RS Aggarwal Class 9 Solutions Chapter 15 Probability Ex 15A

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:
No. of ladies on whom survey was made = 200.
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.

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

RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14F

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:
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14F Q1.1

Question 2.
Solution:
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14F Q2.1

Question 3.
Solution:
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14F Q3.1

Question 4.
Solution:
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14F Q4.1

Question 5.
Solution:
Mean = 8
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14F Q5.1

Question 6.
Solution:
Mean = 28.25
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14F Q6.1
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14F Q6.2

Question 7.
Solution:
Mean = 16.6
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14F Q7.1
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14F Q7.2

Question 8.
Solution:
Mean = 50
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14F Q8.1
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14F Q8.2

Question 9.
Solution:
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14F Q9.1

Question 10.
Solution:
Let assumed mean = 67
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14F Q10.1

Question 11.
Solution:
Here h = 1, Let assumed mean (A) = 21
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14F Q11.1

Question 12.
Solution:
Here h = 400 and let assumed mean (A) = 1000
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14F Q12.1
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14F Q12.2

Hope given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14F 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.

RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14H

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)
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14H Q5.1

Question 6.
Solution:
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14H Q6.1
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14H Q6.2

Question 7.
Solution:
Writing its cumulative frequency table
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14H Q7.1
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14H Q7.2

Question 8.
Solution:
Writing its cumulative frequency table
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14H Q8.1
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:
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14H Q9.1
Here number of terms is 50, which is even
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14H Q9.2

Question 10.
Solution:
Preparing its cumulative frequency table :
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14H Q10.1
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14H Q10.2

Question 11.
Solution:
Preparing its cumulative frequency table we have,
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14H Q11.1
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14H Q11.2

Question 12.
Solution:
Preparing its cumulative frequency table we have,
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14H Q12.1
Hope given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14H 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.

RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14G

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:
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14G Q8.1
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:
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14G Q9.1
Now preparing its cumulative frequency table
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14G Q9.2
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
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14G Q10.1
Now preparing its cumulative frequency table,we find:
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14G Q10.2
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
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14G Q11.1
Now preparing its cumulative frequency table, we find :
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14G Q11.2
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:
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14G Q12.1
Now, preparing its cumulative frequency table.
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14G Q12.2
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.

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

RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14E

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.

RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14D

RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14D

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

Other Exercises

Question 1.
Solution:
(i) First eight natural numbers are 1, 2, 3, 4, 5, 6, 7, 8
∴ Mean = \(\frac { 1+2+3+4+5+6+7+8 }{ 8 } \) = \(\frac { 36 }{ 8 } \) = \(\frac { 9 }{ 2 } \) = 4.5
(ii) First ten odd numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19
∴ Mean = \(\frac { 1+3+5+7+9+11+13+15+17+179 }{ 10 } \) = \(\frac { 100 }{ 10 } \) = 10
(iii) First five prime numbers are 2, 3, 5, 7, 11.
∴ Mean = \(\frac { 2+3+5+7+11 }{ 5 } \) = \(\frac { 28 }{ 5 } \) = 5.6
(iv) First six even numbers are 2, 4, 6, 8, 10, 12
∴ Mean = \(\frac { 2+4+6+8+10+12 }{ 10 } \) = \(\frac { 42 }{ 6 } \) = 7
(v) First seven multiples of 5 are 5, 10, 15, 20, 25, 30, 35
∴ Mean = \(\frac { 5+10+15+20+25+30+35 }{ 7 } \) = \(\frac { 140 }{ 7 } \) = 20
(vi) All the factors of 20 are, 1, 2, 4, 5, 10, 20
∴ Mean = \(\frac { 1+2+4+5+10+20 }{ 6 } \) = \(\frac { 42 }{ 6 } \) = 7

Question 2.
Solution:
No. of families (n) = 10
Sum of children (∑x1) = 2 + 4 + 3 + 4 + 2 + 0 + 3 + 5 + 1 + 6 = 30
∴ Mean \(\left( \overline { x } \right) =\frac { \sum { x1 } }{ n } =\frac { 30 }{ 10 } =3\)

Question 3.
Solution:
Here number of days (n) = 7
Number of books (∑x1) = 105 + 216 + 322 + 167 + 273 + 405 + 346 = 1834
∴ Mean \(\left( \overline { x } \right) =\frac { \sum { x1 } }{ n } =\frac { 1864 }{ 7 } =262 books\)

Question 4.
Solution:
Number of days (n) = 6
Sum of temperature (∑x) = 35.5 + 30.8 + 27.3 + 32.1 + 23.8 + 29.9 = 179.4°F
∴ Mean temperature = \(\frac { \sum { x } }{ n } =\frac { 179.4 }{ 6 } \)=29.9°F

Question 5.
Solution:
Number of students (n) = 12
Sum of marks (∑x) = 64 + 36 + 47 + 23 + 0 + 19 + 81 + 93 + 72 + 35 + 3 + 1 = 474
= \(\frac { \sum { x1 } }{ n } =\frac { 474 }{ 12 } \) = 39.5

Question 6.
Solution:
Here, n = 6
and arithmetic mean =13
Total sum = 13 x 6 = 78.
But sum of 7 + 9+ 11 + 13 + 21=61
Value of x = 78 – 61 = 17
Hence x = 17 Ans.

Question 7.
Solution:
Let x1, x2, x3, … x24 be the 24 numbers
\(\frac { x1+x2+x3+…..+x24 }{ 24 } =35\)
=> x1 + x2 + x3 +….+ x24 = 35 x 24
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14D Q7.1

Question 8.
Solution:
Let x1 + x2 + x3……..x20 be the 20 numbers
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14D Q8.1

Question 9.
Solution:
Let x1, x2, x3 … x15 be the numbers
Mean = \(\frac { x1+x2+x3+…..+x15 }{ 15 } = 27\)
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14D Q9.1

Question 10.
Solution:
Let x1, x2, x3 … x12 be the numbers
Mean = \(\frac { x1+x2+x3+…..+x12 }{ 12 }\) = 40
=> x1+x2+x3+….+x12 = 40 X 12 =480
Now,new numbers are \(\frac { x1 }{ 8 } ,\frac { x2 }{ 8 } ,\frac { x3 }{ 8 } ,..\frac { x12 }{ 8 } \)
Mean = \(\frac { \frac { x1 }{ 8 } +\frac { x2 }{ 8 } +\frac { x3 }{ 8 } +…..+\frac { x12 }{ 8 } }{ 12 } \)
= \(\frac { 1 }{ 8 } \frac { \left( x1+x2+x3+….x12 \right) }{ 12 } \)
= \(\frac { 480 }{ 8X12 } \) = 5
Mean of new numbers =5

Question 11.
Solution:
Let x1, x2, x3,…..x20 are the numbers
Mean = \(\frac { x1+x2+x3+…x20 }{ 20 }\) = 18
=> x1 + x2 + x3 +….+ x20 = 18 X 20 =360
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14D Q11.1

Question 12.
Solution:
Mean weight of 6 boys = 48 kg
Their total weight = 48 x 6 = 288 kg
Weights of 5 boys among them, are 51 kg, 45 kg, 49 kg, 46 kg and 44 kg
Sum of weight of 5 boys = (51 + 45 + 49 + 46 + 44) kg = 235 kg
Weight of 6th boy = 288 kg – 235 kg = 53 kg Ans.

Question 13.
Solution:
Mean of 50 students = 39
Total score = 39 x 50 = 1950
Now correct sum of scores = 1950 – Wrong item + Correct item= 1950 – 23 + 43
= 1950 + 20 = 1970
Correct mean = \(\frac { 1970 }{ 50 } \) = 39.4 Ans.

Question 14.
Solution:
Mean of 100 items = 64
The sum of 100 items = 64 x 100 = 6400
New sum of 100 items = 6400 + 36 + 90 – 26 – 9 = 6526 – 35 = 6491,
Correct mean = \(\frac { 6491 }{ 100 }\) =64.91 = 64.91 Ans.

Question 15.
Solution:
Mean of 6 numbers = 23
Sum of 6 numbers = 23 x 6 = 138
Excluding one number, the mean of remaining 5 numbers = 20
Total of 5 numbers = 20 x 5 = 100
Excluded number = 138 – 100 = 38 Ans.

Hope given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14D 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.

RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14C

RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14C

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

Other Exercises

Question 1.
Solution:
It is clear that the given frequency distribution is in exclusive form, we represent the daily wages (in rupees) along x-axis and no. of workers along y-axis. Then we construct rectangles with class intervals as bases and corresponding frequencies as heights as shown given below:
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14C Q1.1

Question 2.
Solution:
We shall take daily earnings along x-axis and number of stores along y-axis. Then we construct rectangles with the given class intervals as bases and corresponding frequency as height as shown in the graph.
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14C Q2.1

Question 3.
Solution:
We shall take heights along x-axis and number of students along y-axis. Then we shall complete the rectangles with the given class intervals as bases and frequency as height and complete the histogram as shown below :
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14C Q3.1

Question 4.
Solution:
We shall take class intervals along x-axis and frequency along y-axis Then we shall complete the rectangles with given class interval as bases and frequency as heights and complete the histogram as shown below.
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14C Q4.1

Question 5.
Solution:
The given frequency distribution is in inclusive form first we convert it into exclusive form at given below.
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14C Q5.1
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14C Q5.2
Now, we shall take class intervals along x-axis and frequency along y-axis and draw rectangles with class intervals as bases and frequency as heights and complete the histogram as shown.

Question 6.
Solution:
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14C Q6.1

Question 7.
Solution:
In the given distribution class intervals are different size. So, we shall calculate the adjusted frequency for each class. Here minimum class size is 4. We know that adjusted
frequency of the class is \(\frac { max\quad class\quad size }{ class\quad size\quad of\quad this\quad class } \)  x its frequency
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14C Q7.1
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14C Q7.2

Question 8.
Solution:
We shall take two imagined classes one 0-10 at the beginning with zero frequency a id other 70-80 at the end with zero frequency.
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14C Q8.1
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14C Q8.2
Now, plot the points” (5, 0), (15, 2), (25, 5), (35, 12), (45, 19), (55, 9), (65, 4), (75, 0) on the graph and join them its order to get a polygon as shown below.

Question 9.
Solution:
We take Age (in years) along x-axis and number of patients along y-axis. First we complete the histogram and them join the midpoints of their tops in order
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14C Q9.1
We also take two imagined classes. One 0-10 at the beginning and other 70-80 at the end and also join their midpoints to complete the polygon as shown.

Question 10.
Solution:
We take class intervals along x-axis and frequency along y-axis.
First we complete the histogram and then join the midpoints of the tops of adjacent rectangles in order. We shall take two imagined classes 15-20 at the beginning and 50-55 at the end.
We also join their midpoints to complete the polygon as shown
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14C Q10.1

Question 11.
Solution:
We represent class interval on x-axis and frequency on y-axis. First we construct the histogram and then join the midpoints of the tops of each rectangle by line segments. We shall take two imagined class i.e. 560-600 at the beginning and 840-900 at the end. Join their midpoints also to complete
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14C Q11.1

Question 12.
Solution:
We will take the imagined class 11-0 at the beginning and 61-70 at the end. Each with frequency zero. Thus, we have,
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14C Q12.1
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14C Q12.2
Now we will mark the points A (-5.5, 0), B(5.5, 8), C(15.5, 3) D(25.5, 6), E(35.5, 12), F(45.5, 2), G(55.5, 7) and H(65.5, 0) on the graph and join them in order to get the required frequency polygon.

Hope given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14C 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.

RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14B

RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14B

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

Other Exercises

Question 1.
Solution:
We shall take the game along x-axis and number of students along y-axis.
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14B Q1.1

Question 2.
Solution:
We shall take the time on x-axis and temperature (in °C) on y-axis.
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14B Q2.1

Question 3.
Solution:
We shall take name of vehicle on x-axis and velocity (in km/hr) on y-axis
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14B Q3.1

Question 4.
Solution:
We shall take sports on x-axis and number of students on y-axis.
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14B Q4.1

Question 5.
Solution:
We shall take years on x-axis and number of students on y-axis.
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14B Q5.1

Question 6.
Solution:
We shall take years on x-axis and number of students on y-axis.
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14B Q6.1

Question 7.
Solution:
We shall take years on x-axis and number of students on y-axis.
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14B Q7.1

Question 8.
Solution:
We shall take years on x-axis and number of students on y-axis.
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14B Q8.1

Question 9.
Solution:
We shall take years on x-axis and number of students on y-axis.
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14B Q9.1

Question 10.
Solution:
We shall take years on x-axis and number of students on y-axis.
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14B Q10.1

Question 11.
Solution:
We shall take week on x-axis and Rate per 10g (in Rs.) on y-axis.
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14B Q11.1

Question 12.
Solution:
We shall take mode of transport on x-axis and number of students on y-axis.
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14B Q12.1

Question 13.
Solution:
We see from the graph that
(i) It shows the marks obtained by a student in various subjects.
(ii) The student is very well in mathematics.
(iii) The student is very’ poor, in Hindi.
(iv) Average marks
= \(\frac { 60+35+75+50+60 }{ 5 } \) (Here x = 5)
= \(\frac { 280 }{ 5 } \)
= 56 marks
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14B Q13.1

Hope given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14B 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.

RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14A

RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14A

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

Other Exercises

Question 1.
Solution:
Statistics is a science which deals with the collection, presentation, analysis and interpretations of numerical data.

Question 2.
Solution:
(i) Numerical facts alone constitute data
(ii) Qualitative characteristics like intelligence, poverty etc. which cannot be measured, numerically, don’t form data.
(iii) Data are an aggregate of facts. A single observation does not form data.
(iv) Data collected for a definite purpose may not be suited for another purpose.
(v) Data is different experiments are comparable.

Question 3.
Solution:
(i) Primary data : The data collected by the investigator himself with a definite plan in mind are called primary data.
(ii) Secondary data : The data collected by some one other than the investigator are called secondary data. The primary data is more reliable and relevant.

Question 4.
Solution:
(i) Variate : Any character which is capable of taking several different values is called a variate or variable.
(ii) Class interval : Each group into which the raw data is condensed, is called a class interval
(iii) Class size : The difference between the true upper limit and the true lower limit of a class is called class size.
(iv) Class Mark : \(\frac { upper\quad limit+lower\quad limit }{ 2 } \) is called a class mark
(v) Class limits : Each class is bounded by two figures which are called class limits which are lower class limit and upper class limit.
(vi) True class limits : In exclusive form, the upper and lower limits of a class are respectively are the true upper limit and true lower limit but in inclusive form, the true lower limit of a class is obtained by subtracting O.S from lower limit of the class and for true limit, adding 0.5 to the upper limit.
(vii) Frequency of a class : The number of times an observation occurs in a class is called its frequency.
(viii) Cumulative frequency of a class : The cumulative frequency corresponding to a class is the sum of all frequencies upto and including that class.

Question 5.
Solution:
The given data can be represent in form of frequency table as given below:
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14A Q5.1

Question 6.
Solution:
The frequency distribution table of the given data is given below :
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14A Q6.1

Question 7.
Solution:
The frequency distribution table of the
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14A Q7.1

Question 8.
Solution:
The frequency table is given below :
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14A Q8.1

Question 9.
Solution:
The frequency table of given data is given below :
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14A Q9.1

Question 10.
Solution:
The frequency distribution table of the given data in given below :
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14A Q10.1

Question 11.
Solution:
The frequency table of the given data:
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14A Q11.1

Question 12.
Solution:
The cumulative frequency of the given table is given below:
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14A Q12.1

Question 13.
Solution:
The given table can be represented in a group frequency table in given below :
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14A Q13.1

Question 14.
Solution:
Frequency table of the given cumulative frequency is given below :
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14A Q14.1

Question 15.
Solution:
A frequency table of the given cumulative frequency table is given below :
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14A Q15.1

 

Hope given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14A 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.

RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D

RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D.

Other Exercises

Question 1.
Solution:
(i) Radius of sphere = 3.5cm
(a) Volume = \(\frac { 4 }{ 3 } \) πr3
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q1.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q1.2

Question 2.
Solution:
Let r be the radius of the sphere and volume = 38808 cm3
∴\(\frac { 4 }{ 3 } \) πr3 = 38803
=> \(\frac { 4 }{ 3 } \) x \(\frac { 22 }{ 7 } \) r3 = 38803
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q2.1

Question 3.
Solution:
Let r be the radius of the sphere
∴ Volume = \(\frac { 4 }{ 3 } \) πr3
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q3.1

Question 4.
Solution:
Surface area of a sphere = 394.24 m2
Let r be the radius, then 4πr2 = 394.24
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q4.1

Question 5.
Solution:
Surface area of sphere = 576π cm2
Let r be the radius, then 4r2 = 576π
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q5.1

Question 6.
Solution:
Outer diameter of shell = 12cm,
Outer radius (R) = \(\frac { 12 }{ 2 } \) = 6cm
and inner diameter = 8cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q6.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q6.2

Question 7.
Solution:
Length of cuboid of (l) = 12cm
Breadth (b) = 11cm
and height (h) = 9cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q7.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q7.2

Question 8.
Solution:
Radius of sphere (r) = 8cm
Volume = \(\frac { 4 }{ 3 } \)πr3
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q8.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q8.2

Question 9.
Solution:
Radius of solid sphere (R) = 3cm.
Volume = \(\frac { 4 }{ 3 } \)π(R)3 = \(\frac { 4 }{ 3 } \)π(3)3 cm3
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q9.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q9.2

Question 10.
Solution:
Radius of metallic sphere (R) = 10.5cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q10.1

Question 11.
Solution:
Diameter of a cylinder = 8cm
Radius (r) = \(\frac { 8 }{ 2 } \) = 4cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q11.1

Question 12.
Solution:
Diameter of sphere = 6cm
Radius (R) = \(\frac { 6 }{ 2 } \) = 3cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q12.1

Question 13.
Solution:
Diameter of sphere = 18cm
Radius (R) = \(\frac { 18 }{ 2 } \) = 9cm.
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q13.1

Question 14.
Solution:
Diameter of the sphere = 15.6 cm
Radius (R) = \(\frac { 15.6 }{ 2 } \) = 7.8 cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q14.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q14.2

Question 15.
Solution:
Diameter of the canonball = 28cm
Radius (R) = \(\frac { 28 }{ 2 } \) = 14 cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q15.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q15.2

Question 16.
Solution:
Given,
Radius of spherical big ball (R) = 3cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q16.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q16.2

Question 17.
Solution:
Ratio in the radii of two spheres = 1:2
Let radius of smaller sphere = r then,
radius of bigger sphere = 2r
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q17.1

Question 18.
Solution:
Let r1 and r2 be the radii of two spheres
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q18.1

Question 19.
Solution:
Radius of the cylindrical tub = 12cm.
First level of water = 20cm
Raised water level = 6.75cm.
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q19.1

Question 20.
Solution:
Radius of the ball (r) = 9cm.
Volume of ball = \(\frac { 4 }{ 3 } \)πr³
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q20.1

Question 21.
Solution:
Given,
Radius of hemisphere of lead (r) = 9cm.
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q21.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q21.2

Question 22.
Solution:
Given,
Radius of hemispherical bowl (r) = 9cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q22.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q22.2

Question 23.
Solution:
External radius of spherical shell (R) = 9cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q23.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q23.2

Question 24.
Solution:
Inner radius (r) = 4 cm
Thickness of steel used = 0.5
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q24.1

Hope given RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D 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.

RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C

RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C.

Other Exercises

Question 1.
Solution:
Radius of base (r) = 35cm
and height (h) = 84cm.
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q1.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q1.2

Question 2.
Solution:
Height of cone (h) = 6cm
Slant height (l) = 10cm.
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q2.1

Question 3.
Solution:
Volume of right circular cone = (100 π) cm3
Height (h) = 12cm.
Let r be the radius of the cone
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q3.1

Question 4.
Solution:
Circumference of the base = 44cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q4.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q4.2

Question 5.
Solution:
Slant height of the cone (l) = 25cm
Curved surface area = 550 cm2
Let r be the radius
πrl = curved surface area
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q5.1

Question 6.
Solution:
Radius.of base (r) = 35cm.
Slant height (l) = 37cm.
We know that
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q6.1

Question 7.
Solution:
Curved surface area = 4070 cm2
Diameter of the base = 70cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q7.1

Question 8.
Solution:
Radius of the conical tent = 7m
and height = 24 m.
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q8.1

Question 9.
Solution:
Radius of the first cone (r) = 1.6 cm.
and height (h) = 3.6 cm.
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q9.1

Question 10.
Solution:
Ratio in their heights =1:3
and ratio in their radii = 3:1
Let h1,h2 he their height and r1,r2 be their radii, then
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q10.1
The ratio between their volumes is 3:1
hence proved

Question 11.
Solution:
Diameter of the tent = 105m
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q11.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q11.2

Question 12.
Solution:
No. of persons to be s accommodated =11
Area to be required for each person = 4m2
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q12.1

Question 13.
Solution:
Height of the cylindrical bucket (h) = 32cm
Radius (r) = 18cm
Volume of sand filled in it = πr2h
= π x 18 x 18 x 32 cm3
= 10368π cm3
Volume of conical sand = 10368 π cm3
Height of cone = 24 cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q13.1

Question 14.
Solution:
Let h be the height and r be the radius of the cylinder and cone.
Curved surface area of cylinder = 2πrh
and curved surface area of cone = πrl
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q14.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q14.2

Question 15.
Solution:
Diameter of the pillar = 20cm
Radius (r) = \(\frac { 20 }{ 2 } \) = 10cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q15.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q15.2

Question 16.
Solution:
Height of the bigger cone (H) = 30cm
By cutting a small cone from it, then volume of smaller cone = \(\frac { 1 }{ 27 } \) of volume of big cone
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q16.1
Let radius and height of the smaller cone be r and h
and radius and height of the bigger cone be R and H.
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q16.2
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q16.3
Hence at the height of 20cm from the base it was cut off. Ans.

Question 17.
Solution:
Height of the cylinder (h) = 10cm.
Radius (r) = 6cm.
Height of the cone = 10cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q17.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q17.2

Question 18.
Solution:
Diameter of conical vessel = 40cm
Radius (r) = \(\frac { 40 }{ 2 } \) = 20cm
and depth (h) = 24cm.
.’. Volume = \(\frac { 1 }{ 3 } \) πr2h
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q18.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q18.2

Hope given RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C 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.