Correct Answer: (B) 70

Explanation

p*q = p\(^2\) + 2pq - q\(^2\)

when p = -2 and q = -2

p*q = -2\(^2\) + 2(-2)(-2) - (-2)\(^2\)

p*q = 4 + 8 - 4 = 8

when p = -2 and q = -1

p*q = -2\(^2\) + 2(-2)(-1) - (-1)\(^2\)

p*q = 4 + 4 - 1 = 7

when p = -2 and q = 1

p*q = -2\(^2\) + 2(-2)(1) - (1)\(^2\)

p*q = 4 - 4 -1 = -1

when p = -1 and q = -2

p*q = -1\(^2\) + 2(-1)(-2) - (-2)\(^2\)

p*q = 1 + 4 - 4 = 1

when p = -1 and q = -1

p*q = -1\(^2\) + 2(-1)(-1) - (-1)\(^2\)

p*q = 1 + 2 -1 = 2

when p = -1 and q = 2

p*q = -1\(^2\) + 2(-1)(2) - (2)\(^2\)

p*q = 1 - 4 - 4 = -7

when p = 1 and q = -2

p*q = 1\(^2\) + 2(1)(-2) - (-2)\(^2\)

p*q = 1 - 4 - 4 = -7

when p = 1 and q = -1

p*q = 1\(^2\) + 2(1)(-1) - (-1)\(^2\)

p*q = 1 - 2 - 1 = -2

when p = 1 and q = 1

p*q = 1\(^2\) + 2(1)(1) - (1)\(^2\)

p*q = 1 + 2 - 1 = 2

when p = 1 and q = 2

p*q = 1\(^2\) + 2(1)(2) - (2)\(^2\)

p*q = 1 + 4 - 4 = 1

when p = 2 and q = -2

p*q = 2\(^2\) + 2(2)(-2) - (-2)\(^2\)

p*q = 4 - 8 - 4 = -8

when p = 2 and q = -1

p*q = 2\(^2\) + 2(2)(-1) - (-1)\(^2\)

p*q = 4 - 4 - 1 = -1

when p = 2 and q = 1

p*q = 2\(^2\) + 2(2)(1) - (1)\(^2\)

p*q = 4 + 4 - 1 = 7

when p = 2 and q = 2

p*q = 2\(^2\) + 2(2)(2) - (2)\(^2\)

p*q = 4 + 8 - 4 = 8

Correct Answer: D. 3cm

Explanation

hyp = 5cm, adj = \(\frac{8cm}{2}\) 4cm, opp = xcm

Pythagoras theorem:

hyp\(^2\) = opp\(^2\) + adj\(^2\)

5\(^2\) = x\(^2\) + 4\(^2\)

x\(^2\) = 25 - 16

x =√9

x = 3cm

Explanation

(a)

(b) Modal class: This is the class boundary with the highest frequency.

In this case, it is 139.5 and 149.5.

From the histogram, the modal income is 3 +139.5

= 142.5 or N142500 (in thousands of naira)

Correct Answer: C. -1, 3

Explanation

x² - 2x - 3 = 0

x² - 3x + x - 3 = 0

x(x - 3) + (x - 3) = 0

(x + 1)(x - 3) = 0

x = -1 or 3

Explanation

Let P represent the cost ; n, the number of pupils and k and c, constant terms.

\(\therefore P = k + cn\)

\(15705 = k + 50c ... (1)\)

\(13305 = k + 40c .... (2)\)

(1) - (2) :

\(2400 = 10c \implies c = \frac{2400}{10} = 240\)

Put c = 240 in (1), so that

\(15705 = k + (50)(240)\)

\(15705 = k + 12000 \implies k = 3705\)

\(P = 3705 + 240n\) ------ formula connecting price and number of pupils.

(a) When number of pupils, n = 44

\(P = 3705 + 240(44) = 3705 + 10560 = $14265\)

(b) For school not to be at loss:

\(360 \geq k + cn\)

\(360n \geq 3705 + 240n\)

\(360n - 240n \geq 3705 \implies 120n \geq 3705\)

\(n \geq \frac{3705}{120}\)

\(n \geq 30.875 \)

\(n \geq 31\)

The pupils should be 31 at least for the school not to run at loss.

Correct Answer: B. (x + y + z)(x - y - z)

Explanation

\(x^2 - (y + z)^2\)

Using difference of two squares (\(a^2 - b^2 = (a + b)(a - b)|)

⇒ (x + (y + z))(x - (y + z))

⇒ (x + y + z)(x - y - z)

Correct Answer: B. 12 times

Correct Answer: C. 9.32m

Explanation

In right angled \(\bigtriangleup\)QPR

tan 30^o = \(\frac{10}{x + 8}\)

(x + 8) tan 30 = 10

x + 8 = \(\frac{10}{0.5773}\)

x +8 = 17.3

x = 17.3 - 8

x = 9.32

Correct Answer: (d) 0%