Simplify \(\frac{2.25}{0.015}\) leaving your answer in standard form
1.5 x 10-4
1.5 x 10-2
1.5 x 10-3
1.5 x 101
1.5 x 102
Correct answer is E
\(\frac{2.25}{0.015}\)
= \(\frac{2250}{15}\)
= \(150\)
= \(1.5 \times 10^{2}\)
In the diagram above, QRS is a straight line, QP//RT, PRQ = 56°, ∠QPR =84° and ∠TRS = x°. Find x
28°
40°
44°
84°
124°
Correct answer is B
< PQR = 180° - (84° + 56°)
= 40°
\(\therefore\) x = 40° (corresponding angles QP// RT)
Two chords PQ and RS of a circle when produced meet at K. If ∠KPS = 31o and ∠PKR = 42o, find ∠KQR
11o
73o
107o
138o
149o
Correct answer is C
No explanation has been provided for this answer.
In the diagram above, |PQ| = |QR|, |PS| = |RS|, ∠PSR = 30o and ∠PQR = 80o. Find ∠SPQ.
15o
25o
40o
50o
55o
Correct answer is B
Join PR
QRP = QPR
= 180 - 80 = 100/20 = 50o
SRP = SPR
= 180 - 30 = 150/2 = 75o
∴ SPQ = SPR - QPR
= 75 - 50 = 25o
3.3cm
4.4cm
5cm
5.5cm
6.6cm
Correct answer is B
\(\frac{PQ}{RQ} = \frac{RQ}{SQ}\) (Similar triangles)
\(\therefore \frac{PS + 10}{12} = \frac{12}{10}\)
\(10 (PS + 10) = 12 \times 12\)
\(10 PS + 100 = 144 \implies 10 PS = 44\)
\(PS = 4.4 cm\)
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