Correct Answer: A. 8

Explanation

when y = 5; x = ?; y = \(\frac{1}{2}\)x + 1

5 = \(\frac{1}{2}\)x + 1

5 - 1 = \(\frac{1}{2}\)x

4 = \(\frac{1}{2}\)x

x = 4 x 2

x = 8

Correct Answer: D. 66\(\frac{2}{3}\)%

Explanation

Percentage of students with marks ranging from 35 to 50 = \(\frac{f_{35} + f{40} + f{50}}{\sum f}\)

= \(\frac{35 + 40 + 45}{20 + 35 + 40 + 45 + 25 + 15}\) x 100%

= \(\frac{120}{180}\) x 100%

= 66\(\frac{2}{3}\)%

Correct Answer: A. 2

Explanation

5 1 6 2_seven

-2 6 4 4_seven

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2 2 1 5

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the missing number is 2

Correct Answer: D. \(1\frac{1}{2}\)

Explanation

\(\int \frac{4}{x^{3}} \mathrm {d} x = \int 4x^{-3} \mathrm {d} x\)

\(\frac{4x^{-3 + 1}}{-2} = -2x^{-2}|_{1}^{2} = \frac{-2}{x^{2}}|_{1}^{2}\)

= \(\frac{-2}{2^{2}} - \frac{-2}{1^{2}} = -\frac{1}{2} + 2 = 1\frac{1}{2}\)

Correct Answer: D. \(\frac{5}{16}\)

Explanation

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)

= (\(\frac{1}{8} \times \frac{3}{4}\)) + (\(\frac{7}{8} \times \frac{1}{4}\))

= \(\frac{3}{32} + \frac{7}{32}\)

= \(\frac{10}{32}\)

= \(\frac{5}{16}\)

Correct Answer: D. 2\(\frac{3}{4}\)

Explanation

1\(\frac{1}{2} + 2\frac{1}{3} \times \frac{3}{4} - \frac{1}{2}\)

\(\frac{3}{2} + \frac{7}{3} \times \frac{3}{4} - \frac{1}{2} = \frac{3}{2} + \frac{7}{4} - \frac{1}{2}\)

= \(\frac{6 + 7 - 2}{4} = \frac{11}{4}\)

= 2\(\frac{3}{4}\)

Correct Answer: A. 1\(\frac{1}{2}\)

Explanation

(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\(\frac{1}{2}\)

Correct Answer: B. 12.6 years

Explanation

\(Mean = \frac{\text{sum of ages}}{\text{no of pupils}}\)

Let the sum of the ages of the 15 pupils be x and the sum of the 16 pupils be y.

\(\frac{x}{15} = 14.2 \implies x = 14.2 \times 15 = 213\)

\(\frac{y}{16} = 14.1 \implies y = 14.1 \times 16 = 225.6\)

\(\text{Age of new pupil} = 225.6 - 213 = 12.6\)

Correct Answer: C. \(2a^{3} - \frac{2}{a^{3}}\)

Explanation

\(h(x) = x^{3} - \frac{1}{x^{3}}\)

\(h(a) = a^{3} - \frac{1}{a^{3}}\)

\(h(\frac{1}{a}) = (\frac{1}{a})^{3} - \frac{1}{(\frac{1}{a})^{3}} = \frac{1}{a^{3}} - a^{3}\)

\(h(a) - h(\frac{1}{a}) = (a^{3} - \frac{1}{a^{3}}) - (\frac{1}{a^{3}} - a^{3}) = 2a^{3} - \frac{2}{a^{3}}\)

Explanation

\(r = (4t^{3} - t^{2})i - (2t^{2} - t)j\)

\(v = \frac{\mathrm d r}{\mathrm d t} = (12t^{2} - 2t)i - (4t - 1)j\)

\(a = \frac{\mathrm d v}{\mathrm d t} = (24t - 2)i - 4j\)

\(t = 2 : a = (24(2) - 2)i - 4j\)

\(a = 46i - 4j\)

\(F = ma = 3(46i - 4j)\)

= \(138i - 12j)N\)