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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F

February 3, 2023

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F

Other Exercises

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10A
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10B
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10C
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10D
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10E
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F

Question 1.
The 6th term of an A.P. is 16 and the 14th term is 32. Determine the 36th term.
Solution:
Let the first term and common difference of an A.P. be a and d
As, we know that,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F Q1.1

Question 2.
If the third and the 9th terms of an A.P. term is 12 and the last term is 106. Find the 29th term of the A.P.
Solution:
Let the first term and common difference of an A.P. be a and d.
As, we know that,
a_n = a + (n – 1 )d
a₃ = a + (3 – 1 )d = a + 2d
Similarly,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F Q1.2

Question 3.
An A.P. consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term of the A.P.
Solution:
Number of terms in an A.P. = 50
T₃= 12, l = 106
To find T₂₉
Let a be the first term and d be the common difference
=> a + 2d = 12 …(i)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F Q3.1

Question 4.
Find the arithmetic mean of :
(i) – 5 and 41
(ii) 3x – 2y and 3x + 2y
(iii) (m + n)² and (m – n)²
Solution:
(i) Arithmetic mean between – 5 and 41
= \(\\ \frac { -5+41 }{ 2 } \)
= \(\\ \frac { 36 }{ 2 } \)
= 18
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F Q4.1

Question 5.
Find the sum of first 10 terms of the A.P. 4 + 6 + 8 +…..
Solution:
A.P. = 4 + 6 + 8 +…….
Here, a = 4, d = 6 – 4 = 2, n = 10
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F Q5.1

Question 6.
Find the sum of first 20 terms of an A.P. whose first term is 3 and the last term is 60.
Solution:
Sum of first 20 terms of an A.P. in which
a = 3 and a₂₀ = 60
a₂₀ = a + (20 – 1) x d
60 = 3 + 19 x d
19d = 60 – 3
d = \(\\ \frac { 57 }{ 19 } \)
= 3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F Q6.1

Question 7.
How many terms of the series 18 + 15 + 12 +…..when added together will give 45 ?
Solution:
A.P. is 18 + 15 + 12 +…..
Here, a = 18, d = 15 – 18 = – 3
Given : S_n = 45
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F Q7.1

Question 8.
The nth term of a sequence is 8 – 5n. Show that the sequence is an A.P.
Solution:
Given, a_n = 8 – Sn
a₁ = 8 – 5 x (1) = 8 – 5 = 3
a₂ = 8 – 5 x (2) = 8 – 10 = – 2
a₃ = 8 – 5 x (3) = 8 – 15 = – 7
We see that
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F Q8.1

Question 9.
The the general term (nth term) and 23rd term of the sequence 3, 1, – 1, – 3,……
Solution:
The progression 3, 1, – 1, – 3,…..is A.P.
with first term (a) = 3 and common difference (d) = 1 – 3 = – 2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F Q9.1

Question 10.
Which term of the sequence 3, 8, 13,…..is 78 ?
Solution:
Let 78 be the nth term
a = 3, d = 8 – 3 = 5, a_n = 78, n = ?
a + (n – 1)d = a_n
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F Q10.1

Question 11.
Is – 150 a term of 11, 8, 5, 2,….. ?
Solution:
11, 8, 5, 2,….1st term, a = 11
Common difference, d = 8 – 11 = – 3
a_n = – 150
=> a + (n – 1 )d = – 150
=> 11 + (n – 1) ( – 3) = – 150
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F Q11.1

Question 12.
How many two digit numbers are divisible by 3 ?
Solution:
Numbers divisible by 3 are 3, 6, 9, 12,….
Hence, lowest two digit number divisible by 3 = 12
and highest two digit number divisible by 3 = 99
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F Q12.1

Question 13.
How many multiples of 4 lie between 10 and 250 ?
Solution:
Multiples of 4 between 10 and 250 are
12, 16, 20, 24,……, 248
Here, a = 12, d = 4, l = 248
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F Q13.1

Question 14.
The sum of the 4th term and the 8th term of an A.P. is 24 and the sum of 6th term and the 10th term is 44. Find the first three terms of the A.P.
Solution:
In an A.P.
T₄ + T₈ = 24
T₆ + T₁₀ = 44
Let a be the first term and d be the common difference
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F Q14.1

Question 15.
The sum of first 14 terms of an A.P. is 1050 and its 14th term is 140. Find the 20th term.
Solution:
Given a₁₄ = 140
we know, a_n = a + (n – 1) x d
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F Q15.1

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F 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 Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21D

February 3, 2023

RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21D

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 21 Concept of Perimeter and Area Ex 21D

Other Exercises

Question 1.
Solution:
(i) Length (l) = 46 cm
Breadth (b) = 25 cm
Area of rectangle = l x b
= 46 x 25 sq. cm
= 1150 sq.cm.
(ii) Length (l) = 9 m
Breadth (b) = 6 m
Area = l x b = 9 x 6
= 54 sq. metre Ans.
(iii) Length (l) = 14.5 m
Breadth (b) = 6.8 m
Area = l x b
= 14.5 x 6.8 sq. m
= 98.6 sq. m Ans.
(iv) Length (l) = 2m 5cm
= 2x 100 cm + 5 cm
= 200 cm + 5 cm
= 205 cm
Breadth = 60 cm
Area = l x b
= 205 cm x 60 cm
= 12300 cm²
(v) Length (l) = 3.5 km
Breadth (b) = 2 km
Area = l x b
= 3.5 x 2
= 7 sq. km. Ans.

Question 2.
Solution:
Side of a square plot = 14 m
Area = (Side)²
= (14)²
= 196 m²

Question 3.
Solution:
Length of top of table (l) = 2 m, 25 cm = 2.25 m
Breadth (l) = 1 m 20 cm = 1.20 m
Area of the top of the table = l x b
= (2.25 x 1.20) sq. m
= 2.7 sq. m. Ans.

Question 4.
Solution:
Length of carpet (l) = 30 m 75 cm
= 30.75 m
Breadth (b) = 80 cm = 0.80 m
Area of the carpet = l x b
= (30.75 x 0.80) sq. m
= 24.6 sq. m
Cost of one square metre = Rs. 20
Total cost = 24.6 x 150
= Rs. 3690. Ans

Question 5.
Solution:
Length of the sheet of paper 3 m 24 cm
= 300 cm + 24 cm
= 324 cm
Breadth of the sheet of the paper 1 m 72 cm
= 100 cm + 72 cm
= 172 cm
Area of the sheet of paper = (324 x 172) cm²
Also, area of the piece of paper required for an envelope = (18 x 12) cm².
Number of envelopes that can be made
= \(\\ \frac { 324\times 172 }{ 18\times 12 } \)
= 258

Question 6.
Solution:
Length of room (l) = 12.5 m
Breadth (b) = 8 m
Area = l x b
= (12.5 x 8) sq. m
= 100 sq. m
Side of square carpet (a) = 8 m
Area of carpet = a² = (8 x 8) sq. m.
= 64 square metre
Area left without carpet = 100 sq. m – 64 sq. m
= 36 sq. m Ans.

Question 7.
Solution:
Length of the lane = 150 m
Breadth of the lane = 9 m
Area of the lane = (150 x 9) m²
= 1350 m²
Area of the brick = 22.5 cm x 7.5 cm
= 168.75 cm²
RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21D Q7.1

Question 8.
Solution:
Length of room (l) = 13 m
Breadth (b) = 9 m
Area of floor or carpet = l x b
= 13 x 9
= 117 sq. m
RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21D Q8.1

Question 9.
Solution:
Let the length of the rectangular park = 5x metres
and the breadth of the rectangular park = 3x metres
RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21D Q9.1

Question 10.
Solution:
Side of the square plot = 64 m
Perimeter of the square plot = 4 x Side
= 4 x 64
= 256 m
Perimeter of the rectangular plot = Perimeter of the square plot = 256 m
Length of the rectangular plot = 70 m
Perimeter = 2 x (Length + Breadth)
256 = 2 (70 + b)
256 = 140 + 2b
=> 2b = 256 – 140
=> 2b = 116
b = \(\\ \frac { 116 }{ 2 } \) = 58 cm
Area of the rectangular plot = (length x breadth)]
= (70 x 58) m²
= 4060 m²
Area of the square plot = Side x Side
= (64 x 64) m²
= 4096 m².
Square plot has the greater area than that of the rectangular plot by
(4096 – 4060)
= 36 m².

Question 11.
Solution:
Total cost of cultivating the rectangular field = Rs. 71400
Rate of cultivating = Rs. 35 per sq. m
RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21D Q11.1

Question 12.
Solution:
Area of rectangle = 540 sq. cm
Length (l) = 36 cm
RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21D Q12.1

Question 13.
Solution:
Measure of a marble tile = 12cm x 10cm
Area of wall = 4 m x 3 m
= 12 m²
Area of one marble tile
= 12 x 10
= 120 cm²
RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21D Q13.1

Question 14.
Solution:
Area of a rectangle = 600 cm²
Breadth = 25 cm
RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21D Q14.1

Question 15.
Solution:
diagonal of square = 5 √2
RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21D Q15.1

Question 16.
Solution:
(i) We name the given region as shown in the figure
RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21D Q16.1

Question 17.
Solution:
Measures are in cm
(i) In the figure, there are three rectangles and one square
RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21D Q17.1

= 5(6 x 6) cm²
= 180 cm²

Hope given RS Aggarwal Solutions Class 6 Chapter 21 Concept of Perimeter and Area Ex 21D 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.

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25B

February 3, 2023

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25B

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Chapter 25 Probability Ex 25B.

Other Exercises

Question 1.
Nine cards (identical in all respects) are numbered 2 to 10. A card is selected from them at random. Find the probability that the card selected will be :
(i) an even number.
(ii) a multiple of 3 ,
(iii) an even number and a multiple of 3.
(iv) an even number or a multiple of 3.
Solution:
No. of cards = 9
Having numbers marked on it = 2 to 10
∴ Number of possible outcomes = 9
(i) An even number i.e. 2, 4, 6, 8, 10 = 5
∴ Number of even numbers = 5
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25B Q1.1
(ii) A multiple of 3 are 3, 6, 9
Number of multiple of 3 = 3

(iii) An even number and a multiple of 3 are 6
which is one in number

(iv) An even number or a multiple of 3 are 2, 3, 4, 6, 8, 9, 10
which are 7 in numbers
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25B Q1.4

Question 2.
Hundred identical cards are numbered from 1 to 100. The cards are well shuffled and then a card is drawn. Find the probability that the number on the card drawn is :
(i) a multiple of 5.
(ii) a multiple of 6.
(iii) between 40 and 60.
(iv) greater than 85.
(v) less than 48.
Solution:
Number of cards = 100
Marked with numbers from 1 to 100
∴ Number of possible outcome = 100
(i) A multiple of 5 are 5, 10, 15 95, 100
which are 20 in numbers.
∴ Number of favourable outcome = 20
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25B Q2.1
(ii) A multiple of 6 are 6, 12, 18, 24, 90, 96
which are 16 in numbers.
∴ Number of favourable outcome =16

(iii) Between 40 and 60 are 41, 42, ……. , 58, 59,
which are 19

(iv) Greater than 85 are 86 to 100 which are 15 in numbers.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25B Q2.4
(v) Less than 48 are 1 to 47 which are 47 in numbers

Question 3.
From 25 identical cards, numbered, 1, 2, 3, 4, 5 ,…………, 24, 25 ; one card is drawn at random.
Find the probability that the number on the card drawn is a multiple of :
(i) 3
(ii) 5
(iii) 3 and 5
(iv) 3 or 5
Solution:
Number of identical cards = 25
Numbers marked on their are 1 to 25 i.e.
1, 2, 3, 4, 5, …. 21, 22, 23, 24, 25
∴ Number of possible outcome = 25
(i) Multiple of 3 are 3, 6, 9, 12, 15, 18, 21, 24
Which are 8 in numbers.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25B Q3.1
(ii) Multiple of 5 are 5, 10, 15, 20, 25
which are 5 in numbers

(iii) Multiple of 3 and 5 are = 15
which is 1 in number

(iv) Multiple of 3 or 5 are 3, 5, 6, 9, 10, 12, 15, 18, 20, 21, 24, 25
which are 12 in numbers
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25B Q3.4

Question 4.
A die is thrown once. Find the probability of getting a number :
(i) less than 3.
(ii) greater than or equal to 4.
(iii) less than 8
(iv) greater than 6.
Solution:
A die has 6 numbers i.e., 1, 2, 3, 4, 5, 6
∴ Number of possible outcome = 6
(i) Less than 3 are 1, 2 which are 2 in numbers
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25B Q4.1
(ii) Greater than or equal to 4 are 4, 5, 6
which are 3 in number.

(iii) Less than 8 are 1, 2, 3, 4, 5, 6
which are 6 in numbers

(iv) Greater than 6 is nothing on the die

Question 5.
A book contains 85 pages. A page is chosen at random. What is the probability that the sum of the digits on the page is 8 ?
Solution:
Number of pages of the book = 85
which are from 1 to 85
Number of possible outcome = 85
∴ Number of pages whose sum of its page is 8 are : 17, 26, 35, 44, 53, 62, 71, 80 and 8
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25B Q5.1

Question 6.
A pair of dice is thrown. Find the probability of getting a sum of 10 or more if 5 appears on the first die.
Solution:
Numbers marked on each die = 6
∴ Total number of cases = 6 x 6 = 36
∵ Favourable come are (5, 5), (5, 6) as 5 appears on the first, which are 2 in number
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25B Q6.1

Question 7.
If two coins are tossed once, what is the probability of getting :
(i) 2 heads
(ii) at least one head
(iii) both heads or both tails.
Solution:
∵ A coins has two faces Head and Jail or H, T
∴ Two coins are tossed
∴ Number of coins = 2 x 2 = 4
which are HH, HT, TH, TT
(i) When both are Head, then
∴ Number of outcome = 1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25B Q7.1
(ii) At least one head, then
Number of outcomes = 3

(iii) When both head or both tails, then 1
Number of outcomes = 2

Question 8.
Two dice are rolled together. Find the probability of getting :
(i) a total of at least 10.
(ii) a multiple of 2 on one die and an odd number on the other die.
Solution:
∵ A die has 6 faces which are 1, 2, 3, 4, 5,6
∴ On rolling two dice at a time, number of comes = 6 x 6 = 36
∴ Number of possible outcome = 36
(i) a total of atleast 10, the favourable can be (4,6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6) which are 6 in number
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25B Q8.1
(ii) A multiple of 2 on one die and an odd number on the other
∴ Outcome can be (2, 1), (2, 3), (g, 5), (4, 1), (4,3) , (4, 5), (6, 1), (6, 3), (6, 5), (1, 2), (3, 2), (5, 2), (1, 4), (3, 4), (5, 4), (1, 6), (3, 6), 5, 6) which are 18 in numbers.
Number of favourable outcome
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25B Q8.2

Question 9.
A card is drawn from a well-shuffled pack of 52 cards. Find the probability that the card drawn is :
(i) a spade.
(ii) a red card.
(iii) a face card.
(iv) 5 of heart or diamond.
(v) Jack or queen.
(vi) ace and king.
(vii) a red and a king.
(viii) a red or a king.
Solution:
A pack of playing card has 52 cards
∴ Number of possible outcome = 52
(i) A spade
∵ there are 13 cards of spade
∴ Number of favourable outcome = 13
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25B Q9.1
(ii) A red card.
∵ There are 13 + 13 = 26 red card

(iii) A face card.
∵ There are 3 x 4 = 12 faces which are red.

(iv) 5 of heart or diamond.
∴ Number of cards =1 + 1=2

(v) Jack or queen
There are 4 + 4 = 8 such cards
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25B Q9.5
(vi) ace and king.
There is no such card which is ace and king both
∴ P(E) = 0.
(vii) a red and a king
There are 2 such cards which are red kings

(viii) a red or a king
There are 26 cards which are red in which 2 kings are red and 2 more kings which are black = 26 + 2 = 28
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25B Q9.7

Question 10.
A bag contains 16 coloured balls. Six are green, 7 are red and 3 are white. A ball is chosen, without looking into the bag. Find the probability that the ball chosen is :
(i) red
(ii) not red
(iii) white
(iv) not white
(v) green or red
(vi) white or green
(vii) green or red or white.
Solution:
Number of balls in a bag = 16
Green balls = 6
White balls = 3
Red balls = 7
∴ Total possible outcome =16
(i) Red balls = 7
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25B Q10.1
(ii) Not red balls = 16-7 = 9

(iii) White balls = 3

(iv) Not white balls = 16 – 3= 13

(v) Green or red balls = 6 + 7 = 13

(vi) White or green balls = 3 + 6 = 9

(vii) Green or red or white balls =16
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25B Q10.7

Question 11.
A ball is drawn at random from a box .ontammg 12 white. 16 red and 20 green balls. Determine the probability that the ball drawn is:
(i) white
(ii) red
(iii) not green
(iv) red or white.
Solution:
Number of balls in a box
White = 12
Red = 16
Green = 20
Total balls = 12 + 16 +20 = 48
∴ Total possible outcome = 48
(i) White balls = 12
∴ Number of favourable outcome =12
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25B Q11.1
(ii) Red balls = 16
∴ Number of favourable outcome =16

(iii) Not green
Number of balls =12 + 16 = 28
∴ Number of favourable outcome = 28

(iv) Red or white balls =12 + 16 = 28
∴ Number of favourable outcome = 28
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25B Q11.4

Question 12.
A card is drawn from a pack of 52 cards. Find the probability that the card drawn is :
(i) a red card
(ii) a black card
(iii) a spade
(iv) an ace
(v) a black ace
(vi) ace of diamonds
(vii) not a club
(viii) a queen or a jack
Solution:
Number of cards in playing card deck = 52
∴ Number of possible outcome = 52
(i) A red card
There are 13 + 13 = 26 red cards in the deck
∴ Number of favourable outcome = 26
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25B Q12.1
(ii) A black card
There are 13 + 13 = 26 black cards
∴ Number of favourable outcome = 26

(iii) A spade
There are 13 spade cards in the deck
∴ Number of favourable outcome = 13

(iv) An Ace
There use 4 aces in the deck
Number of favourable outcome = 4
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25B Q12.4
(v) A black ace
There are two black aces in a deck
∴ Number of favourable outcome = 2
Number of cards in playing card deck = 52

(vi) Ace of diamonds
∴ There are only one ace of diamonds
∴ Number of favourable of outcome = 1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25B Q12.6
(vii) Not a club
There are 13 x 3 = 39 cards which are not a club
∴ Number of favourable outcome = 39

(viii) A queen or a Jack
There are 4 queen cards and 4 Jack cards
∴ Number of favourable outcome = 4 + 4 = 8
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25B Q12.8

Question 13.
Thirty identical cards are marked with numbers 1 to 30. If one card is drawn at random, find the probability that it is :
(i) a multiple of 4 or 6.
(ii) a multiple of 3 and 5
(iii) a multiple of 3 or 5
Solution:
There are 30 cards which are marks with numbers 1 to 30 and one card is drawn
(i) A multiple of 4 or 6.
∴ There are multiple of 4 or 6 = 4,6, 8, 12, 16, 18, 20, 24, 28, 30 which are 10
∴ Number of favourable outcome = 10
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25B Q13.1
(ii) A multiple of 3 and 5 are 15, 30 which are 2
∴ Number of favourable outcome = 2

(iii) A multiple of 3 or 5
which are 3, 5, 6, 9, 10, 12, 15, 18, 20, 21, 24, 25, 27, 30 which are 14
∴ Number of favourable outcome = 14
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25B Q13.3

Question 14.
In a single throw of two dice, find the probability of :
(i) a doublet
(ii) a number less than 3 on each dice.
(iii) an odd number as a sum.
(iv) a total of at most 10.
(v) an odd number on one dice and a number less than or equal to 4 on the other dice.
Solution:
Number of dice thrown = 2
Each die has 1 to 6 numbers on its faces Number of possible outcomes = 6 X 6 = 36
(i) A doublet : These can be (1, 1), (2, 2),(3, 3), (4, 4), (5, 5) and (6, 6) which are 6
∴ Number of favourable outcome = 6
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25B Q14.1
(ii) A number less than 3 on each die which can be (1, 1), (1, 2), (2, 1), (2, 2) which are 4 in numbers
∴ Number of favourable outcome = 4

(iii) An odd number as a sum which can be (1, 1), (1,3), (1,5), (2, 1), (2,3), (2,5), (3, 1), (3,3), (3,5), (4, 1), (4,3), (4, 5), (5,1),(5,3), (5,5), (6, 1), (6, 3), (6, 5) which are 18 in numbers.
∴ Number of favourable outcome = 18
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25B Q14.3
(iv) Total of at most 10
Which can be = 36 – 3 (which can be (5, 6), (6, 6), (6, 5) = 33
∴ Number of favourable outcome = 33

(v) An odd number on one die and a number less than or equal to 4 on the other die
(1, 1), (1,2), (1,3), (1,4), (3,1), (3,2), (3, 3), (3,4) , (5, 1), (5, 2), (5, 3), (5, 4), (2, 1), (3, 1), (4, 1),(1, 3), (2, 3), (2, 5), (3, 1), (3, 5), (4, 1), (4, 3), (4,5) which are 23 in numbers
∴ Number of favourable outcome = 23
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25B Q14.5

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25B 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 Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21C

February 3, 2023

RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21C

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 21 Concept of Perimeter and Area Ex 21C

Other Exercises

Question 1.
Solution:
1. In the given figure, there are 12 complete squares, so its area is 12 cm².
2. In the given figure, there are 18 complete squares, so its area is 18 cm².
3. In the given figure, there are 14.5 complete squares, so its area is 14.5 cm² .
4. In the given figure, there are 6 complete squares and four half parts of a square, the area of the figure is 8 cm².
5. The given figure contains 9 complete squares and 6 half parts of a square. So the area of the figure is
\(\left( 9+\frac { 6 }{ 2 } \right) \) = 9 + 3 = 12 cm².
6. The given figure contains 16 complete squares. So, its area is 16 cm².
7. The given figure contains 4 complete squares, 8 more than half parts and 4 less than half parts of a square. Neglecting the less than half parts and considering the more than half parts as complete squares, the approximate area of the figure is 12 cm².
8. The given figure contains 7 complete squares and 5 more than half parts and some less than half parts of a square. Neglecting the less than half parts and considering more than half parts as complete square, the area of the figure is 12 cm² approximately.
9. The given figure contains 14 complete squares and four half parts of a square. So, the area of the figure is
\(\left( 14+\frac { 4 }{ 2 } \right) \) cm² = (14 + 2) cm² = 16 cm².

Hope given RS Aggarwal Solutions Class 6 Chapter 21 Concept of Perimeter and Area Ex 21C are helpful to complete your math homework.

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10E

February 3, 2023

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10E

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10E

Other Exercises

Question 1.
Two cars start together in the same direction from the same place. The first car goes at uniform speed of 10 km h^-1. The second car goes at a speed of 8 km h^-1 in the first hour and thereafter increasing the speed by 0.5 km h^-1 each succeeding hour. After how many hours will the two cars meet ?
Solution:
Speed of first car = 10 km/hr
Speed of second car = 8 km/hr in first hour
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10E Q1.1

Question 2.
A sum of Rs 700 is to be paid to give seven cash prizes to the students of a school for their overall academic performance. If the cost of each prize is Rs 20 less than its preceding prize; find the value of each of the prizes.
Solution:
Total amount (S_n) = Rs 700
Cost of each prize is Rs 20 less than its preceding price or d = – 20
d = – 20 and n = 7
\({ S }_{ n }=\frac { n }{ 2 } \left[ 2a+(n-1)d \right] \)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10E Q2.1

Question 3.
An article can be bought by paying Rs 28,000 at once or by making 12 monthly installments. If the first installment paid is Rs 3,000 and every other installment is Rs 100 less than the previous one, find :
(i) amount of installment paid in the 9th month
(ii) total amount paid in the installment scheme.
Solution:
Total price of an article = Rs 28000
No. of installments (n) = 12
First installment (a) = RS 3000
d = Rs 100
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10E Q3.1

Question 4.
A manufacturer of TV sets produces 600 units in the third year and 700 units in the 7th year. Assuming that the production increases uniformly by a fixed number every year, find :
(i) the production in the first year.
(ii) the production in the 10th year.
(iii) the total production in 7 years.
Solution:
A manufacture of TV sets, he produces
No. of units in 3rd year = 600
No. of units in 7th year = 700
Let a be the first term and d be the common difference, then
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10E Q4.1

Question 5.
Mrs. Gupta repays her total loan of Rs 1.18,000 by paying installments every month. If the installment for the first month is Rs 1,000 and it increases by RS 100 every month, what amount will she pay as the 30th installment of loan? What amount of loan she still has to pay after the 30th installment?
Solution:
Total loan to be paid by Mrs. Gupta = Rs 118000
Installment for the first month (a) = Rs 1000
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10E Q5.1

Question 6.
In a school, students decided to plant trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be five times of the class to which the respective section belongs. If there are 1 to 10 classes in the school and each class has three sections, find how many trees were planted by the students?
Solution:
Number of classes = 10
Number of sections of each class = 3
Total number of sections = 10 x 3 = 30
Each section plant tree = 5 times of the class
Each section of 1st class will plant = 1 x 15 = 15
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10E Q6.1

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RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21B

February 3, 2023

RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21B

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 21 Concept of Perimeter and Area Ex 21B

Other Exercises

Question 1.
Solution:
(i) Radius of the circle (r) = 28 cm
Circumference = 2 πr
= 2 x \(\\ \frac { 22 }{ 7 } \) x 28 cm
= 176 cm Ans.
RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21B Q1.1

Question 2.
Solution:
(i) Diameter of the circle (d) = 14 cm
Circumference = πd
= \(\\ \frac { 22 }{ 7 } \) x 14
= 44 cm
RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21B Q2.1

Question 3.
Solution:
Circumference of the circle = 176 cm
Let r be the radius, then
RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21B Q3.1

Question 4.
Solution:
Circumference of a wheel = 264 cm
Let d be its diameter, then
πd = 264
=> \(\\ \frac { 22 }{ 7 } d\)
= 264
RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21B Q4.1

Question 5.
Solution:
Diameter of the wheel (d) = 77 cm
Circumference = πd
RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21B Q5.1

Question 6.
Solution:
Diameter of the wheel = 70 cm
circumference = πd
RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21B Q6.1

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Value Based Questions in Science for Class 10 Chapter 12 Electricity

February 3, 2023

Value Based Questions in Science for Class 10 Chapter 12 Electricity

These Solutions are part of Value Based Questions in Science for Class 10. Here we have given Value Based Questions in Science for Class 10 Chapter 12 Electricity

Question 1.
In the house of Ram, there are 20 incandescent bulbs each of 100 W, three geysers each of 2000 W and 20 tubes each of 40 W. All these appliances work for 5 hours in a day. Every month, he pays heavy amount as electricity bill. His son Sham studying in X standard asked his father to replace all incandescent bulbs with CFL bulb each of 40 W to save electricity.

How much units of electricity are saved per month (30 days) by replacing incandescent bulbs with CFL ?
What values are shown by Sham ?

Answer:

Electric energy consumed by 20 incandescent bulbs in 30 days = P x t = 20 x 100 x 5 x 30
= 300000 Wh = 300 kWh
= 300 units
Electric energy consumed by 20 CFL bulbs = 23 x 40 x 5 x 30 = 120000 Wh
= 120 kWh =120 units
Units of electricity saved = 300 – 120 = 180 units.
Sham helped his father to pay less electricity bill. He also believes that saving energy contributes for the development of our nation.

More Resources

Question 2.
Electricity plays an important role in the development of a country. Ram, a student of class X was studying in the library after school hours. When he left the library, he found that electric fans of all rooms were ‘ON’ although there was no one in the class rooms. He immediately switched ‘OFF’ all the fans and reported the matter to Principal of the school.

Comment on the attitude of Ram.
Why, Ram reported the matter to the Principal ?

Answer:

Ram knows the importance of electricity. He believes that loss of electricity is the loss of school as well as the loss of nation. He is against the misuse of national resources.
He reported the matter to the Principal so that the Principal may ask the peon to ensure that such incident should not occur in future.

Question 3.
A welder was asked by Mr. Sumit to weld an iron grill in his house. He started using electricity by connecting the wires of welding set directly with the transmission wires and not through the energy meter. Sumit’s neighbour objected the action of welder but Sumit sided with the welder. However, the son of Sumit appreciated the neighbour.

Why was the action of welder objected by Sumit’s neighbour ?
Why, Sumit supported the action of the welder ?
Why, Sumit s son appreciated the action of his neighbour ?
Write the commercial unit of electrical energy.

Answer:

It is a crime to steal electrical energy and no honest person can support it.
Sumit thought the welder was doing a favour to him.
Sumit’s son knows that welder is doing wrong.
Kilowatt hour (kWh).

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24D

February 3, 2023

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency (Mean, Median, Quartiles and Mode) Ex 24D

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Chapter 24 Measures of Central Tendency (Mean, Median, Quartiles and Mode) Ex 24D.

Other Exercises

Question 1.
Find the mode of the following data:
(i) 7,9,8,7,7,6,8,10,7 and 6
(ii) 9,11,8,11,16,9,11,5,3,11,17 and 8
Solution:
(i) Mode = 7
because it occurs 4 times
(ii) Mode =11
because it occurs 4 times

Question 2.
The following table shows the frequency distribution of heights of 50 boys:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24D Q2.1
Find the mode of heights.
Solution:
Mode is 122 because it occurs maximum times i.e its., frequency is 18.

Question 3.
Find the mode of following data, using a histogram:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24D Q3.1
Solution:
Mode class = 20 – 30
Mode = 24

We see in the histogram that line AD and CB intersect at P. Draw perpendicular Q to the horizontal x-axis. Which is the value of the mode = 24

Question 4.
The following table shows the expenditure of 60 boys on books. Find the mode of their expenditure:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24D Q4.1
Solution:
Model class is = 30 – 35
and Mode = 34

We see in the histogram that line AD and CB intersect at P. Draw perpendicular Q to the horizontal axis. Which is the value of the mode.

Question 5.
Find the median and mode for the set of numbers 2,2,3,5,5,5,6,8 and 9.
Solution:
Median = \(\frac { 9 +1 }{ 2 }\) = 5th term which is 5
Mode = 5, because it occurs in maximum times.

Question 6.
A boy scored following marks in various class tests during a term, each test being marked out of 20.
15,17,16,7,10,12,14,16,19,12,16.
(i) What are his modal marks ?
(ii) What are his median marks ?
(iii) What are his total marks ?
(iv) What are his mean marks ?
Solution:
Arranging the given data in ascending order : 7, 10,12, 12,14, 15,16,16, 16, 17,19.
(i) Mode = 16 as it occurs in maximum times.
(ii) Median= \(\frac { 11 +1 }{ 2 }\) = 6th term which is 15
(iii) Total marks = 7 + 10+ 12+ 12+ 14+ 15+ 16 + 16+ 16+ 17+ 19= 154
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24D Q6.1

Question 7.
Find the mean, median and mode of the following marks obtained by 16 students in a class test marked out of 10 marks.
0,0,2,2,3,3,3,4,5,5,5,5,6, 6,7,8
Solution:

(ii) Median = Mean of 8th and 9th term
= \(\frac { 4 +5 }{ 2 }\) = \(\frac { 9 }{ 2 }\) = 4.5
(iii) Mode = 5 as it occurs in maximum times.

Question 8.
At a shooting competition the score of a com-petitor were as given below :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24D Q8.1
(i) What was his modal score ?
(ii) What was his median score ?
(iii) What was his total score ?
(iv) What was his mean score ?
Solution:

(i) Modal score =4 as its frequency is 7, the maximum.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24D Q8.3

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10D

February 3, 2023

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10D

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10D

Other Exercises

Question 1.
Find three numbers in A.P. whose sum is 24 and whose product is 440.
Solution:
Let three numbers be a – d, a, a + d
a – d + a + a + d = 24
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10D Q1.1

Question 2.
The sum of three consecutive terms of an A.P. is 21 and the slim of their squares is 165. Find these terms.
Solution:
Let three consecutive numbers in A.P. are
a – d, a, a + d
a – d + a + a + d = 21
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10D Q2.1

Question 3.
The angles of a quadrilateral are in A.P. with common difference 20°. Find its angles.
Solution:
Let the angles of a quadrilateral are
a, a + d, a + 2d, a + 3d
d= 20°
a + a + d + a + 2d + a + 3d = 360°
(Sum of angles of a quadrilateral)
=> 4a + 6d = 360°
=> 4a + 6 x 20° = 360°
=> 4a + 120° = 360°
=> 4a = 360 – 120° = 240°
a = \(\\ \frac { 240 }{ 4 } \) = 60°
Angles are 60°, 80°, 100°, 120°

Question 4.
Divide 96 into four parts which are in A.P. and the ratio between product of their means to product of their extremes is 15 : 7.
Solution:
Number = 96
Let its four parts be a, a + d, a + 2d, a + 3d
a + a + d + a + 2d + a + 3d = 96
=> 4a + 6d = 96
=> 2a + 3d = 48 …(i)
Product of means : Product of extremes = 15 : 7
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10D Q4.1

Question 5.
Find five numbers in A.P. whose sum is \(12 \frac { 1 }{ 2 } \) and the ratio of the first to the last terms is 2 : 3.
Solution:
Let 5 numbers in A.P. be
a, a + d, a + 2d, a + 2d, a + 4d
a + a + d + a + 2d + a + 3d + a + 4d =
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10D Q5.1

Question 6.
Split 207 into three parts such that these parts are in A.P. and the product of the two smaller parts is 4623.
Solution:
Number = 207
Let part be a – d, a, a + d
a – d + a + a + d = 207
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10D Q6.1

Question 7.
The sum of three numbers in A.P. is 15 the sum of the squares of the extreme is 58. Find the numbers
Solution:
Let three numbers in A.P. be a – d, a, a + d
a – d + a + a + d = 15
=> 3a = 15
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10D Q7.1

Question 8.
Find four numbers in A.P. whose sum is 20 and the sum of whose squares is 120.
Solution:
Let four numbers in A.P. be
a – 3d, a – d, a + d, a + 3d
a – 3d + a – d + a + d + a + 3d = 20
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10D Q8.1

Question 9.
Insert one arithmetic mean between 3 and 13.
Solution:
Let A be the arithmetic mean between 3 and 13
\(\left( A=\frac { a+b }{ 2 } \right) \)
A = \(\\ \frac { 3+13 }{ 2 } \)
= \(\\ \frac { 16 }{ 2 } \)
= 8

Question 10.
The angles of a polygon are in A.P. with common difference 5°. If the smallest angle is 120°, find the number of sides of the polygon.
Solution:
Angles of a polygon are in A.P.
and common difference (d) = 5°
Smallest angle (a) = 120°
Let n be the number of sides of the polygon then sum of angles = (2n – 4) x 90°
a = 120° and d = 5°
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10D Q10.1

Question 11.
\(\\ \frac { 1 }{ a } \), \(\\ \frac { 1 }{ b } \) and \(\\ \frac { 1 }{ c } \) are in A.P
S.T : bc, ca and ab are also in A.P
Solution:
\(\\ \frac { 1 }{ a } \), \(\\ \frac { 1 }{ b } \) and \(\\ \frac { 1 }{ c } \) are in A.P
We have to show that
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10D Q11.1

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24C

February 3, 2023

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency (Mean, Median, Quartiles and Mode) Ex 24C

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Chapter 24 Measures of Central Tendency (Mean, Median, Quartiles and Mode) Ex 24C.

Other Exercises

Question 1.
A student got the following marks in 9 questions of a question paper : 3, 5, 7, 3, 8, 0, 1, 4 and 6.
Find the median of these marks.
Solution:
Arranging the given data in descending order, we get:
8, 7, 6, 5,4,3, 3, 1,0
The middle term is 4 which is 5th terms
∴ Median = 4

Question 2.
The weights (in kg) of 10 students of a class are given below :
21, 28.5, 20.5, 24, 25.5, 22, 27.5, 28, 21, 24 Find the median of their weights.
Solution:
Arranging the given data in descending order.
We get 23.5, 28, 27.5, 25.5, 24, 24, 22, 21, 21, 20.5
the middle terms are 24 and 24, 5th and 6th terms.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24C Q2.1

Question 3.
The marks obtained by 19 students of a class are given below :
27, 36, 22, 31, 25, 26, 33, 24, 37, 32, 29, 28, 36, 35, 27, 26, 32, 35, 28. Find :
(i) Media
(ii) Lower quartile
(iii) Upper quartile
(iv) Inter – quartle range
Solution:
(i) Arranging in order say ascending, we get
22, 24, 25, 26, 26, 27, 27, 28, 28, 29, 31, 32, 32, 33, 35, 35,36, 36, 37
Middle term is 10th term i.e. 29
∴ Median = 29
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24C Q3.1

Question 4.
From the following data, find :
(i) Median
(ii) Upper quartile
(iii) Inter-quartile range
25,10, 40, 88, 45, 60, 77, 36,18, 95, 56, 65, 7, 0, 38 and 83.
Solution:
(i) Arrange in ascending order, we get
0,7, 10, 18, 25, 36, 38, 40, 45, 56, 60, 65 ,77, 83, 88, 95
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24C Q4.1

Question 5.
The ages of 37 students in a class are given in the following table :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24C Q5.1
Find the median.
Solution:

Question 6.
The weight of 60 boys are given in the following distribution table :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24C Q6.1
Find : (i) Median
(ii) Lower quartile
(iii) Upper quartile
(iv) Inter-quartile range
Solution:

Question 7.
Estimate the median for the given data by drawing ogive :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24C Q7.1
Solution:

Through mark of 25.5 on the y-axis, draw a line parallel to x-axis which meets the curve at A. From A, draw a perpendicular to x-axis, which meets x-axis at B.
∴ The value of B is the median which is 28.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24C Q7.3

Question 8.
By drawing an ogive; estimate the median for the following frequency distribution :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24C Q8.1
Solution:

Through mark of 28th on the y- axis, draw a line parallel to x-axis which meets the curve at A. From A, draw a perpendicular line segment to x- axis. Which meets it at B.

∴ The value of B is the median which is 18.4

Question 9.
From the following cumulative frequency table, draw ogive and then use it to find :
(i) Median,
(ii) Lower quartile,
(iii) Upper quartile.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24C Q9.1
Solution:

No. of terms = 80
Median = 40th term Through mark of 40 draw a line parallel to x-axis which meets the curve at A. From A, draw a perpendicular to x-axis which meets it at B
(ii) Lower quartile (Q1) = \(\frac { n }{ 4 }\) th term
= \(\frac { 80 }{ 4 }\) th term (Here n = 80 which is even)
= 20th term =18
(iii) Upper quartile (Q1) = \(\frac { 3 }{ 4 }\) nth term =\(\frac { 3 x80 }{ 4 }\) = 60th term = 66 .
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24C Q9.3
∴ Value of B is the median which is 40.

Question 10.
In a school 100 pupils have heights as tabulated below:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24C Q10.1
Find the median height by drawing an ogive.
Solution:

Through mark 50, draw a line parallel to x- axis which meets the curve at A. From A, draw per-pendicular to x-axis which meets x-axis at B is the median which is 148 cm.

P.Q.
Attempt this question on a graph paper. The table shows the distribution of marks gained by a group of 400 students in an examination :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24C Qp1.1

Using a scale of 2 cm to represent 10 marks and 2 cm to represent 50 students, plot these points and draw a smooth curve through the points
Estimate from the graph :
(i) Median marks
(ii) quartile marks. [1997]
Solution:
Plot the points (10, 5), (20, 10), (30, 30), (40, 60), (50,105), (60,180), (70,270), (80, 355), (90, 390), (100, 400) on the graph and join them with free hand to get an ogive (curve) as shown:
(i) Total students = 400
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24C Qp1.2
From 200 on y-axis draw a line parallel to x-axis meeting the curve at P. From P, draw PL perpendicular on x-axis then L is the median which is 62.
(ii) Lower quartile (Q1) = \(\frac { 1 }{ 4 }\) x n = \(\frac { 1 }{ 4 }\) x 400 =100
From 100 ony-axis, draw a line parallel to x-axis meeting the curve at Q, from Q, draw QM ⊥ x-axis.
M is the required lower quartile (Q1) which is 49 3 3
Upper quartile (Q3) = \(\frac { 3 }{ 4 }\) n = \(\frac { 3 }{ 4 }\) x 400 = 300
From 300 on y-axis, draw a line parallel to x-axis meeting the curve at R. From R draw RN perpendicular to x-axis N is the required upper quartile (Q3) which is = 74

P.Q.
Attempt this question on graph paper.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24C Qp2.1
(i) Construct the ‘less than’ cumulative frequency curve for the above data, using 2 cm = 10 years on one axis and 2cm = 10 casualities on the other, (ii) From your graph determine :
(a) Median
(b) Lower quartile. (1995)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24C Qp2.2
No. of terms = 83
∴ Median \(\frac { 83 }{ 2 }\) =41.5 th term .
Through marks 41.5,draw a line segment par allel to x-axis which meets the curve at A. From A, draw a line segment perpendicular to x-axis meeting it at B.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24C Qp2.3

Through 21 on the y-axis draw a line parallel to x-axis meeting the curve at M
From M, draw a perpendicular on x-axis which meets it at N.
∴N is the lower quartile which is 29 (approx)

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24C are helpful to complete your math homework.

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24B

February 3, 2023

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency (Mean, Median, Quartiles and Mode) Ex 24B

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Chapter 24 Measures of Central Tendency (Mean, Median, Quartiles and Mode) Ex 24B.

Other Exercises

Question 1.
The following table gives the ages of 50 students of a class. Find the arithmetic mean of their ages.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24B Q1.1
Solution:

Question 2.
The following table gives the weekly wages of workers in a factory.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24B Q2.1
Calculate the mean by using :
(i) Direct Method
(ii) Short-Cut Method
Solution:
(i) Direct Method:

(ii) Short cut method :

Question 3.
The following are the marks obtained by 70 boys in a class test :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24B Q3.1
Calculate the mean by :
(i) Short-Cut Method
(ii) Step-Deviation Method.
Solution:
(i) Short cut Method :

(ii) Step – Deviation Method:

Question 4.
Find mean by ‘step-deviation method :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24B Q4.1
Solution:

Question 5.
The mean of following frequency distribution is 21\(\frac { 1 }{ 7 }\) Find the value of ‘f ‘.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24B Q5.1
Solution:

Question 6.
Using step-deviation method, calculate the mean marks of the following distribution.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24B Q6.1
Solution:
Let Assumed mean = 72.5

Question 7.
Using the information given in the adjoining histogram; calculate the mean.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24B Q7.1

Question 8.
If the mean of the following observations is 54, find the value of p.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24B Q8.1
Solution:
Mean = 54

⇒ 2106 + 54p = 2370 + 30p
⇒ 54p – 30p = 2370 – 2106 ⇒ 24p = 264
p = \(\frac { 264 }{ 24 }\) = 11
Hence p = 11

Question 9.
The mean of the following distribution is 62.8 and the sum of all the frequencies is 50. Find the missing frequencies f₁ and f₂.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24B Q9.1
Solution:
Mean = 62.8
and sum of frequencies = 50

Question 10.
Calculate the mean of the distribution given below using the short cut method.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24B Q10.1
Solution:

Question 11.
Calculate the mean of the following distribution :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24B Q11.1
Solution:

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