Students can Download Maths Chapter 3 Algebra Ex 3.4 Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

## Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 3 Algebra Ex 3.4

Miscellaneous Practice Problems

Question 1.

6^{2} × 6^{m} = 6^{5}, find the value of ‘m’

Solution:

6^{2} × 6^{m} = 6^{5}

6^{2+m} = 6^{5} [Since a^{m} × a^{n}= a^{m+n}]

Equating the powers, we get

2 + m = 5

m = 5 – 2 = 3

Question 2.

Find the unit digit of 124^{128} × 126^{124}

Solution:

In 124^{128}, the unit digit of base 124 is 4 and the power is 128 (even power).

Therefore, unit digit of 124^{128} is 4.

Also in 126^{124}, the unit digit of base 126 is 6 and the. power is 124 (even power).

Therefore, unit digit of 126^{124} is 6.

Product of the unit digits = 6 × 6 = 36

∴ Unit digit of the 124^{128} × 126^{124} is 6.

Question 3.

Find the unit digit of the numeric expression: 16^{23} + 71^{48} + 59^{61}

Solution:

In 16^{23}, the unit digit of base 16 is 6 and the power is 23 (odd power).

Therefore, unit digit of 16^{23} is 6.

In 71^{48}, the unit digit of base 71 is 1 and the power is 48 (even power).

Therefore, unit digit of 71^{48} is 1.

Also in 59^{61}, the unit digit of base 59 is 9 and the power is 61 (odd power).

Therefore, unit digit of 59^{61} is 9.

Sum of the unit digits = 6 + 1 + 9 = 16

∴ Unit digit of the given expression is 6.

Question 4.

Find the value of

Solution:

Question 5.

Identify the degree of the expression, 2a^{3}be + 3a^{3}b + 3a^{3}c – 2a^{2}b^{2}c^{2}

Solution:

The terms of the given expression are 2a^{3}bc, 3a^{3}b + 3a^{3}c – 2a^{2}b^{2}c^{2}

Degree of each of the terms: 5,4,4,6.

Terms with the highest degree: – 2a^{2}b^{2}c^{2}

Therefore degree of the expression is 6.

Question 6.

If p = -2, q = 1 and r = 3, find the value of 3p^{2}q^{2}r.

Solution:

Given p = -2; q = 1; r = 3

∴ 3p^{2}q^{2}r = 3 × (-2)^{2} × (1)^{2} × (3)

= 3 × (-2 × 1)^{2} × (3) [Since a^{m} × b^{m} = (a × b)^{m}]

= 3 × (-2)^{2} × (3)

= 3 × (-1)^{2} × 2^{2} × 3

= 3^{1+1} × 1 × 4 [Since a^{m} × a^{n} = a^{m+n}]

= 3^{2} × 4 = 9 × 4

∴ 3p^{2}q^{2}r = 36

Challenge Problems

Question 7.

LEADERS is a WhatsApp group with 256 members. Every one of its member is an admin for their own WhatsApp group with 256 distinct members. When a message is posted in LEADERS and everybody forwards the same to their own group, then how many members in total will receive that message?

Solution:

Members of the groups LEADERS = 256

Members is individual groups of the members of LEADERS = 256

Total members who receive the message

= 256 × 256 = 2^{8} × 2^{8}

2^{8+8} = 2^{16}

= 65536

Totally 65536 members receive the message.

Question 8.

Find x such that 3^{x+2} = 3^{x} + 216.

Solution:

Given 3^{x+2} = 3^{x} + 216 ; 3^{x+2} = 3^{x }+ 216

Dividing throught by 3^{x}, we get

Equating the powers of same base

Question 9.

If X = 5x^{2} + 7x + 8 and Y = 4x^{2} – 7x + 3, then find the degree of X + Y.

Solution:

Given x = 5x^{2} + 7x + 8

X + Y = 5x^{2} + 7x + 8 + (4x^{2} – 7x + 3)

= (5x^{2} + 4x^{2}) + (7x – 7x) + (8 + 3)

= x^{2} (5 + 4) + x(7 – 7) + (8 + 3) = 9x^{2 }+ 11

Degree of the expression is 2.

Question 10.

Find the degree of (2a^{2} + 3ab – b^{2}) – (3a^{2} -ab- 3b^{2})

Solution:

(2a^{2} + 3ab – b^{2}) – (3a^{2} – ab – 3b^{2})

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

= 2a^{2} + 3ab – b^{2} – 3a^{2} + ab + 3b^{2}

= 2a^{2} – 3a^{2} + 3ab + ab + 3b^{2} – b^{2}

= 2a^{2} – 3a^{2} + ab (3 + 1) + b^{2}(3 – 1)

= – a^{2} + 4 ab + 2b^{2}

Hence degree of the expression is 2.

Question 11.

Find the value of w, given that x = 4, y = 4, z = – 2 and w = x^{2} – y^{2} + z^{2} – xyz.

Solution:

Given x = 3; y = 4 and z = -2.

w = x^{2} – y^{2} + z^{2} – xyz

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

w = 9 – 16 + 4 + 24

w = 37 – 16

w = 21

Question 12.

Simplify and find the degree of 6x^{2} + 1 – [8x – {3x^{2} – 7 – (4x^{2} – 2x + 5x + 9)}]

Solution:

6x^{2} + 1 – [8x – (3x^{2 }– 7 – (4x^{2 }– 2x + 5x + 9)}]

= 6x^{2} + 1 – [8x – {3x^{2} – 7 – 4x^{2 }– 2x + 5x + 9}]

= 6x^{2} + 1 – [8x – 3x^{2} + 7 + 4x^{2} – 2x + 5x + 9}]

= 6x^{2} – 1 – [8x + 3x^{2} – 7 – 4x^{2} + 2x – 5x – 9]

= 6x^{2} + 3x^{2} – 4x^{2} – 8x + 2x – 5x – 1 – 7 – 9]

= x^{2}(6 + 3 – 4) + x(8 + 2 – 5) – 15

= 5x^{2} – 11x – 15

Degree of the expression is 2.

Question 13.

The two adjacent sides of a rectangle are 2x^{2} – 5xy + 3z^{2} and 4xy – x^{2} – z^{2}. Find the perimeter and the degree of the expression.

Solution:

Let the two adjacent sides of the rectangle as

l = 2x^{2} – 5xy + 3z^{2} and b = 4xy – x^{2}y + 3z^{2}

Perimeter of the rectangle

= 2(l + b) = 2(2x^{2} – 5xy + 3z^{2} + 4xy – x^{2} – z^{2})

= 4x^{2} – 10xy + 6z^{2} + 8xy – 2x^{2} – 2z^{2}

= 4x^{2} – 2x^{2} – 10xy + 8xy + 6z^{2} – 2z^{2}

= x^{2}(4 – 2) + xy (-10 + 8) + z^{2} (6 – 2z^{2})

Perimeter = 2x^{2} – 2xy + 4z^{2}

Degree of the expression is 2.