval : 
0 – 4 
5 – 9 
10 – 14 
15 – 19 
20 – 24 
No. of students: 
2 
6 
12 
5 
3 
Solution
The given frequency of distribution is not continuous so we first make it continuous and prepare the cumulative frequency distribution as under
Now, we mark the upper class limits along xaxis and cumulative frequency along yaxis.
Thus we plot the points (4, 5, 2), (9, 5, 8), (14, 5, 08), (19, 5, 23) and (24, 5, 26)
Cumulative frequency.
4. The monthly profits (in Rs.) of 100 shops are distributed as follows:
Profits per shop: 
0 – 50 
50 – 100 
100 – 150 
150 – 200 
200 – 250 
250 – 300 
No. of shops: 
12 
18 
27 
20 
17 
6 
Draw the frequency polygon for it.
Solution
We have,
5. The following table gives the height of trees:
Height 
No. of trees 
Less than 7 
26 
Less than 14 
57 
Less than 21 
92 
Less than 28 
134 
Less than 35 
216 
Less than 42 
287 
Less than 49 
341 
Less than 56 
360 
Draw ‘less than’ ogive and ‘more than’ ogive.
Solution
Less than method,
It is given that,
Now, we mark the upper class limits along x  axis and cumulative frequency along y  axis.
Thus we plot the points (7, 26) (14, 57) (21, 92) (28, 134) (35, 216) (42, 287) (49, 341) (56, 360)
More than method: we prepare the cf table by more than method as given below:
Now, we mark on x  axis lower class limits, y  axis cumulative frequency
Thus, we plot graph at(0,360)(7, 334)(14,303)(21, 268)(28,226)(35,144)(42, 73)(49, 19)
Profit (in lakhs in Rs) 
Number of shops(frequency) 
More than or equal to 5 
30 
More than or equal to 10 
28 
More than or equal to 15 
16 
More than or equal to 20 
14 
More than or equal to 25 
10 
More than or equal to 30 
7 
More than or equal to 35 
3 
Draw both ogives for the above data and hence obtain the median.
Solution
More than method
Now, we mark on x  axis lower class limits, y  axis cumulative frequency
Thus, we plot the points. (5, 30)(10, 28)(15, 16)(20, 14)(25, 10)(30, 7) and (35, 3)
Less than method.
Now, we mark the upper class limits along x  axis and cumulative frequency along y  axis.
Thus we plot the points, (10, 2) (15, 14)(20, 16)(25, 20)(30, 23)(35, 27)(40, 30)
We find that the two types of curves intersect of point P from point L it is drawn on x  axis.
The value of a profit corresponding to M is 17.5 lakh, Hence median is 17.5 lakh
Daily income (in Rs): 
100 – 120 
120 – 140 
140 – 160 
160 – 180 
180 – 200 
Number of workers: 
12 
14 
8 
6 
10 
Convert the above distribution to a less than type cumula five frequency distribution and draw its ogive.
Solution
We first prepare the cumulative frequency table by less than method as given below.
8. The following table gives production yield per hectare of wheat of 100 farms of a village:
Production yield in kg per hectare: 
50 – 55 
55 – 60 
60 – 65 
65 – 70 
70 – 75 
75 – 80 
Number of farms: 
2 
8 
12 
24 
38 
16 
Draw ‘less than’ ogive and ‘more than’ ogive.
Solution
Less than method:
Cumulative frequency table by less than method.
Now, we mark on x  axis upper class limit, y  axis cumulative frequencies.
We plot the points(50, 100)(55, 98)(60, 90)(65, 78)(70, 54)(75, 16)
9. During the medical check – up of 35 students of a class, their weights were recorded as follows:
Weight (in kg) 
No. of students 
Less than 38 
0 
Less than 40 
3 
Less than 42 
5 
Less than 44 
9 
Less than 46 
14 
Less than 48 
28 
Less than 50 
32 
Less than 52 
35 
Draw a less than type ogive for the given data. Hence, obtain the median weight from the graph and verify the result by using the formula.
Solution
Less than method
It is given that
On x  axis upper class limits. Y  axis cf.
We plot the points (38, 0)(40, 3)(42, 5)(44, 9)(46, 4)(48, 28)(50, 32)(52, 35)
More than method: Cf table
x  axis lower class limits on y  axis  cf
We plot the points(38, 35)(40, 32)(42, 30)(44, 26)(46, 21)(48, 7)(50, 3)
The verification,
We have