Find the inter-quartile range of 1, 3, 4, 5, 8, 9, 10, 11, 12, 14, 16
6
7
8
9
Correct answer is C
\(Q_1 = \frac{1}{4}\) (N + 1)th
\(\frac{1}{4} \times 12^{th}\) no.
= 3rd no (\(\cong\) 4)
\(Q_3 = \frac{3}{4}\) (N + 1)th
= \(\frac{3}{4}\) x 12th no.
= 9th no. (\(\cong\) 12)
Hence, interquartile range
= \(Q_3 - Q_1\)
= 12 - 4
= 8
If 3x\(^o\) 4(mod 5), find the least value of x
1
2
3
4
Correct answer is C
3x \(\equiv\) 4(mod 5)
In modulo 5, multiples of 5 that give solution to the given equation are 5, 20, 35, 50,... but 5 will yield the leaast value of x.
Thus; 3x = 4 + 5 = 9
x = \(\frac{9}{3}\)
x = 3
Find the \(n^{th}\) term of the sequence 2 x 3, 4 x 6, 8 x 9, 16 x 12...
2\(^n\) x 3(n + 1)
2\(^n\) x 3n
2\(^n\) x 3\(^n\)
2\(^n\) x 3\(^n - 1\)
Correct answer is B
2 x 3, 4 x 6, 8 x 9, 16 x 12,...
2\(^1\) x 3 x 1, 2\(^2\) x 3 x 2, 2\(^3\) x 3 x 3, 2\(^4\) x 3 x 4,.... 2\(^n\) x 3n
(3x + 2)(1 - x)
(3x + 2)(2x + 1)
3\((x + 2)^2\)
3(x + 1)(1 - x)
Correct answer is D
\((x + 2)^2\) - \((2x + 1)^2\)
= \((x^2 + 4x + 4) - (4x^2 + 4x + 1)\)
= \(x^2 \) + 4x + 4 - 4 \(x^2 \) - 4x - 1
= -3 \(x^2 \) + 3
= 3 - 3 \(x^2 \)
= 3(1 - \(x^2 \))
= 3(1 + x)(1 - x)
Solved the equation \(2x^2 - x - 6\) = 0
x = \(\frac{-3}{2}\) or 2
x = -2 or \(\frac{3}{2}\)
x = -3 or 2
x = 3 or -2
Correct answer is A
\(2x^2 - x - 6\) = 0
\(2x^2 - 4x + 3x - 6\) = 0
2x(x - 2) + 3(x - 2) = 0
(2x + 3) (x - 2) = 0
Either; 2x + 3 = 0 or x - 2 = 0
x = \(\frac{-3}{2}\) or x = 2