Find the total surface area of a cylinder of base radius 5cm and length 7cm ( π = 3.14)
17.8cm2
15.8cm2
75.4cm2
54.7cm2
\(377.0cm^{2}\)
Correct answer is E
The total surface area of a cylinder = 2πrl + 2πr2
= 2πr(l + r)
= 2 × 3.14 x 5(7+5)
2 × 3.14 × 12 x 5
= 377.1cm (1DP)
X and Y are two sets such that n(X) = 15, n(Y) = 12 and n{X ∩ Y} = 7. Find ∩{X ∪ Y}
21
225
15
20
Correct answer is D
n(X ∪ Y) = n(X) + n(Y) − n(X ∩ Y) = 15 + 12 − 7 ∴ n(X ∪ Y) = 20
(x − 5) 2
(x + 5)(x + 4)
(x – 5)(x + 3)
(x + 3) 2
Correct answer is B
\( x^2 + 9x + 20 \)
Find the two numbers whose product is 20 and its sum is 9}
\( 5x \times 4x = 20x^2\)
\(5x + 4x = 9x\)
\((x^2 + 5x) + (4x + 20)\)
\(
x(x + 5)+ 4(x + 5)
= (x + 5)(x + 4)\)
1 \( \frac{3}{8}\)
2 \( \frac{3}{4}\)
4 \( \frac{3}{8}\)
3 \( \frac{1}{5}\)
Correct answer is C
1¼ ÷ [ 2 ÷ ¼] of 28
Apply BODMAS rules
\( \frac{5}{4}\) ÷ [ 2 ÷ ¼ × 28 ]
\( \frac{5}{4} ÷ \frac{2}{7} \)
\( \frac{5}{4} × \frac{7}{2} \)
\( \frac{35}{8}\)
= \(4 \frac{3}{8}\)
43o
47o
54o
86o
Correct answer is A
Construction: draw a line from Q to point P and another line from S to point P.
< SOQ = 2< QPS (< at centre is twice < on the circumference)
< QPS \(\frac{86}{2} = 43\)
< SQR = < QPS ( < between a chord and tangent = < in the alternate segment)
< SQR = 43o
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