Further Mathematics Questions & Answers - Free Online Practice

376.

P, Q, R, S are points in a plane such that PQ = 8i - 5j, QR = 5i + 7j, RS = 7i + 3j and PS = xi + yj. Find (x, y).

(-6, -15)

(-6, 5)

(20, 5)

(20, 15)

View answer & explanation

Correct answer is C

No explanation has been provided for this answer.

377.

In a class of 50 pupils, 35 like Science and 30 like History. What is the probability of selecting a pupil who likes both Science and History?

0.10

0.30

0.60

0.70

View answer & explanation

Correct answer is B

Let the number of students that like both Science and History = x

Number of Science only = 35 - x

Number of History only = 30 - x

35 - x + 30 - x + x = 50

65 - x = 50

x = 15

P(Science and History) = $\frac{15}{50} = 0.30$

378.

The probabilities that a husband and wife will be alive in 15 years time are m and n respectively. Find the probability that only one of them will be alive at that time.

m + n

m + n - 2mn

1 - mn

View answer & explanation

Correct answer is C

P(only one alive) = P(husband and not wife) + P(wife and not husband)

= m (1 - n) + n ( 1 - m)

= m - mn + n - mn

= m + n - 2mn

379.

The distance s metres of a particle from a fixed point at time t seconds is given by $s = 7 + pt^{3} + t^{2}$ , where p is a constant. If the acceleration at t = 3 secs is $8 ms^{-2}$ , find the value of p.

$\frac{1}{3}$

$\frac{4}{9}$

$\frac{5}{9}$

$1$

View answer & explanation

Correct answer is A

Differentiate distance twice to get the acceleration and then equate to get p.

$s = 7 + pt^{3} + t^{2}$

$\frac{\mathrm d s}{\mathrm d t} = v(t) = 3pt^{2} + 2t$

$\frac{\mathrm d v}{\mathrm d t} = a(t) = 6pt + 2$

$a(3) = 6p(3) + 2 = 8 \implies 18p = 8 - 2 = 6$

$p = \frac{6}{18} = \frac{1}{3}$

380.

The mean age of 15 pupils in a class is 14.2 years. One new pupil joined the class and the mean changed to 14.1 years. Calculate the age of the new pupil.

12.4 years

12.6 years

13.2 years

14.1 years

View answer & explanation

Correct answer is B

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

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

$\frac{x}{15} = 14.2 \implies x = 14.2 \times 15 = 213$

$\frac{y}{16} = 14.1 \implies y = 14.1 \times 16 = 225.6$

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

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