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RS Aggarwal Solutions Class 10 Chapter 13 Constructions Ex 13a

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 13 Constructions Ex 13a.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 13 Constructions Ex 13a
• RS Aggarwal Solutions Class 10 Chapter 13 Constructions Ex 13b
• RS Aggarwal Solutions Class 10 Chapter 13 Constructions Test Yourself

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:

Question 8.
Solution:
BC will be divided in the ratio 3 : 4

Question 9.
Solution:

Question 10.
Solution:

Hope given RS Aggarwal Solutions Class 10 Chapter 13 Constructions Ex 13a are helpful to complete your math homework.

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RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance MCQ

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance MCQ.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance Ex 14a
• RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance MCQ

Question 1.
Solution:

Correct option: (c)
Let AB be the pole and let AC be its shadow.
Now, AB = AC
If θ is the angle of elevation of the sun, then

Question 2.
Solution:

Correct option: (c)
Let AB be the pole and let AC be its shadow.
Let the length of the shadow = x m.
Then, height of the pole = √3x m.
If e is the angle of elevation of the sun, then

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:
Correct option: (d)
Let the length of a shadow of 12-rn-high tree = x m
Now,
Ratio of lengths of objects = Ratio of lengths of their shadow

Question 6.
Solution:

Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:

Question 12.
Solution:

Question 13.
Solution:

Question 14.
Solution:

Question 15.
Solution:

Question 16.
Solution:

Question 17.
Solution:

Question 18.
Solution:

Question 19.
Solution:

Question 20.
Solution:

Question 21.
Solution:

Question 22.
Solution:

Question 23.
Solution:

Question 24.
Solution:

Question 25.
Solution:

Hope given RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance MCQ are helpful to complete your math homework.

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RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance Ex 14a

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance Ex 14a.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance Ex 14a
• RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance MCQ

Question 1.
Solution:
Let AB be the tower standing on a level ground and O be the position of the observer. Then OA = 20 m and ∠OAB = 90° and ∠AOB = 60°

Let AB = h meters
From the right ∆OAB, we have

Question 2.
Solution:
Let OB be the length of the string from the level of ground and O be the point of the observer, then, AB = 75m and ∠OAB = 90° and ∠AOB = 60°, let OB = l meters.
From the right ∆OAB, we have

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:

Question 8.
Solution:
Let SP be the statue and PB be the pedestal. Angles of elevation of S and P are 60° and 45° respectively.
Further suppose AB = x m, PB = h m

Question 9.
Solution:
Let AB be the unfinished tower and let AC be complete tower.
Let O be the point of observation. Then,
OA = 75 m
∠AOB = 30° and ∠AOC = 60°

Question 10.
Solution:
Let AB be the tower and BC be flagpole, Let O be the point of observation.
Then, OA = 9 m, ∠AOB = 30° and ∠AOC = 60°


Hence, height of the tower= 5.196 m and the height of the flagpole = 10.392 m.

Question 11.
Solution:

Question 12.
Solution:

Question 13.
Solution:

Question 14.
Solution:

Question 15.
Solution:

Question 16.
Solution:

Question 17.
Solution:
Let AB and CD be the first and second towers respectively.
Then, CD = 90 m and AC = 60 m.
Let DE be the horizontal line through D.

Draw BF ⊥CD,
Then, BF = AC = 60 m
∠FBD = ∠EDB = 30°

Question 18.
Solution:

Question 19.
Solution:

Question 20.
Solution:

Question 21.
Solution:

Question 22.
Solution:

Question 23.
Solution:
Let AB be the height of the deck and let CD be the cliff..
Let the man be at B, then, AB= 16 m
Let BE ⊥CD and AE ⊥ CD
Then, ∠EBD = 60 and ∠EBC = 30

CE = AB = 16m
Let CD = h meters
Then, ED = (h 16)m
From right ∆BED, we have

Question 24.
Solution:

Question 25.
Solution:

Question 26.
Solution:
Let AB be the tower and let the angle of elevation of its top at C be 30°.
Let D be a point at a distance 150 m from C such that the angle of elevation of the top of tower at D is 60°.
Let h m be the height of the tower and AD = x m

Question 27.
Solution:
Let AB be the light house and let C and D be the positions of the ship.
Let AD =x, CD = y


The distance travelled by the ship during the period of observation = 115.46 m

Question 28.
Solution:

Question 29.
Solution:

Question 30.
Solution:

Question 31.
Solution:

Question 32.
Solution:

Question 33.
Solution:

Hope given RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance Ex 14a are helpful to complete your math homework.

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RS Aggarwal Solutions Class 10 Chapter 15 Probability MCQ

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 15 Probability MCQ.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 15 Probability Ex 15a
• RS Aggarwal Solutions Class 10 Chapter 15 Probability Ex 15b
• RS Aggarwal Solutions Class 10 Chapter 15 Probability MCQ

Question 1.
Solution:
Correct option: (c)
We know that, the probability of an event E will always lie between O and 1, where O is the probability of an impossible event and 1 is the probability of a sure event.

Question 2.
Solution:
Correct option: (b)
Let E be the event.
So, the probability of the event happening will be P(E).
Thus, the probability of the event not happening will be P(E’).
Given that, P(E)= p
We know that, P(E’)+ P(E)= 1
p + P(E’) = 1
=P(E’)= 1 – p

Question 3.
Solution:
Correct option: (b)
The probability of an impossible event is always O.

Question 4.
Solution:
Correct option: (c)
The probability of a sure event is always 1.

Question 5.
Solution:
Correct option: (a)
We know that, the probability of an event E will always lie between O and 1.
Since 1.5>1, it cannot be the probability of an event.

Question 6.
Solution:

Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:
Correct option: (d)
The total number of cards = 9
The primes numbers would be 2, 3, 5, 7, 11.
So, there are 5 numbers.
P(getting a prime number) = 5/9.

Question 11.
Solution:
Correct option: (b)
The total number of tickets = 40
The multiples of 7 between 1 and 40 are 7, 14, 21, 28 and 35.
So, there are 5 numbers,
P(getting a multiple of 7) = 5/40 = 1/8.

Question 12.
Solution:
Correct option: (d)
We know that, the probability of an event E will always lie between O and 1.
Since 7/6 >1, it cannot be the probability of an event.

Question 13.
Solution:

Question 14.
Solution:
Correct option: (d)
An event that cannot occur is called an impossible event.
The probability of an impossible event is 0.

Question 15.
Solution:
Correct option: (b)
The total number of tickets = 20
The multiples of 5 between 1 and 20 are 5, 10, 15 and 20.
So, there are 4 numbers.
P(getting a multiple of 5) = 4/20 = 1/5.

Question 16.
Solution:

Question 17.
Solution:
Correct option: (d)
The total number of bckets = 10
The numbers less than 10 are 6, 7, 8 and 9.
So, there are 4 numbers.
P(getting a number less than 10) = 2/5.

Question 18.
Solution:
Correct option: (a)
The numbers on a die are 1, 2, 3, 4, 5 and 6.
So, there are 6 numbers in total.
The even numbers on the die are 2, 4 and 6.
So, there are 3 numbers.
P(getting an even number) = 3/6 = 1/2.

Question 19.
Solution:
Correct option: (d)
The numbers on a fair die are 1, 2, 3, 4, 5 and 6.
So, there are 6 numbers in total.
The numbers greater than 2 are 3, 4, 5 and 6.
So, there are 4 numbers.
P(getting a number greater than 2) = 4/6 = 2/3.

Question 20.
Solution:
Correct option: (b)
The numbers on a die are 1, 2, 3, 4, 5 and 6.
So, there are 6 numbers in total.
The odd number on a die greater than 3 is 5.
So, there is only 1 number.
P(getting an odd number greater than 3) = 1/6.

Question 21.
Solution:
Correct option: (c)
The numbers on a die are 1, 2, 3, 4, 5 and 6.
So, there are 6 numbers in total.
The prime numbers on the die are 2, 3 and 5.
So, there are 3 numbers.
P(getting a prime number on the die) = 3/6 = 1/2.

Question 22.
Solution:

Question 23.
Solution:
Correct option: (d)
When two coins are tossed the outcomes are:
{HH, HT, TH, TT}
So, there are 4 numbers in total.
P(getting one head) = 1/4.

Question 24.
Solution:

Question 25.
Solution:
Correct option: (d)
When two coins are tossed the simultaneously the outcomes are:
{HH, HT, TH, TT}
So, there are 4 outcomes.
Getting atmost one head means the possible outcomes are:
{HT, TH, TT}
So, there are 3 possible outcomes.
P(getting atmost one head) = 3/4.

Question 26.
Solution:
Correct option: (c)
When three coins are tossed the simultaneously the
outcomes are:
{HHH, HHT, HTH, THH, THT, HTT, TTH and TTT}
So, there are 8 possible outcomes.
P(getting exactly two heads) = 3/8.

Question 27.
Solution:
Correct option: (b)
The number of prizes = 8
The number of blanks = 16
So, the total number of tickets=8+ 16 = 24
P(getting a prize) = 8/24 = 1/3.

Question 28.
Solution:
Correct option: (c)
The number of prizes= 6
The number of blanks = 24
So, the total number of tickets = 6+24=30
P(not getting a prize)= 24/30 = 4/5.

Question 29.
Solution:
Correct option: (c)
The bag contains 3 blue, 2 white and 4 red marbles.
So, the total number of marbles = 3+2+4= 9
Since the marbles cannot be white, it can blue or red.
The number of blue or red marbles= 3+4=7
P(getting a blue or red marble)= 7/9.

Question 30.
Solution:
Correct option: (b)
The bag contains 4 red and 6 black balls.
So, the total number of balls= 4+6= 10
The number of black balls = 6
P(getting a black ball) = 6/10 = 3/5.

Question 31.
Solution:
Correct option: (c)
The bag contains 8 red, 2 black and 5 white balls,
So, the total number of balls= 8+2+5=15
Since the ball should not be black, it can be red or white.
The number of red and white balls = 13
P(getting a red and white ball) = 13/15.

Question 32.
Solution:
Correct option: (c)
The bag contains 3 white, 4 red and 5 black balls.
So, the total number of balls = 3+ 4+ 5= 12
For the ball that is drawn to be neither black nor white, it should be red.
The number of red balls = 4
P(getting a red ball) = 4/12 = 1/3.

Question 33.
Solution:
Correct option: (b)
The total number of cards = 52
The number of black kings = 2
P(getting a black king) = 2/52 = 1/26.

Question 34.
Solution:
Correct option: (a)
The total number of cards = 52
The number of queens = 4
P(getting a queen) = 4/52 = 1/13.

Question 35.
Solution:
Correct option: (c)
The total number of cards = 52
The number of face cards = 12
P(getting a face card) = 12/ 52 = 3/13.

Question 36.
Solution:
Correct option: (b)
The total number of cards = 52
The number of black face cards = 6
P(getting a black face card) = 6/52 = 3/26.

Question 37.
Solution:
Correct option: (c)
The total number of cards = 52
The number of 6 in the deck of cards = 4
P(getting a 6) = 4/52 = 1/13.

Hope given RS Aggarwal Solutions Class 10 Chapter 15 Probability MCQ are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Solutions Class 10 Chapter 15 Probability Ex 15b

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 15 Probability Ex 15b.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 15 Probability Ex 15a
• RS Aggarwal Solutions Class 10 Chapter 15 Probability Ex 15b
• RS Aggarwal Solutions Class 10 Chapter 15 Probability MCQ

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:
Total number of tickets = 100
(i) Even numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98, 100
Total number of even number = 50
P(getting a even number) = 50/100 = 1/2.

(ii) Numbers less than 16 are 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15
Total number of numbers less than 16 is 14
P(getting a number less than 16) = 14/100 = 7/50.

(iii) Numbers which are perfect square are 4, 9, 16, 25, 36, 49, 64, 81, 100
Total number of perfect squares = 9
P(getting a perfect square) = 9/100.

(iv) Prime numbers less than 40 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37
Total number of prime numbers =12
P(getting a prime number less 40) = 12/100 = 3/25.

Question 8.
Solution:
The total number of discs in the box are 80.
The numbers which are perfect squares are 1, 4, 9, 16, 25, 36, 49 and 64.
So, there are 8 possible outcomes.
P(getting a number which is a perfect square) = 8/80 = 1/10.

Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:

Question 12.
Solution:

Question 13.
Solution:

Question 14.
Solution:

Question 15.
Solution:

Question 16.
Solution:
Two dice are thrown simultaneously
Total number of outcomes = 6 6 = 36

(i) Favourable cases are:
(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 6) = 25.
∴ Probability that 5 will not come upon either die = 25/36.

(ii) Favourable cases are:
(1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6) = 11
Probability that 5 will come at least once = 11/36.

(iii) 5 will come up on both dice in 1 case = (5,5)
∴ probability that 5 will come on both dice = = 1/36.

Question 17.
Solution:

Question 18.
Solution:

Question 19.
Solution:

Question 20.
Solution:

Question 21.
Solution:

Question 22.
Solution:

Question 23.
Solution:

Question 24.
Solution:

Question 25.
Solution:

Hope given RS Aggarwal Solutions Class 10 Chapter 15 Probability Ex 15b are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.