Ondo State Joint Promotion Exam 2019 Mathematics Answer

MATHEMATICS OBJ:

.1-10: CDABCDBADB
11-20: CABCADBCAD
21-30: BCADAAADBB
31-40: ADACBDBDBC
41-50: ACAAADCCCB

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MATHEMATICS THEORY:
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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

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(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

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(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)

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