Category: Class 10

Students can download Maths Chapter 8 Statistics and Probability Ex 8.4 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 8 Statistics and Probability Ex 8.4

Question 1.
If P (A) = \(\frac{2}{3}\), P(B) = \(\frac{2}{5}\), P(A ∪ B) = \(\frac{1}{3}\), then find P(A ∩ B).
Answer:
P(A ∪ B) = P (A) + P (B) – P (A ∩ B)
\(\frac{1}{3}=\frac{2}{3}+\frac{2}{5}\) – P (A ∩ B)

Question 2.
A and B are two events such that, P(A) = 0.42, P(B) = 0.48, and P(A∩B)=016. Find (i) P(not A)
(ii) P(not B)
(iii) P(A or B)
Solution:
(a) P(A) = 0.42 ;
P(B) = 0.48
P(A∩B) = 0.16
(i) P(not A) = P(\(\overline{\mathbf{A}}\)) = 1 – P(A) = 1 – 0.42 = 0.58
(ii) P(not B) = P(\(\overline{\mathbf{B}}\)) = 1 – P(B) = 1 – 0.48 = 0.52
(iii) P(A or B) = P(A∪B) = P(A) + P(B) – P(A∩B)
= 0.42 + 0.48 – 0.16
= 0.74

Question 3.
If A and B are two mutually exclusive events of a random experiment and P (not A) = 0.45, P (A ∪ B) = 0.65, then find P(B).
Answer:
P(not A) = 0.45
1 – P (A) = 0.45
P (A) = 1 – 0.45 = 0.55
P(A ∪ B) = P (A) + P (B)
0. 65 = 0.55 + P(B)
0. 65 – 0.55 = P(B)
0.10 = P (B)
P(B) = 0.1

Question 4.
The probability that atleast one of A and B occur is 0.6. If A and B occur simultaneously with probability 0.2, then find P(\(\overline{\mathbf{A}}\)) + P(\(\overline{\mathbf{B}}\)).
Solution:
P(A∪B) = 0.6
P(A∩B) = 0.2
P(A) + P(B) = [1 – P(A∪B)] + [1 – P(A∩B)] = [1 – 0.6] + [1 – 0.2]
= 0.4 + 0.8 = 1.2

Question 5.
The probability of happening of an event A is 0.5 and that of B is 0.3. If A and B are mutually exclusive events, then find the probability of neither A nor B happen.
Answer:
Here P(A) = 0.5, P (B) = 0.3
P(A ∪ B) = P (A) + P(B) [A and B are mutually exclusive]
= 0.5 + 0.3
= 0.8
Probability of neither A nor [P(A ∪ B)’] = 1 – P(A ∪ B) = 1 – 0.8 = 0.2

Question 6.
Two dice are rolled once. Find the probability of getting an even number on the first die or a total of face sum 8.
Answer:
Sample space = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
n (S) = 36
Let A be the event of getting an even number on the first time
A = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
n (A) = 18
\(P(A)=\frac{n(A)}{n(S)}=\frac{18}{36}\)
(ii) Let B be the event of getting a total of face sum 8.
B = {(2, 6) (3, 5) (4, 4) (5, 3) (6, 2)}
n(B) = 5
\(P(B)=\frac{n(B)}{n(S)}=\frac{5}{36}\)
A ∩ B = {(2, 6) (4, 4) (6, 2)}
n(A ∩ B) = 3
P(A ∩ B) = \(\frac{3}{36}\)
P(A ∪ B) = P (A) + P (B) – P (A ∩ B)

The required probability = \(\frac{5}{9}\)

Question 7.
From a well-shuffled pack of 52 cards, a card is drawn at random. Find the probability of its being either a red king or a black queen.
Answer:
n(S) = 52
Let A be the event of getting a red king
n(A) = 2
\(P(A)=\frac{n(A)}{n(S)}=\frac{2}{52}\)
Let B be the event of getting a black Queen king
n(B) = 2
\(P^{\prime}(B)=\frac{n(B)}{n(S)}=\frac{2}{52}\)
It A and B are mutually exclusive
P(A ∪ B) = P(A) + P(B)
\(=\frac{2}{52}+\frac{2}{52}=\frac{4}{52}=\frac{1}{13}\)
The required probability is \(\frac{1}{13}\)

Question 8.
A box contains cards numbered 3, 5, 7, 9,… 35, 37. A card is drawn at random from the box. Find the probability that the drawn card have either multiples of 7 or a prime number.
Answer:
Sample space = {3, 5, 7, 9,…….,35, 37}
n(S) = 18
Let A be the event of getting a multiple of 7
A = {7, 21, 35}
n(A) = 3
\(P(A)=\frac{n(A)}{n(S)}=\frac{3}{18}\)
Let B be the event of getting a prime number
B = {3, 5, 7,11, 13, 17, 19, 23, 29, 31, 37}
n(B) = 11

Probability of getting a multiple of 7 or a prime number = \(\frac{13}{18}\)

Question 9.
Three unbiased coins are tossed once. Find the probability of getting atmost 2 tails or atleast 2 heads.
Answer:
Sample space = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
n(S) = 8
Let A be the event of getting atmost 2 tails.
A = {HTT, THT, TTH, HHT, HTH, THH, HHH}
n(A) = 7

Probability of getting atmost two tails or atleast 2 heads = \(\frac{7}{8}\)

Question 10.
The probability that a person will get an electrification contract is \(\frac{3}{5}\) and the probability that he will not get plumbing contract is \(\frac{5}{8}\). The probability of getting atleast one contract is \(\frac{5}{7}\). What is the probability that he will get both?
Answer:
Let A and B represent the event of getting electrification control and plumbing contract.


Probability of getting both the job is \(\frac{73}{280}\)

Question 11.
In a town of 8000 people, 1300 are over 50 years and 3000 are females. It is known that 30% of the females are over 50 years. What is the probability that a chosen individual from the town is either a female or over 50 years?
Answer:
Total number of people in a town is 8000.
n(S) = 8000
Total number of females = 3000
Let A be the event of getting number of females
n(A) = 3000
\(P(A)=\frac{n(A)}{n(S)}=\frac{3000}{8000}\)
Number of people over 50 years = 1300
Let B be the event of getting number of people over 50 years.
n(B) = 1300
\(P(B)=\frac{n(B)}{n(S)}=\frac{1300}{8000}\)
Given 30% of the females are over 50 years.


Proability of getting either a female or over 50 years = \(\frac{17}{40}\)

Question 12.
A coin is tossed thrice. Find the probability of getting exactly two heads or atleast one tail or two consecutive heads.
Answer:
Sample space = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
n(S) = 8
Let A be the event of getting exactly two heads.
A = {HHT, HTH, THH}
n(A) = 3
\(P(A)=\frac{n(A)}{n(S)}=\frac{3}{8}\)
Let B be the event of getting atleast one tail
B = {HHT, HTH, HTT, THH, THT, TTH, TTT}
n(B) = 7
\(P(B)=\frac{n(B)}{n(S)}=\frac{7}{8}\)
Let C be the event of getting consecutively
C = {HHH, HHT, THH}
n(C) = 3



The probability is 1.

Question 13.
If A, B, C are any three events such that probability of B is twice as that of probability of A and probability of C is thrice as that of probability of A and if P (A ∩ B) = \(\frac{1}{6}\), P(B ∩ C) = \(\frac{1}{4}\), P(A ∩ C) = \(\frac{1}{8}\), P(A ∪ B ∪ C) = \(\frac{9}{10}\) and P (A ∩ B ∩ C) = \(\frac{1}{15}\), then find P(A), P(B) and P(C)?
Answer:
By the given condition,
P(B) = 2 P(A), P(C) = 3 P(A)
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) – P(A ∩ C) + P(A ∩ B ∩ C)

Question 14.
In a class of 35, students are numbered from 1 to 35. The ratio of boys and girls is 4 : 3. The roll numbers of students begin with boys and end with girls. Find the probability that a student selected is either a boy with prime roll number or a girl with composite roll number or an even roll number.
Answer:
Sample space (S) = {1, 2, 3,… ,35}
n(S) = 35
Total number of students = 35
Number of boys = \(\frac{4}{7}\) × 35 = 20 [Boys Numbers = {1, 2, 3,…, 20}]
Number of girls = \(\frac{3}{7}\) × 35 = 15 [Girls Numbers = { 21, 22,…, 35}]
Let A be the event of getting a boy role number with prime number
A = {2, 3, 5, 7, 11, 13, 17, 19}
n(A) = 8
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}\) = \(\frac{8}{35}\)
Let B be the event of getting girls roll number with composite number.
B = {21, 22, 24, 25, 26, 27, 28, 30, 32, 33, 34, 35}
n(B) = 12
P(B) = \(\frac{n(\mathrm{B})}{n(\mathrm{S})}\) = \(\frac{12}{35}\)
Let C be the event of getting an even roll number.
C = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34}
n(C) = 17


Probability of getting roll number is \(\frac{29}{35}\)

Students can download Maths Chapter 8 Statistics and Probability Ex 8.3 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 8 Statistics and Probability Ex 8.3

Question 1.
Write the sample space for tossing three coins using tree diagram.
Answer:
Sample space = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

Question 2.
Write the sample space for selecting two balls from a bag containing 6 balls numbered 1 to 6 (using tree diagram).
Answer:
Sample space = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

Question 3.
If A is an event of a random experiment such that P(A) : P(\(\bar{A}\)) = 17 : 15 and n(s) = 640 then find (i) P(\(\bar{A}\))
(ii) n(A)
Answer:

Question 4.
A coin is tossed thrice. What is the probability of getting two consecutive tails?
Answer:
Sample space = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
n(S) = 8
Let A be the event of getting consecutive tails
A = {HTT, TTH, TTT}
n(A) = 3
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{3}{8}\)

Question 5.
At a fete, cards bearing numbers 1 to 1000, one number on one card are put in a box. Each player selects one card at random and that card is not replaced. If the selected card has a perfect square number greater than 500, the player wins a prize. What is the probability that
(i) the first player wins a prize
(ii) the second player wins a prize if the first has won?
Solution:
22² = 484
31² = 961
23² = 529
32² = 1024
23, 24, 25, 26, 27, 28, 29, 30, 31 has squares below 500 × 1000.
(i) P(first player wins a prize) = \(\frac{9}{1000}\)
(ii) P(second player ins if first has won) = \(\frac{8}{999}\)

Question 6.
A bag contains 12 blue balls and x red balls. If one ball is drawn at random (i) what is the probability that it will be a red ball? (ii) If 8 more red balls are put in the bag, and if the probability of drawing a red ball will be twice that of the probability in (i), then find x.
Answer:
Sample space = 12 + x
n(S) = x + 12
(i) Let A be the event of getting a red ball
n(A) = x
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}\) = \(\left(\frac{x}{x+12}\right)\)
(ii) 8 more red balls are added
Sample space = x + 12 + 8 = x + 20
Number of red balls = x + 8
Probability of drawing red ball = \(\frac{x+8}{x+20}\)
By the given condition
\(\frac{x+8}{x+20}=2\left(\frac{x}{x+12}\right)\)
(x + 8)(x + 12) = 2x(x + 20)
x² + 20x + 96 = 2x² + 40x
x² + 20x – 96 = 0
(x + 24)(x – 4) = 0
x = -24 (or) x = 4
The value of x = 4 (Number of balls will not be negative)
The probability of getting red balls = \(\left(\frac{x}{x+12}\right)=\frac{4}{16}=\frac{1}{4}\)

Question 7.
Two unbiased dice are rolled once. Find the probability of getting
(i) a doublet (equal numbers on both dice)
(ii) the product as a prime number
(iii) the sum as a prime number
(iv) the sum as 1
Answer:
(i) Sample space = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
n(S) = 36
Let A be the event of getting doublet
A = {(1, 1) (2, 2) (3, 3) (4, 4) (5, 5) (6, 6)}
n(A) = 6
\(P(A)=\frac{n(A)}{n(S)}=\frac{6}{36}=\frac{1}{6}\)
(ii) Let B be the event of getting a product is a prime number.
B = {(1, 2) (1, 3) (1, 5) (2, 1) (3, 1) (5, 1)}
n(B) = 6
\(P(B)=\frac{n(B)}{n(S)}=\frac{6}{36}=\frac{1}{6}\)
(iii) Let C be the event of getting a sum is a prime number
C = {(1, 1) (1, 2) (1, 4) (1, 6) (2, 1) (2, 3) (2, 5) (3, 2), (3, 4) (4, 1) (4, 3) (5, 2) (5, 6) (6, 1) (6, 5)}
n(C) = 15
\(P(C)=\frac{n(C)}{n(S)}=\frac{15}{36}=\frac{5}{12}\)
(iv) Let D be the event of getting a sum is 1
n(D) = 0
\(P(D)=\frac{n(D)}{n(S)}=\frac{0}{36}=0\)
Probability of getting a sum is 1 is 0

Question 8.
Three fair coins are tossed together. Find the probability of getting
(i) all heads
(ii) atleast one tail
(iii) atmost one head
(iv) atmost two tails
Solution:
Possible outcomes = {(HHH), (THH), (HTH), (HHT), (TTH), (THT), (HTT), (TTT)}
No. of possible outcomes = 2 × 2 × 2 = 8
(i) Prob(all heads) = \(\frac{1}{8}\)
(ii) Atleast one tail = {(THH), (HTH), (HHT), (TTH), (THT), (HTT), (TTT)}
Prob(atleast one tail) = \(\frac{7}{8}\)
(iii) Atmost one head = {(HTT), (THT), (TTH), (TTT)}
∴ Prob(atmost one head) = \(\frac{4}{8}=\frac{1}{2}\)
(iv) Atmost two tail = {(HHH), (THH), (HTH), (HHT), (TTH), (THT), (HTT)}
∴ Prob(atmost two tail) = \(\frac{7}{8}\)

Question 9.
Two dice are numbered 1, 2, 3, 4, 5, 6 and 1, 1, 2, 2, 3, 3 respectively. They are rolled and the sum of the numbers on them is noted. Find the probability of getting each sum from 2 to 9 separately.
Answer:
1st dice A = {1, 2, 3, 4, 5, 6}
2nd dice B = {1, 1, 2, 2, 3, 3}
Sample Space (S) = {(1, 1), (1, 1), (1, 2), (1, 2), (1, 3), (1, 3), (2, 1), (2, 1), (2, 2), (2, 2), (2, 3), (2, 3), (3, 1), (3, 1), (3, 2), (3, 2), (3, 3), (3, 3), (4, 1), (4, 1), (4, 2), (4, 2), (4, 3), (4, 3), (5, 1), (5, 1), (5, 2), (5, 2), (5, 3), (5, 3),(6, 1), (6, 1), (6, 2), (6, 2), (6, 3), (6, 3)}
n(S) = 36
(i) Let A₁ be the event of getting sum is 2
A1 = {(1, 1) (1, 1)}
n(A₁) = 2
\(P\left(A_{1}\right)=\frac{n\left(A_{1}\right)}{n(S)}=\frac{2}{36}=\frac{1}{18}\)
(ii) Let A₂ be the event of getting a sum is 3.
A₂ = {(1, 2) (1, 2) (2, 1) (2, 1)}
n(A₂) = 4
\(P\left(A_{2}\right)=\frac{4}{36}=\frac{1}{9}\)
(iii) Let A₃ be the event of getting a sum is 4.
A₃ = {(1, 3) (1, 3) (2, 2) (2, 2) (3, 1) (3, 1)}
n(A₃) = 6
\(P\left(A_{3}\right)=\frac{6}{36}=\frac{1}{6}\)
(iv) Let A₄ be the event of getting a sum is 5.
A₄ = {(2, 3) (2, 3) (3, 2) (3, 2) (4, 1) (4, 1)}
n(A₄) = 6
\(P\left(A_{4}\right)=\frac{6}{36}=\frac{1}{6}\)
(v) Let A₅ be the event of getting a sum is 6.
A₅ = {(3, 3) (3, 3) (4, 2) (4, 2) (5, 1) (5, 1)}
n(A₅) = 6
\(P\left(A_{5}\right)=\frac{6}{36}=\frac{1}{6}\)
(vi) Let A₆ be the event of getting a sum is 7.
A₆ = {(4, 3) (4, 3) (5, 2) (5, 2) (6, 1) (6, 1)}
n(A₆) = 6
\(P\left(A_{6}\right)=\frac{6}{36}=\frac{1}{6}\)
(vii) Let A₇ be the event of getting a sum is 8.
A₇ = {(5, 3) (5, 3) (6, 2) (6, 2)}
n(A₇) = 4
\(P\left(A_{7}\right)=\frac{4}{36}=\frac{1}{9}\)
(viii) Let A₈ be the event of getting a sum is 9.
A₈ = {(6, 3) (6, 3)}
n(A₈) = 2
\(P\left(A_{8}\right)=\frac{2}{36}=\frac{1}{18}\)

Question 10.
A bag contains 5 red balls, 6 white balls, 7 green balls, 8 black balls. One ball is drawn at random from the bag. Find the probability that the ball drawn is
(i) white
(ii) black or red
(iii) not white
(iv) neither white nor black
Answer:
Sample space (S) = 5 + 6 + 7 + 8
n(S) = 26
(i) Let A be the event of getting a white ball
n(A) = 6
\(P(A)=\frac{n(A)}{n(S)}\)
\(P(A)=\frac{6}{26}=\frac{3}{13}\)
(ii) Let A be the event of getting a black ball
n(A) = 8
\(P(A)=\frac{n(A)}{n(S)}=\frac{8}{26}\)
Let B be the event of getting a red ball
n(B) = 5
\(P(B)=\frac{n(B)}{n(S)}=\frac{5}{26}\)
Probability of getting black or red ball
P(A ∪ B) = P (A) + P (B)
= \(\frac{8}{26}+\frac{5}{26}=\frac{13}{26}=\frac{1}{2}\)
(iii) Not white probability of getting white ball
P(A) = \(\frac{3}{13}\) from (i)
Probability of not getting white ball P(\(\bar{A}\)) = 1 – P(A)
\(1-\frac{3}{13}=\frac{13-3}{13}=\frac{10}{13}\)
(iv) Probability of getting a white ball.
P(A) = \(\frac{6}{26}\) (from 1)
Let B be the event of getting a black ball
n(B) = 8
\(P(B)=\frac{n(B)}{n(S)}=\frac{8}{26}\)
P(A ∪ B) = P(A) + P(B) = \(\frac{6}{26}+\frac{8}{26}=\frac{14}{26}\)
Probability of neither white nor black P(A ∪ B)’ = 1 – P(A ∪ B)
= \(1-\frac{14}{26}\)
= \(\frac{26-14}{26}=\frac{12}{26}=\frac{6}{13}\)

Question 11.
In a box there are 20 non-defective and some defective bulbs. If the probability that a bulb selected at random from the box found to be defective is \(\frac{3}{8}\) then, find the number of defective bulbs.
Answer:
Let the number of defective bulbs be “x”
Sample space (S) = 20 + x
n(S) = 20 + x
Let A be the event of getting to be defective
n(A) = x
\(P(A)=\frac{n(A)}{n(S)}\)
⇒ \(\frac{3}{8}=\frac{x}{20+x}\)
⇒ 8x = 3(20 + x) = (60 + 3x)
⇒ 8x – 3x = 60
⇒ 5x = 60
⇒ x = \(\frac{60}{5}\)
⇒ x = 12
Number of defective bulbs = 12

Question 12.
The king and queen of diamonds, queen and jack of hearts, jack and king of spades are removed from a deck of 52 playing cards and then well shuffled. Now one card is drawn at random from the remaining cards. Determine the probability that the card is
(i) a clavor
(ii) a queen of red card
(iii) a king of black card
Answer:
King diamond + Queen diamonds = 1 + 1 = 2 …….(1)
Queen hearts + Jack of hearts = 1 + 1 = 2 …….(2)
Jack spade + King of spades =1 + 1 = 2 …….(3)
Remaining number of cards = 52 – (6)
n(S) = 46
(i) Let A be the event of getting a clavor
n (A) = 13
\(P(A)=\frac{n(A)}{n(S)}=\frac{13}{46}\)
(ii) Let B be the event of getting a queen of red card
n(B) = 2
But the above two cards are removed from (1) and (2)
n(B) = 0
\(P(B)=\frac{n(B)}{n(S)}=\frac{0}{46}=0\)
(iii) Let B be the event of getting a king of black card
n(B) = (2 – 1) [from (3) one black card is removed]
n (B) = 1
\(P(B)=\frac{n(B)}{n(S)}=\frac{1}{46}\)

Question 13.
Some boys are playing a game, in which the stone thrown by them landing in a circular region given in the figure is considered as win and landing other than the circular region is considered as a loss. What is the probability to win the game?
Answer:

Area of a rectangle = l × b sq. feet = 3 × 4 sq. feet = 12 sq. feet
sample space (S) = 12
n(S) = 12
Let A be the event of getting the stone landing in a circular region
n(A) = Area of a circle
= πr²
= π × 1 × 1 (radius of a circle = 1 feet)
= π

Probability to win the game = \(\frac{11}{42}\) (or) \(\frac{157}{600}\)

Question 14.
Two customers Priya and Amuthan are visiting a particular shop in the same week (Monday to Saturday). Each is equally likely to visit the shop on any one day as on another day. What is the probability that both will visit the shop on
(i) the same day
(ii) different days
(iii) consecutive days?
Answer:
Sample space (S) = 6 × 6 = 36
n(S) = 36
[priya and Amuthan are visiting a particular shop in any one of 6 days is 6 × 6 = 36]
(i) Let A be the event of getting both are shopping on the same day
A = {(Mon, Mon) (Tue, Tue) (Wed, Wed) (Thu, Thu) (Fri, Fri) (Sat, Sat)}
n(A) = 6
\(P(A)=\frac{n(A)}{n(S)}\)
\(=\frac{6}{36}=\frac{1}{6}\)
(ii) Let B be the event of shopping in different days.
n(B) = 36 – 6 = 30
\(P(B)=\frac{n(B)}{n(S)}\)
\(=\frac{30}{36}=\frac{5}{6}\)
(iii) Let C be the event of shopping consecutive days
C = {(Mon, Tue) (Tue, Wed) (Wed, Thu) (Thu, Fri) (Fri, Sat)}
n(C) = 5
P(C) = \(\frac{n(\mathrm{C})}{n(\mathrm{S})}\) = \(\frac{5}{36}\)

Question 15.
In a game, the entry fee is ₹ 150. The game consists of tossing a coin 3 times. Dhana bought a ticket for entry. If one or two heads show, she gets her entry fee back. If she throws 3 heads, she receives double the entry fees. Otherwise, she will lose. Find the probability that she
(i) gets double entry fee
(ii) just gets her entry fee
(iii) loses the entry fee.
Answer:
Sample space (S) = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
n(S) = 8
(i) Let A be the event of getting double entry fee (only getting 3 heads)
n(A) = 1
\(P(A)=\frac{n(A)}{n(S)}=\frac{1}{8}\)
(ii) Let B be the event of getting her entry fee (one or two heads to show)
n(B) = Probability of one head + Probability of 2 head
= \(\frac{3}{8}+\frac{3}{8}=\frac{6}{8}=\frac{3}{4}\)
(iii) To loss the entry means not getting the head (only tail)
n(C) = 1
\(P(C)=\frac{n(C)}{n(S)}=\frac{1}{8}\)

Students can download Maths Chapter 8 Statistics and Probability Ex 8.2 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 8 Statistics and Probability Ex 8.2

Question 1.
The standard deviation and mean of a data are 6.5 and 12.5 respectively. Find the coefficient of variation.
Answer:
Standard deviation of a data (σ) = 6.5
Mean of the data (\(\bar{x}\)) = 12.5
Coefficient of variation = \(\frac{\sigma}{\bar{x}} \times 100 \%\)
= \(\frac{6.5}{12.5} \times 100 \%=52 \%\)
Coefficient of variation = 52%

Question 2.
The standard deviation and coefficient of variation of a data are 1.2 and 25.6 respectively. Find the value of mean.
Answer:
Standard deviation (σ) = 1.2
Coefficient of variation = 25.6

Question 3.
If the mean and coefficient of variation of a data are 15 and 48 respectively, then find the value of standard deviation.
Answer:
Mean (\(\bar{x}\)) = 15
Co efficient of variation = 48

Question 4.
If n = 5, \(\bar{x}\) = 6, Σx² = 765, then calculate the coefficient of variation.
Answer:

Question 5.
Find the coefficient of variation of 24, 26, 33, 37, 29, 31.
Answer:
Arrange in ascending order we get 24, 26, 29, 31, 33, 37
Assumed mean = 29

Question 6.
The time taken (in minutes) to complete homework by 8 students in a day are given by 38, 40, 47, 44, 46, 43, 49, 53. Find the coefficient of variation.
Answer:
Arrange in ascending order we get, 38, 40, 43, 44, 46, 47, 49, 53.
Assumed mean = 46


= \(\frac{453}{45}\)%
= 10.066
Coefficient of variation = 10.07%

Question 7.
The total marks scored by two students Sathya and Vidhya in 5 subjects are 460 and 480 with standard deviation of 4.6 and 2.4 respectively. Who is more consistent in performance?
Answer:
Total marks scored by sathya = 460
Total marks scored by vidhya = 480
Number of subjects = 5
Mean marks of sathya = \(\frac{460}{5}\)
\(\bar{x}\) = 92%
Given standard deviation, (σ) = 4.6

Vidhya coefficient of variation is less than Sathya.
Vidhya is more consistent.

Question 8.
The mean and standard deviation of marks obtained by 40 students of a class in three subjects Mathematics, Science and Social Science are given below.

Which of the three subjects shows the highest variation and which shows the lowest variation in marks?
Answer:
(i) Mathematics:
Mean (\(\bar{x}\)) = 56
Standard deviation (σ) = 12
Coefficient variation (CV₁) = \(\frac{\sigma}{\bar{x}} \times 100=\frac{12}{56} \times 100\)

Science shows the highest variation
Social science shows the lowest variation

Question 9.
The temperature of two cities A and B in the winter season are given below.

Find which city is more consistent in temperature changes?
Answer:
(i) city A:
Assumed mean = 22




C.V of city A < C.V of city B
City A is more consistent in temperature change.

Students can download Maths Chapter 8 Statistics and Probability Ex 8.1 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 8 Statistics and Probability Ex 8.1

Question 1.
Find the range and coefficient of range of the following data.
(i) 63, 89, 98, 125, 79, 108, 117, 68
(ii) 43.5, 13.6, 18.9, 38.4, 61.4, 29.8
Answer:
(i) Here the largest value (L) = 125
The smallest value (S) = 63
Range = L – S = 125 – 63 = 62

Question 2.
If the range and the smallest value of a set of data are 36.8 and 13.4 respectively, then find the largest value.
Solution:
If the range = 36.8 and
the smallest value = 13.4 then
the largest value = L = R + S
= 36.8 + 13.4 = 50.2

Question 3.
Calculate the range of the following data.

Answer:
Smallest value (S) = 400
Largest value (L) = 650
Range = L – S = 650 – 400 = 250

Question 4.
A teacher asked the students to complete 60 pages of a record notebook. Eight students have completed only 32, 35, 37, 30, 33, 36, 35 and 37 pages. Find the standard deviation of the pages yet to be completed by them.
Answer:
The remaining number of pages to be completed is 60 – 32; 60 – 35; 60 – 37; 60 – 30; 60 – 33; 60 – 36; 60 – 35 and 60 – 37
The pages to be completed are, 28, 25, 23, 30, 27, 24, 25, and 23
Arrange in ascending order we get, 23, 23, 24, 25, 25, 27, 28 and 30

Question 5.
Find the variance and standard deviation of the wages of 9 workers given below:
₹ 310, ₹ 290, ₹ 320, ₹ 280, ₹ 300, ₹ 290, ₹ 320, ₹ 310, ₹ 280.
Answer:
Arrange in ascending order we get,
280, 280, 290, 290, 300, 310, 310, 320 and 320
Assumed mean = 300


Variance = 222.222
Standard deviation = √Variance = √222.222 = 14.907 = 14.91
Variance = 222.22
Standard deviation = 14.91

Question 6.
A wall clock strikes the bell once at 1 o’clock, 2 times at 2 o’clock, 3 times at 3 o’clock and so on. How many times will it strike in a particular day? Find the standard deviation of the number of strikes the bell make a day.
Answer:
Wall clock strikes the bell in 12 hours
1, 2, 3, 4, 5,… ,12
Wall clock strikes in a day (24 hours)
2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24.
Assumed mean = 14



The standard deviation of bell strike in a day is 6.9

Question 7.
Find the standard deviation of the first 21 natural numbers.
Answer:
Here n = 21
The standard deviation of the first ‘n’ natural numbers,

The standard deviation of the first 21 natural numbers = 6.06

Question 8.
If the standard deviation of a data is 4.5 and if each value of the data is decreased by 5, then find the new standard deviation.
Solution:
If the standard deviation of a data is 4.5 and each value of the data decreased by 5, the new standard deviation does not change and it is also 4.5.

Question 9.
If the standard deviation of a data is 3.6 and each value of the data is divided by 3, then find the new variance and new standard deviation.
Answer:
The standard deviation of the data = 3.6
Each value of the data is divided by 3
New standard deviation = \(\frac{3.6}{3}\) = 1.2
New Variance = (1.2)² = 1.44 [∴ Variance = (S.D)²]
New standard Deviation = 1.2
New variance = 1.44

Question 10.
The rainfall recorded in various places of five districts in a week are given below.

Find its standard deviation.
Answer:
Assumed mean = 60

Question 11.
In a study about viral fever, the number of people affected in a town were noted as

Find its standard deviation.
Answer:
Assumed mean = 35

Question 12.
The measurements of the diameters (in cms) of the plates prepared in a factory are given below. Find the standard deviation.

Answer:
Assumed mean = 34.5

Question 13.
The time taken by 50 students to complete a 100 meter race are given below. Find its standard deviation.

Answer:
Assumed mean = 11


Standard deviation (σ) = 1.24

Question 14.
For a group of 100 candidates, the mean and standard deviation of their marks were found to be 60 and 15 respectively. Later on, it was found that the scores 45 and 72 were wrongly entered as 40 and 27. Find the correct mean and standard deviation.
Answer:
Number of candidates = 100
n = 100
Mean (\(\bar{x}\)) = 60
standard deviation (σ) = 15
Mean (\(\bar{x}\)) = \(\frac{\Sigma x}{n} \Rightarrow 60=\frac{\Sigma x}{100}\)
Σx = 6000
Correct total = 6000 + (45 – 40) + ( 72 – 27) = 6000 + 5 + 45 = 6050
Correct mean (\(\bar{x}\)) = \(\frac{6050}{100}\) = 60.5
Given standard deviation = 15


Correct mean = 60.5
Correct standard deviation (σ) = 14. 61

Question 15.
The mean and variance of seven observations are 8 and 16 respectively. If five of these are 2, 4, 10, 12 and 14, then find the remaining two observations.
Answer:
Let the missing two observations be ‘a’ and ‘b’
Arithmetic mean = 8

Students can download Maths Chapter 7 Mensuration Additional Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 7 Mensuration Additional Questions

I. Multiple Choice Questions:

Question 1.
The curved surface area of a right circular cylinder of radius 1 cm and height 1 cm is equal to ______
(1) π cm²
(2) 2π cm²
(3) 3π cm²
(4) 2 cm²
Answer:
(2) 2π cm²
Hint:
C.S.A of a cylinder = 2πrh sq. units = 2 × π × 1 × 1 cm² = 2π cm²

Question 2.
The total surface area of a solid right circular cylinder whose radius is half of its height h is equal to ______ sq. units.
(1) \(\frac{3}{2} \pi h\)
(2) \(\frac{2}{3} \pi h^{2}\)
(3) \(\frac{3}{2} \pi h^{2}\)
(4) \(\frac{2}{3} \pi h\)
Answer:
(3) \(\frac{3}{2} \pi h^{2}\)
Hint:
T.S.A = 2πr(h + r)
[radius is half of the height]
= \(2 \pi \times \frac{h}{2}\left(\frac{h}{2}+h\right)\)
= \(=\pi h\left(\frac{3 h}{2}\right)=\frac{3}{2} \pi h^{2}\) sq. units

Question 3.
Base area of a right circular cylinder is 80 cm². If its height is 5 cm, then the volume is equal to _______
(1) 400 cm³
(2) 16 cm³
(3) 200 cm³
(4) \(\frac{400}{3}\) cm³
Answer:
(1) 400 cm³
Hint:
Volume of a cylinder = πr²h cu. units
[Base area (πr²) = 80 cm² = 80 × 5 cm³ = 400 cm³

Question 4.
If the total surface area of a solid right circular cylinder is 200π cm² and its radius is 5 cm, then the sum of its height and radius is ______
(1) 20 cm
(2) 25 cm
(3) 30 cm
(4) 15 cm
Answer:
(1) 20 cm
Hint:
T.S.A of a cylinder = 200π cm²
2πr (h + r) = 200π
2 × 5 (h + r) = 200
(h + r) = 20 cm

Question 5.
The curved surface area of a right circular cylinder whose radius is a units and height is b units, is equal to ______
(1) πa²b sq.cm
(2) 2πab sq.cm
(3) 2π sq.cm
(4) 2 sq.cm
Answer:
(2) 2πab sq.cm .
Hint:
C.S.A. of a cylinder = 2πrh sq. units = 2 × π × a × b sq. cm = 2πab sq. cm

Question 6.
Radius and height of a right circular cone and that of a right circular cylinder are respectively, equal. If the volume of the cylinder is 120 cm³, then the volume of the cone is equal to _______
(1) 1200 cm³
(2) 360 cm³
(3) 40 cm³
(4) 90 cm³
Answer:
(3) 40 cm³
Hint:
Volume of the cone = \(\frac{1}{3}\) × volume of the cylinder
= \(\frac{1}{3}\) × 120 cm³
= 40 cm³

Question 7.
If the diameter and height of a right circular cone are 12 cm and 8 cm respectively, then the slant height is
(1) 10 cm
(2) 20 cm
(3) 30 cm
(4) 96 cm
Answer:
(1) 10 cm
Hint:
Slant height of a cone

Question 8.
If the circumference at the base of a right circular cone and the slant height are 120π cm and 10 cm respectively, then the curved surface area of the cone is equal to ______
(1) 1200π cm²
(2) 600π cm²
(3) 300π cm²
(4) 600 cm²
Answer:
(2) 600π cm²
Hint:
Circumference (2πr) = 120π cm
Slant height (l) = 10 cm;
Curved surface area of a cone = πrl sq. units
= \(\frac{120 \pi}{2}\) × 10 cm² = 600π cm²

Question 9.
If the volume and the base area of a right circular cone are 48π cm and 12π cm respectively, then the height of the cone is equal to ______
(1) 6 cm
(2) 8 cm
(3) 10 cm
(4) 12 cm
Answer:
(4) 12 cm
Hint:
Volume of a cone = 48π cm³
[Base area (πr²) = 12π]
\(\frac{1}{3}\) πr²h = 48π
\(\frac{1}{3}\) × 12π × h = 48π
[Substitute πr² = 12π]
h = \(\frac{48}{4}\) = 12 cm

Question 10.
If the height and the base area of a right circular cone are 5 cm and 48 sq.cm respectively, then the volume of the cone is equal to _______
(1) 240 cm³
(2) 120 cm³
(3) 80 cm³
(4) 480 cm³
Answer:
(3) 80 cm³
Hint:
Volume of a cone (V) = \(\frac{1}{3}\) πr²h sq. units
Base area (πr²) = 48 sq. cm
V = \(\frac{1}{3}\) × 48 × 5 = 80 cm³

Question 11.
The ratios of the respective heights and the respective radii of two cylinders are 1 : 2 and 2 : 1 respectively. Then their respective volumes are in the ratio _______
(1) 4 : 1
(2) 1 : 4
(3) 2 : 1
(4) 1 : 2
Answer:
(3) 2 : 1
Hint:
h₁ : h₂ = 1 : 2
r₁ : r₂ = 2 : 1
Ratio of their volumes
= \(\frac{1}{3} \pi r_{1}^{2} h_{1}: \frac{1}{3} \pi r_{2}^{2} h_{2}\)
= 22 × 1 : 12 × 2 = 4 : 2 = 2 : 1

Question 12.
If the radius of a sphere is 2 cm, then the curved surface area of the sphere is equal to ________
(1) 8π cm²
(2) 16 cm²
(3) 12π cm²
(4) 16π cm²
Answer:
(4) 16π cm²
Hint:
C.S.A of a sphere = 4πr² sq. units
[radius = 2 cm]
= 4 × π × 22 cm²
= 16π cm²

Question 13.
The total surface area of a solid hemisphere of diameter 2 cm is equal to _______
(1) 12 cm²
(2) 12π cm²
(3) 4π cm²
(4) 3π cm²
Answer:
(4) 3π cm²
Hint:
Radius of a hemisphere = \(\frac{2}{2}\) = 1 cm
Total surface area of a hemisphere = 3πr² sq. units = 3 × π × 12 cm² = 3π cm²

Question 14.
If the volume of a sphere is \(\frac{9}{16} \pi\) cu.cm, then its radius is ________
(1) \(\frac{4}{3}\) cm
(2) \(\frac{3}{4}\) cm
(3) \(\frac{3}{2}\) cm
(4) \(\frac{2}{3}\) cm
Answer:
(2) \(\frac{3}{4}\) cm
Hint:
Volume of the sphere = \(\frac{9}{16} \pi\)

Question 15.
The surface areas of two spheres are in the ratio of 9 : 25. Then their volumes are in the ratio _______
(1) 81 : 625
(2) 729 : 15625
(3) 27 : 75
(4) 27 : 125
Answer:
(4) 27 : 125
Hint:
Ratio of their surface area = 9 : 25

Question 16.
The total surface area of a solid hemisphere whose radius is a units, is equal to ________
(1) 2πa² sq. units
(2) 3πa² sq. units
(3) 3πa sq. units
(4) 3a² sq. units
Answer:
(2) 3πa² sq. units
Hint:
T.S.A. of a solid hemisphere = 3πr² sq. units
= 3 × π × a × a sq.units
= 3πa² sq. units

Question 17.
If the surface area of a sphere is 100π cm², then its radius is equal to ______
(1) 25 cm
(2) 100 cm
(3) 5 cm
(4) 10 cm
Answer:
(3) 5 cm
Hint:
Surface area of a sphere = 100π cm²
4πr² = 100π
r² = 25
r = √25 = 5 cm

Question 18.
If the surface area of a sphere is 36π cm², then the volume of the sphere is equal to _______
(1) 12π cm³
(2) 36π cm³
(3) 72π cm³
(4) 108π cm³
Answer:
(2) 36π cm³
Hint:
Surface area of a sphere = 36π cm²
4πr² = 36π
r² = 9
r = 3 cm
Volume of a sphere = \(\frac{4}{3} \pi r^{3}\) cu. units
= \(\frac{4}{3} \pi\) × 3 × 3 × 3 cm³ = 36π cm³

Question 19.
If the total surface area of a solid hemisphere is 12π cm² then its curved surface area is equal to ______
(1) 6π cm²
(2) 24π cm²
(3) 36π cm²
(4) 8π cm²
Answer:
(4) 8π cm²
Hint:
T.S.A of a hemisphere = 12π cm²
3πr² = 12π
r² = 4
r = 2
Curved surface area of a hemisphere = 2πr² = 2 × π × 4 = 8π cm²

Question 20.
If the radius of a sphere is half of the radius of another sphere, then their respective volumes are in the ratio _____
(1) 1 : 8
(2) 2 : 1
(3) 1 : 2
(4) 8 : 1
Answer:
(1) 1 : 8
Hint:
\(r_{1}=\frac{r_{2}}{2} \Rightarrow \frac{r_{1}}{r_{2}}=\frac{1}{2} \Rightarrow r_{1}: r_{2}=1: 2\)

II. Answer the following questions:

Question 1.
Curved surface area and circumference at the base of a solid right circular cylinder are 4400 sq.cm and 110 cm respectively. Find its height and diameter.
Answer:
Given, Circumference of the base of a cylinder = 110 cm
2πr = 110 ……. (1)
Curved surface area = 4400 cm²
2πrh = 4400 cm² ……. (2)
From (1) & (2), \(\frac{(2)}{(1)} \Rightarrow \frac{2 \pi r h}{2 \pi r}=\frac{4400}{110}=40 \mathrm{cm}\)
Height of the cylinder (h) = 40 cm
From (1), 2πr = 110
2 × \(\frac{22}{7}\) × r = 110
r = \(\frac{35}{2}\)
We know that, diameter (d) = 2 × radius
d = 2 × \(\frac{35}{2}\) = 35 cm
Diameter of the Circular cylinder = 35 cm

Question 2.
A mansion has 12 right cylindrical pillars each having radius 50 cm and height 3.5 m. Find the cost of painting the lateral surface of the pillars at ₹ 20 per square metre.
Answer:
Given, Radius of a cylinder (r) = 50 cm = 0.5 m
Height of a cylinder (h) = 3.5 m
Curved surface area of a pillar = 2πrh sq. units
Curved surface area of 12 pillars = 12 × 2πrh
= 12 × 2 × \(\frac{22}{7}\) × 0.5 × 3.5 m²
= 132 sq. m.
Cost for painting the lateral surface of pillars per metre = ₹ 20
Cost of painting = 132 × ₹ 20 = ₹ 2640

Question 3.
The total surface area of a solid right circular cylinder is 231 cm². Its curved surface area is two thirds of the total surface area. Find the radius and height of the cylinder.
Answer:
Given, Total surface area of a cylinder (T.S. A) = 231 sq.cm
Curved surface area = \(\frac{2}{3}\) × T.S.A = \(\frac{2}{3}\) × 231 = 154 cm²
2πrh = 154 cm² …… (1)
Total surface area = 231 cm²
2πr (h + r) = 231
2πrh + 2πr²= 231
154 + 2πr² = 231 [from (1)]
2πr² = 231 – 154 = 77

Radius of the cylinder = 3.5 cm
Height of the cylinder = 7 cm

Question 4.
The total surface area of a solid right circular cylinder is 1540 cm². If the height is four times the radius of the base, then find the height of the cylinder.
Answer:
Given, Let the radius of the cylinder be ‘r’
Height of a cylinder (h) = 4r (by given condition)
Total surface area = 1540 cm²
2πr(h + r) = 1540 cm²

Height of the cylinder = 4r = 4 × 7 = 28 cm

Question 5.
If the vertical angle and the radius of a right circular cone are 60° and 15 cm respectively, then find its height and slant height.
Answer:
Given, In the figure, OAB is a cone and OC ⊥ AB
∠AOC = \(\frac{60^{\circ}}{2}\) = 30°
In the right ∆OAC, tan 30° = \(\frac{\mathrm{AC}}{\mathrm{OC}}\)

Slant Height of the cone (l) = 15 × 2 = 30 cm

Question 6.
The central angle and radius of a sector of a circular disc are 180° and 21 cm respectively. If the edges of the sector are joined together to make a hollow cone, then find the radius of the cone.
Answer:

Given, Radius of a sector (r) = 21 cm
The angle of the sector (θ) = 180°
Let “R” be the radius of the cone.
Circumference of the base of a cone = Arc length of the sector

Radius of the cone (R) = 10.5 cm

Question 7.
If the curved surface area of a solid hemisphere is 2772 sq.cm, then find its total surface area.
Answer:
Given, the Curved surface area of a solid hemisphere = 2772 cm²
2πr² = 2772

Total surface area = 3πr² sq. units
= 3 × \(\frac{22}{7}\) × 21 × 21
= 4158 sq.cm
Aliter:
C.S.A of a hemisphere = 2772 cm²
2πr² = 2772 cm²
πr² = \(\frac{2772}{2}\) = 1386 cm
T.S.A of a hemisphere = 3πr² sq.units = 3 × 1386 cm² = 4158 cm²

Question 8.
An inner curved surface area of a hemispherical dome of a building needs to be painted. If the circumference of the base is 17.6 m, find the cost of painting it at the rate of ₹ 5 per sq. m.
Answer:
Given, Circumference of the dome = 17.6 m
2πr = 17.6
\(r=\frac{17.6 \times 7}{2 \times 22}=\frac{8.8 \times 7}{22}=2.8 \mathrm{m}\)
Curved surface area of the dome = 2πr² sq. units
= 2 × \(\frac{22}{7}\) × 2.8 × 2.8 m²
= 49.28 m²
Cost of painting for one sq.metre = ₹ 5
Cost of painting the curved surface = 49.28 × ₹ 5 = ₹ 246.40

Question 9.
Volume of a solid cylinder is 62.37 cu.cm. Find the radius if its height is 4.5 cm.
Answer:
Given, Height of a cylinder (h) = 4.5 cm
Volume of a solid cylinder = 62.37 cu. cm

Radius of a cylinder (r) = 2.1 cm

Question 10.
A rectangular sheet of metal foil with dimension 66 cm × 12 cm is rolled to form a cylinder of height 12 cm. Find the volume of the cylinder.
Answer:
After rolling the rectangular sheet into a cylinder

Volume of the cylinder = 4158 cm³

Question 11.
The circumference of the base of a 12 m high wooden solid cone is 44 m. Find the volume.
Answer:
Given, Height of the wooden solid cone (h) = 12 m
Circumference of the base = 44 m
2πr = 44
r = \(\frac{44 \times 7}{2 \times 22}\) = 7 m
Volume of the wooden solid = \(\frac{1}{3} \pi r^{2} h\) cu. units
= \(\frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times 12 \mathrm{m}^{3}\)
= 88 × 7
= 616 m³
Volume of the solid = 616 m³

Question 12.
Find the volume of the largest right circular cone that can be cut out of a cube whose edge is 14 cm.
Answer:
Given, Edge of the cube = 14 cm
The largest circular cone is cut out from the cube.
Radius of the cone (r) = \(\frac{14}{2}\) = 7 cm
Height of the cone (h) = 14 cm
Volume of a cone

Volume of a cone = 718.67 cm³

Question 13.
The thickness of a hemispherical bowl is 0.25 cm. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl. (Take π = \(\frac{22}{7}\))
Answer:
Let r, R and w be the inner and outer radii and thickness of the hemispherical bowl respectively.

Given that r = 5 cm, w = 0.25 cm
R = r + w = 5 + 0.25 = 5.25 cm
Now, outer surface area of the bowl = 2πR²
= 2 × \(\frac{22}{7}\) × 5.25 × 5.25
= 173.25 sq. cm
Thus, the outer surface area of the bowl = 173.25 sq. cm

Question 14.
Volume of a hollow sphere is \(\frac{11352}{7}\) cm3. If the outer radius is 8 cm, find the inner radius of the sphere. (Take π = \(\frac{22}{7}\))
Answer:
Let R and r be the outer and inner radii of the hollow sphere respectively.
Let V be the volume of the hollow sphere.

Hence, the inner radius r = 5 cm

Question 15.
How many litres of water will a hemispherical tank whose diameter is 4.2 m?
Answer:
Radius of the tank = \(\frac{4.2}{2}\) = 2.1 m
Volume of the hemisphere
= \(\frac{2}{3} \pi r^{3}\) cu.units
= \(\frac{2}{3} \times \frac{22}{7} \times 2.1 \times 2.1 \times 2.1 \mathrm{m}^{3}\)
= 19.404 m³
= 19.404 x 1000 lit
= 19,404 litres

III. Answer the following questions.

Question 1.
A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.
Answer:

For cylindrical part:
Radius (r) = 7 cm
Height (h) = 6 cm
Curved surface area = 2πrh = 2 × \(\frac{22}{7}\) × 7 × 6 cm² = 264 cm²
For hemispherical part:
Radius (r) = 7 cm
Surface area (h) = 2πr²
= 2 × \(\frac{22}{7}\) × 7 × 7 cm²
= 308 cm²
Total surface area = (264 + 308) = 572 cm²

Question 2.
A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.
Answer:

Question 3.
From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2.
Answer:

For cylinderical part:
Height (h) = 2.4 cm
Diameter (d) = 1.4 cm
Radius (r) = 0.7 cm
Total surface area of the cylindrical part

For conical part:
Base area (r) = 0.7 cm
Height (h) = 2.4 cm

Question 4.
A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.
Answer:
Diameter of the cylindrical well = 7 m
Radius of the cylinder (r) = \(\frac{7}{2}\) m
Depth of the well (h) = 20 m
Volume = πr²h
= \(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 20 \mathrm{m}^{3}\)
= 22 × 7 × 5 m³
Volume of the earth taken out = 22 × 7 × 5 m³
Now this earth is spread out to form a cuboidal platform having
Length (l) = 22 m
Breadth (b) = 14 m
Let ‘h’ be the height of the platform.
Volume of the platform = 22 × 14 × h m³
Volume of the platform = Volume of the earth taken out
22 × 14 × h = 22 × 7 × 5
\(h=\frac{22 \times 7 \times 5}{22 \times 14}=\frac{5}{2} \mathrm{m}=2.5 \mathrm{m}\)
Thus, the required height of the platform is 2.5 m.

Question 5.
The perimeters of the ends of the frustum of a cone are 207.24 cm and 169.56 cm. If the height of the frustum is 8 cm, find the whole surface area of the frustum. [Use π = 3.14]
Answer:

Let the radii of circular ends are R and r [R > r]
Perimeter of circular ends are 207.24 cm and 169.56 cm
2πR = 207.24 cm

The whole surface area of the frustum = π [(R² + r²) + (R + r) l]
Required whole surface area of the frustum
= 3.14 [33² + 27² + (33 + 27) × 10] cm²
= 3.14 [1089 + 729 + 600] cm²
= 3.14 [2418] cm²
= 7592.52 cm²

Question 6.
A cuboid-shaped slab of iron whose dimensions are 55 cm × 40 cm × 15 cm is melted and recast into a pipe. The outer diameter and thickness of the pipe are 8 cm and 1 cm respectively. Find the length of the pipe. (Take π = \(\frac {22}{7}\))
Answer:

Let h₁ be the length of the pipe
Let R and r be the outer and inner radii of the pipe respectively.
Iron slab:
Volume = lbh = 55 × 40 × 15 cm³
Iron pipe:
Outer diameter, 2R = 8 cm
Outer radius, R = 4 cm
Thickness, w = 1 cm
Inner radius, r = R – w = 4 – 1 = 3 cm
Now, the volume of the iron pipe = Volume of the iron slab

Time is taken by the pipe to empty half of the tank = 3 hours 12 minutes.

Question 7.
The perimeter of the ends of a frustum of a cone are 44 cm and 8.4π cm. If the depth is 14 cm., then find its volume.
Answer:
Given let the radius of the top of the frustum be “R” and the radius of the bottom of the frustum be “r”

Question 8.
A tent is in the shape of a right circular cylinder surmounted by a cone. The total height and the diameter of the base are 13.5 m and 28 m. If the height of the cylindrical portion is 3 m, find the total surface area of the tent.
Answer:

Given, Total height of solid = 13.5 cm
Diameter of the cylinder (d) = 28 m
Height of a cylinder (h) = 3 m
Height of a conical portion = 13.5 – 3 = 10.5 m
From the diagram, Radius of a cone = Radius of a cylinder
Radius (r) = 14 m

Students can download Maths Chapter 7 Mensuration Unit Exercise 7 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 7 Mensuration Unit Exercise 7

Question 1.
The barrel of a fountain-pen cylindrical in shape is 7 cm long and 5 mm in diameter. A full barrel of ink in the pen will be used for writing 330 words on an average. How many words can be written using a bottle of ink containing one-fifth of a litre?
Answer:

Question 2.
A hemispherical tank of radius 1.75 m is full of water. It is connected with a pipe which empties the tank at the rate of 7 litres per second. How much time will it take to empty the tank completely?
Answer:
Radius of the hemispherical tank = 1.75 m
Volume of the tank = \(\frac{2}{3} \pi r^{3}\) cu.units

Time taken = \(\frac{11229.17}{7}\) = 1604.17 seconds = 26.74 minutes = 27 minutes (approximately)

Question 3.
Find the maximum volume of a cone that can be carved out of a solid hemisphere of radius r units.
Answer:
Radius of a cone = Radius of a hemisphere = r unit

Height of a cone = r units
(height of the cone = radius of a hemisphere)
Maximum volume of the cone

Question 4.
An oil funnel of the tin sheet consists of a cylindrical portion 10 cm long attached to a frustum of a cone. If the total height is 22 cm, the diameter of the cylindrical portion by 8 cm and the diameter of the top of the funnel be 18 cm, then find the area of the tin sheet required to make the funnel.
Answer:

Total height of oil funnel = 22 cm
Height of the cylindrical portion = 10 cm
Height of the frustum (h) = 22 – 10 = 12 cm
Radius of the cylindrical portion = 4 cm
Radius of the bottom of the frustum = 4 cm
Top radius of the funnel (frustum) = \(\frac{18}{2}\) = 9 cm
Area of the tin sheet required = C.S.A of the frustum + C.S.A of the cylinder
= π (R + r) l + 2πrh sq. units.
= [π(9 + 4) \(\sqrt{12^{2}+(9-4)^{2}}\) + 2π × 4 × 10] cm²
= π [13 × \(\sqrt{144+25}\) + 25 + 80] cm²
= \(\frac{22}{7}\) [13 × 13 + 80] cm²
= \(\frac{22}{7}\) [169 + 80] cm²
= \(\frac{22}{7}\) × 249 cm²
= 782.57 cm²
Area of sheet required to make the funnel = 782.57 cm²

Question 5.
Find the number of coins, 1.5 cm in diameter and 2 mm thick, to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm.
Answer:
Radius of the cylinder = \(\frac{4.5}{2}\) cm
Height of the cylinder = 10 cm
Volume of the cylinder = πr²h cu. units


Number of coins = 450

Question 6.
A hollow metallic cylinder whose external radius is 4.3 cm and internal radius is 1.1 cm and the whole length is 4 cm is melted and recast into a solid cylinder of 12 cm long. Find the diameter of a solid cylinder.
Answer:
External radius of the hollow cylinder R = 4.3 cm
Internal radius of the hollow cylinder r = 1.1 cm
Length of the cylinder (h) = 4 cm
Length of the solid cylinder (H) = 12 cm
Let the radius of the solid cylinder be “x”
Volume of the solid cylinder = Volume of the hollow cylinder

Diameter of the solid cylinder = 2 × 2.4 = 4.8 cm

Question 7.
The slant height of a frustum of a cone is 4 m and the perimeter of circular ends are 18 m and 16 m. Find the cost of painting its curved surface area at ₹ 100 per sq. m.
Answer:
Slant height of a frustum (l) = 4 m
Perimeter of the top part = 18 m
2πR = 18


Cost of painting = ₹ 100 × 68 = ₹ 6800

Question 8.
A hemispherical hollow bowl has material of volume cubic \(\frac{436 \pi}{3}\) cubic cm. Its external diameter is 14 cm. Find its thickness.
Answer:
External radius of a hemisphere (R) = 7 cm
Volume of a hemi-spherical bowl = \(\frac{436 \pi}{3}\) cm³

Internal radius = 5 cm
Thickness of the hemisphere = (7 – 5) cm = 2 cm

Question 9.
The volume of a cone is 1005\(\frac{5}{7}\) cu. cm. The area of its base is 201\(\frac{1}{7}\) sq. cm. Find the slant height of the cone.
Answer:
Area of the base of a cone = 201\(\frac{1}{7}\) sq. cm

Question 10.
A metallic sheet in the form of a sector of a circle of radius 21 cm has a central angle of 216°. The sector is made into a cone by bringing the bounding radii together. Find the volume of the cone formed.
Answer:

Radius of a cone (r) = 21 cm
Central angle (θ) = 216°
Let “R” be the radius of a cone
Circumference of the base of a cone = arc length of the sector
2πR = \(\frac{\theta}{360} \times 2 \pi r\)

Students can download Maths Chapter 7 Mensuration Ex 7.5 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 7 Mensuration Ex 7.5

Multiple Choice Questions

Question 1.
The curved surface area of a right circular cone of height 15 cm and base diameter 16 cm is
(1) 60π cm²
(2) 68π cm²
(3) 120π cm²
(4) 136π cm²
Solution:
(4) 13671 cm²
Hint:
Here, h = 15 cm, r = 8 cm

C.S.A of a cone = πrl sq. units. = π × 8 × 17 = 136π cm³

Question 2.
If two solid hemispheres of same base radius r units are joined together along with their bases, then the curved surface area of this new solid is
(1) 4πr² sq. units
(2) 6πr² sq. units
(3) 3πr² sq. units
(4) 8πr² sq. units
Answer:
(1) 4πr² sq. units
Hint:
When you joined two hemispheres together, the solid sphere is formed
C.S.A of the new solid = C.S.A of a sphere = 4πr² sq. units.

Question 3.
The height of a right circular cone whose radius is 5 cm and slant height is 13 cm will be
(1) 12 cm
(2) 10 cm
(3) 13 cm
(4) 5 cm
Solution:
(1) 12 cm
Hint:
Here r = 5 cm and l = 13 cm

Question 4.
If the radius of the base of a right circular cylinder is halved keeping the same height, then the ratio of the volume of the cylinder thus obtained to the volume of original cylinder is _________
(1) 1 : 2
(2) 1 : 4
(3) 1 : 6
(4) 1 : 8
Answer:
(2) 1 : 4
Hint:
Let the radius of the cylinder be “r” and the height be “h”
Radius of the new cylinder = \(\frac{r}{2}\) (Height will be same)
Volume of the new cylinder : Volume of the original cylinder
= \(\pi r_{1}^{2} h: \pi r_{2}^{2} h\) (πh is same)
= \(r_{1}^{2}: r_{2}^{2}\)
= \(\left(\frac{r}{2}\right)^{2}: r^{2}\)
= \(\frac{r^{2}}{4}: r^{2}=\frac{1}{4}: 1\)
= 1 : 4

Question 5.
The total surface area of a cylinder whose radius is \(\frac{1}{3}\) of its height is _______
(1) \(\frac{9 \pi h^{2}}{8}\) sq. units
(2) 24πh² sq.units
(3) \(\frac{8 \pi h^{2}}{8}\) sq.units
(4) \(\frac{56 \pi h^{2}}{8}\) sq.units
Answer:
(3) \(\frac{8 \pi h^{2}}{8}\) sq.units
Hint:
Let the height of the cylinder be “h”
Radius of the cylinder = \(\frac{1}{3}\) h
T.S.A of the cylinder = 2πr(h + r)

Question 6.
In a hollow cylinder, the sum of the external and internal radii is 14 cm and the width is 4 cm. If its height is 20 cm, the volume of the material in it is
(1) 5600π cm³
(2) 11200π cm³
(3) 56π cm³
(4) 3600π cm³
Solution:
(2) 112007π cm³
Hint:
Here, let the external radius be “R” and the internal radius be “r”
R + r = 14 ……(1)
Width (R – r) = 4 ……(2)
Height of the hollow cylinder = 20 cm
Volume of the hollow cylinder = πh × (R² – r²)
= πh(R + r) (R – r)
= π × 20 (14) × 4
= π × 1120
= 1120π cm³

Question 7.
If the radius of the base of a cone is tripled and the height is doubled then the volume is ______
(1) made 6 times
(2) made 18 times
(3) made 12 times
(4) unchanged
Answer:
(2) made 18 times
Hint:
Radius of a cone = r
Height of a cone = h
Volume of the cone = \(\frac{1}{3}\) πr²h cu. units
When the radius is increased three-time (tripled) and the height is doubled
Radius is 3r and the height is 2h
Volume of the new cone

Volume is increased 18 times.

Question 8.
The total surface area of a hemisphere is how many times the square of its radius.
(1) π
(2) 4π
(3) 3π
(4) 2π
Solution:
(3) 3π
Hint:
T.S.A of the hemisphere = 3πr²
The square of the radius is 3π times.

Question 9.
A solid sphere of radius x cm is melted and cast into a shape of a solid cone of the same radius. The height of the cone is _______
(1) 3x cm
(2) x cm
(3) 4x cm
(4) 2x cm
Answer:
(3) 4x cm
Hint:
Radius of a sphere = Radius of a cone = x cm
Volume of a cone = Volume of a sphere

Question 10.
A frustum of a right circular cone is of height 16cm with radii of its ends as 8cm and 20cm. Then, the volume of the frustum is
(1) 3328π cm³
(2) 3228π cm³
(3) 3240πcm³
(4) 3340π cm³
Solution:
(1) 3328π cm³
Hint:
Here, h = 16 cm, r = 8 cm, R = 20 cm
Volume of the frustum

Question 11.
A shuttlecock used for playing badminton has the shape of the combination of ______
(1) a cylinder and a sphere
(2) a hemisphere and a cone
(3) a sphere and a cone
(4) frustum of a cone and a hemisphere
Answer:
(4) frustum of a cone and a hemisphere
Hint:

Question 12.
A spherical ball of radius r₁ units is melted to make 8 new identical balls each of radius r₂ units. Then r₁ : r₂ is _______
(1) 2 : 1
(2) 1 : 2
(3) 4 : 1
(4) 1 : 4
Answer:
(1) 2 : 1
Hint:
Volume of the first sphere : Volume of second sphere = 8 : 1

Question 13.
The volume (in cm³) of the greatest sphere that can be cut off from a cylindrical log of wood of base radius 1 cm and height 5 cm is ________
(1) \(\frac{4}{3} \pi\)
(2) \(\frac{10}{3} \pi\)
(3) 5π
(4) \(\frac{20}{3} \pi\)
Answer:
(1) \(\frac{4}{3} \pi\)
Hint:
Radius of the sphere = 1 cm
Volume of the Sphere = \(\frac{4}{3}\) πr³ cu. units
= \(\frac{4}{3}\) × π × 1 × 1 × 1 cm³
= \(\frac{4}{3}\) π cm³

Question 14.
The height and radius of the cone of which the frustum is a part are h₁ units and r₁ units respectively. Height of the frustum is h₂ units and the radius of the smaller base is r₂ units. If h₂ : h₁ = 1 : 2 then r₂ : r₁ is ______
(1) 1 : 3
(2) 1 : 2
(3) 2 : 1
(4) 3 : 1
Answer:
(2) 1 : 2
Hint:
h₂ : h₁ = 1 : 2
h₁ : h₂ = 2 : 1
Ratio of their volumes

Volume is 2 : 1 the ratio of their radius also 2 : 1
r₁ : r₂ = 2 : 1 But r₂ : r₁ = 1 : 2

Question 15.
The ratio of the volumes of a cylinder, a cone and a sphere, if each has the same diameter and same height is
(1) 1 : 2 : 3
(2) 2 : 1 : 3
(3) 1 : 3 : 2
(4) 3 : 1 : 2
Solution:
(4) 3 : 1 : 2
Hint:
Volume of (cylinder : cone : sphere)

Students can download Maths Chapter 7 Mensuration Ex 7.4 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 7 Mensuration Ex 7.4

Question 1.
An aluminium sphere of radius 12 cm is melted to make a cylinder of radius 8 cm. Find the height of the cylinder.
Answer:
Sphere – Radius r₁ = 12 cm
Cylinder – Radius r₂ = 8 cm
h₂ = ?
Volume of cylinder = Volume of sphere melted

∴ Height of the cylinder made = 36 cm.

Question 2.
Water is flowing at the rate of 15 km per hour through a pipe of diameter 14 cm into a rectangular tank which is 50 m long and 44 m wide. Find the time in which the level of water in the tanks will rise by 21 cm.
Answer:
Length of the rectangular tank (l) = 50 m = 5000 cm
Width of the rectangular tank (b) = 44 m = 4400 cm
Level of water in the tank (h) = 21 cm
Volume of the tank = l × b × h cu. units = 5000 × 4400 × 21 cm³
Radius of the pipe (r) = 7 cm
Speed of the water = 15 km/hr.
(h) = 15000 × 100 cm / hr.
Volume of water flowing in one hour

Question 3.
A conical flask is full of water. The flask has base radius r units and height h units, the water poured into a cylindrical flask of base radius x r units. Find the height of water in the cylindrical flask.
Answer:
Radius of the conical flask = r units
Height of the conical flask = h units
Volume of the conical flask = \(\frac{1}{3} \pi r^{2} h\) cu.units
Radius of the cylindrical flask = x r units
Let the height of the cylindrical flask be “H” units
Volume of the cylindrical flask = Volume of the Conical flask

Height of the cylindrical flask = \(\frac{h}{3 x^{2}}\) units

Question 4.
A solid right circular cone of diameter 14 cm and height 8 cm is melted to form a hollow sphere. If the external diameter of the sphere is 10 cm, find the internal diameter.
Answer:
Radius of a cone (V) = 7 cm
Height of a cone (h) = 8 cm
External radius of the hollow sphere (R) = 5 cm
Let the internal radius be “x”
Volume of the hollow sphere = Volume of the Cone


Internal diameter of the Hollowsphere = 2 × 3 = 6 cm.

Question 5.
Seenu’s house has an overhead tank in the shape of a cylinder. This is filled by pumping water from a sump (underground tank) which is in the shape of a cuboid. The sump has dimensions 2 m × 1.5 m × 1 m. The overhead tank has its radius of 60 cm and height 105 cm. Find the volume of the water left in the sump after the overhead tank has been completely filled with water from the sump which has been full, initially.
Answer:
Length of the cuboid tank (l) = 2 cm = 200 cm
Breadth of the cuboid tank (b) = 1.5 cm = 150 cm
Height of the tank (h) = 1 m = 100 cm
Volume of the cuboid = l × b × h cu. units
= 200 × 150 × 100 cm³
= 30,00,000 cm³
Radius of the tank (r) = 60 cm
Height of the tank (h) = 105 cm
Volume of the cylindrical tank = πr²h cu. units
= \(\frac{22}{7}\) × 60 × 60 × 105 cm³
= 22 × 60 × 60 × 15 cm³
= 1188000 cm³
Volume of water left in the sump = Volume of the sump – Volume of the tank
= 3000000 – 1188000 cm³
= 1812000 cm³

Question 6.
The internal and external diameter of a hollow hemispherical shell is 6 cm and 10 cm respectively. If it is melted and recast into a solid cylinder of diameter 14 cm, then find the height of the cylinder.
Answer:
Internal radius of the shell (r) = 3 cm
External radius of the shell (R) = 5 cm
Radius of the cylinder (r) = 7 cm
Let the height of the cylinder be “h”
Volume of the cylinder = Volume of the hemispherical shell

Height of the cylinder = 1.33 cm

Question 7.
A solid sphere of radius 6 cm is melted into a hollow cylinder of uniform thickness. If the external radius of the base of the cylinder is 5 cm and its height is 32 cm, then find the thickness of the cylinder.
Answer:
Radius of a sphere (r) = 6 cm
External radius of the cylinder (R) = 5 cm
Height of the cylinder (h) = 32 cm
Let the internal radius of the cylinder be ‘x’
Volume of the hollow cylinder = Volume of a sphere
πh (R² – r²) = \(\frac{4}{3}\) πr³
π × 32 (5 + x) (5 – x) = \(\frac{4}{3}\) × π × 6 × 6 × 6
32 (25 – x²) = 4 × 2 × 6 × 6
25 – x² = 9
x² = 25 – 9 = 16
x = √16 = 4
Thickness of the cylinder = 5 – 4 = 1 cm.

Question 8.
A hemispherical bowl is filled to the brim with juice. The juice is poured into a cylindrical vessel whose radius is 50% more than its height. If the diameter is same for both the bowl and the cylinder then find the percentage of juice that can be transferred from the bowl into the cylindrical vessel.
Answer:
Let the height of the cylinder be “h”
radius is 50% more than the height


From (1) and (2) we get,
Volume of the cylinder = Volume of the hemisphere
It is possible to transfer the full quantity from the bowl into the cylindrical vessel.
100 % of the juice can be transferred.

Students can download Maths Chapter 7 Mensuration Ex 7.3 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 7 Mensuration Ex 7.3

Question 1.
A vessel is in the form of a hemispherical bowl mounted by a hollow cylinder. The diameter is 14 cm and the height of the vessel is 13 cm. Find the capacity of the vessel.
Answer:
Radius of a hemisphere = Radius of the cylinder

Question 2.
Nathan, an engineering student was asked to make a model shaped like a cylinder with two cones attached at its two ends. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of the model that Nathan made.
Answer:
Radius of the cone = Radius of the cylinder

r = \(\frac{3}{2}\) cm
Height of the cone (H) = 2 cm
Height of the cylinder (h) = 12 – (2 + 2) cm = 8 cm
Volume of the model = Volume of the cylinder + Volume of 2 cones

Question 3.
From a solid cylinder whose height is 2.4 cm and the diameter 1.4 cm, a cone of the same height and same diameter is carved out. Find the volume of the remaining solid to the nearest cm³.
Answer:
Radius of a cylinder = Radius of a cone r = 0.7 cm
Height of a cylinder = Height of a cone (h) = 2.4 cm

Volume of the remaining solid = Volume of the cylinder – Volume of a cone

Volume of the remaining soild = 2.46 cm³

Question 4.
A solid consisting of a right circular cone of height 12 cm and radius 6 cm standing on a hemisphere of radius 6 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of the water displaced out of the cylinder, if the radius of the cylinder is 6 cm and height is 18 cm.

Answer:
Radius of a cone = Radius of a hemisphere = Radius of a cylinder
r = 6 cm
Height of a cone (h) = 12 cm
Volume of the water displaced = Volume of the solid inside = Volume of the cone + Volume of the hemisphere

Volume of water displaced = 905. 14 cm³.

Question 5.
A capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. If the length of the entire capsule is 12 mm and the diameter of the capsule is 3 mm, how much medicine it can hold?
Answer:
Radius of a hemisphere = Radius of a Cylinder

r = \(\frac{3}{2}\) mm = 1.5 mm
Height of the cylinderical portion = 12 mm – (1.5 mm + 1.5 mm) = (12 – 3) mm = 9 mm
Volume of the capsule

Volume of the capsule = 77.8 cu. mm

Question 6.
As shown in figure a cubical block of side 7 cm is surmounted by a hemisphere. Find the surface area of the solid.

Answer:
Side of a cube (a) = 7 cm
Radius of a hemisphere (r) = \(\frac{7}{2}\) cm
Surface area of the solid = T.S.A of the cube + C.S.A of the hemisphere – Area of the base of the hemisphere

Question 7.
A right circular cylinder just encloses a sphere of radius r units. Calculate
(i) the surface area of the sphere
(ii) the curved surface area of the cylinder
(iii) the ratio of the areas obtained in (i) and (ii).
Answer:
(i) Surface area of sphere = 4πr² sq. units

Question 8.
A shuttlecock used for playing badminton has the shape of a frustum of a cone is mounted on a hemisphere. The diameters of the frustum are 5 cm and 2 cm. The height of the entire shuttlecock is 7 cm. Find its external surface area.
Answer:

Students can download Maths Chapter 7 Mensuration Ex 7.2 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 7 Mensuration Ex 7.2

Question 1.
A 14 m deep well with inner diameter 10 m is dug and the earth taken out is evenly spread all around the well to form an embankment of width 5 m. Find the height of the embankment.

Answer:
Radius of the well (r₁) = 5 m
Depth of the well (h) = 14 m
Width of the embankment = 5 m
Outer radius (R) = 5 + 5 = 10 m
Let the height of the embankment be “H”
Volume of Earth in the embankment = Volume of the well
πH(R² – r²) = \(\pi r_{1}^{2} h\)
H(10² – 5²) = 5 × 5 × 14
H (100 – 25) = 5 × 5 × 14
H = \(\frac{5 \times 5 \times 14}{75}\) = 4.67 m
Height of the embankment = 4.67 m

Question 2.
A cylindrical glass with diameter 20 cm has water to a height of 9 cm. A small cylindrical metal of radius 5 cm and height 4 cm is immersed it completely. Calculate the rise of the water in the glass?
Answer:
Radius of the cylindrical glass (r) = 10 cm
Height of the water (h) = 9 cm
Radius of the cylindrical metal (R) = 5 cm
Height of the metal (H) = 4 cm
Let the height of the water raised be “h”
Volume of the water raised in the cylinder = Volume of the cylindrical metal
πr²h = πr²H
10 × 10 × h = 5 × 5 × 4
h = \(\frac{5 \times 5 \times 4}{10 \times 10}\) = 1 cm
Raise of water in the glass = 1 cm

Question 3.
If the circumference of a conical wooden piece is 484 cm then find its volume when its height is 105 cm.
Answer:
Circumference of the wooden piece = 484 cm
2πr = 484
2 × \(\frac{22}{7}\) × r = 484 cm
r = \(\frac{484 \times 7}{2 \times 22}\)
r = 77 cm
Height of the wooden piece (h) = 105 cm
Volume of the conical wooden piece = \(\frac{1}{3} \pi r^{2} h\) cu.units
= \(\frac{1}{3} \times \frac{22}{7} \times 77 \times 77 \times 105 \mathrm{cm}^{3}\)
= 22 × 11 × 77 × 35 cm³
= 652190 cm³
Volume of the wooden piece = 652190 cm³

Question 4.
A conical container is fully filled with petrol. The radius is 10m and the height is 15 m. If the container can release the petrol through its bottom at the rate of 25 cu. meter per minute, in how many minutes the container will be emptied. Round off your answer to the nearest minute.
Answer:
The radius of the conical container (r) = 10 m
Height of the container (h) = 15 m

Question 5.
A right-angled triangle whose sides are 6 cm, 8 cm and 10 cm is revolved about the sides containing the right angle in two ways. Find the difference in volumes of the two solids so formed.

Answer:
Three sides of a triangle are 6 cm, 8 cm and 10 cm.
Case (i): If the triangle is revolved about the side 6 cm, the cone will be formed with radius 6 cm and height 8 cm.
Volume of the cone = \(\frac{1}{3} \pi r^{2} h\) cu. units
= \(\frac{1}{3}\) × π × 6 × 6 × 8 = 96π cm³
Case (ii): If the triangle is revolved about the side 8 cm, the cone will be formed with radius 8 cm and height 6 cm.
Volume of the cone = \(\frac{1}{3}\) × π × 8 × 8 × 6 = 128π cm³
Difference in volume of the two solids = (128π – 96π) cm³ = 32π cm³ = 32 × \(\frac{22}{7}\) cm³ = 100.57 cm³
The difference in the volume of the two solids = 100.57 cm³

Question 6.
The volumes of two cones of same base radius are 3600 cm³ and 5040 cm³. Find the ratio of heights.
Answer:
Let the radius of the two cones be ‘r’
Let the height of the two cones be h₁ and h₂
Ratio of their volumes = 3600 : 5040 (÷ 10)
\(\frac{1}{3} \pi r^{2} h_{1}: \frac{1}{3} \pi r^{2} h_{2}\) = 360 : 504 (÷4)
h₁ : h₂ = 90 : 126 (÷3)
= 30 : 42 (÷3)
= 10 : 14 (÷2)
h₁ : h₂ = 5 : 7
Ratio of heights = 5 : 7

Question 7.
If the ratio of radii of two spheres is 4 : 7, find the ratio of their volumes.
Answer:
Let the ratio of their radii is r₁ : r₂
r₁ : r₂ = 4 : 7
Ratio of their volumes
V₁ : V₂ = \(\frac{4}{3} \pi r_{1}^{3}: \frac{4}{3} \pi r_{2}^{3}\)
= \(r_{1}^{3}: r_{2}^{3}\)
= 43 : 73
Ratio of their volumes = 64 : 343

Question 8.
A solid sphere and a solid hemisphere have an equal total surface area. Prove that the ratio of their volume is 3√3 : 4.
Answer:
Total surface area of a sphere = \(4 \pi r_{1}^{2}\) sq. units
Total surface area of a hemisphere = \(3 \pi r_{2}^{2}\) sq. units
Ratio of Total surface area = \(4 \pi r_{1}^{2}: 3 \pi r_{2}^{2}\)
1 = \(\frac{4 \pi r_{1}^{2}}{3 \pi r_{2}^{2}}\) (Same Surface Area)
1 = \(\frac{4 r_{1}^{2}}{3 r_{2}^{2}}\)

Ratio of their volumes = 3√3 : 4
Hence it is proved.

Question 9.
The outer and the inner surface areas of a spherical copper shell are 576π cm² and 324π cm² respectively. Find the volume of the material required to make the shell.
Answer:
Outer surface area of a spherical shell = 576π cm²
4πR² = 576π
4 × R² = 576
R² = \(\frac{576}{4}\) = 144
R = √144 = 12 cm
Inner surface area of a spherical shell = 324π cm²
4πr² = 324π
4r² = 324
r² = 81
r = √81 = 9
Volume of the material required = Volume of the hollow hemisphere = \(\frac{4}{3}\) π(R³ – r³) cm³

Volume of the material required = 4186.29 cm³

Question 10.
A container open at the top is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends are 8 cm and 20 cm respectively. Find the cost of milk which can completely fill a container at the rate of ₹ 40 per litre.
Answer:
Height of the frustrum (h) = 16 cm
Radius of the upper part (R) = 20 cm
Radius of the lower part (r) = 8 cm
Volume of the frustum

Cost of milk in the container = 10.459 × 40 = ₹ 418.36
Cost of the milk = ₹ 418.36