ML Aggarwal ICSE Solutions Class 10 Math 4th Chapter Linear Inequations
Class 10 Chapter 4 Linear inequationsExercise:
(1) Solve the inequation (3x -11) < 3, where x∈ (1, 2, 3,….10}. Also represent its solution on a number line.
Solution:
Given in equation is (3x -11) <3,
Where x∈ {1, 2, 3, …….., 10}
(3x – 11) <3 = 3x < (3 +11)
3x < 14
x < 14/3
But x∈ = {1,2,3, ……10}
Hence, the solution set is (1, 2, 3, 4).
And, we can represent solution set on a number line as given below.
(2) Solve: 2 (x-3) < 1, where x∈ {1, 2, 3, —–10}
Solution:
Given inequation is 2 (x-3) < 1,
Where x∈ {1, 2, 3, —–10}
= 2 (x-3) < 1 = (2x – 6) < 1
2x < (1+6)
2x < 7
x < 7/2
But, x∈ {1,2,3, —–10}
Hence, the set of solution is {1, 2, 3}.
(3) Solve: (5 – 4x) > (2 – 3x), x∈W Also represent its solution on the number line.
Solution:
Given in equation is (5-4x) > (2-3x)
Where, x∈W= {0, 1, 2, —-}
Then, (5-4x) > (2-3x) = (5-2) > (4x-3x)
3 > (x)
x <3
But, here x∈W = {0, 1, 2, ….}.
Hence, the solution set is {0, 1, 2}.
We can represent the solution set on number line as below:
(4) List the solution set of 30 – 4 (2x-1) < 30, where x is a positive integer.
Solution:
Given inequation is 30-4 (2x-1) < 30,
Where x is a positive integer.
= 30-4 (2x-1) < 30
(30 – 8x + 4) < 30
– 8x < (30 – 30 – 4)
– 8x < – 4 ⇒ 8x => 8x > 4
x > 4/8
But, x is a positive integer.
Then, solution set is {1, 2, 3, 4, ….]
(5) Solve: 2 (x – 2) < (3x – 2), where x∈ {-3, -2, -1, 0, 1, 2, 3}
Solution:
Given equation is 2 (x-2) < (3x-2),
Where x∈ {-3, -2, -1, 0, 1, 2, 3}
Then 2 (x – 2) < (3x – 2)= (2x – 4) < (3x – 2)
(-4 + 2) < (3x – 2x)
(- 2) < x
x > – 2
But, here x∈ {-3, -2, -1, 0, 1, 2, 3}
Hence, the required solution set is {-1, 0, 1, 2, 3}.
(6) If x is a negative integer, find the solution set of 2/3 + 1/3 (x+1) > 0.
Solution:
Given in equation is 2/3 + 1/3 +(x+1) > 0,
Where x is a negative integer.
2 + (x + 1) > 0
(x + 3) > 0
X > -3
But x is a negative integer here.
Hence, the required set solution is (-2, -1).
(7) Solve: x – 3 (2 + x) > 2 (3x – 1), where x∈ {-3, -2, -1, 0, 1, 2, 3} Also represent its solution on the number line.
Solution:
Given inequation is x – 3 (2 + x) > 2 (3x – 1)
Where, x∈ {-3, -2, -1, 0, 1, 2, 3}
Then, x – 3 (2 + x)> 2 (3x-1)
x- (6+3x) > (6x-2)
(x-6 -3x) > (6x-2)
(-2x-6) > (6x-2)
(-6+2) > (6x+2x)
-4 > 8x
-4/8 >x
x < -1/2
But, here, x∈ {-3, -2, -1, 0, 1, 2}
Hence, the required set solution is {-3, -2, -1}.
We can represent the solution set {-3, -2, -1} on number line as follows:
(8) Solve: (x-3) < (2x-1), where x∈ {1, 2, 3, 4, 5, 6, 7, 9}
Solution:
Given that, (x-3) < (2x-1)
Where x∈ {1, 2, 3, 4, 5, 6, 7, 9}
Then, (x-3) < (2x-1)
(x – 3) < (2x – 1)
(x – 2x) < (-1 +3)
(-x) <2
X > – 2
But, here x∈ {1, 2, 3, 4, 5, 6, 7, 9}
Hence, the required solution set is {1, 2, 3, 4, 5, 6, 7, 9).
(9) List the solution set of the in equation (1/2 + 8x) > (5x – 3/2), where x ∈ Z.
Solution:
Given inequation is, (1/2 + 8x) > (5x – 3/2)
Where x ∈ Z.
Then, (1/2 + 8x) > (5x – 3/2)
(8x – 5x) > (-3/2 – 1/2)
(3x) > (-4/2 = -2)
3x > -2
X > -2/3
But, x ∈ Z
Hence, the required set of solution is {0, 1, 2, 3, 4, ….}
(10) List the solution set of (11-2x/5) ≥ (9-3x/8) + (3/4), x ∈ N
Solution:
Given that, (11-2x/5) ≥ (9-3x/8 + 3/4)
(11-2x/5) ≥ (9-3x+6/8)
8 (11-2x) ≥ 5 (9 – 3x + 6)
(88 – 16x) ≥ (45 – 15x + 30)
(15x – 16x) ≥ (45 + 30 – 88)
(x) ≥ (75 – 88)
x ≥ – 13, N = {1, 2, 3, 4, —–}
x ≥ – 13
Or, x ≤ 13 But x ∈ N
Hence, the required solution set is {1, 2, 3, 4, 5, …..13}
(11) Find the values of x, which satisfy the inequation:
(-2 ≤ ½ -2x/3 ≤ 1 5/6), x∈N
Graph the solution on the number line.
Solution:
Given that, (-2 ≤ ½ -2x/3 ≤ 1 5/6), x∈N.
= -2 -1/2 ≤ 1/2 -2x/3 -1/2 < 11/6 –1/2
By subtracting 1/2 on both sides of inequality.
-5/2 ≤ -2x/3 ≤ 8/6
-15 ≤ – 4x ≤ 48
= 15 ≥ 4× ≥ -8
= 15/4 ≥ x ≥ -8/4
15/4 ≥ x ≥ -2
But x∈N = {1, 2, 3, 4, 5 ……}
Hence, the required set of solution is {1, 2, 3}
We can represent the set of solution {1, 2, 3) on number line as below.
(12) If X∈W, find the solution set of 3/5 x – (2x-1)/3 > 1. Also graph the solution set on the number line, if possible.
Solution:
Given that, 3/15 x – (2x-1)/3 > 1, where x∈W
= 9x – 5 (2x – 1) > I5
= 9x – 10x + 5 > I5
= – x > I5 – 5
= -x > 10
x < – 10
But x∈W = {0, 1, 2, 3, ……}
Hence, the required set of solution is empty set (⌀). And hence, we cannot represent it on the number line also.
(14) Given that x∈I, solve the inequation and graph the solution on the number line: 3 ≥ (x-4/2) + x/3 ≥ 2 (2004)
Solution:
Given that, 3 ≥ (x-4/2) + x/3 ≥ 2 (2004), X∈I
= 3 ≥ (3x-12+2x)/6 ≥ 2
3 ≥ (5x-12)/6
= 18 ≥ (5x – 12)
That means, (5x – 12) ≤ 18
5x ≤ (18+12)
5x ≤ 30
x ≤ 6
But X∈I
Hence, required set of solution is {…, -1, 0, 1, 2, 3, 4, 5, 6}
(15) Solve: 1 ≥ 15 – 7 x > 2x-27, x∈N
Solution:
Given that, 1 ≥ (15 – 7x) > (2x – 27) Where x∈N
= 1 ≥ (I5 – 7x) and (15-7x) ≥ (2x-27)
= 7x ≥ (15 – 1) and (-7x – 2x) > (-27 -15)
= 7x ≥ 14 and -9x > -42
= x ≥ 2 and – x > – 42/9
=2 ≤ x and x < 42/9 = 14/3
Thus, here, 2 ≤ x < 14/3
But x∈N
Hence, the required set of solution is {2, 3, 4}
(16) If x∈Z, Solve: (2+4x) < (2x – 5) < 3x. Also, represent its solution on the number line.
Solution:
Given that, (2+4x) < (2x-5) ≤ 3x, where x∈Z
= (2+4x) < (2x-5) and (2x-5) ≤ 3x
= (4x-2x) < (-5 -2) and (2x – 3x) ≤ 5
= 2x < -7 and (-x) ≤ 5
= x < -7/2 and x ≥ -5
= -5 ≤ x < -7/2
But x∈Z
Hence, the required set of solution is {-5, -4}
Thus, we can represent the set of solution {-5, -4} on the number line as bellow.
(17) Solve (4x-10)/3 ≤ (5x-7)/2, where x∈R and represent the solution set on the number line.
Solution:
Given that, (4x-10)/3 ≤ (5x-7)/2, wher x∈R
= 2 (4x – 10) ≤ 3 (5x – 7)
= (8x – 20) ≤ (15x – 21)
= (8x – 15x) ≤ (-21 + 20)
= -7× ≤ -1
= -x ≤ -1/ 7
= X > 1/7
But x∈R
Thus, the required solution set is {x: x∈R, x > 1/7}
We can represent the solution set on number line as below.
(18) Solve: 3x/5 – (2x – 1)/3 > 1, x∈R and represent the solution set on number line.
Solution:
Given that, 3x/5 – (2x-1)/3 > 1, x∈R
= 3 (3x)-5(2x-1) > 15
= 9x-10x+5 > I5
= -x+5> 15
= -x + 5 > 15
= – x > 15 – 5
= x > 10
= x < – 10
But x ∈R
Hence, the required solution set is {x: x∈R, X < -10}
We can represent the solution set on number line as below.
(19) Given that x∈R, solve the following inequation & graph the solution on the number line: -1 ≤ (3+4x) <23.
Solution:
Given that, -1 ≤ (3+4x) < 23, where X∈R
= (-1-3) ≤ 4x < (23-3)
= -4 ≤ 4x < 20
= – 4 ≤ 4x and 4x < 20
= -1 ≤ x and x < 5
= -1 ≤ x < 5
But x∈R
Hence, the required set solution is {x∈R: -1 ≤ x < 5}
We can present the required solution set on number line as bellow:
(20) Solve the following inequation & graph the solution on the number line.
-2 2/3 ≤ (x+1/3) < (3+1/3), where x∈R
Solution:
Given inequation is -2 2/3 ≤ (x+1/3) < (3+1/3), where x∈R
Throughout equation multiply by 3,
We get -8/3 ≤ (x+1/3) < (10/3) —– (1)
(1) × 3 = -8 ≤ (3x+1) < 10
(-8-1) ≤ 3x < (10-1)
-9 ≤ 3x < 9
-3 ≤ x < 3
But x∈R
Thus, the solution set found is {x: x∈R, -3 ≤ x <3}
i.e. (-3, -2, -1, 0, 1, 2)
(21) Solve the following in equation and represent the solution set on the number line. – 3 < (- 1/2 – 2x/3) ≤ 5/6, x∈R
Solution:
-3 < (- 1/2 – 2x/3) ≤ 5/6 is the given inequality, where x∈R
-3 < (-1/2 – 2x/3) and (-1/2 – 2x/3) ≤ 5/6
– (1/2 + 2x/3) > -3 and -2x/-3 ≤ (5/6 + 1/2)
-2x/3 > -3 + 1/2 and -2x/3 ≤ (5-3)/6
= -2x/3 > -5/2 and -2x/3 ≤ 8/6
= 2x/3 < 5/2 and 2/3 x ≥ – 8/6
x < (5/2 × 3/2) and => x ≥ (8/6 × 3/2)
x < 15/4 and => x ≥ -2
Thus, x < 15/4 and x ≥ -2
= – 2 ≤ x < 15/4
But x∈R
Thus, the required solution set is {x: X∈R, -2 ≤ x < 15/4}
i.e., {-2, -1, 0, 1, 2, 3}
The required solution set can be represented on the number line as follows:
(22) Solve the following inequation, write the solution set and represent it on number line -3 (x-7) ≥ (15 – 7x) > (x+1/3), n∈R
Solution:
Given inequation is, -3 (x – 7) ≥ (15 – 7x) > (x+1/3), where n∈R
-3 (x-7) ≥ (15 – 7x) and (15 – 7x) > (x+1/3)
(-3x + 21) ≥ (15 – 7x) and 3 (15 – 7x) > (x+1)
(-3x + 7x) ≥ (15 – 21) and (45 – 21x) > (x+1)
= 4x ≥ – 6 and (45 – 1)> (x + 21x)
= x ≥ – 6/4 and 44 > 22x
= x ≥ – 3/2 and 2 > x
-3/2 ≤ x
Thus, -3/2 ≤ x < 2
But x∈R
Thus, the required solution set is {x: X∈R, -3/2 ≤ x < 2}
The required solution set can be represented on the number line as follows:
(23) Solve the following in equation, write the solution set and represent it on the real number line. (-2 + 10x) ≤ (13x + 10) < (24 + 10x), X∈Z
Solution:
Given inequation is (-2 + 10x) ≤ (13+10) < (24 + 10)X∈Z
(-2+10x) ≤ (13x + 10) and (13x + 10) < (24 + 10x)
(-2-10) ≤ (13x – 10x) and (13x – 10x) < (24-10)
-12 ≤ 3X and 3x < 14
– 4 ≤ x and x < 14/3
X ≥ – 4 and x < 14/3
Thus, – 4 ≤ x < 14/3 But x∈Z
Hence, the required solution set is {x: – 4 ≤ x < 14/3, X∈Z} or [-4, 14/3]
The required solution set can be represented on the number line as follows:
(24) Solve the inequation (2x-5) ≤ (5x + 4) < 11, where X∈I. Also represent the solution set on the number line.
Solution:
Given inequation is (2x-5) ≤ (5x+4) < 11, where X∈I
(2x – 5) ≤ (5x + 4) and (5x + 4) < 11
(2x – 5x) ≤ (4+5) and 5x < 11 – 4
-3x ≤ 9 and 5x < 7
-x < 3 and x < 7/5
X>-3
Thus, -8 < x < 7/5 But x∈I
Thus, the required set of solution is {x: -3 < x < 7/5, X∈I}
The required set can be represented on the number line as follows:
(25) If X∈I, A is the solution set of 2 (x – 1) < (3x – 1) and B is the solution set of (4x – 3) ≤ (8 + x), find A∩B.
Solution:
Given inequation is
2 (x-1) < (3x-1) and (4x-3) ≤ (8+x)
(2x – 2) < (3x – 1)and (4x – x) ≤ (8+3)
(2x – 3x) < (2 – 1)and 3x ≤ 11
– x < 1 and x ≤ 11/3
X > – 1
Thus, -1 < X ≤ 11/3 where, X∈I
Thus, the required two solution sets are as given below:
A = {0, 1, 2, 3, …} and B = {3, 2, 1, 0, -1, …}
Hence, A∩B = {0, 1, 2, 3}
(26) If P is the solution set of (-3x + 4) < (2x – 3), X∈N and Q is the solution set of (4x – 5) < 12, X∈W find
(i) PNQ
(ii) Q – P
Solution:
Given inequations are,
(-3x + 4) < (2x – 3) and (4x – 5) < 12
(- 3x – 2x) < (-3 – 4) and 4x < (12 + 5)
(-5x) < (-7) and 4x < 17
– 5x < -7 and x < 17/4
-5x < -7/5 and x < 17/4
= x > 7/5
But x∈N
Thus, required solution set is P = {0, 3, 4, 5, ….}
(i) P∩Q = {2, 3, 4}
(ii) Q – P = {1, 0}
But x∈W
Thus, required solution set is Q = {4, 3, 2, 1, 0}
(27) 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:
Given set are A = {x: (x-5) > (7x+3), x∈R} and
B = {x: (18x – 9) ≥ (15 + 12x), X∈R}
For Set A | For Set B |
(11x – 5) > (7x + 3) | (18x – 9) ≥ (15 + 12x) |
(11x – 7x) > (3 + 5) | (18x – 12x) ≥ (15 + 9) |
4x > 8 | 6x ≥ 24 |
x > 2, x∈R | x ≥ 4, x∈R |
Thus, the required set solution is found to be
A∩B = {x: x >, 4, x∈R}
Hence, Range of A∩B = {x: x ≥ 4, X∈R} and it can be represented on graph or number line as follows.
(28) Given, P = {x: 5 < 2x – 1 ≤ 11, X∈R} and Q = {x: -1 ≤ 3 + 4x < 23, x∈I} where R- Real numbers, I -> Integers
Represent P & Q on number line. Write down the elements of PNQ.
Solution:
Given that, P = {x: 5 < (2x – 1) ≤ 11, X∈R} and Q = {x: -1 ≤ (3 + 4x) < 23, X∈I}
R => Real numbers and I => Integers
For P
5 < (2x – 1) ≤ 11
= 5 < (2x – 1) and (2x – 1) ≤ 11
(5 + 1) < 2x and (2x) ≤ (11 + 1)
6 < 2x and 2x ≤ 12
3 < x and x ≤ 6
x > 3 and x ≤ 6
=> x > 3 or 3 < X
Hence, the required solution set is 3 < x ≤ 6 = {4, 5, 6}
For Q
– 1 ≤ (3 + 4x) <23
-1 ≤ (3 + 4x) and (3 + 4) < 23
= (-1 – 3) ≤ 4x and 4x < (23 – 3)
– 4 ≤ 4x and 4x < 20
– 1 ≤ x and x < 5
x ≥ -1 => -1 ≤ x < 5 x∈I
The required solution set is found to be –1 ≤ x < 5
i.e. {-1, 0, 1, 2, 3, 4}
P∩Q = {4}
(29) If X∈I, find the smallest value of x, which satisfies the inequation (2x + 5/2) > (5x/3 + 2)
Solution:
Given inequality is (2x + 5/2) > (5x/3 + 2), X∈I
= (2x – 5x/3) >(2 – 5/2)
= (x/3) > (-1/2)
2x > – 3
X > -3/2
But X∈I
Hence, the required smallest value of x is x = -1
(30) Given (20-5x) < 5 (x + 8), find the smallest value of x, When
(i) X∈I
(ii) X∈W
(iii) X∈N
Solution:
Given inequality is (20 – 5x) < 5 (x + 8)
= (20 – 5x) < (5x + 40)
= (- 5X – 5x) < (40 – 20)
= – 10x < 20
= – x <2
= X > – 2
(i) When X∈I, the smallest value of x is found to be x = -1.
(ii) When X∈W, the smallest value of x is found to be x = 0.
(iii) When X∈N, the smallest value of x is found to be x = 1
(32) Solve the given inequation and graph the solution on the number line: (2y – 3) < (y + 1) ≤ (4y + 7), Y∈R.
Solution:
Given inequality is, (2y – 3) < (y+1) ≤ (4y+7), Y∈R
= (2y – 3) < (y + 1) and (y + 1) ≤ (4y + 7)
= (2y – y) < (1 + 3) and 1-7 ≤ (4y – y)
y < 4 and -6 ≤ 3y
-2 ≤ y i.e. y ≥ -2
Thus, 4 > y ≥ – 2 or -2 ≤ y < 4 is the required solution set which can be represented on the number line as below:
(33) Find the greatest integer which is such that if 7 is added to its double, the resulting number becomes greater than three times the integer.
Solution:
Let us consider the greatest integer be ‘x’. From given condition we can write,
(2x + 7) > 3x
= 2x – 3x > – 7
= – x > – 7
= X ≤ 7
Thus, the greatest value of x is found to be x = 7.
(35) One third of a bamboo pole is buried in mud, one-sixth of it is in water and the part above the water greater than or equal to 3 meters. Find the length of the shortest pole.
Solution:
Let us consider the length of shortest pole is ‘x’
Then, from given condition we can write,
Length of pole buried in mud is ‘x/3’ and length of pole which is in water is ‘x/6. According to given conditions,
x – (x/3 + x/6) ≥ 3
= x – (2x + x)/6 ≥ 3
= x – x/2 ≥ 3
X ≥ 6
Thus, the shortest length of the pole is found to be 6 meters.
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