## RS Aggarwal Class 9 Solutions Chapter 15 Probability Ex 15A

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 15 Probability Ex 15A.

Question 1.
Solution:
Number of trials = 500 times
Let E be the no. of events in each case, then
∴No. of heads (E1) = 285 times
and no. of tails (E2) = 215 times
∴ Probability in each case will be
∴(i)P(E1) = $$\frac { 285 }{ 500 }$$ = $$\frac { 57 }{ 100 }$$ = 0.57
(ii) P(E2) = $$\frac { 215 }{ 500 }$$ = $$\frac { 43 }{ 100 }$$ = 0.43

Question 2.
Solution:
No. of trials = 400
Let E be the no. of events in each case, then
No. of 2 heads (E1) = 112
No. of one head (E2) = 160 times
and no. of O. head (E3) = 128 times
∴ Probability in each case will be:
∴ (i)P(E1) = $$\frac { 112 }{ 400 }$$ = $$\frac { 28 }{ 100 }$$ = 0.28
(ii)P(E2) = $$\frac { 160 }{ 400 }$$ = $$\frac { 40 }{ 100 }$$= 0.40
(iii) P(E3) = $$\frac { 128 }{ 400 }$$ = $$\frac { 32 }{ 100 }$$ = 0.32 Ans.

Question 3.
Solution:
Number of total trials = 200
Let E be the no. of events in each case, then
No. of three heads (E1) = 39 times
No. of two heads (E2) = 58 times
No. of one head (E3) = 67 times
and no. of no head (E4) = 36 times
∴ Probability in each case will be .
(i) P(E1) = $$\frac { 39 }{ 200 }$$ = 0.195
(ii) P(E3) = $$\frac { 67 }{ 200 }$$ = 0.335
(iii) P(E4) = $$\frac { 36 }{ 200 }$$ = $$\frac { 18 }{ 100 }$$ = 0.18
(iv) P(E2) = $$\frac { 58 }{ 200 }$$ = $$\frac { 29 }{ 100 }$$ = 0.29

Question 4.
Solution:
Solution No. of trials = 300 times
Let E be the no. of events in each case, then
No. of outcome of 1(E1) = 60
No. of outcome of 2(E2) = 72
No. of outcome of 3(E3) = 54
No. of outcome of 4(E4) 42
No. of outcome of 5(E5) = 39
No. of outcome of 6(E6) = 33
The probability of
(i) P(E3) = $$\frac { 54 }{ 300 }$$ = $$\frac { 18 }{ 100 }$$ = 0.18
(ii) P(E6) = $$\frac { 33 }{ 100 }$$ = $$\frac { 11 }{ 100 }$$= 0.11
(iii) P(E5) = $$\frac { 39 }{ 300 }$$ = $$\frac { 13 }{ 100 }$$ = 0.13
(iv) P(E1) = $$\frac { 60 }{ 300 }$$ = $$\frac { 20 }{ 100 }$$= 0.20 Ans.

Question 5.
Solution:
Let E be the no. of events in each case.
No. of ladies who like coffee (E1) = 142
No. of ladies who like coffee (E2) = 58
Probability of
(1) P(E1) = $$\frac { 142 }{ 200 }$$ = $$\frac { 71 }{ 100 }$$ = 0.71
(ii) P(E2) = $$\frac { 58 }{ 200 }$$ = $$\frac { 29 }{ 100 }$$ = 0.29 Ans.

Question 6.
Solution:
Total number of tests = 6
No. of test in which the students get more than 60% mark = 2
Probability will he
P(E) = $$\frac { 2 }{ 6 }$$ = $$\frac { 1 }{ 3 }$$Ans.

Question 7.
Solution:
No. of vehicles of various types = 240
No. of vehicles of two wheelers = 64.
Probability will be P(E) = $$\frac { 84 }{ 240 }$$ = $$\frac { 7 }{ 20 }$$ = 0.35 Ans.

Question 8.
Solution:
No. of phone numbers are one page = 200
Let E be the number of events in each case,
Then (i) P(E5) = $$\frac { 24 }{ 200 }$$ = $$\frac { 12 }{ 100 }$$ = 0.12
(ii) P(E8) = $$\frac { 16 }{ 200 }$$ = $$\frac { 8 }{ 100 }$$ = 0.08 Ans.

Question 9.
Solution:
No. of students whose blood group is checked = 40
Let E be the no. of events in each case,
Then (i) P(E0) = $$\frac { 14 }{ 40 }$$ = $$\frac { 7 }{ 20 }$$ = 0.35
(ii) P(EAB) = $$\frac { 6 }{ 40 }$$ = $$\frac { 3 }{ 20 }$$ = 0.15 Ans.

Question 10.
Solution:
No. of total students = 30.
Let E be the number of elements, this probability will be of interval 21 – 30
P(E) = $$\frac { 6 }{ 30 }$$ = $$\frac { 1 }{ 5 }$$ = 0.2 Ans.

Question 11.
Solution:
Total number of patients of various age group getting medical treatment = 360
Let E be the number of events, then
(i) No. of patient which are 30 years or more but less than 40 years = 60.
P(E) = $$\frac { 60 }{ 360 }$$ = $$\frac { 1 }{ 6 }$$
(ii) 50 years or more but less than 70 years = 50 + 30 = 80
P(E) = $$\frac { 80 }{ 360 }$$ = $$\frac { 2 }{ 9 }$$
(iii) Less than 10 years = zero
P(E) = $$\frac { 0 }{ 360 }$$ = 0
(iv) 10 years or more 90 + 50 + 60 + 80 + 50 + 30 = 360

Hope given RS Aggarwal Class 9 Solutions Chapter 15 Probability Ex 15A are helpful to complete your math homework.

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## RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14F

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14F.

Other Exercises

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:
Mean = 8

Question 6.
Solution:
Mean = 28.25

Question 7.
Solution:
Mean = 16.6

Question 8.
Solution:
Mean = 50

Question 9.
Solution:

Question 10.
Solution:
Let assumed mean = 67

Question 11.
Solution:
Here h = 1, Let assumed mean (A) = 21

Question 12.
Solution:
Here h = 400 and let assumed mean (A) = 1000

Hope given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14F are helpful to complete your math homework.

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## RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14H

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14H.

Other Exercises

Question 1.
Solution:
Arranging the given data in ascending order :
0, 0, 1, 2, 3, 4, 5, 5, 6, 6, 6, 6
We see that 6 occurs in maximum times.
Mode = 6 Ans.

Question 2.
Solution:
Arranging in ascending order, we get:
15, 20, 22, 23, 25, 25, 25, 27, 40
We see that 25 occurs in maximum times.
Mode = 25 Ans.

Question 3.
Solution:
Arranging in ascending order we get:
1, 1, 2, 3, 3, 4, 5, 5, 6, 6, 7, 8, 9, 9, 9, 9, 9
Here, we see that 9 occurs in maximum times.
Mode = 9 Ans.

Question 4.
Solution:
Arranging in ascending order, we get:
9, 19, 27, 28, 30, 32, 35, 50, 50, 50, 50, 60
Here, we see that 50 occurs in maximum times.
Modal score = 50 scores Ans.

Question 5.
Solution:
Arranging in ascending order, we get:
10, 10, 11, 11, 12, 12, 13, 14,15, 17
Here, number of terms is 10, which is even
∴ Median = $$\frac { 1 }{ 2 } \left[ \frac { 10 }{ 2 } th\quad term+\left( \frac { 10 }{ 2 } +1 \right) th\quad term \right]$$
= $$\frac { 1 }{ 2 }$$ (5th term + 6th term)

Question 6.
Solution:

Question 7.
Solution:
Writing its cumulative frequency table

Question 8.
Solution:
Writing its cumulative frequency table

Here, number of items is 40 which is even.
∴ Median = $$\frac { 1 }{ 2 } \left[ \frac { 40 }{ 2 } th\quad term+\left( \frac { 40 }{ 2 } +1 \right) th\quad term \right]$$
= $$\frac { 1 }{ 2 }$$ (20th term + 21th term)
= $$\frac { 1 }{ 2 }$$ (30 + 30) = $$\frac { 1 }{ 2 }$$ x 60 = 30
Mean= $$\frac { \sum { { f }_{ i }{ x }_{ i } } }{ \sum { { f }_{ i } } }$$ = $$\frac { 1161 }{ 40 }$$ = 29.025
∴Mode = 3 median – 2 mean = 3 x 30 – 2 x 29.025 = 90 – 58.05 = 31.95

Question 9.
Solution:
Preparing its cumulative frequency table we get:

Here number of terms is 50, which is even

Question 10.
Solution:
Preparing its cumulative frequency table :

Question 11.
Solution:
Preparing its cumulative frequency table we have,

Question 12.
Solution:
Preparing its cumulative frequency table we have,

Hope given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14H are helpful to complete your math homework.

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## RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14G

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14G.

Other Exercises

Question 1.
Solution:
Arranging in ascending order, we get:
2,2,3,5,7,9,9,10,11
Here, number of terms is 9 which is odd.
∴ Median = $$\frac { n+1 }{ 2 }$$ th term = $$\frac { 9+1 }{ 2 }$$ th term = 5th term = 7 Ans.
(ii) Arranging in ascending order, we get: 6, 8, 9, 15, 16, 18, 21, 22, 25
Here, number of terms is 9 which is odd.
∴ Median = $$\frac { n+1 }{ 2 }$$ th term = $$\frac { 9+1 }{ 2 }$$ th term = 5th term = 16 Ans.
(iii) Arranging in ascending order, we get: 6, 8, 9, 13, 15, 16, 18, 20, 21, 22, 25
Here, number of terms is 11 which is odd.
∴ Median = $$\frac { 11+1 }{ 2 }$$ th term = $$\frac { 12 }{ 2 }$$ th term = 6th term = 16 Ans.
(iv) Arranging in ascending order, we get:
0, 1, 2, 2, 3, 4, 4, 5, 5, 7, 8, 9, 10
Here, number of terms is 13, which is odd.
Median = $$\frac { 13+1 }{ 2 }$$ th term = $$\frac { 14 }{ 2 }$$ th term = 7th term = 4 Ans.

Question 2.
Solution:
Arranging in ascending order, we get 9, 10, 17, 19, 21, 22, 32, 35
Here, number of terms is 8 which is even
∴Median = $$\frac { 1 }{ 2 } \left[ \frac { 8 }{ 2 } th\quad term+\left( \frac { 8 }{ 2 } +1 \right) th\quad term \right]$$
= $$\frac { 1 }{ 2 }$$ [4th term + 5th term] = $$\frac { 1 }{ 2 }$$ (19 + 21) = $$\frac { 1 }{ 2 }$$ x 40 = 20
(ii) Arranging in ascending order, we get:
29, 35, 51, 55, 60, 63, 72, 82, 85, 91
Here number of terms is 10 which is even
∴Median = $$\frac { 1 }{ 2 } \left[ \frac { 10 }{ 2 } th\quad term+\left( \frac { 10 }{ 2 } +1 \right) th\quad term \right]$$
= $$\frac { 1 }{ 2 }$$ (60 + 63) = $$\frac { 1 }{ 2 }$$ x 123 = 61.5 Ans.
(iii) Arranging in ascending order we get
3, 4, 9, 10, 12, 15, 17, 27, 47, 48, 75, 81
Here number of terms is 12 which is even.
∴Median = $$\frac { 1 }{ 2 } \left[ \frac { 12 }{ 2 } th\quad term+\left( \frac { 12 }{ 2 } +1 \right) th\quad term \right]$$
= $$\frac { 1 }{ 2 }$$ (6th term + 7th term) = $$\frac { 1 }{ 2 }$$ (15 + 17)= $$\frac { 1 }{ 2 }$$ x 32
= 16 Ans.

Question 3.
Solution:
Arranging the given data in ascending order, we get :
17, 17, 19, 19, 20, 21, 22, 23, 24, 25, 26, 29, 31, 35, 40
∴ Median = $$\frac { 15+1 }{ 2 }$$ th term = $$\frac { 16 }{ 2 }$$ th term = 8th term = 23
∴ Median score = 23 Ans.

Question 4.
Solution:
Arranging in ascending order, we get:
143.7, 144.2, 145, 146.5, 147.3, 148.5, 149.6, 150, 152.1
Here, number of terms is 9 which is odd.
Median = $$\frac { 9+1 }{ 2 }$$ th term = $$\frac { 10 }{ 2 }$$ th term = 5th term = 147.3 cm
Hence, median height = 147.3 cm Ans.

Question 5.
Solution:
Arranging in ascending order, we get:
9.8, 10.6, 12.7, 13.4, 14.3, 15, 16.5, 17.2
Here number of terms is 8 which is even
∴Median = $$\frac { 1 }{ 2 } \left[ \frac { 8 }{ 2 } th\quad term+\left( \frac { 8 }{ 2 } +1 \right) th\quad term \right]$$
= $$\frac { 1 }{ 2 }$$[4th term + 5th term]
= $$\frac { 1 }{ 2 }$$ (13.4 + 14.3) = $$\frac { 1 }{ 2 }$$ (27.7) = 13.85
∴ Median weight = 13.85 kg. Ans.

Question 6.
Solution:
Arranging in ascending order, we get:
32, 34, 36, 37, 40, 44, 47, 50, 53, 54
Here, number of terms is 10 which is even.
∴Median = $$\frac { 1 }{ 2 } \left[ \frac { 10 }{ 2 } th\quad term+\left( \frac { 10 }{ 2 } +1 \right) th\quad term \right]$$
= $$\frac { 1 }{ 2 }$$ [5th term + 6th term ] = $$\frac { 1 }{ 2 }$$ (40 + 44) = $$\frac { 1 }{ 2 }$$ x 84 = 42 .
∴ Median age = 42 years.

Question 7.
Solution:
The given ten observations are 10, 13, 15, 18, x + 1, x + 3, 30, 32, 35, 41
These are even
∴Median = $$\frac { 1 }{ 2 } \left[ \frac { 10 }{ 2 } th\quad term+\left( \frac { 10 }{ 2 } +1 \right) th\quad term \right]$$
= $$\frac { 1 }{ 2 }$$ [5th term + 6th term ] = $$\frac { 1 }{ 2 }$$(x + 1 + x + 3) = $$\frac { 1 }{ 2 }$$(2x + 4)
= x + 2
But median is given = 24
∴ x + 2 = 24 => x = 24 – 2 = 22
Hence x = 22.

Question 8.
Solution:
Preparing the cumulative frequency table, we have:

Here, number of terms (n) = 41, which is odd,
Median = $$\frac { 41+1 }{ 2 }$$ th term = $$\frac { 42 }{ 2 }$$ th term = 21st term = 50 (∵ 20th to 28th term = 50)
Hence median weight = 50 kg Ans.

Question 9.
Solution:
Arranging first in ascending order, we get:

Now preparing its cumulative frequency table

Here, number of terms is 37 which is odd.
Median = $$\frac { 37+1 }{ 2 }$$ th term = $$\frac { 38 }{ 2 }$$ th term = 19 th term = 22 (∵18th to 21st = 22)
Hence median – 22 Ans.

Question 10.
Solution:
first arranging in ascending order we get

Now preparing its cumulative frequency table,we find:

Here, number of terms is 43, which if odd.
Median = $$\frac { 43+1 }{ 2 }$$ th term = $$\frac { 44 }{ 2 }$$ th term = 22nd term = 25 25 (∵ 11th to 26th = 25)

Question 11.
Solution:
Arranging in ascending order,we get

Now preparing its cumulative frequency table, we find :

Here, number of terms = 50 which is even
∴Median = $$\frac { 1 }{ 2 } \left[ \frac { 50 }{ 2 } th\quad term+\left( \frac { 50 }{ 2 } +1 \right) th\quad term \right]$$
= $$\frac { 1 }{ 2 }$$ (154 + 155) = $$\frac { 1 }{ 2 }$$ (309) = 154.5 (∵ 22nd to 25th = 154, 26th to 34th= 155)

Question 12.
Solution:
Arranging in ascending order, we get:

Now, preparing its cumulative frequency table.

Here, number of terms is 60 which is even
∴Median = $$\frac { 1 }{ 2 } \left[ \frac { 60 }{ 2 } th\quad term+\left( \frac { 60 }{ 2 } +1 \right) th\quad term \right]$$
= $$\frac { 1 }{ 2 }$$ (30th term + 31st term)
= $$\frac { 1 }{ 2 }$$ (20 + 23) = $$\frac { 1 }{ 2 }$$ x 43 = 21.5 (∵ 18th to 30th term = 20, 31st term to 34th = 23)
Hence median = 21.5 Ans.

Hope given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14G are helpful to complete your math homework.

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## RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14E

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14E.

Other Exercises

Question 1.
Solution:
Mean marks of 7 students = 226
∴Total marks of 7 students = 226 x 1 = 1582
Marks obtained by 6 of them are 340, 180, 260, 56, 275 and 307
∴ Sum of marks of these 6 students = 340 + 180 + 260 + 56 + 275 + 307 = 1418
∴ Marks obtained by seventh student = 1582 – 1418 = 164 Ans.

Question 2.
Solution:
Mean weight of 34 students = 46.5 kg.
∴ Total weight of 34 students = 46.5 x 34 kg = 1581 kg
By including the weight of teacher, the mean weight of 35 persons = 500 g + 46.5 kg
= 46.5 + 0.5 = 47.0 kg
∴ Total weight = 47.0 x 35 = 1645 kg
∴ Weight of the teacher = (1645 – 1581) kg = 64 kg Ans.

Question 3.
Solution:
Mean weight of 36 students = 41 kg
∴ Total weight of 36 students = 41 x 36 kg = 1476 kg
After leaving one student, number of students = 36 – 1 =35
Their new mean = 41.0 – 200g = (41.0 – 0.2) kg = 40.8 kg.
∴ Total weight of 35 students = 40.8 x 35 = 1428 kg
∴ Weight of leaving student = (1476 – 1428) kg = 48 kg Ans.

Question 4.
Solution:
Average weight of 39 students = 40 kg
∴ Total weight of 39 students = 40 x 39 = 1560 kg
By admitting of a new student, no. of students = 39 + 1 =40
and new mean = 40 kg – 200 g = 40 kg – 0.2 kg = 39.8 kg
∴Weight of 40 students = 39.8 x 40 kg = 1592 kg
∴Weight of new student = 1592 – 1560 kg = 32 kg Ans.

Question 5.
Solution:
Average salary of 20 workers = Rs. 7650
∴Their total salary = Rs. 7650 x 20 = Rs. 153000
By adding the salary of the manager, their mean salary = Rs. 8200
∴Their total salary = Rs. 8200 x 21 = Rs. 172200
∴Salary of the manager = Rs. 172200 – Rs. 153000 = Rs. 19200 Ans.

Question 6.
Solution:
Average wage of 10 persons = Rs. 9000
Their total wage = Rs. 9000 x 10 = Rs. 90000
Wage of one person among them = Rs. 8100
and wage of new member = Rs. 7200
∴ New total wage = Rs. (90000 – 8100 + 7200) = Rs. 89100
Their new mean wage = Rs.$$\frac { 89100 }{ 10 }$$ = Rs. 8910

Question 7.
Solution:
Mean consumption of petrol for 7 months of a year = 330 litres
Mean consumption of petrol for next 5 months = 270 litres
∴Total consumption for first 7 months = 7 x 330 = 2320 l
and total consumption for next 5 months = 5 x 270 = 1350 l
Total consumption for 7 + 5 = 12 months = 2310 + 1350 = 3660 l
Average consumption = $$\frac { 3660 }{ 12 }$$ = 305 liters per month.

Question 8.
Solution:
Total numbers of numbers = 25
Mean of 15 numbers = 18
∴Total of 15 numbers = 18 x 15 = 270
Mean of remaining 10 numbers = 13
∴Total = 13 x 10 = 130
and total of 25 numbers = 270 + 130 = 400
∴Mean = $$\frac { 400 }{ 25 }$$ = 16 Ans.

Question 9.
Solution:
Mean weight of 60 students = 52.75 kg.
Total weight = 52.75 x 60 = 3165 kg
Mean of 25 out of them = 51 kg.
∴Their total weight = 51 x 25 = 1275 kg
∴ Total weight of remaining 60 – 25 = 35 students = 3165 – 1275 = 1890 kg 1890
Mean weight = $$\frac { 1890 }{ 35 }$$ = 54 kg Ans.

Question 10.
Solution:
Average increase of 10 oarsman = 1.5 kg.
∴ Total increased weight =1.5 x 10 = 15kg
Weight of out going oarsman = 58 kg .
∴ Weight of new oarsman = 58 + 15 = 73 kg Ans.

Question 11.
Solution:
Mean of 8 numbers = 35
∴Total of 8 numbers = 35 x 8 = 280
After excluding one number, mean of 7 numbers = 35 – 3 = 32
Total of 7 numbers = 32 x 7 = 224
Hence excluded number = 280 – 224 = 56 Ans.

Question 12.
Solution:
Mean of 150 items = 60
Total of 150 items = 60 x 150 = 9000
New total = 9000 + 152 + 88 – 52 – 8 = 9000 + 240 – 60 = 9180
∴New mean = $$\frac { 9180 }{ 150 }$$ = 61.2 Ans.

Question 13.
Solution:
Mean of 31 results = 60
∴Total of 31 results = 60 x 31 = 1860
Mean of first 16 results = 58
∴Total of first 16 results = 58 x 16 = 928
and mean of last 16 results = 62
∴Total of last 16 results = 62 x 16 = 992
∴16th result = (928 + 992) – 1860 = 1920 – 1860 = 60 Ans.

Question 14.
Solution:
Mean of 11 numbers = 42
∴Total of 11 numbers = 42 x 11 = 462
Mean of first 6 numbers = 37
∴Total of first 6 numbers = 37 x 6 = 222
Mean of last 6 numbers = 46
∴Total of last 6 numbers = 46 x 6 = 276
∴6th number = (222 + 276) – 462 = 498 – 462 = 36 Ans.

Question 15.
Solution:
Mean weight of 25 students = 52 kg
∴ Total weight of 25 students = 52 x 25 = 1300 kg
Mean weight of first 13 students = 48 kg
∴ Total weight of first 13 students = 48 x 13 kg = 624 kg
Mean of last 13 students = 55 kg
∴Total of last 13 students = 55 x 13 kg = 715 kg
∴Weight of 13th student = (624 + 715) – 1300 = 1339 – 1300 = 39 kg Ans.

Question 16.
Solution:
Mean of 25 observations = 80
Total of 25 observations = 80 x 25 = 2000
Mean of another 55 observations = 60
∴ Total of these 55 observations = 60 x 55 = 3300
Total number of observations = 25 + 55 = 80
and total of 80 numbers = 2000 + 3300 = 5300
Mean of 80 observations = $$\frac { 5300 }{ 80 }$$ = 66.25 Ans.

Question 17.
Solution:
Marks in English = 36
Marks in Hindi = 44
Marks in Mathematics = 75
Marks in Science = x
∴Total number of marks in 4 subjects = 36 + 44 + 75 + x = 155 + x
Average marks in 4 subjects = 50
∴Total marks = 50 x 4 = 200
∴155 + x = 200
=> x = 200 – 155
=> x = 45
Hence, marks in Science = 45 Ans.

Question 18.
Solution:
Mean of monthly salary of 75 workers = Rs. 5680
Total salary of 75 workers = Rs. 5680 x 75 = Rs. 426000
Mean salary of 25 among them = Rs. 5400
Total of 25 workers = Rs. 5400 x 25 = Rs. 135000
Mean salary of 30 among them = Rs. 5700
∴Total of 30 among them = Rs. 5700 x 30 = Rs. 171000
∴ Total salary of 25 + 30 = 55 workers = Rs. 135000 + 171000 = Rs. 306000
∴Total salary of remaining 75 – 55 = 20 workers = Rs. 426000 – 306000 = Rs. 120000
∴ Mean of remaining 20 workers = Rs. $$\frac { 120000 }{ 20 }$$ = Rs. 6000 Ans.

Question 19.
Solution:
Let distance between two places = 60 km
∴Time taken at the speed of 15 km/h = $$\frac { 60 }{ 15 }$$ = 4 hours
and time taken at speed of 10 km/h for coming back = $$\frac { 60 }{ 10 }$$ = 6 hours
Total the taken = 4 + 6 = 10
hours and distance covered = 60 + 60 = 120 km
∴Average speed = $$\frac { 120 }{ 10 }$$ =12 km/hr.

Question 20.
Solution:
No. of total students = 50
No. of boys = 40
∴ No. of girls = 50 – 40 = 10
Average weight of class = 44kg
∴Total weight of 50 students = 44 x 50 = 2200 kg
Average weight of 10 girls = 40 kg
∴Total weight = 40 x 10 = 400 kg
Total weight of 40 boys = 2200 – 400 = 1800 kg
∴Average weight of 40 boys = $$\frac { 1800 }{ 40 }$$kg = 45 kg Ans.

Hope given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14E are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.