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

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

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

Miscellaneous Practice Problems

Question 1.
6² × 6^m = 6⁵, find the value of ‘m’
Solution:
6² × 6^m = 6⁵
6^2+m = 6⁵ [Since a^m × aⁿ= a^m+n]
Equating the powers, we get
2 + m = 5
m = 5 – 2 = 3

Question 2.
Find the unit digit of 124¹²⁸ × 126¹²⁴
Solution:
In 124¹²⁸, the unit digit of base 124 is 4 and the power is 128 (even power).
Therefore, unit digit of 124¹²⁸ is 4.
Also in 126¹²⁴, the unit digit of base 126 is 6 and the. power is 124 (even power).
Therefore, unit digit of 126¹²⁴ is 6.
Product of the unit digits = 6 × 6 = 36
∴ Unit digit of the 124¹²⁸ × 126¹²⁴ is 6.

Question 3.
Find the unit digit of the numeric expression: 16²³ + 71⁴⁸ + 59⁶¹
Solution:
In 16²³, the unit digit of base 16 is 6 and the power is 23 (odd power).
Therefore, unit digit of 16²³ is 6.
In 71⁴⁸, the unit digit of base 71 is 1 and the power is 48 (even power).
Therefore, unit digit of 71⁴⁸ is 1.
Also in 59⁶¹, the unit digit of base 59 is 9 and the power is 61 (odd power).
Therefore, unit digit of 59⁶¹ 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, 2a³be + 3a³b + 3a³c – 2a²b²c²
Solution:
The terms of the given expression are 2a³bc, 3a³b + 3a³c – 2a²b²c²
Degree of each of the terms: 5,4,4,6.
Terms with the highest degree: – 2a²b²c²
Therefore degree of the expression is 6.

Question 6.
If p = -2, q = 1 and r = 3, find the value of 3p²q²r.
Solution:
Given p = -2; q = 1; r = 3
∴ 3p²q²r = 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 × 2² × 3
= 3¹⁺¹ × 1 × 4 [Since a^m × aⁿ = a^m+n]
= 3² × 4 = 9 × 4
∴ 3p²q²r = 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 = 2⁸ × 2⁸
2⁸⁺⁸ = 2¹⁶
= 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 = 5x² + 7x + 8 and Y = 4x² – 7x + 3, then find the degree of X + Y.
Solution:
Given x = 5x² + 7x + 8
X + Y = 5x² + 7x + 8 + (4x² – 7x + 3)
= (5x² + 4x²) + (7x – 7x) + (8 + 3)
= x² (5 + 4) + x(7 – 7) + (8 + 3) = 9x² + 11
Degree of the expression is 2.

Question 10.
Find the degree of (2a² + 3ab – b²) – (3a² -ab- 3b²)
Solution:
(2a² + 3ab – b²) – (3a² – ab – 3b²)
= (2a² + 3ab – b²) + (- 3a² + ab + 3b²)
= 2a² + 3ab – b² – 3a² + ab + 3b²
= 2a² – 3a² + 3ab + ab + 3b² – b²
= 2a² – 3a² + ab (3 + 1) + b²(3 – 1)
= – a² + 4 ab + 2b²
Hence degree of the expression is 2.

 

Question 11.
Find the value of w, given that x = 4, y = 4, z = – 2 and w = x² – y² + z² – xyz.
Solution:
Given x = 3; y = 4 and z = -2.
w = x² – y² + z² – xyz
w = 3² – 4² + (-2)² – (3)(3)(-2)
w = 9 – 16 + 4 + 24
w = 37 – 16
w = 21

Question 12.
Simplify and find the degree of 6x² + 1 – [8x – {3x² – 7 – (4x² – 2x + 5x + 9)}]
Solution:
6x² + 1 – [8x – (3x² – 7 – (4x² – 2x + 5x + 9)}]
= 6x² + 1 – [8x – {3x² – 7 – 4x² – 2x + 5x + 9}]
= 6x² + 1 – [8x – 3x² + 7 + 4x² – 2x + 5x + 9}]
= 6x² – 1 – [8x + 3x² – 7 – 4x² + 2x – 5x – 9]
= 6x² + 3x² – 4x² – 8x + 2x – 5x – 1 – 7 – 9]
= x²(6 + 3 – 4) + x(8 + 2 – 5) – 15
= 5x² – 11x – 15
Degree of the expression is 2.

 

Question 13.
The two adjacent sides of a rectangle are 2x² – 5xy + 3z² and 4xy – x² – z². Find the perimeter and the degree of the expression.
Solution:
Let the two adjacent sides of the rectangle as
l = 2x² – 5xy + 3z² and b = 4xy – x²y + 3z²

Perimeter of the rectangle
= 2(l + b) = 2(2x² – 5xy + 3z² + 4xy – x² – z²)
= 4x² – 10xy + 6z² + 8xy – 2x² – 2z²
= 4x² – 2x² – 10xy + 8xy + 6z² – 2z²
= x²(4 – 2) + xy (-10 + 8) + z² (6 – 2z²)
Perimeter = 2x² – 2xy + 4z²
Degree of the expression is 2.

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