\(\frac{3}{32}\)
\(\frac{7}{3}\)
\(\frac{5}{3}\)
\(\frac{5}{16}\)
Correct answer is D
Pr. (winning 100m race) = \(\frac{1}{8}\)
Pr. (losing 100m race) = 1 - \(\frac{1}{8}\) = \(\frac{7}{8}\)
Pr. (winning high jump) = \(\frac{1}{4}\)
Pr. (losing high jump ) = 1 - \(\frac{1}{4}\) = \(\frac{3}{4}\)
Pr. (winning only one) = Pr. (Winning 100m race and losing high jump) or Pr.(Losing 100m race and winning high jump)
= (\(\frac{1}{8} \times \frac{3}{4}\)) + (\(\frac{7}{8} \times \frac{1}{4}\))
= \(\frac{3}{32} + \frac{7}{32}\)
= \(\frac{10}{32}\)
= \(\frac{5}{16}\)
The mean of the numbers 2, 5, 2x and 7 is less than or equal to 5. Find the range of the values of x
x \(\leq\) 3
x \(\geq\) 3
x < 3
x > 3
Correct answer is A
mean \(\leq\) 5; \(\frac{2 + 5 + 2x + 7}{4}\) \(\leq\) 5
= \(\frac{14 + 2x}{4} \leq 5\)
= 14 + 2x \(\leq\) 5 x 4
14 + 2x \(\leq\) 20 ; 2x \(\leq\) 20 - 14
2x \(\leq\) 20 - 14
2x \(\leq\) 6
x \(\leq\) \(\frac{6}{2}\)
x \(\leq\) 3
\(\frac{7}{18}\)x
\(\frac{11}{20}\)x
\(\frac{4}{15}\)x
\(\frac{5}{18}\)x
Correct answer is D
x kmh-1 = y ms-1
\(\frac{x km}{1 hr}\) = y ms-1
\(x \times \frac{1km}{1hr}\) = y ms-1
\(x \times \frac{1000m}{60 \times 60s}\) = y ms-1
\(x \times \frac{1000}{3600} \frac{m}{s}\) = y ms-1
\(x \times \frac{5}{18} ms^{-1}\)
\(x \times \frac{5}{18} ms^{-1}\) = y ms-1
y = \(\frac{5}{18}\)x
If x2 + kx + \(\frac{16}{9}\) is a perfect square, find the value of k
\(\frac{8}{3}\)
\(\frac{7}{3}\)
\(\frac{5}{3}\)
\(\frac{2}{3}\)
Correct answer is A
x2 + kx + \(\frac{16}{9}\); Perfect square
But, b2 - 4ac = 0, for a perfect square
where a - 1; b = k; c = \(\frac{16}{9}\)
k2 - 4(1) x \(\frac{16}{9}\) = 0
k2 - \(\frac{64}{9}\) = 0
k2 = \(\frac{64}{9}\)
k = \(\sqrt{\frac{64}{9}}\)
k = \(\frac{8}{3}\)
If the sum of the roots of the equation (x - p)(2x + 1) = 0 is 1, find the value of x
1\(\frac{1}{2}\)
\(\frac{1}{2}\)
-\(\frac{3}{2}\)
-1\(\frac{1}{2}\)
Correct answer is A
(x - p)(2x + 1) = 0
2x2 + x - 2px - p = 0
2x2 + x (1 - 2p) - p = 0
2x2 - (2p - 1)x - p = 0
divide through by 2
x2 - \(\frac{(2p - 1)}{2}\)x - \(\frac{p}{2}\) = 0
compare to x2 - (sum of roots)x + product of roots = 0
sum of roots = \(\frac{2p - 1}{2}\)
But sum of roots = 1
Given; \(\frac{2p - 1}{2}\) = 1
2p - 1 = 2 x 1
2p - 1 = 2
2p = 2 + 1 = 3
p = \(\frac{3}{2}\)
p = 1\(\frac{1}{2}\)