Given that p:q = \(\frac{1}{3}\):\(\frac{1}{2}\) and q:r = \(\frac{2}{5}\), find p:r
4:105
7:15
20:21
2:35
3:20
Correct answer is B
\(p : q = \frac{1}{3} : \frac{1}{2}\)
\(\frac{p}{q} = \frac{2}{3}\)
\(2q = 3p ... (1)\)
\(q : r = \frac{2}{5} : \frac{4}{7}\)
\(\frac{q}{r} = \frac{2}{5} \times \frac{7}{4} = \frac{7}{10}\)
\(10q = 7r ... (2)\)
Eliminating q, we have
(1) : \(2q = 3p \)
\(10q = 15p\)
\(\implies 15p = 7r\)
\(\therefore \frac{p}{r} = \frac{7}{15}\)
\(p : r = 7 : 15\)
42cm2
3cm2
21cm2
24cm2
12cm2
Correct answer is C
Length of arc = \(\frac{\theta}{360}\) x 2\(\pi\)r = 6
\(\theta\) x 2\(\pi\)r = 360 x 6
\(\theta\) = \(\frac{360 \times 6}{2\pi r}\)
Area of the sector = \(\frac{\theta}{360}\) x \(\pi\)r2
\(\frac{360 \times 6}{2\pi r}\) x \(\frac{1}{360}\) x \(\pi\)r2 = r
= 3 x 7
= 21cm2
If f(x - 2) = 3x2 + 4x + 1. Find the area of the sector
8
40
7
24
32
Correct answer is B
f(x - 2) = 3x2 + 4x + 1
f(1) will be f(3 - 2)
When x = 3, f(1) = 3(3)2 + 4 x 3 = 1
27 + 12 +1 = 40
Find the median of the set of numbers 110, 116, 113, 119, 118, 127, 118, 117, 113
117.5
118
117
116
113
Correct answer is C
Re-arrange in ascending order: 110, 113, 113, 116, |117|, 118, 119, 127 a= 117
25%
11%
20%
80%
30%
Correct answer is C
Loss% = \(\frac{\text{actual loss}}{\text{cost price}}\) x 100%
\(\frac{N150 - N120}{N150} \times 100%\)
\(\frac{N30}{N150} \times 100%\)
= 20%