AP State Syllabus AP Board 7th Class Maths Solutions Chapter 10 Algebraic Expressions Ex 3 Textbook Questions and Answers.

## AP State Syllabus 7th Class Maths Solutions 10th Lesson Algebraic Expressions Exercise 3

Question 1.

Find the length of the line segment PR in the following figure in terms of ’a’.

Solution:

The length of PR = \(\overline{\mathrm{PQ}}+\overline{\mathrm{QR}}\)

= 3a + 2a

= 5a units

Question 2.

(i) Find the perimeter of the following triangle.

(ii) Find the perimeter of the following rectangle.

Solution:

i) The perimeter of the triangle = \(\overline{\mathrm{AB}}+\overline{\mathrm{BC}}+\overline{\mathrm{CA}}\)

= 2x + 6x + 5x

= 13x units

ii) The perimeter of the rectangle ABCD = \(\overline{\mathrm{AB}}+\overline{\mathrm{BC}}+\overline{\mathrm{CD}}+\overline{\mathrm{DA}}\)

or

= 2 (l + b)

= 2 (3x + 2x)

= 2 × 5x

= 10x units

Question 3.

Subtract the second terni from first term.

(i) 8x, 5x

(ii) 5p, 11p

(iii) 13m^{2}, 2m^{2}

Solution:

i) 8x – 5x = 3x

ii) 5p – 11p = -6p

iii) 13m^{2} – 2m^{2} = 11m^{2}

Question 4.

Find the value of following monomials, if x =1.

(i) -x

(ii) 4x

(iii) -2x^{2}

Solution:

1) If x = 1 ⇒ – x = – (1)= – 1

ii) If x = 1 ⇒ 4x = 4 x 1 = 4

iii) If x= 1 ⇒ – 2x^{2} = – 2(1)^{2} = – 2 × 1= – 2

Question 5.

Simplify and find the value of 4x + x – 2x^{2} + x – 1 when x = -1.

Solution:

Given expression = 4x + x – 2x^{2} + x – 1

= 4x + x + x – 2x^{2} – 1

= – 2x^{2} + 6x – 1

If x = – 1 then, the value of – 2x^{2} + 6x – 1 = – 2 (-1)2 + 6( – 1) – 1

= – 2(1) – 6 – 1

= – 2 – 6 – 1

= – 9

Question 6.

Write the expression 5x^{2} – 4 – 3x^{2} + 6x + 8 + 5x – 13 in its simplified form. Find its value when x = -2

Solution:

The given algebraic expression = 5x^{2} – 4 – 3x^{2} + 6x + 8 + 5x – 13.

= (5x^{2} – 3x^{2}) (6x + 5x) + (-4 – 13)

= 2x^{2} + 11x – 17

When x = – 2. then the value of 2x^{2}+ 11x – 17 = 2(- 2)^{2} + 11(-2) – 17

= 2 × 4 – 22 – 17

= 8 – 39

= – 31

Question 7.

If x = 1; y = 2 find the values of the following expressions

(i) 4x – 3y + 5

(ii) x^{2} + y^{2}

(iii) xy + 3y – 9

Solution:

i) 4x – 3y+5

If x = 1, y = 2 then the value of 4x – 3y + 5 = 4(1) – 3(2) + 5

= 4 – 6 + 5

= 9 – 6 = 3

ii) x^{2} + y^{2}

If x = 1, y = 2, the value of x^{2} + y^{2}= (1)^{2} + (2)^{2}

=(1 × 1) + (2 × 2)

= 5

iii) xy + 3y – 9

If x = 1; y = 2,the value of xy + 3y – 9 = 1 × 2 + 3(2) – 9

= 2 + 6 – 9

= 8 – 9

Question 8.

Area of a rectangle is given by A = l × b . If l = 9cm, b = 6cm, find its area?

Solution:

Formula for area of a rectangle (A) = l × b

If l = 9cm, b = 6cm

then the area of the rectangle = l × b

= 9 × 6

= 54 cm^{2}

Question 9.

Simple interest is given by I = \(\frac{P T R}{100}\) . If P = ₹ 900, T =2 years; and R =5%, find the simple interest.

Solution:

Given that the formula for simple interest (I) = \(\frac{P T R}{100}\)

P = Rs.900, T = 2years, R = 5%

∴ The required simple interest (I) = \(\frac{\mathrm{PTR}}{100}=\frac{900 \times 2 \times 5}{100}\) = 9 x 2 x 5 = Rs. 90

Question 10.

The relationship between speed (s), distance (d) and time (t) is given by S = – . Find the

value of s, if d = 135 meters and t = 10 seconds.

Solution:

Given that s = \(\frac{\mathrm{d}}{\mathrm{t}}\) and d = 135 meters, t = 10 seconds

∴ The value of s = \(\frac{\mathrm{d}}{\mathrm{t}}=\frac{135}{10}\)

∴ speed (s) = 13.5 m/sec.