25
21
16
3
Correct answer is B
Number of people playing both rugby and cricket = 4 Number that play cricket only = 11 - 4 = 7 Number that play rugby only = 14 - 4 = 10. Number that play for at least one of the teams = 4 + 7 + 10 = 21.
Simplify \(\frac{\sqrt{12} - \sqrt{3}}{\sqrt{12} + \sqrt{3}}\)
\(\frac{1}{3}\)
9
16cm
3
Correct answer is A
\(\frac{\sqrt{12} - \sqrt{3}}{\sqrt{12} + \sqrt{3}}\)
\(\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}\)
\(\therefore \frac{\sqrt{12} - \sqrt{3}}{\sqrt{12} + \sqrt{3}} = \frac{2\sqrt{3} - \sqrt{3}}{2\sqrt{3} + \sqrt{3}}\)
= \(\frac{\sqrt{3}}{3\sqrt{3}}\)
= \(\frac{1}{3}\)
Evaluate \(\frac{(81)^{\frac{3}{4}} - (27)^{\frac{1}{3}}}{3 \times 2^3}\)
27
1
\(\frac{1}{3}\)
\(\frac{1}{8}\)
Correct answer is B
\(\frac{(81)^{\frac{3}{4}} - (27)^{\frac{1}{3}}}{3 \times 2^3}\)
= \(\frac{(3^{4})^{\frac{3}{4}} - (3^{3})^{\frac{1}{3}}}{3 \times 2^{3}}\)
= \(\frac{3^{3} - 3^{1}}{24}\)
= \(\frac{27 - 3}{24} = 1\)
Express the product of 0.0014 and 0.011 in standard form
1.54 x 10-2
1.54 x 10-3
1.54 x 10-4
1.54 x 10-5
Correct answer is D
\(0.0014 = 1.4 \times 10^{-3}\)
\(0.011 = 1.1 \times 10^{-2}\)
\(\therefore 0.0014 \times 0.011 = 1.4 \times 10^{-3} \times 1.1 \times 10^{-2}\)
= \(1.54 \times 10^{-5}\)
N28,800
N29,040
N31,200
N31,944
Correct answer is B
Present salary = N24,000.
Increment after first year = \(\frac{10}{100} \times N24,000 = N2,400\)
Salary at the beginning of second year = N(24,000 + 2,400)
= N26,400.
Increment after second year = \(\frac{10}{100} \times N26,400 = N2,640\)
Salary at the beginning of third year = N(26,400 + 2,640)
= N29,040.
\(\therefore\) His salary at the beginning of the third year = N29,040.