MATHEMATICS OBJ:

.1-10: CDABCDBADB

11-20: CABCADBCAD

21-30: BCADAAADBB

31-40: ADACBDBDBC

41-50: ACAAADCCCB

====================

MATHEMATICS THEORY:

===========================

PLEASE TAKE NOTE OF THE FOLLOWING SIGNS AND THEIR MEANING.

* means multiplication

– means minus/subtraction

[SECTION A] ANSWER ALL THE FIVE (5) QUESTION IN THIS SESSION

(1a)

123x = 2710

1² 2¹ 30x= 2710 (Converting the LHS to base ten)

(1 * x²) + (2 * x¹) +(3 * x0) = 27

x² + 2x + 3 = 27 (Since x0 = 1 )

x² + 2x + 3 – 27 = 0

x² + 2x – 24 = 0

By Factorization;

x² + 6x – 4x – 24 = 0

x ( x + 6) – 4 (x + 6) = 0

(x-4) (x+6) = 0

x-4 = 0 or x + 6 = 0.

x = 4 or -6

Since x must be a positive value, the x = 4

=======================

(2a)

¼ of 64x = 163x

From the law of indices

ab * ac = ab+c

:- ¼ * 64x = 163x

4-1 * 43x = 42(3x)

4-1 * 43x = 46x

Also; pa = pb imply a=b

3x – 1 =6x

-1 =6x – 3x

-1 =3x; 3x = -1

x = -1 /3

========================

================================

(4a)

Let x rep. the man’s age.

Let y rep. the son’s age.

Four years ago;

x – 4 = 4 (y – 4) – -> ; eqn i

x – y = 30 – -> ; eqn ii

Solve simultaneously; using substitution method.

From eqn ii; x = 30 + y

Substitute into eqn i.

30 + y – 4 = 4 (y – 4)

26 + y = 4y – 16

26 + 16 = 4y – y

42 = 3y; y = 42/3 = 14

x = 30 + 14; x = 44

:- the man’s age = 44 While

the son’s age = 14

(4b)

======================

===============================

[SECTION B] ANSWER ANY FIVE (5) QUESTION IN THIS SESSION

===============================

==============================

================================

==========================

===========================

============================

## Leave a Reply