# OP Malhotra Class 9 ICSE Maths Solutions Chapter 15

## OP Malhotra Class 9 ICSE Maths Solutions Chapter 15 Mean Median and Frequency Polygon

Welcome to NCTB Solutions. Here with this post we are going to help 9th class students for the Solutions of OP Malhotra Class 9 ICSE Math Book, Chapter 15, Mean, Median and Frequency Polygon. Here students can easily find step by step solutions of all the problems for Mean, Median and Frequency Polygon, Exercise 15a, 15b and 15c Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 15 solutions. Here in this post all the solutions are based on ICSE latest Syllabus.

Mean, Median and Frequency Polygon Exercise 15(a) Solution :

Question no – (1)

Solution :

(i) Frist 5 natural numbers

Mean of 1st 5 natural numbers

= 1 + 2 +3 + 4 + 5/5

= 15/5

= 3

(ii) Frist 7 whole numbers

Mean of 1st 7 whole numbers are

= 0 + 1 + 2 + 3 + 4 + 5 + 6/7

= 21/7

= 3

(iv) Frist 4 prime numbers

Mean of 1st 4 prime No is

= 2 + 3 + 5 + 7/4

= 17/4

= 4.24

(iv) 1.3 cm, 5.7 cm, 89,8 cm, 6.4 cm, 6.9 cm

Mean = 1.3 + 5.7 + 9.8 + 6.4 + 6.9/5

= 30.1/5

= 6.02 cm

(v) Rs 7, Rs 19, Rs 31, Rs 43, Rs 70

Mean = (7 + 19 + 31 + 43 + 70/5) Rs

= 170/5

= 70

Question no – (2)

Solution :

As per the question, Seema’s scores are, 80, 85, 90, 71, 60, 100.

Seema Mean Score,

= 80 + 85 + 90 + 71 + 60 + 100/6

= 486/6

= 81

Therefore, her mean score will be 81.

Question no – (3)

Solution :

According to the question,

Shaleen’s last six batting scores were 138, 144, 155, 142, 167, 172

Mean of batting score,

= 13.8 + 144 + 155 + 142 + 167 + 172/6

= 918/6

= 153

Therefore, his mean score will be 153.

Question no – (4)

Solution :

As per the question we know,

Madhu worked 2 hours on Monday,

2 1/2 hours on Monday,

3 1/4 hrs on Tuesday,

2 3/4 hrs on Wednesday

Mean of working hour,

= 2 1/2 + 3 1/4 + 2 3/4/3

= 7 + (1/23 + 1/4 + 3/4)/3

= 7 + (2 + 1 + 3/4)/3

= 7 + 6/4

= 7 6/4/3

= 7 2/3/3

= 17/2 / 3

= 17/6

= 2 5/6

Therefore, she worked 2 5/6 mean number of hours.

Question no – (5)

Solution :

Mean of Ayushree score,

= 68 + 75 + 70 + 45 + 57 + 77/6

= 392/6

= 65.33%

Mean of Aanaya score,

= 52 + 87 + 64 + 53 + 74 + 81 + 86/7

= 497/7

= 71%

Therefore, Ananya has highest mean score.

Question no – (6)

Solution :

As we know that,

First 10 odd natural numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19

Mean of 1st 10 odd natural numbers,

= 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19/10

= 100/10

= 10
Thus, the Mean of 1st 10 odd natural numbers is 10

Question no – (7)

Solution :

Sum of 5 marks given,

= 16 + 14 + x + 23 + 20

= 73 + x

Total of 5 no is = 5 × 18 = 90

73 + x = 90

= x = 90 – 73

= 17

Therefore, the value will be 17

Question no – (8)

Solution :

Here, total t terms,

= 24 × 5

= 120

Sum of Mean,

= x + x + 2 + x + 4 + x + 6 + x + 8

= 5x + 20

5x + 20 = 120

= 5x = 120 – 20

= 100

= x = 100/5

= 20

Therefore, x will be 20

Question no – (9)

Solution :

From the given question we know,

45 minutes,

30 minutes,

60 minutes,

50 minutes

20 minutes

Mean practice time,

= 45 + 30 + 60 + 50 + 20/5

= 205/5

= 41 min

Hence, her mean practice time will be 41 min.

Question no – (10)

Solution :

Let, marks = x

= 73 + 86 + 78 + 75 + x/5

= 312 + x/5

Here Mean score is 80

Now, 312 + x/5 = 80

= 31 + x = 400

= x = 400 – 312

= 88 marks

Therefore, the least number of marks Nisha can secure will be 88 marks.

Question no – (11)

Solution :

Here, Total runs,

= (60 × 10) = 600 runs of runs of 11 innings

= 11 × 62

= 682 runs

Number of runs scored in 11th innings,

= 682 – 600

= 82 runs

Therefore, 88 runs are to be scored in the eleventh innings.

Question no – (12)

Solution :

 Weight (x) Number of students (f) (f x0) 30 8 240 31 10 310 32 15 480 33 8 264 34 9 306 Total 50 1600

∴ Mean,

= 1600/50

= 32 kg

Therefore, the mean of the given frequency distributions will be 32 kg.

Question no – (13)

Solution :

 X F F×x 2 5 7 8 Total 15 90

∴ Mean,

= 90/15

= 6

Therefore, the mean of the given frequency distribution will be 6.

Question no – (14)

Solution :

According to the question,

 X F Fx 0.1 20 2 0.2 60 12 0.3 20 6 0.4 40 16 0.5 10 5 0.6 50 30 Total 200 71

∴ Mean,

= 200/71

= 0.355

Therefore, the mean of the given frequency distribution will be 0.355.

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Updated: June 20, 2023 — 8:05 am