SCERT AP 7th Class Maths Solutions Pdf Chapter 6 Algebraic Expressions Unit Exercise Questions and Answers.

## AP State Syllabus 7th Class Maths Solutions 6th Lesson Algebraic Expressions Unit Exercise

Question 1.

Fill in the blanks:

(i) The constant term in the expression a + b + 1 is ………………..

Answer:

1.

(ii) The variable in the expression 3x – 8 is ………………..

Answer:

x

(iii) The algebraic term in the expression 2d – 5 is ………………..

Answer:

2d

(iv) The number of terms in the expression ……………….. p^{2} – 3pq + q is

Answer:

3

(v) The numerical coefficient of the term – ab is …………………..

Answer:

– 1

Question 2.

Write below statements are True or False:

(i) \(\frac{3 x}{9 y}\) is a binomial.

Answer:

False

(ii) The coefficient of b in – 6abc is – 6a.

Answer:

False

(iii) 5pq and – 9qp are like terms.

Answer:

True

(iv) The sum of a + b and 2a + 7 is 3a + 7b.

Answer:

False

(v) When x = – 2, then the value of x + 2 is 0.

Answer:

True.

Question 3.

Identify like terms among the following:

3a, 6b, 5c, – 8a, 7c, 9c, – a, \(\frac{2}{3}\)b, \(\frac{7 c}{9}\), \(\frac{a}{2}\).

Answer:

Given terms are

3a, 6b, 5c, – 8a, 7c, 9c, – a, \(\frac{2}{3}\)b, \(\frac{7 c}{9}\), \(\frac{a}{2}\)

Like terms: 3a, – 8a, – a, \(\frac{a}{2}\)

6b, \(\frac{2}{3}\)b

5c, 7c, 9c, \(\frac{7 c}{9}\)

Question 4.

Arjun and his friend George went to a stationary shop. Arjun bought 3 pens and 2 pencils whereas George bought one pen and 4 pencils. If the price of each pen and pencil is ₹ x and ₹ y respectively, then find the total bill amount in x and y.

Answer:

Given cost of each pen is ₹ x and cost of each pencil is ₹ y .

Arjun bought 3 pens and 2 pencils. Cost of 3 pens = 3 × ₹ x = ₹ 3x

Cost of 2 pencils = 2 × ₹y = ₹ 2y

Amount paid by Arjun = 3x + 2y = ₹ (3x + 2v)

George bougth one pen and 4 pencils.

Cost of 1 pen = 1 × ₹ x = ₹ x

Cost of 4 pencils = 4 × ₹ y = ₹ 4y

Amount paid by George = x + 4y = ₹(x + 4y)

Total bill amount

= Arjun amout + George amount

= (3x + 2y) + (x + 4y)

= 3x + 2y + x + 4y

= 3x + x + 2y + 4y

= (3 + 1)x + (2 +4)y

∴ Total bill amount = ₹ (4x + 6y)

Question 5.

Find the errors and correct the following :

(i) 7x + 4y = 11xy

Answer:

7x and 4y are not like terms and different variables x, y.

So, we should not add the coefficients.

(ii) 8a^{2} + 6ac = 14a^{3}c

Answer:

8a^{2} and 6ac are not like terms. So, we should not add the coefficients.

(iii) 6pq^{2} – 9pq^{2} = 3pq2

Answer:

6pq^{2} – 9pq^{2} = (6 – 9)pq^{2}

= – 3pq^{2}

(iv) 15mn – mn = 15

Answer:

15mn – mn = (15 – 1) mn

= 14 mn

(v) 7 – 3a = 4a

Answer:

7 and 3a are not like terms.

So, we should not subtract the coefficient 7 and 3.

Question 6.

Add the expressions

(i) 9a + 4, 2 – 3a

Answer:

Given expressions are 9a + 4; 2 – 3a

Write the given expressions in standard form.

9a + 4, – 3a + 2.

The sum = (9a + 4) + (- 3a + 2)

= 9a + 4 – 3a + 2

= (9a – 3a) + (4 + 2)

= (9 – 3)a + 6

= 6a + 6

(ii) 2m – 7n, 3n + 8m, m + n

Answer:

Given expressions are

2m – 7n, 3n + 8m, m + n

Write the given expressions in standard form.

2m – 7n, 8m + 3n, m + n

The sum

= (2m 7n) + (8m + 3n) + (m + n)

= 2m – 7n + 8m + 3n + m + n

= (2m + 8m + m) + (- 7n + 3n + n)

= (2 + 8 + 1)m + (- 7 + 3 + 1)n

= 11 m + (- 3)n

= 11 m – 3n

Question 7.

Subtract:

(i) – y from y

Answer:

– y from y = y – (-y) = y + y = 2y

(ii) 18 pq from 25pq

Answer:

18pq from 2 5pq

= 25 pq – 18 pq

= (25 – 18) pq = 7pq

(iii) 6t + 5 from 1 – 9t

Answer:

Given expressions are (6t + 5), (1 – 9t)

Write the given expressions in the standard form (6t + 5); (- 9t + 1)

(6t + 5) from (- 9t + 1)

= (- 9t + 1) – (6t + 5)

= – 9t + 1 – 6t – 5

= (- 9t – 6t) + (1 – 5)

= (- 9 – 6) t + (- 4) = – 15t – 4

Question 8.

Simplify the following :

(i) t + 2 + t + 3 + t + 6- t- 6 + t

Answer:

Given t + 2 + t + 3+ t + 6 – t – 6 + t

= 3t + 5

(ii) (a + b + c) + (2a + 3b – c) – (4a + b – 2c)

Answer:

Given (a + b + c) + (2a + 3b – c) – (4a + b – 2c)

= a + b + c + 2a + 3b – c – 4a – b + 2c

= (a + 2a – 4a) + (b + 3b – b) + (c – c + 2c)

= (1 + 2 – 4)a + (1 + 3 – 1)b + (1 – 1 + 2)c

= (- 1) a + 3b + 2c

= – a + 3b + 2c

(iii) x + (y + 1) + (x + 2) + (y + 3) + (x + 4) + (y + 5)

Answer:

Given x + (y + 1) + (x + 2) + (y + 3) + (x + 4) + (y + 5)

= x + y + 1 + x + 2 + y + 3 + x + 4 + y + 5

= (x + x + x) + (y + y + y) + (1 + 2 + 3 + 4 + 5)

= 3x + 3y + 15

Question 9.

The perimeter of a triangle is 8x^{2} + 7x – 9 and two of its sides are x^{2} – 3x + 4, 2x^{2} + x – 9 respectively, then find third side.

Answer:

Let the sides of triangle are A, B, C.

A = x^{2} – 3x + 4; B = 2x^{2} + x – 9 ; C = ?

Perimeter = 8x^{2} + 7x – 9

Perimeter of the triangle = A + B + C

To get the third side (C). subtract sum of A and B from the perimeter.

∴ C = Perimeter – (A + B)

So,

A + B = (x^{2} – 3x + 4) + (2x^{2} + x – 9)

= x^{2} – 3x + 4 + 2x^{2} + x – 9

= x^{2} + 2x^{2} – 3x + x +4 – 9

= (1 + 2) x^{2} + (- 3 + 1) x – 5

A + B = 3x^{2} – 2x – 5

Additive inverse of A + B is – (A + B)

– (A + B) = – (3x^{2} – 2x – 5) .

– (A + B) = – 3x^{2} + 2x + 5

C = perimeter + [- (A + B)]

= (8x^{2} + 7x – 9) + (- 3x^{2} + 2x + 5)

= 8x^{2} + 7x – 9 – 3x^{2} + 2x + 5

= 8x^{2} – 3x^{2} + 7x + 2x – 9 + 5

= (8 – 3) x^{2} + (7 + 2) x – 4

C = 5x^{2} + 9x – 4

∴ Third side is 5x^{2} + 9x – 4.

Question 10.

The perimeter of a rectangle is 2a^{3} – 4a^{2} – 12a + 10, if length is 3a^{2} – 4, find its breadth.

Answer:

Given length of rectangle l = 3a^{2} – 4 and breadth b = ?

Perimeter of a rectangle = 2a^{3} – 4a^{2} – 12a + 10

Perimeter of a rectangle

= 2(l + b)

= 2a^{3} – 4a^{2} – 12a + 10

= \(\frac{1}{2}\) × 2(a^{3} – 2a^{2} – 6a + 5)

⇒ l + b = (a^{3} – 2a^{2} – 6a + 5)

⇒ l + b – l = (a^{3} – 2a^{2} – 6a + 5) – l

∴ b = (a^{3} – 2a^{2} – 6a + 5) – l

Additive inverse of 1 is – 1 = – (3a^{2} – 4)

∴ – l = – 3a^{2} + 4 .

b = (a^{3} – 2a^{2} – 6a + 5) + (- 1)

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

= a^{3} – 2a^{2} – 6a + 5 – 3a^{2} + 4

= a^{3} – 2a^{2} – 3a^{2} – 6a + 5 + 4

= a^{3} + (- 2 – 3) a^{2} – 6a + 9

= a^{3} + (- 5) a^{2} – 6a + 9

∴ Breadth of rectangle

= a^{3} – 5a^{2} – 6a + 9