The DAV Maths Book Class 7 Solutions Pdf and **DAV Class 7 Maths Chapter 7 Worksheet 2** Solutions of Linear Equations in One Variable offer comprehensive answers to textbook questions.

## DAV Class 7 Maths Ch 7 WS 2 Solutions

Question 1.

Adding 4 to twice a number yields \(\frac{25}{2}\). Find the number.

Answer:

Let the number be x

∴ its twice = 2x

Now, 2x + 4 = \(\frac{25}{2}\)

2x = \(\frac{25}{2}\) – 4

2x = \(\frac{25-24}{2}\)

2x = \(\frac{1}{6}\)

x = \(\frac{1}{6×2}\)

∴ x = \(\frac{1}{12}\)

Hence, the required number is \(\frac{1}{12}\).

Question 2.

A number when added to its two-thirds is equal to 55. Find the number.

Answer:

Let the number be x

its two-third = \(\frac{2}{3}\)x

x + \(\frac{2}{3}\)x = 55

⇒ \(\frac{3x+2x}{3}\) = 55

⇒ \(\frac{5x}{3}\) = 55

⇒ x = 55 × \(\frac{3}{5}\)

⇒ x = 11 × 3 = 33

x = 33

Hence, the requires number is 33.

Question 3.

A number when multiplied by 4 exceeds itself by 45. Find the number.

Answer:

Let the number be x

4x – x = 45

3x = 45

Hence, the required number is 15.

Question 4.

Find the number which when multiplied by 8 and then reduced by 9 is equal to 47.

Answer:

Let the number be x

8x – 9 =47

8x = 9 + 47

⇒ 8x = 56

x = \(\frac{56}{8}\) = 7

Hence, the required number is 7.

Question 5.

The sum of two numbers is 72. If one of the numbers is 6 more than the other number, find the numbers.

Solution:

Let one number be x

Other number = 72 – x

Now, x = 72 – x + 6

⇒ x + x = 72 + 6

⇒ 2x = 78

⇒ x = \(\frac{78}{2}\) = 39

Hence, first number is 39 and the other number = 72 – 39 = 33.

Question 6.

The sum of two numbers is 99. If one exceeds the other by 9, find the numbers.

Answer:

Let one number be x

∴ Other number = (99 – x)

Now, x – (99 – x) = 9

⇒ x – 99 + x = 9

⇒ 2x = 99 + 9

⇒ 2x = 108

⇒ x = \(\frac{108}{2}\) = 54

Hence, one number is 54 and other number is 99 – 54 = 45.

Question 7.

One number is 10 more than the other. If their sum is 52, find the numbers.

Answer:

Let one number be x

Other number = x + 10

Now, x + x + 10 =52

⇒ 2x = 52 -10

⇒ 2x = 42

⇒ x = \(\frac{42}{2}\) = 21

Hence, one number = 21 and the other number = 52 – 21 = 31.

Question 8.

The length of a rectangle is 20 cm more than its breadth. If the perimeter is 100 cm, find the dimensions of the rectangle.

Answer:

Let the breadth of the rectangle be x cm

its length = (x + 20) cm and

Perimeter of the rectangle = 2 [l + b]

∴ 2[x + 20 + x] = 100

⇒ 2[2x + 20] = 100

⇒ 4x + 40 = 100

⇒ 4x = 100-40

⇒ 4x = 60

⇒ x = \(\frac{60}{4}\) = 15

Hence, breadth = 15 cm and length = 15 + 20 = 35 cm.

Question 9.

The length of a rectangle is 3 times its breadth. If the perimeter is 84 cm, find the length of the rectangle.

Answer:

Let the breadth be x cm

∴ its length = 3x cm

Perimeter = 2[l + b]

∴ 2[3x + x] = 84

⇒ 2 x 4x = 84

⇒ 8x = 84

⇒ x = \(\frac{84}{8}\) = 10.5 cm

Hence, length = 10.5 × 3 = 31.5 cm.

Question 10.

Two equal sides of an isosceles triangle are each 2 cm more than thrice the third side. If the perimeter of triangle is 67 cm, find the lengths of its sides.

Answer:

Let the third side be x cm

∴ length of other two sides = (3x + 2) cm each

Perimeter of triangle = Sum of three sides

∴ x + 3x + 2 + 3x + 2 = 67

⇒ 7x + 4 = 67

⇒ 7x = 67 – 4

⇒ 7x = 63

⇒ x = \(\frac{63}{7}\) = 9

Hence, the length of sides are 9cm, 3 × 9 + 2 = 29 cm, 3 × 9 + 2 = 29 cm.

Question 11.

Length of a rectangle is 16cm less than twice its breadth. If the perimeter of the rectangle is 100 cm, find its length and breadth.

Answer:

Let the breadth be x cm

∴ length = (2x -16) cm

Perimeter = 2[l + b]

∴ 2[l + b] = 100

⇒ 2 [2x -16 + x] = 100

⇒ 2(3x -16) = 100

⇒ 6x – 32 = 100

⇒ 6x = 100 + 32

⇒ 6x = 132

⇒ x = \(\frac{136}{6}\) = 22

Hence, breadth = 22cm

length = (2 × 22 -16) = (44 – 16) = 28 cm

Question 12.

Find two consecutive positive integers whose sum is 63.

Answer:

Let the two consecutive integers be x and x + 1

x + x + 1 = 63

⇒ 2x = 63 – 1

⇒ 2x = 62

⇒ x = \(\frac{62}{2}\) = 31

Hence, the two consecutive integers are 31 and 31 + 1 = 32.

Question 13.

A sum oft 800 is in the form of denominations of ₹ 10 and ₹ 20. If the total number of notes is 50, find the number of notes of each type.

Answer:

Let the number of t 10 notes be x

∴ The number of 120 notes = (50 – x)

∴ 10x + 20(50 – x) = 800

⇒ 10x + 1000 – 20x = 800

⇒ -10x = 800 – 1000

⇒ -10x = -200

⇒ x = \(\frac{-200}{-10}\) = 20

Hence, number of 110 notes = 20 and number of t 20 notes = 50 – 20 = 30.

Question 14.

In a class of 49 students, number of girls is y of the boys. Find the number of the boys in the class.

Answer:

Let number of boys be x

Number of girls = (49 – x)

49 – x = \(\frac{2}{5}\)x

⇒ 5(49 – x) = 2x

⇒ 245 – 5x = 2x

⇒ -5x – 2x = -245

⇒ -7x = -245

⇒ x = \(\frac{-245}{-7}\) = 35

Hence, number of boys = 35 and number of girls = 49 – 35 = 14.

Question 15.

Leena has ₹ 117 in the form of 5 rupee coins and 2 rupee coins. The number of 2 rupee coins is 4 times that of 5 rupee coins. Find the number of coins of each denomination.

Answer:

Let the number of 5-rupee coins be x

2-rupee coins = 4x

∴ 5x + 2 (4x) = 117

⇒ 5x + 8x = 117

⇒ 13x = 117

∴ x = \(\frac{117}{13}\) = 9

Hence, number of 5-rupee coin = 9 and number of 2-rupee coin = 9 × 4 = 36.

Question 16.

A total of ₹ 80,000 is to be distributed among 200 persons as prizes. A prize is either of ₹ 500 or ₹ 100. Find the number of each prize.

Answer:

Let number of ₹ 500 prize be x

∴ Number of 1100 prize = (200 – x)

∴ 500x + 100(200 – x) = 80,000

⇒ 500x + 20000 – 100x = 80,000

⇒ 400x = 80,000-20,000

⇒ 400x = 60,000

∴ x = = 150

Hence, number of ₹ 500 prize = 150 and number of ₹ 100 prize = 200 -150 = 50

Question 17.

When \(\frac{1}{3}\) is subtracted from a number and difference is multiplied by 4, the result is 28, find the number.

Answer:

Let the required number be x

4(x – \(\frac{1}{3}\)) = 28

⇒ 4x = 28 + \(\frac{4}{3}\)

⇒ 4x – \(\frac{4}{3}\) = 28

⇒ 4x = \(\frac{84+4}{3}\)

⇒ 4x = \(\frac{88}{3}\)

x = \(\frac{88}{3 \times 4}=\frac{88}{12}=\frac{22}{3}\)

Hence, the required number is \(\frac{22}{3}\)

Question 18.

Sudesh is twice as old as Seema. If six years is subtracted from Seema’s age and 4 years are added to Sudesh’s age, Sudesh will be four times Seema’s age. How old were they 3 years ago?

Solution:

Let Seema’s age be x years

Sudesh’s age = 2x years

If Seema’s age = (x – 6) years

and Sudesh’s age = (2x + 4) years

2x + 4 = 4 (x – 6)

⇒ 2x + 4 = 4x – 24

⇒ 2x – 4x = – 4 – 24

⇒ – 2x = – 28

⇒ x = 14

Seema’s present age = 14 years and Sudesh’s present age = 2 × 14 = 28 years.

Hence, 3 years ago, Seema’s age = 14 – 3 = 11

Question 19.

The ages of Leena and Heena are in the ratio 7 : 5. Ten years hence, the ratio of their ages will be 9 : 7. Find their present ages.

Answer:

Let their present ages be 7x years and 5x years respectively.

∴ After 10 years, Leena’s age = (7x + 10) years

After 10 years, Heena’s age = (5x +10) years .

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

⇒ 7(7x + 10) = 9(5x + 10)

⇒ 49x + 70 = 45x + 90.

⇒ 49x – 45x = 90-70

⇒ 4x = 20

⇒ x = \(\frac{20}{4}\) = 5

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

and the present age of Heena = 5 × 5 = 25 years.

Question 20.

Vikas is three years older than Deepika. Six years ago, Vikas’s age was four times Deepika’s age. Find the age of Deepika and Vikas.

Answer:

Let Deepika’s age be x years

∴ Vikas’s age = 3x years

Six years ago, Deepika’s age was = (x – 6) years

Six years ago Vikas’s age was = (3x – 6) years

∴ 4(x – 6) = (3x – 6)

⇒ 4x – 24 = 3x – 6

⇒ 4x – 3x = 24 – 6

⇒ x = 18

Hence Deepika’s age = 18 years and Vikas’s age = 18 × 3 = 54 years.

## DAV Class 7 Maths Chapter 7 Value Based Questions

Question 1.

Aman is an avid reader. He has a good collection of books. In his summer vacation he read a book of 300 pages in four days.

(i) If the number of pages he read on second, third and fourth day is half the number of pages he read on previous day, find the number of pages he read on first day.

(ii) Discuss the importance of reading books.

Answer:

(i) Let the number of pages Aman read on first day be x.

∴ Number of pages Aman read on second day = \(\frac{x}{2}\)

Number of pages Aman read on third day = \(\frac{1}{2}\left(\frac{x}{2}\right)=\frac{x}{4}\)

Number of pages Aman read on fourth day = \(\frac{1}{2}\left(\frac{x}{2}\right)=\frac{x}{4}\)

Now, x + \(\frac{x}{2}+\frac{x}{4}+\frac{x}{8}\) = 300

⇒ \(\frac{8 x+4 x+2 x+x}{8}\) = 300

⇒ \(\frac{15 x}{8}\) = 300

⇒ x = \(\frac{8 \times 300}{15}\)

⇒ x = 8 × 20

∴ x = 160

So, the number of pages he read on first day were 160.

(ii) Importance of reading books:

- Reading books improves vocabulary.
- Through reading, we expose ourselves to new things, new information, new ways to solve a problem, and new ways to achieve one thing.
- Through reading, we begin to have a greater understanding on a topic.
- Reading improves understanding.
- Reading books boost imagination and creativity.

Question 2.

Students of Rotary Club prepared ‘Tie and Dye Dupattas’, ‘Wall Hangings’ and ‘Hand Bags’ for sale in an Exhibition to help the poor people. They earned a total profit oft 2000.

(i) If profit on Dupattas is 130 less than two times the profit on Wall Hangings and profit on Hand Bags is 180 more than twice the profit on Wall Hangings. Find the profit on Wall Hangings,

(ii) What value is exhibited by the students?

Answer:

Let the profit on Wall Hangings be ₹ x.

Now, profit on Dupattas = 2x – 30

and profit on Hand Bags = 2x + 80

Now x + 2x – 30 + 2x + 80 = 2000

⇒ 5x + 50 = 2000

⇒ 5x = 1950

⇒ x = \(\frac{1950}{5}\) = 390

Therefore, the profit on Wall Hangings is ₹ 390.

(ii) Values: Students are helping and caring towards poor people.