A binary operation Δ is defined by a Δ b = a + 3b + 2.
Find (3 Δ 2) Δ 5
35
59
28
87
Correct answer is C
a Δ b = a + 3b + 2 (3 Δ 2) Δ 5 = (3 + 3(2) + 2) Δ 5 = 11 Δ 5 = 11 + 3(5) + 2 = 28
N21,850
N18,780
N27,400
N32,500
Correct answer is C
Education = \(\frac{15}{100} \times N 50,000\)
= N 7,500
Food = N 13,600
Electricity = \(\frac{3}{100} \times N 50,000\)
= N 1,500
Leftover : N (50,000 - (7,500 + 13,600 + 1,500))
= N 27,400
Find the equation of the locus of a point A(x, y) which is equidistant from B(0, 2) and C(2, 1)
4x + 2y = 3
4x - 3y = 1
4x - 2y = 1
4x + 2y = -1
Correct answer is C
Since A(x, y) is the point of equidistance between B and C, then
AB = AC
(AB)\(^2\) = (AC)\(^2\)
Using the distance formula,
(x - 0)\(^2\) + (y - 2)\(^2\) = (x - 2)\(^2\) + (y - 1)\(^2\)
x\(^2\) + y\(^2\) - 4y + 4 = x\(^2\) - 4x + 4 + y\(^2\) - 2y + 1
x\(^2\) - x\(^2\) + y\(^2\) - y\(^2\) + 4x - 4y + 2y = 5 - 4
4x - 2y = 1
N20,000
N28,000
N31,200
N41,000
Correct answer is A
Let the number of days worked by the assistant = t
∴∴ The bricklayer worked (t + 10) days.
1500(t + 10) + 500(t) = N 95,000
1500t + 15,000 + 500t = N 95,000
2000t = N 95,000 - N 15,000
2000t = N 80,000
t = 40 days
∴∴ The assistant worked for 40 days and received N (500 x 40)
= N 20,000
If \(25^{1 - x} \times 5^{x + 2} \div (\frac{1}{125})^{x} = 625^{-1}\), find the value of x.
x = -4
x = 2
x = -2
x = 4
Correct answer is A
\(25^{1 - x} \times 5^{x + 2} \div (\frac{1}{125})^{x} = 625^{-1}\)
\((5^2)^{(1 - x)} \times 5^{(x + 2)} \div (5^{-3})^x = (5^4)^{-1}\)
\(5^{2 - 2x} \times 5^{x + 2} \div 5^{-3x} = 5^{-4}\)
\(5^{(2 - 2x) + (x + 2) - (-3x)} = 5^{-4}\)
Equating bases, we have
\(2 - 2x + x + 2 + 3x = -4\)
\(4 + 2x = -4 \implies 2x = -4 - 4\)
\(2x = -8\)
\(x = -4\)