OP Malhotra Linear Inequations in One Variable S.Chand Class-10 ICSE Maths Solutions Ch-4. We Provide Step by Step Answer of Exe-4 Banking with Self Evaluation and Revision of S Chand OP Malhotra Maths . Visit official Website CISCE for detail information about ICSE Board Class-10.
OP Malhotra Linear Inequations in One Variable S.Chand Class-10 ICSE Maths Solutions Ch-4
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OP Malhotra Linear Inequations in One Variable S.Chand Class-10 ICSE Maths Solutions Ch-4
Question. 1:
On a bargain counter, the shopkeeper puts labels on various goods showing their prices, Rs. P, where P is real numbers. Write a mathematical sentences for each of the following labels
(a) more than Rs. 7.50
(b) not less than Rs. 10
(c) not more than Rs. 22
(d) less than Rs. 11
Answer:
Since the price on label is Rs P, then
(a) more than Rs. 7.50 ⇒ P > 7.50
(b) not less than Rs. 10 ⇒ P ≮ 10 or P ≥ 10
(c) not more than Rs. 22 ⇒ P ≯ 22 or P ≤ 10
(d) less than Rs. 11 ⇒ P < 11
Question 2
You are given the following numbers : -2.6, 5.1, -3, 0.4, 1.2, -3.1, 4.7
Fill in the blanks
(a) A = {x : x ≥ -3 } = {……..}
(b) B = {x : x ≤ 1 } = {……..}
Answer:
You are given the following numbers : -2.6, 5.1, -3, 0.4, 1.2, -3.1, 4.7
It mean, you are given a replacement set as U = {-3.1, -3, -2.6, 0.4, 1.2, 4.7, 5.1}.
(a) Solution of A = {x : x ≥ -3 }
= {-3, -2.6, 0.4, 1.2, 4.7, 5.1}
(b) Solution of B = {x : x ≤ 1 }
= {-3.1, -3, -2.6, 0.4}
Question 3
If the replacement set is {-2, -1, +1, +2, +4, +5, +9}, what is the solution set of each of the following mathematical senetences.
(a) x + 3/2 > 5/2
(b) x – 4 = -3
(c) 2x – 5 ≥ 10
(d) 3y/2 ≤ 5/2
(e) 4x2 = 16.
Answer:
Since the replacement set is {-2, -1, +1, +2, +4, +5, +9}, then
(a) Solution set of x + 3/2 > 5/2
⇒ x > 5/2 – 3/2
⇒ x > 1 is {+2, +4, +5, +9}.
(b) Solution set of x – 4 = -3
⇒ x = -3 + 4
⇒ x = 1 is {+1}.
(c) Solution set of 2x – 5 ≥ 10
⇒ 2x ≥ 15
⇒ x ≥ 7.5 is {+9}.
(d) Solution set of 3y/2 ≤ 5/2
⇒ 3y ≤ 5
⇒ y ≤ 5/3
⇒ y ≤ 1.66 is {-2, -1, +1}.
(e) Solution set of 4x2 = 16
⇒ x2 = 4
⇒ x = +2 or -2 is {-2, +2}.
Question 4
List the solution set of 30 – 4(2x – 1) < 30, given that x is a positive integer.
Answer:
Since the replacement set is a positive integer, then
Solution set of
30 – 4(2x – 1) < 30
⇒ -4(2x – 1) < 30 – 30
⇒ -4(2x – 1) < 0
⇒ (2x – 1) > 0
⇒ x > 1/2
⇒ x > 0.5
is x = {1, 2, 3, 4, 5,…}
Question 5 OP Malhotra Linear Inequations in One Variable
If the replacement set is {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, what is the solution set of each of the following mathematical senetences.
(a) x + 4/3 = 7/3
(b) 2x + 1 < 3
(c) x – 6 > 10 – 6
(d) x + 5 = 20
(e) 2x + 3 ≥ 17
Answer:
Since the replacement set is {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, then
(a) Solution set of x + 4/3 = 7/3
⇒ x = 3/3
⇒ x = 1 is {1}.
(b) Solution set of 2x + 1 < 3
⇒ 2x < 2
⇒ x < 1 is {0}.
(c) Solution set of x – 6 > 10 – 6
⇒ x > 10 is φ.
(d) Solution set of x + 5 = 20
⇒ x = 15 is φ.
(e) Solution set of 2x + 3 ≥ 17
⇒ 2x ≥ 14
⇒ x ≥ 7 is {7, 8, 9}.
Question 6
(a) x ∈ {2, 4, 6, 9} and y ∈ {4, 6, 18, 27, 54}. Form all ordered pairs (x, y) such that x is a factor of y and x < y.
(b) Find the truth set of the inequality
x > y + 2, where (x, y) ∈ {(1, 2), (2, 3), (5, 1), (7, 3), (5, 6)}
Answer:
(a) x ∈ {2, 4, 6, 9} and y ∈ {4, 6, 18, 27, 54}.
(x, y) = x is a factor of y and x < y.
2 is a factor of 4, 6, 18, 54 and 2 < 4, 2 < 6, 2 < 18, 2 < 54, then (x, y) forms the ordered pairs as (2, 4), (2, 6), (2, 18), (2, 54).
4 is a factor of 4 but 4 = 4, then (4, 4) is not in (x, y).
6 is a factor of 18, 54 and 6 < 18, 6 < 54, then (x, y) forms the ordered pairs as (6, 18), (6, 54).
9 is a factor of 18, 27, 54 and 9 < 18, 9 < 27, 9 < 54 then (x, y) forms the ordered pairs as (9, 18), (9, 27), (9, 54).
Thus, the solution set of
(x, y) is
{(2, 4), (2, 6), (2, 18), (2, 54), (6, 18), (6, 54), (9, 18), (9, 27), (9, 54)}.
(b) Given inequality is x > y + 2. It means x is greater than y + 2.
In {(1, 2), (2, 3), (5, 1), (7, 3), (5, 6)},
we see that (5, 1) and (7, 3) are two pairs which satisfy the given inequality.
Thus, the solution set of (x, y) is {(5, 1), (7, 3)}.
Question 7
Find out the truth set of the following open sentences replacement sets are given
(i) 5/x > 7 ; {1, 2}
(ii) 5/x > 2 ; {1, 2, 3, 4, 5, 6}
(iii) x2 = 9 ; {-3, -2, -1, 1, 2, 3}
(iv) x + 1/x = 2 ; {0, 1, 2, 3}
(v) 3x2 < 2x ; {-4, -3, -2, -1, 0, 1, 2, 3, 4}
(vi) 2(x – 3) < 1 ; {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
Answer:
(i) Given replacement set is {1, 2}.
Now given inequality is 5/x > 7 ⇒ x < 5/7 ⇒ x < 0.71
Thus, the solution set of the given inequality is x ∈ φ.
(ii) Given replacement set is {1, 2, 3, 4, 5, 6}.
Now given inequality is 5/x > 2 ⇒ x < 5/2 ⇒ x < 2.5
Thus, the solution set of the given inequality is x ∈ {1, 2}.
(iii) Given replacement set is {-3, -2, -1, 1, 2, 3}.
Now given equation is x2 = 9 ⇒ x = 3 or -3.
Thus, the solution set of the given equation is x ∈ {-3, 3}.
(iv) Given replacement set is {0, 1, 2, 3}.
Now given equation is x + 1/x = 2
⇒ (x2 + 1) = 2x
⇒ (x2 -2x + 1) = 0
⇒ (x – 1)2 = 0
⇒ (x – 1) = 0
⇒ x = 1.
Thus, the solution set of the given equation is x ∈ {1}.
(v) Given replacement set is {-4, -3, -2, -1, 0, 1, 2, 3, 4}.
Now given inequality is 3x2 < 2x
⇒ 3x2 – 2x < 0
⇒ x(3x – 2) < 0
⇒ 0 < x < 2/3
⇒ 0 < x < 0.66
Thus, the solution set of the given inequality is x ∈ φ.
(vi) Given replacement set is {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.
Now given inequality is 2(x – 3) < 1
⇒ x – 3 < 0.5
⇒ x < 3.5.
Thus, the solution set of the given inequality is
x ∈ {1, 2, 3}.
Question 8 OP Malhotra Linear Inequations in One Variable
Statement: The sum of the lengths of any two sides of a triangle is always greater than the length of its third side.
Let x, x+1, x+2 be the lengths of the three sides of a triangle.
(i) Write down the three inequations in x, each of which represents the given statement.
(ii) List the set of possible values of x which satisfy all the three inequations obtained in your answer to part (i) above, given that x is an integer.
Answer:
The lengths of the three sides of a triangle are x, x+1, x+2.
(i)
Since the sum of the lengths of any two sides of a triangle is always greater than the length of its third side, we have three inequality as below.
x + (x + 1) > (x + 2)
⇒ 2x + 1 > x + 2
⇒ x > 1 …(1)
x + (x + 2) > (x + 1)
⇒ 2x + 2 > x + 1
⇒ x > -1 …(2)
(x + 1) + (x + 2) > x
⇒ 2x + 3 > x
⇒ x > -3 …(3)
On combining (1), (2), (3), we get x > 1.
(ii) Since replacement set of x is an integer, means x ∈ {…, -3, -2, -1, 0, 1, 2, 3, …}.
The solution set of x > 1 is x ∈ { 2, 3, 4, …}.
Question 9
Answer True or False
(a) If x + 10 = y + 14, then x > y
(b) |-4| – 4 = 8
(c) If 10 – x > 3, then x < 7
(d) If p = q + 2, then p > q
(e) If a and b are two negative integers such that a < b, then 1/a < 1/b
(f) 3 ∈ {x : 3x – 2 ≥ 5}
Answer:
(a) x + 10 = y + 14
⇒ x – y = 14 – 10
⇒ x – y = 4
⇒ x – y > 0
⇒ x > y.
Thus, the given statement “If x + 10 = y + 14, then x > y” is TRUE.
(b) |-4| – 4 = 4 – 4 = 0
Thus, the given statement “|-4| – 4 = 8” is FALSE.
(c) 10 – x > 3
⇒ 10 – 3 > x
⇒ 7 > x
⇒ x > 7
Thus, the given statement “If 10 – x > 3, then x < 7” is TRUE.
(d) p = q + 2
⇒ p – q = 2
⇒ p – q > 0
⇒ p > q
Thus, the given statement “If p = q + 2, then p > q” is TRUE.
(e) a < b
⇒ 1/a > 1/b
Thus, the given statement “If a and b are two negative integers such that a < b, then 1/a < 1/b” is FALSE.
(f) 3x – 2 ≥ 5
⇒ 3x ≥ 7 ⇒ x ≥ 7/3
⇒ x ≥ 2.3
Thus, the given statement “3 ∈ {x : 3x – 2 ≥ 5}” is TRUE.
Question 10
Find the solution of the inequation 2 ≤ 2p – 3 ≤ 5. Hence graph the solution set on the number line.
Answer:
2 ≤ 2p – 3 ≤ 5
⇒ 5 ≤ 2p ≤ 8
⇒ 5/2 ≤ p ≤ 4
⇒ 2.5 ≤ p ≤ 4
Question 11
If x is a negative integer, find the solution set of 2/3 + (x + 1)/3 > 0.
Answer:
2/3 + (x + 1)/3 > 0
⇒ (x + 1)/3 > -2/3
⇒ x + 1 > -2
⇒ x > -3 …(1)
Since the replacement set of x is negative integer, it means x ∈ {…, -5, -4, -3, -2, -1}.
From (1), x ∈ {-2, -1}.
Question 12
Write open mathematical sentences, using x for the variable whose graphs would be
Answer:
(i)
From the graph, we see that the shaded arrow is left of -2 (included) and all the real numbers are taken up to -2.
Thus, required mathematical sentence is
{x : x ≤ -2 and x ∈ R}.
(ii)
From the graph, we see that the shaded arrow is left of 4 (included) and all the real numbers are taken up to 4.
Thus, required mathematical sentence is
{x : x ≤ 4 and x ∈ R}.
(iii)
From the graph, we see that the solid circles are at 4 and 5 and only natural numbers are taken.
Thus, required mathematical sentence is
{x : 4 ≤ x ≤ 5 and x ∈ N}.
(iv)
From the graph, we see that the solid circles are at 1, 3 and 5 and only natural numbers are taken.
Thus, required mathematical sentence is
{x : 1 ≤ x ≤ 5 and x ∈ N}.
(v)
From the graph, we see that the shaded arrow is right of -2 (excluded) and all the real numbers are ahead of -2.
Thus, required mathematical sentence is
{x : x > -2 and x ∈ R}.
Question 13 OP Malhotra Linear Inequations in One Variable
Answer TRUE or FALSE:
(i) If 2 – x < 0, then x > 2.
(ii) The graph of the inequations y ≤ 2x includes the origin.
Answer:
(i) 2 – x < 0 ⇒ 2 < x.
Thus, the given statement is TRUE.
(ii) On putting x = 0 and y = 0 in y ≤ 2x, we get that 0 = 0, which is true.
Thus, the given statement is TRUE.
Question 14
If 25 – 4x ≤ 16, then find
(i) the smallest value of x when x is real number.
(ii) the smallest values of x when x is an integer.
Answer:
25 – 4x ≤ 16 ⇒ 25 – 16 ≤ 4x ⇒ 9 ≤ 4x ⇒ x ≥ 9/4.
(i) Thus, the smallest value of x is 9/4 = 2 (1/4.)
(ii) The smallest integer greater than x ≥ 9/4 is 3.
Question 15
Given x ∈ {1, 2, 3, 4, 5, 6, 7, 9}, find the values of x for which -3 < 2x – 1< x + 4
Answer:
Given replacement set is x ∈ {1, 2, 3, 4, 5, 6, 7, 9}.
Given inequation is -3 < 2x – 1 < x + 4.
On solving -3 < 2x – 1 ⇒ 2x > -3 + 1
⇒ 2x > -2
⇒ x > -1 … (1)
On solving 2x – 1 < x + 4
⇒ 2x – x < 1 + 4
⇒ x < 5 …(2)
From (1) and (2), we get -1 < x < 5.
Thus, the solution set of the given inequation is x ∈ {1, 2, 3, 4}.
Question 16
Solve the inequality 2x – 10 < 3x – 15
Answer:
Given inequality is 2x – 10 < 3x – 15
⇒ 2x – 3x < 10 – 15
⇒ -x < -5
⇒ x > 5
Question 17 OP Malhotra Linear Inequations in One Variable
Solve the inequation 3 – 2x ≥ x – 12, given that x ∈ N.
Answer:
Given inequation is 3 – 2x ≥ x – 12
⇒ -x – 2x ≥ -3 – 12
⇒ -3x ≥ -15
⇒ x ≤ 5
Since the replacement of x is natural number, then the solution set of the given inequation is
x ∈ {1, 2, 3, 4, 5}.
Question 18
x ∈ {real numbers} and -1 < 3 – 2x ≤ 7, evaluate x and represent it on number line.
Answer:
Given inequation is -1 < 3 – 2x ≤ 7
From -1 < 3 – 2x,
⇒ -1 < 3 – 2x ⇒ -1 – 3 < – 2x
⇒ -4 < -2x
⇒ 2x < 4
⇒ x < 2 …(1)
and 3 – 2x ≤ 7
⇒ 3 – 7 ≤ 2x
⇒ -4 ≤ 2x
⇒ -2 ≤ x …(2)
On combining (1) and (2), we get -2 ≤ x < 2.
The graph of this inequation is
Question 19
Find the range of values of x which satisfies and graph these values of x on number line.
Answer:
Given inequation is
On simplifying, we get
Now the graph of these values of x is as below.
Question 20
Solution and graph the solution set of
(a) (b)
Answer:
(a) Given inequation is
On solving first inequation, we get 6 ≥ 2 – x
⇒ x ≥ 2 – 6
⇒ x ≥ -4 … (1)
On solving second inequation, we get x/3 + 2 < 3
⇒ x/3 < 3 – 2
⇒ x/3 < 1
⇒ x < 3 … (2)
On combining (1) and (2), we get -4 ≤ x < 3.
The graph of this inequation is
(b) Given inequation is
On solving first inequation, we get
x/2 < (6 – x)/4
⇒ 2x < 6 – x
⇒ 3x < 6
⇒ x < 2… (1)
On solving second inequation, we get
(2 – x)/6 < (7 – x)/9
⇒ 3(2 – x) < 2(7 – x)
⇒ 6 – 3x < 14 – 2x
⇒ x > -8 … (2)
On combining (1) and (2), we get -8 < x < 2.
The graph of this inequation is
Question 21
Find the range of values of x which satisfy
Graph these values of x on real number line.
Answer:
(1.) Given inequation is (2.) On simplifying,we get
Graph of these values of x on number line is as below.
Question 22 OP Malhotra Linear Inequations in One Variable
Write down the range of real values of x for which the inequation x > 3 and -2 ≤ x < 5 are both true.
Answer:
Graphing the given inequations x > 3 and -2 ≤ x < 5 on same number line, we get
Thus, we get 3 < x < 5.
Question 23
Solve and graph the solution set of 3x – 4 > 11 or 5 – 2x ≥ 7; x ∈ R.
Answer:
On solving the first inequation, we get 3x – 4 > 11
⇒ 3x > 15
⇒ x > 5 …(1)
On solving the second inequation, we get 5 – 2x ≥ 7
⇒ – 2x ≥ 2
⇒ x ≤ 1 …(2)
On combining (1) and (2), we get x ≤ 1 or x > 5.
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OP Malhotra Linear Inequations in One Variable S.Chand Class-10 ICSE Maths Solutions Ch-4
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