# ML Aggarwal Class-8 Linear Equations and Inequalities ICSE Maths

ML Aggarwal Class-8 Linear Equations and Inequalities in One Variable ICSE Mathematics Solutions Chapter-12. We provide step by step Solutions of Exercise / lesson-12 Linear Equations and Inequalities in One Variable Class-8th ML Aggarwal ICSE Mathematics.

Our Solutions contain all type Questions with Exe-12.1 , Exe-12.2, Exe-12.3  Objective Type Questions (including Mental Maths Multiple Choice Questions Value Based Questions , HOTS),  and Check Your Progress to develop skill and confidence. Visit official Website for detail information about ICSE Board Class-8 Mathematics.

## ML Aggarwal Class-8 Linear Equations and Inequalities in One Variable ICSE Mathematics Solutions Chapter-12

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Exercise 12.1 ,

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### Ex 12.1 , ML Aggarwal Class-8 Linear Equations and Inequalities in One Variable ICSE Mathematics Solutions

Solve the following equations (1 to 12):

Question 1.
(i) 5x – 3 = 3x – 5
(ii) 3x – 7 = 3(5 – x)

(i) 5x – 3 = 3x – 5
⇒ 5x – 3x = -5 + 3
⇒ 2x = — 2
⇒ x = $\frac{-2}{2}$ = -1

(ii) 3x – 7 = 3(5 – x)
⇒ 3x – 7 = 15 – 3x
⇒ 3x + 3x = 15 + 7
⇒ 6x = 22
⇒ x = $\frac{22}{6}=\frac{11}{3}$

Question 2.
(i) 4(2x + 1) = 3(x – 1) + 7
(ii) 3(2p – 1) = 5 – (3p – 2)

(i) 4(2x + 1) = 3(x – 1) + 7
⇒ 8x + 4 = 3x – 3 + 7
⇒ 8x + 4 = 3x + 4
⇒ 8x – 3x = 0
⇒ x = 0

(ii) 3(2p – 1) = 5 – (3p – 2)
⇒ 6p – 3 = 5 – 3p + 2
⇒ 6p + 3p = 5 + 3 + 2
⇒ 9p = 10
⇒ p = $\frac{10}{9}=1 \frac{1}{9}$

Question 3.
(i) 5y – 2[y – 3(y – 5)] = 6
(ii) 0.3(6 – x) = 0.4(x + 8)

(i) 5y – 2[y – 3(y – 5)] = 6
⇒ 5y – 2[y – 3y + 15] = 6
⇒ 5y – 2[-2y + 15] = 6
⇒ 5y + 4y – 30 = 6
⇒ 9y = 6 + 30
⇒ 9y = 36
⇒ y = $\frac{36}{9}$
⇒ y = 4

(ii) 0.3 (6 – x) = 0.4 (x + 8)
⇒ 1.8 – 0.3x = 0.4x + 3.2
⇒ 0.3x – 0.4x = 3.2 – 1.8
⇒ -0.7x = 1.4
⇒ x = $\frac{-1 \cdot 4}{0 \cdot 7}$
⇒ x = $\frac{-14}{7}$
⇒ x = -2

Question 4.
(i) $\frac{x-1}{3}=\frac{x+2}{6}+3$
(ii) $\frac{x+7}{3}=1+\frac{3 x-2}{5}$

⇒ x – 4 = 6 × 3
⇒ x – 4 = 18
⇒ x = 18 + 4
⇒ x = 22

⇒ -4x + 41 = 15
⇒ -4x = 15 – 41
⇒ -4x = -26
⇒ x = $\frac{-26}{-4}$
⇒ x = $\frac{13}{2}$
⇒ x = $6 \frac{1}{2}$

Question 5.

⇒ 3 (-y + 5) = 6 (1 + 2y)
⇒ -3y +15 = 6+ 12y
⇒ -3y – 12y = 6 – 15
⇒ -15y = -9

⇒ 47p + 480 = 55 × 60
⇒ 47p + 480 = 3300
⇒ 47p = 3300 – 480
⇒ 47p = 2820
⇒ p = $\frac{2820}{47}$
⇒ p = 60

Question 6.

⇒ 3 (n + 1) = 2(5 – n)
⇒ 3n + 3 = 10 – 2n
⇒ 3n + 2n = 10 – 3
⇒ 5n = 7 ⇒ n = $\frac{7}{5}=1 \frac{2}{5}$

⇒ (6t – 4) + (6t + 9) = 6t + 7
⇒ 12t + 5 = 6t + 7
⇒ 12t – 6t = 7 – 5
⇒ 6t = 2 ⇒ t = $\frac{2}{6}=\frac{1}{3}$

Question 7.
(i) 4(3x + 2) – 5(6x – 1) = 2(x – 8) – 6(7x – 4)
(ii) 3(5x + 7) + 5(2x – 11) = 3(8x – 5) – 15

(i) 4(3x + 2) – 5(6x – 1) = 2(x – 8) – 6(7x – 4)
⇒ 12x + 8 – 30x + 5 = 2x – 16 – 42x + 24
⇒ -18x + 13 = -40x + 8
⇒ -18x + 40x = 8 – 13
⇒ 22x = -5 ⇒ x = $\frac{-5}{22}$

(ii) 3(5x + 7) + 5(2x – 11) = 3(8x – 5) – 15
⇒ 15x + 21 + 10x – 55 = 24x – 15 – 15
⇒ 25x – 34 = 24x – 30
⇒ 25x – 24x = -30 + 34
⇒ x = 4

Question 8.

(i) $\frac{3-2 x}{2 x+5}=-\frac{3}{11}$
⇒ 11(3 – 2x) = -3(2x + 5)
⇒ 33 – 22x = -6x – 15
⇒ -22x + 6x = -15 – 33
⇒ -16x = -48
⇒ x = $\frac{48}{16}$ = 3

(ii) $\frac{5 p+2}{8-2 p}=\frac{7}{6}$
⇒ 6(5p + 2) = 7(8 – 2)p
⇒ 30p + 12 = 56 – 14p
⇒ 30p +14p = 56 – 12
⇒ 44p = 44
⇒ p =$\frac{44}{44}$ = 1

Question 9.

5 (x – 4) = 7x
⇒ 5x – 20 = 7x
⇒ 5x – 7x = 20
⇒ -2x = 20

⇒ 4(x + 4) = 5(2x + 3)
⇒ 4x + 16 = 10x + 15
⇒ -6x = -1
⇒ x = $\frac{-1}{-6}=\frac{1}{6}$

Question 10.

⇒ 2x2 – 7x – 5 = x (2x – 2)
⇒ 2x2 – 7x – 5 = 2x2 – 2x
⇒ -7x – 5 = -2x
⇒ -7x + 2x = 5
⇒ -5x = 5
⇒ x = $\frac{5}{-5}$
⇒ x = $\frac{-5}{5}$
⇒ x = -1

⇒ 3x(1 – 15x) = 15x (3x – 1)
⇒ 3(1 – 15x) = 15 (3x – 1)
⇒ 3 – 45x = 45x – 15
⇒ -45x – 45x = -15 – 3
⇒ -90x = -18
⇒ x = $\frac{-18}{-90} \Rightarrow x=\frac{1}{5}$

Question 11.

⇒ (2x – 3)(3x + 1) = (3x – 1) (2x – 1)
⇒ 6x2 + 2x – 9x – 3 = 6x2 – 3x – 2x + 1
⇒ 6x2 – 7x – 3 = 6x2 – 5x + 1
⇒ 6x2 – 7x – 6x2 + 5x = 1 + 3
⇒ -2x = 4 ⇒ x = $\frac{4}{-2}$ = -2

(ii) $\frac{2 y+3}{3 y+2}=\frac{4 y+5}{6 y+7}$
⇒ (2y + 3) (6y + 7) = (4y + 5) (3y + 2)
⇒ 12y2 + 14y + 18y + 21 = 12y2 + 8y + 15y + 10
⇒ 32y + 21 = 23y + 10
⇒ 32y – 23y = 10 – 21
⇒ 9y = -11 ⇒ y = $\frac{-11}{9}$

Question 12.

If x = p + 1, find the value of p from the equation $\frac{1}{2}$ (5x – 30) – $\frac{1}{3}$ (1 + 7p) = $\frac{1}{4}$

Given x =p + 1 …(i)
Also $\frac{1}{2}$(5x – 30) – $\frac{1}{3}$(1 + 7p) = $\frac{1}{4}$ …(ii)
Putting the value of x from (i) in (ii), we get,

⇒ 4 × (p – 77) = 1 × 6
⇒ 4p – 308 = 6
⇒ 4p = 6 + 308
⇒ 4p = 314

Question 13.
Solve $\frac{x+3}{3}-\frac{x-2}{2}=1$, Hence find p if $\frac{1}{x}+p$ = 1.

### ML Aggarwal Class-8 Linear Equations and Inequalities in One Variable ICSE Mathematics Solutions  Ex 12.2

Question 1.

Three more than twice a number is equal to four less than the number. Find the number.

Let the number = x
Twice the number = 2x
According to problem, 3 + 2x = x – 4
⇒ 3 + 2x + 4 = x
⇒ 7 = x – 2x
⇒ 7 = -x
⇒ -x = 7
⇒ x = -7
Hence, the number = -7

Question 2.
When four consecutive integers are added, the sum is 46. Find the integers.

Let x be the first integer, then the next
three consecutive integers are x + 1, x + 2 and x + 3
According to problem,
x + (x + 1) + (x + 2) + (x + 3) = 46
⇒ x + x + 1 + x + 2 + x + 3 = 46
⇒ 4x + 6 = 46
⇒ 4x = 46 – 6
⇒ 4x = 40
⇒ x = $\frac{40}{4}$ = 10
Hence four consecutive integers are 10, (10 + 1), (10 + 2) and (10 + 3)
i.e. 10, 11, 12 and 13

Question 3.
Manjula thinks a number and subtracts $\frac{7}{3}$ from it. She multiplies the result by 6. The result now obtained is 2 less than twice the same number she thought of. What is the number?

Let a number thought by Manjula = x
According to the condition,
$\left(x-\frac{7}{3}\right)$ × 6 = 2x – 2
⇒ 6x – 14 = 2x – 2
⇒  6x – 2x = -2 + 14 = 12
⇒  4x = 12
⇒ x = $\frac{12}{4}$ = 3
Flence required number = 3

Question 4.
A positive number is 7 times another number. If 15 is added to both the numbers, then one of the new number becomes $\frac{5}{2}$ times the other new number. What are the numbers?

Let the required number = x
Then another number = $\frac{x}{7}$
According to the condition,

Hence the numbers are 35 and 5

Question 5.
When three consecutive even integers are added, the sum is zero. Find the integers.

Let the first even integer be x,
then next two consecutive even integers are (x + 2) and (x + 4)
According to given problem,
x + (x +2) + (x + 4) = 0
⇒ x + x + 2 + x + 4 =0
⇒ 3x + 6 = 0
⇒ 3x = – 6
⇒ x = $\frac{-6}{3}$
⇒ x = — 2
Hence three consecutive integers are -2, – 2 + 2, – 2 + 4 i.e. – 2, 0, 2

Question 6.

Find two consecutive odd integers such that two-fifth of the smaller exceeds two-ninth of the greater by 4.

Let the first odd integer be x,
then next consecutive odd integers is (x + 2)
According to given problem,

Hence two consecutive odd integers are
25 and (x + 2) = (25 + 2) = 27

Question 7.
The denominator of a fraction is 1 more than twice its numerator. If the numerator and denominator are both increased by 5, it becomes $\frac{3}{5}$. Find the original fraction.

Let the numerator of the original fraction be x
Then, its denominator = 2x + 1
∴ The fraction = $\frac{x}{2 x+1}$
According to given problem,

⇒ 5 (x + 5) = 3 (2x + 6)
⇒ 5x + 25 = 6x + 18
⇒ 5x – 6x = 18 – 25
⇒ -x = -7
⇒ x = 7
Hence, the original fraction = $\frac{x}{2 x+1}=\frac{7}{2 \times 7+1}=\frac{7}{15}$

Question 8.
Find two positive numbers in the ratio 2 : 5 such that their difference is 15.

Let the two numbers be 2x and 5x
[∵ ratio of these two numbers = $\frac{2 x}{5 x}=\frac{2}{5}$ = 2 : 5 ]
According to given problem,
5x – 2x = 15
⇒ 3x = 15
⇒ x = $\frac{15}{3}$
⇒ x = 5
Hence the numbers are 2 × 5 and 5 × 5 i.e. 10 and 25

Question 9.
What number should be added to each of the numbers 12, 22, 42 and 72 so that the resulting numbers may be in proportion ?

Let the required number be x
Then according to given problem,
12 + x, 22 + x, 42 + x and 72 + x are in proportion
⇒ $\frac{12+x}{22+x}=\frac{42+x}{72+x}$
⇒ (12 + x) (72 + x) = (42 + x) (22 + x)
⇒ 12 (72 + x) + x (72 + x) = 42 (22 + x) + x (22 + x)
⇒ 864 + 12x + 72x + x2 = 924 + 42x + 22x + x2
⇒ 864 + 84x + x2 = 924 + 64x + x2
⇒ 864 + 84x + x2 – 924 – 64x – x2 = 0
⇒ 864 + 84x – 64x – 924 = 0
⇒ 84x – 64x = 924 – 864
⇒ 20x = 60
⇒ x = $\frac{60}{20}$
⇒ x = 3
Hence, the required number is 3.

Question 10.
The digits of a two-digit number differ by 3. If the digits are interchanged and the resulting number is added to the original number, we get 143. What can be the original number?

Let one’s digit of a 2-digit number = x
Then ten’s digit = x + 3
∴ Number = x + 10(x + 3) = x + 10x + 30 = 11x + 30
By interchanging the digits,
One’s digit of new number = x + 3
and ten’s digit = x
∴ Number = x + 3 + 10x = 11x + 3
According to the condition,
11x + 30+ 11x + 3 = 143
⇒ 22x + 33 = 143
⇒ 22x = 143-33 = 110
⇒ x = $\frac{110}{22}$ = 5
∴ Original number = 11x+ 30 = 11 × 5 + 30 = 55 + 30 = 85

Question 11.
Sum of the digits of a two-digit number is 11. When we interchange the digits, it is found that the resulting new number is greater than the original number by 63. Find the two-digit number.

Sum of two digits of a 2-digit number = 11
Let unit’s digit of a 2-digit number = x
Then ten’s digit = 11 – x
∴ Number = x + 10(11 – x) = x + 110 – 10x = 110 -9x
By interchanging the digit,
One’s digit of new number = 11 – x
and ten’s digit = x
∴ Number = 11 – x + 10x = 11 + 9x
According to the condition,
11 + 9x – (110 – 9x) = 63
11 + 9x – 110 + 9x = 63
18x = 63 – 11 + 110 = 162
x = $\frac{162}{18}$ = 9
∴ Original number = 110 – 9x = 110 – 9 × 9 = 110 – 81 = 29

Question 12.
Ritu is now four times as old as his brother Raju. In 4 years time, her age will be twice of Raju’s age. What are their present ages?

Let the age of Raju = x years
then the age of Ritu = 4 × x years = 4x years
In 4 years time,
age of Raju = (x +4) years
age of Ritu = (4x + 4) years
According to given problem,
4x + 4 = 2 (x + 4)
⇒ 4x + 4 = 2x + 8
⇒ 4x – 2x = 8 – 4
⇒ 2x = 4 ⇒ x = $\frac{4}{2}$
⇒ x = 2
Hence, the age of Raju = 2 years
and the age of Ritu = 4 × 2 years = 8 years.

Question 13.
A father is 7 times as old as his son. Two years ago, the father was 13 times as old as his son. How old are they now?

Let the present age of son = x years
Then, age of his father = 7 × x years = 7x years
Two years ago age of son = (x – 2) years
Two years ago age of his father = (7x – 2) years
According to given problem,
7x – 2 = 13 (x – 2)
⇒ 7x – 2 = 13x – 26
⇒ 7x – 13x = -26 + 2
⇒ -6x = -24
⇒ x = $\frac{-24}{-6}$
⇒ x = 4
Hence, age of son = 4 years
and age of his father = 7 × 4 years = 28 years.

Question 14.
The ages of Sona and Sonali are in the ratio 5 : 3. Five years hence, the ratio of their ages will be 10 : 7. Find their present ages.

Given ratio of ages of Sona and Sonali = 5 : 3
let the present ages of Sona and Sonali is 5x and 3x years
five years hence, the age of Sona = 5x + 5
and five years to hence the age of Sonali = 3x + 5
According to given problem,
$\frac{5 x+5}{3 x+5}=\frac{10}{7}$
⇒ 7(5x + 5) = 10(3x + 5)
⇒ 35x + 35 = 30x + 50
⇒ 35x – 30x = 50 – 35
⇒ 5x = 15 ⇒ x = $\frac{15}{5}$
⇒ x = 3
Hence, the present age of Sona and Sonali is 5 × 3 and 3 × 3 years
i.e. 15 and 9 years.

Question 15.
An employee works in a company on a contract of 30 days on the condition that he will receive ₹200 for each day he works and he will be fined ₹20 for each day he is absent. If he receives ₹3800 in all, for how many days did he remain absent?

Period of contract = 30 days
If an employees works a day, he will get ₹200
If he is absent, he will be fined ₹20 per day
At the end of contract period, he get ₹3800
Let he remained absent for x days
Then he worked for = (30 – x) days
According to the condition,
(30 – x) × 200 – x × 20 = 3800
⇒ 6000 – 200x – 20x = 3800
⇒  220x = 6000 – 3800 = 2200
⇒  x = $\frac{2200}{220}$ = 10
He remained absent for 10 days.

Question 16.
I have a total of ₹300 in coins of denomination ₹1, ₹2 and ₹5. The number of coins is 3 times the number of ₹5 coins. The total number of coins is 160. How many coins of each denomination are with me?

Amount of coins = ₹300
and total coins = 160
Let number of coins of ₹5 = x
Then number of coins of ₹2 = 3x
and number of coins of ₹1 = 150 – (x + 3x) = 150 – 4x
According to the condition,
(160 – 4x) × 1 + 3x × 2 + x × 5 = 300
⇒ 160 – 4x + 6x + 5x = 300 ⇒ 160 + 7x = 300
⇒ 7x = 300 – 160 = 140
⇒ x = $\frac{140}{7}$ = 20
∴ 5 rupee coins = 20
2 rupee coins = 3 × 20 = 60
and 1 rupee coins = 160 – 60 – 20 = 80

Question 17.
A local bus is carrying 40 passengers, some with ₹5 tickets and the remaining with ₹7.50 tickets. If the total receipts from these passengers is ₹230, find the number of passengers with ₹5 tickets.

Let the number of passengers with ₹5 tickets = x
Then, the number of passengers with ₹7.50 tickets = (40 – x)
According to given problem,
5 × x + (40 – x) × 7.50 = 230
⇒ 5x + 300 – 7.5x = 230
⇒ 5x – 7.5x = 230 – 300
⇒ -2.5x = -70
⇒ x = $\frac{70}{2.5}$ = 28
Hence, the number of passengers with ₹5 tickets = 28.

Question 18.
On a school picnic, a group of students agree to pay equally for the use of a full boat and pay ₹10 each. If there had been 3 more students in the group, each would have paid ₹2 less. How many students were there in the group ?

Let, the number of students in a group = x
when 3 students are more then,
the total number of students in the group = x + 3
According to given problem,
⇒ 10 × x = (x + 3) × (10 – 2)
⇒ 10x = (x + 3) × 8
⇒ 10x = 8 (x + 3)
⇒ 10x = 8x + 24
⇒ 10x – 8x = 24
⇒ 2x = 24
⇒ x = $\frac{24}{2}$
⇒ x = 12
Hence, the number of students in the group = 12

Question 19.
Half of a herd of deer are grazing in the field and three-fourths of the remaining are playing nearby. The rest 9 are drinking water from the pond. Find the number of deer in the herd.

Let total deer in the herd = x
Number of deer grazing in the field = $\frac{x}{2}$
Remaining = x – $\frac{x}{2}=\frac{x}{2}$
$\frac{3}{4}$ of the remaining deer playing
$\frac{3}{4} \times \frac{1}{2} x=\frac{3}{8} x$
Rest of deer = $\frac{x}{2}-\frac{3}{8} x=\frac{1}{8} x$
∴ $\frac{1}{8}$x = 9
⇒ x = 9 × 8 = 72
∴ Total number of deer = 72

Question 20.
Sakshi takes some flowers in a basket and visits three temples one by one. At each temple, she offers one half of the flowers from the basket. If she is left with 6 flowers at the end, find the number of flowers she had in the beginning.

Let total flowers in a basket = x
Flowers offered in the first temple = $\frac{x}{2}$
Remaining flowers = x – $x-\frac{x}{2}=\frac{x}{2}$
Flowers offered in the second temple
$\frac{x}{2} \times \frac{1}{2}=\frac{x}{4}$
Remaining flower = $\frac{x}{2}-\frac{x}{4}=\frac{x}{4}$
Flowers offered in the third temple = $\frac{x}{4} \times$ $\frac{1}{2}=\frac{x}{8}$
Remaining flowers = $\frac{x}{4}-\frac{x}{8}=\frac{x}{8}$
∴ Total , number of flowers = 48

Question 21.
Two supplementary angles differ by 50°. Find the measure of each angle.

Let the angle be x
Then, its supplementary angle = 180° – x
According to given problem,
x – (180° – x) = 50°
⇒ x – 180° + x = 50°
⇒ 2x – 180° = 50°
⇒ 2x = 180° + 50°
⇒ 2x = 230°
⇒ x = $\frac{230^{\circ}}{2}$
⇒ x = 115°
Hence, the measurement of each angle be 115° and (180° – 115°)
i.e. 115° and 65°.

Question 22.
If the angles of a triangle are in the ratio 5 : 6 : 7, find the angles.

Let the angles of a triangle are 5x, 6x, and 7x
Then, we know that,
5x + 6x + 7x = 180°
⇒ 18x = 180°
⇒ x = $\frac{180^{\circ}}{18}$
⇒ x= 10°
Hence, the angle of a triangle are 5 × 10°, 6 × 10°,
and 7 × 10° i.e. 50°, 60° and 70°.

Question 23.
Two equal sides of an isosceles triangle are 3x – 1 and 2x + 2 units. The third side is 2x units. Find x and the perimeter of the triangle.

Two equal sides of an isosceles triangle are 3x – 1 and 2x + 2
i. e. 3x – 1 = 2x + 2
⇒ 3x – 2x = 2 + 1
⇒ x = 3
Third side of triangle = 2x = 2 × 3 = 6 units
equal sides of an triangle = 3 × 3 – 1= 9 – 1 = 8 units
∴ Perimeter of the triangle = (8 + 8 + 6) units = 22 units

Question 24.
If each side of a triangle is increased by 4 cm, the ratio of the perimeters of the new triangle and the given triangle is 7 : 5. Find the perimeter of the given triangle.

Let the perimeter of original triangle = x cm
By increasing each side by 4 cm The perimeter will be
= x + 4 × 3 = (x + 12) cm
Now Ratio of perimeter of new triangle and given triangle = 7 : 5
⇒ $\frac{x+12}{x}=\frac{7}{5}$
⇒ 5x + 60 = 7x (By corss multiplication)
⇒ 7x – 5x = 60
⇒ 2x = 60
⇒ x = $\frac{60}{2}$ = 30
∴ Perimeter of given triangle = 30 cm

Question 25.
The length of a rectangle is 5 cm less than twice its breadth. If the length is decreased by 3 cm and breadth increased by 2 cm, the perimeter of the resulting rectangle is 72 cm. Find the area of the original rectangle.

Let, the breath of the original rectangle = x cm
Then, length of the original rectangle = (2x – 5) cm
When, length is decreased by 3 cm then new length
= [(2x – 5) – 3)] cm = (2x – 8) cm
When breadth is increased by 2 cm,
then new length = (x + 2) cm
New perimeter=2(new length + new breadth)
= 2 [(2x – 8) + (x + 2)]
= 2 [2x – 8 + x + 2]
= 2 (3x – 6) = 6x – 12
According to the given problem,
6x – 12 = 72
⇒ 6x = 72 + 12
⇒ 6x = 84
⇒ x = $\frac{84}{6}$
⇒ x = 14
Breadth of the original rectangle = 14 cm
and length of the original rectangle = (2 × 14 – 5) cm = 23 cm
Area of the original rectangle = Length × Breadth
= 23 × 14 cm2 = 322 cm2

Question 26.

A rectangle is 10 cm long and 8 cm wide. When each side of the rectangle is increased by x cm, its perimeter is doubled. Find the equation in x and hence find the area of the new rectangle.

Length of rectangle (l) = 10 cm
and width (b) = 8 cm
Perimeter = 2(l + b) = 2(10 + 8) cm = 2 × 18 = 36 cm
By increasing each side by x cm
Then perimeter = 2[10 + x + 8 + x]
= 2(18 + 2x) = (36 + 4x) cm
According to the condition,
36 + 4x = 2(36)
⇒ 36 + 4x = 72
⇒ 4x = 72 – 36 = 36
⇒ x = $\frac{36}{4}$
⇒ x = 9
Length of new rectanlge = l + x = 10 + 9 = 19 cm
and breadth = b + x = 8 + 9 = 17 cm
Area = Length x Breadth = 19 × 17 cm= 323 cm2

Question 27.

A steamer travels 90 km downstream in the same time as it takes to travel 60 km upstream. If the speed of the stream is 5 km/hr, find the speed of the streamer in still water.

Let, the speed of the streamer in still water be x km/hr,
Then, the speed downstream = (x + 5) km/hr
And, the speed upstream = (x – 5) km/r.
According to given problem
$\frac{90}{x+5}=\frac{60}{x-5}$
⇒ 90(x – 5) = 60(x + 5)
⇒ 90x – 450 = 60x + 300
⇒ 90x – 60x = 300 + 450
⇒ 30x = 750
⇒ x = $\frac{750}{30}$
⇒ x = 25
Hence, the speed of the streamer in still water be 25 km/hr.

Question 28.
A steamer goes downstream and covers the distance between two ports in 5 hours while it covers the same distance upstream in 6 hours. If the speed of the stream is 1 km/h, find the speed of the streamer in still water and the distance between two ports.

Speed of the stream in still water = 1 km/h
Let speed of streamer = x km/h
∴ It down speed = (x + 1) km/h
and up speed = (x – 1) km/h
According to the condition,
(x + 1) × 5 = (x – 1) × 6
⇒ 5x + 5 = 6x – 6
⇒ 6x – 5x = 5 + 6
⇒ x = 11
∴ Speed of streamer in still water = 11 km/h
and distance between two points = (11 + 1) × 5 = 60 km/h

Question 29.
Distance between two places A and B is 350 km. Two cars start simultaneously from A and B towards each other and the distance between them after 4 hours is 62 km. If the speed of one car is 8 km/h less than the speed of other cars, find the speed of each car.

Distance between two places A and B = 350 km
Let speed of car C1 = x km/h
The speed of car C2 = (x – 8) km/h

In 4 hours, there will be 62 km distance between these two cars.
∴ x × 4 + (x – 8) × 4 = 350 – 62
⇒ 4x + 4x – 32 = 288
⇒ 8x = 288 + 32 = 320
⇒ x = $\frac{320}{8}$ = 40
Speed of one car C1 = 40 km/h
and speed of car C2 = 40 – 8 = 32 km/h

### ML Aggarwal Solutions Exe- 12.3 Class-8 Linear Equations and Inequalities in One Variable ICSE Mathematics Solutions

Question 1.
If the replacement set = {-7, -5, -3, – 1, 3}, find the solution set of:
(i) x > – 2
(ii) x < – 2
(iii) x > 2
(iv) -5 < x ≤ 5
(v) -8 < x < 1
(vi) 0 ≤ x ≤ 4

Given replacement set = {-7, -5, -3, -1, 3}
(i) Solution set of x > – 2 is {-1,0, 1,3}
(ii) Solution set of x < – 2 is { -7, -5, -3} m
(iii) Solution set of x >2 is {3}
(iv) Solution set of – 5 < x ≤ 5 is {-3, -1, o, 1, 3}
(v) Solution set of – 8 < x < 1 is { -7, -5, -3, -1, 0}
(vi) Solution set of 0 ≤ x ≤ 4 is {0, 1, 3}

Question 2.

Represent the solution of the following inequalities graphically:
(i) x ≤ 4, x ϵ N
(ii) x < 5, x ϵ W
(iii) -3 ≤ x < 3, x ϵ I

(i) Given x < 4, x ϵ N
The solution set = {1, 2, 3, 4}
These four numbers are shown by thick dots on the number line.

(ii) Given x < 5, x ϵ W
The solution set = {0, 1, 2, 3, 4}
These five numbers are shown by thick dots on the number line.

(iii) Given – 3 ≤ x < 3, x ϵ I
The solution set = {-3, -2, -1, 0, 1, 2}
These six numbers are shown by thick dots on the number line.

Question 3.
If the replacement set is {-6, -4, -2, 0, 2, 4, 6}, then represent the solution set of the inequality – 4 ≤ x < 4 grahically.

The given replacement set is {-6, -4, -2, 0, 2, 4, 6}
and, given inequality – 4 ≤ x < 4 Solution set is {-4, -2, 0, 2}
Graphically representation of solution set is as under.

Question 4.
Find the solution set of the inequality x < 4 if the replacement set is
(i) {1, 2, 3, ………..,10}
(ii) {-1, 0, 1, 2, 5, 8}
(iii) {-5, 10}
(iv) {5, 6, 7, 8, 9, 10}

The given inequation x < 4
(i) replacement set is {1, 2, 3, -, 10} for this set, solution set is {1, 2, 3}
(ii) replacement set is {- 1, 0, 1, 2, 5, 8) for this set, solution set is {-1, 0, 1, 2}
(iii) replacement set is {-5, 10} for this set, solution set is {-5}
(iv) repalcement set is {5, 6, 7, 8, 9, 10} for this set, solution set is ϕ

Question 5.
If the replacement set = {-6, -3, 0, 3, 6, 9, 12}, find the truth set of the following.:
(i) 2x – 3 > 7
(ii) 3x + 8 ≤ 2
(iii) -3 < 1 – 2x

The given replacement set = {-6, -3, 0, 3, 6, 9, 12}
(i) 2x – 3 > 7
⇒ 2x > 7 + 3
⇒ 2x > 10
⇒ x > $\frac{10}{2}$
⇒ x > 5
Its solution set is {6, 9, 12}

(ii) 3x + 8 ≤ 2
⇒ 3x ≤ 2 – 8
⇒ 3x < – 6
⇒ x ≤ $-\frac{6}{3}$
⇒ x ≤ – 2
Its solution set is {-6, -3}

(iii) -3 < 1 – 2x
⇒ 2x – 3 < 1
⇒ 2x < 1 +3
⇒ 2x < 4
⇒ x < $\frac{4}{2}$
⇒ x < 2
Its solution set is {-6, -3, 0}

Question 6.
Solve the following inequations:
(i) 4x + 1 < 17, x ϵ N
(ii) 4x + 1 ≤ 17, x ϵ W
(iii) 4 > 3x – 11, x ϵ N
(iv) -17 ≤ 9x – 8, x 6ϵ Z

(i) 4x + 1 < 17
⇒ 4x < 17 – 1
⇒ 4x < 16
⇒ x < $\frac{16}{4}$
⇒ x < 4
As x ϵ N, the solution set is {1, 2, 3}

(ii) 4x + 1 ≤ 17
⇒ 4x ≤ 17- 1
⇒ 4x ≤ 16
⇒ x ≤ $\frac{16}{4}$
⇒ x < 4. As x ϵ W, the solution set is {0, 1,2, 3,4}

(iii) 4 > 3x – 11
⇒ 4 + 11 > 3x
⇒ 15 > 3x
⇒ $\frac{15}{3}$ >x
⇒ 5 > x
⇒ x > 5
As x ϵ N, the solution set is {1, 2, 3, 4}

(iv) 17 ≤ 9x – 8
⇒ -17 + 8 ≤ 9x
⇒ -9 ≤ 9x
⇒ $\frac{-9}{9}$ ≤ x
⇒ -1 ≤ x
⇒ x > – 1
As x ϵ Z, the solution set is {-1, 0, 1, 2, ……..}

Question 7.
Solve the following inequations :

(i) $\frac{2 y-1}{5} \leq 2$
⇒ 2y- 1 ≤ 10
⇒ 2y ≤ 10 + 1
⇒ 2y ≤ 11
⇒ y ≤ $\frac{11}{2}$
As y ϵ N, the solution set is {1, 2, 3, 4, 5}

⇒ 2y + 4 ≤ 9
⇒ 2y ≤ 9 – 4
⇒ 2y ≤ 5
⇒ y ≤ $\frac{5}{2}$
As y ϵ N, the solution set is {0, 1, 2}

(iii) $\frac{2}{3}$P + 5 < 9
⇒ $\frac{2}{3}$p < 9 – 5
⇒ $\frac{2}{3}$p < 4
⇒ 2p < 4 × 3
⇒ 2p < 12
⇒ P < $\frac{12}{2}$ ⇒ p < 6
As p ϵ W, the solution set is {0, 1, 2, 3, 4, 5}

(iv) -2 (p + 3) > 5
⇒ -2p – 6 > 5
⇒ -2p > 5 + 6
⇒ -2p > 11
⇒ p < $\frac{11}{(-2)}$
⇒ p < $\frac{-11}{2}$
As p ϵ I, the solution set is {… -8, -7, -6}

Question 8.
Solve the following inequations:
(i) 2x – 3 < x + 2, x ϵ N
(ii) 3 – x ≤ 5 – 3x, x ϵ W
(iii) 3 (x – 2) < 2 (x -1), x ϵ W
(iv) $\frac{3}{2}-\frac{x}{2}$ > -1, x ϵ N

(i) 2x – 3 < x + 2
⇒ 2x – x < 2 + 3
⇒ x < 5
As x ϵ N, the solution set is {1, 2, 3, 4}

(ii) 3 – x ≤ 5 – 3x
⇒ -x + 3x ≤ 5 – 3
⇒ 2x ≤ 2
⇒ x ≤ $\frac{2}{2}$
⇒ x ≤ 1
As x ϵ W, the solution set is {0, 1}

(iii) 3 (x – 2) < 2 (x – 1)
⇒ 3x – 6 < 2x – 2
⇒ 3x – 2x < – 2 + 6
⇒ x < 4
As x ϵ W, the solution set is {0, 1, 2, 3}

As x ϵ N, the solution set is {1, 2, 3, 4}

Question 9.
If the replacement set is {-3, -2, -1,0, 1, 2, 3} , solve the inequation $\frac{3 x-1}{2}<2$. represent its solution on the number line.

The given replacement set is {-3, -2, -1, 0, 1, 2, 3}
And inequation, $\frac{3 x-1}{2}<2$
⇒ 3x – 1 < 4
⇒ 3x < 4 + 1
⇒ 3x < 5
⇒ x < $\frac{5}{3}$
Hence, solution set is {-3, -2, -1,0, 1}
Graphical representation of this solution set is

Question 10.
Solve $\frac{x}{3}+\frac{1}{4}<\frac{x}{6}+\frac{1}{2}$, x ϵ W. Also represent its solution on the number line.
Solution:

As x ϵ W, the solution set is {0, 1}
Graphical representation of this solution set is

Question 11.
Solve the following inequations and graph their solutions on a number line
(i) -4 ≤ 4x < 14, x ϵ N
(ii) -1 < $\frac{x}{2}$ + 1 ≤ 3, x ϵ I

(i) Given – 4 ≤ 4x < 14
Dividing by 4

As x ϵ N, then its solution set is {1, 2, 3}
Its graphical representation is

As x ϵ I, then its solution set is {-3, -2, -1, 0, 1, 2, 3, 4}
Its Graphical representation is

### Chapter-12 Objective Type Questions, ML Aggarwal Class-8 Linear Equations and Inequalities in One Variable ICSE Mathematics Solutions

#### Mental Maths

Question 1.
Fill in the blanks:
(i) An equation of the type ax + b = 0 where a ≠ 0 is called a …………. in variable x.
(ii) Any value of the variable which satisfies the equation is called a …………. of the equation.
(iii) The process of finding all the solutions of an equation is called ………….
(iv) We can add the …………. to both sides of an equation.
(v) We can divide both sides of an equation by the same …………. number.
(vi) The solution set of the inequality 3x ≤ 10, x ϵ N is ………….

(i) An equation of the type ax + b = 0
where a ≠ 0 is called a linear equation in variable x.
(ii) Any value of the variable which satisfies
the equation is called a solution of the equation.
(iii) The process of finding all the solutions of
an equation is called solving the equation.
(iv) We can add the same number to both sides of an equation.
(v) We can divide both sides of an equation
by the same non-zero number.
(vi) The solution set of the inequality 3x ≤ 10, x ϵ N is (1, 2, 3).

Question 2.
State whether the following statements are true (T) or false (F):
(i) An equation is a statement that two expressions are equal.
(ii) A term may be transposed from-one side of the equation to the other side, but its sign will not change.
(iii) We cannot subtract the same number from both sides of an equation.
(iv) 3x + 2 = 4(x + 7) + 9 is a linear equation in variable x.
(v) x = 1 is the solution of equation 4(x + 5) = 24.

(i) An equation is a statement that two expressions are equal. True
(ii) A term may be transposed from one side of
the equation to the other side, but its sign will not change. False
Correct:
The sign will change.
(iii) We cannot subtract the same number from
both sides of an equation. False
Correct:
We can subtract.
(iv) 3x + 2 = 4(x + 7) + 9 is a linear equation in variable x. True
(v) x = 1 is the solution of equation 4(x + 5) = 24. True
4(1 + 5) = 24 ⇒ 4 × 6 = 24

#### MCQs,

Multiple Choice Questions

Choose the correct answer from the given four options (3 to 16):
Question 3.
Which of the following is not a linear equation in one variable?
(a) 3x + 2 = 0
(b) 2y – 4 = y
(c) x + 2y = 7
(d) 2(x – 3) + 7 = 0

x + 2y = 7 is not a linear equation in one variable
as there are two variables x and y. (c)

Question 4.
The solution of the equation $\frac{2}{3} x+1=\frac{15}{9}$ is
(a) 1
(b) $\frac{3}{2}$
(c) 2
(d) $\frac{2}{3}$

The solution of the equation

Question 5.
The solution of the equation 4z + 3 = 6 + 2z is
(a) 1
(b) $\frac{3}{2}$
(c) 2
(d) 3

Solution of equation 4z + 3 = 6 + 2z
⇒ 4z – 2z = 6 – 3 ⇒ 2z = 3 ⇒ z = $\frac{3}{2}$ (b)

Question 6.
The solution of the equation $\frac{3 x}{5}+1=\frac{4 x}{15}$ is +7 is
(a) 12
(b) 14
(c) 16
(d) 18

Question 7.
The solution of the equation $\frac{x}{2}-\frac{1}{5}=\frac{x}{3}+$$\frac{1}{4}$ is
(a) 2.7
(b) 1.8
(c) 2.9
(d) 1.7

The solution of the equation

Question 8.
The solution of the equation $\frac{8 x-3}{3 x}=2$ is

Question 9.
If we subtract $\frac{1}{2}$ from a number and multiply the result by $\frac{1}{2}$, we get $\frac{1}{8}$, then the number is

Let number be x, then

Question 10.
Fifteen years from now Ravi’s age will be four times his present age. What is Ravi’s present age?
(a) 4 years
(b) 5 years
(c) 6 years
(d) 3 years

Let present age of Ravi = x years
After 15 years, his age will be = (x + 5) years
∴ x + 15 = 4x
⇒ 15 = 4x – x = 3x
⇒ x = $\frac{15}{3}$ = 5
∴ His present age = 5 years (b)

Question 11.
If the sum of three consecutive integers is 51, then the largest integer is
(a) 16
(b) 17
(c) 18
(d) 19

Let first integers = x
Then next two integers = x + 1, x + 2
∴ x + x + 1 + x + 2 = 51
⇒ 3x + 3 = 51
⇒ 3x = 51 – 3 = 48
⇒ x = $\frac{48}{3}$ = 16
∴ First integer = 16
and other two integer = 17, 18
Largest integers =18 (c)

Question 12.
If the perimeter of a rectangle is 13 cm and its Width is $2 \frac{3}{4}$ cm, then its length is

Perimeter of a rectangle = 13 cm

Question 13.
What should be added to twice the rational number $\frac{-7}{3}$ to get $\frac{3}{7}$ ?

According to the condition,

Question 14.
Sum of digits of a two digit number is 8. If the number obtained by reversing the digits is 18 more than the original number, then the original number is
(a) 35
(b) 53
(c) 26
(d) 62

Sum of digits of a two digit number = 8
Let unit digit = x
Then tens digit = 8 – x
∴ Number = x + 10(8 – x) = x + 80 – 10x = 80 – 9x
By reversing the digits,
Unit digit = 8 – x
and tens digit = x
∴ Number = 8 – x + 10x = 8 + 9x
∴ 8 + 9x = 80 – 9x + 18
⇒ 9x + 9x = 80 + 18 – 8
⇒ 18x = 90
⇒ x = $\frac{90}{18}$ =5
∴ Number = 80 – 9x = 80 – 9 × 5 = 80 – 45 = 35 (a)

Question 15.
Arjun is twice as old as Shriya. If five years ago his age was three times Shriya’s age, then Arjun’s present age is
(a) 10 years
(b) 15 years
(c) 20 years
(d) 25 years

Let Shriya’s age = x years
Then Arjun’s age = 2x
5 years ago,
Age of Shriya was = (x – 5) years
and age of Arjun’s = (2x – 5) years
∴ 2x – 5 = 3(x – 5)
⇒ 2x – 5 = 3x – 15
⇒ 3x – 2x = 15 – 5 = 10
⇒ x = 10
∴ Arjun’s present age = 2x = 2 × 10 = 20 years (c)

Question 16.
If the replacement set is {-5, -3, -1,0, 1, 3}, then the solution set of the inequation -3 < x < 3 is
(a) {-2,-1, 0, 1, 2}
(b) {-1, 0, 1, 2}
(c) {-3,-1, 0, 1, 3}
(d) {-1,0, 1}

Replacement set = {-5, -3, -1, 0, 1,3}
-3 < x < 3
∴ x = {-2, -1, 0, 1, 2} from the replacement set,
Solution set x = {-1, 0, 1} (d)

#### Value Based Questions

Question 1.
Seema is habitual of saving her pocket money. She collected some 50 paise and 25 paise coins in her piggy bank. If she collected ₹25 and number of 50 paise coins is double the number of 25 paise coins. How many coins of each type did she collect? What values are being promoted? Is saving a good habit?

Seema has 50 paise and 25 paise coins in his piggy bank.
Total amount = ₹25
Let 25 paise coins = x
Then 50 paise coins = 2x

∴ 25-paise coins = 20
and 50 paise coins = 20 × 2 = 40
This is a good habit to save some money,
we can solve any financial problem with its help at any time.

Question 2.
Ramesh gave one-fourth of his property to his two sons in equal shares and rest to his wife Sunita. Sunita gave one-third of her share to an orphanage. If the amount given by Sunita to the orphanage was ₹20000, find the total value of the Ramesh’s property and the amount each person got? What value is shown by the Sunita?

Let Ramesh’s property = x
$\frac{1}{4}$th part of property was given to two sons equally = $\frac{x}{4}$
So, each son’s share = $\frac{x}{4} \times \frac{1}{2}=\frac{x}{8}$
Rest to his wife Sunita = $x-\frac{1}{4} x=\frac{3}{4} x$
Sunita gave one third of her share to orphanages
Property given to orphanage
$\frac{1}{3} \text { of } \frac{3}{4} x=\frac{1}{4} x$
∴ $\frac{1}{4}$x = ₹20000
∴ Total value of Ramesh property = ₹20000 × $\frac{4}{1}$ = ₹80000
Each son will get = ₹$\frac{x}{8}$ × 80000 = ₹ 10000
and wife will get = ₹80000 × $\frac{3}{4}$ = ₹60000
Sunita done a good deed to help the orphanage
where needy person are living and they need your help and support.

#### (HOTS)

Higher Order Thinking Skills

Question 1.

A man covers a distance of 24 km in $3 \frac{1}{2}$ hours partly on foot at the speed of 4.5 km/h and partly on bicycle at the speed of 10 km/h. Find the distance covered on foot.

Total distance = 24 km
Time taken = $3 \frac{1}{2}$ hours = $\frac{7}{2}$ hours
Let a man travels x km on foot at the speed of 4.5 km
and (24 – x) km on bicycle at the speed of 10 km/hr

∴ He travelled 9 km on foot.

Question 2.
The perimeter of a rectangle is 240 cm. If its length is decreased by 10% and breadth is increased by 20% we get the same perimeter. Find the original length and breadth of the rectangle.

Perimeter of a rectangle = 240 cm
∴ Length + breadth = $\frac{240}{2}$ = 120 cm
Let length = x cm
Then breadth = (120 – x) cm
By decreasing length by 10%
and increasing breadth by 20%, we get

According to the condition,

∴ Length = 80 cm
and breadth = 120 – 80 = 40 cm

Question 3.
A person preparing a medicine wants to convert 15% alcohol solution into 32% alcohol solution. Find how much pure alcohol he should mix in 400 mL of 15% alcohol solution to obtain required solution?

15% of alcohol mixture = 400 mL
∴ Alcohol = $\frac{15}{100}$ × 400 = 60 mL
and other solution = 400 – 60 = 340 mL
In new mixture alcohol = 32%
Other solution = 100 – 32 = 68%
In 86 mL, alcohol = 32
and in 340 mL, alcohol will be = $\frac{32 \times 340}{68}$ = 160 mL
∴ More alcohol required = 160 – 60 = 100 mL

Question 4.
Rahul covers a distance from P to Q on bicycle at 10 km/h and returns back at 9 km/h. Anuj covers the distance from P to Q and Q to P both at 12 km/h. On calculating we find that Anuj took 10 minutes less than Rahul. Find the distance between P and Q.

Let distance between P and Q = x km
Speed of Rahul from P to Q = 10 km/h
and back Q to P = 9 km/h

Question 5.
Solve:

Solution:

Question 1.

Solve the following equations:

⇒ x = $\frac{24}{-3}$ = 8
⇒ x = 8
(iii) 5x – 3(4x – 3) = 2(4 – 5x) + 10
⇒ 5x – 12x + 9 = 8 – 10x + 10
⇒ 5x – 12x + 10x = 8 + 10 – 9
⇒ 3x = 9
⇒ x = $\frac{9}{3}$ = 3
∴ x = 3
(iv) $\frac{3 x+2}{5 x+4}=\frac{3}{4}$
By cross multiplication
12x + 8 = 15x + 12
⇒ 12x – 15x = 12 – 8
⇒ – 3x = 4

By cross multiplication
(2x – 3) (5x + 20) = (2x + 1) (5x – 4)
⇒ 10x2 + 40x – 15x – 60 = 10x2 – 8x + 5x – 4
⇒ 25x – 60 = -3x -4
⇒ 25x + 3x = -4 + 60
⇒ 28x = 56
⇒ x = $\frac{56}{28}$ = 2
∴ x = 2

⇒ 2x2 – 7x – 5 = x (2x – 2)
⇒ 2x2 – 7x – 5 = 2x2 – 2x
⇒ 2x2 – 7x – 2x2 – 5 + 2x = 0
⇒ -7x – 5 + 2x = 0
⇒ -5x – 5 = 0
⇒ -5x = 5
⇒ x = $\frac{5}{-5}$
⇒ x = – 1

Question 2.
The sum of three consecutive multiples of 11 is 363. Find these multiples.

Let first multiple of 11 = 11x
Then second multiple = 11x + 11
and third multiple = 11x + 22
∴ 11x + 11x+ 11 + 11x + 22 = 363
⇒ 33x + 33 = 363
⇒ x + 1 = 11
⇒ x = 11 – 1 = 10
∴ Multiples are 11 × 10 = 110
110 + 11 = 121
110 + 22 = 132
Hence number are 110, 121, 132

Question 3.
Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.

Sum of two numbers = 95
Let first number = x
Then second number = 95 – x
According to the condition,
x – (95 – x) = 15
⇒ x – 95 + x = 15
⇒ 2x = 15 + 95 = 110
⇒ x = $\frac{110}{2}$ = 55
∴ First number = 55
and second = 95 – 55 = 40
Hence numbers are 55, 40

Question 4.
One-half of a number is equal to one-third of its succeeding number. Find the first number.

Let, the first number be x
According to given problem,
$\frac{x}{2}=\frac{1}{3}(x+1)$
⇒ 3x = 2 (x + 1)
⇒ 3x = 2x + 2
⇒ 3x – 2x = 2 x = 2
Hence, the first number be 2.

Question 5.
The numerator of a rational number is 8 less than its denominator. If the numerator is increased by 2 and denominator is decreased by 1, the number obtained is $\frac{1}{2}$. Find the number.

Let denominator of a rational number = x
Then its numerator = x – 8
and fraction = $\frac{x-8}{x}$
According to the condition,
$\frac{x-8+2}{x-1}=\frac{1}{2} \Rightarrow \frac{x-6}{x-1}=\frac{1}{2}$
⇒ (x – 6) × 2 = x – 1
⇒ 2x – 12 = x – 1
⇒ 2x – x = 12 – 1
⇒ x = 11
∴ Fraction = $\frac{11-8}{11}=\frac{3}{11}$

Question 6.
The present ages of Rohit and Mayank are in the ratio 11 : 8. 8 years later the sum of their ages will be 54 years. What are their present ages?

Ratio in the present ages of Rohit and Mayank = 11 : 8
Let Rohit’s age = 11x years
and Mayank’s age = 8x years
8 years later,
Rohit’s age = 11x + 8
and Mayank’s age = 8x + 8
According to the condition,
⇒ 11x + 8 + 8x + 8 = 54
⇒ 19x = 54 – 8 – 8 = 54 – 16 = 38
x = $\frac{38}{19}$
∴ Rohit’s present age =11 × 2 = 22 years
and Mayank’s age = 8 × 2=16 years

Question 7.
A father’s age is 3 times the sum of ages of his two sons. Five years later he will be twice the sum of ages of his two sons. Find the present age of the father.

Let sum of ages of two sons = x years
Then fathers age = 3x years
5 years later
Sum of ages of two sons = x + 5 + 5 = x + 10 years
and father’s age = (3x + 5) years
According to the condition,
3x + 5 = 2(x + 10)
⇒ 3x + 5 = 2x + 20
⇒ 3x – 2x = 20 – 5
⇒ x = 15
∴ Father’s age = 3 × 15 = 45 years

Question 8.
The digits of a two-digit number differ by 7. If the digits are interchanged and the resulting number is added to the original number we get 121. Find the original number.

Let unit’s digit = x
Then ten’s digit = x – 7
Number = x + 10(x – 7) = x + 10x – 70 = 11x – 70
After interchanged the digits,
Unit’s digit = x – 7
and ten’s digit = x
∴ Number = x – 7 + 10x = 11x – 7
According to the condition,
11x – 70 + 11x – 7 = 121
⇒ 22x – 77 = 121
⇒ 22x = 121 + 77 = 198
x = $\frac{198}{22}$ = 9
∴ Original number = 11x – 70 = 11 × 9 – 70 = 99 – 70 = 29
Hence number is 29 or 92

Question 9.
The ten’s digit of a two-digit number exceeds its unit’s digit by 5. When digits are reversed, the new number added to the original number becomes 99. Find the original number.

Let, the digit at unit’s place = x
And, digit at ten’s place = x + 5
Number = 10 × (x + 5) + 1 × 5
= 10 (x + 5) + x
= 10x + 50 + x
= 11x + 50
Reversing the number = 1 × (x + 5) + 10 × x
= x + 5 + 10x = 11x + 5
According to given problem,
(11x + 50) + (11x + 5) = 99
⇒ 11x + 50 + 11x + 5 = 99
⇒ 22x + 55 = 99
⇒ 22x = 99 – 55
⇒ 22x = 44
⇒ x = $\frac{44}{22}$
⇒ x = 2
Hence, the number = 11 × 2 + 50
= 22 + 50 = 72

Question 10.
Sonia went to a bank with ₹2,00,000. She asked the cashier to give her ₹500 and ₹2000 currency notes in return. She got 250 currency notes in all. Find the number of each kind of currency notes.

Total amount = ₹2,00,000
and total number of currency notes = 250
Let 500 rupees notes = x
Then 2000’s rupee notes = 250 – x
According to the condition,
x × 500 + (250 – x) × 2000 = 2,00,000
⇒ 500x + 5,00,000 – 2000x = 2,00,000
⇒ -1500x = 2,00,000 – 5,00,000
⇒ -1500x = -3,00,000
⇒ x = $\frac{-3,00,000}{-1500}$ = 200
∴ 500 rupees notes = 200
and 2000 rupees notes = 250 – 200 = 50

Question 11.

Ajay covers a distance of 240 km in $4 \frac{1}{4}$ hours. Some part of the journey was covered at the speed of 45 km/h and the remaining at 60 km/h. Find the distance covered by him at the rate of 60 km/h.

Let the distance cohered by Ajay at the rate of 60 km/hr = x km.
Then, remaining distance covered by Ajay at the rate of 45 km/hr = (240 – x) km.
∵ Time taken by Ajay to cover the distance at the rate of 60 km/hr. = $\frac{x}{60}$ hr [ time = $\frac{\text { Distance }}{\text { time }}$
Also, time taken by Ajay to cover the distance at the rate of 45 km/hr = $\frac{(240-x)}{45}$ hr
According to given problem,
Total time to cover the distance

Hence, the distance covered by Ajay at the rate of 60 km/hr = 195 km.

Question 12.
If x ϵ {even integers), represent the solution set of the inequation -5 ≤ x < 5 on a number line.

Given inequation -5 ≤ x < 5
As x ϵ {even integers) then, solution is {-4, -2, 0, 2, 4}
Graphical representation on the number line

Question 13.
Solve the following inequality and graph its solution on a number line:

As x ϵI, then its solution set is {-4, -3, -2, -1, 0, 1, 2}
Graphical representation is

As x ϵ I, then its solution set is {-4, -3, -2, -1, 0, 1, 2}
Graphical representation is

—:  End of  ML Aggarwal Class-8 Linear Equations and Inequalities in One Variable Solutions :–