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Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression

November 10, 2023May 26, 2022 by Veerendra

Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression

Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression

Arithmetic Progression Exercise 10A – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
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Solution:

Question 2.
The nth term of sequence is (2n – 3), find its fifteenth term.
Solution:
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Question 3.
If the pth term of an A.P. is (2p + 3), find the A.P.
Solution:
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Question 4.
Find the 24th term of the sequence:
12, 10, 8, 6,……
Solution:
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Question 5.
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Solution:

Question 6.
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Solution:

Question 7.
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Solution:

Question 8.
Is 402 a term of the sequence :
8, 13, 18, 23,………….?
Solution:
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Question 9.
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Solution:

Question 10.
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Solution:

Question 11.
Which term of the A.P. 1 + 4 + 7 + 10 + ………. is 52?
Solution:
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Question 12.
If 5th and 6th terms of an A.P are respectively 6 and 5. Find the 11th term of the A.P
Solution:
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Question 13.

Solution:
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Question 14.
Find the 10th term from the end of the A.P. 4, 9, 14,…….., 254
Solution:
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Question 15.
Determine the arithmetic progression whose 3rd term is 5 and 7th term is 9.
Solution:
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Question 16.
Find the 31st term of an A.P whose 10th term is 38 and 10th term is 74.
Solution:
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Question 17.
Which term of the services :
21, 18, 15, …………. is – 81?
Can any term of this series be zero? If yes find the number of term.
Solution:
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Question 18.
An A.P. consists of 60 terms. If the first and the last terms be 7 and 125 respectively, find the 31^stterm.
Solution:
For a given A.P.,
Number of terms, n = 60
First term, a = 7
Last term, l = 125
⇒ t₆₀ = 125
⇒ a + 59d = 125
⇒ 7 + 59d = 125
⇒ 59d = 118
⇒ d = 2
Hence, t₃₁ = a + 30d = 7 + 30(2) = 7 + 60 = 67

Question 19.
The sum of the 4^th and the 8^th terms of an A.P. is 24 and the sum of the sixth term and the tenth term is 34. Find the first three terms of the A.P.
Solution:
Let ‘a’ be the first term and ‘d’ be the common difference of the given A.P.
t₄ + t₈ = 24 (given)
⇒ (a + 3d) + (a + 7d) = 24
⇒ 2a + 10d = 24
⇒ a + 5d = 12 ….(i)
And,
t₆ + t₁₀ = 34 (given)
⇒ (a + 5d) + (a + 9d) = 34
⇒ 2a + 14d = 34
⇒ a + 7d = 17 ….(ii)
Subtracting (i) from (ii), we get
2d = 5

Question 20.
If the third term of an A.P. is 5 and the seventh terms is 9, find the 17^th term.
Solution:
Let ‘a’ be the first term and ‘d’ be the common difference of the given A.P.
Now, t₃ = 5 (given)
⇒ a + 2d = 5 ….(i)
And,
t₇ = 9 (given)
⇒ a + 6d = 9 ….(ii)
Subtracting (i) from (ii), we get
4d = 4
⇒ d = 1
⇒ a + 2(1) = 5
⇒ a = 3
Hence, 17^th term = t₁₇ = a + 16d = 3 + 16(1) = 19

Arithmetic Progression Exercise 10B – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
In an A.P., ten times of its tenth term is equal to thirty times of its 30th term. Find its 40th term.
Solution:
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Question 2.
How many two-digit numbers are divisible by 3?
Solution:
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Question 3.
Which term of A.P. 5, 15, 25 ………… will be 130 more than its 31st term?
Solution:
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Question 4.
Find the value of p, if x, 2x + p and 3x + 6 are in A.P
Solution:
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Question 5.
If the 3rd and the 9th terms of an arithmetic progression are 4 and -8 respectively, Which term of it is zero?
Solution:
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Question 6.
How many three-digit numbers are divisible by 87?
Solution:
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Question 7.
For what value of n, the nth term of A.P 63, 65, 67, …….. and nth term of A.P. 3, 10, 17,…….. are equal to each other?
Solution:
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Question 8.
Determine the A.P. Whose 3rd term is 16 and the 7th term exceeds the 5th term by 12.
Solution:
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Question 9.
If numbers n – 2, 4n – 1 and 5n + 2 are in A.P. find the value of n and its next two terms.
Solution:
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Question 10.

Solution:
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Question 11.
If a, b and c are in A.P show that:
(i) 4a, 4b and 4c are in A.P
(ii) a + 4, b + 4 and c + 4 are in A.P.
Solution:
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Question 12.
An A.P consists of 57 terms of which 7th term is 13 and the last term is 108. Find the 45th term of this A.P.
Solution:
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Question 13.
4th term of an A.P is equal to 3 times its first term and 7th term exceeds twice the 3rd time by I. Find the first term and the common difference.
Solution:
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Question 14.
The sum of the 2nd term and the 7th term of an A.P is 30. If its 15th term is 1 less than twice of its 8th term, find the A.P
Solution:
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Question 15.
In an A.P, if mth term is n and nth term is m, show that its rth term is (m + n – r)
Solution:
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Question 16.
Which term of the A.P 3, 10, 17, ………. Will be 84 more than its 13th term?
Solution:
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Arithmetic Progression Exercise 10C – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Find the sum of the first 22 terms of the A.P.: 8, 3, -2, ………..
Solution:
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Question 2.
How many terms of the A.P. :
24, 21, 18, ……… must be taken so that their sum is 78?
Solution:
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Question 3.
Find the sum of 28 terms of an A.P. whose nth term is 8n – 5.
Solution:
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Question 4(i).
Find the sum of all odd natural numbers less than 50
Solution:
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Question 4(ii).
Find the sum of first 12 natural numbers each of which is a multiple of 7.
Solution:
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Question 5.
Find the sum of first 51 terms of an A.P. whose 2nd and 3rd terms are 14 and 18 respectively.
Solution:
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Question 6.
The sum of first 7 terms of an A.P is 49 and that of first 17 terms of it is 289. Find the sum of first n terms
Solution:
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Question 7.
The first term of an A.P is 5, the last term is 45 and the sum of its terms is 1000. Find the number of terms and the common difference of the A.P.
Solution:
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Question 8.
Find the sum of all natural numbers between 250 and 1000 which are divisible by 9.
Solution:
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Question 9.
The first and the last terms of an A.P. are 34 and 700 respectively. If the common difference is 18, how many terms are there and what is their sum?
Solution:
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Question 10.
In an A.P, the first term is 25, nth term is -17 and the sum of n terms is 132. Find n and the common difference.
Solution:
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Question 11.
If the 8th term of an A.P is 37 and the 15th term is 15 more than the 12th term, find the A.P. Also, find the sum of first 20 terms of A.P.
Solution:
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Question 12.
Find the sum of all multiples of 7 between 300 and 700.
Solution:
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Question 13.

Solution:
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Question 14.
The fourth term of an A.P. is 11 and the term exceeds twice the fourth term by 5 the A.P and the sum of first 50 terms
Solution:
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Arithmetic Progression Exercise 10D – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Find three numbers in A.P. whose sum is 24 and whose product is 440.
Solution:
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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:
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Question 3.
The angles of a quadrilateral are in A.P. with common difference 20°. Find its angles.
Solution:
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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:
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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:
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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:
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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:
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Question 8.
Find four numbers in A.P. whose sum is 20 and the sum of whose squares is 120.
Solution:
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Question 9.
Insert one arithmetic mean between 3 and 13.
Solution:

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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:
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Question 11.
\(\frac{1}{a}, \frac{1}{b} \text { and } \frac{1}{c}\) are in A.P. Show that : be, ca and ab are also in A.P.
Solution:
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Question 12.
Insert four A.M.s between 14 and -1.
Solution:
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Question 13.
Insert five A.M.s between -12 and 8.
Solution:
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Question 14.
Insert six A.M.s between 15 and -15.
Solution:
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Arithmetic Progression Exercise 10E – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Two cars start together in the same direction from the same place. The first car goes at uniform speed of 10 km hr^-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:
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Question 2.
A sum of ₹ 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 ₹ 20 less than its preceding prize; find the value of each of the prizes.
Solution:
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Question 3.
An article can be bought by paying ₹ 28,000 at once or by making 12 monthly instalments. If the first instalment paid is ₹ 3,000 and every other instalment is ₹ 100 less than the previous one, find :
(i) amount of instalment paid in the 9th month
(ii) total amount paid in the instalment scheme.
Solution:
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Question 4.
A manufacturer of TV sets produces 600 units in the third year and 700 units in the 7^th 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 10^th year.
(iii) the total production in 7 years.
Solution:
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Question 5.
Mrs. Gupta repays her total loan of ₹ 1.18,000 by paying instalments every month. If the instalment for the first month is ₹ 1,000 and it increases by ₹ 100 every month, what amount will she pay as the 30^th instalment of loan? What amount of loan she still has to pay after the 30^th instalment?
Solution:
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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:
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Arithmetic Progression Exercise 10F – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
The 6^th term of an A.P. is 16 and the 14^th term is 32. Determine the 36^th term.
Solution:
Let ‘a’ be the first term and ‘d’ be the common difference of the given A.P.
Now, t₆ = 16 (given)
⇒ a + 5d = 16 ….(i)
And,
t₁₄ = 32 (given)
⇒ a + 13d = 32 ….(ii)
Subtracting (i) from (ii), we get
8d = 16
⇒ d = 2
⇒ a + 5(2) = 16
⇒ a = 6
Hence, 36^th term = t₃₆ = a + 35d = 6 + 35(2) = 76

Question 2.
If the third and the 9^th terms of an A.P. term is 12 and the last term is 106. Find the 29^th term of the A.P.
Solution:
For an A.P.,a
t₃ = 4
⇒ a + 2d = 4 … (i)
t₉ = -8
⇒ a + 8d = -8 …. (ii)
Subtracting (i) from (ii), we get
6d = -12
⇒ d = -2
Substituting d = -2 in (i), we get
a = 2(-2) = 4
⇒ a – 4 = 4
⇒ a = 8
⇒ General term = t_n = 8 + (n – 1)(-2)
Let p^th term of this A.P. be 0.
⇒ 8 + (0 – 1) (-2) = 0
⇒ 8 – 2p + 2 = 0
⇒ 10 – 2p = 0
⇒ 2p = 10
⇒ p = 5
Thus, 5^th term of this A.P. is 0.

Question 3.
An A.P. consists of 50 terms of which 3^rd term is 12 and the last term is 106. Find the 29^th term of the A.P.
Solution:
For a given A.P.,
Number of terms, n = 50
3^rd term, t₃ = 12
⇒ a + 2d = 12 ….(i)
Last term, l = 106
⇒ t₅₀ = 106
⇒ a + 49d = 106 ….(ii)
Subtracting (i) from (ii), we get
47d = 94
⇒ d = 2
⇒ a + 2(2) = 12
⇒ a = 8
Hence, t₂₉ = a + 28d = 8 + 28(2) = 8 + 56 = 64

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:
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Question 5.
Find the sum of first 10 terms of the A.P. 4 + 6 + 8 + ………
Solution:
Here,
First term, a = 4
Common difference, d = 6 – 4 = 2
n = 10
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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:
Here,
First term, a = 3
Last term, l = 57
n = 20
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Question 7.
How many terms of the series 18 + 15 + 12 + ……. when added together will give 45 ?
Solution:
Here, we find that
15 – 18 = 12 – 15 = -3
Thus, the given series is an A.P. with first term 18 and common difference -3.
Let the number of term to be added be ‘n’.
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⇒ 90 = n[36 – 3n + 3]
⇒ 90 = n[39 – 3n]
⇒ 90 = 3n[13 – n]
⇒ 30 = 13n – n²
⇒ n² – 13n + 30 = 0
⇒ n² – 10n – 3n + 30 = 0
⇒ n(n – 10) – 3(n – 10) = 0
⇒ (n – 10)(n – 3) = 0
⇒ n – 10 = 0 or n – 3 = 0
⇒ n = 10 or n = 3
Thus, required number of term to be added is 3 or 10.

Question 8.
The n^th term of a sequence is 8 – 5n. Show that the sequence is an A.P.
Solution:
t_n = 8 – 5n
Replacing n by (n + 1), we get
t_n+1 = 8 – 5(n + 1) = 8 – 5n – 5 = 3 – 5n
Now,
t_n+1 – t_n = (3 – 5n) – (8 – 5n) = -5
Since, (t_n+1 – t₂) is independent of n and is therefore a constant.
Hence, the given sequence is an A.P.

Question 9.
The the general term (n^th term) and 23^rd term of the sequence 3, 1, -1, -3, ……
Solution:
The given sequence is 1, -1, -3, …..
Now,
1 – 3 = -1 – 1 = -3 – (-1) = -2
Hence, the given sequence is an A.P. with first term a = 3 and common difference d = -2.
The general term (n^th term) of an A.P. is given by
t_n = a + (n – 1)d
= 3 + (n – 1)(-2)
= 3 – 2n + 2
= 5 – 2n
Hence, 23^rd term = t₂₃ = 5 – 2(23) = 5 – 46 = -41

Question 10.
Which term of the sequence 3, 8, 13, …….. is 78 ?
Solution:
The given sequence is 3, 8, 13, …..
Now,
8 – 3 = 13 – 8 = 5
Hence, the given sequence is an A.P. with first term a = 3 and common difference d = 5.
Let the n^th term of the given A.P. be 78.
⇒ 78 = 3 + (n – 1)(5)
⇒ 75 = 5n – 5
⇒ 5n = 80
⇒ n = 16
Thus, the 16th term of the given sequence is 78.

Question 11.
Is -150 a term of 11, 8, 5, 2, ……… ?
Solution:
The given sequence is 11, 8, 5, 2, …..
Now,
8 – 11 = 5 – 8 = 2 – 5 = -3
Hence, the given sequence is an A.P. with first term a = 11 and common difference d = -3.
The general term of an A.P. is given by
t_n = a + (n – 1)d
⇒ -150 = 11 + (n – 1)(-5)
⇒ -161 = -5n + 5
⇒ 5n = 166
⇒ n =\(\frac{166}{5}\)
The number of terms cannot be a fraction.
So, clearly, -150 is not a term of the given sequence.

Question 12.
How many two digit numbers are divisible by 3 ?
Solution:
The two-digit numbers divisible by 3 are as follows: 12, 15, 18, 21, …….. 99
Clearly, this forms an A.P. with first term, a = 12
and common difference, d = 3
Last term = n^th term= 99
The general term of an A.P. is given by
t_n = a + (n – 1)d
⇒ 99 – 12 + (n – 1)(3)
⇒ 99 – 12 + 3n-3
⇒ 90 – 3n
⇒ n = 30
Thus, 30 two-digit numbers are divisible by 3.

Question 13.
How many multiples of 4 lie between 10 and 250 ?
Solution:
Numbers between 10 and 250 which are multiple of 4 are as follows: 12, 16, 20, 24,……, 248
Clearly, this forms an A.P. with first term a = 12,
common difference d= 4 and last term l = 248
l – a + (n – 1)d
⇒ 248 – 12 + (n – 1) × 4
⇒ 236 – (n – 1) × 4
⇒ n – 1 = 59
⇒ n = 60
Thus, 60 multiples of 4 lie between 10 and 250.

Question 14.
The sum of the 4^th term and the 8^th term of an A.P. is 24 and the sum of 6^th term and the 10^th term is 44. Find the first three terms of the A.P.
Solution:
Given, t₄ + t₈ = 24
(a + 3d) + (a + 7d) = 24
= 2a+ 10d = 24
> a + 5d = 12 ….(i)
And,
t₆₂ + t₁₀ = 44
= (a + 5d) + (a + 9d) = 44
= 2a+ 14d = 44
= a + 7 = 22 …(ii)
Subtracting (i) from (ii), we get
2d = 10
= d = 5
Substituting value of din (i), we get
a + 5 × 5 = 12
= a + 25 = 12
= a = -13 = 1^st term
a + d = -13 + 5 = -8 = 2^nd term
a + 2d = -13 + 2 × 5 = -13 + 10= -3 = 3^rd term
Hence, the first three terms of an A.P. are – 13,- 8 and -5.

Question 15.
The sum of first 14 terms of an A.P. is 1050 and its 14^th term is 140. Find the 20^th term.
Solution:
Let ‘a’ be the first term and ‘d’ be the common difference of the given A.P.
Given,
S₁₄= 1050
\(\frac{14}{2}[2 a+(14-1) d]=1050\)
⇒ 7[2a + 13d] = 1050
⇒ 2a + 13d = 150
⇒ a + 6.5d = 75 ….(i)
And, t₁₄ = 140
⇒ a + 13d = 140 ….(ii)
Subtracting (i) from (ii), we get
6.5d = 65
⇒ d = 10
⇒ a + 13(10) = 140
⇒ a = 10
Thus, 20^th term = t₂₀ = 10 + 19d = 10 + 19(10) = 200

Question 16.
The 25^th term of an A.P. exceeds its 9^th term by 16. Find its common difference.
Solution:
n^th term of an A.P. is given by tn= a + (n – 1) d.
⇒ t₂₅ = a + (25 – 1)d = a + 24d and
t₉ = a + (9 – 1)d = a + 8d
According to the condition in the question, we get
t₂₅ = t₉ + 16
⇒ a + 24d = a + 8d + 16
⇒ 16d = 16
⇒ d = 1

Question 17.
For an A.P., show that:
(m + n)^th term + (m – n)^th term = 2 × m^thterm
Solution:
Let a and d be the first term and common difference respectively.
⇒(m + n)^th term = a + (m + n – 1)d …. (i) and
(m – n)^th term = a + (m – n – 1)d …. (ii)
From (i) + (ii), we get
(m + n)^th term + (m – n)^th term
= a + (m + n – 1)d + a + (m – n – 1)d
= a + md + nd – d + a + md – nd – d
= 2a + 2md – 2d
= 2a + (m – 1)2d
= 2[ a + (m – 1)d]
= 2 × m^th term
Hence proved.

Question 18.
If the n^th term of the A.P. 58, 60, 62,…. is equal to the n^th term of the A.P. -2, 5, 12, …., find the value of n.
Solution:
In the first A.P. 58, 60, 62,….
a = 58 and d = 2
t_n = a + (n – 1)d
⇒ t_n = 58 + (n – 1)2 …. (i)
In the first A.P. -2, 5, 12, ….
a = -2 and d = 7
t_n = a + (n – 1)d
⇒ t_n = -2 + (n – 1)7 …. (ii)
Given that the n^th term of first A.P is equal to the n^th term of the second A.P.
⇒58 + (n – 1)2 = -2 + (n – 1)7 … from (i) and (ii)
⇒58 + 2n – 2 = -2 + 7n – 7
⇒ 65 = 5n
⇒ n = 15

Question 19.
Which term of the A.P. 105, 101, 97 … is the first negative term?
Solution:
Here a = 105 and d = 101 – 105 = -4
Let an be the first negative term.
⇒ a_2n < 0
⇒ a + (n – 1)d < 0
⇒ 105 + (n – 1)(-4)

Question 20.
How many three digit numbers are divisible by 7?
Solution:
The first three digit number which is divisible by 7 is 105 and the last digit which is divisible by 7 is 994.
This is an A.P. in which a = 105, d = 7 and t_n = 994.
We know that n^th term of A.P is given by
t_n = a + (n – 1)d.
⇒ 994 = 105 + (n – 1)7
⇒ 889 = 7n – 7
⇒ 896 = 7n
⇒ n = 128
∴ There are 128 three digit numbers which are divisible by 7.

Question 21.
Divide 216 into three parts which are in A.P. and the product of the two smaller parts is 5040.
Solution:
Let the three parts of 216 in A.P be (a – d), a, (a + d).
⇒a – d + a + a + d = 216
⇒ 3a = 216
⇒ a = 72
Given that the product of the two smaller parts is 5040.
⇒ a(a – d ) = 5040
⇒ 72(72 – d) = 5040
⇒ 72 – d = 70
⇒ d = 2
∴ a – d = 72 – 2 = 70, a = 72 and a + d = 72 + 2 = 74
Therefore the three parts of 216 are 70, 72 and 74.

Question 22.
Can 2n² – 7 be the n^th term of an A.P? Explain.
Solution:
We have 2n² – 7,
Substitute n = 1, 2, 3, … , we get
2(1)² – 7, 2(2)² – 7, 2(3)² – 7, 2(4)² – 7, ….
-5, 1, 11, ….
Difference between the first and second term = 1 – (-5) = 6
And Difference between the second and third term = 11 – 1 = 10
Here, the common difference is not same.
Therefore the nth term of an A.P can’t be 2n² – 7.

Question 23.
Find the sum of the A.P., 14, 21, 28, …, 168.
Solution:
Here a = 14 , d = 7 and t_n = 168
t_n = a + (n – 1)d
⇒ 168 = 14 + (n – 1)7
⇒ 154 = 7n – 7
⇒ 154 = 7n – 7
⇒ 161 = 7n
⇒ n = 23
We know that,
Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression image - 81
Therefore the sum of the A.P., 14, 21, 28, …, 168 is 2093.

Question 24.
The first term of an A.P. is 20 and the sum of its first seven terms is 2100; find the 31^st term of this A.P.
Solution:
Here a = 20 and S₇ = 2100
We know that,
Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression image - 82
To find: t₃₁ =?
t_n = a + (n – 1)d

Therefore the 31st term of the given A.P. is 2820.

Question 25.
Find the sum of last 8 terms of the A.P. -12, -10, -8, ……, 58.
Solution:
First we will reverse the given A.P. as we have to find the sum of last 8 terms of the A.P.
58, …., -8, -10, -12.
Here a = 58 , d = -2
Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression image - 84
Therefore the sum of last 8 terms of the A.P. -12, -10, -8, ……, 58 is 408.

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Selina Concise Mathematics Class 10 ICSE Solutions Shares and Dividends

November 10, 2023May 26, 2022 by Prasanna

Selina Concise Mathematics Class 10 ICSE Solutions Shares and Dividends

Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividends

Shares and Dividends Exercise 3A – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
How much money will be required to buy 400, ₹ 12.50 shares at a premium of ₹ 1?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Shares and Dividends - 1

Question 2.
How much money will be required to buy 250, ₹ 15 shares at a discount of ₹ 1.50?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Shares and Dividends - 2

Question 3.
A person buys 120 shares at a nominal value of ₹ 40 each, which he sells at ₹ 42.50 each. Find his profit and profit percent.
Solution:
Nominal value of 120 shares = ₹ 40 × 120= ₹ 4,800
Market value of 120 shares = ₹ 42.50 × 120= ₹ 5,100
His profit = ₹ 5,100 – ₹ 4,800 = ₹ 300
profit = \(\frac { 300 }{ 4800 }\) × 100% = 6.25%

Question 4.
Find the cost of 85 shares of ₹ 60 each when quoted at ₹ 63.25.
Solution:
Market value of 1 share = ₹ 63.25
Market value of 85 shares = ₹ 63.25 × 85 = ₹ 5,376.25

Question 5.
A man invests ₹ 800 in buying ₹ 5 shares and when they are selling at a premium of ₹ 1.15, he sells all the shares. Find his profit and profit percent.
Solution:
Nominal value of 1 share = ₹ 5
Market value 1 share = ₹ 5 + ₹ 1.15 = ₹ 6.15
Total money invested = ₹ 800
No of shares purchased = \(\frac { 800 }{ 5 }\) = 160
Market value of 160 shares = 160 × 6.15= ₹ 984
His profit = ₹ 984 – ₹ 800 = ₹ 184
profit = \(\frac { 184 }{ 800 }\) × 100% = 23%

Question 6.
Find the annual income derived from 125, ₹ 120 shares paying 5% dividend.
Solution:
Nominal value of 1 share = ₹ 60
Nominal value 250 shares= ₹ 60 x 250= ₹ 15,000
Dividend = 5% of ₹ 15,000
= \(\frac { 5 }{ 100 }\) × 15,000 = ₹ 750

Question 7.
A man invests ₹ 3,072 in a company paying 5% per annum, when its ₹ 10 share can be bought for ₹ 16 each. Find :
(i) his annual income
(ii) his percentage income on his investment.
Solution:
Market value of 1 share = ₹ 16
Nominal value of 1share = ₹ 10
Money invested = ₹ 3,072
Selina Concise Mathematics Class 10 ICSE Solutions Shares and Dividends - 3

Question 8.
A man invests ₹ 7,770 in a company paying 5% dividend when a share of nominal value of ₹ 100 sells at a premium of ₹ 5. Find:
(i) the number of shares bought;
(ii) annual income;
(iii) percentage income.
Solution:
Total money invested = ₹ 7,770
Nominal value of 1 share = ₹ 100
Market value of 1 share = ₹ 100 + ₹ 5 = ₹ 105
Selina Concise Mathematics Class 10 ICSE Solutions Shares and Dividends - 4

Question 9.
A man buys ₹ 50 shares of a company, paying 12% dividend, at a premium of ₹ 10. Find:
(i) the market value of 320 shares;
(ii) his annual income;
(iii) his profit percent.
Solution:
Nominal value of 1 share = ₹ 50
Market value of 1 share = ₹ 50 + ₹ 10 = ₹ 60
Market value of 320 shares = 320 x 60 = ₹ 19,200
Nominal value of 320 shares = 320 x 5 = ₹ 16,000
Selina Concise Mathematics Class 10 ICSE Solutions Shares and Dividends - 5

Question 10.
A man buys ₹ 75 shares at a discount of ₹ 15 of a company paying 20% dividend. Find:
(i) the market value of 120 shares;
(ii) his annual income;
(iii) his profit percent.
Solution:
Nominal value of 1 share = ₹ 75
Market value of 1 share = ₹ 75 – ₹ 15 = ₹ 60
Market value of 120 shares = 120 × 60 = ₹ 7,200
Nominal value of 120 shares = 120 × 75 = ₹ 9,000
Selina Concise Mathematics Class 10 ICSE Solutions Shares and Dividends - 6

Question 11.
A man has 300, ₹ 50 shares of a company paying 20% dividend. Find his net income after paying 3% income tax.
Solution:
Nominal value of 1 share = ₹ 50
Nominal value of 300 shares = 300 × 50 = ₹ 15,000
Selina Concise Mathematics Class 10 ICSE Solutions Shares and Dividends - 7
His net income = ₹ 3,000 – ₹ 90 = ₹ 2,910

Question 12.
A company pays a dividend of 15% on its ten-rupee shares from which it deducts income tax at the rate of 22%. Find the annual income of a man who owns one thousand shares of this company.
Solution:
Nominal value of 1 share = ₹ 10
Nominal value of 1000 shares = 1000 × 10 = ₹ 10,000
Selina Concise Mathematics Class 10 ICSE Solutions Shares and Dividends - 8
His net income = ₹ 1,500 – ₹ 330 = ₹ 1,170

Question 13.
A man invests ₹ 8,800 in buying shares of a company of face value of rupees hundred each at a premium of 10%. If he earns ₹ 1,200 at the end of the year as dividend, find:
(i) the number of shares he has in the company.
(ii) the dividend percent per share.
Solution:
Total investment = ₹ 8,800
Nominal value of 1 share = ₹ 100
Market value of 1 share = ₹ 110
∴ No of shares purchased = \(\frac { 8800 }{ 110 }\) = 80
Nominal value of 80 shares = 80 × 100= ₹ 8,000
Let dividend% = y%
then y% of ₹ 8,000 = ₹ 1,200
⇒ \(\frac { y }{ 100 }\) × 8,000 = 1,200
⇒ y = 15%

Question 14.
A man invests ₹ 1,680 in buying shares of nominal value ₹ 24 and selling at 12% premium. The dividend on the shares is 15% per annum. Calculate:
(i) the number of shares he buys;
(ii) the dividend he receives annually.
Solution:
Nominal value of 1 share = ₹ 24
Market value of 1 share = ₹ 24+ 12% of ₹ 24
= ₹ 24+ ₹ 2.88= ₹ 26.88
Total investment = ₹ 1,680
∴ No of shares purchased = \(\frac { 1680 }{ 26.88 }\) = 62.5
Nominal value of 62.5 shares = 62.5 x 24= ₹ 1,500
Dividend = 15% of ₹ 1,500
= \(\frac { 15 }{ 100 }\) × 1,500 = ₹ 225

Question 15.
By investing ₹ 7,500 in a company paying 10 percent dividend, an annual income of ₹ 500 is received. What price is paid for each of ₹ 100 share ?
Solution:
Total investment = ₹ 7,500
Nominal value of 1 share = ₹ 100
No. of shares purchased = y
Nominal value of y shares = 100 x y = ₹ (100y)
Dividend% = 10%
Dividend = ₹ 500
Selina Concise Mathematics Class 10 ICSE Solutions Shares and Dividends - 9

Shares and Dividends Exercise 3B – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
A man buys 75, ₹ 100 shares of a company which pays 9 percent dividend. He buys shares at such a price that he gets 12 percent of his money. At what price did he buy the shares ?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Shares and Dividends - 10

Question 2.
By purchasing ₹ 25 gas shares for ₹ 40 each, a man gets 4 percent profit on his investment. What rate percent is the company paying? What is his dividend if he buys 60 shares?
Solution:
Nominal value of 1 share = ₹ 25
Market value of 1 share = ₹ 40
Profit% on investment = 4%
Then profit on 1 share = 4% of ₹ 40= ₹ 1.60
∴ Dividend% = \(\frac { 1.60 }{ 25 }\) × 100% = 6.4%
No. of shares purchased= 60
Then dividend on 60 shares = 60 × ₹ 1.60 = ₹ 96

Question 3.
Hundred rupee shares of a company are available in the market at a premium of ₹ 20. Find the rate of dividend given by the company, when a man’s return on his investment is 15%.
Solution:
Nominal value of 1 share = ₹ 100
Market value of 1 share = ₹ 100 + ₹ 20 = ₹ 120
Profit% on investment of 1 share =15%
Then profit= 15% of ₹ 120 = ₹ 18
∴ Dividend% = \(\frac { 18 }{ 100 }\) × 100% = 18%

Question 4.
₹ 50 shares of a company are quoted at a discount of 10%. Find the rate of dividend given by the company, the return on the investment on these shares being 20 percent.
Solution:
Nominal value of 1 share = ₹ 50
Market value of 1 share = ₹ 50 – 10% of ₹ 50
= ₹ 50 – ₹ 5 = ₹ 45
Profit % on investment = 20%
Then profit on 1 share = 20% of ₹ 45 = ₹ 9
∴ Dividend% = \(\frac { 9 }{ 50 }\) × 100% = 18%

Question 5.
A company declares 8 percent dividend to the share holders. If a man receives ₹ 2,840 as his dividend, find the nominal value of his shares.
Solution:
Dividend% = 8%
Dividend = ₹ 2,840
Let nominal value of shares = ₹ y
then 8% of y = ₹ 2,840
⇒ \(\frac { 8 }{ 100 }\) × y = ₹ 2,840
⇒ y = ₹ 35000

Question 6.
How much should a man invest in ₹ 100 shares selling at ₹ 110 to obtain an annual income of ₹ 1,680, if the dividend declared is 12%?
Solution:
Nominal value of 1 share = ₹ 100
Market value of 1 share = ₹ 110
Let no. of shares purchased = n
Then nominal value of n shares = ₹ (100n)
Dividend% = 12%
Dividend = ₹ 1,680
Selina Concise Mathematics Class 10 ICSE Solutions Shares and Dividends - 12
Then market value of 140 shares= 140 × 110 = ₹ 15,400

Question 7.
A company declares a dividend of 11.2% to all its share-holders. If its ₹ 60 share is available in the market at a premium of 25%, how much should Rakesh invest, in buying the shares of this company, in order to have an annual income of ₹ 1,680?
Solution:
Nominal value of 1 share = ₹ 60
Market value of 1 share = ₹ 60+ 25% of ₹ 60
= ₹ 60 + ₹ 15 = ₹ 75
Let no. of shares purchased = n
Then nominal value of n shares = ₹ (60n)
Dividend% = 11.2%
Dividend = ₹ 1,680
Selina Concise Mathematics Class 10 ICSE Solutions Shares and Dividends - 13
Then market value of 250 shares = 250 × 75 = ₹ 18,750

Question 8.
A man buys 400, twenty-rupee shares at a premium of ₹ 4 each and receives a dividend of 12%. Find:
(i) the amount invested by him.
(ii) his total income from the shares.
(iii) percentage return on his money.
Solution:
Nominal value of 1 share = ₹ 20
Market value of 1 share = ₹ 20 + ₹ 4 = ₹ 24
No. of shares purchased = 400
Nominal value of 400 shares = 400 × 20 = ₹ 8,000
(i) Market value of 400 shares = 400 × 24 = ₹ 9,600
Selina Concise Mathematics Class 10 ICSE Solutions Shares and Dividends - 14

Question 9.
A man buys 400, twenty-rupee shares at a discount of 20% and receives a return of 12% on his money. Calculate:
(i) the amount invested by him.
(ii) the rate of dividend paid by the company.
Solution:
Nominal value of 1 share = ₹ 20
Market value of 1 share = ₹ 20 – 20% of ₹ 20
= ₹ 20 – ₹ 4 = ₹ 16
No. of shares purchased = 400
Nominal value of 400 shares = 400 x 20 = ₹ 8,000
(i) Market value of 400 shares = 400 x 16 = ₹ 6,400
(ii) Return%= 12%
Income = 12% of ₹ 6,400
Selina Concise Mathematics Class 10 ICSE Solutions Shares and Dividends - 15

Question 10.
A company, with 10,000 shares of ₹ 100 each, declares an annual dividend of 5%.
(i) What is the total amount of dividend paid by the company?
(ii) What should be the annual income of a man who has 72 shares in the company?
(iii) If he received only 4% of his investment, find the price he paid for each share.
Solution:
Nominal value of 1 share = ₹ 100
Nominal value of 10,000 shares = 10,000 x ₹ 100 = ₹ 10,00,000
(i) Dividend% = 5%
Dividend = 5% of ₹ 10,00,000
= \(\frac { 5 }{ 100 }\) × 10,00,000 = ₹ 50,000
(ii) Nominal value of 72 shares= ₹ 100 x 72 = ₹ 7,200
Dividend = 5% of ₹ 7,200
= \(\frac { 5 }{ 100 }\) × 7,200 = ₹ 360
(iii) Let market value of 1 share = ₹ y
Then market value of 10,000 shares = ₹ (10,000y)
Return% = 4%
then 4% of ₹ 10,000y = ₹ 50,000
⇒ \(\frac { 4 }{ 100 }\) × 10,000y = ₹ 50,000
⇒ y = ₹ 125

Question 11.
A lady holds 1800, ₹ 100 shares of a company that pays 15% dividend annually. Calculate her annual dividend. If she had bought these shares at 40% premium, what is the return she gets as percent on her investment. Give your answer to the nearest integer.
Solution:
Nominal value of 1 share = ₹ 100
Market value of 1 share = ₹ 100 + 40% of ₹ 100
= ₹ 100 + ₹ 40 = ₹ 140
No. of shares purchased = 1800
Nominal value of 1800 shares = 1800 × 100 = ₹ 1,80,000
Market value of 1800 shares= 1800 × 140 = ₹ 2,52,000
(i)Dividend% = 15%
Dividend = 15% of ₹ 1,80,000
Selina Concise Mathematics Class 10 ICSE Solutions Shares and Dividends - 16

Question 12.
A man invests ₹ 11,200 in a company paying 6 percent per annum when its ₹ 100 shares can be bought for ₹ 140. Find:
(i) his annual dividend
(ii) his percentage return on his investment.
Solution:
Nominal value of 1 share = ₹ 100
Market value of 1 share = ₹ 140
Total investment = ₹ 11,200
No of shares purchased = \(\frac { 11,200 }{ 140 }\) = 80 shares
Then nominal value of 80 shares= 80 × 100= ₹ 8,000
(i) Dividend% = 6%
Dividend = 6% of ₹ 8,000
Selina Concise Mathematics Class 10 ICSE Solutions Shares and Dividends - 17

Question 13.
Mr. Sharma has 60 shares of nominal value ₹ 100 and decides to sell them when they are at a premium of 60%. He invests the proceeds in shares of nominal value ₹ 50, quoted at 4% discount, and paying 18% dividend annually. Calculate :
(i) the sale proceeds
(ii) the number of shares he buys and
(iii) his annual dividend from the shares.
Solution:
1st case
Nominal value of 1 share = ₹ 100
Nominal value of 60 shares = ₹ 100 × 60= ₹ 6,000
Market value of 1 share = ₹ 100 + 60% of ₹ 100
= ₹ 100+ ₹ 60 = ₹ 160
Market value of 60 shares = ₹ 160 × 60 = ₹ 9,600 Ans.
(ii) Nominal value of 1 share = ₹ 50
Market value of 1 share= ₹ 50 – 4% of ₹ 50
= ₹ 50 – ₹ 2 = ₹ 48
No of shares purchased = \(\frac { 9,600 }{ 48 }\) = 200 shares
(iii) Nominal value of 200 shares = ₹ 50 × 200 = ₹ 10,000
Dividend% = 18%
Dividend = 18% of ₹ 10,000
= \(\frac { 18 }{ 100 }\) × 10,000 = ₹ 1800

Question 14.
A company with 10,000 shares of nominal value ₹ 100 declares an annual dividend of 8% to the share-holders.
(i) Calculate the total amount of dividend paid by the company.
(ii) Ramesh had bought 90 shares of the company at ₹ 150 per share. Calculate the dividend he receives and the percentage of return on his investment.
Solution:
(i) Nominal value of 1 share = ₹ 100
Nominal value of 10,000 shares = ₹ 100 × 10,000 = ₹ 10,00,000
Dividend% = 8%
Dividend = 8% of ₹ 10,00,000
= \(\frac { 8 }{ 100 }\) × 10,00,000 = ₹ 80,000
(ii) Market value of 90 shares = ₹ 150 × 90 = ₹ 13,500
Nominal value of 90 shares = ₹ 100 × 90 = ₹ 9,000
Dividend = 8% of ₹ 9,000
= \(\frac { 8 }{ 100 }\) × 9,000 = ₹ 720
Selina Concise Mathematics Class 10 ICSE Solutions Shares and Dividends - 18

Question 15.
Which is the better investment :
16% ₹ 100 shares at 80 or 20% ₹ 100 shares at 120?
Solution:
1st case
16% of ₹ 100 shares at 80 means;
Market value of 1 share = ₹ 80
Nominal value of 1 share = ₹ 100
Dividend = 16%
Income on ₹ 80= 16% of ₹ 100 = ₹ 16
Income on ₹ 1 = \(\frac { 16 }{ 80 }\) = ₹ 0.20
2nd case
20% of ₹ 100 shares at 120 means;
Market value of 1 share = ₹ 120
Nominal value of 1 share = ₹ 100
Dividend = 20%
Income on ₹ 120 = 20% of ₹ 100= ₹ 20
Income on ₹ 1 = \(\frac { 20 }{ 120 }\) = ₹ 0.17
Then 16% ₹ 100 shares at 80 is better investment.

Question 16.
A man has a choice to invest in hundred-rupee shares of two firms at ₹ 120 or at ₹ 132. The first firm pays a dividend of 5% per annum and the second firm pays a dividend of 6% per annum. Find:
(i) which company is giving a better return.
(ii) if a man invests ₹ 26,400 with each firm, how much will be the difference between the annual returns from the two firms.
Solution:
(i) 1st firm
Market value of 1 share = ₹ 120
Nominal value of 1 share = ₹ 100
Dividend = 5%
Income on ₹ 120 = 5% of ₹ 100 = ₹ 5
Income on ₹ 1 = \(\frac { 5 }{ 120 }\) = ₹ 0.041
2nd firm
Market value of 1 share = ₹ 132
Nominal value of 1 share = ₹ 100
Dividend = 6%
Income on ₹ 132 = 6% of ₹ 100 = ₹ 6
Income on ₹ 1 = \(\frac { 6 }{ 132 }\) = ₹ 0.045
Then investment in second company is giving better return.
(ii) Income on investment of ₹ 26,400 in fi₹ t firm
= \(\frac { 5 }{ 120 }\) × 26,400 = ₹ 1,100
Income on investment of ₹ 26,400 in second firm
= \(\frac { 6 }{ 132 }\) × 26,400 = ₹ 1,200
∴ Difference between both returns = ₹ 1,200 – ₹ 1,100 = ₹ 100

Question 17.
A man bought 360, ten-rupee shares of a company, paying 12% per annum. He sold the shares when their price rose to ₹ 21 per share and invested the proceeds in five-rupee shares paying 4.5 percent per annum at ₹ 3.50 per share. Find the annual change in his income.
Solution:
1st case
Nominal value of 1 share = ₹ 10
Nominal value of 360 shares = ₹ 10 × 360 = ₹ 3,600
Market value of 1 share = ₹ 21
Market value of 360 shares = ₹ 21 × 360 = ₹ 7,560
Dividend% = 12%
Dividend = 12% of ₹ 3,600
= \(\frac { 12 }{ 100 }\) × 3,600 = ₹ 432
2nd case
Nominal value of 1 share= ₹ 5
Market value of 1 share= ₹ 3.50
∴ No of shares purchased = \(\frac { 7,560 }{ 3.50 }\) = 2,160 shares
Nominal value of 2160 shares=₹ 5 × 2160= ₹ 10,800
Dividend%= 4.5%
Dividend= 4.5% of ₹ 10,800
= \(\frac { 4.5 }{ 132 }\) × 10,800 = ₹ 486
Annual change in income = ₹ 486 – ₹ 432
= ₹ 54 increase

Question 18.
A man sold 400 (₹ 20) shares of a company, paying 5% at ₹ 18 and invested the proceeds in (₹ 10) shares of another company paying 7% at ₹ 12. How many (₹ 10) shares did he buy and what was the change in his income?
Solution:
1st case
Nominal value of 1 share = ₹ 20
Nominal value of 400 shares = ₹ 20 x 400= ₹ 8,000
Market value of 1 share = ₹ 18
Market value of 400 shares = ₹ 18 x 400= ₹ 7,200
Dividend% = 5%
Dividend = 5% of ₹ 8,000
= \(\frac { 5 }{ 100 }\) × 8,000 = ₹ 400
2nd case
Nominal value of 1 share = ₹ 10
Market value of 1 share = ₹ 12
∴ No of shares purchased = \(\frac { 7,200 }{ 12 }\) = 600 shares
Nominal value of 600 shares = ₹ 10 x 600 = ₹ 6,000
Dividend% = 7%
Dividend = 7% of ₹ 6,000
= \(\frac { 7 }{ 100 }\) × 6,000 = ₹ 420
Annual change in income = ₹ 420 – ₹ 400
= ₹ 20 increase

Question 19.
Two brothers A and B invest ₹ 16,000 each in buying shares of two companies. A buys 3% hundred-rupee shares at 80 and B buys ten-rupee shares at par. If they both receive equal dividend at the end of the year, find the rate per cent of the dividend received by B.
Solution:
For A
Total investment = ₹ 16,000
Nominal value of 1 share = ₹ 100
Market value of 1 share = ₹ 80
∴ No of shares purchased = \(\frac { 16,000 }{ 80 }\) = 200 shares
Nominal value of 200 shares = ₹ 100 × 200 = ₹ 20,000
Dividend% = 3%
Dividend = 3% of ₹ 20,000
= \(\frac { 3 }{ 100 }\) × 20,000 = ₹ 600
For B
Total investment= ₹ 16,000
Nominal value of 1 share= ₹ 10
Market value of 1 share= ₹ 10
∴ No of shares purchased = \(\frac { 16,000 }{ 10 }\) = 1600 shares
Nominal value of 1600shares= 10 × 1600= ₹ 16,000
Dividend received by B = Dividend received by A = ₹ 600
Selina Concise Mathematics Class 10 ICSE Solutions Shares and Dividends - 19

Question 20.
A man invests ₹ 20,020 in buying shares of nominal value ₹ 26 at 10% premium. The dividend on the shares is 15% per annum. Calculate :
(i) the number of shares he buys.
(ii) the dividend he receives annually.
(iii) the rate of interest he gets on his money.
Solution:
Total investment = ₹ 20,020
Nominal value of 1 share = ₹ 26
Market value of 1 share = ₹ 26+ 10% of ₹ 26
= ₹ 26+ ₹ 2.60 = ₹ 28.60
∴ No of shares purchased = \(\frac { 20,020 }{ 28.60 }\) = 700 shares
Nominal value of 700 shares= ₹ 26 x 700 = ₹ 18,200
Dividend% = 15%
Dividend = 15% of ₹ 18,200
= \(\frac { 15 }{ 100 }\) × 18,200 = ₹ 2,730
Selina Concise Mathematics Class 10 ICSE Solutions Shares and Dividends - 20

Shares and Dividends Exercise 3C – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
By investing ₹ 45,000 in 10% ₹ 100 shares, Sharad gets ₹ 3,000 as dividend. Find the market value of each share.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Shares and Dividends - 21

Question 2.
Mrs. Kulkarni invests ₹ 1, 31,040 in buying ₹ 100 shares at a discount of 9%. She sells shares worth Rs.72,000 at a premium of 10% and the rest at a discount of 5%. Find her total gain or loss on the whole.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Shares and Dividends - 22

Question 3.
A man invests a certain sum on buying 15% ₹ 100 shares at 20% premium. Find :
(i) His income from one share
(ii) The number of shares bought to have an income, from the dividend, ₹ 6480
(iii) Sum invested
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Shares and Dividends - 23

Question 4.
Gagan invested ₹ 80% of his savings in 10% ₹ 100 shares at 20% premium and the rest of his savings in 20% ₹ 50 shares at ₹ 20% discount. If his incomes from these shares is ₹ 5,600 calculate:
(i) His investment in shares on the whole
(ii) The number of shares of first kind that he bought
(iii) Percentage return, on the shares bought on the whole.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Shares and Dividends - 24

Question 5.
Ashwarya bought 496, ₹ 100 shares at ₹ 132 each, find :
(i) Investment made by her
(ii) Income of Ashwarya from these shares, if the rate of dividend is 7.5%.
(iii) How much extra must ashwarya invest in order to increase her income by ₹ 7,200.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Shares and Dividends q5

A company pays a dividend of 15% on its ₹ 100 shares from which income tax at the rate of 20% is deducted. Find :
(i) The net annual income of Gopal who owns 7,200 shares of this company
(ii) The sum invested by Ramesh when the shares of this company are bought by him at 20% premium and the gain required by him(after deduction of income tax) is ₹ 9,000
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Shares and Dividends - 25

Mr. Joseph sold some ₹ 100 shares paying 10% dividend at a discount of 25% and invested the proceeds in ₹ 100 shares paying 16% dividend at a discount of 20%. By doing so, his income was increased by ₹ 4,800. Find the number of shares originally held by Mr. Joseph.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Shares and Dividends - 26

Question 6.
Gopal has some ₹ 100 shares of company A, paying 10% dividend. He sells a certain number of these shares at a discount of 20% and invests the proceeds in ₹ 100 shares at ₹ 60 of company B paying 20% dividend. If his income, from the shares sold, increases by ₹ 18,000, find the number of shares sold by Gopal.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Shares and Dividends - 27

Question 7.
A man invests a certain sum of money in 6% hundred-rupee shares at ₹ 12 premium. When the shares fell to ₹ 96, he sold out all the shares bought and invested the proceed in 10%, ten-rupee shares at ₹ 8. If the change in his income is ₹ 540, Find the sum invested originally
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Shares and Dividends - 28

Question 8.
Mr. Gupta has a choice to invest in ten-rupee shares of two firms at ₹ 13 or at ₹ 16. If the first firm pays 5% dividend and the second firm pays 6% dividend per annum, find:
(i) which firm is paying better.
(ii) if Mr. Gupta invests equally in both the firms and the difference between the returns from them is ₹ 30, find how much, in all, does he invest.
Solution:
(i) 1st firm
Nominal value of 1 share = ₹ 10
Market value of 1 share = ₹ 13
Dividend% = 5%
Dividend = 5% of ₹ 10 = ₹ 0.50
Selina Concise Mathematics Class 10 ICSE Solutions Shares and Dividends - 29
2nd firm
Nominal value of 1 share = ₹ 10
Market value of 1 share = ₹ 16
Dividend% = 6%
Dividend = 6% of ₹ 10 = ₹ 0.60

Then first firm is paying better than second firm.
(ii) Let money invested in each firm = ₹ y
Selina Concise Mathematics Class 10 ICSE Solutions Shares and Dividends - 31
Total money invested in both firms = ₹ 31,200 × 2
= ₹ 62,400

Question 9.
Ashok invested Rs. 26,400 in 12%, Rs. 25 shares of a company. If he receives a dividend of Rs. 2,475, find the :
(i) number of shares he bought.
(ii) market value of each share.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Shares and Dividends q9

Question 10.
A man invested ₹ 45,000 in 15% Rs100shares quoted at ₹ 125. When the market value of these shares rose to ₹ 140, he sold some shares, just enough to raise ₹ 8,400. Calculate:
(i) the number of shares he still holds;
(ii) the dividend due to him on these remaining shares.
Solution:
(i) Total investment = ₹ 45,000
Market value of 1 share = ₹ 125
∴ No of shares purchased = \(\frac { 45,000 }{ 125 }\) = 360 shares
Nominal value of 360 shares = ₹ 100 × 360= ₹ 36,000
Let no. of shares sold = n
Then sale price of 1 share = ₹ 140
Total sale price of n shares = ₹ 8,400
Then n = \(\frac { 8,400 }{ 140 }\) = 60 shares
The no. of shares he still holds = 360 – 60 = 300
(ii) Nominal value of 300 shares = ₹ 100 × 300 = ₹ 30,000
Dividend% = 15%
Dividend = 15% of ₹ 30,000
= \(\frac { 15 }{ 100 }\) × 30,000 = ₹ 4,500

Question 11.
Mr.Tiwari. invested ₹ 29,040 in 15% Rs100 shares quoted at a premium of 20%. Calculate:
(i) the number of shares bought by Mr. Tiwari.
(ii) Mr. Tiwari’s income from the investment.
(iii) the percentage return on his investment.
Solution:
Total investment = ₹ 29,040
Nominal value of 1 share = ₹ 100
Market value of 1 share = ₹ 100+ 20% of ₹ 100
= ₹ 100 + ₹ 20 = ₹ 120
∴ No of shares purchased = \(\frac { 29,040 }{ 120 }\) = 242 shares
Nominal value of 242 shares = ₹ 100 x 242 = ₹ 24,200
Dividend% = 15%
Dividend = 15% of ₹ 24,200
= \(\frac { 15 }{ 100 }\) × 24,200 = ₹ 3,630
Selina Concise Mathematics Class 10 ICSE Solutions Shares and Dividends - 32

Question 12.
A dividend of 12% was declared on ₹ 150 shares selling at a certain price. If the rate of return is 10%, calculate:
(i) the market value of the shares.
(ii) the amount to be invested to obtain an annual dividend of ₹ 1,350.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Shares and Dividends - 33

Question 13.
Divide ₹ 50,760 into two parts such that if one part is invested in 8% ₹ 100 shares at 8% discount and the other in 9% ₹ 100 shares at 8% premium, the annual incomes from both the investments are equal.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Shares and Dividends - 34

Question 14.
Mr. Shameem invested 33 1/3% of his savings in 20% ₹ 50 shares quoted at ₹ 60 and the remainder of the savings in 10% ₹ 100 share quoted at ₹ 110. If his total income from these investments is ₹ 9,200; find :
(i) his total savings
(ii) the number of ₹ 50 share
(iii) the number of ₹ 100 share.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Shares and Dividends - 36

Question 15.
Vivek invests ₹ 4,500 in 8%, ₹ 10 shares at ₹ 5. He sells the shares when the price rises to ₹ 30, and invests the proceeds in 12% ₹ 100 shares at ₹ 125. Calculate :
(i) the sale proceeds
(ii) the number of ₹ 125 shares he buys.
(iii) the change in his annual income from dividend.
Solution:
1st case
Total investment = ₹ 4,500
Market value of 1 share = ₹ 15
∴ No of shares purchased = \(\frac { 4,500 }{ 15 }\) = 300 shares
Nominal value of 1 share = ₹ 10
Nominal value of 300 shares = ₹ 10 × 300= ₹ 3,000
Dividend = 8% of ₹ 3,000
= \(\frac { 8 }{ 100 }\) × 3,000 = ₹ 240
Sale price of 1 share = ₹ 30
Total sale price= ₹ 30 × 300= ₹ 9,000
(ii) new market price of 1 share= ₹ 125
∴ No of shares purchased = \(\frac { 9,000 }{ 125 }\) = 72 shares
(iii) New nominal value of 1 share= ₹ 100
New nominal value of 72 shares = ₹ 100 × 72 = ₹ 7,200
Dividend% = 12%
New dividend = 12% of ₹ 7,200
= \(\frac { 12 }{ 100 }\) × 7,200 = ₹ 864
Change in annual income = ₹ 864 – ₹ 240 = ₹ 624

Question 16.
Mr.Parekh invested ₹ 52,000 on ₹ 100 shares at a discount of ₹ 20 paying 8% dividend. At the end of one year he sells the shares at a premium of ₹ 20. Find:
(i) The annual dividend
(ii) The profit earned including his dividend.
Solution:
Rate of dividend = 8%
Investment = ₹ 52000
Market Rate = ₹ 100 – 20 = ₹ 80
No. of shares purchased = \(\frac { 52000 }{ 80 }\) = 650
(i) Annual dividend = 650 × 8 = ₹ 5200
(ii) On selling, market rate = ₹ 100+20 = ₹ 120
⇒ Sale price = 650 × 120 = ₹ 78000
Profit = ₹ 78000 – ₹ 52000 = ₹ 26000
⇒ Total gain = 26000 + 5200 = ₹ 31200

Question 17.
Salman buys 50 shares of face value ₹ 100 available at ₹ 132.
(i) What is his investment?
(ii) If the dividend is 7.5%, what will be his annual income?
(iii) If he wants to increase his annual income by ₹ 150, how many extra shares should he buy?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Shares and Dividends - 37

Question 18.
Salman invests a sum of money in ₹ 50 shares, paying 15% dividend quoted at 20% premium. If his annual dividend is ₹ 600, calculate :
(i) The number of shares he bought.
(ii) His total investment.
(iii) The rate of return on his investment.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Shares and Dividends - 38

Question 19.
Rohit invested ₹ 9,600 on ₹ 100 shares at ₹ 20 premium paying 8% dividend. Rohit sold the shares when the price rose to ₹ 160. He invested the proceeds (excluding dividend) in 10% ₹ 50 shares at ₹ 40. Find the :
(i) Original number of shares.
(ii) Sale proceeds.
(iii) New number of shares.
(iv) Change in the two dividends.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Shares and Dividends - 39

Question 20.
How much should a man invest in Rs. 50 shares selling at Rs. 60 to obtain an income of Rs. 450, if the rate of dividend declared is 10%. Also find his yield percent, to the nearest whole number.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Shares and Dividends ex 3c q20

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Selina Concise Mathematics Class 10 ICSE Solutions Linear Inequations (in one variable)

November 10, 2023May 26, 2022 by Prasanna

Selina Concise Mathematics Class 10 ICSE Solutions Linear Inequations (in one variable)

Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 4 Linear Inequations (in one variable)

Linear Inequations in One Variable Exercise 4A – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.

Solution:

Question 2.
State, whether the following statements are true or false:
(i) a < b, then a – c < b – c (ii) If a > b, then a + c > b + c
(iii) If a < b, then ac > bc
(iv) If a > b, then \(\frac { a }{ c } <\frac { b }{ c }\)
(v) If a – c > b – d, then a + d > b + c
(vi) If a < b, and c > 0, then a – c > b – c
Where a, b, c and d are real numbers and c ≠ 0.
Solution:
(i) a < b ⇒ a – c < b – c The given statement is true.
(ii) If a > b ⇒ a + c > b + c
The given statement is true.
(iii) If a < b ⇒ ac < bc The given statement is false.
(iv) If a > b ⇒ \(\frac { a }{ c } >\frac { b }{ c }\)
The given statement is false.
(v) If a – c > b – d ⇒ a + d > b + c
The given statement is true.
(vi) If a < b ⇒ a – c < b – c (Since, c > 0)
The given statement is false.

Question 3.
If x ∈ N, find the solution set of inequations.
(i) 5x + 3 ≤ 2x + 18
(ii) 3x – 2 < 19 – 4x
Solution:
(i) 5x + 3 ≤ 2x + 18
5x – 2x ≤ 18 – 3
3x ≤ 15
x ≤ 5
Since, x ∈ N, therefore solution set is {1, 2, 3, 4, 5}.
(ii) 3x – 2 < 19 – 4x
3x + 4x < 19 + 2
7x < 21
x < 3
Since, x ∈ N, therefore solution set is {1, 2}.

Question 4.
Selina Concise Mathematics Class 10 ICSE Solutions Linear Inequations (in one variable) - 3
Solution:
(i) x + 7 ≤ 11
x ≤ 11 – 7
x ≤ 4
Since, the replacement set = W (set of whole numbers)
⇒ Solution set = {0, 1, 2, 3, 4}
(ii) 3x – 1 > 8
3x > 8 + 1
x > 3
Since, the replacement set = W (set of whole numbers)
⇒ Solution set = {4, 5, 6, …}
(iii) 8 – x > 5
– x > 5 – 8
– x > -3
x < 3
Since, the replacement set = W (set of whole numbers)
⇒ Solution set = {0, 1, 2}
Selina Concise Mathematics Class 10 ICSE Solutions Linear Inequations (in one variable) - 4
Since, the replacement set = W (set of whole numbers)
∴ Solution set = {0, 1, 2}

Since, the replacement set = W (set of whole numbers)
∴ Solution set = {0, 1}
(vi) 18 ≤ 3x – 2
18 + 2 ≤ 3x
20 ≤ 3x
x ≥ \(\frac { 20 }{ 3 }\)
Since, the replacement set = W (set of whole numbers)
∴ Solution set = {7, 8, 9, …}

Question 5.
Solve the inequation:
3 – 2x ≥ x – 12 given that x ∈ N.
Solution:
3 – 2x ≥ x – 12
-2x – x ≥ -12 – 3
-3x ≥ -15
x ≤ 5
Since, x ∈ N, therefore,
Solution set = {1, 2, 3, 4, 5}

Question 6.
If 25 – 4x ≤ 16, find:
(i) the smallest value of x, when x is a real number,
(ii) the smallest value of x, when x is an integer.
Solution:
25 – 4x ≤ 16
-4x ≤ 16 – 25
-4x ≤ -9
x ≥ \(\frac { 9 }{ 4 }\)
x ≥ 2.25
(i) The smallest value of x, when x is a real number, is 2.25.
(ii) The smallest value of x, when x is an integer, is 3.

Question 7.
Selina Concise Mathematics Class 10 ICSE Solutions Linear Inequations (in one variable) - 6
Solution:

Question 8.

Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Linear Inequations (in one variable) - 9
Thus, the required smallest value of x is -1.

Question 9.
Find the largest value of x for which
2(x – 1) ≤ 9 – x and x ∈ W.
Solution:
2(x – 1) ≤ 9 – x
2x – 2 ≤ 9 – x
2x + x ≤ 9 + 2
3x ≤ 11
x ≤ \(\frac { 11 }{ 3 }\)
x ≤ 3.67
Since, x ∈ W, thus the required largest value of x is 3.

Question 10.

Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Linear Inequations (in one variable) - 11

Question 11.
Given x ∈ {integers}, find the solution set of:
-5 ≤ 2x – 3 < x + 2
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Linear Inequations (in one variable) - 12

Question 12.
Given x ∈ {whole numbers}, find the solution set of:
-1 ≤ 3 + 4x < 23

Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Linear Inequations (in one variable) - 13

Linear Inequations in One Variable Exercise 4B – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Selina Concise Mathematics Class 10 ICSE Solutions Linear Inequations (in one variable) - 15
Solution:

Question 2.
Selina Concise Mathematics Class 10 ICSE Solutions Linear Inequations (in one variable) - 16
Solution:

Question 3.
Selina Concise Mathematics Class 10 ICSE Solutions Linear Inequations (in one variable) - 18
Solution:

Question 4.
Selina Concise Mathematics Class 10 ICSE Solutions Linear Inequations (in one variable) - 20
Solution:

Question 5.
x ∈ {real numbers} and -1 < 3 – 2x ≤ 7, evaluate x and represent it on a number line.
Solution:
-1 < 3 – 2x ≤ 7
-1 < 3 – 2x and 3 – 2x ≤ 7
2x < 4 and -2x ≤ 4
x < 2 and x ≥ -2
Solution set = {-2 ≤ x < 2, x ∈ R}
Thus, the solution can be represented on a number line as:
Selina Concise Mathematics Class 10 ICSE Solutions Linear Inequations (in one variable) - 23

Question 6.
List the elements of the solution set of the inequation
-3 < x – 2 ≤ 9 – 2x; x ∈ N.
Solution:
-3 < x – 2 ≤ 9 – 2x
-3 < x – 2 and x – 2 ≤ 9 – 2x
-1 < x and 3x ≤ 11
-1 < x ≤ \(\frac { 11 }{ 3 }\)
Since, x ∈ N
∴ Solution set = {1, 2, 3}

Question 7.
Selina Concise Mathematics Class 10 ICSE Solutions Linear Inequations (in one variable) - 24
Solution:

Question 8.
Selina Concise Mathematics Class 10 ICSE Solutions Linear Inequations (in one variable) - 26
Solution:

Question 9.
Given x ∈ {real numbers}, find the range of values of x for which -5 ≤ 2x – 3 < x + 2 and represent it on a number line.
Solution:
-5 ≤ 2x – 3 < x + 2
-5 ≤ 2x – 3 and 2x – 3 < x + 2
-2 ≤ 2x and x < 5
-1 ≤ x and x < 5
Required range is -1 ≤ x < 5.
The required graph is:
Selina Concise Mathematics Class 10 ICSE Solutions Linear Inequations (in one variable) - 28

Question 10.
If 5x – 3 ≤ 5 + 3x ≤ 4x + 2, express it as a ≤ x ≤ b and then state the values of a and b.
Solution:
5x – 3 ≤ 5 + 3x ≤ 4x + 2
5x – 3 ≤ 5 + 3x and 5 + 3x ≤ 4x + 2
2x ≤ 8 and -x ≤ -3
x ≤ 4 and x ≥ 3
Thus, 3 ≤ x ≤ 4.
Hence, a = 3 and b = 4.

Question 11.
Solve the following inequation and graph the solution set on the number line:
2x – 3 < x + 2 ≤ 3x + 5, x ∈ R.
Solution:
2x – 3 < x + 2 ≤ 3x + 5
2x – 3 < x + 2 and x + 2 ≤ 3x + 5
x < 5 and -3 ≤ 2x
x < 5 and -1.5 ≤ x
Solution set = {-1.5 ≤ x < 5}
The solution set can be graphed on the number line as:
Selina Concise Mathematics Class 10 ICSE Solutions Linear Inequations (in one variable) - 29.

Question 12.
Solve and graph the solution set of:
(i) 2x – 9 < 7 and 3x + 9 ≤ 25, x ∈ R (ii) 2x – 9 ≤ 7 and 3x + 9 > 25, x ∈ I
(iii) x + 5 ≥ 4(x – 1) and 3 – 2x < -7, x ∈ R
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Linear Inequations (in one variable) - 70

Question 13.
Solve and graph the solution set of:
(i) 3x – 2 > 19 or 3 – 2x ≥ -7, x ∈ R
(ii) 5 > p – 1 > 2 or 7 ≤ 2p – 1 ≤ 17, p ∈ R
Solution:
(i) 3x – 2 > 19 or 3 – 2x ≥ -7
3x > 21 or -2x ≥ -10
x > 7 or x ≤ 5
Graph of solution set of x > 7 or x ≤ 5 = Graph of points which belong to x > 7 or x ≤ 5 or both.
Thus, the graph of the solution set is:
Selina Concise Mathematics Class 10 ICSE Solutions Linear Inequations (in one variable) - 31
(ii) 5 > p – 1 > 2 or 7 ≤ 2p – 1 ≤ 17
6 > p > 3 or 8 ≤ 2p ≤ 18
6 > p > 3 or 4 ≤ p ≤ 9
Graph of solution set of 6 > p > 3 or 4 ≤ p ≤ 9
= Graph of points which belong to 6 > p > 3 or 4 ≤ p ≤ 9 or both
= Graph of points which belong to 3 < p ≤ 9
Thus, the graph of the solution set is:
Selina Concise Mathematics Class 10 ICSE Solutions Linear Inequations (in one variable) - 32

Question 14.
Selina Concise Mathematics Class 10 ICSE Solutions Linear Inequations (in one variable) - 33
Solution:
(i) A = {x ∈ R: -2 ≤ x < 5}
B = {x ∈ R: -4 ≤ x < 3}
(ii) A ∩ B = {x ∈ R: -2 ≤ x < 5}
It can be represented on number line as:

B’ = {x ∈ R: 3 < x ≤ -4}
A ∩ B’ = {x ∈ R: 3 ≤ x < 5}
It can be represented on number line as:
Selina Concise Mathematics Class 10 ICSE Solutions Linear Inequations (in one variable) - 35

Question 15.
Use real number line to find the range of values of x for which:
(i) x > 3 and 0 < x < 6
(ii) x < 0 and -3 ≤ x < 1
(iii) -1 < x ≤ 6 and -2 ≤ x ≤ 3
Solution:
(i) x > 3 and 0 < x < 6
Both the given inequations are true in the range where their graphs on the real number lines overlap.
The graphs of the given inequations can be drawn as:
Selina Concise Mathematics Class 10 ICSE Solutions Linear Inequations (in one variable) - 36
From both graphs, it is clear that their common range is
3 < x < 6
(ii) x < 0 and -3 ≤ x < 1
Both the given inequations are true in the range where their graphs on the real number lines overlap.
The graphs of the given inequations can be drawn as:
Selina Concise Mathematics Class 10 ICSE Solutions Linear Inequations (in one variable) - 37
From both graphs, it is clear that their common range is
-3 ≤ x < 0
(iii) -1 < x ≤ 6 and -2 ≤ x ≤ 3
Both the given inequations are true in the range where their graphs on the real number lines overlap.
The graphs of the given inequations can be drawn as:
Selina Concise Mathematics Class 10 ICSE Solutions Linear Inequations (in one variable) - 38
From both graphs, it is clear that their common range is
-1 < x ≤ 3

Question 16.
Illustrate the set {x: -3 ≤ x < 0 or x > 2, x ∈ R} on the real number line.
Solution:
Graph of solution set of -3 ≤ x < 0 or x > 2
= Graph of points which belong to -3 ≤ x < 0 or x > 2 or both
Thus, the required graph is:
Selina Concise Mathematics Class 10 ICSE Solutions Linear Inequations (in one variable) - 39

Question 17.
Given A = {x: -1 < x ≤ 5, x ∈ R} and B = {x: -4 ≤ x < 3, x ∈ R}
Represent on different number lines:
(i) A ∩ B
(ii) A’ ∩ B
(iii) A – B
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Linear Inequations (in one variable) - 40

Question 18.
P is the solution set of 7x – 2 > 4x + 1 and Q is the solution set of 9x – 45 ≥ 5(x – 5); where x ∈ R. Represent:
(i) P ∩ Q
(ii) P – Q
(iii) P ∩ Q’
on different number lines.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Linear Inequations (in one variable) - 41

Question 19.
Selina Concise Mathematics Class 10 ICSE Solutions Linear Inequations (in one variable) - 42
Solution:

Question 20.
Given: A = {x: -8 < 5x + 2 ≤ 17, x ∈ I}, B = {x: -2 ≤ 7 + 3x < 17, x ∈ R}
Where R = {real numbers} and I = {integers}. Represent A and B on two different number lines. Write down the elements of A ∩ B.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Linear Inequations (in one variable) - 44

Question 21.
Solve the following inequation and represent the solution set on the number line 2x – 5 ≤ 5x +4 < 11, where x ∈ I
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Linear Inequations (in one variable) - 45

Question 22.
Selina Concise Mathematics Class 10 ICSE Solutions Linear Inequations (in one variable) - 46
Solution:

Question 23.
Given:
A = {x: 11x – 5 > 7x + 3, x ∈ R} and
B = {x: 18x – 9 ≥ 15 + 12x, x ∈ R}.
Find the range of set A ∩ B and represent it on number line.
Solution:
A = {x: 11x – 5 > 7x + 3, x ∈ R}
= {x: 4x > 8, x ∈ R}
= {x: x > 2, x ∈ R}
B = {x: 18x – 9 ≥ 15 + 12x, x ∈ R}
= {x: 6x ≥ 24, x ∈ R}
= {x: x ≥ 4, x ∈ R}
A ∩ B = {x: x ≥ 4, x ∈ R}
It can be represented on number line as:
Selina Concise Mathematics Class 10 ICSE Solutions Linear Inequations (in one variable) - 48

Question 24.
Selina Concise Mathematics Class 10 ICSE Solutions Linear Inequations (in one variable) - 49
Solution:

Question 25.
Selina Concise Mathematics Class 10 ICSE Solutions Linear Inequations (in one variable) - 51
Solution:

Question 26.
Selina Concise Mathematics Class 10 ICSE Solutions Linear Inequations (in one variable) - 53
Solution:

Question 27.
Find three consecutive largest positive integers such that the sum of one-third of first, one-fourth of second and one-fifth of third is atmost 20.
Solution:
Let the required integers be x, x + 1 and x + 2.
According to the given statement,
Selina Concise Mathematics Class 10 ICSE Solutions Linear Inequations (in one variable) - 55
Thus, the largest value of the positive integer x is 24.
Hence, the required integers are 24, 25 and 26.

Question 28.
Solve the given inequation and graph the solution on the number line.
2y – 3 < y + 1 ≤ 4y + 7, y ∈ R
Solution:
2y – 3 < y + 1 ≤ 4y + 7, y ∈ R
⇒ 2y – 3 – y < y + 1 – y ≤ 4y + 7 – y
⇒ y – 3 < 1 ≤ 3y + 7
⇒ y – 3 < 1 and 1 ≤ 3y + 7
⇒ y < 4 and 3y ≥ 6 ⇒ y ≥ – 2
⇒ – 2 ≤ y < 4
The graph of the given equation can be represented on a number line as:
Selina Concise Mathematics Class 10 ICSE Solutions Linear Inequations (in one variable) - 56

Question 29.
Solve the inequation:
3z – 5 ≤ z + 3 < 5z – 9, z ∈ R.
Graph the solution set on the number line.
Solution:
3z – 5 ≤ z + 3 < 5z – 9
3z – 5 ≤ z + 3 and z + 3 < 5z – 9
2z ≤ 8 and 12 < 4z
z ≤ 4 and 3 < z
Since, z R
∴ Solution set = {3 < z ≤ 4, x ∈ R }
It can be represented on a number line as:
Selina Concise Mathematics Class 10 ICSE Solutions Linear Inequations (in one variable) - 57

Question 30.
Selina Concise Mathematics Class 10 ICSE Solutions Linear Inequations (in one variable) - 58
Solution:

Question 31.
Selina Concise Mathematics Class 10 ICSE Solutions Linear Inequations (in one variable) - 60
Solution:

Question 32.
Selina Concise Mathematics Class 10 ICSE Solutions Linear Inequations (in one variable) - 61
Solution:

Question 33.
Selina Concise Mathematics Class 10 ICSE Solutions Linear Inequations (in one variable) - 64
Solution:

Question 34.
Solve the following in equation and write the solution set:
13x – 5 < 15x + 4 < 7x + 12, x ∈ R
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Linear Inequations (in one variable) - 67

Question 35.
Solve the following inequation, write the solution set and represent it on the number line.
-3(x – 7) ≥ 15 – 7x > x+1/3, x R.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Linear Inequations (in one variable) q36

Question 36.
Solve the following inequation and represent the solution set on a number line.

Solution:

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Selina Concise Mathematics Class 10 ICSE Solutions Solving Simple Problems (Based on Quadratic Equations)

November 10, 2023May 26, 2022 by Prasanna

Selina Concise Mathematics Class 10 ICSE Solutions Solving Simple Problems (Based on Quadratic Equations)

Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Simple Problems (Based on Quadratic Equations)

Solving Simple Problems (Based on Quadratic Equations) Exercise 6A – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
The product of two consecutive integers is 56. Find the integers.
Solution:
Let the two consecutive integers be x and x + 1.
From the given information,
x(x + 1) = 56
x² + x – 56 = 0
(x + 8) (x – 7) = 0
x = -8 or 7
Thus, the required integers are – 8 and -7; 7 and 8.

Question 2.
The sum of the squares of two consecutive natural numbers is 41. Find the numbers.
Solution:
Let the numbers be x and x + 1.
From the given information,
x² + (x + 1)² = 41
2x² + 2x + 1 – 41 = 0
x² + x – 20 = 0
(x + 5) (x – 4) = 0
x = -5, 4
But, -5 is not a natural number. So, x = 4.
Thus, the numbers are 4 and 5.

Question 3.
Find the two natural numbers which differ by 5 and the sum of whose squares is 97.
Solution:
Let the two numbers be x and x + 5.
From the given information,
x² + (x + 5)² = 97
2x² + 10x + 25 – 97 = 0
2x² + 10x – 72 = 0
x² + 5x – 36 = 0
(x + 9) (x – 4) = 0
x = -9 or 4
Since, -9 is not a natural number. So, x = 4.
Thus, the numbers are 4 and 9.

Question 4.
The sum of a number and its reciprocal is 4.25. Find the number.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Solving Simple Problems (Based on Quadratic Equations) - 1

Question 5.
Two natural numbers differ by 3. Find the numbers, if the sum of their reciprocals is \(\frac { 7 }{ 10 }\)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Solving Simple Problems (Based on Quadratic Equations) - 2

Question 6.
Divide 15 into two parts such that the sum of their reciprocals is \(\frac { 3 }{ 10 }\)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Solving Simple Problems (Based on Quadratic Equations) - 3

Question 7.
The sum of the square of two positive integers is 208. If the square of larger number is 18 times the smaller number, find the numbers.
Solution:
Let the two numbers be x and y, y being the bigger number. From the given information,
x²+ y²= 208 ….. (i)
y² = 18x ….. (ii)
From (i), we get y²=208 – x². Putting this in (ii), we get,
208 – x²= 18x
⇒ x² + 18x – 208 = 0
⇒ x² + 26X – 8X – 208 = 0
⇒ x(x + 26) – 8(x + 26) = 0
⇒ (x – 8)(x + 26) = 0
⇒ x can’t be a negative number , hence x = 8
⇒ Putting x = 8 in (ii), we get y² = 18 x 8=144
⇒ y = 12, since y is a positive integer
Hence, the two numbers are 8 and 12.

Question 8.
The sum of the squares of two consecutive positive even numbers is 52. Find the numbers.
Solution:
Let the consecutive positive even numbers be x and x + 2.
From the given information,
x² + (x + 2)² = 52
2x² + 4x + 4 = 52
2x² + 4x – 48 = 0
x² + 2x – 24 = 0
(x + 6) (x – 4) = 0
x = -6, 4
Since, the numbers are positive, so x = 4.
Thus, the numbers are 4 and 6.

Question 9.
Find two consecutive positive odd numbers, the sum of whose squares is 74.
Solution:
Let the consecutive positive odd numbers be x and x + 2.
From the given information,
x² + (x + 2)² = 74
2x² + 4x + 4 = 74
2x² + 4x – 70 = 0
x² + 2x – 35 = 0
(x + 7)(x – 5) = 0
x = -7, 5
Since, the numbers are positive, so, x = 5.
Thus, the numbers are 5 and 7.

Question 10.
The denominator of a fraction is one more than twice the numerator. If the sum of the fraction and its reciprocal is 2.9; find the fraction.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Solving Simple Problems (Based on Quadratic Equations) - 4

Question 11.
Three positive numbers are in the ratio 1/2 : 1/3 : 1/4. Find the numbers if the sum of their squares is 244.
Solution:
Given, three positive numbers are in the ratio 1/2 : 1/3 : 1/4 = 6 : 4 : 3
Let the numbers be 6x, 4x and 3x.
From the given information,
(6x)² + (4x)² + (3x)² = 244
36x² + 16x² + 9x² = 244
61x² = 244
x² = 4
x = ± 2
Since, the numbers are positive, so x = 2.
Thus, the numbers are 12, 8 and 6.

Question 12.
Divide 20 into two parts such that three times the square of one part exceeds the other part by 10.
Solution:
Let the two parts be x and y.
From the given information,
x + y = 20 ⇒ y = 20 – x
3x² = (20 – x) + 10
3x² = 30 – x
3x² + x – 30 = 0
3x² – 9x + 10x – 30 = 0
3x(x – 3) + 10(x – 3) = 0
(x – 3) (3x + 10) = 0
x = 3, -10/3
Since, x cannot be equal to -10/3, so, x = 3.
Thus, one part is 3 and other part is 20 – 3 = 17.

Question 13.
Three consecutive natural numbers are such that the square of the middle number exceeds the difference of the squares of the other two by 60.
Assume the middle number to be x and form a quadratic equation satisfying the above statement. Hence; find the three numbers.
Solution:
Let the numbers be x – 1, x and x + 1.
From the given information,
x² = (x + 1)² – (x – 1)² + 60
x² = x² + 1 + 2x – x² – 1 + 2x + 60
x² = 4x + 60
x² – 4x – 60 = 0
(x – 10) (x + 6) = 0
x = 10, -6
Since, x is a natural number, so x = 10.
Thus, the three numbers are 9, 10 and 11.

Question 14.
Out of three consecutive positive integers, the middle number is p. If three times the square of the largest is greater than the sum of the squares of the other two numbers by 67; calculate the value of p.
Solution:
Let the numbers be p – 1, p and p + 1.
From the given information,
3(p + 1)² = (p – 1)² + p² + 67
3p² + 6p + 3 = p² + 1 – 2p + p² + 67
p² + 8p – 65 = 0
(p + 13)(p – 5) = 0
p = -13, 5
Since, the numbers are positive so p cannot be equal to -13.
Thus, p = 5.

Question 15.
A can do a piece of work in ‘x’ days and B can do the same work in (x + 16) days. If both working together can do it in 15 days; calculate ‘x’.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Solving Simple Problems (Based on Quadratic Equations) - 5

Question 16.
One pipe can fill a cistern in 3 hours less than the other. The two pipes together can fill the cistern in 6 hours 40 minutes. Find the time that each pipe will take to fill the cistern.
Solution:
Let one pipe fill the cistern in x hours and the other fills it in (x – 3) hours.
Given that the two pipes together can fill the cistern in 6 hours 40 minutes, i.e.,
Selina Concise Mathematics Class 10 ICSE Solutions Solving Simple Problems (Based on Quadratic Equations) - 6

So, x = 15.
Thus, one pipe fill the cistern in 15 hours and the other fills in (x – 3) = 15 – 3 = 12 hours.

Question 17.
A positive number is divided into two parts such that the sum of the squares of the two parts is 20. The square of the larger part is 8 times the smaller part. Taking x as the smaller part of the two parts, find the number.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Solving Simple Problems (Based on Quadratic Equations) - 8

Solving Simple Problems (Based on Quadratic Equations) Exercise 6B – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
The sides of a right-angled triangle containing the right angle are 4x cm and (2x – 1) cm. If the area of the triangle is 30 cm²; calculate the lengths of its sides.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Solving Simple Problems (Based on Quadratic Equations) - 9

Question 2.
The hypotenuse of a right-angled triangle is 26 cm and the sum of other two sides is 34 cm. Find the lengths of its sides.
Solution:
Hypotenuse = 26 cm
The sum of other two sides is 34 cm.
So, let the other two sides be x cm and (34 – x) cm.
Using Pythagoras theorem,
(26)² = x² + (34 – x)²
676 = x² + x² + 1156 – 68x
2x² – 68x + 480 = 0
x² – 34x + 240 = 0
x² – 10x – 24x + 240 = 0
x(x – 10) – 24(x – 10) = 0
(x – 10) (x – 24) = 0
x = 10, 24
When x = 10, (34 – x) = 24
When x = 24, (34 – x) = 10
Thus, the lengths the three sides of the right-angled triangle are 10 cm, 24 cm and 26 cm.

Question 3.
The sides of a right-angled triangle are (x – 1) cm, 3x cm and (3x + 1) cm. Find:
(i) the value of x,
(ii) the lengths of its sides,
(iii) its area.
Solution:
Longer side = Hypotenuse = (3x + 1) cm
Lengths of other two sides are (x – 1) cm and 3x cm.
Using Pythagoras theorem,
(3x + 1)² = (x – 1)² + (3x)²
9x² + 1 + 6x = x² + 1 – 2x + 9x²
x² – 8x = 0
x(x – 8) = 0
x = 0, 8
But, if x = 0, then one side = 3x = 0, which is not possible.
So, x = 8
Thus, the lengths of the sides of the triangle are (x – 1) cm = 7 cm, 3x cm = 24 cm and (3x + 1) cm = 25 cm.
Area of the triangle = ½ × 7 cm × 24 cm = 84 cm²

Question 4.
The hypotenuse of a right-angled triangle exceeds one side by 1 cm and the other side by 18 cm; find the lengths of the sides of the triangle.
Solution:
Let one hypotenuse of the triangle be x cm.
From the given information,
Length of one side = (x – 1) cm
Length of other side = (x – 18) cm
Using Pythagoras theorem,
x² = (x – 1)² + (x – 18)²
x² = x² + 1 – 2x + x² + 324 – 36x
x² – 38x + 325 = 0
x² – 13x – 25x + 325 = 0
x(x – 13) – 25(x – 13) = 0
(x – 13) (x – 25) = 0
x = 13, 25
When x = 13, x – 18 = 13 – 18 = -5, which being negative, is not possible.
So, x = 25
Thus, the lengths of the sides of the triangle are x = 25 cm, (x – 1) = 24 cm and (x – 18) = 7 cm.

Question 5.
The diagonal of a rectangle is 60 m more than its shorter side and the larger side is 30 m more than the shorter side. Find the sides of the rectangle.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Solving Simple Problems (Based on Quadratic Equations) - 10
Let the shorter side be x m.
Length of the other side = (x + 30) m
Length of hypotenuse = (x + 60) m
Using Pythagoras theorem,
(x + 60)² = x² + (x + 30)²
x² + 3600 + 120x = x² + x² + 900 + 60x
x² – 60x – 2700 = 0
x² – 90x + 30x – 2700 = 0
x(x – 90) + 30(x – 90) = 0
(x – 90) (x + 30) = 0
x = 90, -30
But, x cannot be negative. So, x = 90.
Thus, the sides of the rectangle are 90 m and (90 + 30) m = 120 m.

Question 6.
The perimeter of a rectangle is 104 m and its area is 640 m². Find its length and breadth.
Solution:
Let the length and the breadth of the rectangle be x m and y m.
Perimeter = 2(x + y) m
∴ 104 = 2(x + y)
x + y = 52
y = 52 – x
Area = 640 m²
∴ xy = 640
x(52 – x) = 640
x² – 52x + 640 = 0
x² – 32x – 20x + 640 = 0
x(x – 32) – 20 (x – 32) = 0
(x – 32) (x – 20) = 0
x = 32, 20
When x = 32, y = 52 – 32 = 20
When x = 20, y = 52 – 20 = 32
Thus, the length and breadth of the rectangle are 32 cm and 20 cm.

Question 7.
A footpath of uniform width runs round the inside of a rectangular field 32 m long and 24 m wide. If the path occupies 208 m², find the width of the footpath.
Solution:
Let w be the width of the footpath.
Selina Concise Mathematics Class 10 ICSE Solutions Solving Simple Problems (Based on Quadratic Equations) - 11
Area of the path = Area of outer rectangle – Area of inner rectangle
∴ 208 = (32)(24) – (32 – 2w)(24 – 2w)
208 = 768 – 768 + 64w + 48w – 4w²
4w² – 112w + 208 = 0
w² – 28w + 52 = 0
w² – 26w – 2w + 52 = 0
w(w – 26) – 2(w – 26) = 0
(w – 26) (w – 2) = 0
w = 26, 2
If w = 26, then breadth of inner rectangle = (24 – 52) m = -28 m, which is not possible.
Hence, the width of the footpath is 2 m.

Question 8.
Two squares have sides x cm and (x + 4) cm. The sum of their area is 656 sq. cm. Express this as an algebraic equation in x and solve the equation to find the sides of the squares.
Solution:
Given that, two squares have sides x cm and (x + 4) cm.
Sum of their area = 656 cm²
∴ x² + (x + 4)² = 656
x² + x² + 16 + 8x = 656
2x² + 8x – 640 = 0
x² + 4x – 320 = 0
x² + 20x – 16x – 320 = 0
x(x + 20) – 16(x + 20) = 0
(x + 20) (x – 16) = 0
x = -20, 16
But, x being side, cannot be negative.
So, x = 16
Thus, the sides of the two squares are 16 cm and 20 cm.

Question 9.
The dimensions of a rectangular field are 50 m and 40 m. A flower bed is prepared inside this field leaving a gravel path of uniform width all around the flower bed. The total cost of laying the flower bed and gravelling the path at Rs 30 and Rs 20 per square metre, respectively, is Rs 52,000. Find the width of the gravel path.
Solution:
Let the width of the gravel path be w m.
Length of the rectangular field = 50 m
Breadth of the rectangular field = 40 m
Let the length and breadth of the flower bed be x m and y m respectively.
Therefore, we have:
x + 2w = 50 … (1)
y + 2w = 40 … (2)
Also, area of rectangular field = 50 m 40 m = 2000 m²
Area of the flower bed = xy m²
Area of gravel path = Area of rectangular field – Area of flower bed = (2000 – xy) m²
Cost of laying flower bed + Gravel path = Area x cost of laying per sq. m
52000 = 30 xy + 20 (2000 – xy)
52000 = 10xy + 40000
xy = 1200
Using (1) and (2), we have:
(50 – 2w) (40 – 2w) = 1200
2000 – 180w + 4w² = 1200
4w² – 180w + 800 = 0
w² – 45w + 200 = 0
w² – 5w – 40w + 200 = 0
w(w – 5) – 40(w – 5) = 0
(w – 5) (w – 40) = 0
w = 5, 40
If w = 40, then x = 50 – 2w = -30, which is not possible.
Thus, the width of the gravel path is 5 m.

Question 10.
An area is paved with square tiles of a certain size and the number required is 128. If the tiles had been 2 cm smaller each way, 200 tiles would have been needed to pave the same area. Find the size of the larger tiles.
Solution:
Let the size of the larger tiles be x cm.
Area of larger tiles = x² cm²
Number of larger tiles required to pave an area is 128.
So, the area needed to be paved = 128 x² cm² …. (1)
Size of smaller tiles = (x – 2)cm
Area of smaller tiles = (x – 2)² cm²
Number of larger tiles required to pave an area is 200.
So, the area needed to be paved = 200 (x – 2)² cm² …. (2)
Therefore, from (1) and (2), we have:
128 x² = 200 (x – 2)²
128 x² = 200x² + 800 – 800x
72x² – 800x + 800 = 0
9x² – 100x + 100 = 0
9x² – 90x – 10x + 100 = 0
9x(x – 10) – 10(x – 10) = 0
(x – 10)(9x – 10) = 0
Selina Concise Mathematics Class 10 ICSE Solutions Solving Simple Problems (Based on Quadratic Equations) - 12
Hence, the size of the larger tiles is 10 cm.

Question 11.
A farmer has 70 m of fencing, with which he encloses three sides of a rectangular sheep pen; the fourth side being a wall. If the area of the pen is 600 sq. m, find the length of its shorter side.
Solution:
Let the length and breadth of the rectangular sheep pen be x and y respectively.
From the given information,
x + y + x = 70
2x + y = 70 … (1)
Also, area = xy = 600
Using (1), we have:
x (70 – 2x) = 600
70x – 2x² = 600
2x² – 70x + 600 = 0
x² – 35x + 300 = 0
x² – 15x – 20x + 300 = 0
x(x – 15) – 20(x – 15) = 0
(x – 15)(x – 20) = 0
x = 15, 20
If x = 15, then y = 70 – 2x = 70 – 30 = 40
If x = 20, then y = 70 – 2x = 70 – 40 = 30
Thus, the length of the shorter side is 15 m when the longer side is 40 m. The length of the shorter side is 20 m when the longer side is 30 m.

Question 12.
A square lawn is bounded on three sides by a path 4 m wide. If the area of the path is \(\frac { 7 }{ 8 }\) that of the lawn, find the dimensions of the lawn.
Solution:
Let the side of the square lawn be x m.
Area of the square lawn = x² m²
The square lawn is bounded on three sides by a path which is 4 m wide.
Selina Concise Mathematics Class 10 ICSE Solutions Solving Simple Problems (Based on Quadratic Equations) - 13
Area of outer rectangle = (x + 4) (x + 8) = x² + 12x + 32
Area of path = x² + 12x + 32 – x² = 12x + 32
From the given information, we have:

Since, x cannot be negative. So, x = 16 m.
Thus, each side of the square lawn is 16 m.

Question 13.
The area of a big rectangular room is 300 m². If the length were decreased by 5 m and the breadth increased by 5 m; the area would be unaltered. Find the length of the room.
Solution:
Let the original length and breadth of the rectangular room be x m and y m respectively.
Area of the rectangular room = xy = 300
⇒ y = \(\frac { 300 }{ x }\) …..(1)
New length = (x – 5) m
New breadth = (y + 5) m
New area = (x – 5) (y + 5) = 300 (given)
Using (1), we have:
Selina Concise Mathematics Class 10 ICSE Solutions Solving Simple Problems (Based on Quadratic Equations) - 15
But, x cannot be negative. So, x = 20.
Thus, the length of the room is 20 m.

Solving Simple Problems (Based on Quadratic Equations) Exercise 6C – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
The speed of an ordinary train is x km per hr and that of an express train is (x + 25) km per hr.
(i) Find the time taken by each train to cover 300 km.
(ii) If the ordinary train takes 2 hrs more than the express train; calculate speed of the express train.
Solution:
(i) Speed of ordinary train = x km/hr
Speed of express train = (x + 25) km/hr
Distance = 300 km
Selina Concise Mathematics Class 10 ICSE Solutions Solving Simple Problems (Based on Quadratic Equations) - 16
(ii) Given that the ordinary train takes 2 hours more than the express train to cover the distance.
Therefore,

But, speed cannot be negative. So, x = 50.
∴ Speed of the express train = (x + 25) km/hr = 75 km/hr

Question 2.
If the speed of a car is increased by 10 km per hr, it takes 18 minutes less to cover a distance of 36 km. Find the speed of the car.
Solution:
Let the speed of the car be x km/hr.
Distance = 36 km
Selina Concise Mathematics Class 10 ICSE Solutions Solving Simple Problems (Based on Quadratic Equations) - 18
From the given information, we have:

But, speed cannot be negative. So, x = 30.
Hence, the original speed of the car is 30 km/hr.

Question 3.
If the speed of an aeroplane is reduced by 40 km/hr, it takes 20 minutes more to cover 1200 km. Find the speed of the aeroplane.
Solution:
Let the original speed of the aeroplane be x km/hr.
Selina Concise Mathematics Class 10 ICSE Solutions Solving Simple Problems (Based on Quadratic Equations) - 20
But, speed cannot be negative. So, x = 400.
Thus, the original speed of the aeroplane is 400 km/hr.

Question 4.
A car covers a distance of 400 km at a certain speed. Had the speed been 12 km/h more, the time taken for the journey would have been 1 hour 40 minutes less. Find the original speed of the car.
Solution:
Let x km/h be the original speed of the car.
We know that,

It is given that the car covers a distance of 400 km with the speed of x km/h.
Thus, the time taken by the car to complete 400 km is
t = \(\frac { 400 }{ x }\)
Now, the speed is increased by 12 km.
∴ Increased speed = (x + 12) km/hr.
Also given that, increasing the speed of the car will decrease the time taken by 1 hour 40 minutes.
Selina Concise Mathematics Class 10 ICSE Solutions Solving Simple Problems (Based on Quadratic Equations) - 22

Question 5.
A girl goes to her friend’s house, which is at a distance of 12 km. She covers half of the distance at a speed of x km/hr and the remaining distance at a speed of (x + 2) km/hr. If she takes 2 hrs 30 minutes to cover the whole distance, find ‘x’.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Solving Simple Problems (Based on Quadratic Equations) - 23

Question 6.
A car made a run of 390 km in ‘x’ hours. If the speed had been 4 km/hour more, it would have taken 2 hours less for the journey. Find ‘x’.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Solving Simple Problems (Based on Quadratic Equations) - 24

Question 7.
A goods train leaves a station at 6 p.m., followed by an express train which leaved at 8 p.m. and travels 20 km/hour faster than the goods train. The express train arrives at a station, 1040 km away, 36 minutes before the goods train. Assuming that the speeds of both the train remain constant between the two stations; calculate their speeds.
Solution:
Let the speed of goods train be x km/hr. So, the speed of express train will be (x + 20) km/hr.
Distance = 1040 km
Selina Concise Mathematics Class 10 ICSE Solutions Solving Simple Problems (Based on Quadratic Equations) - 25
It is given that the express train arrives at a station 36 minutes before the goods train. Also, the express train leaves the station 2 hours after the goods train. This means that the express train arrives at the station

before the goods train.
Selina Concise Mathematics Class 10 ICSE Solutions Solving Simple Problems (Based on Quadratic Equations) - 27
Since, the speed cannot be negative. So, x = 80.
Thus, the speed of goods train is 80 km/hr and the speed of express train is 100 km/hr.

Question 8.
A man bought an article for Rs x and sold it for Rs 16. If his loss was x per cent, find the cost price of the article.
Solution:
C.P. of the article = Rs x
S.P. of the article = Rs 16
Loss = Rs (x – 16)
We know:
Selina Concise Mathematics Class 10 ICSE Solutions Solving Simple Problems (Based on Quadratic Equations) - 28
Thus, the cost price of the article is Rs 20 or Rs 80.

Question 9.
A trader bought an article for Rs x and sold it for Rs 52, thereby making a profit of (x – 10) per cent on his outlay. Calculate the cost price.
Solution:
C.P. of the article = Rs x
S.P. of the article = Rs 52
Profit = Rs (52 – x)
We know:
Selina Concise Mathematics Class 10 ICSE Solutions Solving Simple Problems (Based on Quadratic Equations) - 29
Since, C.P. cannot be negative. So, x = 40.
Thus, the cost price of the article is Rs 40.

Question 10.
By selling a chair for Rs 75, Mohan gained as much per cent as its cost. Calculate the cost of the chair.
Solution:
Let the C.P. of the chair be Rs x
S.P. of chair = Rs 75
Profit = Rs (75 – x)
We know:
Selina Concise Mathematics Class 10 ICSE Solutions Solving Simple Problems (Based on Quadratic Equations) - 30
But, C.P. cannot be negative. So, x = 50.
Hence, the cost of the chair is Rs 50.

Solving Simple Problems (Based on Quadratic Equations) Exercise 6D – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
The sum S of n successive odd numbers starting from 3 is given by the relation n(n + 2). Determine n, if the sum is 168.
Solution:
From the given information, we have:
n(n + 2) = 168
n² + 2n – 168 = 0
n² + 14n – 12n – 168 = 0
n(n + 14) – 12(n + 14) = 0
(n + 14) (n – 12) = 0
n = -14, 12
But, n cannot be negative.
Therefore, n = 12.

Question 2.
A stone is thrown vertically downwards and the formula d = 16t² + 4t gives the distance, d metres, that it falls in t seconds. How long does it take to fall 420 metres?
Solution:
From the given information,
16t² + 4t = 420
4t² + t – 105 = 0
4t² – 20t + 21t – 105 = 0
4t(t – 5) + 21(t – 5) = 0
(4t + 21)(t – 5) = 0
t = -21/4, 5
But, time cannot be negative.
Thus, the required time taken is 5 seconds.

Question 3.
The product of the digits of a two digit number is 24. If its unit’s digit exceeds twice its ten’s digit by 2; find the number.
Solution:
Let the ten’s and unit’s digit of the required number be x and y respectively.
From the given information,
Selina Concise Mathematics Class 10 ICSE Solutions Solving Simple Problems (Based on Quadratic Equations) - 31

Question 4.
The product of the digits of a two digit number is 24. If its unit’s digit exceeds twice its ten’s digit by 2; find the number.
Solution:
The ages of two sisters are 11 years and 14 years.
Let in x number of years the product of their ages be 304.
Selina Concise Mathematics Class 10 ICSE Solutions Solving Simple Problems (Based on Quadratic Equations) - 32
But, the number of years cannot be negative. So, x = 5.
Hence, the required number of years is 5 years.

Question 5.
One year ago, a man was 8 times as old as his son. Now his age is equal to the square of his son’s age. Find their present ages.
Solution:
Let the present age of the son be x years.
∴ Present age of man = x² years
One year ago,
Son’s age = (x – 1) years
Man’s age = (x² – 1) years
It is given that one year ago; a man was 8 times as old as his son.
∴ (x² – 1) = 8(x – 1)
x² – 8x – 1 + 8 = 0
x² – 8x + 7 = 0
(x – 7) (x – 1) = 0
x = 7, 1
If x = 1, then x² = 1, which is not possible as father’s age cannot be equal to son’s age.
So, x = 7.
Present age of son = x years = 7 years
Present age of man = x² years = 49 years

Question 6.
The age of the father is twice the square of the age of his son. Eight years hence, the age of the father will be 4 years more than three times the age of the son. Find their present ages.
Solution:
Let the present age of the son be x years.
Present age of father = 2x² years
Eight years hence,
Son’s age = (x + 8) years
Father’s age = (2x² + 8) years
It is given that eight years hence, the age of the father will be 4 years more than three times the age of the son.
2x² + 8 = 3(x + 8) +4
2x² + 8 = 3x + 24 +4
2x² – 3x – 20 = 0
2x² – 8x + 5x – 20 = 0
2x(x – 4) + 5(x – 4) = 0
(x – 4) (2x + 5) = 0
x = 4, -5/2
But, the age cannot be negative, so, x = 4.
Present age of son = 4 years
Present age of father = 2(4)² years = 32 years

Question 7.
The speed of a boat in still water is 15 km/hr. It can go 30 km upstream and return downstream to the original point in 4 hours 30 minutes. Find the speed of the stream.
Solution:
Let the speed of the stream be x km/hr.
∴ Speed of the boat downstream = (15 + x) km/hr
Speed of the boat upstream = (15 – x) km/hr
Selina Concise Mathematics Class 10 ICSE Solutions Solving Simple Problems (Based on Quadratic Equations) - 50
But, x cannot be negative, so, x = 5.
Thus, the speed of the stream is 5 km/hr.

Question 8.
Mr. Mehra sends his servant to the market to buy oranges worth Rs 15. The servant having eaten three oranges on the way. Mr. Mehra pays Rs 25 paise per orange more than the market price.
Taking x to be the number of oranges which Mr. Mehra receives, form a quadratic equation in x. Hence, find the value of x.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Solving Simple Problems (Based on Quadratic Equations) - 34

Question 9.
Rs 250 is divided equally among a certain number of children. If there were 25 children more, each would have received 50 paise less. Find the number of children.
Solution:
Let the number of children be x.
It is given that Rs 250 is divided amongst x students.
Selina Concise Mathematics Class 10 ICSE Solutions Solving Simple Problems (Based on Quadratic Equations) - 35
Since, the number of students cannot be negative, so, x = 100.
Hence, the number of students is 100.

Question 10.
An employer finds that if he increased the weekly wages of each worker by Rs 5 and employs five workers less, he increases his weekly wage bill from Rs 3,150 to Rs 3,250. Taking the original weekly wage of each worker as Rs x; obtain an equation in x and then solve it to find the weekly wages of each worker.
Solution:
Original weekly wage of each worker = Rs x
Original weekly wage bill of employer = Rs 3150
Selina Concise Mathematics Class 10 ICSE Solutions Solving Simple Problems (Based on Quadratic Equations) - 36
Since, wage cannot be negative, x = 45.
Thus, the original weekly wage of each worker is Rs 45.

Question 11.
A trader bought a number of articles for Rs 1,200. Ten were damaged and he sold each of the remaining articles at Rs 2 more than what he paid for it, thus getting a profit of Rs 60 on whole transaction.
Taking the number of articles he bought as x, form an equation in x and solve it.
Solution:
Number of articles bought by the trader = x
It is given that the trader bought the articles for Rs 1200.
Selina Concise Mathematics Class 10 ICSE Solutions Solving Simple Problems (Based on Quadratic Equations) - 37
Number of articles cannot be negative. So, x = 100.

Question 12.
The total cost price of a certain number of identical articles is Rs 4800. By selling the articles at Rs 100 each, a profit equal to the cost price of 15 articles is made. Find the number of articles bought.
Solution:
Let the number of articles bought be x.
Total cost price of x articles = Rs 4800
Cost price of one article = Rs \(\frac { 4800 }{ x }\)
Selling price of each article = Rs 100
Selling price of x articles = Rs 100x
Given, Profit = C.P. of 15 articles
∴ 100x – 4800 = 15 × \(\frac { 4800 }{ x }\)
100x² – 4800x = 15 4800
x² – 48x – 720 = 0
x² – 60x + 12x – 720 = 0
x(x – 60) + 12(x – 60) = 0
(x – 60) (x + 12) = 0
x = 60, -12
Since, number of articles cannot be negative. So, x = 60.
Thus, the number of articles bought is 60.

Solving Simple Problems (Based on Quadratic Equations) Exercise 6E – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
The distance by road between two towns A and B is 216 km, and by rail it is 208 km. A car travels at a speed of x km/hr and the train travels at a speed which is 16 km/hr faster than the car. Calculate:
(i) the time taken by the car to reach town B from A, in terms of x;
(ii) the time taken by the train to reach town B from A, in terms of x.
(iii) If the train takes 2 hours less than the car, to reach town B, obtain an equation in x and solve it.
(iv) Hence, find the speed of the train.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Solving Simple Problems (Based on Quadratic Equations) - 38

Question 2.
A trader buys x articles for a total cost of Rs 600.
(i) Write down the cost of one article in terms of x.
If the cost per article were Rs 5 more, the number of articles that can be bought for Rs 600 would be four less.
(ii) Write down the equation in x for the above situation and solve it for x.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Solving Simple Problems (Based on Quadratic Equations) - 39

Question 3.
A hotel bill for a number of people for overnight stay is Rs 4800. If there were 4 people more, the bill each person had to pay, would have reduced by Rs 200. Find the number of people staying overnight.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Solving Simple Problems (Based on Quadratic Equations) - 40

Question 4.
An Aero plane travelled a distance of 400 km at an average speed of x km/hr. On the return journey, the speed was increased by 40 km/hr. Write down an expression for the time taken for:
(i) the onward journey;
(ii) the return journey.
If the return journey took 30 minutes less than the onward journey, write down an equation in x and find its value.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Solving Simple Problems (Based on Quadratic Equations) - 41

Question 5.
Rs 6500 was divided equally among a certain number of persons. Had there been 15 persons more, each would have got Rs 30 less. Find the original number of persons.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Solving Simple Problems (Based on Quadratic Equations) - 42

Question 6.
A plane left 30 minutes later than the schedule time and in order to reach its destination 1500 km away in time, it has to increase its speed by 250 km/hr from its usual speed. Find its usual speed.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Solving Simple Problems (Based on Quadratic Equations) - 43

Question 7.
Two trains leave a railway station at the same time. The first train travels due west and the second train due north. The first train travels 5 km/hr faster than the second train. If after 2 hours, they are 50 km apart, find the speed of each train.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Solving Simple Problems (Based on Quadratic Equations) - 44

Question 8.
The sum S of first n even natural numbers is given by the relation S = n(n + 1). Find n, if the sum is 420.
Solution:
S = n(n + 1)
Given, S = 420
n(n + 1) = 420
n² + n – 420 = 0
n² + 21n – 20n – 420 = 0
n(n + 21) – 20(n + 21) = 0
(n + 21) (n – 20) = 0
n = -21, 20
Since, n cannot be negative.
Hence, n = 20.

Question 9.
The sum of the ages of a father and his son is 45 years. Five years ago, the product of their ages (in years) was 124. Determine their present ages.
Solution:
Let the present ages of father and his son be x years and (45 – x) years respectively.
Five years ago,
Father’s age = (x – 5) years
Son’s age = (45 – x – 5) years = (40 – x) years
From the given information, we have:
(x – 5) (40 – x) = 124
40x – x² – 200 + 5x = 124
x² – 45x +324 = 0
x² – 36x – 9x +324 = 0
x(x – 36) – 9(x – 36) = 0
(x – 36) (x – 9) = 0
x = 36, 9
If x = 9,
Father’s age = 9 years, Son’s age = (45 – x) = 36 years
This is not possible.
Hence, x = 36
Father’s age = 36 years
Son’s age = (45 – 36) years = 9 years

Question 10.
In an auditorium, seats were arranged in rows and columns. The number of rows was equal to the number of seats in each row. When the number of rows was doubled and the number of seats in each row was reduced by 10, the total number of seats increased by 300. Find:
(i) the number of rows in the original arrangement.
(ii) the number of seats in the auditorium after re-arrangement.
Solution:
Let the number of rows in the original arrangement be x.
Then, the number of seats in each row in original arrangement = x
Total number of seats = x × x = x²
From the given information,
2x(x – 10) = x² + 300
2x² – 20x = x² + 300
x² – 20x – 300 = 0
(x – 30) (x + 10) = 0
x = 30, -10
Since, the number of rows or seats cannot be negative. So, x = 30.
(i) The number of rows in the original arrangement = x = 30
(ii) The number of seats after re-arrangement = x² + 300 = 900 + 300 = 1200

Question 11.
Mohan takes 16 days less than Manoj to do a piece of work. If both working together can do it in 15 days, in how many days will Mohan alone complete the work?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Solving Simple Problems (Based on Quadratic Equations) - 45

Question 12.
Two years ago, a man’s age was three times the square of his son’s age. In three years time, his age will be four times his son’s age. Find their present ages.
Solution:
Let the age of son 2 years ago be x years.
Then, father’s age 2 years ago = 3x² years
Present age of son = (x + 2) years
Present age of father = (3x² + 2) years
3 years hence:
Son’s age = (x + 2 + 3) years = (x + 5) years
Father’s age = (3x² + 2 + 3) years = (3x² + 5) years
From the given information,
3x² + 5 = 4(x + 5)
3x² – 4x – 15 = 0
3x² – 9x + 5x – 15 = 0
3x(x – 3) + 5(x – 3) = 0
(x – 3) (3x + 5) = 0
x = 3,
Since, age cannot be negative. So, x = 3.
Present age of son = (x + 2) years = 5 years
Present age of father = (3x² + 2) years = 29 years

Question 13.
In a certain positive fraction, the denominator is greater than the numerator by 3. If 1 is subtracted from the numerator and the denominator both, the fraction reduces by. Find the fraction.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Solving Simple Problems (Based on Quadratic Equations) - 46

Question 14.
In a two digit number, the ten’s digit is bigger. The product of the digits is 27 and the difference between two digits is 6. Find the number.
Solution:
Given, the difference between two digits is 6 and the ten’s digit is bigger than the unit’s digit.
So, let the unit’s digit be x and ten’s digit be (x + 6).
From the given condition, we have:
x(x + 6) = 27
x² + 6x – 27 = 0
x² + 9x – 3x – 27 = 0
x(x + 9) – 3(x + 9) = 0
(x + 9) (x – 3) = 0
x = -9, 3
Since, the digits of a number cannot be negative. So, x = 3.
Unit’s digit = 3
Ten’s digit = 9
Thus, the number is 93.

Question 15.
Some school children went on an excursion by a bus to a picnic spot at a distance of 300 km. While returning, it was raining and the bus had to reduce its speed by 5 km/hr and it took two hours longer for returning. Find the time taken to return.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Solving Simple Problems (Based on Quadratic Equations) - 47

Question 16.
Rs.480 is divided equally among ‘x’ children. If the number of children were 20 more, then each would have got Rs.12 less. Find ‘x’.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Solving Simple Problems (Based on Quadratic Equations) - 48

Question 17.
A bus covers a distance of 240 km at a uniform speed. Due to heavy rain its speed gets reduced by 10 km/h and as such it takes two hrs longer to covers the total distance. Assuming the uniform speed to be ‘x’ km/h, form an equation and solve it to evaluate ‘x’.

Solution:

Question 18.
The sum of the ages of Vivek and his younger brother Amit is 47 years. The product of their ages in years is 550. Find their ages.
Solution:
Given that he sum of the ages of Vivek and his younger brother Amit is 47 years.
Let the age of Vivek = x
⇒ the age of Amit = 47 – x
The product of their ages in years is 550 …. given
⇒ x(47 – x) = 550
⇒ 47x – x² = 550
⇒ x² – 47x + 550 = 0
⇒ x² – 25x – 22x + 550 = 0
⇒ x(x – 25) – 22(x – 25) = 0
⇒ (x – 25) (x – 22) = 0
⇒ x = 25 or x = 22
Given that Vivek is an elder brother.
∴ x = 25 years = age of Vivek and
age of Amit = 47 – 25 = 22 years

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Selina Concise Mathematics Class 10 ICSE Solutions Matrices

November 10, 2023May 26, 2022 by Prasanna

Selina Concise Mathematics Class 10 ICSE Solutions Matrices

Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices

Matrices Exercise 9A – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
State, whether the following statements are true or false. If false, give a reason.
(i) If A and B are two matrices of orders 3 × 2 and 2 × 3 respectively; then their sum A + B is possible.
(ii) The matrices A_2 × 3 and B_2 × 3 are conformable for subtraction.
(iii) Transpose of a 2 × 1 matrix is a 2 × 1 matrix.
(iv) Transpose of a square matrix is a square matrix.
(v) A column matrix has many columns and one row.
Solution:
(i) False
The sum A + B is possible when the order of both the matrices A and B are same.
(ii) True
(iii) False
Transpose of a 2 1 matrix is a 1 2 matrix.
(iv) True
(v) False
A column matrix has only one column and many rows.

Question 2.
Selina Concise Mathematics Class 10 ICSE Solutions Matrices - 1
Solution:
If two matrices are equal, then their corresponding elements are also equal. Therefore, we have:
x = 3,
y + 2 = 1 ⇒ y = -1
z – 1 = 2 ⇒ z = 3

Question 3.
Selina Concise Mathematics Class 10 ICSE Solutions Matrices image - 2
Solution:
If two matrices are equal, then their corresponding elements are also equal.
(i)
a + 5 = 2 ⇒ a = -3
-4 = b + 4 ⇒ b = -8
2 = c – 1 ⇒ c = 3
(ii) a= 3
a – b = -1
⇒ b = a + 1 = 4
b + c = 2
⇒ c = 2 – b = 2 – 4 = -2

Question 4.
If A = [8 -3] and B = [4 -5]; find: (i) A + B (ii) B – A
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Matrices image - 3

Question 5.
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Solution:

Question 6.
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Question 7.
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Question 8.
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Question 9.
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Solution:

Question 10.
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Solution:

Question 11.
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Solution:

Matrices Exercise 9B – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
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Solution:

Question 2.
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Solution:

Question 3.
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Solution:

Question 4.
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Solution:

Question 5.
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Solution:

Question 6.
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Solution:

Question 7.
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Solution:

Question 8.
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Solution:

Question 9.
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Solution:

Question 10.
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Solution:

Question 11.
Selina Concise Mathematics Class 10 ICSE Solutions Matrices image - 39
Solution:

Matrices Exercise 9C – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Selina Concise Mathematics Class 10 ICSE Solutions Matrices image - 41
Solution:

The number of columns in the first matrix is not equal to the number of rows in the second matrix. Thus, the product is not possible.

Question 2.
Selina Concise Mathematics Class 10 ICSE Solutions Matrices image - 43
Solution:

Question 3.
Selina Concise Mathematics Class 10 ICSE Solutions Matrices image - 46
Solution:

Question 4.
Selina Concise Mathematics Class 10 ICSE Solutions Matrices image - 48
Solution:

Question 5.
Selina Concise Mathematics Class 10 ICSE Solutions Matrices image - 50
Solution:

Question 6.
Selina Concise Mathematics Class 10 ICSE Solutions Matrices image - 52
Solution:

(iii) Product AA (=A²) is not possible as the number of columns of matrix A is not equal to its number of rows.

Question 7.
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Solution:

Question 8.
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Solution:

Question 9.
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Solution:

Question 10.
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Solution:

Question 11.
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Solution:

Question 12.
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Solution:

Question 13.
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Solution:

Question 14.
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Solution:

Question 15.
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Solution:

Question 16(i).
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Solution:

Question 16(ii).
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Solution:

Question 16(iii).
Selina Concise Mathematics Class 10 ICSE Solutions Matrices image - 76
Solution:

Question 17.
Selina Concise Mathematics Class 10 ICSE Solutions Matrices image - 78
Solution:
We know, the product of two matrices is defined only when the number of columns of first matrix is equal to the number of rows of the second matrix.

Question 18.
Selina Concise Mathematics Class 10 ICSE Solutions Matrices image - 81
Solution:

Question 19.
Selina Concise Mathematics Class 10 ICSE Solutions Matrices image - 83
Solution:

Question 20.
If A and B are any two 2 x 2 matrices such that AB = BA = B and B is not a zero matrix, what can you say about the matrix A?
Solution:
AB = BA = B
We know that AI = IA = I, where I is the identity matrix.
Hence, B is the identity matrix.

Question 21.
Selina Concise Mathematics Class 10 ICSE Solutions Matrices image - 85
Solution:

Question 22.
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Solution:

Question 23.
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Solution:

Question 24.
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Solution:

Question 25.
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Solution:

Question 26.
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Solution:

Question 27.
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Solution:

Question 28.
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Solution:

Question 29.
Selina Concise Mathematics Class 10 ICSE Solutions Matrices image - 101
Solution:

Question 30.
Selina Concise Mathematics Class 10 ICSE Solutions Matrices image - 103
Solution:

Question 31.
State, with reason, whether the following are true or false. A, B and C are matrices of order 2 x 2.
(i) A + B = B + A
(ii) A – B = B – A
(iii) (B. C). A = B. (C. A)
(iv) (A + B). C = A. C + B. C
(v) A. (B – C) = A. B – A. C
(vi) (A – B). C = A. C – B. C
(vii) A² – B² = (A + B) (A – B)
(viii) (A – B)² = A² – 2A. B + B²
Solution:
(i) True.
Addition of matrices is commutative.
(ii) False.
Subtraction of matrices is commutative.
(iii) True.
Multiplication of matrices is associative.
(iv) True.
Multiplication of matrices is distributive over addition.
(v) True.
Multiplication of matrices is distributive over subtraction.
(vi) True.
Multiplication of matrices is distributive over subtraction.
(vii) False.
Laws of algebra for factorization and expansion are not applicable to matrices.
(viii) False.
Laws of algebra for factorization and expansion are not applicable to matrices.

Matrices Exercise 9D – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Selina Concise Mathematics Class 10 ICSE Solutions Matrices image - 105
Solution:

Question 2.
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Solution:

Question 3.
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Solution:

Question 4.
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Question 5.
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Question 6.
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Question 7.
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Question 8.
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Question 9.
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Question 10.
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Question 11.
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Question 12.
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Question 13.
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Question 14.
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Question 15.
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Solution:

Question 16.
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Solution:

Question 17.
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Question 18.
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Question 19.
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Solution:

Question 20.
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Solution:

Question 21.
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Solution:

Question 22.

Solution:

Question 23.

Solution:

Question 24.

Solution:

Question 25.

Solution:

More Resources for Selina Concise Class 10 ICSE Solutions

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Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression

Arithmetic Progression Exercise 10A – Selina Concise Mathematics Class 10 ICSE Solutions

Arithmetic Progression Exercise 10B – Selina Concise Mathematics Class 10 ICSE Solutions

Arithmetic Progression Exercise 10C – Selina Concise Mathematics Class 10 ICSE Solutions

Arithmetic Progression Exercise 10D – Selina Concise Mathematics Class 10 ICSE Solutions

Arithmetic Progression Exercise 10E – Selina Concise Mathematics Class 10 ICSE Solutions

Arithmetic Progression Exercise 10F – Selina Concise Mathematics Class 10 ICSE Solutions

Selina Concise Mathematics Class 10 ICSE Solutions Shares and Dividends

Shares and Dividends Exercise 3A – Selina Concise Mathematics Class 10 ICSE Solutions

Shares and Dividends Exercise 3B – Selina Concise Mathematics Class 10 ICSE Solutions

Shares and Dividends Exercise 3C – Selina Concise Mathematics Class 10 ICSE Solutions

Selina Concise Mathematics Class 10 ICSE Solutions Linear Inequations (in one variable)

Linear Inequations in One Variable Exercise 4A – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1. Solution:

Linear Inequations in One Variable Exercise 4B – Selina Concise Mathematics Class 10 ICSE Solutions

Selina Concise Mathematics Class 10 ICSE Solutions Solving Simple Problems (Based on Quadratic Equations)

Solving Simple Problems (Based on Quadratic Equations) Exercise 6A – Selina Concise Mathematics Class 10 ICSE Solutions

Solving Simple Problems (Based on Quadratic Equations) Exercise 6B – Selina Concise Mathematics Class 10 ICSE Solutions

Solving Simple Problems (Based on Quadratic Equations) Exercise 6C – Selina Concise Mathematics Class 10 ICSE Solutions

Solving Simple Problems (Based on Quadratic Equations) Exercise 6D – Selina Concise Mathematics Class 10 ICSE Solutions

Solving Simple Problems (Based on Quadratic Equations) Exercise 6E – Selina Concise Mathematics Class 10 ICSE Solutions

Selina Concise Mathematics Class 10 ICSE Solutions Matrices

Matrices Exercise 9A – Selina Concise Mathematics Class 10 ICSE Solutions

Matrices Exercise 9B – Selina Concise Mathematics Class 10 ICSE Solutions

Matrices Exercise 9C – Selina Concise Mathematics Class 10 ICSE Solutions

Matrices Exercise 9D – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.

Solution: