7\(\frac{1}{2}\) years
10 years
5 years
12 years
14 years
Correct answer is B
\(A = P(1 + \frac{R}{100})^{T}\)
\(285.20 = P(1 + \frac{5}{100})^{3}\)
\(285.20 = P(1.05)^{3} \implies 1.16P = 285.20\)
\(P = \frac{285.20}{1.16} = $245.86\)
Given Amount = $434.00
Interest = Amount - Principal
Interest = $434.00 - $245.86 = $188.14
\(T = \frac{100I}{PR}\)
\(T = \frac{100 \times 188.14}{245.86 \times 7.5}\)
\(T = \frac{18814}{1843.95} = 10.2\)
Approximately 10 years.
13years
12 years 7\(\frac{1}{2}\) months
13 years 5 months
13 years 10 months
11 years
Correct answer is D
Total age of the 7 pupils = 7 x 12 = 84
Total age of the 25 pupils = 25 x 14 = 350
Total age of the 6 pupils = 6 x 11 = 66
7 pupils leaving a class of 25 pupils given 18 pupils
Total age of the 18 pupils = 350 - 84 = 266
6 pupils joining 18 pupils = 24
Total age of 24 pupils = 266 + 66 = 332
Average age of 24 pupils now in class \(\frac{332}{24}\)
= 13yrs 10 months
If \(\sin x° = \frac{a}{b}\), what is \(\sin (90 - x)°\)?
\(\frac{\sqrt{b^2 - a^2}}{b}\)
1\(\frac{-a}{b}\)
\(\frac{b^2 - a^2}{b}\)
\(\frac{a^2 - b^2}{b}\)
\(\sqrt{b^2 - a^3}\)
Correct answer is A
\(\sin x = \frac{a}{b}\)
\(\sin^{2} x + \cos^{2} x = 1\)
\(\sin^{2} x = \frac{a^{2}}{b^{2}}\)
\(\cos^{2} x = 1 - \frac{a^{2}}{b^{2}} = \frac{b^{2} - a^{2}}{b^{2}}\)
\(\therefore \cos x = \frac{\sqrt{b^{2} - a^{2}}}{b}\)
\(\sin (90 - x) = \sin 90 \cos x - \cos 90 \sin x\)
= \((1 \times \frac{\sqrt{b^{2} - a^{2}}}{b}) - (0 \times \frac{a}{b})\)
= \(\frac{\sqrt{b^{2} - a^{2}}}{b}\)
6
5
8
7
10
Correct answer is D
| x | 5 | 6 | 7 | 8 | 9 |
| f | 6 | 4 | 8 | 7 | 5 |
Mode is the No having the highest frequency = 7
6
5
8
7
10
Correct answer is B
Add all the No.s together to give 100
\(\sum\)xf = 100
N = 20
x = \(\frac{\sum xf}{N}\)
= \(\frac{100}{20}\)
= 5