Mean Median and Frequency Polygon Class 9 OP Malhotra Exe-15A ICSE Maths Solutions Ch-15. Step by Step Solutions / Answer of Questions of OP Malhotra Maths. Visit official Website CISCE for detail information about ICSE Board Class-9 Mathematics
Mean Median and Frequency Polygon Class 9 OP Malhotra Exe-15A ICSE Maths Solutions Ch-15
| Board | ICSE |
| Publications | S Chand |
| Subject | Maths |
| Class | 9th |
| Chapter-15 | Mean Median and Frequency Polygon |
| Writer | OP Malhotra |
| Exe-15B | How to Calculate Mean |
| Edition | 2025-2026 |
How to Calculate Mean
Mean Median and Frequency Polygon Class 9 OP Malhotra Exe-15A ICSE Maths Solutions Ch-15.
Que-1: Find the mean of each of the following set of numbers :
(a) First 5 natural numbers
(b) First 7 whole numbers
(c) First 4 prime numbers
(d) 1.3 cm, 5.7 cm, 9.8 cm, 6.4 cm, 6.9 cm
(e) Rs. 7, Rs. 19, Rs. 31, Rs. 43, Rs. 70
Sol: (a) First 5 natural numbers are 1, 2, 3, 4, 5
∴ Mean = (1+2+3+4+5)/5 = 15/5 {x¯=∑xi/n}
= 3
(b) First 7 whole numbers are 0, 1,2, 3, 4, 5, 6
∴ Mean = (0+1+2+3+4+5+6)/7 {x¯ = ∑xi/n}
= 21/7 = 3
(c) First 4 prime numbers are 2, 3, 5, 7
∴ Mean = (2+3+5+7)/4 = 17/4 {x¯ = ∑xi/n}
= 4.25
(d) 1.3 cm, 5.7 cm, 9.8 cm, 6.4 cm, 6.9 cm
∴ Mean = (1.3+5.7+9.8+6.4+6.9)/5 {x¯ = ∑xi/n}
= 30.1/5
= 6.02 cm
(e) Rs. 7, Rs. 19, Rs. 31, Rs. 43, Rs. 70
∴ Mean = (7+19+31+43+70)/5 {x¯ = ∑xi/n}
= Rs. 170/5
= Rs. 34
Que-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 ?
Sol: Score obtained are 80, 85, 90, 71, 60, 100
∴ Mean = (80+85+90+71+60+100)/6 {x¯ = ∑xi/n}
= 486/6 = 81
Que-3: Shaleen’s last six batting scores were 138, 144, 155, 142, 167, 172.
What was his mean score ?
Sol: Last 6 batting scores 138, 144, 155, 142, 167, 172
∴ Mean score = (138+144+155+142+167+172)/6 {x¯ = ∑xi/n}
= 918/6 = 153
Que-4: Madhu worked 2*(1/2) hours on Monday, 3*(1/4) hrs. on Tuesday, and 2*(3/4) hrs. on Wednesday. What is the mean number of hours she worked on these three days ?
Sol: Working hours from Monday to Wednesday for 3 days = 2*(1/2), 3*(1/4) and 2*(3/4) (hours)
∴ Mean hours = [{2*(1/2)}+{3*(1/4)}+{2*(3/4)}/3] = {8*(1/2)}/3
= 17/(2×3) = 17/6 hours = 2*(5/6) hours
Que-5: Ayushree sat for six tests and Ananya sat for seven tests. Their percentage scores were:
Who has the higher mean score?
Sol: Ayushree’s score (in percent) = 68, 75, 70, 45, 57, 77
∴ Her mean = (68+75+70+45+57+77)/6 {x¯=∑xi/n}
= 392/6 = 65*(1/3)%
and Ananya’s score (in percent) = 52, 87, 64, 53,74,81,86
∴ Her mean = (52+87+64+53+74+81+86)/7
= 497/7 = 71%
It is clear that Ananya’s score is better
Que-6: Find the mean of first ten odd natural numbers.
Sol: First 10 odd natural numbers are 1,3,5, 7, 9, 11, 13, 15, 17, 19
∴ Mean
= (1+3+5+7+9+11+13+15+17+19)/10 {x¯=∑xi/n}
= 100/10 = 10
Que-7: If the mean of 16, 14, x, 23, 20 is 18 find the value of x.
Sol: Mean of 5 terms = 18
∴ Then total = 18 x 5 = 90
But sum of given numbers is
= 16 + 14 + x + 23 + 20 = 73 + x
∴ 73 + x = 90 ⇒ x = 90 – 73 = 17
Hence x = 17
Que-8: If the mean of x, x + 2, x + 4, x + 6, x + 8 is 24, find x.
Sol: Mean of 5 terms = 24
∴ Their total = 24 x 5 = 120
Now sum of mean = x + x + 2 + x + 4 + x + 6 + x + 8
= 5x + 20
∴ 5x + 20 = 120 ⇒ 5.x = 120 – 20
⇒ 5x = 100 ⇒ x = 100/5 = 20
∴ x = 20
Que-9: Madhu practiced on her sitar 45 minutes, 30 minutes, 60 minutes, 50 minutes and 20 minutes. What was her mean practice time?
Sol: Madhu practiced on her sitar for 45 minutes, 30 minutes, 60 minutes, 50 minutes and 20 minutes
∴ Her mean practice
= (45+30+60+50+20)/5 {x¯=∑xi/n}
= 205/5 = 41 minutes
Que-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?
Sol: Marks secured by Nisha m 4 tests are = 73, 86, 78, 75
Let she secured marks in the next (fifth) test Then mean marks secured will be
= (73+86+78+75+x)/5 = (312+x)/5
But her mean score = 80 marks
∴ (312+x)/5 = 80 ⇒ 312 + x = 400
⇒ x = 400 – 312 = 88
Hence her score in the fifth test = 88 marks
Que-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 to raise the mean score to 62.
Sol: Mean score of a cricketer = 60 run
No. of innings (H) = 10
∴ Total runs = 60 x 10 = 600 runs
Mean score in 11 innings = 62
∴ Total runs = 62 x 11 = 682 run
∴ No. of runs scored in 11 th innings
= 682 – 600 = 82 runs
Que-12: Find the mean of the following frequency distributions:
Sol: Mean (x) = ∑fi.xi / ∑fi
= 1600/50 = 32 kg
Que-13:
Sol: Mean = ∑fixi / ∑fi
= 90/15 = 6
Que-14:
Sol: Mean = ∑fi.xi / ∑fi
= 71/200 = 0.355
Que-15: The makes scored in a test by a class of 25 boys are as follows:
Draw a frequency table and calculate the mean.
Sol: No. of boys (n) = 25
Highest score = 25
Lowest score = 16
∴ Range = 25 – 169
Now taking class intervals
Mean = ∑fi.xi / ∑fi
= 533/25 = 21.32
∴ Mean marks = 21.32 marks
–: Mean Median and Frequency Polygon Class 9 OP Malhotra Exe-15A ICSE Maths Ch-15. Solutions :—
Return to :– OP Malhotra S Chand Solutions for ICSE Class-9 Maths
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