Find at which rate per annum simple interest N525 will amount to N588 in 3 years.
3%
2%
5%
4%
Correct answer is D
I = A − P
= N588 − N525
∴I = N63
I = PRT ÷ 100
R = [100I ÷PT]
R = [(100 × 63 ) ÷ (525 × 3)]
= (6300 ÷ 1575) = 4
∴ The rate = 4 %
Make x the subject of the equation
s = 2 + \(\frac{t}{5} \)(x + ⅗y)
x = 5[(s − 2) ÷ t] - 3/5y
x = 25[(s − 2) ÷ t] − 3ty
x = [1 ÷ (s − 2)3ty]
x = [5(s − 2) 2 ÷ 3ty] × t
Correct answer is A
No explanation has been provided for this answer.
N132.50K
N136.30K
N125.40K
N257.42K
Correct answer is B
\( A = P \left(1 + \frac{r}{100}\right)^n \)
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
Find the equation of a line which is form origin and passes through the point (−3, −4)
y = \( \frac{3x}{4} \)
y = \( \frac{4x}{3} \)
y = \( \frac{2x}{3} \)
y = \( \frac{x}{2} \)
Correct answer is B
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 \(y = \frac{4}{3} x\)
If x + y = 90 simplify \((sinx + siny)^2\)−2sinxsiny
1
0
2
-1
Correct answer is A
Given: \(x + y = 90° ... (1)\)
\((\sin x + \sin y)^{2} - 2\sin x \sin y = \sin^{2} x + \sin^{2} y + 2\sin x \sin y - 2\sin x \sin y\)
= \(\sin^{2} x + \sin^{2} y ... (2)\)
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