The DAV Class 7 Maths Solutions and **DAV Class 7 Maths Chapter 6 Brain Teasers** Solutions of Algebraic Expressions offer comprehensive answers to textbook questions.

## DAV Class 7 Maths Ch 6 Brain Teasers Solutions

Question 1.

A. Tick (S) the correct option:

(i) (3p^{2} – 14pq + 2r) – (14pq + 3p^{2} + 2r^{2}) is a:

(a) monomial

(b) binomial

(c) trinomial

(d) none

Answer:

(3p^{2} – 14pq + 2r) – (14pp + 3p^{2} + 2r^{2})

= 3p^{2} – 14pq + 2r – 14pq – 3p^{2} – 2r^{2}

= -2 r^{2} – 28 pq + 2r

As this algebraic expression have three terms so it is a trinomial.

Hence, (c) is the correct answer.

(ii) H.C.F. of the terms of the expression (4p^{3}q^{2}r – 12pq^{2}r^{2} + 16p^{2}q^{2}r^{2}) is:

(a) 4pq^{2}r

(b) -4pq^{2}r

(c) 16p^{3}q^{2}r^{2}

(d) -16p^{3}q^{2}r^{2}

Answer:

We know HCF shall be the product of the highest common coefficient and highest common powers of the variables respectively.

HCF of 4p^{3}q^{2}r, -12pq^{2}r^{2} and 16p^{2}q^{2}r^{2} is 4pq^{2}r.

Hence, (a) is the correct option.

(iii) For what value of’t’ the expressions (2x^{2} – 5x + 10) amd (2x^{2} – tx + 2t) are equal:

(a) 2

(b) 5

(c) 3

(d) 4

Answer:

The given expressions are (2x^{2} – 5x + 10) and (2x^{2} – tx + 2t).

Now, comparing the coefficients of x, we have -5 = -t

⇒ t = 5

Also, comparing the constant terms in the given expressions, we have

10 = 21

⇒ t = \(\frac{10}{2}\) = 5

Hence, (b) is the correct option.

(iv) Value of ‘p’ if the expression z^{2} + 3z – p equals to 3 for z = 2 is:

(a) 6

(b) 5

(c) 7

(d) 4

Answer:

On putting the value of z = 2 in the expression z^{2} + 3z – p, we have

(2)2 + 3(2) – p = 3

4 + 6 – p = 3

⇒ 10 – p = 3

⇒ p = 10 – 3

⇒ p = 7

Hence, (c) is the correct option.

(v) Factors of 2x^{3} + 5x – 6x^{2} – 15 are:

(a) (2x^{2} + 5) and (x + 3)

(b) (2x^{2} – 5) and (x + 3)

(c) (2x^{2} + 5) and (x – 3)

(d) (2x^{2} – 5) and (x + 3)

Answer:

= (2x^{3} – 6x^{2}) + (5x – 15) = 2x^{2} (x – 3) + 5(x – 3)

= (x – 3)(2x^{2} + 5)

= (2x^{2} + 5) (x – 3)

Hence, (c) is the correct option.

B. Answer the following questions:

(i) In the given figure, ABCD is a rectangle with length (p^{2}q – 5) and breadth (q^{2} + 4p). Find the area of shaded triangle PAB.

Answer:

Draw PQ perpendicular to AB

Here, base (AB) = (p2q – 5) and altitude (PQ) = (q2 + 4p)

∴ Area of shaded triangle PAB = \(\frac{1}{2}\) × Base × Altitude

= \(\frac{1}{2}\) × AB × PQ

= \(\frac{1}{2}\) × (p^{2}q – 5) × (q^{2} + 4p)

= \(\frac{1}{2}\) (p^{2}q – 5) (q^{2} + 4P)

= \(\frac{1}{2}\)[p^{2}q[q^{2} + 4p) – 5(q^{2} + 4p)]

= \(\frac{1}{2}\) (p^{2}q^{3} + 4p^{3}q – 5q^{2} – 20p) sq. units

(ii) By how much does the expression 72z^{2} – 45z + 4 exceed the expression 30z – 42z^{2} – 17?

Answer:

The expression (72z^{2} – 45z + 4) exceeds the expression (30z – 42z^{2} -17) by (72z^{2} – 45z + 4) – (30z – 42z^{2} – 17).

Now, subtracting by horizontal method, we have (72z^{2} – 45z + 4) – (30z – 42z^{2} – 17)

= 72z^{2} – 45z + 4 – 30z + 42z^{2} + 17

= (72z^{2} + 42z^{2}) + (-45z – 30z) + (4 + 17)

= 114z^{2} – 75z + 21

So, the expression (72z^{2} – 45z + 4) exceeds the expression (30z – 42z^{2} – 17) by (114z^{2} – 75z + 21).

(iii) The perimeter of a triangle is (x^{2}y + 10) units. One of the sides is of length (x^{2}y – 4) units and another side is (3 – 2x^{2}y) units. Find the length of the third side.

Answer:

Perimeter of a triangle = Sum of three sides of the triangle.

∴ (x^{2}y + 10) = (x^{2}y – 4) + (3 – 2x^{2}y) + Length of the third side

⇒ (x^{2}y + 10) = (-x^{2}y -1) + length of the third side

⇒ Length of the third side = (x^{2}y + 10) – (-x^{2}y – 1)

= x^{2}y + 10 + x^{2}y + 1

= 2 x^{2}y + 11 units

So, the length of the third side is given by the algebraic expression (2x^{2}y + 11) units.

(iv) Find the HCF of the terms of the expression

2b^{3}c + 4b^{4}c^{2} + 16b^{2}c^{2}

Answer:

HCF of 2b^{3}c, 4b^{4}c^{2} and 16b^{2}c^{2} is 2b^{2}c.

∴2b^{3}c + 4b^{4}c^{2} + 16b^{2}c^{2}

= 2b^{2}c (b + 2b^{2}c + 8c)

So, factors of the expression 2b^{3}c + 4b^{4}c^{2} + 16b^{2}c^{2} are 2b^{2}c and (b + 2b^{2}c + 8c).

(v) If area of a rectangle with length (a – b) and breadth (2a + b) is the same as the algebraic expression Kab + 2a^{2} – b^{2}, find K.

Answer:

Length of rectangle = (a – b)

Breadth of rectangle = (2a + b)

∴ Area of rectangle = Length x Breadth

= (a – b)(2a + b)

= a(2a + b) – b(2a + b)

= 2a^{2} + ab – 2ab – b^{2}

= 2a^{2} – ab – b^{2}

Now, according to the question, we have

2 a^{2} – a b – b^{2}

= Kab + 2 a^{2} – b^{2}

Comparing the coefficients of term ‘ab’, we have

K = -1

So, the value of K is -1.

Questions 2.

Find the following products.

(i) (13ax – 4) (5ay + 1)

Answer:

(13ax – 4) (5ay + 1) = 13ax (5ay + 1) – 4 (5ay + 1)

= 13ax × 5ay + 1 × 13ax – 4 × 5ay – 4 × 1 .

= 65a^{2}xy + 13 ax – 20ay – 4

(ii) (3x^{2} + 5x – 7) (x + 5y)

Answer:

(3x^{2} + 5x – 7) (x + 5y)

= 3x^{2} (x + 5y) + 5x (x + 5y) – 7 (x + 5y)

= 3x^{3} + 15x^{2}y + 5x^{2} + 25xy – 7x – 35y

(iii) -6a^{2} b (a^{4} + b^{4} – 3a^{2} b^{2})

Answer:

– 6a^{2}b (a^{4} + b^{4} – 3a^{2}b^{2})

= – 6a^{2}b × a4 – 6a^{2}b × b^{4} – 6a^{2}b × – 3a^{2}b^{2}

= – 6a^{6}b – 6a^{2}b^{5} + 18a^{4}b^{3}

(iv) (\(\frac{3}{5}\) x – \(\frac{2}{9}\)y)(15x – 9y)

Answer:

(\(\frac{3}{5}\) x – \(\frac{2}{9}\)y) (15x – 9y) = \(\frac{3}{5}\)x (15x – 9y) – \(\frac{2}{9}\)y (15x – 9y)

= \(\frac{3}{5}\)x × 15x – \(\frac{3}{5}\)x × 9y – \(\frac{2}{9}\)y × 15x – \(\frac{2}{9}\)y × – 9y

= 9x^{2} – \(\frac{27}{5}\)xy – \(\frac{10}{3}\)y – xy + 2y^{2}

= 9x^{2} – \(\left(\frac{27}{5}+\frac{10}{3}\right)\)xy + 2y^{2}

= 9x^{2} – \(\frac{131}{15}\)xy + 2y^{2}

(v) (x^{9}) (- x^{10}) (x^{11}) (-x^{12})

Answer:

(x^{9}) (- x^{10}) (x^{11}) (- x^{12})

= (-)( -) x^{9+10+11+12}

= x^{42}

(vi) 0.9 p^{3} q^{3} (10p – 20q)

Answer:

0.9p^{3}q^{3} (10p – 20q)

= 0.9p^{3}q^{3} × 10p – 0.9p^{3}q^{3} × 20q

= 9p^{4}q^{3} – 18p^{3}q^{4}

(vii) (7x^{2} – 11x + 10) (x^{3} – x^{2})

Answer:

(7x^{2} – 11x + 10) (x^{3} – x^{2})

= 7x^{2}(x^{3} – x^{2}) – 11x (x^{3} – x^{2}) + 10(x^{3} – x^{2})

= 7x^{5} – 7x^{4}– 11x^{4} + 11 x^{3} + 10x^{3} – 10x^{2}

= 7x^{5} – 18x^{4} + 21x^{3} – 10x^{2}

(viii) (\(\frac{10}{9}\)a^{5}b^{6}c^{6}) (-\(\frac{3}{2}\)b^{2}c) (\(\frac{6}{5}\)c^{3}d^{4})(-a^{3}d^{5})

Answer:

(\(\frac{10}{9}\)a^{5}b^{6}c^{6}) (-\(\frac{3}{2}\)b^{2}c) (\(\frac{6}{5}\)c^{3}d^{4})(-a^{3}d^{5})

= \(\) × -1 × a^{5} × a^{3} × b^{6} × b^{2} × c^{6} × c × c^{3} × d^{4} × d^{5}

= 2 a^{8}b^{8}c^{10}d^{9}

(ix) (0.7x^{3} – 0.5y^{3}) (0.7x^{3} + 0.5y^{3})

Answer:

(0.7x^{3} – 0.5y^{3}) (0.7x^{3} + 0.5y^{3})

= 0.7x^{3}(0.7x^{3} + 0.5y^{3}) – 0.5y^{3} (0.7x^{3} + 0.5y^{3})

= 0.7x^{3} x 0.7x^{3} + 0.7x^{3} x 0.5 y^{3} – 0.5y^{3} x 0.7x^{3} – 0.5y^{3} x 0.5 y^{3}

= 0.49x^{6} + 0.35x^{3}y^{3} – 0.35x^{3}y^{3} – 0.25y^{6}

= 0.49x^{6} – 0.25y^{6}

(x) (64a^{2} – 56ab + 49b^{2}) (8a + 7b)

Answer:

(64a^{2} – 56ab + 49b^{2}) (8a + 7b)

= 64a^{2}(8a + 7b) – 56ab (8a + 7b) + 49b^{2} (8a + 7b)

= 64a^{2} x 8a + 64a^{2} x 7b – 56ab x 8a – 56ab x 7b + 49b^{2} x 8a + 49b^{2} x 7b

= 512a^{3} + 448a^{2}b – 448a^{2}b – 392ab^{2} + 392ab^{2} + 343b^{3}

= 512a^{3} + 343b^{3}

Question 3.

Find the variable part in the product of:

(i) 127x^{3}y^{9}, 255x^{6}y^{5} and 313y^{3}z^{4}

(ii) \(\frac{61}{17}\)a^{2}b^{4}, \(\frac{97}{43}\)b^{3} , \(\frac{29}{41}\)a^{6}b and \(\frac{111}{123}\)

Solution:

(i) Variable part in 127x^{3} y^{9} = x^{3} y^{9}

Variable part in 255x^{6} y^{5} = x^{6} y^{5}

Variable part in 313y^{2} z^{4} = y^{2} z^{4}

(ii) Variable part in \(\frac{61}{17}\)a^{2} b^{4} = a^{2} b^{4}

Variable part in \(\frac{97}{43}\)b^{3} = b^{3}

Variable part in \(\frac{29}{41}\)a^{6} b = a^{6} b

Variable part in \(\frac{111}{123}\) = nil

Question 4.

Express (5x^{6})(12x^{2}y) (\(\frac{3}{20}\)xy^{2}) as a monomial and then evaluate at x = 1, y = 2.

Answer:

(5x^{6})(12x^{2}y) (\(\frac{3}{20}\)xy^{2}) = 5 × 12 × x^{6} × x^{2}y × xy^{2}

= 9x^{9}y^{3}

Put x = 1, y = 2

= 9(1 )^{9} (2)^{3}

= 9 × 1 × 8 = 72

Question 5.

Express 1.5a^{2}(10ab – 4b^{2}) as a binomial and then evaluate at a = – 2, b = 3.

Answer:

1.5a^{2}(10ab – 4b^{2}) = 1.5a^{2} × 10ab – 1.5a^{2} × 4b^{2}

= 15a^{3}b – 6a^{2}^{2}

Put a = – 2, b = 3

= 15(-2)^{3}(3) – 6 (- 2)^{2}(3)^{2}

= 15 × -8 × 3 – 6 × 4 × 9

= – 360 – 216

– 576

Question 6.

Simplify and then verify the result for the given values:

(i) (3x – 4y) (4x^{2}y + 3 xy^{2}); x = 2, y = -1

Answer:

(i) (3x – 4y) (4x^{2}y + 3xy^{2}); x = 2, y = – 1

= 3x(4x^{2}y + 3xy^{2}) – 4y (4x^{2}y + 3xy^{2})

= 12x^{3}y + 9x^{2}y^{2} – 16x^{2}y^{2} – 12xy^{3}

= 12x^{3}y – 7x^{2}y^{2} – 12xy^{3}

Verification:

L.H.S = (3x – 4y) (4x^{2}y + 3xy^{2})

Putting x = 2, y = -1

= [3(2) – 4(-1)] [4(2)^{2} (-1) + 3(2)(-1)^{2}]

= (6 + 4) (-16 + 6)

= (10)(-10) = -100

R.H.S = 12x^{3}y – 7x^{2}y^{2} – 12xy^{3}

Putting x = 2, y = -1

= 12(2)^{3}(-1) – 7(2)^{2}(-1)^{2} – 12(2)(-2)^{3}

= -96 – 28 + 24

= -100

Hence verified

L.H.S = R.H.S

(ii) (\(\frac{1}{4}\)a^{2} + \(\frac{5}{9}\)b^{2})(a + b + ab); a = 2, b = 3

Answer:

(\(\frac{1}{4}\)a^{2} + \(\frac{5}{9}\)b^{2})(a + b + ab); a = 2, b = 3

= \(\frac{1}{4}\)a^{2}(a + b + ab) + \(\frac{5}{9}\)b^{2}(a + b + ab)

= \(\frac{1}{4}\)a^{2} × a + \(\frac{1}{4}\)a^{2} × b + \(\frac{1}{4}\)a^{2} × ab + \(\frac{5}{9}\)b^{2} × a + \(\frac{5}{9}\)b^{2} × b + \(\frac{5}{9}\)b^{2} × ab

= \(\frac{1}{4}\)a^{3} + \(\frac{1}{4}\)a^{2}b + \(\frac{1}{4}\)a^{3}b + \(\frac{5}{9}\)ab^{2} + \(\frac{5}{9}\)b^{3} + \(\frac{5}{9}\)ab^{3}

Verification:

L.H.S = (\(\frac{1}{4}\)a^{2} + \(\frac{5}{9}\)b^{2})(a + b + ab)

Putting a= 2, b = 3

= (\(\frac{1}{4}\)(2)^{2} + \(\frac{5}{9}\)(3)^{2})(2 + 3 + 2 × 3)

= [\(\frac{1}{4}\) × 4 + \(\frac{5}{9}\) × 9] [5 + 6]

= (1 + 5)(11) = 77

R.H.S = \(\frac{1}{4}\)a^{3} + \(\frac{1}{4}\)a^{2}b + \(\frac{1}{4}\)a^{3}b + \(\frac{5}{9}\)ab^{2} + \(\frac{5}{9}\)b^{3} + \(\frac{5}{9}\)ab^{3}

= \(\frac{1}{4}\)(2)^{3} + \(\frac{1}{4}\)(2)^{2}(3) + \(\frac{1}{4}\)(2)^{3}(3) + \(\frac{5}{9}\)(2)(3)^{2} + \(\frac{5}{9}\)(3)^{3} + \(\frac{5}{9}\)(2)(3)^{3}

= \(\frac{1}{4}\) × 8 + \(\frac{1}{4}\) × 4 × 3 + \(\frac{1}{4}\) × 8 × 3 + \(\frac{5}{9}\) × 2 × 9 + \(\frac{5}{9}\) × 27 + \(\frac{5}{9}\) × 2 × 27

= 2 + 3 + 6 + 10 + 15 + 30 = 77

Hence verified L.H.S. = R.H.S

(iii) (x^{3}y – y^{2})(x^{3}y + y^{2});x = -1, y = -2

Answer:

(x^{3}y – y^{2})(x^{3}y + y^{2});x = -1, y = -2

= x^{3}y(x^{3}y + y^{2}) – y^{2} (x^{3}y + y^{2})

= x^{3}y × x^{3}y + x^{3}y × y^{2} – y^{2} × x^{3}y – y^{2}× y^{2}

= x^{6}y^{2} + x^{3}y^{3} – x^{3}y^{3} – y^{4}

Verification:

L.H.S = (x^{3}y – y^{2})(x^{3}y + y^{2})

= [(-1)^{3} (- 2) – (- 2)^{2}] [(-1)^{3}(- 2) + (- 2)^{2}]

= [- 1 × -2 – 4][-1 × -2 + 4]

= (2 – 4) (2 + 4)

= – 2 × 6

= – 12

R.H.S = x^{6}y^{2} + x^{3}y^{3} – x^{3}y^{3} – y^{4}

Putting x = -1, y = -2

= (-1)^{6}(-2)^{2} – (-2)^{4}

= 1 × 4 – 16

= 4 – 16

= -12

Hence verified

L.H.S = R.H.S

(iv) (2p + 3q) (4p^{2} + 12pq + 9q^{2}); p = \(\frac{1}{2}\),q = \(\frac{1}{3}\)

= (2p + 3q) (4p^{2} + 12pq + 9q^{2})

= 2p (4p^{2} + 12pq + 9q^{2}) + 3q (4p^{2} + 12pq + 9q^{2})

= 8p^{3} + 24 p^{2}q + 18 pq^{2} + 12p^{2}q + 36 pq^{2} + 27q^{3}

= 8p^{3} + 36 p^{2}q + 54 pq^{2} + 27 q^{3}

Verification

L.H.S = (2p + 3q) (4p^{2} + 12pq + 9q^{2})

Putting p = \(\frac{1}{2}\), q = \(\frac{1}{3}\)

= [2(\(\frac{1}{2}\)) + 3(\(\frac{1}{3}\))] [4(\(\frac{1}{2}\))^{2} + 12(\(\frac{1}{2}\))(\(\frac{1}{3}\)) + 9(\(\frac{1}{3}\))^{2}]

= (1 + 1) (4 × \(\frac{1}{4}\) + 12 × \(\frac{1}{6}\) + 9 × \(\frac{1}{9}\))

= 2(1 + 2 + 1)

= 2 × 4 = 8

R.H.S = 8p^{3} + 36p^{2}q + 54 pq^{2} + 27 q^{3}

= 8(\(\frac{1}{2}\))3 – 36(\(\frac{1}{2}\))2(\(\frac{1}{3}\)) + 54(\(\frac{1}{2}\)) (\(\frac{1}{3}\))^{2} + 27(\(\frac{1}{3}\))^{3}

= 8 × \(\frac{1}{8}\) + 36 × \(\frac{1}{4}\) \(\frac{1}{3}\) + 54 × \(\frac{1}{2}\) × \(\frac{1}{9}\) + 27 × \(\frac{1}{27}\)

= 1 + 3 + 3 + 1

= 8

Hence verified L.H.S. = R.H.S.

(v) (m^{2} + mn + n^{2})(m – n); m = 4, n = 3

Answer:

(m^{2} + mn + n^{2}) (m – n); m = 4, n = 3

= m^{2}(m – n) + mn(m – n) + n^{2} (m – n)

= m^{3} – m^{2}n + m^{2}n – mn^{2} + mn^{2} – n^{3}

= nr-nr

Verification:

L.H.S. = (m^{2} + mn + n^{2}) (m – n)

Putting m = 4, n = 3

= [(4)^{2} + 4 × 3 + (3)^{2}](4 – 3)

= (16 + 12 + 9) (1) = 37

R.H.S. = m^{3} – n^{3}

= (4)^{3} – (3)^{3}

= 64 – 27 = 37

Hence verified L.H.S. = R.H.S.

Question 7.

Simplify:

(i) 3x^{2} (3y^{2} + 2) – x (x – 2xyz) + y (2x^{2}y – 2y)

(ii) (2x + 7)(5x + 9)-(4x^{2} + 1)(x – 3)

(iii) (y^{2} – 7y + 4) (3y^{2} – 2) + (y + 1) (y^{2} + 2y)

Answer:

(i) 3x^{2}(3y^{2} + 2) – x(x – 2xy^{2}) + y (2x^{2}y – 2y)

= 3x^{2} × 3y^{2} + 3x^{2} × 2 – x × x – x – 2xy^{2} + y × 2x^{2}y – 2y × y

= 9x^{2}y^{2} + 6x^{2} – x^{2} + 2x^{2}y^{2} + 2x^{2}y^{2} – 2 y^{2}

= 13x^{2}y^{2} + 5x^{2} – 2y^{2}

(ii) (2x + 7) (5x + 9) – (4x^{2} +1) (x – 3)

= [2x (5x + 9) + 7(5x + 9)] – [4x^{2} (x – 3) + 1 (x – 3)]

= (10x^{2} + 18x + 35x + 63) – (4x^{3} – 12x^{2} + x – 3)

= 10x^{2} + 53x + 63 – 4x^{3} + 12x^{2} – x + 3

= – 4x^{3} + 22x^{2} + 52x + 66

(iii) (y^{2} – 7y + 4) (3y^{2} – 2) + (y + 1) (y^{2} + 2y)

= [y^{2} (3y^{2} – 2) – 7y(3y^{2} – 2) + 4(3y^{2} – 2)] + [y(y^{2} + 2y) + 1 (y^{2} + 2y)]

= 3y^{4} – 2y^{2} – 21y^{3} + 14y + 12y^{2} – 8 + y^{3} + 2y^{2} + y^{2} + 2y

= 3y^{4} – 20y^{3} + 13y^{2} + 16y – 8

Question 8.

Find the HCF of the terms in the expression 3a^{2}b^{2} + 6ab^{2}c^{2} + 12a^{2}b^{2}c^{2}.

Answer:

3a^{2}b^{2} + 6ab^{2}c^{2} + 12a^{2}b^{2}c^{2}

= 3 × a × a × b × b + 2 × 3 × a × b × b × c × c + 2 × 2 × 3 × a × a × b × b × c × c

H.C.F. = 3 × a × b × b

= 3 ab^{2}

Question 9.

Factorise:

(a) ab^{2} – bc^{2} – ab + c^{2}

Answer:

= (ab^{2} – ab) – (bc^{2} – c^{2})

= ab(b -1) – c^{2}(b -1)

= (b – 1) (ab – c^{2})

(b) 4(p + q) (3a -b) + 6 (p + q) (2b – 3a)

Answer:

4(p + q)(3a – b) + 6(p + q)(2b – 3a)

= 2(p + q) [2(3a – b) + 3(2b – 3a)]

= 2(p + q) (6a -2b + 6b- 9a)

= 2(p + q) (4b – 3a)

(c) axy + bc xy – az – bcz

Answer:

axy + be xy – az – b c z

= xy (a + bc) – z(a + bc)

= xy (a + bc) – z(a + bc)

= (a + bc) (xy – z)

### DAV Class 7 Maths Chapter 6 Enrichment Questions

Question 1.

Sum of first n natural numbers is given by the expression , \(\left(\frac{n^2+n}{2}\right)\)

Sume of first five natural numbers (i.e., 1+2 + 3 + 4 + 5) will be

\(\frac{5^2+5}{2}=\frac{30}{2}\) = 15

Sum of first seven natural numbers (i.e., 1+2 + 3 + 4 + 5 + 6 + 7) will be

\(\frac{7^2+7}{2}=\frac{56}{2}\) = 28

Now, write down the sum of first ten natural numbers.

Answer:

Sum of first n natural numbers is given by the expression \(\left(\frac{n^2+n}{2}\right)\).

Therefore, putting the value of n = 10 in the above expression and solving it, we will have the sum of first 10 natural numbers.

∴ Sum of first 10 natural numbers = \(\left(\frac{10^2+10}{2}\right)\)

= \(\left(\frac{100+10}{2}\right)\)

= \(\frac{110}{2}\)

= 55

So, the sum of first 10 natural numbers is 55.

Additional Questions

Question 1.

Find the product:

(i) (0.7x + 0.9y) (0.49x^{2} – 0.63xy + 0.81y^{2})

Solution:

(i) (0,7x + 0.9y) (0.49x^{2} – 0.63xy + 0.81y^{2})

= 0.7x (0.49x^{2} – 0.63xy + 0.81y^{2}) + 0.9y (0.49x^{2} – 0.63xy + 0.81y^{2})

= 0.343x^{3} + 0.729y^{3}

(ii) (\(\frac{1}{2}\)x – \(\frac{1}{2}\)y) (\(\frac{1}{2}\)x^{2} + \(\frac{1}{2}\)y^{2})

Hence verified L.H.S. = R.H.S.

Question 2.

Simplify the following and verify the result for x = 1, y = – 1 and z = 1.

(i) (\(\frac{3}{2}\)x^{2} – \(\frac{1}{5}\)y) (\(\frac{1}{2}\)x + \(\frac{3}{4}\)xy – \(\frac{1}{3}\)y)

Answer:

= \(\frac{3}{2}\)x^{2}(\(\frac{1}{2}\)x + \(\frac{3}{4}\)xy – \(\frac{1}{3}\)y) – \(\frac{1}{5}\)y(\(\frac{1}{2}\)x + \(\frac{3}{4}\)xy – \(\frac{1}{3}\)y)

= \(\frac{3}{4}\)x^{3} + \(\frac{9}{8}\)x^{3}y – \(\frac{1}{2}\)x^{2}y – \(\frac{1}{10}\)xy – \(\frac{3}{20}\)xy^{2} + \(\frac{1}{15}\)y^{2}

Verification:

L.H.S = (\(\frac{3}{2}\)x^{2} – \(\frac{1}{5}\)y) (\(\frac{1}{2}\)x + \(\frac{3}{2}\)xy – \(\frac{1}{3}\)y)

Put x = 1 and y = -1

Hence verified L.H.S = R.H.S

(ii) (x^{2} + y^{2} + z^{2} – xy – yz – zx) (x + y + z)

= x^{2}(x + y + z) + y^{2}(x + y + z) + z^{2} (x + y + z) – xy(x + y + z) – yz (x + y + z) – zx(x + y + z)

= x^{3} + y^{3} + z^{3} – 3xyz

Verification:

L.H.S = (x^{2} + y^{2} + z^{2} – xy – yz – zx) (x + y + z)

= [(1)^{2} + (-1)^{2} + (1)^{2} – (1) (- 1) – (-1) (1) – (1) (1)] [(1) + (-1) + (1)]

= [1 + 1 + 1 + 1 + 1 – 1] [1 – 1 + 1]

= 4 × 1 = 4

R.H.S = x^{3} + y^{3} + z^{3} – 3 xyz

Put x = 1, y = -1, z = 1

= (1)^{3} + (- 1)^{3} + (1)^{3} – 3(1) (- 1) (1)

= 1 – 1 + 1 + 3 = 4

Hence verified L.H.S. = R.H.S.

Question 3.

Simplify the following and verify the result also.

(i) (m^{3} + n^{3}) (2m – 3n); m = -2, n = 0

Answer:

(m^{3} + n^{3}) (2m – 3n)

= m^{3} (2m – 3n) + n^{3} (2m – 3n)

= m^{3} × 2m – m^{3} × 3n + n^{3} × 2m – n^{3} × 3n

= 2m^{3} – 3m^{3}n + 2mn^{3} – 3n^{3}

Verification:

L.H.S. = (m^{3} + n^{3}) (2m – 3n)

Putting m = – 2, n = 0

= [(- 2)^{3} + (0)^{3}] [2(-2) – 3 × (0)]

= -8 × -4

= 32

R.H.S. = 2m^{4} – 3m^{3}n + 2mn^{3} – 3n^{3}

Putting m = -2,n = 0

= 2(- 2)^{4} – 3 (- 2)^{3} (0) + 2(- 2) (0)^{3} – 3 (0)^{3}

= 2 × 16 – 0 + 0 – 0

= 32

Hence verified L.H.S. = R.H.S.

(ii) (x^{2}y – xy^{2}) (2x + 1); x = 1, y = 2

Answer:

(x^{2}y – xy^{2}) (2x + 1)

= x^{2}y (2x + 1) – xy^{2}(2x + 1)

= x^{2}y × 2x + x^{2}y × 1 – xy^{2} × 2x – xy^{2} × 1

= 2x^{3}y + x^{2}y- 2x^{2} y^{2} – xy^{2}

Verification:

L.H.S = (x^{2}y – xy^{2}) (2x +1)

Putting x = 1, y = 2

= [(1)^{2} (2) – (1) (2)^{2}] [2(1) + 1]

= (2 – 4) (2 + 1)

= -2 × 3 = -6

R.H.S = 2x^{3}y + x^{2}y – 2x^{2}y^{2} – xy^{2}

= 2(1 )^{3} (2) + (1)^{2} (2) – 2(1)^{2} (2)^{2} – (1) (2)^{2}

= 2× 1 × 2 + 1 × 2 – 2 × 1 × 4 – 1 × 4

= 4 + 2 – 8 – 4

= 6 – 12

= -6

Hence verified L.H.S. – R.H.S.

(iii) (9b^{2}c – 4c^{2}) (2bc^{2} + 5c); b = 0, c = 2

Answer:

(9b^{2}c – 4c^{2}) (2bc^{2} + 5c)

= 9b^{2}c (2bc^{2} + 5c) – 4c^{2} (2bc^{2} + 5c)

= 9b^{2}c x 2bc^{2} + 9b^{2}c x 5c – 4c^{2} x 2bc^{2} – 4c^{2} x 5c

= 18b^{3}c^{3} + 45b^{2}c^{2} – 8bc^{4} – 20c^{3}

Verification:

L.H.S. = (9b^{2}c – 4c^{2}) (2bc^{2} + 5c)

Putting b = 0 and c = 2

= [9(0)^{2} (2) – 4(2)^{2}] [2 × (0) (2)^{2} + 5(2)]

= [0 – 16] [0 + 10]

= -16 × 10

= – 160

R.H.S. = 18b^{3}c^{3} + 45b^{2} – 8bc^{4} – 20c^{3}

Putting b = 0,c = 2

= 18(0)^{3} (2)^{3} + 45(0)^{2} (2)^{2} – 8(0) (2)^{4} – 20(2)^{3}

= 0 + 0 – 0 – 20 × 8

= -160

Hence verified L.H.S. = R.H.S.

Question 4.

Find the common factors of the following terms:

(i) 3x^{3}y^{2},15 xy^{3}, 60x^{2}y^{2}

Answer:

3x^{3}y^{2},15xy^{3}, 60x^{2}y^{2}

Factors of 3x^{3}y^{2} = 3 × x × x × x × y × y

Factors of 15xy^{3} = 3 × 5 × x × y × y × y

Factors of 60x^{2}y^{2} = 2 × 2 × 3 × 5 × x × x × y × y

Common factors = 3 × x × y × y = 3xy^{2}

(ii) 4a^{3}b, 12a^{2}b^{4},16a^{4}b^{2}

Answer:

4a^{3}b, 12a^{2}b^{4},16a^{4}b^{2}

Factors of 4 a3b = 2 × 2 × a × a × a × b

Factors of 12 a2b4 = 2 × 2 × 3 × a × a × b × b × b × b

Factors of 16fl4fr2 = 2 × 2 × 2 × 2 × a × a × a × a × b × b

Common factors are 2 × 2 × a × a × b = 4 a2b

(iii) 8x^{2}(x – y), 12x(x – y), 24x^{3}y^{2} (x – y)

Answer:

8x^{2}(x – y), 12x(x – y), 24x^{3}y^{2} (x – y)

Factors of 8x^{2}(x – y) = 2 × 2 × 2 × x × x × x – y)

Factors of 12x(x – y) =2 × 2 × 3 × x × (x – y)

Factors of 24x^{3}y^{2} (x – y) = 2 × 2 × 2 × 3 × x × x × x × y × y × (x – y)

Common factors are 2 × 2 × x × (x – y) = 4x(x – y)

(iv) 21x^{3}y^{7}z^{2}, 35x^{2} y^{6}z^{3}, 42x^{5}y^{3}z^{4}

Answer:

21x^{3}y^{7}z^{2}, 35x^{2} y^{6}z^{3}, 42x^{5}y^{3}z^{4}

Factors of 21x^{3}y^{7}z^{2} = 3 × 7 × x × x × x × y × y × y × y × y × y × y × z × z

Factors of 35x^{2} y^{6}z^{3} = 5 × 7 × x × x × y × y × y × y × y × y × z × z × z

Factors of 42x^{5}y^{3}z^{4} = 2 × 3 × 7 × x × x × x × x × x × y × y × y × z × z × z × z

Common Factors are 7 × x × x × y × y × y × z × z = 7x^{2}y^{3}z^{2}

Question 5.

Find H.C.F. of the terms and factorise the following expressions.

(i) 21x^{4}y^{5} + 28x^{5}y^{3} – 35x^{3}y^{7}

Answer:

21x^{4}y^{5} + 28x^{5}y^{3} – 35x^{3}y^{7}

Common factors = 7x^{3}y^{3}

H.C.F. of the terms = 7x^{3}y^{3}

∴ 21x^{4}y^{5} + 28x^{5}y^{3} – 35 x^{3}y^{7} = 7x^{3}y^{3}(3xy^{3} + 4x^{2} – 5y^{4})

(ii) 5p^{3}q – 20pq^{3} + 3p^{2}q^{2}

Answer:

5p^{3}q – 20pq^{3} + 3p^{2}q^{2}

Common factor = pq

∴ H.C.F. of the terms = pq

∴ 5p^{3}q – 20pq^{3} + 3p^{2}q^{2} = pq(5p^{2} + 4x^{2} – 5y^{4})

Question 6.

Factorise the following expressions:

(i) (3x – 5y)^{2} – 15x^{2}y + 25xy^{2}

Answer:

(3x – 5y)^{2} – 15x^{2}y + 25xy^{2}

= (3x – 5y)^{2} – 5xy (3x – 5y)

= (3x – 5y) (3x – 5y – 5xy)

(ii) 9x^{2} – 16y^{2} + 18x^{3}y – 12xy^{3}

Answer:

9x^{2} – 16y^{2} + 18x^{3}y – 12xy^{3}

= (3x)^{2} – (4y)^{2} + 6xy (3x – 4y)

= (3x – 4y) (3x + 4y) + 6xy (3x – 4y)

= (3x – 4y) (3x + 4y + 6xy)

Question 7.

Simplify the following and verify the result for the given values:

(i) (2a – 3b) (4a^{2} + 6ab + 9b^{2}); a = 1, b = 2

Answer:

(2a – 3b) (4a^{2} + 6ab + 9b^{2}) = 2a(4a^{2} + 6ab + 9b^{2}) – 3b(4a^{2} + 6ab + 9b^{2})

= 8a^{3} – 27b^{3}

Verification:

L.H.S. = (2a – 3b) (4a^{2} + 6ab + 9b^{2})

Put a = 1,b = 2

= [2(1)-3 (2)] [4 (1)^{2} + 6(1)(2) + 9 (2)^{2}]

= (2 – 6) (4 + 12 + 36)

= – 4 × 52 = – 208

= 8a^{3} – 27b^{3}

R.H.S

Put x = 1, b = 2

= 8(1)^{3} – 27 (2)^{3}

= 8 – 27 × 8

= 8 – 216

= -208

Hence verified L.H.S = R.H.S

(ii) (\(\frac{1}{3}\)p^{2} – \(\frac{2}{5}\)q^{2}) (p + q – pq); p = -1, q = 1

Answer:

(\(\frac{1}{3}\)p^{2} – \(\frac{2}{5}\)q^{2}) (p + q – pq)

= \(\frac{1}{3}\)p^{2}(p + q – pq) + \(\frac{2}{5}\)q^{2}(p + q – pq)

= \(\frac{1}{3}\)p^{3} + \(\frac{1}{3}\)p^{2}q – \(\frac{1}{3}\)p^{3}q + \(\frac{2}{5}\)pq^{2} + \(\frac{2}{5}\)q^{3} – \(\frac{2}{5}\)pq^{3}

Verification:

L.H.S. = (\(\frac{1}{3}\)p^{2} – \(\frac{2}{5}\)q^{2})

Put p = -1, q = 1

= [\(\frac{1}{3}\)(- 1)2 + \(\frac{2}{5}\)(1)2] [-1 + 1 – (- 1) (1)]

= \(\left(\frac{1}{3}+\frac{2}{5}\right)\)(-1 + 1 + 1)

= \(\left(\frac{5+6}{15}\right)\)(1)

= \(\frac{11}{15}\)

R.H.S.

= \(\frac{1}{3}\)p^{3} + \(\frac{1}{3}\)p^{2}q – \(\frac{1}{3}\)p^{3}q + \(\frac{2}{5}\)pq^{2} + \(\frac{2}{5}\)q^{3} – \(\frac{2}{5}\)pq^{3}

Put p = – 1 and q = 1

Hence verified L.H.S. = R.H.S.

Question 8.

Factorise:

(i) pxy + qr xy – pz – qrz

Answer:

pxy + qrxy – pz – qrz = xy (p + qr) – z(p + qr)

= (p + qr) (xy – z)

(ii) 8(x + y)(3p – q) + 12(x + y)(q – 3p)

Answer:

8(x + y) (3p – q) + 12(x + y)(q – 3p)

= 8(x + y) (3p – q) – 12(x + y) (3p – q)

= – 4(x + y) (3p – q)