The DAV Class 8 Maths Solutions and **DAV Class 8 Maths Chapter 9 Worksheet 2 **Solutions of Linear Equations in One Variable offer comprehensive answers to textbook questions.

## DAV Class 8 Maths Ch 9 WS 2 Solutions

Question 1.

The present ages of A and B are in the ratio 7 : 5. Ten years later, their ages will be in the ratio 9 : 7. Find their present ages.

Solution:

Let the present ages of A and B be 7x years and 5x years respectively.

After ten years, the age of A will be (7x + 10) years.

After ten years, the age of B will be (5x + 10) years.

As per the condition,

\(\frac{7 x+10}{5 x+10}=\frac{9}{7}\)

⇒ 7 (7x + 10) = 9 (5x + 10)

⇒ 49x + 70 = 45x + 90

⇒ 49x – 45x = 90 – 70

⇒ 4x = 20

∴ x = 5

Hence, the present age of A = 7 × 5 = 35 years

and that of B = 5 × 5 = 25 years.

Question 2.

Two years ago, father was three times as old as his son and two years hence, twice his age will be equal to five times that of his son. Find their present ages.

Solution:

Let the present age of son be x years

∴ Present age of father = 3x years

After two years the son’s age = (x + 2) years

After two years the father’s age = (3x + 2) years

As per the conditions,

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

⇒ 6x + 4 = 5x + 10

⇒ 6x – 5x = 10 – 4

⇒ x = 6

Hence, the present age of the son is 6 years

and the present age of father = 6 × 3 = 18 years.

Question 3.

The sum of the digits of a two-digit number is 8. The number obtained by interchanging the digits exceeds the given number by 18. Find the given number.

Solution:

Let unit place digit be x.

∴ Ten’s place digit = (8 – x)

∴ Number = x + 10 (8 – x)

= x + 80 – 10x

= – 9x + 80

Number obtained by enterchanging the digits = 10x + 8 – x = 9x + 8

As per the condition,

(9x + 8) – (- 9x + 80) = 18

⇒ 9x + 8 + 9x – 80 = 18

⇒ 18x – 72 = 18

⇒ 18x = 18 + 72

⇒ 18x = 90

∴ x = 5

Unit place digit = 5

and ten’s place digit = 8 – 5 = 3

Hence the required number = 5 + 3 × 10 = 35

Question 4.

The ones digit of a two digit number is twice the tens digit. When the number formed by reversing the digits is added to the original number, the sum is 99. Find the original number.

Solution:

Let the ten’s place be x.

∴ One’s place will be 2x

Number formed = 2x + 10 × x = 12x

Number obtained by reversing the digits = x + 10

∴ 2x = 21x

As per the condition,

12x + 21x = 99

⇒ 33x = 99

⇒ x = \(\frac{99}{33}\) = 3

Unit place = 2 × 3 = 6

and Ten’s place = 3

Hence, the required number = 6 + 3 × 10 = 36.

Question 5.

The sum of the digits of a 2-digit number is 11. The number obtained by interchanging the digits exceeds the original number by 27. Find the number.

Solution:

Let the unit place digit be x.

∴ Ten’s place digit = (11 – x)

Original number = x + 10 (11 – x)

= x + 110 – 10x

= – 9x + 110

New number obtained by reversing the digits = 10x + 11 – x = 9x + 11

As per the condition,

(9x + 11) – (- 9x + 110) = 27

⇒ 9x + 11 + 9x – 110 = 27

⇒ 18x – 99 = 27

⇒ 18x = 99 + 27

⇒ 18x = 126

⇒ x = 7

Unit place digit = 7

and Ten’s place digit = 11 – 7 = 4

Hence, the required number = 7 + 4 × 10 = 47.

Question 6.

The sum of three consecutive multiples of 7 is 777. Find these multiples,

Solution:

Let the three consecutive multiples of 7 be x, x + 7 and x + 14

∴ x + x + 7 + x + 14 = 777

⇒ 3x + 21 = 777

⇒ 3x = 777 – 21

⇒ 3x = 756

⇒ x = \(\frac{756}{3}\) = 252

Hence, the required multiples are 252, 252 + 7 = 259 and 259 + 7 = 266.

Question 7.

The sum of three consecutive multiples of 9 is 999. Find these multiples.

Solution:

Let the three consecutive multiples of 9 be x, x + 9, x + 18

∴ x + x + 9 + x + 18 = 999

⇒ 3x + 27 = 999

⇒ 3x = 999 – 27

⇒ 3x = 972

⇒ x = \(\frac{972}{3}\) = 324

Hence, the required multiples of 9 are 324, 324 + 9 = 333 and 333 + 9 = 342

Question 8.

The denominator of a rational number is greater than its numerator by 7. If 3 is subtracted from the numerator and 2 is added to the denominator, the new number becomes \(\frac{1}{5}\). Find the rational number.

Solution:

Let the numerator be x

∴ Denominator = x + 7

As per condition, the new numerator = x – 3

and the new denominator = x + 7 + 2 = x + 9

∴ New rational number = \(\frac{x-3}{x+9}\)

⇒ \(\frac{x-3}{x+9}=\frac{1}{5}\)

⇒ 5x – 15 = x + 9

⇒ 5x – x = 15 + 9

⇒ 4x = 24

⇒ x = 6

Numerator = 6 and

the denominator = 6 + 7 = 13

Hence, the new rational number = \(\frac{6}{13}\).

Question 9.

The numerator and denominator of a rational number are in the ratio 3 : 4. If the

denominator is increased by 3, the ratio becomes 3 : 5. Find the rational number.

Solution:

Let the numerator be 3x and denominator be 4x.

As per conditions,

\(\frac{3 x}{4 x+3}=\frac{3}{5}\)

⇒ 15x = 3 (4x + 3)

⇒ 15x=12x+9

⇒ 15x – 12x = 9

⇒ 3x = 9

∴ x = 3

Numerator = 3 × 3 = 9

Denominator = 4 × 3 = 12

Hence, the required rational number = \(\frac{9}{12}\).

Question 10.

A motor boat goes downstream and covers a distance in 4 hours, while it covers the same distance upstream in 5 hours. If the speed of the stream is 3 km/hr, find the speed of the motor boat in still water.

Solution:

Let the speed of the motor boat be x km/hr.

Speed of the motor boat in daystream = (x + 3) km/hr

Speed of the motor boat in upstream = (x – 3) km/hr

Distance covered by motor boat in 4 hrs = 4 (x + 3) km

Distance covered motor boat is 5 hrs = 5 (x – 3) km

As per the condition,

⇒ 4 (x + 3) = 5 (x – 3)

⇒ 4x + 12 = 5x – 15

⇒ 4x – 5x = – 12 – 15

⇒ x = – 27

∴ x = 27

Hence, the speed of the motor boat = 27 km/hr.

Question 11.

A streamer, going downstream in a river, covers the distance between two towns in 15 hours. Coming back upstream, it covers this distance in 20 hours. The speed of the water is 3 km / hr. Find the distance between the two towns.

Solution:

Let the speed of the streamer be x km/hr

Speed of the streamer in down stream = (x + 3) km/hr

Speed of the streamer in upstream = (x – 3) km/hr

Distance covered in 15 hours = 15(x + 3) km

Distance covered in 20 hours = 20(x – 3) km

As per the condition,

15 (x + 3) = 20 (x – 3)

⇒ 15x – 20x = – 45 – 60

⇒ – 5x = – 105

⇒ x = 21

Hence, the distance between the towns = 15 (21 + 3)

= 15 × 24 = 360 km.

Question 12.

The distance between two towns is 300 km. Two cars start simultaneously from these towns and move towards each other. The speed of one car is more than the other by 7 km/hr. If the distance between the cars after 2 hours is 34 km, find the speed of the cars.

Solution:

Let the speed of one car be x km/hr.

∴ Speed of the other car = (x + 7) km/hr

Distance travelled by 1st car in 2 hours = 2x km

and the distance travelled by 2nd car in 2 hours = 2 (x + 7) km

Total distance covered by the two cars = 2x + 2 (x + 7) km

As per the conditions,

2x + 2 (x + 7) = 300 – 34

⇒ 4x = 266 – 14

⇒ 4x = 252

⇒ x = \(\frac{252}{4}\) = 63

Hence the speed of 1st car = 63 km/hr

and the speed of 2nd car = 63 + 7 = 70 km/hr.

Question 13.

The length of a rectangle is greater than the breadth by 18 cm. If both length and breadth are increased by 6 cm, then the area increases by 168 cm^{2}. Find the length and breadth of the rectangle.

Solution:

Let the breadth of a rectangle be x cm.

Its length = (x + 18) cm

As per the conditions,

New length = (x + 18 + 6) = (x + 24) cm

and new breadth = (x + 6) cm

∴ New area = (x + 24) (x + 6) cm^{2}

and original area = x (x + 18)

∴ (x + 24) (x + 6) = x (x + 18) = 168

x^{2} + 30x + 144 – x^{2} – 18x = 168

⇒ 12x + 144 = 168

⇒ 12x = 24

∴ x = 2

Hence, the required length = 2 + 18 = 20 cm

and breadth = 2 cm.

Question 14.

The length of a rectangle is greater than the breadth by 3 cm. If the length is increased by 9 cm and the breadth is reduced by 5 cm, the area remains the same. Find the dimensions of the rectangle.

Solution:

Let the breadth be x cm

Length = (x + 3) cm

New length = (x + 3 + 9) = (x + 12) cm

New breadth = (x – 5) cm

∴ New area = Length × Breadth

= (x + 12) (x – 5) cm^{2}

Original area = (x + 3) x cm^{3}

As per the condition,

(x + 12) (x – 5) = (x + 3) x

x^{2} + 12x – 5x – 60 = x^{2} + 3x

12x – 5x – 3x = 60

4x = 60

x = 15

Hence breadth = 15 cm

and length = (15 + 3) = 18 cm.

Question 15.

The difference between two positive integers is 30. The ratio of these integers is 2 : 5. Find the integers.

Solution:

Let the two integers be 2x and 5x.

∴ 5x – 2x = 30

⇒ 3x = 30

⇒ x = 10

Hence, the required integers are 2 × 10 = 20 and 5 × 10 = 50.

Question 16.

The sum of two positive integers is 105. The integers are in the ratio 2 : 3. Find the integers.

Solution:

Let the two integers be 2x and 3x.

∴ 2x + 3x = 105

⇒ 5x = 105

⇒ x = 21

The required intergers are 2 × 21 = 42 and 3 × 21 = 63.

Question 17.

A money box contains one-rupee and two-rupee coins in the ratio 5 : 6. If the total value of the coins in the money box is ₹ 85, find the number of two-rupee coins.

Solution:

Let number of one-rupee coins be 5x

and the number of two-rupee coins = 6x

As per the condition,

\(\frac{5 x}{100}+\frac{6 x}{50}\) = 85

⇒ \(\frac{5 x+12 x}{100}\) = 85

⇒ 17x = 85 × 100

⇒ x = \(\frac{85 \times 100}{17}\)

⇒ x = 500

Hence the number of two-rupee coins = 6 × 500 = 3000.

Question 18.

A purse has only one-rupee coins and two-rupee coins in it. The number of two rupee coins is one-third the number of one-rupee coins. If the purse has only ₹ 115, find the number of two-rupee coins.

Solution:

Let the number of one-rupee coins be x

∴ The number of two rupee coins = \(\frac{x}{3}\)

As per the condition,

\(\frac{x}{100}+\frac{x}{3 \times 50}\) = 115

⇒ \(\frac{x}{100}+\frac{x}{150}\) = 115

⇒ \(\frac{3 x+2 x}{300}\) = 115

⇒ 5x = 300 × 115

⇒ x = \(\frac{300 \times 115}{5}\)

⇒ x = 6900

Hence the number of two-rupee coins = \(\frac{6900}{3}\) = 2300.

### DAV Class 8 Maths Chapter 9 Value Based Questions

Question 1.

If a scooterist drives at the rate of 24 km/hr from his home, he reaches his work place 5 minutes late. But if he drives at the rate of 30 km/hr, he reaches his work place 4 minutes early.

(i) Find the distance of his work place from his home.

(ii) Is it advisable to drive at high speed? Give reason.

Solution:

(i) Let the distance of work place from home be x km.

Case I.

Speed of scooterist = 24 km/hr

Let the time taken be t_{1} hr.

∴ Speed = \(\frac{\text { distance }}{\text { time }}\)

⇒ 24 = \(\frac{x}{t_1}\)

⇒ t_{1} = \(\frac{x}{24}\)

Case II.

Speed of scooterist = 30 km/hr

Let the time taken be t_{2} hr.

Speed = \(\frac{\text { distance }}{\text { time }}\)

⇒ 30 = \(\frac{x}{t_2}\)

⇒ t_{2} = \(\frac{x}{30}\)

ATQ,

t_{1} – t_{2} = 9 minutes = \(\frac{9}{60}\) hr

⇒ \(\frac{x}{24}-\frac{x}{30}=\frac{9}{60}\)

⇒ \(\frac{5 x-4 x}{120}=\frac{9}{60}\)

⇒ x = \(\frac{9 \times 120}{60}\)

⇒ x = 18 km

Thus, the distance of work place from home of scooterist is 18 km.

(ii) No, it may lead to trouble.

Question 2.

DAV school wants to give 15 prizes to its students on the values of discipline, politeness and punctuality. If the number of prizes for politeness is live-sixth of the number of prizes for discipline and the number of prizes for punctuality is four-fifths that of number of prizes of politeness, then

(i) Find the number of prizes for each value.

(ii) Apart from the above three values, write one more value for which the prize can be given to students.

Solution:

(i) Let the number of prizes for discipline be x.

Then, the number of prizes for politeness is \(\frac{5 x}{6}\)

Also, the number of prizes for politeness is \(\frac{4}{5} \times \frac{5 x}{6}=\frac{2 x}{3}\)

Now, total number of prizes = 15

⇒ \(x+\frac{5 x}{6}+\frac{2 x}{3}\) = 15

⇒ \(\frac{18 x+15 x+12 x}{18}\) = 15

⇒ \(\frac{45 x}{18}\) = 15

⇒ x = \(\frac{15 \times 18}{45}\)

⇒ x = 6

Thus, the number of prizes for discipline = 6

The number of prizes for politeness = \(\frac{5 \times 6}{6}\) = 5

The number of prizes for punctuality = \(\frac{2 \times 6}{3}\) = 4

(ii) Compassion, social justice, honesty or any other value.