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

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6A.

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6C
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6D
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6E

Question 1.
The product of two consecutive integers is 56. Find the integers.
Solution:
Let the first integer = x
Then second integer = x + 1
Now according to the condition given
x (x + 1) = 56
⇒ x² + x – 56 = 0
⇒ x² + 8x – 7x – 56 = 0
⇒ x (x + 8) – 7 (x + 8) = 0
⇒ (x + 8) (x – 7) = 0
Either x + 8 = 0, then x = – 8
or x – 7 = 0, then x = 7
(i) If x = -8, then
first integer = -8
and second integer = – 8 + 1 = – 7
(ii) If x = 7, then
first integer = 7
and second integer = 7 + 1 = 8
Integers are 7, 8 or -8, -7

Question 2.
The sum of the squares of two consecutive natural numbers is 41. Find the numbers.
Solution:
Let the first natural number = x
Then second natural number = x + 1
Now according to the condition given,
(x)² + (x + 1)² = 41
⇒ x² + x² + 2x + 1 = 41
⇒ 2x² + 2x + 1 – 41 = 0
⇒ 2x² + 2x – 40 = 0
⇒ x² + x – 20 = 0 (Dividing by 2)
⇒ x² + 5x – 4x – 20 = 0
⇒ x (x + 5) – 4 (x + 5) = 0
⇒ (x + 5) (x – 4) = 0
Either x + 5 = 0 then x = – 5 But it is not a natural number
or x – 4 = 0, Then x = 4
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 first natural number = x
Then second natural number = x + 5
Now according to the given condition,
(x)² + (x + 5)² = 97
⇒ x² + x² + 10x + 25 – 97 = 0
⇒ 2x² + 10x – 72 = 0
⇒ x² + 5x – 36 = 0 (Dividing by 2)
⇒ x² + 9x – 4x – 36 = 0
⇒ x (x + 9) – 4 (x + 9) = 0
⇒ (x + 9) (x – 4) = 0
Either x + 9 = 0, then x = – 9 But it is not a natural number
or x – 4 = 0, then x = 4
First number = 4
and second number = 4 + 5 = 9

Question 4.
The sum of a number and its reciprocal is 4.25. Find the number.
Solution:
Let the number = x
Now according to the condition,

⇒ 4x (x – 4) – 1 (x – 4) = 0
⇒ (x – 4) (4x – 1) = 0
Either x – 4 = 0, then x = 4
or 4x – 1 = 0, 4x = 1 then x = \(\frac { 1 }{ 4 }\)
Number is 4 or \(\frac { 1 }{ 4 }\)

Question 5.
Two natural numbers differ by 3. Find the numbers, if the sum of their reciprocals is \(\frac { 7 }{ 10 }\)
Solution:
Let the first natural number = x
Then second natural number = x + 3
Now according to the given condition,

But it is not a natural number
or x – 2 = 0, then x = 2
First number = 2
and second number = 2 + 3 = 5

Question 6.
Divide 15 into two parts such that the sum of their reciprocals is \(\frac { 3 }{ 10 }\).
Solution:
Let first part = x
Then second part = 15 – x (sum = 15)
Now according to the given condition,

⇒ x (x – 5) – 10 (x – 5) = 0
⇒ (x – 5) (x – 10) = 0
Either x – 5 = 0, then x = 5
or x – 10 = 0, then x = 10
If x = 5, then first part = 15 – 5 = 10
If x = 10, then second part = 15 – 10 = 5
Parts are 5, 10

Question 7.
The sum of the squares of two positive integers is 208. If the square of the larger number is 18 times the smaller number, find the numbers.
Solution:
Let x be the larger number and y be the smaller number, then
According to the conditions x2 = 18 ….(i)
and x² + y² = 208 ….(ii)
⇒ 18y + y² = 208 [From (i)]
⇒ y² + 18y – 208 = 0
⇒ y² + 26y – 8y – 208 = 0
⇒ y (y + 26) – 8 (y + 26) = 0
⇒ (y + 26) (y – 8) = 0
Either y + 26 = 0, then y = -26
But it is not possible as it is not positive
or y – 8 = 0, then y = 8
Then x² = 18y ⇒ y² = 18 x 8 ⇒x² = 144 = (12)² ⇒ x = 12
Number are 12, 8

Question 8.
The sum of the squares of two consecutive positive even numbers is 52. Find the numbers.
Solution:
Let first even number = 2x
and second even number = 2x + 2
Now according to the given condition,
(2x)² + (2x + 2)² = 52
⇒ 4x² + 4x² + 8x + 4 = 52
⇒ 4x² + 4x² + 8x + 4 – 52 = 0
⇒ 8x² + 8x – 48 = 0
⇒ x² + x – 6 = 0 (Dividing by 8)
⇒ x² + 3x – 2x – 6 = 0
⇒ x (x + 3) – 2 (x + 3) = 0
⇒ (x + 3) (x – 2) = 0
Either x + 3 = 0, then x = – 3 But it is not possible because it is not positive.
or x – 2 = 0, then x = 2
First even number = 2 x 2 = 4
Second number = 4 + 2 = 6
Hence numbers are 4, 6

Question 9.
Find two consecutive positive odd numbers, the sum of whose squares is 74.
Solution:
Let first odd number = 2x -1
Second odd number = 2x – 1 + 2 = 2x + 1
Now according to the given condition,
(2x – 1)² + (2x + 1)² = 74
⇒ 4x² – 4x + 1 + 4x² + 4x + 1 = 74
⇒ 8x² + 2 – 74 = 0
⇒ 8x² – 72 = 0
⇒ x² – 9 = 0 (Dividing by 8)
⇒ (x + 3) (x – 3) = 0
Either x + 3 = 0, then x = -3 But it is not possible because it is not positive.
or x – 3 = 0, then x = 3
First odd number = 2x – 1 = 2 x 3 – 1 = 6 – 1 = 5
and second odd number = 5 + 2 = 7
Number are 5, 7

Question 10.
The denominator of a positive fraction is one more than twice the numerator. If the sum of the fraction and its reciprocal is 2.9; find the fraction.
Solution:
Let numerator of a fraction = x
Then denominator = 2x + 1

Question 11.
Three positive numbers are in the ratio \(\frac { 1 }{ 2 }\) : \(\frac { 1 }{ 3 }\) : \(\frac { 1 }{ 4 }\) Find the numbers; if the sum of their squares is 244.
Solution:
Multiply each ratio by L.C.M. of de-nominators i.e. by 12.
We get, 6 : 4 : 3
Let first positive number = 6x
Then second number = 4x
and third number = 3x
According to the given condition,
(6x)² + (4x)² + (3x)² = 244
⇒ 36x² + 16x² + 9x² = 244
⇒ 61x² = 244
⇒ x² – 4 = 0
⇒ (x + 2) (x – 2) = 0
Either x + 2 = 0, then x = -2 But it is not possible because it is not positive
or x – 2 = 0, then x = 2
First number = 6x = 6 x 2 = 12
Second number = 4x = 4 x 2 = 8
and third number = 3x = 3 x 2 = 6
Numbers are 12, 8, 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 first part = x
Then second part = 20 – x (Sum = 20)
Now, according to the given condition,
3(x)² – (20 – x) = 10
⇒ 3x² – 20 + x – 10 = 0
⇒ 3x² + x – 30 = 0
⇒ 3x² + 10x – 9x – 30 = 0
⇒ x (3x + 10) – 3 (3x + 10) = 0
⇒ (3x + 10) (x – 3) = 0
Either 3x + 10 = 0, then 3x = -10 ⇒ x = \(\frac { -10 }{ 3 }\)
But it is not possible.
or x – 3 = 0 then x = 3
Then first part = 3
and second part = 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 middle number = x
Then, first number = x – 1
and third number = x + 1
Now according to the condition,
(x)²= [(x + 1)² – (x – 1)²] + 60
⇒ x² = [x² + 2x + 1 – x² + 2x – 1 ] + 60
⇒ x² = 4x + 60
⇒ x² – 4x – 60 = 0
⇒ x² – 10x + 6x – 60 = 0
⇒ x (x – 10) + 6 (x – 10) = 0
⇒ (x – 10) (x + 6) = 0
Either x – 10 = 0 then x = 10
or x + 6 = 0 then x = -6. But it is not a natural number.
Middle number = 10
First number = 10 – 1 = 9
and third number = 10 + 1 = 11
Hence numbers are 9, 10, 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:
Middle number = p
then smallest number = p – 1
and greatest number = p + 1
Now, according to the condition.
3 (p + 1)² – (p – 1)² – p² = 67
⇒ 3 (p² + 2p + 1) – (p² – 2p + 1) – p² = 67
⇒ 3p² + 6p + 3 – p² + 2p – 1 – p² – 67 = 0
⇒ p² + 8p + 2 – 67 = 0
⇒ p² + 8p – 65 = 0
⇒ p² + 13p – 5p – 65 = 0
⇒ p (p + 13) – 5 (p + 13) = 0
⇒ (p + 13) (p – 5) = 0
Either p + 13 = 0, then p = -13 But it is not possible.
or p – 5 = 0, then p = 5
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:

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 first pipe can fill the cistern in = x hrs.
Second pipe will fill the cistern in = x – 3 hrs.

But it is not possible.
x = 15
First pipe can fill in 15 hrs.
and second pipe in 15 – 3 = 12 hrs.

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 A as the smaller part of the two parts. Find the number. (2010)
Solution:
Let larger part = y
and smaller part = x
x² + y² = 20 ….(i)
and y² = 8x ….(ii)
Substituting the value of y² in „
x² + 8x = 20
⇒ x² + 8x – 20 = 0
⇒ x² + 10x – 1x – 20 = 0 {-20 = 10 x (-2), 8 = 10 – 2}
⇒ x (x + 10) – 2(x + 10) = 0
⇒ (x + 10) (x – 2) = 0
Either x + 10 = 0, then x = – 10 which is not possible because it is negative
or x – 2 = 0, then x = 2
Smaller number = 2
and larger number = y² = 8x = 8 x 2 = 16
⇒ y² = 16 = (4)²
⇒ y = 4
Number = x + y = 2 + 4 = 6

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6A are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D.

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5C
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D

Solve each of the following equations :
Question 1.

Solution:

Question 2.
(2x + 3)² = 81
Solution:
(2x + 3)2 = 81
⇒ 4x² + 12x + 9 = 81
⇒ 4x² + 12x + 9 – 81 = 0
⇒ 4x² + 12x – 72 = 0
⇒ x² + 3x – 18 = 0 (Dividing by 4)
⇒ x² + 6x – 3x – 18 = 0
⇒ x (x + 6) – 3 (x + 6) = 0
⇒ (x + 6) (x – 3) = 0
Either x + 6 = 0, then x = -6
or x – 3 = 0, then x = 3
x = 3, – 6

Question 3.
a² x² – b² = 0
Solution:
a² x² – b² = 0
⇒ (ax)² – (b)² =0
⇒ (ax + b) (ax – b) =
Either ax + b = 0, then x = \(\frac { -b }{ a }\)
or ax – b = 0. then x = \(\frac { b }{ a }\)
x = \(\frac { b }{ a }\) , \(\frac { -b }{ a }\)

Question 4.

Solution:

Question 5.
x + \(\frac { 4 }{ x }\) = – 4; x ≠ 0
Solution:
x + \(\frac { 4 }{ x }\) = -4
⇒ x² + 4 = -4x
⇒ x² + 4x + 4 = 0
⇒ (x + 2)² = 0
⇒ x + 2 = 0
⇒ x = – 2

Question 6.
2x⁴ – 5x² + 3 = 0
Solution:
2x⁴ – 5x² + 3 = 0
⇒ 2(x²)² – 5x² + 3 = 0
⇒ 2(x²)² – 3x² – 2x² + 3 = 0
⇒ 2x⁴ – 3x² – 2x² + 3 = 0
⇒ x² (2x² – 3) – 1 (2x² – 3) = 0
⇒ (2x² – 3) (x² – 1) = 0

Question 7.
x⁴ – 2x² – 3 = 0
Solution:
x⁴ – 2x² – 3 = 0
⇒ (x²)² – 2x² – 3 = 0
⇒ (x²)² – 3x² + x² – 3 = 0
⇒ x² (x² – 3) + 1 (x² -3) = 0
⇒ (x² – 3) (x² + 1) = 0
Either x² – 3 = 0, then x² = 3 ⇒ x = √3
or x² + 1 = 0, then x² = – 1 In this case roots are not real
x = ±√3 or √3 , – √3

Question 8.

Solution:

Question 9.

Solution:

Question 10.

Solution:

Question 11.
(x² + 5x + 4)(x² + 5x + 6) = 120
Solution:
Let x² + 5x + 4 = y then x² + 5x + 6 = y + 2
Now (x² + 5x + 4) (x² + 5x + 6) = 120
⇒ y (y + 2) – 120 = 0
⇒ y² + 2y – 120 = 0
⇒ y² + 12y – 10y – 120 = 0
⇒ y (y + 12) – 10 (y + 12) = 0
⇒ (y + 12) (y – 10) = 0
Either y + 12 = 0, then y = – 12
or y – 10 = 0, then y = 10
(i) when y = -12, then x² + 5x + 4 = -12
⇒ x² + 5x + 4 + 12 = 0
⇒ x² + 5x + 16 = 0
Here a = 1, b = 5, c = 16
D = b² – 4ac = (5)² – 4 x 1 x 16 = 25 – 64 = -39
D < 0, then roots are not real
(ii) When y = 10, then x² + 5x + 4 = 10
⇒ x² + 5x + 4 – 10 = 0
⇒ x² + 5x – 6 = 0
⇒ x² + 6x – x – 6 = 0
⇒ x (x + 6) – 1 (x + 6) = 0
⇒ (x + 6) (x – 1) = 0
Either x + 6 = 0, then x = – 6
or x – 1 = 0, then x = 1
x = 1, -6

Question 12.
Solve each of the following equations, giving answer upto two decimal places:
(i) x² – 5x – 10 = 0 [2005]
(ii) 3x² – x – 7 = 0 [2004]
Solution:
(i) Given Equation is : x² – 5x – 10 = 0
On comparing with, ax² + bx + c = 0
a = 1, b = -5 , c = -10

Question 13.

Solution:

Question 14.
Solve:
(i) x² – 11x – 12 = 0; when x ∈ N
(ii) x² – 4x – 12 = 0; when x ∈ I
(iii) 2x² – 9x + 10 = 0; when x ∈ Q.
Solution:
(i) x² – 11x – 12 = 0
⇒ x² – 12x + x – 12 = 0
⇒ x (x – 12) + 1 (x – 12) = 0
⇒ (x – 12) (x + 1) = 0
Either x – 12 = 0, then x = 12
or x + 1 = 0, then x = -1
x ∈ N
x = 12
(ii) x² – 4x – 12 = 0
⇒ x² – 6x + 2x – 12 = 0
⇒ x (x – 6) + 2 (x – 6)=0
⇒ (x – 6) (x + 2) = 0
Either x – 6 = 0, then x = 6
or x + 2 = 0, then x = -2
x ∈ I
x = 6, -2
(iii) 2x² – 9x + 10 = 0
⇒ 2x² – 4x – 5x + 10 = 0
⇒ 2x (x – 2) – 5 (x – 2) = 0
⇒ (x – 2) (2x – 5) = 0
Either x – 2 = 0, then x = 2
or 2x – 5 = 0, then 2x = 5 ⇒ x = \(\frac { 1 }{ 2 }\)
x ∈ Q
x = 2, \(\frac { 5 }{ 2 }\) or 2, 2.5

Question 15.
Solve: (a + b)² x² – (a + b) x – 6 = 0, a + b ≠ 0.
Solution:
(a + b)² x² – (a + b) x – 6 = 0
Let (a + b) x = y, then y² – y – 6 = 0
⇒ y² – 3y + 2y – 6 = 0
⇒ y (y – 3) + 2 (y – 3) = 0
⇒ (y – 3) (y + 2) = 0
Either y – 3 = 0, then y = 3
or y + 2 = 0, then y = – 2
(i) If y = 3, then

Question 16.

Solution:


Either x + p = 0, then x = -p
or x + q = 0, then x = -q
Hence x = -p, -q

Question 17.

Solution:
(i) x (x + 1) + (x + 2) (x + 3) = 42
⇒ x² + x + x² + 3x + 2x + 6 – 42 = 0
⇒ 2x² + 6x – 36 = 0
⇒ x² + 3x – 18 = 0
⇒ x² + 6x – 3x – 18 = 0
⇒ x (x + 6) – 3(x + 6) = 0
⇒ (x + 6) (x – 3) = 0
Either x + 6 = 0, then x = -6
or x – 3 = 0, then x = 3
Hence x = 3, -6

Question 18.
For each equation, given below, find the value of ‘m’ so that the equation has equal roots. Also, find the solution of each equation:
(i) (m – 3) x² – 4x + 1 = 0
(ii) 3x² + 12x + (m + 7) = 0
(iii) x² – (m + 2) x + (m + 5) = 0
Solution:

Question 19.
Without solving the following quadratic equation, find the value of ‘p’ for which the roots are equal. px² – 4x + 3 = 0.
Solution:
px² – 4x + 3 = 0 …..(i)
Compare (i) with ax² + bx + c = 0
Here a = p, b = -4, c = 3
D = b² – 4ac = (-4)² – 4.p.(3) = 16 – 12p
As roots are equal, D = 0
16 – 12p = 0
⇒ \(\frac { 16 }{ 12 }\) = p
⇒ p = \(\frac { 4 }{ 3 }\)

Question 20.
Without solving the following quadratic equation, find the value of m for which the given equation has real and equal roots : x² + 2 (m – 1) x + (m + 5) = 0.
Solution:
x² + 2 (m – 1) x + (m + 5) = 0.
Here, a = 1, b = 2 (m – 1), c = m + 5
So, discriminant, D = b² – 4ac
= 4(m – 1)² – 4 x 1 (m + 5)
= 4m² + 4 – 8m – 4m – 20
= 4m² – 12m – 16
For real and equal roots D = 0
So, 4m² – 12m – 16 = 0
⇒ m² – 3m – 4 = 0 (Dividingby4)
⇒ m² – 4m + m – 4 = 0
⇒ m (m – 4) + 1 (m – 4) = 0
⇒ (m – 4) (m + 1) = 0
⇒ m = 4 or m = -1

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 4 Linear Inequations (In one variable) Ex 4B

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 4 Linear Inequations Ex 4B.

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 4 Linear Inequations Ex 4A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 4 Linear Inequations Ex 4B

Question 1.
Represent the following inequalities on real number lines:
(i) 2x – 1 < 5
(ii) 3x + 1 ≥ – 5
(iii) 2 (2x – 3) ≤ 6
(iv) -4 < x < 4
(v) -2 ≤ x < 5 (vi) 8 ≥ x > -3
(vii) -5 < x ≤ -1
Solution:

Question 2.
For each graph given below, write an inequation taking x as the variable :

Solution:

Question 3.
For the following inequations, graph the solution set on the real number line :
(i) – 4 ≤ 3x – 1 < 8
(ii) x – 1 < 3 – x ≤ 5
Solution:

Question 4.
Represent the solution of each of the following inequalities on the real number line :
(i) 4x – 1 > x + 11
(ii) 7 – x ≤ 2 – 6x
(iii) x + 3 ≤ 2x + 9
(iv) 2 – 3x > 1 – 5x
(v) 1 + x ≥ 5x – 11
(vi) \(\frac { 2x + 5 }{ 2 }\) > 3x – 3
Solution:

Question 5.
x ∈ {real numbers} and -1 < 3 – 2x ≤ 7, evaluate x and represent it on a number line.
Solution:

Question 6.
List the elements of the solution set of the inequation – 3 < x – 2 ≤ 9 -2x ; x ∈ N.
Solution:

Question 7.
Find the range of values of x which satisfies

Graph these values of x on the number line.
Solution:

Question 8.
Find the values of x, which satisfy the inequation:

Graph the solution on the number line. (2007)
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 real number line.
Solution:

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:
Here in, 5x – 3 ≤ 5 + 3x ≤ 4x + 2
⇒ 5x – 3 ≤ 5 + 3x and 5 + 3x ≤ 4x + 2
⇒ 5x – 3x ≤ 5 + 3 and 3x – 4x ≤ 2 – 5
⇒ 2x ≤ 8 and – x ≤ – 3
⇒ x ≤ 4 and x ≥ 3
Solution is 3 ≤ x ≤ 4
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:

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:
(i) 2x – 9 < 7 and 3x + 9 ≤ 25; x ∈ R.

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:

Question 14.
The diagram represents two inequations A and B on real number lines :

(i) Write down A and B in set builder notation.
(ii) Represent A ∩ B and A ∩ B’ on two different number lines.
Solution:

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:

Question 16.
Illustrate the set {x : -3 ≤ x < 0 or x > 2 ; x ∈ R} on a real number line.
Solution:

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:

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:

Question 19.
If P = {x : 7x – 4 > 5x + 2, x ∈ R} and Q = {x : x – 19 ≥ 1 – 3x , x ∈ R}: find the range of set P ∩ Q and represent it on a number line.
Solution:

Question 20.
Find the range of values of x, which satisfy:

Graph, in each of the following cases, the values of x on the different real number lines:
(i) x ∈ W
(ii) x ∈ Z
(iii) x ∈ R.
Solution:

Question 21.
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 an B is on two different number lines. Write down the elements of A ∩ B.
Solution:

Question 22.
Solve the following inequation and represent the solution set on the number line 2x – 5 ≤ 5x + 4< 11, where x ∈ I.
Solution:

Question 23.
Given that x ∈ I, solve the inequation and graph the solution on the number line:

Solution:

Question 24.
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 a number line. (2005)
Solution:

Question 25.
Find the set of values of x, satisfying:

Solution:

Question 26.

Solution:

Question 27.
Solve the inequation:

Graph the solution set on the number line.
Solution:

Question 28.
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 first positive integer = x
Then, second integer = x + 1
and third integer = x + 2
According to the condition,

Question 29.
Solve the given inequation and graph the solution on the number line.
2y – 3 < y + 1 < 4y + 7; y ∈ R (2008)
Solution:

Question 30.
Solve the inequation:
3z – 5 ≤ z + 3 < 5z – 9; z ∈ R.
Graph the solution set on the number line.
Solution:

Question 31.
Solve the following inequation and represent the solution set on the number line.

Solution:

Question 32.
Solve the following inequation and represent the solution set on the number line:

Solution:

Question 33.
Solve the following inequation, write the solution set and represent it on the number

Solution:

Question 34.

Solution:

Question 35.
Solve the following inequation and write the solution set:
13x – 5 < 15x + 4 < 7x + 12, x ∈ R
Represent the solution on a real number line. (2015)
Solution:

Question 36.
Solve the following inequation, write the solution set and represent it on the number line.
-3 (x – 7) ≥ 15 – 7x > \(\frac { x + 1 }{ 3 }\), x ∈ R. (2016)
Solution:

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 4 Linear Inequations Ex 4B are helpful to complete your math homework.

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 4 Linear Inequations (In one variable) Ex 4A

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 4 Linear Inequations Ex 4A.

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 4 Linear Inequations Ex 4A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 4 Linear Inequations Ex 4B

Question 1.
State, true or false :

Solution:

Question 2.
State whether the following statements are true or false:
(i) If 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 \(\frac { a }{ b }\) < \(\frac { b }{ c }\) (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) True
(ii) True
(iii) False
(iv) False
(v) True
(vi) False

Question 3.
If x ∈ N, find the solution set of inequations,
(i) 5x + 3 ≤ 2x + 18
(ii) 3x – 2 < 19 – 4x
Solution:

x = {1, 2}

Question 4.
If the replacement set is the set of whole numbers, Solve:
(i) x + 7 ≤ 11
(ii) 3x – 1 > 8
(iii) 8 – x > 5
(iv) 7 – 3x ≥ – \(\frac { 1 }{ 2 }\)
(v) x – \(\frac { 3 }{ 2 }\) < \(\frac { 3 }{ 2 }\) – x
(vi) 18 ≤ 3x – 2
Solution:

Question 5.
Solve the inequation :
3 – 2x ≥ x – 12 given that x ∈ N. [1987]
Solution:
3 – 2x ≥ x – 12
⇒ – 2x – x ≥ – 12 – 3
⇒ – 3x ≥ -15
⇒ – x ≥ – 5
⇒ x ≤ 5
Solution Set= {1, 2, 3, 4, 5} or {x ∈ N : x ≤ 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
⇒ 4x ≥ 9
x ≥ \(\frac { 9 }{ 4 }\)
(i) The smallest value of x, when x is a real number \(\frac { 9 }{ 4 }\) or 2.25
(ii) The smallest value of x, when x is an integer 3.

Question 7.
If the replacement set is the set of real numbers, solve:

Solution:

Question 8.
Find the smallest value of x for which 5 – 2x < 5\(\frac { 1 }{ 2 }\) – \(\frac { 5 }{ 3 }\) x, where x is an integer.
Solution:

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\(\frac { 2 }{ 3 }\)
x ∈ W and value of x is largest x = 3

Question 10.
Solve the inequation:
12 + 1\(\frac { 5 }{ 6 }\) x ≤ 5 + 3x and x ∈ R. (1999)
Solution:

Question 11.
Given x ∈ (integers), find the solution set of: -5 ≤ 2x – 3 < x + 2.
Solution:
-5 ≤ 2x – 3 < x + 2
(i) -5 ≤ 2x – 3
⇒ -2x ≤ -3 + 5
⇒ -2x ≤ 2
⇒ x ≤ -1
⇒ -1 ≤ x
(ii) 2x – 3 < x + 2
⇒ 2x – x < 2 + 3
⇒ x < 5
From (i) and (ii),
-1 ≤ x < 5
x = {-1, 0, 1, 2, 3, 4}

Question 12.
Given x ∈ (whole numbers), find the solution set of: -1 ≤ 3 + 4x < 23.
Solution:
-1 ≤ 3 + 4x < 23
(i) -1 ≤ 3 + 4x
⇒ -1 – 3 ≤ 4x
⇒ -4 < 4x
⇒ -1 ≤ x
(ii) 3 + 4x < 23
⇒ 4x < 23 – 3
⇒ 4x < 20
⇒ x < 5
From (i) and (ii)
-1 ≤ x < 5 and x ∈ W
Solution set = {0, 1, 2, 3, 4}

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 4 Linear Inequations Ex 4A are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.