ML Aggarwal Linear Equations and Inequalities in One Variable Exe-12.2 Class 8 ICSE Ch-12 Maths Solutions. We Provide Step by Step Answer of Exe-12.2 Questions for Linear Equations and Inequalities in One Variable as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-8.

**ML Aggarwal Linear Equations and Inequalities in One Variable Exe-12.2 Class 8 ICSE Maths Solutions**

Board | ICSE |

Publications | Avichal Publishig Company (APC) |

Subject | Maths |

Class | 8th |

Chapter-12 | Linear Equations and Inequalities in One Variable |

Writer | ML Aggarwal |

Book Name | Understanding |

Topics | Solution of Exe-12.2 Questions |

Edition | 2023-2024 |

**Linear Equations and Inequalities in One Variable Exe-12.2**

ML Aggarwal Class 8 ICSE Maths Solutions

Page-213

**Question 1. Three more than twice a number is equal to four less than the number. Find the number.**

**Answer:**

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.**

**Answer:**

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 = 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 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?**

**Answer:**

Let a number thought by Manjula = x

According to the condition,

(x – 7/3) × 6 = 2x – 2

⇒ 6x – 14 = 2x – 2

⇒ 6x – 2x = -2 + 14 = 12

⇒ 4x = 12

⇒ x = 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 5/2 times the other new number. What are the numbers?**

**Answer:**

Let the required number be x

Then the other number = x / 7

x + 15 = 5 / 2 {(x / 7) + 15}

2 (x + 15) = (5x / 7) + (5 × 15)

2x + 30 = (5x / 7) + 75

2x – (5 / 7) x = 75 – 30

{(14 – 5) / 7}x = 45

9x / 7 = 45

x = 45 × 7 / 9

We get,

x = 35

One number = 35

Other number = 35 / 7 = 5

Therefore, the numbers are 35 and 5

**(ML Aggarwal Linear Equations and Inequalities in One Variable Exe-12.2 Class 8 )**

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

**Answer:**

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 = -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.**

**Answer:**

Let the first odd integer = x

Then next consecutive odd integers = (x + 2)

(2 / 5) (x) = (2 / 9) (x + 2) + 4

2x / 5 = {2 (x + 2)} / 9 + 4

(2x / 5) – {2 (x + 2)} / 9 = 4

{18x – 10 (x + 2)} / 45 = 4

(18x – 10x – 20) / 45 = 4

(8x – 20) / 45 = 4

8x – 20 = 4 × 45

8x – 20 = 180

8x = 180 + 20

8x = 200

x = 200 / 8

x = 25

So, two consecutive odd integers are x = 25 and

(x + 2) = (25 + 2) = 27

Hence, two consecutive odd integers are 25 and 27

**(ML Aggarwal Linear Equations and Inequalities in One Variable Exe-12.2 Class 8 )**

**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 3/5. Find the original fraction.**

**Answer:**

Let the numerator of the original fraction = x

Then, its denominator = 2x + 1

Hence,

The fraction = x / (2x + 1)

(x + 5) / {(2x + 1) + 5} = 3 / 5

(x + 5) / (2x + 1 + 5) = 3 / 5

5 (x + 5) = 3 (2x + 6)

5x + 25 = 6x + 18

5x – 6x = 18 – 25

-x = – 7

x = 7

Original fraction = x / (2x + 1)

= 7 / {2 (7) + 1}

= 7 / (14 + 1)

= 7 / 15

Therefore, the original fraction is 7 / 15

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

**Answer:**

Let the two numbers be 2x and 5x

Because the ratio of these two numbers = 2x / 5x

= 2 / 5

= 2: 5

5x – 2x = 15

3x = 15

x = 15 / 3

x = 5

So, the numbers are 2x = 2 × 5 = 10 and 5x = 5 × 5 = 25

Hence, the required numbers are 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 ?**

**Answer:**

Let x be the required number

(12 + x), (22 + x), (42 + x) and (72 + x) are in proportion

(12 + x) / (22 + x) = (42 + x) / (72 + x)

On cross multiplication,

(12 + x) (72 + x) = (42 + x) (22 + x)

12 (72 + x) + x (72 + x) = 42 (22 + x) + x (22 + x)

864 + 12x + 72x + x^{2} = 924 + 42x + 22x + x^{2}

864 + 84x + x^{2} = 924 + 64x + x^{2}

864 + 84x + x^{2} – 924 – 64x – x^{2} = 0

864 + 84x – 64x – 924 = 0

84x – 64x = 924 – 864

20x = 60

x = 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?**

**Answer:**

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 = 110/22 = 5

∴ Original number = 11x+ 30 = 11 × 5 + 30 = 55 + 30 = 85

**(ML Aggarwal Linear Equations and Inequalities in One Variable Exe-12.2 Class 8 )**

**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.**

**Answer:**

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 = 162/18 = 9

∴ Original number = 110 – 9x = 110 – 9 × 9 = 110 – 81

= 29

**Linear Equations and Inequalities in One Variable Exe-12.2**

**ML Aggarwal Class 8 ICSE Maths Solutions**

Page-214

**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?**

**Answer:**

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 = 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?**

**Answer:**

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 = -24/-6

⇒ x = 4

Hence, age of son = 4 years

and age of his father = 7 × 4 years = 28 years.

**(ML Aggarwal Linear Equations and Inequalities in One Variable Exe-12.2 Class 8 )**

**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.**

**Answer:**

Ratio of ages of Sona and Sonali = 5: 3

Let us consider the present age of Sona and Sonali be 5x and 3x years respectively

Five years hence,

The age of Sona = 5x + 5 and

The age of Sonali = 3x + 5

(5x + 5) / (3x + 5) = 10 / 7

On cross multiplication,

7 (5x + 5) 10 (3x + 5)

35x + 35 = 30x + 50

35x – 30x = 50 – 35

5x = 15

x = 15 / 5

x = 3

Present age of Sona = 5x = 5 × 3 = 15 years and

Present age of Sonali = 3x = 3 × 3 = 9 years

Hence, the present age of Sona and Sonali is 15years 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?**

**Answer:**

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 = 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?**

**Answer:**

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 = 140/7 = 20

∴ 5 rupee coins = 20

2 rupee coins = 3 × 20 = 60

and 1 rupee coins = 160 – 60 – 20 = 80

**(ML Aggarwal Linear Equations and Inequalities in One Variable Exe-12.2 Class 8 )**

**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.**

**Answer:**

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 = 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 ?**

**Answer:**

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 = 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.**

**Answer:**

Let the number of deer in the herd = x

Number of deer grazing in the field = x / 2

Remaining = x – (x / 2)

= x / 2

Given that the (3 / 4) of the remaining deer are playing

= (3 / 4) × (1 / 2) x

= (3 / 8) x

Rest of deer = (x / 2) – (3 / 8) x

= (1 / 8) x

(1 / 8) x = 9

x = 9 × 8

x = 72

Therefore, total number of deer in the herd = 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.**

**Answer:**

Let the total number of flowers in the basket = x

Flowers offered in first temple = x / 2

Remaining flowers = x – (x / 2)

= (x / 2)

Flowers offered in the second temple

(x / 2) × (1 / 2) = x / 4

Remaining flowers = (x / 2) – (x / 4)

= (x / 4)

Flowers offered in the third temple = (x / 4) × (1 / 2)

= (x / 8)

Remaining flowers = (x / 4) – (x / 8)

= (x / 8)

(x / 8) = 6

x = 6 × 8

x = 48

Hence, number of flowers she had in the beginning = 48

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

**Answer:**

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 = 230°/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.**

**Answer:**

Let the angles of a triangle are 5x, 6x, and 7x

Then, we know that,

5x + 6x + 7x = 180°

⇒ 18x = 180°

⇒ x = 180°/18

⇒ x= 10°

Hence, the angle of a triangle are 5 × 10°, 6 × 10°,

and 7 × 10° i.e. 50°, 60° and 70°.

**(ML Aggarwal Linear Equations and Inequalities in One Variable Exe-12.2 Class 8 )**

**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.**

**Answer:**

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.**

**Answer:**

Let the perimeter of original triangle be x cm

If each side of a triangle is increased by 4, then,

The perimeter will be = x + 4 × 3

= (x + 12) cm

Ratio of perimeter of new triangle and given triangle = 7: 5

(x + 12) / x = 7 / 5

On cross multiplication,

5 (x + 12) = 7x

5x + 60 = 7x

7x – 5x = 60

2x = 60

x = 60 / 2

x = 30

Hence, the perimeter of the given triangle is 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.**

**Answer:**

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 = 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 cm^{2} = 322 cm^{2}

**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.**

**Answer:**

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 = 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^{2 }= 323 cm^{2}

**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.**

**Answer:**

Let the speed of the steamer = x km/ h

The speed downstream = (x + 5) km/h and

The speed upstream = (x – 5) km/ h

90 / (x + 5) = 60 / (x – 5)

On cross multiplication, we get,

90 (x – 5) = 60 (x + 5)

90x – 450 = 60x + 300

90x – 60x = 300 + 450

30x = 750

x = 750 / 30

x = 25

Hence, the speed of the streamer in still water is 25 km/ h

**(ML Aggarwal Linear Equations and Inequalities in One Variable Exe-12.2 Class 8 )**

**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.**

**Answer:**

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

**Linear Equations and Inequalities in One Variable Exe-12.2**

ML Aggarwal Class 8 ICSE Maths Solutions

Page-215

**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.**

**Answer:**

Distance between two places A and B = 350 km

Let the speed of car C_{1} = x km/h and

Speed of car C_{2} = (x – 8) km/h

After 4 hours, the distance between two cars is 62 km

Hence,

x × 4 + (x – 8) × 4 = 350 – 62

4x + 4x – 32 = 288

8x = 288 + 32

8x = 320

x = 320 / 8

x = 40

Hence, speed of car C_{1} = 40 km/h

Speed of car C_{2} = (x – 8) = (40 – 8) = 32 km/h

**Question 30. Nine person went to a hotel to take their dinner. Eight of them spent ₹60 each for the dinner and the ninth person spent ₹40 more than the averages expenditure of all the nine. How much money was spent by all of them?**

**Answer:**

8 persons spent Rs. 60 each for their meals.

So the total money spent by 8 persons = 8*60 = 480

Suppose that the ninth person spent = x

So the total money spent by 9 persons = 480+x

Average of the money spent by 9 persons = (480+x)/9

As per question, the ninth one spent Rs., 40 more than the average expenditure of all the nine. So…

x = (480+x)/9+40

x = 480/9+x/9+40

x-x/9 = 480/9+40

(9x-x)/9 = (480+360)/9

8x/9 = 840/9

x = (840/9)*(9/8)

x = 105

So the total money spent by 9 persons = 480+x

= 480+105

= 585

— End of Linear Equations and Inequalities in One Variable **Exe-12.2 **Class 8 ICSE Maths Solutions :–

Return to : **– **ML Aggarwal Maths Solutions for ICSE Class -8

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