Correct Answer: B. -1

Explanation

p - 2g + 1 = g + 3p.........(1)

p - 2 = 0 .........(2)

From (2), p = 2; put p = 2 into (1);

2 - 2g + 1 = g + 3(2)

3 - 2g = g + 6

-2g - g = 6 - 3

-3g = 3

g = \(\frac{3}{-3}\)

g = -1

Correct Answer: C. \(x^{2} + 2x - 3 = 0\)

Explanation

Given the roots of an equation such that you can find the sum and product of the roots, the equation can be given as:

\(x^{2} - (\alpha + \beta)x + (\alpha \beta) = 0 \)

\(\alpha + \beta = -3 + 1 = -2\)

\(\alpha \beta = -3 \times 1 = -3\)

Equation: \(x^{2} - (-2)x + (-3) = 0 \implies x^{2} + 2x - 3 = 0\)

Correct Answer: B. 2/15

Explanation

P(X \(\cap\) Y) = \(\frac{1}{3} \times \frac{2}{5}\)

= \(\frac{2}{15}\)

Explanation

(a) \(< ADB = < ACB = 20°\) (Angle in the same segment).

In \(\Delta ACY\),

\(\hat{C} = < ACB\)

(i) \(\hat{C} + \hat{A} + \hat{Y} = 180°\)

\(20° + \hat{A} + 60° = 180°\)

\(\hat{A} = 180° - 80° = 100°\)

\(< CAY = \hat{A} = 100°\)

\(< CAX = 180° - 100° = 80°\)

(ii) In \(\Delta ACB\),

\(\hat{B} = \hat{C} = \frac{180° - 20°}{2} \)

= \(\frac{160°}{2}\)

= \(80°\)

In \(\Delta ABY\),

\(< ABY = \hat{B} = 180° - 80°\)

= \(100°\)

(b) \((m + 1)(m - 3) \equiv m^{2} - km + c\)

\(m^{2} - 3m + m - 3 = m^{2} - 2m - 3\)

\(m^{2} - 2m - 3 \equiv m^{2} - km + c\)

\(\implies k = 2 ; c = -3\)

Explanation

The Chief Examiner reported that in part (a), some Candidates' carried out the requirement of BODMAS correctly tile few couldn't. However, in part (b), most Candidates' could not sketch the diagram correctly and hence failed to earn the marks allocated.

In part (a), expressing as improper fraction we have (\(\frac{1}{4} \times 9\frac{1}{7} + \frac{2}{5} (\frac{2}{2} + \frac{3}{4})) \div (\frac{2}{5} - \frac{1}{4})\) Next, using BODMAS to get (\(\frac{16}{7} + \frac{2}{5} \times \frac{17}{12}) \div \frac{17}{12}\). Simplifying and inverting yields \(\frac{1198}{420} \times \frac{20}{3} = 19\frac{1}{63}\)

In part (b), they were expected to sketch

Then, using trigonometric ratio to have sin 48\(^o\) = \(\frac{x}{250}\) or cos 42\(^o\) = \(\frac{x}{250}\) and simplifying the vertical distance moved to be 186 m.

Correct Answer: C. 0.53 square units

Explanation

Area = \(\frac{1}{2} \times 1.33 \times 0.8\)

= 0.53 square unit

Correct Answer: B. 150°

Correct Answer: C. 8.8 x 10^-3

Correct Answer: D. 24cm

Explanation

Considering the smaller and larger triangle, these two are similar triangles. Hence,

If the height of the smaller triangle = h,

\(\therefore \frac{h}{6} = \frac{h + 12}{12}\)

\(12h = 6h + 72 \implies 6h = 72\)

\(h = 12 cm\)

\(\therefore\) The height of the cone = 12 + 12 = 24 cm

Explanation

a. Let the numbers be p, r and q respectively

Given:

p + r + q = 81 --- (i)

r = 2p --- (ii)

q = r + 6 --- (iii)

Substitute (2p) for r in equation (iii)

q = 2p + 6 --- (iv)

Substitute (2p) for r and (2p + 6) for q in equation (i)

⇒ p + (2p) + (2p + 6) = 81

⇒ 5p + 6 = 81

⇒ 5p = 81 - 6

⇒ 5p = 75

p = \(\frac{75}{5}\) = 15.

From equation (ii)

⇒ r = 2(15) = 30

From equation (iv)

⇒ q = 2(15) + 6

⇒ q = 30 + 6 = 36

∴ The numbers are 15, 30 and 36

b. \(m_1 = \frac{ y_2 - y_1}{ x_2 - x_1}\)

Given points are P( 3, 5) and Q( -5, 7) = P(\( x_1 , y_1)\) , Q(\( x_2 , y_2)\)

the slope of line PQ = \((\frac{ 7 - 5}{ -5 - 3}) = \frac{2}{-8} = -\frac{1}{4}\)

Given that R( x, y) is the midpoint of the line formed by the points P( 3, 5) and Q( -5, 7)

R(x, y) = \(\frac{3 + (-5)}{2}, \frac{5 + 7}{2} = \frac{ 3 - 5}{2} , \frac{ 5+7}{2}\)

= -\(\frac{2}{2} , \frac{12}{2}\) = (-1, 6)

but the line passes through the point R( -1, 6) and is perpendicular to LINE PQ

\(m_2 = \frac{1}{m_1} = -\frac{1}{\frac{-1}{4}}\) = 4

Equation of the line can be obtained using \(y - y_1 = m_2( x - x_1)\)

y - 6 = 4( x - (-1)) = y - 6 = 4( x + 1 )

y - 6 = 4x + 4

y = 4x + 4 + 6

Therefore, y = 4x + 10