RS Aggarwal Chapter 15 Class 9 Maths Exercises 15.3 (ex 15c) Solutions

RS Aggarwal Chapter 15 Class 9 Maths Exercise 15.3 Solutions: Wherever we look, usually we see solids. So far, in all our studies, we have been dealing with figures that can be easily drawn on our notebooks or blackboards. These are called plane figures. We have understood what rectangles, squares, and circles are, what we mean by their perimeters and areas, and how we can find them. We have learned these in earlier classes.

It would be interesting to see what happens if we cut out many of these plane figures of the same shape and size from cardboard sheets and stack them up in a vertical pile. Through this process, we shall obtain some solid figures (briefly called solids) such as a cuboid, a cylinder, etc. Know more about this chapter here.

Download RS Aggarwal Chapter 15 Class 9 Maths Exercise 15.3 Solutions

 


EXERCISE – 15C

Access The RS Aggarwal Chapter 15 Class 9 Maths Exercise 15.3 Solutions

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) = 285500 = 57100 = 0.57
(ii) P(E2) = 215500 = 43100 = 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) = 112400 = 28100 = 0.28
(ii)P(E2) = 160400 = 40100= 0.40
(iii) P(E3) = 128400 = 32100 = 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) = 39200 = 0.195
(ii) P(E3) = 67200 = 0.335
(iii) P(E4) = 36200 = 18100 = 0.18
(iv) P(E2) = 58200 = 29100 = 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) = 54300 = 18100 = 0.18
(ii) P(E6) = 33100 = 11100= 0.11
(iii) P(E5) = 39300 = 13100 = 0.13
(iv) P(E1) = 60300 = 20100= 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) = 142200 = 71100 = 0.71
(ii) P(E2) = 58200 = 29100 = 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) = 26 = 13Ans.

Question 7.
Solution:
No. of vehicles of various types = 240
No. of vehicles of two wheelers = 64.
Probability will be P(E) = 84240 = 720 = 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) = 24200 = 12100 = 0.12
(ii) P(E8) = 16200 = 8100 = 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) = 1440 = 720 = 0.35
(ii) P(EAB) = 640 = 320 = 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) = 630 = 15 = 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) = 60360 = 16
(ii) 50 years or more but less than 70 years = 50 + 30 = 80
P(E) = 80360 = 29
(iii) Less than 10 years = zero
P(E) = 0360 = 0
(iv) 10 years or more 90 + 50 + 60 + 80 + 50 + 30 = 360

Important Definition for RS Aggarwal Chapter 15 Class 9 Maths Ex 15c Solutions

Name of the Solid Figure

Formulas

Cuboid

LSA: 2h(l + b)
TSA: 2(lb + bh + hl)
Volume: l × b × h

l = length,
b = breadth,
h = height

Cube

LSA: 4a2
TSA: 6a2
Volume: a3

a = sides of a cube

Right Circular Cylinder

LSA: 2(π × r × h)
TSA: 2πr (r + h)
Volume: π × r2 × h

r = radius,
h = height

Right Pyramid

LSA: ½ × p × l
TSA: LSA + Area of the base
Volume: ⅓ × Area of the base × h

p = perimeter of the base,
l = slant height, h = height

Prism

LSA: p × h
TSA: LSA × 2B
Volume: B × h

p = perimeter of the base,
B = area of base, h = height

Right Circular Cone

LSA: πrl
TSA: π × r × (r + l)
Volume: ⅓ × (πr2h)

r = radius,
l = slant height,
h = height

Hemisphere

LSA: 2 × π × r2
TSA: 3 × π × r2
Volume: ⅔ × (πr3)

r = radius

Sphere

LSA: 4 × π × r2
TSA: 4 × π × r2
Volume: 4/3 × (πr3)

r = radius

Know more at the official website.

FAQs on RS Aggarwal Chapter 15 Class 9 Maths Exercise 15.3

What do you Understand by Surface Area and Volume? Write the Crucial Topics Involved in Chapter 15, Maths Class 9 RS Aggarwal.

Total Surface area refers to the sum of the areas of all the faces of a solid. Lateral surface area refers to the sum of areas of all the sides of a solid, excluding its top and base. 

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