2018 Paper 3 — Theory | BECE Mathematics

Question 1

(a) If a plastic cup costs N80.00 and \(\frac{1}{2}\) dozen of it cost N420.00, find the discount on \(\frac{1}{2}\)dozen.

Solution

1 plastic cup = ₦80

6 plastic cups at ₦80 per cup = 6 x ₦80

= ₦480

½ dozen price = ₦420

discount on ½ dozen = ₦480 – ₦420

= ₦60

Note

1 dozen = 12

½ dozen = 6

(b) Find the value of Y if Y_two  x  101_two = 110111_two

Solution

Y_two  x  101_two = 110111_two

convert every term to base ten

(Y x 2 ⁰) x (1 x 2² + 0 x 2¹ + 1 x 2⁰)= (1 x 2⁵ + 1 x 2⁴ + 0 x 2³ + 1 x 2² + 1 x 2¹ + 1 x 2⁰)

= Y x (4 + 0 + 1) = (32 + 16 + 0 + 4 + 2 + 1)

= Y x 5 = 55

5Y = 55

divide both sides by 5

\(\frac{5Y}{5} = \frac{55}{5}\)

Y = 11_ten

Next step is to convert 11_ten  to base 2

= 1011_two

(c) the product of three numbers is 3150. Two of the numbers are 14 and 15. Find the third number.

Solution

Let the third number = x

1st number = 14

2nd number = 15

3rd number = x

Product of the numbers = 3150

\(\scriptsize 14 \: \times \: 15 \: \times \: x = 3150\)

x = \(\frac{3150}{14 \: \times \: 15}\)

x = \(\frac{3150}{210}\)

x = 15

∴ The thrid number = 15

Question 2

(a) Factorize 28a² – 63b²

Solution

7(4a² – 9b²)

= 7(2²a² – 3²b²)

= 7(2a + 3b)(2a – 3b) (difference of two squares)

(b) Solve the equation

\(  \normalsize \frac{4x\: +\: 5}{7} \scriptsize \: + \: 3x = \normalsize \frac{x}{7} \: – \: \frac{1}{9} \)

Solution

Multiply through by the L.C.M which is 63

\(  \normalsize \frac{4x\: +\: 5}{7} \scriptsize \: \times \: 63 \: + \: 3x \: \times \: 63 = \normalsize \frac{x}{7}\scriptsize \: \times \: 63 \: – \: \normalsize \frac{1}{9} \scriptsize  \: \times \: 63  \)

= 9(4x + 5) + 189x = 9x – 7

Open the brackets

36x + 45 + 189x = 9x – 7

Collect like terms

36x + 189x – 9x = -7 – 45

216x = -52

Divide both sides by 216

\(\frac{216x}{216} = \frac{-52}{216}\)

\(\scriptsize x = \normalsize \frac{-52}{216}\)

\(\scriptsize x = \normalsize \frac{-13}{54}\)