Time: 40 mins

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Question 1 (a)

Find the values of x, y and z in the figure below.

Solution

30º + yº + 64º = 180º (sum of angles on a straight line = 180º)

⇒ yº + 94º = 180º

⇒ yº = 180º – 94º

∴ yº = 86º

xº + 135º = 180º (sum of angles on a straight line = 180º)

⇒ xº = 180º – 135º

∴ xº =  45º

xº + yº + zº = 180 (sum of angles in a triangle = 180º)

⇒ 45º + 86º + zº = 180º

⇒  zº + 131º = 180º

⇒  zº = 180º – 131º

∴ zº = 49º

Question 1 (b)

Calculate the mean of the data below, correct to one decimal place.

Score (x) 9 10 12 13 15
f 3 5 2 2 3

Solution

x f fx
9 3 27
10 5 50
12 2 24
13 2 26
15 3 45
Σf = 15 Σfx = 172

Mean x = \( \frac{\sum fx}{\sum f} \\ = \frac{172}{15} \\ = \scriptsize 11.4666 \\ = \scriptsize 11.5 \: to \: 1 \:d.p \)

Question 2 (a)

Evaluate 1101two + 1111two

Solution

The rules of binary addition are:

0 + 0 = 0
0 + 1 = 1
1 + 0 = 1
1 + 1 = 0 with a Carry of 1
1 + 1 + 1 = 1 with a Carry of 1

Question 2 (b)

⇒ \( \frac{2x\:-\:1}{3\:-\:4x}= \frac{2}{5} \)

Solution

cross multiply

⇒ \( \scriptsize 5(2x\:-\:1) = 2(3\:-\:4x)\)

open brackets

⇒ \(\scriptsize 10x\:-\:5 = 6\:-\:8x \)

collect like terms

⇒ \( \scriptsize 10x\:+\:8x = 6\:+\:5 \)

⇒ \( \scriptsize 18x = 11 \)

divide both sides by 18

∴  \(\scriptsize x = \normalsize \frac{11}{18} \)

Question 3

Find the rate per annum, if ₦36,000.00 is paid as simple interest on ₦120,000 deposited in the bank for 3 years.

Solution

S.I = \( \frac{PRT}{100} \)

where;

  • P = Principal
  • R= Rate
  • T = Time

Given;

  • S.I = ₦36,000
  • P =  ₦120,000
  • T = 3 years
  • Rate = ?

Make R the subject of the formula: S.I = \( \frac{PRT}{100} \)

⇒ \( \scriptsize 100 \: \times \: S.I = PRT \)

or \( \scriptsize  PRT = 100 \: \times \: S.I \)

∴ \( \scriptsize  R = \normalsize \frac{100 \: \times \: S.I}{PT} \)

Substitute in the given values

⇒ \( \scriptsize  R = \normalsize  \frac{100 \: \times \: 36,000}{120,000 \: \times \: 3} \)

⇒ \( \scriptsize R = \normalsize  \frac{3,600,000}{360,000} \)

⇒ \( \scriptsize R = 10 \%\)

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