Pedagogy of Mathematics (Part II) - Algebra, Algebra, Maths, IX std Maths, Samacheerkalvi maths, II year B.Ed., Pedagogy, Mathematics, Consistency and Inconsistency of linear Equation in two variables,

1. PEDAGOGY OF
MATHEMATICS – PART II
BY
Dr. I. UMA MAHESWARI
Principal
Peniel Rural College of Education,Vemparali,
Dindigul District
iuma_maheswari@yahoo.co.in

2. STD IX
CHAPTER 3 – ALGEBRA
Ex – 3.14 & 3.15

11. Solution:
Let the ten’s digit be x and the unit digit be
y.
The number is 10x + y
If the digits are interchanged
The new number is 10y + x
By the given first condition
10x + y + 10y + x = 110
11x + 11y = 110
x + y = 10 → (1) (Divided by
11)

12. Again by the given second condition
10x + y – 10 = 5(x + y ) + 4
10x + y – 10 = 5x + 5y + 4
5x – 4y = 14 → (2)
(1) × 5 ⇒ 5x + 5y = 50 → (3)
(2) × 1 ⇒ 5x – 4y = 14 → (2)
(3) – (2) ⇒ 9y = 36
y = 36/9
= 4
Substitute the value of y = 4 in
(1)
x + y = 10
x + 4 = 10
x = 10 – 4
= 6
∴ The number is (10 × 6 + 4) =
64

14. ABCD is a cyclic quadrilateral ∠A + ∠C =
180°
(Sum of the opposite angles of a cyclic
quadrilateral is 180°)
(4y + 20)° + (4x)° = 180°
Solution:
4x + 4y = 180 – 20
4x + 4y = 160
x + y = 40 → (1) (divided by 4)
∠B + ∠D = 180° (Sum of the
opposite angles of a cyclic
quadrilateral)
(3y – 5)° + (7x + 5)° = 180°
3y – 5 + 7x + 5 = 180

15. (1) × 3 ⇒ 3x + 3y = 120 →
(3)
(3) – (2) ⇒ -4x = – 60
4x = 60
x = 60/4
Substitute the value of x =
15 in (1)
15 + y = 40
y = 40 – 15 = 25
∠A = 4y + 20 = 4(25) + 20
= 100 + 20 = 120°
∠B = 3y – 5 = 3(25) – 5 = 75 – 5 = 70
∴ ∠B = 70°
∠C = 4x = 4(15) = 60
∴ ∠C = 60°
∠D = 7x + 5 = 7(15) + 5
∠D = 105 + 5 = 110°
∴ ∠A= 120°, ∠B = 70°, ∠C = 60° and ∠D = 110°

16. Solution:
Let the cost price of the TV be Rs “x” and the cost price of the fridge be Rs “y”.
By the given condition
Multiply by 20
x + 2y = 40000 → (1)
Again by the given second condition

18. Solution:
Let the two numbers be x and y.
By the given first condition
x : y = 5 : 6
6x = 5y (Product of the extreme is equal to the product of the means)
6x – 5y = 0 → (1)
Again by the given second condition
x – 8 : y – 8 = 4 : 5
5(x – 8) = 4(y – 8)

19. 5x – 40 = 4y – 32
5x – 4y = – 32 + 40
5x – 4y = 8 → (2)
(1) × 4 ⇒ 24x – 20y = 0 → (3)
(2) × 5 ⇒ 25x – 20y = 40 → (4)
(3) – (4) ⇒ – x + 0 = -40
∴ x = 40
Substitute the value of x = 40 in (1)
The two numbers are 40 and 48
[∴ The ratio of the number = 40 : 48 are 5
: 6]