Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.4

Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.4

Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.4

Miscellaneous Practice Problems

Question 1.
Subtract – 3ab – 8 from 3ab – 8. Also subtract 3ab + 8 from -3ab – 8.
Solution:
Subtracting -3ab – 8 from 3ab + 8
= 3ab + 8 – (-3ab – 8) = 3ab + 8 + (3ab + 8)
= 3ab + 8 + 3ab + 8 = (3 + 3) ab + (8 + 8)
= 6ab + 16
Also subtracting 3 ab + 8 from – 3ab – 8
= – 3ab – 8 – (3ab + 8) = – 3ab – 8 + (-3ab – 8) = – 3ab – 8 – 3 ab – 8
= [(-3) + (- 3)] ab + [(-8) + (-8)] = – 6ab + (- 16)
= -6ab – 16

Question 2.
Find the perimeter of a triangle whose sides are x + 3y, 2x + y, x – y.
Solution:
Perimeter of a triangle = Sum of three sides
= (x + 3y) + (2x + y) + (x – y)
= x + 3y + 2x + y + x – y
= (1 + 2 + 1)x + (3 + 1 + (-1))y = 4x + 3y
∴ Perimeter of the triangle = 4x + 3y

Question 3.
Thrice a number when increased by 5 gives 44. Find the number.
Solution:
Let the required number be x.
Thrice the number = 3x.
Thrice the number increased by 4 = 3x + 5
Given 3x + 5 = 44
3x + 5 – 5 = 44 – 5
3x = 39
3x/3=39/3
x = 13
∴ The required number = 13

Question 4.
How much smaller is 2ab + 4b – c than 5ab – 3b + 2c.
Solution:
To find the answer we have to find the difference.
Here greater number 5ab – 3ab + 2c.
∴ Difference = 5ab – 3b + 2c – (2ab + 4b – c) = 5ab – 3b + 2c + (- 2ab -4b + c)
= 5ab – 3b + 2c – 2ab – 4b + c
= (5 – 2) ab + (-3 – 4) b + (2 + 1) c = 3ab + (-7)b + 3c
= 3ab – 7b + 3c
It is 3ab – 7b + 3c smaller.

Question 5.
Six times a number subtracted from 40 gives – 8. Find the number.
Solution:
Let the required number be x. Six times the number = 6x.
Given 40 – 6x = – 8
-6x + 40 – 40 = -8 – 40
– 6x = – 48
−6x/−6=−48/−6
x = 8
∴ The required number is 8.

Challenge Problems

Question 6.
From the sum of 5x + 7y -12 and 3x – 5y + 2, subtract the sum of 2x – 7y – 1 and – 6x + 3y + 9.
Solution:
Sum of 5x + 7y – 12 and 3x – 5y + 2 .
= 5x + 7y- 12 + 3x – 5y + 2 = (5 + 3) x + (7 – 5) y + ((- 12) + 2)
= 8x + 2y – 10.
Again Sum of 2x – 7y – 1 and – 6x + 3y + 9
= 2x – 7y – 1 + (- 6x + 3y + 9) = 2x – 7y – 1 – 6x + 3y + 9
= (2 – 6) x + (- 7 + 3) y + (- 1 + 9)
= – 4x – 4y + 8
Now 8x + 2y – 10 – (-4x – 4y + 8)
= 8x + 2y – 10 + (4x + 4y – 8)
= 8x + 2y – 10 + 4x + 4y – 8
= (8 + 4) x + (2 + 4) y + ((- 10) + (- 8))
= 12x + 6y – 18

Question 7.
Find the expression to be added with 5a – 3b – 2c to get a – 4b – 2c?
Solution:
To get the required expression we must subtract 5a – 3b + 2c from a – 4b – 2c.
∴ a – 4b – 2c – (5a – 3b + 2c) = a – 4b – 2c + (- 5a + 3b – 2c)
= a – 4b – 2c – 5a + 3b -2c
= (1 – 5) a + (- 4 + 3) b + (- 2 – 2) c
= – 4a – b – 4c.
∴ -4a – b – 4c must be added.

Question 8.
What should be subtracted from 2m + 8n + 10 to get – 3m + 7n + 16?
Solution:
To get the expression we have to subtract – 3m + 7n + 16 from 2m + 8n + 10.
(2m + 8n + 10) – (-3m + 7n + 16) = 2m + 8n + 10 + 3m – 7n – 16
= (2 + 3) m + (8 – 7) n + (10 – 16)
= 5m + n – 6

Question 9.
Give an algebraic equation for the following statement:
“The difference between the area and perimeter of a rectangle is 20”.
Solution:
Let the length of a rectangle = l and breadth = b then Area = lb; Perimeter = 2(1 + b)
Area – Perimeter = 20
∴ lb – 2(l + b)