SOLUTIONS CLASS

These Solved Solutions are powered by: writeshub.com

WELCOME TO SOLUTIONS CLASS !!!

www.facebook.com/ekemini.ekpo

writestable@gmail.com
Ekemini Ekpo - Founder, The Writeshub Group

N/B: The following free solution sets are referenced to original Post UTME Past Questions for the under listed schools. If you do not have the original question sets, please click here to get all the Past Questions before proceeding.

Need personal tutor? Get an Online Tutor here [SSCE, WAEC, NECO, JAMB, Post UTME, Higher Education]

Subjects Covered: SSCE Science, Arts & Commercial Subjects.
Tutors: University Graduates/Undergraduates [with assistance from Lecturers/Teachers]
Fee: FREE

Below is a Session for one school/Class. . . .More Schools/Subjects to be added later in the day.

UNIUYO MATHEMATICS POST UTME SOLUTIONS

Some Needful updates:

Nature of UNIUYO Post UTME Exams

UNIUYO Post UTME Cut off mark

How to get UNIUYO Post UTME Forms, Register and Check Result, Admission List

N/B: UNIUYO Mathematics Post UTME Past Questions contain twenty-five (25) questions per Year. We have provided for you solutions to the first five (5) questions per year free of charge. If you need the complete solutions to all the years (over 150 questions/department) with detailed answers, simply drop your email address in the comments box below.

Qst 1: Given that (1x03)₄ = 115₁₀ find x.

Solution

We can solve the above Number base problem using either of two methods;

Method 1: Equate both sides of the equation. First, convert the L.H.S of the Equation to base ten (10) just as the R.H.S.

Converting (1x03)₄ to base 10,

1 * 4³ + x * 4³ + 0 * 4¹ + 3 * 4⁰

64 + 16x + 0 + 3

(67 + 16x)₁₀

Now, equate the L.H.S with the R.H.S.  So we have,

67₁₀ + 16x₁₀ = 115₁₀ [same base]

16x₁₀ = 115₁₀ - 67₁₀

            = 48 [ignoring the bases, or cancelling them out]

16_x = 48,

X = 48/16 = 3

Therefore, x = 3. [Answer, OPTION C]

Method 2

The above problem may also be solved by comparing the place values of the L.H.S with that of the R.H.S. To do this, first convert the R.H.S to base 4 using the long division method, thus:

4	115	R
4	28	3
4	7	0
4	1	3
	0	1

Reading the values up, the normal way, we have that 115₁₀ = 1303₄

Now, comparing both sides of the equation,

1x03₄ = 1303₄

As you can see, the place value of “x” in the L.H.S equals the Place value of “3” in the R.H.S

Hence, x = 3. [Answer, OPTION C]

Qst 2: A motorist starts from point O and moves on a bearing of 030⁰ for 48km. He then moves 10km toward the west. How far is he now from O?

 Solution

To solve the above problem, we need a small trigonometric sketch of the event described. Thus,

OY/48 = Sin 30⁰

OY = 48Sin 30⁰ [given that Sin 30⁰ = 1/2 ]

            48 * ½ = 48/2 = 24.

[Answer, OPTION C]

Qst 3: If 7 and -3 are the roots of the quadratic equation x² + kx – 21 = 0, what is the value of k?

Solution

Let’s say that the roots of the quadratic equation are a and b.

a = 7; b = -3

Standard format of a quadratic equation by when broken down into factors is (x - a)(x - b).

Substituting the values of a, b into equation above we have:

(x - a)(x - b)

(x - 7)(x – (-3))

(x - 7)(x + 3)

Expanding the brackets,

x (x + 3) – 7 (x + 3)

x² + 3x – 7x – 21

x² – 4x – 21

Comparing the derived quadratic equation with the one given in question we have;

x² – 4x – 21 = x² + kx – 21

-          4x = kx

Hence k = - 4

[Answer, OPTION B]

Qst 4: If the function f(x) = x³ + 2x² + kx – 6 is divisible by x, find k.

Solution

Given that f(x) = x³ + 2x² + kx – 6,

Using simple Polynomial trial and error method, if the function f(x) is divisible by x, then the divisor would be in the form, x – 1 = 0.

Check: if x – 1 = o, then x = o + 1 = 1

Substituting x = 1 into the original equation, we have:

F(x) = x³ + 2x² + kx – 6, [x = 1]

f (1) = [1]³ + 2[1]² + k [1] – 6 = 0,

            k – 3 = 0

Therefore, k = 3.

[Answer, OPTION A]

Qst 5: A woman is 3 times as old as her daughter. 8 years ago, the product of their ages was 112. Find the mother’s age.

Solution

Let daughter’s present age = x

Mum’s present age = 3 times daughter’s present age = 3x

Daughter’s age 8 yrs ago = x - 8

Mum’s age 8 yrs ago = 3x – 8

The product of their ages (mum’s and daughter’s) 8 years ago = 112

i.e.,

(x - 8)(3x – 8) = 112

Expanding this,

x (3x - 8) – 8 (3x - 8)

3x² - 8x + 21x + 64 = 112

3x² – 32x = 48

3x² – 32x - 48 = 0

We now evaluate the above using the Quadratic formula to solve for x.

You may use completing the squares method or any other method to still get the same answer. I choose to use the almighty formula given by;

Therefore, x = 12, - 1.33 [since age cannot be a negative value, the correct age is 12years]

Let daughter’s present age = x = 12years

Mum’s age = 3 times daughter’s present age = 3x = 3 * 12years = 36years

[Answer, OPTION B]

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

Thank you for following our brief maths solution series. To get the complete solutions to all the years (over 150 questions/department) with detailed answers, simply drop your email address in the comments box below.

• Click here to get all the Mathematics + Other Post UTME Past Questions for your school.
• Get an Online Tutor here [SSCE, WAEC, NECO, JAMB, Post UTME, Higher Education]

Above is a Session for one school/Class. . . .More Schools/Subjects to be added later in the day.

CLASS IS CURRENTLY ON BREAK. WE SHALL RESUME SHORTLY WITH TUTORIALS ON OTHER SUBJECTS AND SCHOOLS. . . . .

Drop a comment below. . .

Need to contact us? Click here to contact us

Search Below for More Information