S Chand ICSE Mathematics Class 9 Solution Fifteenth Chapter Mean Median and Frequency Polygon Exercise 15A (O.P. Malhotra, S.K. Gupta, Anubhuti Gangal)

S Chand ICSE Mathematics Class 9 Solution Fifteenth Chapter Mean, Median and Frequency Polygon  Exercise 15A

(1) Find the Mean of each of the following set of numbers:

(i) First 5 natural numbers

Solution: Fist 5 natural numbers are 1,2,3,4,5

∴ Mean = 1 + 2 + 3 + 4 + 5 / 5

= 15 / 5

= 3

(ii) First 7 whole numbers

Solution: first 7 whole numbers are 0,1,2,3,4,5,6

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

= 21 / 7

= 3

(iii) Fist 4 prime numbers

Solution: first 4 prime numbers are = 2, 3, 5, 7

∴ Mean = 2 + 3 + 5 + 7 / 4

= 17 / 4

= 4.25

(iv) 1.3 cm, 5.7 cm, 9.8 cm, 6.4 cm, 6.9 cm

Solution: 1.3 + 5.7 + 9.8 + 6.4 + 6.9

= 30.1 / 5

= 6.02 cm

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

Solution: Mean = 7 + 19 + 31 + 43 + 70 / 5

= 170 / 5

= 34

(2) Seema obtained the following scores (out of 100) on a set of spelling tests:

80, 85, 90, 71,60, 100.

What is her mean score?

Solution: Mean score of Seema = 80 + 85 + 90 + 71 + 60 + 100 / 6

= 486 / 6

= 81

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

What was his mean score?

Solution: Shaleen’s mean score = 138 + 144 + 155 + 142 + 167 + 172 / 6

= 918 / 6

= 153

(4) Madhu worked 2 and ½ hours on Monday, 3 and ¼ hrs. on Tuesday and 2 and ¾ hrs. on Wednesday. What is the mean number of hours she worked on these three days?

Solution:

(5) Ayushree sat for six tests and Ananya sat for seven tests. Their percentage scores were:

Ayushree 68 75 70 45 57 77

 

Ananya 52 87 64 53 74 81 86

Who has the higher mean score?

Solution: Ayushree’s mean score = 68 + 75 + 70 + 45 + 57 + 77 / 6

= 392 / 6

= 65.33

Ananya’s mean score = 52 + 87 + 64 + 53 + 74 + 81 + 86 / 7

= 497 / 7

= 71

So, Ananya has higher score.

(7) If the mean of 16, 14, x, 23, 20 is 18, find the value of x

(8) If the mean of x, x + 2, x + 4, x + 6, x + 8 is 24 find x

Solution:

(9) Madhu Practiced on her Sitar 45 minutes, 30 minutes, 60 minutes, 50 minutes, and 20 minutes. What was her mean practice time?

Solution: Mean = 45 + 30 + 60 + 50 + 20 / 5

= 205 / 5

= 41 minutes.

(10) Nisha secured 73, 86, 78 and 75 marks in four tests. What is the least number of marks she can secure in her next test, if she has to have a mean score of 80 marks in five tests?

Solution: according to the question, 73 + 86 + 78 + 75 + x = 80 x 5

Or, 312 + x = 400

Or, x = 400 – 312

Or, x = 88

(11) A cricketer has a mean score of 60 runs in ten innings. Find out how many runs are to be scored in the eleventh innings tyo raise the mean score to 62.

Solution: Total run in 10 innings = 60 x 10 = 600

Total run in 11 innings = 62 x 11 = 682

∴ Run in 11th innings = 682 – 600 = 82 runs.

(12) Find the mean of the following frequency distributions:

Weight (kg) 30 31 32 33 34
Number of students 8 10 15 8 9

Solution: Mean = 30 x 8 + 31 x 10 + 32 x 15 + 33 x 8 + 34 x 9 / 50

= 240 + 310 + 480 + 264 + 306 / 50

= 1600 / 50

= 32 kg

(13)

X 2 5 7 8
f 2 4 6 3

Solution: 2 x 2 + 5 x 4 + 7 x 6 + 8 x 3 / 15

= 4 + 20 + 42 + 24 / 15

= 90 / 15

= 6

(14)

x 0.1 0.2 0.3  0.4 0.5 0.6
 f 20 60 20 40 10 50

Solution: Mean = 0.1 x 20 + 0.2 x 60 + 0.3 x 20 + 0.4 x 40 + 0.5 + 10 + 0.6 x 50

= 2 + 12 + 6 + 16 + 5 + 30 / 200

= 71 / 200

= 0.355

(15) The marks scored in a test by a class of 25 boys are as follows:

24 25 23 20 20 19  
22 20 24 22 18 23  
23 18 20 16 25 24  
17 18 23 22 23 20 24

Draw a frequency table and calculate the mean.

Solution:

____ x _____

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  1. Please give solutions for all chapters

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