SCERT AP 7th Class Maths Solutions Pdf Chapter 9 Algebraic Expressions Ex 9.4 Textbook Exercise Questions and Answers.

## AP State Syllabus 7th Class Maths Solutions 9th Lesson Algebraic Expressions Ex 9.4

Question 1.

Find the value of the expression

2x^{2} – 4x + 5 when

(i) x = 1

(ii) x = – 2

(iii) x = 0.

Answer:

Given expression is 2x^{2} – 4x + 5

(i) When x = 1, then

= 2(1)^{2} – 4(1) + 5

= 2 × 1 – 4 + 5

= 2 – 4 + 5 = 3

When x = 1, then 2x^{2} – 4x + 5 = 3

(ii) When x = – 2, then

= 2(- 2)^{2} – 4(- 2) + 5

= 2(4) + 8 + 5

= 8 + 8 + 5 = 21

When x = – 2, then 2x^{2} – 4x + 5 = 21

(iii) When x = 0, then

= 2(0)^{2} – 4(0) + 5

= 2(0) – 0 + 5

= 0 – 0 + 5 = 5

When x = 0, then 2x^{2} – 4x + 5 = 5

Question 2.

Find the value of Expressions when m = 2, n = – 1.

(i) 2m + 2n

Answer:

Given expression is 2m + 2n

If m = 2, n = – 1, then

2m + 2n = 2(2) + 2(- 1) = 4 – 2 = 2

∴ If m = 2, n = – 1, then 2m + 2n = 2

(ii) 3m – n

Answer:

Given expression is 3m – n

If m = 2, n = – 1, then

3m – n = 3(2) -(-1) = 6 + 1 = 7

∴ If m = 2, n = – 1, then 3m – n = 7

(iii) mn – 2.

Answer:

Given expression is mn – 2

If m = 2, n = – 1, then

mn – 2 = 2 × (-1)-2

= – 2 – 2 = – 4

∴ If m = 2, n = – 1, then mn – 2 = – 4

Question 3.

Simplify and find the value of the expression 5x^{2} – 4 – 3x^{2} + 6x + 8 + 5x – 13 when x = – 2.

Answer:

Given expression is

5x^{2} – 4 – 3x^{2} + 6x + 8 + 5x – 13

= (5x^{2} – 3x^{2}) + (6x + 5x) + (-4 + 8 – 13)

= (5 – 3)x^{2} + (6 + 5)x + (- 9)

= 2x^{2} + 11x – 9

If x = – 2, then 2x^{2} + 11x – 9

= 2(- 2)^{2} + 11 (- 2) – 9

= 2(4) – 22 – 9

= 8 – 22 – 9

∴ If x = – 2, then 2x^{2} + 11x – 9 = – 23

Question 4.

Find the length of the line segment PQ when a = 3 cm.

Answer:

From the figure,

Given PR = 3a and RQ = 2a

PQ = PR + RQ

= 3a + 2a

= (3 + 2)a

PQ = 5a

If a = 3 cm, then PQ = 5(3)

∴ PQ = 15 cm

Question 5.

The area of a square field of side ‘s’ meters is s2 sq. m. Find the area of square field, when

(i) s = 5m

(ii) s =12m

(iii) s = 6.5m

Answer:

From the figure,

Area of square = s2 sq.m.

(i) If s = 5m, then

s^{2} = (5)^{2} = 5 × 5 = 25 sq.m

(ii) If s = 12 m, then

s^{2} = (12)^{2} = 12 × 12 = 144 sq.m

(iii) If s = 6.5m, then

s^{2} = (6.5)^{2} = 6.5 × 6.5 = 42.25 sq.m

Question 6.

The area of triangle is given by \(\frac{1}{2}\) ∙ b ∙ h and if b = 12 cm, h = 8 cm, then find the area of triangle.

Answer:

Given area of the triangle = \(\frac{1}{2}\) ∙ b ∙ h

If b = 12 cm, h = 8 cm

= 6 × 8 = 48 sq.cm.

Question 7.

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

Answer:

Given simple interest I = \(\frac{\text { PTR }}{100}\)

If P = ₹ 900, T= 2 and R = 5% then

= 9 × 2 × 5 = 90

Question 8.

Find the errors and correct them in the following:

The value of following when a = – 3.

(i) 3 – a = 3 – 3 = 0

Answer:

3 – a = 3 – (- 3) (when a = – 3)

= 3 + 3 = 6

(Error is – (- 3) = – 3)

(ii) a^{2} + 3a = (- 3)^{2} + 3(- 3) = 9 + 0 = 9

Answer:

a^{2} + 3a = (- 3)^{2} + 3(- 3) (when a = – 3)

= (- 3 × – 3) – 9

(Error is 3(- 3) = 0)

= 9 – 9 = 0

(iii) a^{2} – a – 6 = (- 3)^{2} – (- 3) – 6 = 9 – 3 – 6 = 0

Answer:

a^{2} – a – 6

= (- 3)^{2} – (- 3) – 6 (when a = – 3)

= (- 3 × – 3) + 3 – 6

(Error is – (- 3) = – 3)

= 9 + 3 – 6

= 12 – 6 = 6

(iv) a^{2} + 4a + 4 = (- 3)^{2} + 4(-3) + 4 = 9 + 1 + 4 = 14

Answer:

a^{2} + 4a + 4

= (- 3)^{2} + 4(- 3) + 4 (when a = – 3)

= (- 3 × – 3)- 12 + 4

(Error is 4 (- 3) = 1)

= 9 – 12 + 4 = 13 – 12 = 1

(v) a^{3} – a^{2} – 3 = (- 3)^{3} – (-3)^{2} – 3 = – 9 + 6 – 3 = – 6

Answer:

a^{3} – a^{2} – 3

= (- 3)^{3} – (- 3)^{2} – 3 (when a = – 3)

= (- 3 × – 3 × – 3) – (- 3 × – 3) – 3

= – 27 – (9) – 3

(Error is (- 3)^{3} = – 9 and (- 3)^{2} = 6)

= – 27 – 9 – 3 = – 39