WAEC Past Questions and Answers - Page 9

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41.

P(3,4) and Q(-3, -4) are two points in a plane. Find the gradient of the line that is normal to the line PQ. 

A.

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

B.

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

C.

\(\frac{-3}{4}\)

D.

\(\frac{-4}{3}\)

Correct answer is B

Gradient = \(\frac{-4 - 4}{-3 - 3}\)

= \(\frac{-8}{6}\)

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

Normal = -(\(\frac{1}{\frac{1}{4}}\))

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

42.

The distance(s) in metres covered by a particle in motion at any time, t seconds, is given by S =120t - 16t\(^2\). Find in metres, the distance covered by the body before coming to rest.

A.

220

B.

222

C.

223

D.

225

Correct answer is D

s = 120t - 16t\(^2\) - 16t\(^2\)

\(\frac{ds}{dt} = 120 - 32t\)

120 - 32t = 0

\(\frac{32t}{32}\) = \(\frac{120}{32}\)

t = 3.75

s = 120(3.75) - 16(3.75)\(^2\)

= 450 - 225 

= 225

43.

A bag contains 5 red and 5 blue identical balls. Three balls are selected at random without replacement. Determine the probability of selecting balls alternating in color.

A.

\(\frac{7}{18}\)

B.

\(\frac{5}{18}\)

C.

\(\frac{5}{36}\)

D.

\(\frac{1}{36}\)

Correct answer is C

probability of selecting three balls with alternating colours = probability of selecting a blue ball (without replacement) x probability of selecting a red ball (without replacement) x probability of selecting a blue ball (without replacement)

In this case, we will be using the first instance (both would give the same answer since the number of blue and red balls are equal)

first: probability of selecting a red ball (without replacement) = \(\frac{5}{10}\) = \(\frac{1}{2}\)

second: the probability of selecting a blue ball (without replacement) = \(\frac{5}{9}\)

third: probability of selecting a red ball (without replacement) = \(\frac{4}{8}\) = \(\frac{1}{2}\)

Therefore, the probability of selecting three balls with alternating colours = \(\frac{1}{2}\) x \(\frac{5}{9}\) x \(\frac{1}{2}\)

= \(\frac{5}{36}\)

44.

The variables x and y are such that y =2x\(^3\) - 2x\(^2\) - 5x + 5. Calculate the corresponding change in y and x changes from 2.00 to 2.05.

A.

0.58

B.

0.95

C.

1.48

D.

1.95

Correct answer is A

When x = 2

y = 2(2)\(^3\) - 2(2)\(^2\) 

-5(2) + 5

y = 16 - 8 - 10 + 5

y = 3 

 

when x = 2.05 

y = 2(2.105)\(^3\) - 2(2.05)\(^2\)

- 5(2.05) + 5

y = 17.23 - 8.405 - 10.25 + 5

y = 3.575 - 3

Difference = 3.575 - 3

= 0.58