## ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 12 Linear Equations and Inequalities in One Variable Ex 12.2

Question 1.

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

Solution:

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.

Solution:

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?

Solution:

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?

Solution:

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.

Solution:

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.

Solution:

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.

Solution:

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.

Solution:

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 ?

Solution:

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 + 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 = \(\frac{60}{20}\)

⇒ 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?

Solution:

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.

Solution:

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?

Solution:

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?

Solution:

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.

Solution:

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?

Solution:

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?

Solution:

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.

Solution:

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 ?

Solution:

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.

Solution:

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.

Solution:

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.

Solution:

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.

Solution:

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.

Solution:

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.

Solution:

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.

Solution:

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 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.

Solution:

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^{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.

Solution:

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.

Solution:

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.

Solution:

Distance between two places A and B = 350 km

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

The speed of car C_{2} = (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 C_{1} = 40 km/h

and speed of car C_{2} = 40 – 8 = 32 km/h