The ages of students in a small primary school were recorded in the table below.
| Age | 5 - 6 | 7 - 8 | 9 -10 |
| Frequency | 29 | 40 | 38 |
Estimate the mean
7.7
7.5
7.8
7.6
Correct answer is A
| Class Interval | Class Mark | Frequency (f) | fx |
| 5 - 6 | 5.5 | 29 | 5.5 x 29 = 159.5 |
| 7 - 8 | 7.5 | 40 | 7.5 x 40 = 300 |
| 9 - 10 | 9.5 | 38 | 9.5 x 38 = 361 |
| \(\sum f = 107\) | \(\sum fx = 820.5\) |
Mean = \(\frac{\sum fx}{\sum f} = \frac {820.5}{107}\) = 7.7 (1 d.p)
Evaluate the following limit: \(lim_{x\to2} \frac {x^2 + 4x - 12}{x^2 - 2x}\)
4
8
0
2
Correct answer is A
\(lim_{x\to2} \frac {x^2 + 4x - 12}{x^2 - 2x}\) = \(lim_{x\to2} \frac {(x - 2)(x + 6)}{x(x - 2)}\)
\(lim_{x\to2} \frac {x + 6}{x}\)
\(\frac {2 + 6}{2} = \frac {8}{2}\) = 4
15 years
25 years
20 years
30 years
Correct answer is C
Amount (A) = Principal (P) + Interest (I) i.e. A = P + I
1 = \(\frac {PTR}{100}\)
A = 3P; T = 10 years (Given)
Since A = P + I
\(\implies 3P = P + \frac {P\times 10 \times R}{100}\)
\(\implies 3P = P + \frac {10PR}{100}\)
\(\implies 3P = P + \frac {PR}{10}\)
\(\implies 3P - P = \frac {PR}{10}\)
\(\implies 2P = \frac {PR}{10}\)
\(\implies \frac {2P}{1} = \frac {PR}{10}\)
\(\implies\) PR = 20P
\(\therefore\) R = 20%
Since the rate stays the same
A = 5P; R = 20%;T =?; A =P + I
\(\implies 5P = P + \frac {p \times T \times 20}{100}\)
\(\implies 5P - P = \frac {2PT}{10}\)
\(\implies 4P = \frac {2PT}{10}\)
\(\implies 2P = \frac {PT}{10}\)
\(\implies\) PT = 20P
\(\therefore\) T = 20 years
Find the value of t, if the distance between the points P(–3, –14) and Q(t, –5) is 9 units.
3
2
-3
-2
Correct answer is C
Let the given points be:
P(-3, -14) = (x\(_1\), y\(_1\))
Q(t, -5) = (x\(_2\), y\(_2\))
PQ = 9 units (given)
Using the distance formula,
d = √ [ \((x_2 - x_1)^2 + (y_2 - y_1)^2\)]
PQ = √ [ \((t - (-3))^2 + (-5 + 14)^2\)]
\(\implies\) √ [ \((t + 3)^2 + 81\)] = 9
Squaring on both sides,
⇒ (t + 3)\(^2\) + 81 = 81
⇒ (t + 3)\(^2\) = 0
⇒ t + 3 = 0
∴ t = -3
If \(\frac {3 - \sqrt 3}{2 + \sqrt 3} = a + b\sqrt 3\), what are the values a and b?
a = 9, b = -5
a = 5, b = 9
a = 9, b = 5
a = -5, b = 9
Correct answer is A
\(\frac {3 - \sqrt 3}{2 + \sqrt 3} = a + b\sqrt 3\)
Rationalize
= \(\frac {3 - \sqrt 3}{2 + \sqrt 3} \times \frac {2 - \sqrt 3}{2 - \sqrt 3}\)
= \(\frac {(3 - \sqrt 3)}{(2 + \sqrt 3)} \frac {(2 - \sqrt 3)}{(2 - \sqrt 3)}\)
= \(\frac {6 - 3 \sqrt 3 - 2 \sqrt 3 + (\sqrt 3)^2}{4 - 2 \sqrt 3 + 2 \sqrt 3 - (\sqrt 3)^2}\)
= \(\frac {6 - 5 \sqrt 3 + 3}{4 - 3}\)
= \(\frac {9 - 5 \sqrt 3}{1} = 9 - 5 \sqrt 3\)
= 9 + (-5) \(\sqrt 3\)
\(\therefore a = 9, b = - 5\)