State Board Commerce (XI-XII) - Test Papers
How to prepare for NTSE Stage 2 in the last few days
NTSE or The National Talent Search Examination is conducted by the National Council of Educational Research and Training (NCERT) nation-wide to award scholarships to aspiring students. So, if you have successfully passed their stage I exam, it is time for you to buckle up for their Stage II exam to be held on May, 14, 2017.
The last few days before the exam is a crucial time where you need to concentrate on revising and revising only. First and foremost fix yourself a daily routine for the remnant days so that you have a goal for the entire day when you wake up in the morning. Try to clock in 8 to 9 hours of intense studying during the last few days. Waking up early is another key point to be followed. Also, you are advised to do some morning exercise that will help you keep your body as well as mind fresh and alert. Avoid studying in the afternoon post lunch and try to squeeze in a power nap here.
Try to concentrate on important topics as mentioned below:
1. MAT (Mental Ability Test): Maximum weightage is given to this section in the exam. Analytical and application-based questions, clocks and calendar are what you should be thorough with. Also, logical Venn Diagrams, Coding-Decoding, Number Series, Missing Terms, Seating Arrangements are some other important topic to look into.
Mental Ability Test (MAT): Since this is an important section of the exam, let’s delve deeper into it. This section has 50 questions all together. You need to get 20 questions correct to make it to the cutoff score. This is clearly the most difficult section and all you get is just 45 minutes to attempt this section. This section comprises of some of the trickiest questions for testing your logical reasoning ability. Try to incorporate shortcut formulas and tricks to solve questions to save time. Try to give maximum time to practice this section.
- Geography is yet another important subject from the exam score point of view. Questions based on maps should be revised over and over again.
- History: This subject might feel too boring for students but try to revise this as well. Topics like Nazism and Rise of Hitler, Nationalism in India and the Rise of Nationalism in Europe need special attention.
- Social Studies: Industries, Climate, Natural Vegetation and Wildlife, Physical Division of India, Globalization and Indian Economy, Union Executive and Legislature are some of the important topics to be covered in this subject.
- Physics: This is a subject that demands you to understand the basics. Concentrate on topics like Motion, Light, Gravitation, Electricity and Magnetism.
- Mathematics: Constant practice of these topics is the only way to score in Maths. Area & Volumes, Number System, Triangles, Algebraic Expressions, and Equations and Regular Polygons.
- Chemistry: Your last minute topics of interest should be Solutions, Metals and Non-Metals, Acids, Bases, and Salts, Chemical Reactions, State of Matter and Gas Law.
- Language Test: Brush up your sentence completion and comprehension reading skills. Make sure you also time yourself during practicing the comprehensions.Last but not the least; solve as many mock test papers as possible on a daily basis as this will hone your time management skill. This way you will also get familiar with the NTSE Stage-II exam pattern and help you determine as to how to attempt the paper, how to allocate time to each section and which section needs more of your attention.
So, get going and start following these easy tips to secure your scholarship!
NCERT Solutions for Class 10 Maths – Probability
Exercise – 13.1
Q.1: Q and R are events such as P(Q) = 0.6 , P(R) = 0.3, P(Q∩R) = 0.2. Determine the value of P(QR) and P(RQ)
Sol:
According to the Q., it is given that P(Q) = 0.6 , P(R) = 0.3, P (Q∩R) = 0.2
P(QR) = P(Q∩R)P(R) = 0.20.3 = 23
P(RQ) = P(Q∩R)P(Q) = 0.20.6 = 13
Q.2: Determine the value of P (QR) if Q & R are the events such that P(Q) = 0.5 and P(Q∩R) = 0.32
Sol:
According to the Q., it is given that P(Q) = 0.5 , P(Q∩R) = 0.32.
P(QR) = P(Q∩R)P(R)
= 0.320.5 = 1625
Q.3: If Q and R are events such that P(Q) = 0.8 , P(R) = 0.5 , P(RQ) = 0.4. Determine the value of
(a) P(Q∩R)
(b) P(QR)
(c) P(Q∪R)
Sol:
According to the Q, it is given that:
P(Q) = 0.8 , P(R) = 0.5 , P(RQ) = 0.4
(a) P(RQ) = P(Q∩R)P(Q)
i.e. 0.4 = P(Q∩R)0.8
Therefore, P(Q∩R) = 0.32
(b) P(QR) = P(Q∩R)P(R) = 0.320.5 = 0.64
(c) P(Q∪R)=P(Q)+P(R)–P(Q∩R)
= P(Q∪R)=0.8+0.5–32 = 0.98
Q.4: Determine the value of P(Q∩R) if 2P(Q) = P(R) = 513 & P(QR)=25
Sol:
According to the Q, it is given that:
2P(Q) = P(R) = 513 & P(QR)=25
P(Q) = 526
P(R) = 513
P(QR)=25
= Q∩RR = 25
= P(Q∩R) = 25×513
= 213
We know that:
P(Q∪R)=P(Q)+P(R)–P(Q∩R)
= Q∩R = 526+513–213
= Q∩R = 5+10–426
= Q∩R = 1126
Q.5: If Q and R are events such that P(Q) = 611 , P(R) = 511, P(Q∪R = 711. Determine the value of
(a) P(Q∩R)
(b) P(QR)
(c) P(RQ)
Sol:
According to the Q., it is given that:
P(Q) = 611 , P(R) = 511 & P(Q∩R) = 711
(a) P(Q∪R)=P(Q)+P(R)–P(Q∩R)
= 711 = 611 + 511 – P(Q∩R)
= P(Q∩R) = 1111 –711
= P(Q∩R) = 411
(b) P(QR) = P(Q∩R)P(R)
= 411511 = 45
(c) P(RQ) = P(Q∩R)P(Q)
= 411611 = 23
Q.6: An experiment consists of tossing up of a coin three times. Determine the following:
(a) Q: obtaining heads on third toss & R: obtaining heads from the first consecutive two tosses.
(b) Q: obtaining at least two heads & R: obtaining at most two heads.
(c) Q: obtaining at most two tails & R: obtaining at most one tail.
Sol:
According to the Q., a coin is tossed thrice in an order to conduct an experiment.
Sample space (S) = {TTT , TTH ,THT , THH , HTT , HTH , HHT , HHH }
(a) Number of favourable outcomes of event Q = { TTH , THH ,HTH , HHH }
Number of favourable outcomes of event F = {HHT , HHH}
Q∩R = { HHH }
P(R) = 28=14
P(QR) = P(Q∩R)P(R) = 1814 = 12
(b) Number of favourable outcomes of event Q = { THH ,THH ,HHT , HHH }
Number of favourable outcomes of event F = {TTT , TTH , THT , THT , THH , HTT , HTH , HHT}
Q∩R = { HHT , HTH , HHT }
P(R) = 78
P(QR) = P(Q∩R)P(R) = 3878 = 37
(c) Number of favourable outcomes of event Q = {TTH , THT , THH , HTH , HTT ,HHT , HHH }
Number of favourable outcomes of event F = {TTT , TTH , THT , THT , THH , HTT , HHT}
Q∩R = {TTH , THT , THH , HTH , HTT , HTT , HHT }
P(R) = 78
P(QR) = P(Q∩R)P(R) = 6878 = 67
Q.7: An experiment consists of tossing up of a coin where:
(a) Q: obtaining tail on one coin & R: head appears in one coin
(b) Q: tail does not appear & R: head does not appears
Sol:
An experiment consists of tossing up of a coin only once.
Sample space (S) = {TT, TH, HT, HH}
(a) Number of favourable outcomes of event Q = {TH, HT}
Number of favourable outcomes of event F = {TH, HT}
Q∩R = { TH , HT }
P(R) = 28=28=14
P(QR) = P(Q∩R)P(R)
= 22 = 1
(b) Number of favourable outcomes of event Q = { HH }
Number of favourable outcomes of event F = {TT }
Q∩R = { ϕ }
P(R) = 1 =
P(QR) = P(Q∩R)P(R) = 01 = 0
Q.8: An experiment consists of throwing up of a dice three times .Find:
Q: the number 4 appears during the third toss
R: 6 & 5 appears consecutively during the first two tosses
Sol:
According to the Q., it is given that a die is being tossed thrice.
Total number of elements in the sample space = 216
Favorable outcomes of event Q =
{ (6,1,4) , (6,2,4) , (6,3,4) , (6,4,4) , (6,5,4) , (6,6,4)
(5,1,4) , (5,2,4) , (5,3,4) , (5,4,4) , (5,5,4) , (5,6,4)
(4,1,4) , (4,2,4) , (4,3,4) , (4,4,4) , (4,5,4) , (4,6,4)
(3,1,4) , (3,2,4) , (3,3,4) , (3,4,4) , (3,5,4) , (3,6,4)
(2,1,4) , (2,2,4) , (2,3,4) , (2,4,4) , (2,5,4) ,(2,6,4)
(1,1,4) , (1,2,4) , (1,3,4) , (1,4,4) , (1,5,4) , (1,6,4) }
Favorable outcomes of event R = { (6,5,6) , (6,5,5) , (6,5,4) , (6,5,3) , (6,5,2) , (6,5,1) }
Q∩R = { (6,5,4) }
P(R) = 6216
Q∩R = 1216
P(QR) = P(Q∩R)P(R) = 12166216 = 16
Q.9: An experiment consists of taking into account the line up all the members in a family namely the father, mother and the child.
Q: the son is placed in one of the end
R: the position of the father is in the middle
Sol:
An experiment consists of taking into account all the members of a family namely the father , mother and the child.
Let us consider the father , mother and the child are denoted by F , M and S respectively.
Sample space = { SFM , SMF , FSM , FMS , MSF , MFS }
Favourable outcomes of event Q = { SFM , SMF , FMS , MFS }
Favourable outcomes of event R = { SFM , MFS }
P(R) = 26=13
Q∩R = 26=13
P(QR) = P(Q∩R)P(R) = 1313 = 1
Q.10: An experiment consists of a black and a red die which are rolled.
(a) Determine the conditional probability of getting a sum of numbers greater than 9 such that the black die results in a 5
(b) Determine the conditional probability of getting a sum of numbers 8 such that red die results in numbers less than 4
Sol:
According to the Q., it is given that the a black and a red die are rolled
Total number of elements in the sample space = 36
(a) Favourable outcomes of event Q = { (4,6) , (5,5) , (5,6) , (6,4) ,(6,5) ,(6,6) }
Favourable outcomes of event R = { (5,1) , (5,2) , (5,3) , (5,4) , (5,5) , (5,6) }
Q∩R = { (5,6) , (5,5) }
Conditional probability of getting a sum greater than 9 , so that the black die results in 5 .
P (QR) = P(Q∩R)P(R)
= 2361836 = 19
Q.11: An experiment consists of rolling up of a dice including events such as Q = { 5,3,1 } , R = { 3,2} , S = { 5,4,3,2}. Determine:
(a) P(QR) and P(RQ)
(b) P(QS) and P(SQ)
(c) P(Q∪R)S and P(Q∪S)S
Sol:
According to the Q., it is given that a die is rolled in which events namely Q, R and S are recorded.
Sample space = { 6 , 5 , 4 , 3 , 2 , 1}
Favourable outcomes of event Q = { 5 , 3 , 1 }
Favourable outcomes of event R = { 3 , 2 }
Favourable outcomes of event S = { 5 , 4 , 3 , 2 }
Therefore , P(Q) = 36=12
P(R) = 26=13
P(S) = 46=23
(a) Q∩R = {3}
P (Q∩R) = 16
P(QR)=Q∩RR=1613=12 P(RQ)=Q∩RQ=1612=13
(b) Q∩S = {5 , 3}
P (Q∩R) = 16
P(QS)=Q∩SS=1323=12 P(SQ)=Q∩SQ=1312=23
(c) Q∪R = {5 , 3 , 2 , 1}
(Q∪R)∩S=5,3,2,1∩5,4,3,2 = {5,3,2}
Q∩R = {3}
(Q∩R)∩S=3∩5,4,3,2 = { 3 }
P(Q∪S)=46=23 P[(Q∪R)∩S]=36=12 P(Q∩R)=16 P[(Q∪R)S]=P[(Q∪R)∩S]P(S)=1223=34 P[(Q∩R)S]=P[(Q∩S)∩S]P(S)=1623=14
Q.12: Assume that each of the children born in a family can either be a boy or a girl. If a family having 2 children, determine the conditional probability that both of them are girls given:
(a) The youngest child is a girl
(b) At least one is a girl.
Sol:
According to the Q., it is given that a family is having 2 children where both of them are girls.
Let us represent boy and the girl child with the letter (b) and (g) respectively.
Sample space = {(g, g), (g, b), (b, g), (b, b)}
Let us consider Q be the event which indicates that both of the child born to a family are girls.
Q = {(g, g)}
(a) Let us consider R be the event that the youngest child born in the family is a girl.
R = {(g, g), (b, g)}
Q∩R=(g,g)
P (Q) = 24=12
P (Q∩R=14
According to the Q., both of the children are girls and the youngest being a girl child.
P (QR)=P(Q∩R)P(R)=1412=12
The required probability is 12
(b) Let us consider S be the event that at least one child born in the family is a girl.
S = {(g, g), (b, g), (g, b)}
Q∩S=(g,g)
P(S) = 34
P (Q∩S=14
According to the Q., both of the children are girls and the youngest being a girl child.
P (QS)=P(Q∩S)P(S)=1434=13
The required probability is 13
Q.13: A Q. bank consisting of 500 easy multiple choice Q.s , 400 difficult multiple choice Q.s , 300 easy False / True Q.s & 200 difficult False / True Q.s . Determine the probability that a Q. chosen from this Q. bank will be an easy multiple choice Q…
Sol:
| False / True | Multiple choice | Sum total | |
| Difficult | 200 | 400 | 600 |
| Easy | 300 | 500 | 800 |
| Sum total | 500 | 900 | 1400 |
Let us consider:
Easy Q. = E
Difficult Q. = D
Multiple choice Q. = M
False / True Q. = T
Sum total of all Q.s in the Q. ban k = 1400
Number of multiple choice Q.s = 900
Number of False / True Q.s = 500
Probability of getting an easy multiple choice Q. in the Q. bank =
P (E∩M)=5001400=514
Probability of selecting a multiple choice Q. be it easy or difficult
P (M) = 9001400=914
The conditional probability of getting an easy multiple choice Q. from the Q. bank = P (EM)=P(E∩M)P(M)=5494=59
The required probability is 59
Q.14: On rolling two dice simultaneously two different numbers appears on both the faces of the dice. Find the probability that the sum of two different numbers is 4.
Sol:
Total number of elements in the sample space = 36
Let Q be the event that the sum of two different numbers is 4 & R be the event that two numbers appearing on both the faces of the dice are different.
Q = {(3, 1), (2, 2), (1, 3)}
R = { (6 , 1) , (6 , 2) , (6 , 3) , (6 , 4) , (6 , 5) , (6 , 6)
(5 , 1) , (5 , 2) , (5 , 3) , (5 , 4) , (5 , 5) , (5 , 6)
(4 , 1) , (4 , 2) , (4 , 3) , (4 , 4) , (4 , 5) , (4 , 6)
(3 , 1) , (3 , 2) , (3 , 3) , (3 , 4) , (3 , 5) , (3 , 6)
(2 , 1) , (2 , 2) , (2 , 3) , (2 , 4) , (2 , 5) , (2 , 6)
(1 , 2) , (1 , 3) , (1 , 4) , (1 , 5) , (1 , 6) }
Q∩R=(1,3),(3,1))
P(R) = 3036=56
Q∩R=236=118
Let P(QR) = P(Q∩R)P(R) = 11856 = 115
The required probability is 115
Q.15: A die is rolled and if any number multiples of 3 come up then it is rolled again. This time if any other number appears then a coin is tossed. Determine the conditional probability of the event ‘tail appears ‘given that ‘at least one die shows a 3 ‘.
Sol:
Sample space of the above conducted experiment = { (6 , 6) , (6 , 5) , (6 , 4) , (6 , 3) , (6 , 2) , (6 , 1) , (5 , T) , (5 , H) , (4 , T) , (4 , H) , (3, 6) , (3 , 5) ,(3 , 4) , (3 , 3) , (3 , 2) , (3 , 1) , (2 , T) , (2 , H) , (1, T) , (1, H) }
Let Q be the event that a tail appears
R be the event at least one die shows 3
Q = {(1, T), (2, T), (4, T), (5, T)}
R = { (3, 6) , (3 , 5) ,(3 , 4) , (3 , 3) , (3 , 2) , (3 , 1) , (6 , 3) }
Q∩R=ϕ P(Q∩R)=0
Then,
P (R) ={ { P (3, 6) , P (3 , 5) ,P (3 , 4) , P (3 , 3) , P (3 , 2) , P (3 , 1) , P (6 , 3) }}
= 136 + 136 + 136 + 136 + 136 + 136 + 136 = 736
Probability that the die shows a tail given that at least one die shows 3
= P (QR) = 0736 = 0
Q.16: If P (Q) = 12
P (R) = 0,
Then P (QR) =?
(i) 0 (ii) 12 (iii) not defined (iv) 1
Sol:
It is given that:
P (Q) = 12
P (R) = 0
P (QR)
= Q∩R0
= not defined
Hence, the correct answer is C
Q.17: If Q and Rare events such that P( QR) = P( RQ) , then :
(a) Q⊂R , Q ≠ R
(b) Q = R
(c) Q∩R=ϕ
(d) P (Q) = P(R)
Sol:
According to the Q., it is given that:
P( QR) = P( RQ)
= P(Q∩R)P(R) = P(Q∩R)P(Q)
= P (Q) = P(R)
The correct answer is D
Exercise – 13.2
Q.1: If P (Q) = 35 & P (R) = 15 . Considering events Q and R are independent.
Sol:
Given, P (Q) = 35 & P (R) = 15
Q and R are the events which are independent in nature.
P(Q capR)=P(Q)×P(R)
= 35×15 = 325
Q.2: 2 cards are drawn at random from a well shuffled deck of 52 cards. Determine the probability that the drawn cards are black.
Sol:
In a well shuffled deck of 52 cards there are only 26 black cards (13 spade cards and 13 clubs cards)
Let Q be the event black card is drawn from a deck of cards.
P (Q) = 2652=12
Let R be the probability a black card on the second draw.
P (R) = 2552
Thus, the probability of getting both black cards = (2652)(2552) = 25102
Q.3: A bag full of oranges is inspected by selecting 3 random oranges without replacement. If it is seen that all the 3 drawn oranges are good then the box is sent for sale, otherwise it is sent for rejection. Determine the probability if a box containing 15 oranges out of which 12 of which are good and rest are bad will be approved for sale or not?
Sol:
Let us consider events Q, R and S be the events of drawing first, second and third drawing of orange.
P (Q) be the event of first drawing of orange = 1215
P(R) be the event of second drawing of orange = 1114
P(S) be the event of third drawing of orange = 1013
Thus , getting the probability of all good apples = (1215)(1114)(1013)
= 4491
Thus, the probability of getting approved for sale is 4491
Q.4: An unbiased coin is tossed. An EVENT Q and R be the head appears on the coin & 3 appears on the face of the die. Find out whether Q and R are independent events or not?
Sol:
Sample space = { (T, 6), (T, 5), (T, 4), (T, 3), (T, 2), (T, 1), (H, 6), (H, 5), (H, 4), (H, 3), (H, 2), (H, 1) }
Let Q be the event of head appearing on the face of the coin
Q = { (H , 6) , (H, 5) , (H , 4) , (H , 3) , (H , 2) , (H , 1) }
P (Q) = 612=12
Let R be the event that the number 3 appears on the face of the die
R = {(T, 3), (H, 3)}
P (R) = 212=16
Therefore,
Q∩R = {(H, 3)}
P (Q∩R) = 112
P(Q)×P(R) = 12×16 = P(Q)∩P(R)
Therefore, Q & R are independent events.
Q.5: An experiment consists of marking a die 1, 2, 3 in red and 4, 5, 6 in green and then rolled. Let Q be the event that the number is even and R be the event that color of the face of the die is red. Determine the probability that both of the events are independent.
Sol:
Sample space (S) when a die is rolled = {6, 5, 4, 3, 2, 1}
According to the Q., it is given that:
Q = Event that the number appears is even
R = the colour of the face of the die is red.
Q = {6, 4, 2}
P (Q) = 36=12
R = {3, 2, 1}
P (R) = 36=12
Therefore, Q∩R=2
P(QR)=P(Q∩R)=16 P(Q)⋅P(R)=12×12=14≠16
Hence, events Q and R are not independent events.
Q.6: Q and R be the events such that P (Q) = 35 , P (R) = 310 & Q∩R=15 . Determine whether events Q and R are independent or not /
Sol:
According to the Q., it is given that:
P (Q) = 35 , P (R) = 310 & Q∩R=15 .
P(Q)⋅P(R)=35×310=950≠15
Hence, events Q and R are not independent events.
Q.7: Q and R be the events such that P (Q) = 12 , Q∩R=35 , P (R) = p . Determine the value of p if Q and R are:
(a) Mutually exclusive
(b) Independent.
Sol:
(a) Let us consider Q and R are mutually exclusive.
Q∩R=ϕ, P (R) = p
P(Q∩R)=0
We know, that:
P(Q∩R)=P(Q)+P(R)–P(Q∪R)
= 35=12+p–0
= p = 110
(b) Considering events Q and R are independent events.
P(Q∩R)=P(Q)×P(R)=12p P(Q∩R)=P(Q)+P(R)–P(Q∪R)
= 35=12+p–12p
= p = 210=15
Q.8: Let Q and R be the independent events such that P (Q) = 0.3, P (R) = 0.4
Determine:
(a) P (P(Q∩R))
(b) P (P(Q∪R))
(c) P(QR)
(d) P(RQ)
Sol:
(a) Considering the events Q and R independent events,
P(Q∩R)=P(Q)⋅P(R)=(0.3)(0.4)=0.12
(b) P(Q∩R)=P(Q)+P(R)–P(Q∪R)
P (P(Q∪R)) = 0.3 + 0.4 – 0.12 = 0.58
(c) P(QR)=P(Q∩R)P(R)
= P(QR)=0.120.4 = 0.3
(d) P(RQ)=P(Q∩R)P(Q)
= P(RQ)=0.120.3 = 0.4
Q.9: Let Q and R be the two events such that P(Q)=12, P(R)=12, P(Q∩R)=18. Determine the value of P (neither Q nor R)
Sol:
According to the Q., it is given that:
P(Q)=12, P(Q∩R)=18, P(R)=12
P (neither Q or R) = P (Q′∩R′)
P (neither Q or R) = P (Q∩R)′
= 1 – P(Q∩R)
= 1 – P(Q)+P(R)–P(Q∩R)
= 1 – 14+12–18
= 1 – 58 = 38
Q.10: Q and R are events such that P(Q) = 12 , P(R) = 12 & P(neither Q nor R) = 14
Sol:
P(Q′∪R′)=14
= P((Q∩R)′)=14
= 1 – P(Q∩R)=14
= P(Q∩R)=34
= P(Q∩R)=P(Q)⋅P(R)=(12)(712)=724
34≠724
Hence, Q & R are not independent events.
Q.11: Q and R are two independent events such that P (Q) 0.3 & P(R) = 0.6. Determine:
(a) P (Q and R)
(b) P (Q and not R)
(c) P (Q or R)
(d) P (neither Q nor R)
Sol:
(a) P (Q and R) = P(Q∩R)=P(Q)⋅P(R)=(0.3)(0.6)=0.18
(b) P (Q and not R)
= P(Q∩R′)
= P(Q)–P(Q∩R)
= 0.3 – 0.18 = 0.12
(c) P (Q or R)
= (Q∩R)
= P(Q)+P(R)–P(Q∩R)
= 0.3 + 0.6 – 0.18 = 0.72
(d) P (neither Q nor R)
= P(Q′∩R′)
= P((Q∩R)′)
= 1 – P(Q∩R)
= 1 – 0.72 = 0.28
Q.12: An experiment consists of tossing up of die thrice. Determine the probability of getting an odd number at least once.
Sol:
Probability of getting an odd number in a single throw = 36=12
Probability of getting an even number = 36=12
Probability of getting an even number = 12×12×12 = 18
According to the Q.,
Determining the probability of getting an odd number at least once
= 1 – Probability of getting an odd number
= 1 – probability of getting an even number thrice
= 1 – 18 = 78
Q.13: From a bag of 10 black balls and 8 red balls 2 balls are drawn at a random without random. Determine the probability:
(a) Both the drawn balls are red
(b) First and the second drawn ball are black & red respectively.
(c) One of them is black and the other one is red.
Sol:
According to the Q., it is given that:
The bag contains a total number of number balls of = 18
No. of red balls = 8
No. of black balls = 10
(a) Probability that a red ball appears in the first draw = 818=419
The ball so obtained is replaced.
Probability that a red ball appears in the second draw = 818=419
Probability that both the red balls appear = 49×49=1681
(b) Probability that a black ball appears in the first draw = 1018=519
The ball so obtained is replaced.
Probability that a black ball appears in the second draw = 818=419
Probability that both the black balls appear = 59×49=2081
(c) Probability that a red ball appears in the first draw = 818=419
The ball so obtained is replaced.
Probability that a black ball appears in the second draw = 818=59
Probability that both the red balls appear = 49×59=2081
Hence, the probability of getting one black and one red ball =
Probability of getting first red ball and then black ball + Probability of getting first black ball and then red ball
= 2081+2081 = 4081
Q.14: The probability of solving particular problems by methods Q and R are 12 and 13 respectively. Considering that the problem is solved independently, determine the probability:
(a) The problem is solved
(b) At least one of them solves the problem correctly.
Sol:
Probability that the problem is solved by Q, P (Q) = 12
Probability that the problem is solved by R, P (R) = 13
According to the Q., it is given that both problems are solved independently by Q and R:
= P (QR) = P(Q)×P(R)=12×13=16
P (Q’) = 1 – P (Q) = 1 – 12=12
P (R’) = 1 – P(R) = 1 – 13=23
(a) Probability that the problem:
= P(Q∩R)
= P (Q) + P(R) – P (QR)
= 12+13–16 = 23
(b) Probability that exactly one solves the above mentioned problem:
= P(A)×P(B′)+P(B)×P(A′)
= (12×23)+(12×13)
= 13+16
= 12
Q.15: An experiment consists of drawing up of a card at random from a well shuffled deck of 52 cards. Events Q and R associated with the following experiment are independent in nature
(a) Q: the drawn card is a spade
R: the drawn card is an ace
(b) Q: it is a black card
R: the drawn card is a king
(c) Q: the drawn card can either be a king or queen
R: the drawn card is either a queen or jack.
Sol:
(a) Total number of spades = 13
Total number of ace cards in a pack of cards = 4
Probability of drawing a spade card P (Q) = 1352=14
Probability of drawing an ace card P(F) = 14
In a pack of cards, there is only card which is an ace of spades. Hence probability of getting an ace of spades = 152
= P(Q)×P(R)
= 14×113 = P (QR)
P (QR) = P(Q)×P(R)
Hence, events Q and R are independent in nature.
(b) In a pack of 52 cards, there are 26 black cards and 4 cards which are kings.
P (Q) = P (the drawn card is black) = 2652=12
P (R) = P (the drawn card is a king) = 452=113
In a pack of 52 cards, there are 2 black cards which are kings.
P (QR) = P (the drawn card is black in colour as well as a king) = 252=126
= P(Q)×P(R)
= 12×113=126 = P (QR)
Hence, events Q and R are independent in nature.
(c) In a pack of 52 cards, there are 4 kings, 4 jacks and 4 queens.
P (Q) = P (the drawn card can either be a king or queen) = 852=213
P (R) = P (the drawn card is a queen or a jack) = 852=213
There are 4 cards which are king or queen and queen or jack = 452=113
= P(Q)×P(R)
= 213×213≠4169 = P (QR)
Hence, events Q and R are not independent in nature.
Q.16: Among the students residing in a hostel, 60% of them read Hindi newspaper, 20% of them read both Hindi and English newspapers. A student is selected as a random:
(a) Determine the probability the students reads neither Hindi nor English newspapers.
(b) Determine the probability that the student reads Hindi newspapers, but also reads English newspaper too.
(c) If the students read English newspapers, find the probability that they reads Hindi newspaper too.
Sol:
According to the Q. it is given that Q be the event who reads Hindi newspapers and R be the event who reads English newspapers.
P (Q) = 60% = 60100=35
P (R) = 40% = 40100=25
P (Q∩R) = 20% = 20100=15
(a) Probability that the student reads Hindi or English newspapers:
= (Q∪R)′
= 1 – (Q∪R)
= 1 – { P(Q) + P(R) – (Q∪R) }
= 1 – { 35 + 25 – 15 }
= 1 – 45 = 15
(b) Probability that a guy chose to read English newspaper , if they already reads Hindi newspaper = P(QR)
= Q∩RR = 1535 = 13
(c) Probability that a guy chose to read Hindi newspaper, if they already reads English newspaper = P(RQ)
= Q∩RQ = 1525 = 12
Q.17: Find the probability of obtaining an even number, when two dice are rolled at once:
(i) 0
(ii) 13
(iii) 112
(iv) 136
Sol:
Total number of outcomes when two dice are rolled at once = 36
2 is the only even prime number.
Let Q be the event of getting an even prime number.
Therefore, Q = {2, 2}
P (Q) = 136
Hence, the correct answer to the above mentioned Q. is (iv)
Q.18: 2 events Q and R will be independent, such as:
(i) Q and R are mutually exclusive.
(ii) P (Q’R’) = [1 – P (Q) ][1 – P(R) ]
(iii) P(Q) = P (R)
(iv) P (Q) + P (R) = 1
Sol:
2 events Q and R are said to be independent, if
P (QR) = P(Q)×P(R)
= P (Q’R’) = [ 1 – P(Q)][1 – P(R)]
= P(Q′∩R′) = 1–P(Q)–P(R)+P(Q)⋅P(R)
= 1- P(Q′∩R′) = P(Q)–P(R)+P(Q)⋅P(R)
= P(Q∪R) = P(Q)–P(R)+P(Q)⋅P(R)
= P (Q) + P (R) – P (QR) = P(Q)–P(R)+P(Q)⋅P(R)
= P (QR) = P(R)+P(Q)⋅P(R)
This implies that event Q and R are independent event,
P (Q’R’) = [1 – P(Q)][1 – P(R)]
Let P (Q) = a
P (R) = b
0 ˂ a, b ˂ 1
Q and R are both mutually exclusive events.
Therefore,
= P(Q∪R)=ϕ
= P (QR) = 0
Therefore, P(Q) × P (R) = ab ≠ 0
Therefore, P(Q)×P(R)≠P(QR)
(ii) Event of getting an even number = {6, 4, 2}
P (R) = 36 = 12
= P(Q∪R)=ϕ
= P(Q)×P(R)=14≠0
(iii) Let Q be the event of getting a odd number = {5 , 3, 1 }
P (Q) = 36 = 12
(iv) P(Q) + P(R) = 12 + 12 = 1
From the above solution it cannot be concluded that Q and R are independent events.
Therefore, the correct answer to the above mentioned Q. is (ii).
Exercise – 13.3
Q.1: An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red?
Answer I:
The urn contains 5 red and 5 black balls.
Let a red ball be drawn in the first attempt.
So, P (drawing a red ball) = 5/10 = 1/2
If two red balls are added to the urn, then the urn contains 7 red and 5 black balls.
P (drawing a red ball) = 7/12
Let a black ball be drawn in the first attempt.
P (drawing a black ball in the first attempt) = 5/10 = 1/ 2
If two black balls are added to the urn, then the urn contains 5 red and 7 black balls.
P (drawing a red ball) = 5/12
Therefore, probability of drawing second ball as red is 12×712+12×512=12[712+512]=12×1=12
- 2: A bag contains 4 red and 4 black balls; another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.
Answer 2:
Let E1 and E2 be the events of selecting first bag and second bag respectively.
P(E1) = P(E2) = 1/2
Let A be the met of getting a red ball
P(A
E1) = P (drawing a red ball from first bag) = 4/8 = 1/2
P(AlE2) = P(drawing a red ball from second bag) = 2/8 = 1/4
The probability of drawing a ball from the first bag, given that it is red, is given by P(E2
A)
By using Bayes’ theorem, we obtain
P(E1
A) = P(E1).P(A
E2)P(E1).P(A
E1)+P(E2).O(A
E2)
= 12×1212×12+12×14
= 1414+18 = 1438 = 23
Q.3: Of the students in a college, it is known that 6 0% reside in hostel and 40% are day scholars (not residing in hostel) Previous year results report that 30% of all students who reside in hostel attain A grade and 20% of day scholars attain A grade in their annual examination At the end of the year, one student is chosen at random from the college and he has an A grade, what is the probability that the student is hostler?
Answer 3:
Let E1 and E2 be the events that the student is a hostler and a day scholar respectively and A be the event that the chosen student gets grade A.
P(E1) = 60% = 0.6
P(E2) = 40% = 0.4
P(A
E1))= P(student getting an A grade is a hostler)= 30% = 0.3
P(A
E2)= P(student getting an A grade is a day scholar)= 20%= 0.2
The probability that a randomly chosen student is a hostler, given that he has an A grade, is given by P(E1
A)
By using Bayes’ theorem, we obtain
P(E1
A)=P(E1.P(A
E1)P(E!.P(A
E1)+P(E2).P(A
E2)
= 0.6×0.30.6×0.3+0.4×0.2
= 0.180.26 = 1826 = 913
Q.4: In answering a Q. on a multiple choice test, a student either knows the answer or guesses. Let 3/4 be the probability that he knows the answer and ¼ be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability 1/4. What is the probability that the student knows the answer given that he answered it correctly?
Answer 4:
Let E1 and E2 be the respective events that the student knows the answer and he guesses the answer.
Let A be the event that the answer is correct.
P(E1) = 3/4
P(E2) = 1/4
The probability that the student answered correctly, given that he knows the answer, is 1
P(A
E1) = 1
Probability that the student answered correctly, given that he guessed, is 1/4.
P(A
E2) = 1/4
The probability that the student knows the answer, given that he answered it correctly, is given by P(E1
A)
By using Bayes’ theorem, we obtain
P(E1
A)=P(E1.P(A
E1)P(E!.P(A
E1)+P(E2).P(A
E2)
= 34×134×1+1414
= 3434+116 = 341316 = 1213
Q.5: A laboratory blood test is 99% effective in detecting a certain disease when it is in fact, present. However, the test also yields a false positive result for 0 5% of the healthy person tested (that is, if a healthy person is tested, then, with probability 0.005, the test will imply he has the disease) If 0.1 percent of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive?
Answer 5:
Let E1 and E2 be the respective events that a person has a disease and a person has no disease.
Since E1 and E2 are events complimentary to each other,
P (E1) + P (E2) = 1
P(E2) = 1 -P(E1)= 1 – 0.001= 0.999
Let A be the event that the blood test result is positive
P(E1) = 0.1% = 1/100= 0.001
P(AlE1) = P(result is positive given the person has disease) = 99% = 0.99
P(AlE2) = P(result is positive given that the person has no disease) = 0.5% = 0.005
Probability that a person has a disease, given that his test result is positive, is given by
P(E1
A)
By using Bayes’ theorem, we obtain
P(E1
A)=P(E1.P(A
E1)P(E!.P(A
E1)+P(E2).P(A
E2)
= 0.001×0.990.001×0.99+0.000×0.005
= 0.000990.005985 = 110665 = 22133
Q.6: There are three coins. One is two headed coin (having head on both faces), another is a biased coin that comes up heads 73% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin?
Answer 6:
Let E1, E2, and E3 be the respective events of choosing a two headed coin, a biased coin, and an unbiased coin
P (E1) + P (E2) + P (E3) = 1/3.
Let A be the event that the coin shows heads.
A two-headed coin will always show heads
P (AlE1) = P(coin showing heads, given that it is a two-headed coin) = 1
Probability of heads coming up, given that it is a biased coin= 75%
P(AlE2)= P(coin showing heads. given that it is a biased coin) = 3/4
Since the third coin is unbiased, the probability that it shows heads is always 1/2
P(AlE3) = P(coin showing heads, given that it is an unbiased coin) = 1/2.
The probability that the coin is two-headed, given that it shows heads, is given by
P(E1
A)
By using Bayes’ theorem, we obtain
P(E1
A)=P(E1.P(A
E1)P(E!.P(A
E1)+P(E2).P(A
E2)
= 13×113×1+1334+13×12
= 1313(1+34+12
= 194 = 49
Q.7: An insurance company insured 2000 scooter drivers, 4000 Car drivers and 6000 truck drivers. The probability of accidents are 0 01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?
Answer 7:
Let E1, E2, and E3 be the respective events that the driver is a scooter driver, a car driver, and a truck driver.
Let A be the event that the person meets with an accident.
There are 2000 scooter drivers, 4000 car drivers, and 6000 truck drivers.
Total number of drivers = 2000 + 4000 + 6000 = 12000
P (E1) = P (driver is a scooter driver) = 2000/ I2000 = 1/6
P (E2) = P (driver is a car driver) = 4000/12000 = 1/3
P (E3) = P (driver is a truck driver) = 6000/1 2000 = 1/2
P(AlE1) = P(scooter driver met with an accident) = 0.01= 1/100
P(A
E2)= P (car driver met with an accident) = 0.03 = 3/100
P(A
E3) = P(truck driver met with an accident) = 0.15 = 15/100
The probability that the coin is two-headed, given that it shows heads, is given by P(E1
A)
By using Bayes’ theorem, we obtain
P(E1
A)=P(E1.P(A
E1)P(E1.P(A
E1)+P(E2).P(A
E2)+P(E3).P(A
E3)
= 16×110016×1100+133100+12×15100
= 161100110016+1+152
= 1610412 = 152
Q.8: A factory has two machines A and B. Past record shows that machine A produced 60% of the Items of output and machine B produced 40% of the items. Further, 2% of the items produced by machine A and 1% produced by machine B were defective. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that was produced by machine B?
Answer 8:
Let E1 and E2 be the respective events of items produced by machines A and B. Let X be the event that the produced item was found to be defective.
Probability of items produced by machine A, P (E1) = 60% = 3/5
Probability of items produced by machine B, P (E2) = 40% = 2/5
Probability that machine A produced defective items, P (X
E1) = 2% = 2/100
Probability that machine B produced defective items. P (X
E2) = 1% = 1/100
The probability that the randomly selected item was from machine B, given that it is defective, is given by P (E2
X)
By using Bayes’ theorem, we obtain
P(E1
A)=P(E2.P(X
E2)P(E1.P(X
E1)+P(E2).P(X
E2)
= 25×110035×2100+251100
= 25006500+2500
= 28 = 14
Q.9: Two groups are competing for the position on the board of directors of a corporation. The probabilities that the first and the second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability is 0 3 if the second group wins. Find the probability that the new product introduced was by the second group.
Answer 9:
L.et E1 and E2 be the respective events that the first group and the second group win the competition. Let A be the event of introducing a new product.
P (E1) = Probability that the first group wins the competition = 0.6
P (E2) = Probability that the second group wins the competition = 0.4
P (A
E1) = Probability of introducing a new product if the first group wins =0.7
P(A
E2) = Probability of introducing a new product if the second group wins =0.3
The probability that the new product is introduced by the second group is given by P(E2
A)
By using Hayes’ theorem, we obtain
P(E2
A)=P(E2.P(A
E2)P(E1.P(A
E1)+P(E2).P(A
E2)
= 0.4×0.330.6×0.7+0.4×0.3
= 0.120.42+0.12 = 0.1254 = 29
Q.10: Suppose a girl throws a die if she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1, 2, 3 or 4, she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1, 2, 3 or 4 with the die?
Answer 10:
Let E1 be the event that the outcome on the die is 5 or 6 and E2 be the event that the outcome on the die is 1, 2, 3, or 4.
P(E1) = 2/6 = 1/3 and P(E2)= 4/6 = 2/3.
Let A be the event of getting exactly one head.
P (A
E1) =Probability of getting exactly one head by tossing the coin three times if she gets 5 or 6 = 3/8
P (A
E2) = Probability of getting exactly one head in a single throw of coin if she gets 1, 2, 3, or 4 =1/2.
The probability that the girl threw 1, 2, 3, or 4 with the die, if she obtained exactly one head, is given by P (E2
A)
By using Bayes’ theorem, we obtain
P(E2
A)=P(E1.P(A
E1)P(E!.P(A
E1)+P(E2).P(A
E2)
= 23×1213×38+2312
= 1313(1+38
= 1118 = 811
Q.11: A manufacturer has thee machine operators A, B and C. The first operator A produces 1% defective items, whereas the other two operators B and C produce 5% and 7% defective items respectively. A is on the Job for 50% of the time, B is on the job for 30% of the time and C is on the Job for 20% of the time A defective item is produced. What is the probability that was produced by A?
Answer 11:
Let E1, E2, and E3 be the respective events of the time consumed by machines A, B, and C for the job
P(E1)= 50% = 50/100 = 1/2
P(E2) = 30% = 30/100 = 3/10
P(E3) = 20% = 20/100 = 1/5
Let A be the event of getting exactly one head.
P (A
E1) =1% = 1/100
P (A
E2) = 5% = 5/100
P (A
E3) = 7% = 7/100
The probability that the defective item was produced by A is given by P (E1
A)
By using Bayes’ theorem, we obtain
P(E1
A)=P(E1.P(A
E1)P(E1.P(A
E1)+P(E2).P(A
E2)+P(E3).P(A
E3)
= 12×110012×1100+3105100+15×7100
= 1100121100(12+32+75
= 12175 = 534
Q.12: A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.
Answer 12:
Let E1 and E2 be the respective events of choosing a diamond card and a card which is not diamond.
Let A denote the lost card
Out of 52 cards, 13 cards are diamond and 39 cards are not diamond
P(E1) = 13/52 = 1/4
P(E2) = 39/52 = 3/4
When one diamond card is lost, there we 12 diamond cards out of 51 cards
Two cards can be drawn out of 12 diamond cards in 12C2 ways.
Similarly, 2 diamond cards can be drawn out of 51 cards in 51C2 ways. The probability of getting two cards, when one diamond card is lost, is given by P(A
E1).
P(A
E1)=12C251C2=12!2!×10!×2!×49!50×51!=11×1250×51=22425
When the lost card is not a diamond, there are 13 diamond cards out of 51 cards.
Two cards can be drawn out of 13 diamond cards in 13C2 ways whereas 2 cards can be drawn out of 51 cards in 51C2 ways.
The probability of getting two cards, when one card is lost which is not diamond, is given by P (E1
A)
By using Bayes’ theorem, we obtain
P(E1
A)=P(E1.P(A
E1)P(E1.P(A
E1)+P(E2).P(A
E2)
= 14×2242514×22425+3426425
= 14252242514(22425)+3426425
= 11225 = 1150
Q.13: Probability that A speaks troth is 4/5. A coin is tossed a reports that a head appears. The probability that actually there was head 5 is:
(A) 4/5
(B) ½
(C) 1/5
(D) 2/5
Answer 13:
Let E1 and E2 be the events such that
E1: A speaks truth
E2: A speaks false
Let X be the event that a head appears
P (E1) = 4/5
P(E2) =
1 – 4
E1) = P(X
E2)= 1/2
The probability that there is actually a head is given by P (E1
X)
P(E1
X)=P(E1.P(X
E1)P(E1.P(X
E1)+P(E2).P(X
E2)
= 45×1245×12+1512
= 124512(45+15) = 451 = 45
Therefore, the correct answer is A.
Q.14; If A and B are two events such that A⊂BandP(B)≠0, then which of the following is correct?
(A) P(A
B)=P(B)P(A)
(B) P(A
B) < P(A)
(C) P(A
B) >= P(A)
(D) None of these
If A⊂B,then,A∩B=A.
⇒P(A∩B)=P(A)
Also, P(A) < P(B)
Consider, P(A
B)=P(A∩B)P(B)=P(a)P(B)≠P(B)P(A) . . . . (1)
Consider, P(A
B)=P(A∩B)P(B)=P(A)P(B) . . . . . . . . . (2)
It is known that P(B) <= 1
⇒1P(B)≥1 ⇒P(A)P(B)≥P(A)
From (2), we obtain
P(A
B)≥P(A) . . . . . . . . . (3)
Hence, P(A
B) is not less than P(A)
Thus, from (3), it can be concluded that the relation given in alternative C is correct.
Exercise – 13.4
Q-1: Check whether the following are the probability distributions of a random variable or not. Also, justify your answer.
(i)
| X | 0 | 1 | 2 |
| P(x) | 0.3 | 0.5 | 0.2 |
(ii)
| X | 0 | 1 | 2 | 3 | 4 |
| P(X) | 0.2 | 0.6 | 0.3 | -0.2 | 0.3 |
(iii)
| Y | 0 | 1 | 2 |
| P(Y) | 0.7 | 0.2 | 0.3 |
(iv)
| Z | 3 | 2 | 1 | 0 | -1 |
| P(Z) | 0.4 | 0.3 | 0.5 | 0.2 | 0.06 |
Solution:
We know that the sum of all of the probabilities in the probability distribution should be one.
(i)
| X | 0 | 1 | 2 |
| P(x) | 0.3 | 0.5 | 0.2 |
Sum of the probabilities = 0.3 + 0.5 + 0.2 = 1.0
Hence, this given table is a probability distribution of the random variables.
(ii)
| X | 0 | 1 | 2 | 3 | 4 |
| P(X) | 0.2 | 0.6 | 0.3 | -0.2 | 0.3 |
Here, from the table we can see that for X = 3, P(X) = -0.2
We know that, the probability for any of the observation cannot be negative.
Hence, the table given is not a probability distribution for the random variables.
(iii)
| Y | 0 | 1 | 2 |
| P(Y) | 0.7 | 0.2 | 0.3 |
Sum of all the probabilities given in the Q. = 0.7 + 0.2 + 0.3 = 1.2 ≠ 1
Hence, the table given is not a probability distribution for the random variables.
(iv)
| Z | 3 | 2 | 1 | 0 | -1 |
| P(Z) | 0.4 | 0.3 | 0.5 | 0.2 | 0.06 |
Here, sum of all of the probabilities given in the Q. = 0.4 + 0.3 + 0.5 + 0.2 + 0.06 = 1.46 ≠ 1
Hence, the table given is not a probability distribution for the random variables.
Q.2: There are 5 red balls and 2 black balls. From that, two balls are drawn at random. Let X represents the number of the black balls. Find the values for X. Check whether X is a random variable or not?
Solution:
Let, the two balls drawn randomly are represented as RR, RB, BR, and BB, where R is for a black ball and B is for a red ball.
According to Q,
X represents the number of the black balls.
Thus,
X (RR) = 0
X (RB) = 1
X (BR) = 1
X (BB) = 2
Hence, the possible values of X are 0, 1 and 2.
Therefore, C is a random variable.
Q.3: Let us consider X which represents the difference between the total number of tails and the number of heads which can be obtained when a single coin is tossed for 5 times. Find the possible values for X.
Solution:
As per the data given in the Q,
A coin will be tossed for 5 times and every time the result is observed. Also, X represents the difference between the total number of heads and tails observed after each toss.
Thus,
X (5H, 0T) =
5 – 0
= 5
X (4H, 1T) =
4 – 1
= 3
X (3H, 2 T) =
3 – 2
= 1
X(2H, 3T) =
2 – 3
= 1
X(1H, 4T) =
= 3
X(0H, 5T) =
0 – 5
= 5
Hence, the possible values for X are 5, 3 and 1.
Q.4: What will be the probability distribution for?
(i) Number of tails after tossing a coin for twice.
(ii) Number of heads after simultaneously tossing a coin for three times.
(iii) Number of tails after tossing a coin for four times.
Solution:
(i) Once a coin is tossed for two times, the chances are:
{TT, TH, HT, HH}, where H is for heads and T is for tails.
Let, the total number of tails be represented by X.
Then,
X(TT) = 2, X(TH) = 1, X(HT) = 1 and X(HH) = 0.
Hence, X can take the values of 0, 1 and 2.
We know that,
P(TT) = P(TH) = P(HT) = P(HH) = 14
P(X = 0) = P(HH) = 14
P(X = 1) = P (HT) + P(TH) = 14+14 = 1+14=24=12
P(X = 2) = P(TT) = 14
Hence, the probability distribution which is required as per the Q.’s demand is given below:
| X | 0 | 1 | 2 |
| P(X) | 14 | 12 | 14 |
(ii) Once a coin is tossed for three times, the chances are:
{TTT, TTH, THH, HHH, HHT, HTT, HTH, THT}, where H is for heads and T is for tails.
Let, the total number of tails be represented by X.
Then,
X(HHH) = 3, X(HTH) = 2, X(HHT) = 2, X(THH) = 2, X(TTH) = 1, X(HTT) = 1, X(THT) = 1 and X(TTT) = 0.
Hence, X can take the values of 0, 1, 2 and 3.
Now,
P(X = 0) = P(TTT) = 18
P(X = 1) = P (TTH) + P(HTT) + P(THT) = 18+18+18 = 1+1+18=38
P(X = 2) = P (HTH) + P(THH) + P(HHT) = 18+18+18 = 1+1+18=38
P(X = 3) = P(HHH) = 18
Hence, the probability distribution which is required as per the Q.’s demand is given below:
| X | 0 | 1 | 2 | 3 |
| P(X) | 18 | 38 | 38 | 18 |
(iii) Once a coin is tossed for four times, the chances are:
{TTTT, TTTH, TTHT, TTHH, THTT, THTH, THHT, THHH, HTTT, HTTH, HTHH, HTHT, HHTT, HHTH, HHHT, HHHH}, where H is for heads and T is for tails.
Let, the total number of tails be represented by X.
Then,
X(TTTT) = 4, X(TTTH) = 3, X(TTHT) = 3, X(THTT) = 3, X(HTTT) = 3, X(TTHH) = 2, X(THTH) = 2, X(THHT) = 2, X(HTTH) = 2, X(HTHT) = 2, X(HHTT) = 2, X(THHH) = 1, X(HTHH) = 1, X(HHTH) = 1, X(HHHT) = 1 and X(HHHH) = 0.
Hence, X can take the values of 0, 1, 2, 3 and 4.
Now,
P(X = 0) = P(HHHH) = 116
P(X = 1) = P (THHH) + P(HTHH) + P(HHTH) + P(HHHT) = 116+116+116+116 = 1+1+1+116=416=14
P(X = 2) = P (TTHH) + P(THTH) + P(THHT) + P(HTTH) + P(HTHT) + P(HHTT) = 116+116+116+116+116+116 = 1+1+1+1+1+116=616=38
P(X = 3) = P (TTTH) + P(TTHT) + P(THTT) + P(HTTT) = 116+116+116+116 = 1+1+1+116=416=14
P(X = 0) = P(TTTT ) = 116
Hence, the probability distribution which is required as per the Q.’s demand is given below:
| X | 0 | 1 | 2 | 3 | 4 |
| P(X) | 116 | 14 | 38 | 14 | 116 |
Q.5: What will be the probability distribution of the numbers for the two success tosses of a die, where a success will be defined as?
(i) Obtained number which is greater than 3.
(ii) On the die, six appears for at least once.
Solution:
When we will toss a die for two times, we will obtain at least (6 × 6) = 36 number of the observations.
Let us consider X which is a random variable, which represents total number of successes.
(i) In this case, numbers which is greater than 3 is referred to as success.
P(X = 0) = P (Number which are less than or equal to 3 after having both the tosses) = 36×36=14
P(X = 1) = P(Numbers which are less than or equal to 3 in first toss and greater than 3 in the second toss) + P(Numbers which are greater than 3 in first toss and less than or equal to 3 in the second toss)
= 36×36+36×36
= 12×12+12×12
= 14+14
= 12
P(X = 2) = P(Numbers which are greater than 3 after both toss) = 36×36
= 12×12
= 14
Hence, the probability distribution which is required as per the Q.’s demand is given below:
| X | 0 | 1 | 2 |
| P(X) | 14 | 12 | 14 |
(ii) In this case, success will be when six appeared at least for once after tossing the dice simultaneously.
P(Y = 0) = P(six doesn’t appeared on either of the dice) = 56×56=2536
P(Y = 1) = P(six appeared at least on one of the dice) = 16×56+56×16=536+536=1036=518
Hence, the probability distribution which is required as per the Q.’s demand is given below:
| Y | 0 | 1 |
| P(Y) | 2536 | 518 |
Q.6: Consider a lot of 34 bulbs among which 10 bulbs are defective. From that lot, a sample of 3 bulbs is drawn at random with having replacement. What is the probability distribution for the number of the defective bulbs?
Solution:
As per the data given in the Q., we have
A lot of 34 bulbs among which 10 bulbs are defective.
So, the number of non- defective bulbs = 34 – 10 = 24 bulbs
Now,
A sample of 3 bulbs is drawn at random from the lot with having replacement.
Let us consider X be a random variable which denotes the total number of defective bulbs among the sample drawn.
P(X = 0) = P (3 non- defective and 0 defective) = 3C0. 34×34×34 = 3!0!3!×2764 = 1×2764=2764
P(X = 1) = P (2 non- defective and 1 defective) = 3C1. 14×34×34 = 3!1!2!×964 = 3×2!2!×964=3×964=2764
P(X = 2) = P (1 non- defective and 2 defective) = 3C2. 14×14×34 = 3!2!1!×364 = 3×2!2!×364=3×364=964
P(X = 3) = P (0 non- defective and 3 defective) = 3C3. 14×14×14 = 3!3!0!×164 = 1×164=164
Hence, the probability distribution which is required as per the Q.’s demand is given below:
| X | 0 | 1 | 2 | 3 |
| P(X) | 2764 | 2764 | 964 | 164 |
Q.7: Consider a situation where a coin will be biased in such a manner that the tail will occur likely 4 times the head. What will be the probability distribution of the number of heads, by considering the situation that the coin is tossed for two times?
Solution:
Let us consider that the probability of getting the head in the biased coin is x.
Then,
P(H) = x
⟹ P(T) = 4x
For the biased coin, P(H) + P(T) = 1
⟹ x + 4x = 1
⟹ 5x = 1
⟹ x = 15
Hence, P(H) = 15 and P(T) = 45
When the coin will be tossed for two times, then the sample space will be { TT, TH, HT, HH }
Let, X be the random variable which represents the number of heads.
Then,
P(X = 0) = P(no heads) = P(T) × P(T) = 45×45=1625
P(X = 1) = P(one heads) = P(HT) + P(TH) = 15×45+45×15=425+425=825
P(X = 2) = P(two heads) = P(H) × P(H) = 15×15=125
Hence, the probability distribution which is required as per the Q.’s demand is given below:
| X | 0 | 1 | 2 |
| P(X) | 1625 | 825 | 125 |
Q.8: The following probability distribution is for the random variable X.
| X | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| P(X) | 0 | k | 2k | 2k | 3k | k2 | 2k2 | 7k2 + k |
Find:
(i) The value of k
(ii) P(X < 4)
(iii) P(X > 5)
(iv) P (0 < X < 2)
Solution:
We know that, the sum of all of the probabilities of the probability distribution of a random variable is 1.
Thus,
k + 2k + 2k + 3k + k2 + 2k2 + 7k2 + k = 1
⟹ 10k2 + 9k = 1
⟹ 10k2 + 9k – 1 = 0
⟹ 10k2 + 10k – k – 1 = 0
⟹ 10k(k + 1) – 1(k + 1) = 0
⟹ (k + 1)(10k – 1) = 0
⟹ k = -1, 110
Now,
k = -1 is not possible because the probability for any of the event cannot ever be negative.
So, k = 110
(ii) P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
= o + k + 2k + 2k = 5k
= 5×110 = 12
(iii) P(X > 5) = P(X = 6) + P(X = 7)
= 2k2 + 7k2 + k
= 9k2 + k
= k (9k + 1)
= 110(9×110 + 1)
= 110(910+1)
= 110 (9+1010)
= 110(1910)
= 19100
(iv) P(0 < X < 2) = P(X = 1)
= k = 110
Q.9: Consider P(Y) be the probability distribution of the random variable Y for the following forms, where some number is.
P(Y) =
2a, if a = 0
3a, if a = 1
4a, if a = 2
0, otherwise
(i) Find the value of a.
(ii) Determine P(Y < 2), P(Y ≥ 2), and P (Y ≤ 2).
Solution:
(i) As we know that the sum for all of the probabilities of the probability distribution of random variables is given by 1.
Thus,
2a + 3a + 4a + 0 = 1
⟹ 9a = 1
⟹ a = 19
(ii) P(Y < 2) = P(Y = 0) + P(Y = 1)
= 2a + 3a
= 5a = 5×19 = 59
P(Y ≥ 2) = P(Y = 2) + P(Y > 2)
= 4a + 0
= 4a = 4×19 = 49
P(Y ≤ 2) = P(Y = 2) + P(Y = 1) + P(Y = 0)
= 4a + 3a + 2a
= 9a = 9×19 = 1
Q.10: After tossing a fair coin, what will be the mean number for tails?
Solution:
Let us consider X which denotes the chances of success to get tails.
Thus,
For this case, the sample space will be
S = {TTT, TTH, THT, THH, HTT, HTH, HHT, HHH}
Let, the total number of tails be represented by X.
Then,
X(HHH) = 0, X(HTH) = 1, X(HHT) = 1, X(THH) = 1, X(TTH) = 2, X(HTT) = 2, X(THT) = 2 and X(TTT) = 3.
Here, we can see that X will take values 0, 1, 2 or 3.
Now,
P(X = 0) = P(HHH)
= 12×12×12
= 18
P(X = 1) = P(HTH) + P(HHT) + P(THH)
= 12×12×12 + 12×12×12 + 12×12×12
= 18+18+18
= 38
P(X = 2) = P(TTH) + P(HTT) + P(THT)
= 12×12×12 + 12×12×12 + 12×12×12
= 18+18+18
= 38
P(X = 0) = P(TTT)
= 12×12×12 = 18
Hence, the probability distribution which is required as per the Q.’s demand is given below:
| X | 0 | 1 | 2 | 3 |
| P(X) | 18 | 38 | 38 | 38 |
Mean of X E(X), µ = ∑XiP(Xi)
= 0×18+1×38+2×38+3×18
= 38+68+38
= 128
= 32 = 1.5
Q.11: Take two dices which are thrown simultaneously. If Y denotes the number of times we will get sixes, then what will be the expectation of Y?
Solution:
As per the data given in the Q., Y represents the total number of sixes which will be obtained after throwing two dice simultaneously.
Thus, Y will take the values of 0, 1 or 2.
Hence,
P(Y = 0) = P (no sixes will be obtained in either of the throw) = 56×56=2536
P(Y = 1) = P (six at the first dice and other number rather than 6 on the second dice) + P(other number rather than six on the first dice and six on the second dice)
= 2 (56+16)
= 2×536
= 518
P(Y = 2) = P (sixes will be obtained in both of the throw) = 136
Hence, the probability distribution which is required as per the Q.’s demand is given below:
| Y | 0 | 1 | 2 |
| P(Y) | 2536 | 1036 | 136 |
Therefore,
Expectation of X = E(X) = ∑XiP(Xi)
= 0×2536+1×1036+2×136
= 0+518+118
= 618 = 13
Q.12: From the starting six positive integers, two numbers will be selected at random, without any replacement. Let, Y denotes the larger number among those two numbers obtained. Find E(Y).
Solution:
As per the Q.’s demand, from the starting six positive integers, two numbers will be selected, without having replacement which will be done in 6 × 5 = 30 ways.
Let us consider Y which represents the two numbers obtained which are larger. Thus,
Y will take values 2, 3, 4, 5 or 6.
For Y = 2, the possible observations will be (1, 2) and (2, 1).
Then,
P(Y = 2) = 230=115
For Y = 3, the possible observations will be (1, 3), (2, 3), (3, 1) and (3, 2)
Then,
P(Y = 3) = 430=215
For Y = 4, the possible observations will be (1, 4), (2, 4), (3, 4), (4, 1), (4, 2) and (4, 3)
Then,
P(Y = 4) = 630=15
For Y = 5, the possible observations will be (1, 5), (2, 5), (3, 5), (4, 5), (5, 4), (5, 3), (5, 2), (5, 1)
Then,
P(Y = 5) = 830=415
For Y = 6, the possible observations will be (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 5), (6, 4), (6, 3), (6, 2), (6, 1)
Then,
P(Y = 6) = 1030=13
Hence, the probability distribution which is required as per the Q.’s demand is given below:
| X | 2 | 3 | 4 | 5 | 6 |
| P(X) | 115 | 215 | 15 | 415 | 13 |
Therefore,
Expectation of X = E(X) = ∑XiP(Xi)
= 2×115+3×215+4×15+5×415+6×13
= 215+615+4×315+2015+6×515
= 215+615+1215+2015+3015
= 7015 = 143
Q.13: Y denotes the sum of all the numbers which are obtained when two fair dice are rolled. What will be the variance and the standard deviation of Y.?
Solution:
Number of observations which are obtained when two dice is rolled is 6 × 6 = 36
P(Y = 2) = P(1, 1) = 136
P(Y = 3) = P(1, 2) + P(2, 1) = 236
P(Y = 4) = P(1, 3) + P(2, 2) + P(3, 1) = 336
P(Y = 5) = P(1, 4) + P(2, 3) + P(3, 2) + P(1, 4) = 436
P(Y = 6) = P(1, 5) + P(2, 4) + P(3, 3) + P(4, 2) + P(5, 1) = 536
P(Y = 7) = P(1, 6) + P(2, 5) + P(3, 4) + P(4, 3) + P(5, 2) + P(6, 1) = 636
P(Y = 8) = P(2, 6) + P(3, 5) + P(4, 4) + P(5, 3) + P(6, 2) = 536
P(Y = 9) = P(3, 6) + P(4, 5) + P(5, 4) + P(3, 6) = 436
P(Y = 10) = P(4, 6) + P(5, 5) + P(6, 4) = 336
P(Y = 11) = P(5, 6) + P(6, 5) = 236
P(Y = 12) = P(6, 6) = 136
Hence, the probability distribution which is required as per the Q.’s demand is given below:
| Y | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
| P(Y) | 136 | 236 | 336 | 436 | 536 | 636 | 536 | 436 | 336 | 236 | 136 |
Therefore,
Expectation of X = E(X) = ∑XiP(Xi)
= 2×136+3×236+4×336+5×436+6×536+7×636+8×536+9×436+10×336+11×236+12×136
= 236+636+1236+2036+3036+4236+4036+3636+3036+2236+1236
= 25236 = 7
Expectation of X2 = E(X2) = ∑X2iP(Xi)
= 4×136+9×236+16×336+25×436+36×536+49×636+64×536+81×436+100×336+121×236+144×136
= 436+1836+4836+10036+18036+29436+32036+32436+30036+24236+14436
= 197436 = 54.8333
Then,
Var(X) = E(X2) – [ E(X)2 ]
= 54.8333 – 49 = 5.8333
Hence,
Standard derivation = Var(X)−−−−−−−√
= 5.8333−−−−−√ = 2.415
Q.14: There is a class having 15 students with ages 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19 and 20 years. One student will be selected in such a way that each of them have the same chance of being chosen and Y is recorded as the age of the selected student. Find the probability distribution for the random variable Y. What will be the mean, variance and standard derivation of Y?
Solution:
A class has 15 students with ages 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19 and 20 years.
Every student has the same chance of being selected.
Hence, the probability for every student being selected is 115
The above information will be compiled in the frequency table which is given below:
| Y | 14 | 15 | 16 | 17 | 18 | 19 | 20 | 21 |
| f | 2 | 1 | 2 | 3 | 1 | 2 | 3 | 1 |
P(X = 14) = 215
P(X = 15) = 115
P(X = 16) = 215
P(X = 17) = 315
P(X = 18) = 115
P(X = 19) = 215
P(X = 20) = 315
P(X = 21) = 115
Hence, the probability distribution which is required as per the Q.’s demand is given below:
| Y | 14 | 15 | 16 | 17 | 18 | 19 | 20 | 21 |
| P(Y) | 215 | 115 | 215 | 315 | 115 | 215 | 315 | 115 |
Thus,
Expectation of X = E(X) = ∑XiP(Xi)
= 14×215+15×115+16×215+17×315+18×115+19×215+20×315+21×115
= 2815+1515+3215+5115+1815+3815+6015+2115
= 26315 = 17.53
Expectation of X2 = E(X2) = ∑X2iP(Xi)
= (14)2×215+(15)2×115+(16)2×215+(17)2×315+(18)2×115+(19)2×215+(20)2×315+(21)2×115
= 39215+22515+51215+86715+32415+72215+120015+44115
= 468315 = 312.2
Then,
Var(X) = E(X2) – [ E(X)2 ]
= 312.2 – (17.53)2 = 4.78
Hence,
Standard derivation = Var(X)−−−−−−−√
= 4.78−−−−√ = 2.186
Q.15: 70% members are in the favour and 40% are opposing a certain proposal in a meeting. A member will be randomly selected and we will take Y = 0, if the person opposed, and X = 1 if the person favours. What will be E(Y) and var(Y) in such a situation?
Solution:
As per the data given in the Q., we have
P(Y = 0) = 40% = 40100 = 0.4
P(Y = 1) = (100 – 40)% = 60% = 60100 = 0.6
Hence, the probability distribution which is required as per the Q.’s demand is given below:
| X | 0 | 1 |
| P(X) | 0.4 | 0.6 |
Thus,
Expectation of X = E(X) = ∑XiP(Xi)
= 0 × 0.4 + 1 × 0.6 = 0.6
Expectation of X2 = E(X2) = ∑X2iP(Xi)
= 02 × 0.4 + 12 × 0.6 = 0.6
We know that,
Var(X) = E(X2) – [ E(X)2 ]
= 0.6 – (0.6)2 = 0.6 – 0.36 = 0.24
Q.16: The mean of the numbers which will be obtained after throwing a die on which 1 is written on two of the faces, 2 on three faces and 5 on one face is
(a) 1
(b) 2
(c) 5
(d) 136
Solution:
Let us consider Y which is a random variable representing any number on the die.
Since, it’s a die, then the total number of observations will be six.
Thus, P(Y = 1) = 26=13
P(Y = 2) = 36=12
P(Y = 5) = 16
Hence, the probability distribution which is required as per the Q.’s demand is given below:
| Y | 1 | 2 | 5 |
| P(Y) | 13 | 12 | 16 |
Thus,
Expectation of X = E(X) = ∑XiP(Xi)
= 1×13+2×12+5×16
= 13+1+56 = 2+6+56 = 136
Hence, the correct answer is (d).
Q.17: Let Y be the number of aces obtained. Let any of the two cards are drawn at random from the deck of the cards. Then, the value of E(X) will be
(a) 37221
(b) 513
(c) 213
(d) 113
Solution:
Let, Y be the number of aces obtained.
Therefore, Y will take any of the values of 0, 1, or 2
As we know, there are 52 cards in a deck. Among them 4 cards are aces.
Hence, there are 48 non- ace cards.
P(Y = 0) = P(2 non- ace cards and 0 ace card) = 48C2×4C052C2=11281326
P(Y = 1) = P(1 non- ace cards and 1 ace card) = 48C1×4C152C2=1921326
P(Y = 2) = P(0 non- ace cards and 2 ace card) = 48C0×4C252C2=61326
Hence, the probability distribution which is required as per the Q.’s demand is given below:
| Y | 0 | 1 | 2 |
| P(Y) | 11281326 | 1921326 | 61326 |
Thus,
Expectation of X = E(X) = ∑XiP(Xi)
= 0×11281326+1×1921326+2×61326
= 1921326+121326 = 192+121326 = 2041326 = 213
Hence, the correct answer is (c).
EXERCISE – 13.5
Q.1: A die will be thrown for 6 times. Let us consider that, “getting any odd number “will be the success, then what the probability is of
(a) Getting 5 successes? (b) Getting at least 5 successes? (c) Getting at most 5 successes?
Solution:
Odd numbers on the die are 1, 3 and 5.
When a die is tossed repeatedly, it is Bernoulli trials. Let us consider Y which represents the total number of successes of getting any odd numbers in the experiment of 6 toss trials.
In a single throw of the die, the probability of getting any odd number is a = 36=12
Thus,
b = 1 – A = 12
Y has the binomial distribution.
Hence, P(Y = r) = nCrbn–rar, where n = 0, 1, 2, 3, …………, n
= 6Cr126–r12r
= 6Cr126
(a) Probability of getting 5 successes
P(5 successes) = P(Y = 5)
= 6C5126
= 6!5!×1!126
= 6×164
= 332
(b) Probability of getting at least 5 successes
P(at least 5 successes) = P(Y ≥ 5)
= P(Y = 5) + P(Y = 6)
= 6C5126+6C6126
= 6!5!×1!126+6!6!×0!126
= 6×164+1×164
= 764
(c) Probability of getting at most 5 successes
P(at most 5 successes) = P(Y ≤ 5)
= 1 – P(Y > 5)
= 1 – P(Y = 6)
= 1 – 6C6126
= 1 – 6!6!×0!126
= 1 – 1×164
= 1 – 164 = 6364
Q.2: A pair of dice will be thrown for 4 times. Let us consider that, “getting a doublet” will be the success, then what is the probability for getting two successes?
Solution:
When a die is tossed repeatedly, it is Bernoulli trials. Let us consider Y which represents the total number of successes of getting any doublets in the experiment of 4 toss trials.
In a single throw of the die, the probability of getting any odd number is a = 636=16
Thus,
b = 1 – A = 1–16=6–16=56
Y has the binomial distribution with n = 4.
Hence, P(Y = r) = nCrbn–rar, where n = 0, 1, 2, 3, …………, n
= 4Cr564–r16r
4Cr(54–r64)
Probability of getting 2 successes
P(2 successes) = P(Y = 2)
= 4C2(54–264)
= 4!2!×2!521296
= 4×3×2!2!×2!×251296
= 122×251296
= 6×251296 = 25216
Q.3: In a large bulk of different items, there are 5% of the defective items. Find the probability for the sample having 10 items, which will include at most one defective item only.
Solution:
As the drawing of items will be done with replacement, so the trial will be Bernoulli trials.
Let us consider X which denotes the total number of defective items of the sample from which 10 items are drawn successively.
Thus,
p = 5100=120
⟹ q = 1 – p
⟹ q = 1 – 120=1920
Y has the binomial distribution with n = 4 and p = 120.
Hence, P(Y = r) = nCrbn–rar, where n = 0, 1, 2, 3, …………, n
= 4Cr19204–r120r
P(at most 1 defective item) = P(X ≤ 1)
= P(X = 0) + P(X = 1)
= 10C0(1920)10–0.(120)0+10C1(1920)10–1.(120)1
= 10!10!0!(1920)10.×1+10!9!1!(1920)9.(120)
= 1×(1920)10.×1+10×9!9!1!(1920)9.(120)
= (1920)10×1+10×(1920)9.(120)
= (1920)9[1920+1020]
= (1920)9(2920) = (2920).(1920)9
Q.4: From a deck of 52 cards, well- shuffled, five cards will be drawn successively with having replacement. Find the probability that-
(a) All the five cards chosen will be shades?
(b) Among the five selected cards, only three of them will be spade?
(c) None of them is a spade?
Solution:
As the drawing of cards will be done with replacement, so the trial will be Bernoulli trials.
Let us consider X which denotes the number of spade cards in the deck from which 5 cards are drawn successively.
We know that in a deck of cards, there are 52 cards and among them 13 cards are spade.
Thus,
a = 1352=14
⟹ b = 1 – a
⟹ b = 1 – 14=34
Y has the binomial distribution.
Hence, P(Y = r) = nCrbn–rar, where n = 0, 1, 2, 3, …………, n
= 4Cr19204–r120r
(a) P(all the five cards chosen will be spades) = P(Y = 5) having n = 5 and p = 14
= 5C5.(34)5–5.(14)5
= 5!5!0!.(34)0.(14)5
= 1×1×11024
= 11024
(b) P(among them 3 cards chosen will be spades) = P(Y = 3) having n = 5 and p = 14
= 5C3.(34)5–3.(14)3
= 5!3!2!.(34)2.(14)3
= 5×2×916×164
= 45512
(iii) P(none of them will be spades) = P(Y = 0) having n = 5 and p = 14
= 5C0.(34)5–0.(14)0
= 5!0!5!.(34)5×1
= 1×2431024
= 2431024
Q.5: If the probability that a bulb which is produced by the factory will fuse at least after 150 days of its use, is 0.10. What will be the probability that out of 5 such bulbs,
(a) None
(b) Not more than one
(c) More than 1
(d) At least 1
Bulb will fuse after 150 days of its use.
Solution:
As the drawing of bulbs will be done with replacement, so the trial will be Bernoulli trials.
Let us consider X which denotes the number of bulbs which will fuse at least after 150 days of its use in the experiment on 5 trial bulbs.
As per the data in the Q,
a = 0.1
⟹ b = 1 – 0.1 = 0.9
Y has the binomial distribution.
Hence, P(Y = r) = nCrbn–rar, where n = 0, 1, 2, 3, …………, n
Here, n = 5 and p = 0.1.
Thus,
5Cr(0.9)5–r(0.1)r
(a) P(none) = P(Y = 0)
= 5C0(0.9)5–0(0.1)0
= 5!0!5!(0.9)5×1
= (0.9)5 = 0.59049 = 0.59
(b) P(not more than 1) = P(Y ≤ 1)
= P(Y = 0) + P(Y = 1)
= 5C0(0.9)5–0(0.1)0+5C1(0.9)5–1(0.1)1
= 5!0!5!(0.9)5×1+5!1!4!(0.9)4×0.1
= (0.9)5 + 5×4!1!4!(0.9)4×0.1
= (0.9)5 + 5 × (0.9)4 × 0.1
= 0.59049 + 0.5 × (0.9)4
= 0.59049 + 0.32805 = 0.91854
(c) P(more than 1) = P(Y ≥ 1)
= 1 – P(Y ≤ 1) = 1 – 0.91854 = 0.08146
(d) P(at least one) = P(Y > 1)
= 1 – P(Y < 1) = 1 – P(Y = 0) = 1 – 0.59 = 0.41
Q.6: There are 10 balls in the bag, each marked with any one of the digit from 0 to 9. Find the probability that none of them will be marked with the digit 0, if the four balls will be drawn successively from the bag, with replacement?
Solution:
As the drawing of balls will be done with replacement, so the trial will be Bernoulli trials.
Let us consider X which denotes the number of balls which will be marked with any one of the digits from 0 to 9 from which 4 balls are drawn successively from the bag.
As per the data in the Q.,
a = 110
⟹ b = 1 – 110=910
Y has the binomial distribution.
Hence, P(Y = r) = nCrbn–rar, where n = 0, 1, 2, 3, …………, n
Here, n = 4 and p = 110.
Thus,
4Cr9104–r110r
Now,
P(none of them is marked with 0) = P(Y = 0)
= 4C0(910)4–0.(110)0
= 4!0!4!.(910)4×1
= 1×(910)4×1 = (910)4
Q.7: There are 20 true and false type Q.s which are asked in the Q. paper in an examination. Let us consider a situation that a student picks up his option by tossing a fair coin every time. If its heads on the coin, the student answered true for the Q., and if its tails then the student answered false. What will be the probability that he can answer at least 12 Q.s.
Solution:
Let, Y represents the total number of correctly answered Q.s from those 20 Q.s.
As the picking of answer from the repeated tossing will be done with replacement, so the trial will be Bernoulli trials.
As per the student’s observation, whenever there is head on the coin it represents that the answer is true and whenever there is tails on the coin it represents that the answer is false.
Thus,
a = 12
⟹ b = 1 – p = 1 – 12 = 12
Y will have the binomial distribution with n = 20, a = 12 and b = 12
Thus,
P(Y = r) = nCrbn–rar, where n = 0, 1, 2, 3, …………, n
= 20Cr(12)20–r(12)r
= 20Cr(12)20
P(at least 12 Q.s from those 20 Q.s) = P(Y ≥ 12)
= P(Y = 12) + P(Y = 13) + P(Y = 14) + …….. + P(Y = 20)
= 20C12(12)20 + 20C13(12)20 + 20C14(12)20 + …………………. + 20C20(12)20
= (12)2.[20C12+20C13+20C14+………+20C20]
Q.8: Let us consider Y being a binomial distribution R(6, 12). Prove that, Y = 3 will be the most likely outcome.
Solution:
Here, as per the data given in the Q.,
Y is the binomial distribution with having binomial distribution as R(6, 12).
Hence,
n = 6 and a = 12
⟹ b = 1 – p = 1 – 12 = 12
Y will have the binomial distribution with n = 20, a = 12 and b = 12
Thus,
P(Y = r) = nCrbn–rar, where n = 0, 1, 2, 3, …………, n
= 6Cr(12)6–r(12)r
= 6Cr(12)6
Here, we can see that P(Y = r) can be the maximum if, 6Cr can be maximum.
Thus,
6C0=6C6=6!0!6!=1 6C1=6C5=6!1!5!=6×5!1!5!=6 6C2=6C4=6!2!4!=6×5×4!2!4!=3×5=156C3=6C3=6!3!3!=6×5×4×3!3!3!=2×5×2=15
Here, we can see that the value of 6C3 is maximum.
Hence, for y = 3, P(Y = r) is maximum.
Therefore, Y = 3 is the most likely outcome.
Q.9: In a multiple choice Q.s examination which having three possible answers for every five Q.s, find the probability for the candidate that he will get 4 or more correct answers by having a guess every time.
Solution:
Let, Y represents the total number of correctly answered Q.s among those 5 multiple choice Q.s by having a guess.
As the picking of answer from the repeated guessing for the correct answer from the multiple choices will be done with replacement, so the trial will be Bernoulli trials.
Now,
Probability of getting the correct answer is given by,
a = 13
⟹ b = 1 – a = 1 – 13 = 23
Hence,
Y will have the binomial distribution with n = 5, a = 13 and b = 23
Thus,
P(Y = r) = nCrbn–rar, where n = 0, 1, 2, 3, …………, n
= 5Cr(23)5–r(13)r
P(guessing at least 4 correct answer) = P(Y ≥ 4)
= P(Y = 4) + P(Y = 5)
= 5C4(23)5–4(13)4 + 5C5(23)5–5(13)5
= 5C4(23)1(13)4 + 5C5(23)0(13)5
= 5C4.23.(13)4 + 5C5.1.(13)5
= 5!4!1!.23.(13)4 + 5!5!0!.1.(13)5
= 5×4!4!.23.(13)4 + 1.1.(13)5
= 5.23.(13)4 + 1.1.(13)5
= 10243+1243 = 11243
Q.10: A person purchased a lottery ticket from 50 different lotteries, among each of which his chance to win a prize is 1100. Find the probability that the person will win a prize
(i) Exactly once
(ii) at least once
(iii) at least twice?
Solution:
Let us consider Y which represents the number of winning prizes among 50 lotteries.
As the picking of winner will be done with replacement, so the trial will be Bernoulli trials.
Now,
Y is a binomial distribution with n = 50 and a = 1100
⟹ b = 1 – a = 1 – 1100 = 99100
Thus,
P(Y = r) = nCrbn–rar, where n = 0, 1, 2, 3, …………, n
= 50Cr(99100)50–r(1100)r
(a) P (winning exactly once) = P(X = 1)
= 50C1(99100)50–1(1100)1
= 50!1!49!(99100)49(1100)1
= 50(1100)(99100)49
= 12(99100)49
(ii) P(winning at least once) = P(Y ≥ 1)
= 1 – P(Y < 1)
= 1 – P(X = 0)
= 1 – 50Cr(99100)50
= 1 – 50!0!50!.(99100)50
= 1 – 1.(99100)50
= 1 – (99100)50
(iii) P (at least twice) = P(Y ≥ 2)
= 1 – P(Y < 2)
= 1 – P(Y ≤ 1)
= 1 – [P(Y = 0) + P(Y = 1)]
= [1 – P(Y = 0)] – P(Y = 1)
= 1 – (99100)49. [99100+12]
= 1 – (99100)49. (99+50100)
= 1 – (99+50100) . (99100)49
Q.11: What will be the probability of getting 5 exactly twice in 7 throws of a die?
Solution:
As the repeatedly tossing of a die will be done with replacement, so the trial will be Bernoulli trials.
Let us consider Y which represents the number of times to get 5 in 7 throws of the die.
As per the data given in the Q., a = 16
⟹ b = 1 – a = 1 – 16 = 56
Y is the probability distribution with n = 7, a = 16 and b = 56
Then,
P(Y = r) = nCrbn–rar, where n = 0, 1, 2, 3, …………, n
= 7Cr(56)7–r(16)r
P (getting 5 exactly twice) = P (Y = 2)
= 7C2(56)7–2(16)2
= 7!2!5!(56)2(16)2
= 7×6×5!2×5!(56)2(16)2
= 7×62(56)2(136)
= 21(56)2(136)
= (56)2(712)
Q.12: 10% of the certain articles which are manufactured will be defective. Find the probability for the samples having 12 such articles among which 9 of them will be defective.
Solution:
As the repeated selection of the articles in a random sample space will be done with replacement, so the trial will be Bernoulli trials.
Let us consider Y which represents the number of times to select the defective articles in the random sample space of 12 articles.
As per the data given in the Q., a = 10% = 10100=110
⟹ b = 1 – a = 1 – 110 = 910
Y is the probability distribution with n = 12, a = 110 and b = 910
Then,
P(Y = r) = nCrbn–rar, where n = 0, 1, 2, 3, …………, n
= 12Cr(910)12–r(110)r
P(selecting 9 defective articles) = 12C9(910)12–9(110)9
= 12!9!3!(910)3(110)9
= 12×11×10×9!9!×3×2×1(910)3(110)9
= 12×11×103×2×1(910)3(110)9
= 220.(910)3(110)9
= 22×931011
Q.13: Among the 100 bulbs in the box, 10 bulbs are defective. The probability that out of the sample of 5 bulbs, none of the bulbs are defective is,
(a) 10-1
(b) (12)5
(c) (910)
(d) (910)5
Solution:
As the repeated selection of the defective bulbs in the box will be done with replacement, so the trial will be Bernoulli trials.
Let us consider Y which represents number of the defective bulbs in a random sample of 5 bulbs.
As per the data given in the Q., a = 10100=110
⟹ b = 1 – a = 1 – 110 = 910
Y is the probability distribution with n = 5, a = 110 and b = 910
Then,
P(Y = r) = nCrbn–rar, where n = 0, 1, 2, 3, …………, n
= 5Cr(910)5–r(110)r
P(none of the bulbs are defective) = P(Y = 0)
5C0(910)5–0(110)0
= 5!5!0!(910)5×1
= 1×(910)5×1
= (910)5
Hence, the correct option is (d).
Q.14: The probability that the student will not be a swimmer is 15. The probability that from a group of 5 students, 4 of them will be a swimmer is
(a) (45)415
(b) 5C4(45)415
(c) 5C1(45)415
(d) None of these
Solution:
As the repeated selection of the students in the group who all are swimmers will be done with replacement, so the trial will be Bernoulli trials.
Let us consider Y which represents number of the students from the group of 5 students, who all are swimmers.
As per the data given in the Q., b = 15
⟹ a = 1 – b = 1 – 15 = 45
Y is the probability distribution with n = 5, a = 45 and b = 15
Then,
P(Y = r) = nCrbn–rar, where n = 0, 1, 2, 3, …………, n
= 5Cr(15)5–r(45)r
P(none of the bulbs are defective) = P(Y = 4)
= 5C4(15)5–4(45)4
= 5C4(15)(45)4
Hence, the correct answer is (b).
Related Links
NCERT Solutions for Class 10 Maths – Surface Areas and Volume
Short answer questions:
- Three metallic solid cubes, whose edges are 3 cm, 4cm, and 5cm, are melted and formed into a single cube. Find the edges of the cube so formed.
Sol. Here , edge of there metallic cubes are 3 cm, 4cm, and 5cm.
Volume of single cube = 33+43+53
= 27 +64 +125
= 216 cm3
Therefore,edgeofsinglecube=216−−−√3=6
- The rainwater from a roof of dimensions 22 m * 20m drains into a cylindrical vessel having diameter of base 2 m and height 3.5 m. If the rainwater collected from the roof just fill the cylindrical vessel, then find the rainfall in cm.
Sol. Let the required rainfall in cm be x
therefore according to the statement of the question, we have
Volume of rain = volume of cylinder
x100×22×20=227×1×1×3.5
x= 227×1×1×3.5×10022×20
=2.5m
- A cone having radius 8 cm and height 12 cm is divide into two part by a plane through the mid-point of axis parallel to its base. Find the ratio of the volumes of two parts.
Sol. Here, △ADE∼△ABCbyAAsimilarityruler4=612
∴radiusofsmallerconeradiusoflargercone=HeightofsmallerconeHeightoflargercone
⇒ r=2cm
now,volumeofupperpart(cone)volumeoflowerpart(frustumofcone)
=13πr2h13πh(r2+R2+R)
=2×2×66(4+16+8)
=428
=17
So that required ratio is 1:7.
- Marbles of diameter 1.4 cm are dropped into a cylindrical beaker of diameter 7 cm containing some water. Find the number of marble that should be dropped into the beaker so that the water level rises by 5.6 cm.
Sol. Let us assume that the marbles are spherical in shape,
Hence,Volume of each marble
=43×π×0.7×0.7×0.7
Let n be the number of marbles that are required to raise the water level upto 5.6cm of the cylindrical beaker
therefore n times volume of each marble = Volume of cylindrical beaker of height 5.6cm.
n×43×π×0.7×0.7×0.7 =π×72×72×5.6
n=72×72×5.6×34×10.7×10.7×10.7
n= 150.
- Find the number of metallic circular disc with 1.5 cm base diameter and of height 0.2 cm. to be melted to from a right circular cylinder of diameter 4.5 cm and height 10 cm.
Sol. Volume of each circular disc
=π×1.52×1.52×0.2cm3
Volume of right circular cylinder
=π×4.52×4.52×10cm3
Let the number of metallic circular disc be n
∴n×Volumeofeachcirculardisc=Volumeofcylinder
∴n×π×1.52×1.52×0.2=π×4.52×4.52×10
n=4.52×4.52×10×21.5×21.5×10.2
=450
- A heap of rice is in the form of a cone of diameter 9 m and height 3.5 m. find the volume of the rice. How much canvas cloth is required to just cover the heap?
Sol. Radius of conical heap of rice (r) = 92=4.5m
Height of conical heap of rice (h) = 3.5m
Therefore, Volume of the rice =13×π×r2×h
=13×227×4.5×4.5×3.5
=74.25 m3
Therefore the Slant height of the heap =(4.5)2+(3.5)2−−−−−−−−−−−√
=20.25+12.25−−−−−−−−−−−√
=32.5−−−−√=5.7m
Area of the canves cloth required =227×4.5×5.7
=80.6m3
- Two cones with same base radius 8 cm and height 15 cm are joined together along their bases. Find the surface area of the shape so formed.
Sol.
Slant height of cone = 82+152−−−−−−−√
=64+225−−−−−−−√
=289−−−√
=17cm
Total surface area of the shape so formed
= 2×Curvedsurfaceoftheoriginalconeused
=2×227×8×17
=854.86cm2
- An ice-cream cone full of ice cream having radius 5cm and height 10cm as shown in the figure. Calculate the volume of ice-cream, provided that its 16 part is left unfilled with ice-cream.
Sol. Radius of the cone (r) = 5cm
Total height of ice-cream cone = 10cm
Height of cone = 10 – 5=5cm
Volume of the ice-cream = 13πr2h+23πr3
=13πr2(h+2r)
=13×227×25(5+10)
=13×227×25×15
=392.86cm3
Volume of ice-cream = (1−16)of392.86
=56×392.86
=327.38cm3
- How many spherical lead shots each of diameter 4.2 cm can be obtained from a solid rectangular lead piece with dimensions 66cm, 42cm, and 21cm?
Sol. Let n be the number of spherical lead shots.
n×volumeofeachsphericalleadshot=volumeoftherectangularleadpiece
n=43×227×4.22×4.22×4.22=66×42×21
=×66×42×21×3×7×84×22×4.2×4.2×4.2
=1500
Hence, the required number of spherical lead shots are 1500.
- A well 24m long, 0.4m thick and 6m high is constructed with the bricks each of dimensions 25cm×16cm×10cm. IF the motor occupies 110 th of the well, then find the number of bricks used in constructing the well.
Sol. Volume of the wall = 24m×0.4m×6m
=2400×40×600cm3
=57600000cm3
Volume of the wall occupied by motor = 110×5700000
=5760000cm3
Volume of the wall constructed with bricks = 57600000−5760000cm3
=51840000
Let n be the number of bricks required
∴n×25×16×10 =51840000
⇒n=5184000025×16×10=12960
- A solid metallic hemisphere of radius 8 cm is melted and re-casted into a right circular cone of base radius 6 cm. Determine the height of the cone.
Sol. Let h be the height of the cone.
Volume of the cone = Volume of hemisphere
13πr2h=23πr3
⇒6×6×h=2×8×8×8
h=2×8×8×86×6
h= 28.44
Hence, the height of cone is 28.44cm
- How many cubic centimeters of iron is required to construct an open box whose external dimension are 36 cm, 25cm and 16.5 cm provided the thickness of the iron weights 7.5g, Find the weight of the box.
Sol. The Volume of the external box = 36×25×16.5
=14850
Volume of the internal box=33×22×15
=10890cm3
Therefore length =36-1.5-1.5 =33cm
breadth = 25-1.5-1.5 =22cm
Height =16.5-1.5 =15cm
Volume of iron used =14850-10890
=3960 cubic cm
Weight of the box = 3960×7.5
=29700g
=29.7kg
- Water flows at the rate of 10m/ minute through a cylindrical pipe 5 mm in diameter. How long would it take to fill a conical vessel whose diameter at the base is 40cm and depth 24 cm?
Sol. Radius of cylindrical pipe =52mm
=52cm
Length per min = 10m
=1000cm
thereforeVolumeofwaterflowingperminute=πr2h
=227×520×520×1000
Here, Radius of the conical vessel = 20cm
Depth of the conical vessel =24cm
Now Volume of the conical vessel=13×227×20×20×24
Therefore Time required to the conical vessel
=13×227×20×20×24227×520×520×1000
=51minutes 12 second
- A factory manufactures 120000 pencils daily. All pencils are cylindrical in shape each of length 25cm , circumference of base as 1.5cm. Determine the cost of coloring the curved surface of the pencils manufactured in one day at the rate of 0.05 per dm3.
Sol. Given that, Circumference of base of pencil =1.5cm
2πr=1.5cm
r=1.52πcm
So, Curved surface area of each pencil =2π×1.52π×25
=37.5cm2
Total Curved surface area of 120000 pencils
=0.375×120000=45000dm2
Total cost of coloring the C.S.A is 120000 pencils
=0.05×45000=2250
- water is following at the rate of 15 km/h through a pipe diameter 14cm, a cuboidal pond which is 50 m long and 44m wide. IN what time will the level of water in pond rise by 21cm?
Sol. Hence, Volume of the pound =×50m×44m×21100m
=462m3
Radius of pipe (r) =142×1100
=0.07m
Length of the pipe per hour= 15km
=15000m
Therefore Water collected per hour =227×0.07×0.07×15000
=231m3}
Time required to rise the water level by 21cm
=462231
=2hours
- A milk container of height 16cm is made of metal sheet in the from of a frustum of a cone with radii of its lower and upper ends as 8cm and 200cm 22 Rs. per liter which the container canhold.
Sol. Height of the frustum of cone(h)= 16 cm
Radii of lower and upper end are
R=8cm and R = 20cm
Therefore Volume of the milk =13πh(R2+r2+Rr)
=13×227×16(202+82+20(8))
=22×1621×(400+64+160)
=22×1621×624
=10459.43cm3
=10459.431000litres
=10.459litres
Total cost of the milk =22×10.459Rs
=230.10Rs
- A rocket is in the from of a right circular cylinder close at the lower end surmounted by a cone with the same radius as that of the cylinder .The diameter and height of the cylinder are 6cm and 12 cm respectively. If the slant height of the conical portion is 5cm, find the total surface area and volume of the rocket. [use π=3.14]
Sol. Height of the cone =52−32−−−−−−√
=25−9−−−−−√=16−−√=4
Volume of the rocket =πr2H+13πr2h
=πr2(H+13h)
=3.14×9(12+13×4)
=3.14×19×403
=376.8cm3
Total surface area =πr2+2πrh+πrl
=πr(r+2h+l)
=3.14×3(3+24+5)
=3.14×3×32
=301.44cm2
- A hemispherical bowl of internal radius 9 cm is full of liquid. The liquid is to be filled into cylindrical shaped bottle
Sol. Volume of liquid in hemispherical bowl =23×227×9×9×9cm3
Volume of each of cylindrical bottle = 227×1.5×1.5×4cm3
Let n be the number of cylindrical shaped bottles.
therefore,n×227×1.5×1.5×4 n=2×9×9×93×1.5×1.5×4
=54
Hence, the required number of bottles needed to empty the blow are 54.
- A pen stand made of wood is in the shape of a cuboid with four conical depression and cubical depression to hold the cuboid are 10cm, 5cm and 4cm.The radius of each of the conical depression is 0.5 cm. find the volume of the wood in the enter stand.
Sol. Required volume of wood in stand = Volume of cuboid – 4 (Volume of one conical depression ) – Volume of cubical depression
=10×5×4−13×227×0.5×0.5×2.1−3×3×3
=200-2.2-27
=170.8
Exercise 13.1
- Two cubes each having volume 64 cm3 are connected end to end. Calculate the surface area of the cuboid formed.
Ans- Volume of each cube=64cm3
If “x” is the side of cube, x3=64
⇒x=64−−√3⇒x=4cm.
Thereforesurfaceareaofacube=6x2=6(4)2=6(16)=96cm2
thereforesurfaceareaofresultingcuboid
=2(surfaceareaofeachcube)−(surfaceareaofoneside)
=(2×96)−2x2=192−2(4)2=192−32=160cm2.
- A container which is in the shape of a hollow hemisphere is covered from the top by a hollow cylinder with diameter of 14 cm and the total height of the container is 13 cm. Calculate the inner surface area of the container.
Ans.-Radius = 7 cm
Height of cylindrical portion = 13 – 7 = 6 cm
Curved surface are of cylindrical portion can be calculated as follows:
=2πrh =2×22÷7×7×6 =264cm2
Curved surface area of hemispherical portion can be calculated as follows:
=2πr2 =2×22÷7×7×7 =308cm2
Total surface are = 308 + 264 = 572 sq cm
3) A toy in a shape of a cone having radius 3.5 cm, mounted on a hemisphere with same radius. The total height of the toy is 15.5 cm. Calculate the total surface area of the toy.
Ans.-Radius of cone = 3.5 cm, height of cone = 15.5 – 3.5 = 12 cm
Slant height of cone can be calculated as follows:
a=h2+r2−−−−−−√ a=122+3.52−−−−−−−−√ a=144+12.5−−−−−−−−√ a=156.25−−−−−√=12.5cm
Curved surface area of cone can be calculated as follows:
=πra =22÷7×3.5×12.5 137.5cm2
Curved surface area of hemispherical portion can be calculated as follows:
=2πr2 =2×22÷7×3.5×3.5 77cm2
Hence, total surface area = 137.5 + 77 = 214.5 sq cm.
- A cubical block with side 7 cm is surmounted by a hemisphere. Identify the greatest diameter the hemisphere which can be formed? Determine the surface area of the solid.
Ans.-The greatest diameter = side of the cube = 7 cm
Surface Area of Solid = Surface Area of Cube – Surface Area of Base of Hemisphere + Curved Surface Area of hemisphere
Surface Area of Cube = 6 x Side2
= 6 x 7 x 7 = 294 sq cm
Surface Area of Base of Hemisphere
=πr2
22÷7×3.52=38.5cm2
Curved Surface Area of Hemisphere = 2 x 38.5 = 77 sq cm
Total Surface Area = 294 – 38.5 + 77 = 332.5 sq cm
- A hemispherical dint is removed by cutting out from one side of a cubical wooden block in such a way that the diameter ‘d’ of the hemisphere becomes equal to the edge of the cube. Calculate the surface area of the remaining solid.
Ans.- This question can be solved like previous question. Here the curved surface of the hemisphere is a dint, unlike a projection in the previous question-:
Total Surface Area-:
6a2−π(a÷2)2π+2π(a÷2)2 =6a2+π(a2)2 =14a2×(π+24)
- A medicine capsule having the shape of a cylinder consisting two hemispheres stuck to each of its ends. The length of the total capsule is 14 mm with diameter of 5 mm. Calculate its surface area.Ans.- Height of Cylinder = 14 – 5 = 9 cm, radius = 2.5 cm
Curved Surface Area of Cylinder
2πrh
=2π×2.5×9
=45πcm2
Curved Surface Area of two Hemispheres
=4πr2=4π×2.52
=25πcm2
Total Surface Area
=45π+25π
=70π=220cm2
- A tent having the shape of a cylinder is covered by a conical top. The height of the cylindrical part is 2.1 m with a diameter of 4 m having a slant height of 2.8 m, calculate the area of the fabric used for making the tent. Also, calculate the cost of the fabric of the tent at the rate of Rs 500 per m2. (Note that the bottom of the tent can’t be covered with fabric).
Ans.- Radius of cylinder = 2 m, height = 2.1 m and slant height of conical top = 2.8 m
Curved Surface Area of cylindrical portion
=2πrh =2π×2×2.5 =8.4πm2
Curved Surface Area of conical portion
=πrl =π×2×2.8=5.6πm2
Total CSA
=8.4π+5.6π =14×227=44m2
Cost of fabric = Rate x Surface Area
= 500 x 44 = Rs. 22000
- From a solid cylinder having height and diameter of 2.4 cm and 1.4 cm respectively, a conical cavity with same height and same diameter is hollowed out. Calculate the total surface area of the remaining solid to the closest cm2.
Ans.- Radius = 0.7 cm and height = 2.4 cm
Total Surface Area of Structure = Curved Surface Area of Cylinder + Area of top of cylinder + Curved Surface Area of Cone
Curved Surface Area of Cylinder
= 2πrh
=2π×0.7×2.4=3.36πcm2
Area of top
=πr2 =π×0.72 =0.49πcm2
Slant height of cone can be calculated as follows:
=l=r2+h2−−−−−−√ =2.42+0.72−−−−−−−−−√ =5.76+0.49−−−−−−−−−√ =6.25−−−−√=2.5cm
Curved Surface Area of Cone
=πrl =π×0.7×2.5 =1.75πcm2
Hence, remaining surface area of structure
=3.36π+0.49π+1.75π =5.6π=17.6cm2 =18cm2(approx)
- A wooden article has been made by removing out a hemisphere from each end of a solid cylinder, as given in figure. If the height and radius of base of the cylinder is 10 cm and 3.5 cm respectively, then calculate the complete surface area of the article
Ans.- Radius = 3.5 cm, height = 10 cm
Total Surface Area of Structure = CSA of Cylinder + CSA of two hemispheres
Curved Surface Area of Cylinder
=2πrh =2π×3.5×10 =70πcm2
Surface Area of Sphere
=4πr2 =4π×3.52 =49π
Total Surface Area
=70π+49π=119π =119×227=374cm2
Exercise 13.2
1: A solid cone placed on a hemisphere with both their radii whose value is equal to 1 cm and the height of the cone is equal to its radius. Calculate the volume of the solid in terms of π.
Ans.- radius = 1 cm, height = 1 cm
Volume of hemisphere
=23πr3 =23π×13 =23πcm2
Volume of cone
=13πr2h =13π×12×1=13πcm3
Total volume
=23π+13π=πcm3
2: Harish, who is student, he was asked to make a model of shape similar to a cylinder which contains two cones connected at its two ends by using a metal sheet. The diameter and the height of the model are 3 cm and 12 cm respectively. If both the cone has a height of 2 cm, calculate the volume of air present in the model that Harish made. (Assume that both the inner and outer dimension of the model is almost same).
Ans.- Height of cylinder = 12 – 4 = 8 cm, radius = 1.5 cm, height of cone = 2 cm
Volume of cylinder
=πr2h =π×1.52×8=18πcm3
Volume of cone
=13πr2h =13π×1.52×2 =1.5πcm3
Total volume
=1.5π+1.5π+18π =21π=66cm3
3: A sweet, which contains sugar syrup up to about 30% of its volume. Calculate approximately how much syrup will be available in 45 sweets, each shaped like a cylinder having two hemispherical ends with length 5 cm and 2.8 cm diameter.
Ans.- Length of cylinder = 5 – 2.8 = 2.2 cm, radius = 1.4 cm
Volume of cylinder
=πr2h =π×1.42×2.2 =4.312πcm3
Volume of two hemispheres
=43πr3
=43π×1.43 =10.9763πcm3
Total volume
=4.312π+10.9763π
Volume of syrup = 30% of total volume
=π(4.312+10.9763)×30100 =23.9123×30100×227=7.515cm3
Volume of syrup in 45 sweets = 45 x 7.515 = 338.184 cm3
4: A flower pot that is made of wood having the shape of a cuboid with four conical depressions to hold flowers. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Calculate the volume of wood in the entire pot.
Ans.- Dimensions of cuboid is 15 cm x 10 cm x 3.5 cm, radius of cone is 0.5 cm, depth of cone is 1.4 cm
As we know Volume of cuboid = length x width x height
So, 15×10×3.5=525cm3
Now volume of cone =13πr2h
=13×227×0.52×1.4=1130cm3
therefore Hence volume of wood =Volume of cuboid – 6 x volume of cone
=525−6×1130 =525−115=522.8cm3
5: A container is in the shape of an inverted cone. The height and the radius at the top (which is open) of the container are 8 cm and 5 cm respectively. The container is filled with water up to the upper edge. When lead shots, each of which is a sphere of radius 0.5 cm are dropped inside the vessel, quarter quantity of the water flows out. Calculate the number of lead shots dropped in the vessel.
Ans.- Given, radius of cone is 5 cm, height of cone is 8 cm, radius of sphere is 0.5 cm
So volume of cone=13πr2h
13π×52×8 2003πcm3
Similarly volume of lead shot=43πr3
43π×0.53 16πcm3
Now number of lead shots will be
=2003π×14÷16π
=50π3×6π=100
6: A solid metal pole consisting of a cylinder of height and base diameter as 220 cm and 24 cm respectively is surmounted by another cylinder of height and radius as 60 cm and 8 cm respectively. Calculate the mass of the pole, given that 1 cm3 of iron has approximately 8g mass.
Ans.- Here radius of bigger cylinder = 12 cm, height of bigger cylinder = 220 cm
Similarly, radius of smaller cylinder = 8 cm, height of smaller cylinder = 60 cm
As we know volume of bigger cylinder=πr2h
=π×122×220
=31680πcm3
Now volume of small cylinder=πr2h
=π×82×60
=3840πcm3
Therefore total volume=31680π+3840π
=35520πcm3
Hence, Mass= Density. Volume
=8×35520π=892262.4gm= 892.3kg
7: A solid containing of a right circular cone of height and radius are 120 cm and 60 cm respectively standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of liquid such that it touches the bottom. Calculate the volume of liquid left in the cylinder, if the radius and height of the cylinder are 60 cm and 180 cm respectively.
Ans.- Here, Radius of cone = 60 cm, height of cone = 120 cm
Radius of hemisphere = 60 cm
Radius of cylinder = 60 cm, height of cylinder = 180 cm
So as we know that volume of cone is =13πr2h
=13π×602×120
=144000πcm3
Volume of hemisphere
=43πr3
23π×603=144000cm3
Similarly, volume of solid=(144000+144000)π
=288000πcm3
Now volume of cylinder=πr2h
=π×602×180=648000πcm3
Hence the total volume of liquid left in the cylinder is
=(648000−288000)π=360000π
=1130400cm3
8: A spherical glass vessel having a cylindrical neck of height 8 cm and diameter 2 cm; the radius of the spherical part is 4.25 cm. By determining the amount of water it holds, a child calculates its volume to be 345 cm3. Identify whether he is correct, taking the above as the inside measurements, and π = 3.14.
Ans.- Here, Radius of cylinder = 1 cm, height of cylinder = 8 cm, radius of sphere = 4.25 cm
As we know that volume of cylinder=πr2h
=π×12×8
=8πcm3
Similarly volume of sphere=43πr3
=43π×(8.52)3
=6141256000πcm3
therefore,the total volume will be
=(6141256000+8)π
=(614125+480006000)π=346.51cm3
Exercise 13.3
1: A metallic sphere having radius of 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Calculate the height of the cylinder.
Ans.- Here, Radius of sphere is 4.2 cm, radius of cylinder is 6 cm
As we know that volume of sphere
=43πr3
=43π×4.23
Similarly volume of cylinder
=πr2h=π×62×h
Since volume of cylinder = Volume of sphere
Hence, height of cylinder.
2: Metallic spheres having radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Calculate the radius of the resulting sphere.
Ans.- Here, Radii of spheres = 6 cm, 8 cm, 10 cm
As we know that volume of sphere=43πr3
So total volume of three spheres is
=43π(63+83+103)
=43π(616+512+1000)
=43π×1728
Hence radius of biggest sphere is
=1728−−−−√3=12cm.
3: A 20 m deep well having diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Calculate the height of the platform.
Ans.- Here, Radius of well = 3.5 m, depth of well = 20 m
So the dimension of rectangular platform =22m×14m
Now the volume of earth dug out
=πr2h
=π×3.52×20=770m3
Area of top of platform = Area of Rectangle – Area of Circle
(as the circular portion of mouth of well is open)
=22×14−π×3.52
=308−38.5=269.5m2
So as we know, height=volume/area
=770269.5=2.85m
4: A well is dug with diameter 3 m and a depth of 14 m. The earth taken out of it has been spread in a same quantity surrounding the well forming a shape of circular ring of width 4 m to form an embankment. Calculate the height of the embankment.
Ans.- Here, Radius of well = 1.5 m, depth of well = 14 m, width of embankment = 4 m
So the radius circular embankment=4+1.5=5.5m
Now the volume of earth dug out,
=πr2h
=π×1.52×14=31.5πm3
Area of top of platform = (Area of bigger circle – Area of smaller circle)
So as we know, height=volume/area
=31.5π28π=1.125m
5: A vessel shaped like a right circular cylinder with diameter 12 cm and height 15 cm is full of ice cream. The ice cream needs to be filled into cones having height 12 cm and diameter 6 cm, forming a hemispherical shape on the top. Calculate the number of such cones which can be filled with ice cream.
Ans.- Here, Radius of cylinder = 6 cm, height of cylinder = 15 cm
Similarly, Radius of cone = 3 cm, height of cone = 12 cm
Given, Radius of hemispherical top on ice cream = 3 cm
Now as we know that, volume of cylinder
= πr2h
= π×62×15=540πcm3
Volume of cone is
=13×π×32×12
=36πcm3
Now, volume of hemisphere is
=23πr3
= 23×π×33=18πcm3
So volume of ice-cream will be
= (36+18)π=54πcm3
Hence the number of ice creams = Volume of cylinder/Volume of ice cream
540π54π=10
6: How many silver coins, 1.75 cm in diameter and of thickness 2 mm, can be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm?
Ans.- Here, Radius of coin = 0.875 cm, height = 0.2 cm
Dimensions of cuboid = 5.5 cm x 10 cm x 3.5 cm
As we know that the volume of coin is
=πr2h
= π×0.8752×0.2= 0.48125cm3
Similarly, volume of cuboid is= 5.5×10×3.5=192.5cm3
Therefore Number of coins
= 192.50.48125=400
7: A cylindrical bucket with 32 cm high and having a radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, calculate the radius and slant height of the heap.
Ans.- Given, Radius of cylinder = 18 cm, height = 32 cm
Height of cone = 24 cm
As we know that the volume of cylinder is
= πr2h
= π×182×32
As, Volume of cone = Volume of cylinder
So, volume of cone is
= 13πr2×24
Hence, the radius of cone can be calculated as follows:
r2=3×π×182×32π×24
Or, r2=182×22
Or, r = 36cm
So now the slant height of conical heap can be calculated as follows:
l = h2+r2−−−−−−√
= 242+362−−−−−−−−√
= 576+1296−−−−−−−−−√=1872−−−−√
= 3613−−√cm
8: Water in a canal with 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?
Ans.- Given, Depth = 1.5 m, width = 6 m, height of standing water = 0.08 m
In 30 minutes, length of water column = 5 km = 5000 m
Volume of water in 30 minutes = 1.5 x 6 x 5000 = 45000 cubic m
So as we know that,
Area = Volume/Height
= 450000.08=562500m2
9: A farmer connects a pipe with internal diameter of 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If the water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?
Ans.- Given, Radius of pipe = 10 cm = 0.1 m, length = 3000 m/h
Radius of tank = 5 m, depth = 2 m
So the volume of water in 1 hr. through pipe is
= πr2h
= π×0.12×3000= 30πm3
Similarly, the volume of tank is
= πr2h
= π×52×2=50πm3
Hence the time taken to fill the tank is = Volume of tank/Volume of water in 1 hr.
= 50π30π=1 hr. 40min
Exercise 13.4
1: A drinking glass with the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Calculate the capacity of the glass.
Ans.- Given, R = 2, r = 1 cm and h = 14 cm
We know that, volume of frustum is
=13πh(R2+r2+Rr) =13π×14(22+12+2) =13×227×14×7 =10223cm3
2: The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Calculate the curved surface area of the frustum.
Ans.- Given, Slant height l = 4 cm, perimeters = 18 cm and 6 cm
Radii can be calculated as follows:
We know that, Radius= Perimeter2π
= 182π=9π
And radius = 62π=3π
Therefore the curved surface area of frustum is
= π(R+r)l
= π(9π+3π)4= 48cm2
3: A fez, the cap used by the Turks, is shaped like the frustum of a cone. If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, calculate the area of material used for making it.
Ans.- Given, R = 10 cm, r = 4 cm, slant height = 15 cm
We know that, Curved surface area of frustum is
= π(R+r)l
= π(10+4)15
= 227×14×15=660
Now area of upper base is
= πr2=π×42
= 16π=5027
So the total surface area is
660+5027=71027cm2
4: A vessel, opened from the top and made up of an aluminium sheet, is in the form of a frustum of a cone having height, radii of its lower and upper ends are 16cm, 8 cm and 20 cm, respectively. Calculate the cost of the milk which can completely fill the vessel, at the rate of Rs 20 per litre. Also calculate the cost of aluminium sheet used to make the vessel, if it costs Rs 8 per 100 cm2.
Ans.- Given, Height of frustum = 16 cm, R = 20 cm, r = 8 cm
We know that, volume of frustum is
= 13πh(R2+r2+Rr)
= 13π×16(202+82+160)
= 13×227×16(400+64+160)
= 13×227×16×604=10449.92cm3
Cost of milk @ Rs. 20 per 1000 cubic cm
10.44992 x 20 = Rs. 208.99
For calculating surface area, we need to find slant height which can be calculated as follows:
l=h2+(R−r)2−−−−−−−−−−−√
= 162+(20−8)2−−−−−−−−−−−−√
= 256+144−−−−−−−−√=400−−−√=20cm
Now the surface area of frustum is
= π(R+r)l+πr2
= π[(R+r)l+r2]
= π[(20+8)20+82]
= π(560+64)
= 227×624=1959.36
Therefore Cost of metal sheet @ Rs. 8 per 100 sq cm = 19.5936 x 8 = Rs. 156.75
5: A metallic right circular cone of 20 cm high and whose vertical angle is 60∘ is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter 1/16 cm, calculate the length of the wire.
Ans.- Here volume of frustum will be equal to the volume of wire and by using this relation we can calculate the length of the wire.
In the given figure; AO = 20 cm and hence height of frustum LO = 10 cm
In triangle AOC we have angle CAO = 30 (half of vertical angle of cone BAC)
Therefore;
tan30∘=OCAO
Or, 13√=OC20
Or, OC= 203√
Using similarity cirteria in triangles AOC and ALM it can be shown that LM = 10/√3 (because LM bisects the cone through its height)
Similarly, LO = 10 cm
Volume of frustum can be calculated as follows:
V = 13π×10(4003+1003+2003)
= 70009πcubcm
Volume of cylinder is given as follows:
πr2h
Or, π×(132)2×h
= 70009π
Or, h= 70009×1024=796444.44cm
= 7964.4m
Exercise 13.5
1: A copper wire having a diameter of 3mm, is wound about a cylinder whose length is 12 cm, and diameter 10 cm, so as to cover the curved surface of the cylinder. Calculate the length and mass of the wire, assuming the density of copper to be 8.88 g per cm3.
Ans.-Given, For copper wire: Diameter is = 3 mm = 0.3 cm
For cylinder; length h is = 12 cm, d = 10 cm
Density of copper is = 8.88 gm cm3
So now the curved surface area of cylinder can be calculated as follows:
We know that-: Length = AreaWidth
= 120π0.3=400π
= 1256cm
Similarly, volume of wire can be calculated as follows:
V = πr2h
= 3.14×0.152×1256=88.7364cm3
Hence, mass can be calculated as-: Mass = Density*Volume
8.88×88.7364=788g(approx)
2: A right angled triangle, whose sides with 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Calculate the volume and surface area of the double cone so formed.
Ans.- In the given triangle ABC;
AC2=AB2+BC2
Or, AC2=32+42=9+16=25
Or, AC=5CM
In triangle ABC and triangle BDC;
∠ABC=∠BDC ∠BAC=∠DBC
Hence, △ABC∼△BDC
So, now we get following equations:
ABAC=BDBC
Or, 35=BD4
thereforeBD=3×45=2.4
In triangle BDC;
DC2=BC2−BD2
Or, DC2=42−2.42=16−5.76=10.24
thereforeDC=3.2
= 13×3.14×2.42×1.8=10.85184cm3
Upper Cone: r = 2.4 cm, l = 3 cm, h = 1.8 cm
We know that, volume of cone is
= 13πr2h
Now, Curved surface area of cone
=πrl
= 3.14×2.4×3=22.608cm3
Similarly, for lower cone: r = 2.4 cm, l = 4 cm, h = 3.2 cm
As we know that, volume of cone is
= 13πr2h
= 13×3.14×2.42×3.2=19.2916cm3
Now, Curved surface area of cone is
=πrl
= 3.14×2.4×4=30.144cm2
Total volume = 19.29216 + 10.85184=30.144cm3
thereforeTotalsurfacearea=30.144+22.608=52.752cm2
3: A cistern measuring 150 cm x 120 cm x 110 cm has 129600 cm3 water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one seventeenth of its own volume of water. Identify how many bricks can be put in without overflowing the water, each brick being 22.5 cm x 7.5 cm x 6.5 cm?
Ans.- Here volume of cistern is = length x width x depth
= 150 x 120 x 110
= 1980000 cm3
Vacant space = Volume of cistern – Volume of water
= 1980000 – 129600 = 1850400 cm3
Volume of brick = length x width x height
= 22.5 x 7.5 x 6.5
Since the brick absorbs one seventeenth its volume hence the remaining volume will be equal to 16/17 the volume of brick
Remaining volume is
= 22.5×7.5×6.5×1617
therefore Number of bricks = Remaining volume of cistern/remaining volume of brick
= 1850400×1722.5×7.5×6.5×16
= 3145680017550=1792
5: An funnel made of a metallic sheet consisting of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height and the diameter of the cylindrical portion are 22 cm and 8 cm respectively and diameter of the top of the funnel is 18 cm, calculate the area of the metallic sheet required to make the funnel.
Ans.- Curved surface area of cylinder
Slant height of frustum can be calculated as follows:
Curved surface area of frustum
therefore Total curved surface area is
= 169π+80π=249+3.14
= 781.86cm2
Exercise 13.5
1. A matchbox measures 5cm × 1cm × 3.5cm. Determine the volume of a packet containing 14 such boxes?
Ans.- Given the dimension of the matchbox = 5cm × 1cm × 3.5cm
Let us assume, l = 5cm, b = 1cm, h = 3.5cm
As we know that, Volume of one matchbox = (l × b × h)
= (5×1×3.5)cm3=17.5cm3
∴Thevolumeofapacketcontaining14suchboxes=(17.5×14)cm3=245cm3
2. A cuboidal water tank is 10 m long, 2 m wide and 7.5 m deep. How many litres of water can it hold? (1m3 = 1000 l)
Ans.- Dimensions of water tank = 10m × 2m × 7.5m
Let us assume, l = 10m, b = 2m, h = 7.5m
Therefore Volume of the tank = (l×b×h)m3
= (10×2×7.5)m3=150m3
Hence, the tank can hold = 150 x 1000 litres = 150000 litres of water.
3. A cuboidal vessel is 25 m long and 12 m wide. Determine the height that must be made to hold 400 cubic metres of a liquid?
Ans.– Given, Length = 25 m , Breadth = 12 m and Volume = 400m3
As we know, Volume of cuboid = Length x Breadth x Height
Therefore, Height = Volume of cuboid/(Length × Breadth)
= 40025×12m=1.33m
4. Find the value of digging a cuboidal pit 15 m long, 5 m broad and 3 m deep at the rate of Rs.50 per m3.
Ans.- Here, length = 15m, breadth = 5m and height = 3m
As we know that, Volume of the pit = (l×b×h)m3
= (15×5×3)m3=225m3
The rate of digging is = Rs.50 per m3
∴ The total value of digging the pit = Rs.(225 x 50)
= Rs.11250
5.The capacity of a cuboidal tank is 2,10000 litres of water. Calculate the breadth, given the length is 3.5m and depth is 20m.
Ans.- Given, length = 3.5m, depth = 15m and volume = 30000 litres
As we know that, 1m3=1000litres
∴210000litres=2100001000m3= 2101m3
Breadth = volumeofcuboidlength×depth
= 210(3.5×20)m
= 3m
- A village, with a population of 6000, requires 200 litres of water per head per day. It has a tank measuring 30m × 25m × 8m. Justify the number of days that will take to empty the water tank ?
Ans.- Given, the dimension of the tank = 30m × 25m × 8m
So, l = 30m, b = 25m and h = 8m
As we know that, the total capacity of the tank = (30×25×8)m3=6000m3
Water required for a single person per day = 200 litres
The requirement of water for 6000 person in a single day = (6000 x 200) litres
= (6000×200)1000=1200m3
Hence, the number of days the water will last = (the capacity of the tank /water required per day) = (60001200)=5
∴ The water lasts for 5 days.
- A warehouse measures 50m × 35m × 25m. Calculate the maximum number of wooden box each measuring 2.5m × 1.5m × 1m that can be stored in the warehouse.
Ans.- Given the dimensions of the warehouse = 50m x 35m x 25m
As we know that, the volume of the warehouse will be = (lbh)m3
= (50×35×25)m3=43750m3
Now, the dimension of box = 2.5m x 1.5m x 1m
Similarly, volume of 1 box = (2.5×1.5×1)m3=3.75m3
Hence, Number of box that can be stored = volume of warehouse / volume of 1 box = 437503.75=11666.666=11666
- A solid cuboid having side 20 cm is cut into 16 cubes of equal volume. Calculate the side of the new cuboid and also calculate the ratio between their surface areas.
Ans.- Here the edge of the cube = 20cm
So, Volume of the cuboid = (edge)3cm3
= (20×20×20)cm3=8000cm3
Now, The number of smaller cube = 16
So, the volume of 1 small cube = 800016cm3=500cm3
Let us assume the side of small cube as ‘p’
p3=500⇒p=7.937(approx)
Hence, the surface area the cube = 7.937(side)2
Therefore, the ratio of their surface area
= (7.937 x 20 x 20)/(7.937 x 7.937 x 7.937)
= 401.585 = 40 :1.585
- A river 5 m deep and 60 m wide is flowing at a rate of 6 km per hour. Estimate the amount of water that will fall into the sea in a minute?
Ans.- Given, Depth (h) = 5m
Width (b) = 60m
So, the rate of flow of water (l) = 6km per hour= (600060)mperminute=100mperminute
Therefore, the volume of water flowing into the sea in a minute = lbhm3
(100×60×5)m3=30000m3
Related Links
NCERT Solutions for Class 10 Maths – Areas Related to Circles
1. If perimeter of a circle is equal to that of a square, What would be the ratio of their areas.
Here, perimeter of circle = 2πr and perimeter of square = 4a
∴2πr = 4a ⇒r=2aπ
∴ Ratio of their areas = πr2a2 = πa2×(2aπ)2
= πa2×(4a2π2) = 4π = 422×7 = 1411
2. Find the areas of circle that can be inscribed in a square of side 8 cm.
Diameter of the circle = Side of the square
∴ Diameter = 8 cm
⇒ Radius = 82 = 4 cm
∴ Area of the circle = π×4×4 cm2
= 9π cm2
3. If sum of the areas of two circles with radii R1andR2 is equal to area of a circle radius R, derive the relation among their radii.
Since sum of areas of two circles with radii R1andR2 = Area of circle with radii R.
⇒ πR21+πR22=πR2
⇒ R21+R22=R2 or R = R21+R22−−−−−−−√
4. If sum of the circumference of two circles with radii radii R1andR2 is equal to circumference of a circle radius R, derive the relation among their radii.
Since sum of circumference of two circles with radii R1andR2 = Circumference of circle with radii R.
⇒ 2πR1+2πR2=2πR
⇒ R1+R2=R
5. If the circumference of a circle and the perimeter of a square are equal, then write the relation between their radii.
Since sum of circumference of a circle = Perimeter of a square
⇒ 2πr=4×side
⇒ r=2π×side
∴ Area of circle = π×4π2×(side)2
= 4π(side)2
= 2822(side)2
Area of square = (side)2
Since 2822(side)2 > (side)2
6. To build a single circular park equal in area to the sum of areas of two circular parks of diameter 8 m and 6 m in a locality. What would be the radius of new park?
Here, area of single park = Area of the park with (d= 8m) + Area of the park with (d= 6m)
⇒ πR2=π(4)2+π(3)2
⇒ πR2=π(16+9)
⇒ R2=25
⇒ R = 5 m.
7. Find the area of a square that can be inscribed in a circle of radius 6 cm.
Since square is inscribed in a circle
⇒ Diagonal of square = Diametre of circle
⇒ 2–√×side = 12
Side = 122√ cm
Now, area of the square = 122√×122√ = 72 cm2
8. Find the radius of a circle whose circumference is equal to the sum of the circumference of two circles of diameter 24 cm and 16 cm.
From the above, we obtain
2πR=2πR1+2πR2
⇒ R=R1+R2
⇒ R = 12 + 8 = 20 cm
9. Find the diameter of a circle whose area is equal to the sum of the areas of the two circles of radii 20 cm and 15 cm.
From the above, we obtain
πR2=πR21+πR22
⇒ R2=R21+R22
⇒ R2=(20)2+(15)2 = 400 + 225 = 625
⇒ R = 25 cm
10. Find the area of the largest triangle that can be inscribed in a semicircle of radius r.
Here, among all altitudes, radius is the largest altitude
∴ Areaof△PQR=12×2r×r
= r2sq.units
Hence, area of the largest triangle inscribed in a semicircle r is r2sq.units.
11. In the given figure , a square of diagonal 14 cm is inscribed in a circle. Find the area of the shaded region.
Here, diagonal of the square = 14 cm
⇒ 2–√×side = 14
⇒ side = 142√ cm
∴ Area of the square = 82√×82√ = 98 sq.cm
Diameter of the circle = 14 cm
∴ Radius of the circle = 7 cm.
Area of the circle = π(4)2 = 227×49 = 154 sq.cm
Thus, area of the shaded region = 154 – 98
= 56 sq.cm
12. The wheel of a motor cycle is of radius 28cm. How many revolutions per minitue must the wheel make so as to keep a speed of 60 Km/h?
Here, speed of the motor cycle = 60 Km/h
= 6000060 m/min
= 1000 m/min
Radius of the wheel (r) = 28 cm = 28100 m.
Circumference of the wheel = 2×227×28100 m
= 1.76 m
Number of revolutions per minute = 10001.76
= 568.18
13. A cow is tied with a rope of length 21 m at the corner of a rectangular field of dimensions 27 m × 23 m. Find the area of the field in which the cow can graze.
Length of the rope = 21 m
∴ Area of the field in which the cow can graze
= 90°360°×227×21×21
= 346.5 m2
14. In the given fig, arcs are drawn by taking vertices P,Q and R of an equilateral triangle of side 12 cm to intersect the sides QR, RR and PQ at their respective mid-points X, Y and Z. Find the area of the shaded region.
Side of an equilateral m2△ PQR = 12 cm
Since, arcs are drawn by taking vertices P,Q and R to intersect the sides QR, RR and PQ at their mid-points X, Y and Z.
∴ Radii of each sector (r) = 6 cm
Also, m2△ PQR is an equilateral, therefore
∠P=∠Q=∠R=60°
Required area of the shaded region
= 3×θ360°×πr2
= 3×60°360°×3.14×62
= 3×16×3.14×36
= 56.52 cm2
15. In the given fig, arcs have been drawn with radii 7 cm each and with centers A, B and C.
Find the area of the shaded region.
Here, radii of each sector at A, B and C is 7 cm
Let ∠A=θ1°∠B=θ2°,∠C=θ3°
Required area of shaded region
= θ1°360°πr2+θ2°360°πr2+θ3°360°πr2
= πr2360°(θ1°+θ2°+θ3°)
= 22×7×77×360°×180° [Sinceθ1°+θ2°+θ3°=180°]
= 77 cm2
16. A circular park is surrounded by a road 20 m wide. If the radius of the park is 100 m, find the area of the road.
Here, radius of the circular park (r) = 100 m
Width of the road = 20 m
∴ Radius of the outer circle (R) = 100 + 20 = 120 m
Required area of road
= πR2+πr2
= π (R + r) (R – r)
= 227 (120 + 100) (120 – 100)
= 227×220×20 = 13816 m2
17. A piece of wire 22 cm long is bent into the form of an arc of a circle subtending an angle of 90° at its center. Find the radius of the circle.
Here, length of wire (arc of required circle) = 22 cm
And central angle(θ) = 90°
Let r be the radius of the required circle.
∴θ360°×2πr = length of the arc
∴60°360°×2×227r = 20
⇒r=22×7×360°60°×2×22 = 21 cm
18. The length of the minute hand of a clock is 7 cm. Find the area swept by the minute hand during the time period 6:05 A.M and 6:40 A.M.
Angle swept by the minute hand in one minute = 6°
∴Angle swept by the minute hand in 35 minute = (6×35)°=210°
Length of the minute hand = 7 cm
Area swept by the minute hand = 210°360°×227×7×7
= 89.83 cm2
19. Find the area of the sector of a circle of radius 7 cm, if the corresponding arc length is 4 cm.
Radius of the circle (r) = 7 cm
Length of the arc = 3.5 cm
∴ θ360°×2πr = 4 cm
⇒θ360°×2×227×7 = 4
⇒θ360° = 4×72×22×7 = 111
Now, area of the corresponding arc
= 111×227×7×7
= 14 cm2
20. A circular pond is of diameter 20 m. It is surrounded by a 2 m wide path. Find the cost of constructing the path at the rate of Rs.30 per m2.
Radius of the circular pond r = 202 = 10 m
Width of the path = 2 m
∴ Radius of the outer circle (R) = 10 + 2 = 12 m
Area of the path = πR2−πr2
= π (R + r) (R – r)
= 227 (12 + 10) (12 – 10)
= 227×22×2 = 138.16 m2
Cost of constructing the path @ Rs.30 per m2
= Rs.4144.8
21. Find the number of revolutions made by a circular tier of area 1.32 m2 in rolling a distance of 165 m.
Area of a circular wheel = 1.32 m2
πr2 = 1.32
r2 = 1.32×722 = 0.42
r = 0.42−−−−√ = 0.648 m
Distance traveled in one rotation = 2πr
= 2×227×0.648
= 4.07 m
Total distance to be travelled = 165
∴Number of rotations = 1654.07 = 40.54
22. Find the difference of the areas of sectors of angle 90° and its corresponding major sector of a circle of radius 14 cm.
Radius of the circle (r) = 14 cm
Angle of the sector θ=90°
∴ Area of the sector = θ360°×πr2
= 90°360°×227×14×14
= 154 cm2
Area of major sector
= 360°–90°360°×227×14×14
= 462 cm2
∴ The required distance = 308 cm²
23. Find the area of the shaded region in the give figure.
Here, unshaded region is combination of two semicircles and one rectangle.
∴ Area of unshaded region = Area of rectangle with dimensions (26-3-3-) cm by (12-4-4) cm + 2 area of semicircle with radii (12−4−42)
= 16×4+2×12×π×2×2
= (64+4π) cm2
Required area of shaded region
= 26×12−(64+4π)
= 312 – 64 – 4π
= (248 – 4π) cm2
24. Find the area of the minor segment of a circle of radius 28 cm, when the angle of the corresponding sector is 90°.
Radius of the circle (r) = 28 cm
Central angle (θ) = 90°
Area of minor segment
= Area of sector PRQ – Area of △PRQ
= 90°360°×227×28×28−12×28×28×sin90° 90°
= 616 – 392
= 224 cm2
25. The diameter of front and rear wheels of a tractor 70 cm and 2 m respectively. Find the number of rotations that rear wheel will make in covering a distance in which the wheel makes 1200 rotations.
Radius of the front wheel (r) = 35 cm
Number of rotations = 1200
∴ Total distance covered by front wheel = 1200 2πr
= 1200×2×227×35
= 264000 cm
Radius of the rear wheel (R) = 2002 = 100 cm
Distance covered by the rear wheel in one rotation = 2×227×100
26. Number of rotations = 264000×72×22×100 = 420Sides of a triangular field are 15 m, 16 m and 17 m. With the three corners of the field a cow, a buffalo and a horse are tied separately with ropes of length 7 m each to graze in the field. Find the area of the field which cannot be grazed by the three animals.
Length of the rope = 7 m
Sides of the triangular field are AB = 16 m, BC = 17 m and AC = 15 m
Let cow, buffalo and horse be tied at vertices A, B and C respectively.
∴ Area of △ ABC = 24(24−15)(24−16)(24−17)−−−−−−−−−−−−−−−−−−−−−−−√
= 24×9×8×7−−−−−−−−−−−√ = 2421−−√ m2
Hence, area of the field which cannot be grazed by three animals = (2421−−√−77) m2
27. Find the area of the flower bed ( with semi- circular ends) as shown in the figure.
Here, the flower bed is a combination of two semicircular ends and rectangle.
Radius of semicircular ends = 7 cm
Length and breadth of rectangular field are 35 cm and 8 cm
∴ Area of the shaded region
= 2×12πr2+l×b
= 227×7×7+35×8
= 154 + 280
= 434 cm2
Hence, area of the flower bed is 434 cm2.
28. On a square cardboard sheet of area 196 cm2, four congruent circular plates of maximum size are placed such that each circular plate touches the other two plates and each side of the square sheet is tangent to two circular plates. Find the area of the square sheet not covered by circular plates.
Area of the square PQRS = 196 cm2
Side of the square = 196−−−√ = 14 cm
∴ Diameter of each circular plate = 7 cm
And radius of each circular plate = 3.5 cm
Area of four circular plates = 4×227×3.5×3.5
= 154 cm2
Area of the square sheet not covered by the circular plate = 196 – 154 = 42 cm2.
29. The central angles of two sectors of circles of radii 7 cm and 21 cm are respectively 120° and 40°. Find the areas of two sectors as well as the lengths of those corresponding arcs.
Here, radii of two circles are r1=7cm and r2=21cm
Central angles are Θ1=120° and Θ2=40°
∴ Areas of two sectors are
120°360°×227×7×7 and 40°360°×227×21×21
⇒51.33cm2and154cm2
Lengths of corresponding arcs are
120°360°×227×7×2 and 40°360°×227×21×2
\Rightarrow 14.67 cm and 14.67 cm
Arc lengths of two sectors of two different circles may be equal (14.67 cm), but their areas need not be equal.
30. Floor of a room is of dimension 8 m × 5 m and it is covered with circular tiles of diameters 40 cm each as shown in figure. Find the area of floor that remains uncovered with tiles.
Radii of each circular tile = 20 cm
∴ Number of circular tiles along its length = 20
Number of circular tiles along its breadth = 12.5
Total number of circular tiles =20× 12.5 = 250
Area of each circular tile =3.14×20×20
= 1256 cm²
Area of 250 circular tile = 250 × 1256
= 314000 cm²
Area of the rectangle floor = 800 × 500
= 400000 cm²
Thus, area of the floor that remains uncovered with tiles = 400000 – 314000
= 86000 cm²
= 8.6 cm²
Related Links
NCERT Solutions for Class 10 Maths – Constructions
Exercise 11.1
1) Draw a line segment of length 7.6 cm and divide it in 5: 8 ration. In addition, find the measure of two parts.
Solution:
Procedure for construction:
- Draw any ray AX, making an acute angle with AB.
- Locate 13(= 5 + 8) points A1, A2, A3 ……… A13 on AX so that AA1 = A1A2 Al2 A13.
- Join BA13.
- Through the point A5(m = 5), draw a line parallel to BA13 (by making an angle equal to L AA13 B at A5 intersecting AB at C. Then AC: CB = 5 : 8)
2) Construct a triangle with sides 4 cm, 5 cm and 6 cm and then a similar triangle to it whose sides are 2/3 of the corresponding sides of the first one.
Solution:
Procedure for construction:
- Draw a line segment BC with length 5 cm.
- With B as centre and radius of 4 cm draw an arc.
- With C as centre and radius of 6 cm draw an arc.
- Join AB and AC. Then, ∆ABC is the required triangle.
- Below BC, make an acute angle ∠CBX
- Along BX, mark up three points B1, B2, B3 such that BB1 = B1 B2 = B2B3
- Join B3C
- From B2, draw B2C’llB3c, meeting BC at C’
- From C’ draw C’ All CA, meeting BA at A’
- Then ∆A’BC’ is the required triangle, each of whose sides is two-third of the corresponding sides of ∆ABC.
3) Construct a triangle with side lengths 5 cm, 6 cm and 7 cm and then another triangle whose sides are of the corresponding sides of the first triangle.
Solution:
Procedure for construction :
- Draw a line segment BC with length 6 cm.
- With B as centre and keeping radius as 5 cm, draw an arc.
- With C as centre and keeping radius as 7 cm, draw another arc, intersecting the previously drawn arc at Point A.
- Join AB and AC. Then, ∆ABC is the required triangle.
- Below BC, make an acute angle∠CBX.
- Along BX, mark up seven points B1, B2, B3….. B7 such that BB1 = B1,B2, B6B7.
- Join B5 to C (5 being smaller of 5 and 7 in7/5) and draw a line through B7 parallel to B5C, intersecting the extended line segment BC at C’.
- Draw a line through C’ parallel to CA intersecting the extended line segment BA at A’. Then A’BC’ is the required triangle.
4) Draw a triangle ABC with sides BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are ¾ of the corresponding sides of ∆ABC.
Solution:
Procedure for construction:
(i) Draw a triangle ABC with BC = 6 cm, AB = 5 cm and ∠ABC = 60°.
(ii) Draw any ray BX making an acute angle with BC on the side opposite to the vertex X.
(iii) Locate 4(the greater of 3 and 4 in ¾) points B1, B2, B3, B4 on BX so that BB1 = B1B2 = B2B3 = B3B4.
(iv) Join B4C and draw a line through B3(the 3rd point, 3 being smaller of 3 and 4 in ¾) parallel to B4C to intersect BC at C’.
(v) Draw a line through C’ parallel to the line CA to intersect BA at A’. Then ∆A’BC’ is the required triangle.
Justification of construction
∆ABC ~ ∆A’BC’ , Therefore,
ABA′B=ACA′C′=BCBC′
But, BCBC′=BB3BB4=34
So, ABA′B=ACA′C′=BCBC′=34
5) Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are4/3 times the corresponding sides of ∆ABC.
Solution:
Procedure for construction :
(i) Draw a triangle ABC with BC = 7cm, ∠B = 45° and ∠A = 105°.
(ii) Draw any ray BX making an acute angle with BC on the side opposite to the vertex X.
(iii) Locate 4(the greater of 3 and 4 in 4/3) points B1, B2, B3, B4 on BX so that 3 BB1 = B1 B2 = B2B3 = B3B4.
(iv) Join NC’ and draw a line through B3(the 3rd point, 3 being smaller of 3 and 4 in 1) parallel to NC’ to intersect BC’ at C. 3
(v) Draw a line through C’ parallel to the line CA to intersect BA at A’. Then A NBC’ is the required triangle.
6) Construct a triangle of isosceles type, whose base is 8 cm and height 4 cm and then another triangle whose sides are 1.5 times the corresponding sides of the isosceles triangle.
Solution:
Given:An isosceles triangle whose base is 8 cm and height 4 cm. Scale factor: 1 =
Required: To construct a similar triangle to above whose sides are 1.5 times the above triangle.
Procedure for construction:
(i) Draw a line segment BC = 8 cm.
(ii) Draw a perpendicular bisector AD of BC.
(iii) Join AB and AC we get a isosceles ∆ABC.
(iv) Construct an acute angle∠CBX downwards.
(v) On BX make 3 equal parts.
(vi) Join C to B2 and draw a line through B3 parallel to B2C intersecting the extended line segment BC at C’.
(vii) Again draw a parallel line C’A’ to AC cutting BP at A’.
(viii) ∆A’BC’ is the required triangle.
Exercise 11.2
7) Draw a circle with radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.
Solution:
Procedure for construction:
- Draw a line segment of length AB = 10 cm. Bisect AB by constructing a perpendicular bisector of AB. Let M be the mid-point of AB.
- With M as centre and AM as radius, draw a circle. Let it intersect the given circle at the points P and Q.
- Join PB and QB. Thus, PB and QB are the required two tangents.
Justification: Join AP. Here ∠APB is an angle in the semi-circle. Therefore, ∠APB = 90°. Since AP is a radius of a circle, PB has to be a tangent to a circle. Similarly, QB is also a tangent to a circle.
In a Right ∆APB, AB2 = AP2 + PB2 (By using Pythagoras Theorem)
PB2 = AB2 – AP2 = 102 — 62 = 100 – 36 = 64
PB = 8 cm.
8) Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also, verify the measurement by actual calculation.
Solution:
Procedure for construction:
- Draw a line segment of length OA = 4 cm. With O as centre and OA as radius, draw a circle.
- With O as centre draw a concentric circle of radius 6 cm(0B).
- Let C be any point on the circle of radius 6 cm, join OC.
- Bisect OC such that M is the mid point of OC.
- With M as centre and OM as radius, draw a circle. Let it intersect the given circle of radius 4 cm at the points P and Q.
- Join CP and CQ. Thus, CP and CQ are the required two tangents.
Justification:
Join OP. Here ∠OPC is an angle in the semi-circle. Therefore, ∠OPC = 90°. Since OP is a radius of a circle, CP has to be a tangent to a circle. Similarly, CQ is also a tangent to a circle.
In ∆COP, ∠P = 90°
CO2=CP2+OP2 CP2=CO2–OP2
=62–42
CP=25–√cm
9) Draw a circle with radius 3 cm. On one of its extended diameter, take two points P and Q each at a distance of 7 cm from its centre. From two points P and Q, draw tangents to the circle.
Solution:
Given:
Two points P and Q on the diameter of a circle with radius 3 cm OP = OQ = 7 cm.
Aim:
To construct the tangents to the circle from the given points P and Q.
Procedure for construction:
- Draw a circle with radius 3 cm with centreO.
- Extend its diameter both the sides and cut OP = OQ = 7 cm.
- Bisect OP and OQ.Let mid-points of OP and OQ be M and N.
- With M as centre and OM as radius, draw a circle. Let it intersect (0, 3) at two points A and B. Again taking N as centre ON as radius draw a circle to intersect circle(0, 3) at points C and D.
- Join PA, PB, QC and QD. These are the required tangents from P and Q to circle (0, 3).
10) Draw a pair of tangents to a circle which is of radius 5 cm, such that they are inclined to each other at an angle of 60°.
Solution:
To determine: To draw tangents at the ends of two radius which are inclined to each other at 120°
Procedure for construction :
- Keeping O as centre, draw a circle of radius 5 cm.
- Take a point Q on the circle and join it to O.
- From OQ, Draw∠QOR = 120°.
- Take an external point P.
- Join PR and PQ perpendicular to OR and OQ respectively intersecting at P.
The required tangents are RP and QP.
Related Links
NCERT Solutions for Class 10 Maths -Circles
1. In the figure, PQ is a chord of a circle and PT is the tangent at P such that ∠QPT = 60°. Then find the measure of ∠PRQ.
Solution.
Since OP is perpendicular to PT.
∠OPT = 90°
∠OPQ = 90°- ∠QPT
∠OPQ = 90 – 60 = 30°.
In ΔOPQ, OP= OQ = r ( Radius of the circle )
∠OPQ= ∠OQP = 30.
And,
∠POQ = 180 – ∠OPQ- ∠OQP
= 180° – 30° – 30° = 120°
Also, reflex ∠POQ = 360° – 120° = 240°
Now, ∠PRQ = reflex ∠POQ
= 12x 240° = 120°
2. If the angle between two radii of a circle is 130°, then find the degree measure of the angle between the tangents at the ends of the radii.
Solution.
It is already known that angle between two radii and the angle between the tangents at the ends of the radii are supplementary.
Hence, Angle between the tangents at the ends of the radii is 180° – 130° , i.e., 50°.
3. In the figure, if ∠AOB = 125°, then find the degree measure of ∠
Solution.
It is already known that the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
∠AOB + ∠COD = 180°.
125° + ∠COD = 180°
∠COD = 180° – 125° = 55°.
4. In the figure, AB is a chord of the circle and AOC is its diameter such that ∠ACB = 50°. If AT is the tangent to the circle at the point A, then find the measure of ∠BAT.
Solution.
∠BAT= 50°.
5. At one end A of a diameter AB of a circle of radius 5cm, tangent XAY is drawn to the circle. Find the length of the chord CD parallel to XY and at a distance 8cm from A.
Solution.
Since XAY is a tangent through one end A of a diameter of a circle,
AB is perpendicular to XY
And also, CD is parallel to XAY ⇒ AB is perpendicular to CD
Since OM is perpendicular from centre O to the chord CD.
⤑ OM is perpendicular bisector of chord CD.
That is., CM = MD = 12CD.
Now, AM =8cm (given)
⇒ OM = AM – AO = 8-5 = 3cm.
In OMC, we obtain
CM = CM=OC2−OM2−−−−−−−−−−√
= 52−32−−−−−−√
= 16−−√
= 4 cm.
Hence, CD = 2cm = 2 x 4 = 8 cm.
6. In the given figure, AT is a tangent to the circle with centre O such that OT = 4 cm and ∠OTA = 30°. Find the length of the segment AT.
Solution.
In tight triangle ∆OTA, ∠OTA = 30°.
OAOT=sin30∘
OA=12OT=12×4=2cm
Again, AT=OT2−OA2−−−−−−−−−−√
AT = 42−22−−−−−−√
=16−4−−−−−√ 12−−√
= 23–√cm.
7. In the figure, if O is the centre of a circle, PQ is a chord and the tangent PR at P makes an angle of 50° with PQ, find the measure of ∠POQ.
Solution.
Since OP is perpendicular to PR
∠OPR = 90°
∠OPQ + ∠QPR = 90°
∠OPQ + 50° = 90°
→ ∠OPQ = ∠OQP = 40°
Again,
∠POQ + ∠OPQ + ∠OQP = 180°
→ ∠POQ + 40° + 40° = 180°
→ ∠POQ = 180° – 40° – 40°
= 100°
8. In the figure, if PA and PB are tangents to the circle with centre O such that ∠APB = 50°, find the degree measure of ∠OAB.
Solution.
Since OA is perpendicular to PA and also, OB is perpendicular to PB
∠APB + ∠AOB = 180°
50°+ ∠AOB = 180°
∠AOB = 180° – 50° = 130°
In △AOB,
OA = OB = radii of same circle
∠OAB = ∠OBA = x ( say )
Again, ∠OAB + ∠OBA + ∠AOB = 180°
x +x + 130° = 180°
2x = 180° – 130° = 50°
X = 25°
Hence, ∠OAB =25°
9. In the figure, if PQR is the tangent to a circle at Q whose centre is O, AB is a chord parallel to PR and ∠BQR = 70°, find the measure of ∠AQB.
Solution.
Since AB is parallel to PQR
∠B =∠BQR
→ ∠B =70°
Also, ∠A = ∠BQR [∠s in the corresponding alternate segments are equal ]
→ ∠A = 70°
Now, in △ABQ, we have
∠A + ∠B + ∠AQB = 180°
70° + 70° + ∠AQB = 180°
∠AQB=180° – 70° – 70°
=40°
Hence, ∠AQB = 40°
10. In the figure, DE and DF are tangents from an external point D to a circle with centre A. If DE=5cm and DE is perpendicular DF, find the radius of the circle.
Solution.
Join AE and AF → AE is perpendicular to DE and AF is perpendicular to DF.
AEDF is a square of side 5cm [DE = 5cm(given)]
Hence, radius of the circle is 5 cm.
11. If two tangents inclined at an angle of 60° are drawn to a circle of radius 3cm, find the length of each tangent.
Solution.
.
Here, let PA and PB be two tangents to a circle with centre O.
OA is perpendicular to AP.
Since OP bisects ∠APB
∠OPA = 30°
Now, in △OAP, we have
APOA=cot30∘
→ AP3=3–√
→ AP=33–√cm
12. From a point P which is at a distance of 13cm from the centre O of a circle of radius 5cm, a pair of tangents PQ and PR to the circle are drawn. Find the area of the quadrilateral PQOR.
Solution.
In △OPQ, we have
OQ perpendicular to PQ
OP2=PQ2+OQ2 132=PQ2+52
PQ2=169–25 = 144
PQ = 12cm
Now, (△OPQ) = x PQxOQ
= x12x5 = 30cm2
Thus, (quad, PQOR) = 2 x ar(△OPQ)
= 2×30 cm2
= 60cm2
