WAEC Mathematics Past Questions & Answers - Page 194

966.

Evaluate Cos 45o Cos 30o - Sin 45o Sin 30o leaving the answer in surd form

A.

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

B.

\(\frac{\sqrt{3}-\sqrt{2}}{4}\)

C.

\(\frac{\sqrt{6}-\sqrt{2}}{2}\)

D.

\(\frac{\sqrt{6}-\sqrt{2}}{4}\)

Correct answer is D

\(cos45^o \times cos30^o - sin45^o \times sin30^o\\
\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}}\times \frac{1}{2}\\
\frac{\sqrt{3}}{2\sqrt{2}}-\frac{1}{2\sqrt{2}}; = \frac{\sqrt{3}-1}{2\sqrt{2}}=\frac{\sqrt{6}-\sqrt{2}}{4}\)

967.

Find the equation whose roots are \(-\frac{2}{3}\) and 3

A.

3x2+11x-6=0

B.

3x2+7x+6=0

C.

3x2-11x-6=0

D.

3x2-7x-6=0

Correct answer is D

\(x = -\frac{2}{3} \implies x + \frac{2}{3} = 0\)

\(x = 3 \implies x - 3 = 0\)

\(\implies (x - 3)(x + \frac{2}{3}) = 0\)

\(x^2 - 3x + \frac{2}{3}x - 2 = 0\)

\(x^2 - \frac{7}{3}x - 2 = 0\)

\(3x^2 - 7x - 6 = 0\)

968.

The locus of points equidistant from two intersecting straight lines PQ and PR is

A.

a circle centre P radius Q.

B.

a circle centre P radius PR

C.

the point of intersection of the perpendicular bisectors of PQ and PR

D.

the bisector of angle QPR

Correct answer is C

No explanation has been provided for this answer.

969.

The probabilities of a boy passing English and Mathematics test are x and y respectively. Find the probability of the boy failing both tests

A.

1-(x-y)+xy

B.

1-(x+y)-xy

C.

1-(x+y)+xy

D.

1 - (x - y) + x

Correct answer is C

Prob (passing English) = x
Prob (passing Maths) = Y
Prob (failing English) = 1 - x
Prob (failing Maths) = 1 - y
Prob (failing both test) = Prob(failing English) and Prob(failing Maths) = (1 - x)(1 - y)
=1 - y - x + xy
=1 - (y + x) + xy

970.

Given that p varies as the square of q and q varies inversely as the square root of r. How does p vary with r?

A.

p varies as the square of r

B.

p varies as the square root of r

C.

p varies inversely as the square of r

D.

p varies inversely as r

Correct answer is D

\(p \propto q^2\)

\(q\propto\frac{1}{\sqrt{r}}\)

\(p = kq^2\)

\(q = \frac{c}{\sqrt{r}}\)

where c and k are constants.

\(q^2 = \frac{c^2}{r}\)

\(p = \frac{kc^2}{r}\)

If k and c are constants, then kc\(^2\) is also a constant, say z.

\(p = \frac{z}{r}\)

p varies inversely as r.