RD Sharma Class 10 Solutions Chapter 7 Exercise 7.4 (Updated for 2024)

RD Sharma Class 10 Solutions Chapter 7 Exercise 7.4: The median is the point in a distribution that is in the middle. Students will experience finding the median of a discrete and grouped frequency distribution in this exercise. The RD Sharma Class 10 Solutions is the best location to look for quick access to solutions. Download the RD Sharma Solutions for Class 10 Maths Chapter 7 Exercise 7.4 PDF for further details on this exercise.

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RD Sharma Class 10 Solutions Chapter 7 Exercise 7.4

Access answers to RD Sharma Solutions Class 10 Maths Chapter 7 Exercise 7.4- Important Question with Answers

1. Following are the lives in hours of 15 pieces of the components of aircraft engine. Find the median: 

715, 724, 725, 710, 729, 745, 694, 699, 696, 712, 734, 728, 716, 705, 719.

Solution:

Arranging the given data in ascending order, we have

694, 696, 699, 705, 710, 712, 715, 716, 719, 721, 725, 728, 729, 734, 745

As the number of terms is an old number, i.e., N = 15

We use the following procedure to find the median.

Median = (N + 1)/2 ^th term

= (15 + 1)/2 ^th term

= 8^th term

So, the 8^th term in the arranged order of the given data should be the median.

Therefore, 716 is the median of the data.

2. The following is the distribution of height of students of a certain class in a certain city:

Height (in cm):	160 – 162	163 – 165	166 – 168	169 – 171	172 – 174
No of students:	15	118	142	127	18

Find the median height.

Solution:

Class interval (exclusive)	Class interval  (inclusive)	Class interval frequency	Cumulative frequency
160 – 162	159.5 – 162.5	15	15
163 – 165	162.5 – 165.5	118	133(F)
166 – 168	165.5 – 168.5	142(f)	275
169 – 171	168.5 – 171.5	127	402
172 – 174	171.5 – 174.5	18	420
		N = 420

Here, we have N = 420,

So, N/2 = 420/ 2 = 210

The cumulative frequency just greater than N/2 is 275 then 165.5 – 168.5 is the median class such that

L = 165.5, f = 142, F = 133 and h = (168.5 – 165.5) = 3

= 165.5 + 1.63

= 167.13

3. Following is the distribution of I.Q. of 100 students. Find the median I.Q.

I.Q:	55 – 64	65 – 74	75 – 84	85 – 94	95 – 104	105 – 114	115 – 124	125 – 134	135 – 144
No of students:	1	2	9	22	33	22	8	2	1

Solution:

Class interval (exclusive)	Class interval  (inclusive)	Class interval frequency	Cumulative frequency
55 – 64	54.5 – 64-5	1	1
65 – 74	64.5 – 74.5	2	3
75 – 84	74.5 – 84.5	9	12
85 – 94	84.5 – 94.5	22	34(F)
95 – 104	94.5 – 104.5	33(f)	67
105 – 114	104.5 – 114.5	22	89
115 – 124	114.5 – 124.5	8	97
125 – 134	124.5 – 134.5	2	98
135 – 144	134.5 – 144.5	1	100
		N = 100

Here, we have N = 100,

So, N/2 = 100/ 2 = 50

The cumulative frequency just greater than N/ 2 is 67 then the median class is (94.5 – 104.5) such that L = 94.5, F = 33, h = (104.5 – 94.5) = 10

= 94.5 + 4.85

= 99.35

4. Calculate the median from the following data:

Rent (in Rs):	15 – 25	25 – 35	35 – 45	45 – 55	55 – 65	65 – 75	75 – 85	85 – 95
No of houses:	8	10	15	25	40	20	15	7

Solution:

Class interval	Frequency	Cumulative frequency
15 – 25	8	8
25 – 35	10	18
35 – 45	15	33
45 – 55	25	58(F)
55 – 65	40(f)	98
65 – 75	20	118
75 – 85	15	133
85 – 95	7	140
	N = 140

Here, we have N = 140,

So, N/2 = 140/ 2 = 70

The cumulative frequency just greater than N/ 2 is 98 then the median class is 55 – 65 such that L = 55, f = 40, F = 58, h = 65 – 55 = 10

= 55 + 3 = 58

5. Calculate the median from the following data:

Marks below:	10 – 20	20 – 30	30 – 40	40 – 50	50 – 60	60 – 70	70 – 80	85 – 95
No of students:	15	35	60	84	96	127	198	250

Solution:

Marks below	No. of students	Class interval	Frequency	Cumulative frequency
10	15	0 – 10	15	15
20	35	10 – 20	20	35
30	60	20 – 30	25	60
40	84	30 – 40	24	84
50	96	40 – 50	12	96(F)
60	127	50 – 60	31(f)	127
70	198	60 – 70	71	198
80	250	70 – 80	52	250
			N = 250

Here, we have N = 250,

So, N/2 = 250/ 2 = 125

The cumulative frequency just greater than N/ 2 is 127, then the median class is 50 – 60 such that L = 50, f = 31, F = 96, h = 60 -50 = 10

= 50 + 9.35

= 59.35

6. Calculate the missing frequency from the following distribution, it being given that the median of the distribution is 24.

Age in years:	0 – 10	10 – 20	20 – 30	30 – 40	40 – 50
No of persons:	5	25	?	18	7

Solution:

Let the unknown frequency be taken as x,

Class interval	Frequency	Cumulative frequency
0 – 10	5	5
10 – 20	25	30(F)
20 – 30	x (f)	30 + x
30 – 40	18	48 + x
40 – 50	7	55 + x
	N = 170

It’s given that

Median = 24

Then, median class = 20 – 30; L = 20, h = 30 -20 = 10, f = x, F = 30

4x = 275 + 5x – 300

4x – 5x = – 25

– x = – 25

x = 25

Therefore, the Missing frequency = 25

7. The following table gives the frequency distribution of married women by age at marriage.

Age (in years)	Frequency	Age (in years)	Frequency
15 – 19	53	40 – 44	9
20 – 24	140	45 – 49	5
25 – 29	98	45 – 49	3
30 – 34	32	55 – 59	3
35 – 39	12	60 and above	2

Calculate the median and interpret the results.

Solution:

Class interval (exclusive)	Class interval (inclusive)	Frequency	Cumulative frequency
15 – 19	14.5 – 19.5	53	53 (F)
20 – 24	19.5 – 24.5	140 (f)	193
25 – 29	24.5 – 29.5	98	291
30 – 34	29.5 – 34.5	32	323
35 – 39	34.5 – 39.5	12	335
40 – 44	39.5 – 44.5	9	344
45 – 49	44.5 – 49.5	5	349
50 – 54	49.5 – 54.5	3	352
55 – 54	54.5 – 59.5	3	355
60 and above	59.5 and above	2	357
		N =357

Here, we have N = 357,

So, N/2 = 357/ 2 = 178.5

The cumulative frequency just greater than N/2 is 193, so then the median class is (19.5 – 24.5) such that l = 19.5, f = 140, F = 53, h = 25.5 – 19.5 = 5

Median = 23.98

This means nearly half the women were married between the ages of 15 and 25

8. The following table gives the distribution of the life time of 400 neon lamps:

Life time: (in hours)	Number of lamps
1500 – 2000	14
2000 – 2500	56
2500 – 3000	60
3000 – 3500	86
3500 – 4000	74
4000 – 4500	62
4500 – 5000	48

Find the median life.

Solution:

Life time	Number of lamps fi	Cumulative frequency (cf)
1500 – 2000	14	14
2000 – 2500	56	70
2500 – 3000	60	130(F)
3000 – 3500	86(f)	216
3500 – 4000	74	290
4000 – 4500	62	352
4500 – 5000	48	400
	N = 400

It’s seen that the cumulative frequency just greater than n/2 (400/2 = 200) is 216, and it belongs to the class interval 3000 – 3500, which becomes the Median class = 3000 – 3500

Lower limits (l) of median class = 3000 and,

Frequency (f) of median class = 86

Cumulative frequency (cf) of class preceding median class = 130

And the Class size (h) = 500

Thus, by calculating the median by the formula, we get

= 3000 + (35000/86)

= 3406.98

Thus, the median lifetime of lamps is 3406.98 hours

9. The distribution below gives the weight of 30 students in a class. Find the median weight of students:

Weight (in kg):	40 – 45	45 – 50	50 – 55	55 – 60	60 – 65	65 – 70	70 – 75
No of students:	2	3	8	6	6	3	2

Solution:

Weight (in kg)	Number of students fi	Cumulative frequency (cf)
40 – 45	2	2
45 – 50	3	5
50 – 55	8	13
55 – 60	6	19
60 – 65	6	25
65 – 70	3	28
70 – 75	2	30

It’s seen that the cumulative frequency just greater than n/ 2 (i.e. 30/ 2 = 15) is 19, belongs to class interval 55 – 60.

So, it’s chosen that

Median class = 55 – 60

Lower limit (l) of median class = 55

Frequency (f) of median class = 6

Cumulative frequency (cf) = 13

And, Class size (h) = 5

Thus, by calculating the median by the formula, we get

= 55 + 10/6 = 56.666

So, the median weight is 56.67 kg.

10. Find the missing frequencies and the median for the following distribution if the mean is 1.46

No. of accidents:	0	1	2	3	4	5	Total
Frequencies (no. of days):	46	?	?	25	10	5	200

Solution:

No. of accidents (x)	No. of days (f)	fx
0	46	0
1	x	x
2	y	2y
3	25	75
4	10	40
5	5	25
	N = 200	Sum = x + 2y + 140

It’s given that N = 200

⇒ 46 + x + y + 25 + 10 + 5 = 200

⇒ x + y = 200 – 46 – 25 – 10 – 5

⇒ x + y = 114 —- (i)

And also given, Mean = 1.46

⇒ Sum/ N = 1.46

⇒ (x + 2y + 140)/ 200 = 1.46

⇒ x + 2y = 292 – 140

⇒ x + 2y = 152 —- (ii)

Subtract equation (i) from equation (ii), and we get

x + 2y – x – y = 152 – 114

⇒ y = 38

Now, on putting the value of y in equation (i), we find x = 114 – 38 = 76

Thus, the table becomes:

No. of accidents (x)	No. of days (f)	Cumulative frequency
0	46	46
1	76	122
2	38	160
3	25	185
4	10	195
5	5	200
	N = 200

It’s seen that,

N = 200 N/2 = 200/2 = 100

So, the cumulative frequency just more than N/2 is 122

Therefore, the median is 1.

We have provided complete details of RD Sharma Class 10 Solutions Chapter 7 Exercise 7.4. If you have any queries related to CBSE Class 10, feel free to ask us in the comment section below.

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