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

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