Question Library

\( x^2 + 9x + 20 \) Find the two numbers whose product is 20 and its sum is 9} \( 5x \times 4x = 20x^2\) \(5x + 4x = 9x\) \((x^2 + 5x) + (4x + 20)\) \( x(x + 5)+ 4(x + 5) = (x + 5)(x + 4)\)

Total of their ages: 18 + 20 + 22 = 60 The second staff will get \(\frac{20}{60} \times 120,000\) = N40,000.00K

n(X ∪ Y) = n(X) + n(Y) − n(X ∩ Y) = 15 + 12 − 7 ∴ n(X ∪ Y) = 20

Starting from 3x − z \(\leq\) 9 x = 0, substitute the value of x (i.e. x = 0) into the equation 3x − z \(\leq\) 9 3(0) − z \(\leq\) 9 z \(\leq\) − 9 If z = 0 Then, 3x − 0 \(\leq\) 9 3x \(\leq\) 9 x \(\leq\) 3 Intercept of x and z i.e (x,z) = (3, -9)

log101.5 + 3 log102 − log100.3 log101.5 + log1023 − log100.3 [log10(1.5 × 23) ÷ log100.3] [log10(15/10 × 8 ) ÷ log10( 3/10)] log10(15/10 × 8 × 10/3 ) log1040

The total surface area of a cylinder = 2\pirl + 2\pir2 = 2\pir(l + r) = 2 \times 3.14 x 5(7+5) 2 \times 3.14 \times 12 x 5 = 377.1cm (1DP)

Given: \(x + y = 90^\circ ... (1)\) \((\sin x + \sin y)^{2} - 2\sin x \sin y = \sin^{2} x + \sin^{2} y + 2\sin x \sin y - 2\sin x \sin y\) = \(\sin^{2} x + \sin^{2} y ... (2)\) Recall: \(\sin x = \cos (90 - x) ... (a)\) From (1), \(y = 90 - x ... (b)\) Putting (a) and (b) in (2), we have \(\sin^{2} x + \sin^{2} y \equiv \cos^{2} (90 - x) + \sin^{2} (90 - x)\) = 1

His annual salary = N2000 His allowances = N600 So his taxable income = Annual salary − allowance = N2000 − N600 = N1400 He pay at 5% Then, his allowance income tax 5/100 \times 1400 = N70

Using the quadratic formula, we have \(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\) From the equation \(3x^{2} - 4x - 5 = 0\), a = 3, b = -4 and c = -5. \(\therefore x = \frac{-(-4) \pm \sqrt{(-4)^{2} - 4(3)(-5)}}{2(3)}\) = \(\frac{4 \pm \sqrt{16 + 60}}{6}\) = \(\frac{4 \pm \sqrt{76}}{6}\) = \(\frac{4 \pm 8.72}{6}\) = \(\frac{4 + 8.72}{6} \text{ or } \frac{4 - 8.72}{6}\) = \(\frac{12.72}{6} \text{ or } \frac{-4.72}{6}\) \(x = \text{2.12 or -0.79}\)

nth term of a linear sequence (AP) = a+(n − 1)d first term = 6, last term = 10 sum − 40 i.e. a = 6, l = 10, S = 40 \(S_n\)= n/2(2a + (n − 1)d or Sn = \div2 (a + l) \(S_n\) = n/2(a + l) 40 = n/2(6 + 10) 40 = 8n 8n = 40 8n = 40 n = 40/8 = 5 The number of terms = 5

The slope of the line from (0, 0) passing through (-3, -4) = \(\frac{-4 - 0}{-3 - 0}\) = \(\frac{4}{3}\) Equation of a line is given as \(y = mx + b\), where m = slope and b = intercept. To get the value of b, we use a point on the line, say (0, 0). \(y = \frac{4}{3} x + b\) \(0 = \frac{4}{3}(0) + b\) \(b = 0\) The equation of the line is \(y = \frac{4}{3} x\)

\( A = P \left(1 + \frac{r}{100}\right)^n \) Where P = 126, r = 4,n = 2 A=126 \( \left(1 + \frac{4}{100}\right)^2 \text{Using LCM} \) =126 \( \left(\frac{100+4}{100}\right)^2 = 126 \left(\frac{104}{100}\right)^2 \) =126 \( \left(1.04^2 \right) \) = 126 * 1.04 * 1.04 =136.28 A = 136.30 (approx.) The Amount A, = N136.30k

No expanded working has been attached for this question yet. The verified answer is Option A.

log520 = x 5x = 20(Take log10 of both sides) log5x = log20 xlog5 = log20 x= [log20 ÷ log5] [1.30103 ÷ 0.69897] x = 1.861

I = A − P = N588 − N525 ∴I = N63 I = PRT ÷ 100 R = [100I ÷PT] R = [(100 × 63 ) ÷ (525 × 3)] = (6300 ÷ 1575) = 4 ∴ The rate = 4 %