AP State Syllabus AP Board 7th Class Maths Solutions Chapter 3 Simple Equations Ex 3 Textbook Questions and Answers.

## AP State Syllabus 7th Class Maths Solutions 3rd Lesson Simple Equations Exercise 3

Question 1.

Write the information given in the picture in the form of an equation. Also, find ‘x’ in the following figure.

Solution:

From the figure x + 11 = 15

∴ x = 15 – 11 (transposing +11)

∴ x = 4cm

Question 2.

Write the information given in the picture in the form of an equation. Also, find ‘y in the following figure.

Solution:

From the figure y + 8 = 13

y = 13 – 8 (transposing 8)

∴ y = 5cm

Question 3.

If we add 7 to twice a number, we get 49. Find the number.

Solution:

Let the number be x

Then twice the number = 2x

Onaddingi = 2x + 7

By problem, 2x + 7 = 49

2x = 49 – 7 (transposing + 7)

2x = 42

x = \(\frac { 42 }{ 2 }\) (transposing x 2)

x = 21

Question 4.

If we subtract 22 from three times a number, we get 68. Find the number.

Solution:

Let the number be x

Then three times the number = 3x

On subtracting 22 ⇒ 3x – 22

By problem, 3x – 22 = 68

3x = 68 + 22 (transposing -22)

3x = 90

x = \(\frac { 90 }{ 3 }\)(transposingx3)

x = 30

∴ The required number 30

Question 5.

Find a number which when multiplied by 7 and then reduced by 3 is equal to 53.

Solution:

Let the number be x

Multiplied by 7 ⇒ 7x

Then reduced by 3 ⇒ 7x – 3

By problem, 7x – 3 = 53

7x = 53 + 3 (transposing – 3)

7x = 56

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

x = 8 .

∴ The required number = 8

Question 6.

Sum of two numbers is 95. 1f one exceeds the other by 3, find the numbers.

Solution:

Let one number be = x.

Then the other number x – 3

Sumofthenumbers x + x – 3 = 2x – 3

By problem, 2x – 3 = 95

2x = 95 + 3 (transposing – 3)

2x = 98

x = \(\frac { 98 }{ 2 }\) (transposing x 2)

x = 49

∴if one number x 49 then the other number x – 3 = 49 – 3 = 46

Question 7.

Sum of three consecutive integers is 24. Find the integers.

Solution:

Let the three integers be = x, x + 1, x 2

Sumoftheintegers = x + x + 1 + x + 2 = 3x + 3

By problem, 3x + 3 = 24’

3x = 24 – 3 (Transposing + 3)

3x = 21

x = \(\frac { 21 }{ 3 }\) (transposing x 3)

x = 7

∴ The integers x = 7

x + 1 = 7 + 1 = 8

x + 2 = 7 + 2 = 9

Question 8.

Find the length and breadth of the rectangle given below if its perimeter is 72m.

Solution:

Length of the rectangle = 5x + 4

Breadth of the rectangLe = x – 4

Perimeter of the rectangle = 2 x (length + breadth)

= 2x[(5x + 4)+(x – 4)]

= 2[5x + 4 + x – 4]

= 2(6x)

= 12x

Byproblem, 12x = 72

x = \(\frac { 72 }{ 12 }\) (transposing x 12)

x = 6

∴ Length of the rectangle = 5x + 4 = 5 x 6.4 = 34cm

Breadth of the rectangÌe = x – 4 = 6 – 4 = 2cm

Question 9.

Length of a rectangle exceeds its breadth by 4 m. 1f the perimeter of the rectangle is 84 m, find its length and breadth.

Solution:

Let the breadth be x

Then its length = x + 4

Perimeter of the rectangle = 2 x (length + breadth)

=2[x + 4 + x]

= 2 (2x + 4)

= 4x + 8

By problem, 4x + 8 = 84

4x = 84 – 8 (transposing + 8)

4x = 76

x = \(\frac{76}{4}\) (transposing x 4)

x = 19

∴ Length of the rectangle =x+4= 19+4 = 23m

Breadth of the rectangle = 19m

Question 10.

After 15 years, Hema’s age will become four times that of her present age. Find her present age.

Solution:

Let the preser age of Hema be x years

After 15 years Hema Age = 4x

By problem, x + 15 = 4x

x + 15 – 4x = 4x – 4x (Subtracting 4x from both sides)

– 3x + 15 = 0

– 3x = – 15

x = \(\frac{-15}{-3}\) transposing x ( – 3)]

x = 5

∴ Her present age is 5 years.

∴ Her present age is 5 years.

Question 11.

A sum of ₹.3000 is to be given in the form of 63 prizes. Ifthe prize money is either ₹. 100 or.25. Find the number of prizes of each type.

Solution:

Let the number of ₹ 100 prizes be x

Then the number of ₹ 25 prizes be = 63 – x

Value of the prizes = 100x + (63 – x) x 25

= 100x+ 1575 – 25x

= 75x + 1575

By problem, 75x + 1575 = 3000

75x = 3000 – 1575

75x = 1425

x = \(\frac { 1425 }{ 75 }\)

x = 19

∴ ₹ 100 prizes = 19

₹25 prizes= 63 – x =63 – 19 = 44

Question 12.

A number is divided into two parts such that one part is 10 more than the other. Ifthe two parts are in the ratio 5:3, find the number and the two parts.

Solution:

Let one part be x

Then the other part = x + 10

Ratio of these two parts = x + 10 : x

Byproblem, x + 10 : x=5:3

∴ \(\frac{x+10}{x}=\frac{5}{3}\)

3(x+ 10)=5x

3x + 30 = 5x

30 = 5x – 3x

2x = 30

x = \(\frac{30}{2}\)

x = 15

If one part is 15 then the other part must be x + 10 = 15 + 10 = 25

∴ The number is 15 + 25 40

Question 13.

Suhana said, “multiplying my number by 5 and adding 8 to it gives the same answer as subtracting my number from 20”. Find Suhana’s numbers.

Solution:

Let Suhana’s number be x

Muhtplying by S and adding 8 to that number 5x + 8

Subtracting that number from 20 = 20 – x

By problem above two are equal.

i.e. 5x + 8 = 20 – x

5x + x = 20 – 8

6x = 12

x = \(\frac{12}{6}\) = 2

x = 2

∴ Suhanas number is 2

Question 14.

The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest mark is 87. What is the lowest mark?

Solution:

Let the Lowest mark of the class = x

Twice the least mark = 2x

On adding 7 = 2x + 7

By problem; 2x + 7 = 87

2x = 87 – 7

2x = 80

x = \(\frac{80}{2}\) = 40

x = 40

∴ The lowest mark = 40

Question 15.

In adjacent figure find the magnitude of each of the three angles formed?

(Hint: Sum of all angles at a point on a line is 180°)

Solution:

(Hint : Sum of all angles at a point on a line is 180°)

We know sum of angles at a point = 180°

∴ x° + 2x° + 3x° = 180

6x° = 180°

x = \(\frac{180}{6}\) = 30

∴ The angles are x = 30°

2x = 2 x 30 = 60°

3x° = 3 x 30° = 90°

Question 16.

Solve the following riddle:

I am a number

Tell my identity

Take me two times over

And add a thirty six.

To reach a century

Solution:

Let the number be x

Two times the number = 2x

On adding 36 = 2x 36

By problem, 2x + 36 = 100 – 4

2x 36 = 96

2x = 96 – 36

2x = \(\frac{60}{2}\)

x = 30