If 23x = 325, find the value of x
7
6
5
4
Correct answer is A
23x = 325
2 \(\times x^1 + 3 \times x^0 = 3 \times 5^1 + 2 \times 5^0\)
= 2x + 3 = 15 + 2
2x + 3 = 17
2x = 17 - 3
2x = 14
x = \(\frac{14}{2}\)
x = 7
Simplify 10\(\frac{2}{5} - 6 \frac{2}{3} + 3\)
6\(\frac{4}{15}\)
6\(\frac{11}{15}\)
7\(\frac{4}{15}\)
7\(\frac{11}{15}\)
Correct answer is B
10\(\frac{2}{5} - 6 \frac{2}{3} + 3\)
\(\frac{52}{5} - \frac{20}{3} + \frac{3}{1}\)
= \(\frac{156 - 100 + 45}{15}\)
\(\frac{156 + 45 - 100}{15}\)
= \(\frac{201 - 100}{15}\)
= \(\frac{101}{15}\)
= 6\(\frac{11}{15}\)
Using the venn diagram, find n(x \(\cap\) y1)
2
3
4
6
Correct answer is A
(x \(\cap\) y2) = [a, b]
n(x \(\cap\) y2) = 2
In the diagram, ST//PQ reflex angle SRQ = 198o and < RQp = 72o. Find the value of y
18o
54o
87o
92o
Correct answer is B
R +198 = 360(angles at a point)
R = 360 - 198 = 162o
C = 72(alternate angle), b + c = 180(angles or straight line)
b = 190 - c = 180 - 72 = 108
a + b = R = 162
a = R - b = 162 - 108
= 54, but a = y(alternate angles)
y = 54o
In the diagram, PQ is a straight line. Calculate the value of the angle labelled 2y
130o
120o
110o
100o
Correct answer is D
Sum of angles on straight line = 180o
80o + 50o + x + 10o = 180o
140o + x = 180o
x = 180o - 140o
x = 40o
total sum of all angles = 360o
2yo + 2xo + 80o + 50o + x + 10o = 360o
2yo + 2(40)o + 80o + 50o + 40 + 10 = 360o
2yo + 80 + 180o = 360o
2yo + 260o = 360o
2yo = 360o - 260o
2yo = 100o
angle 2yo = 100o
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