WAEC Mathematics Past Questions & Answers - Page 20

96.

Evaluate \((101_{two})^3\)

A.

\(111101_{two}\)

B.

\(11111101_{two}\)

C.

\(1111101_{two}\)

D.

\(11001_{two}\)

Correct answer is C

\((101_{two})^3\) = \((5_{ten})^3\)

5\(^3\) = 125

And 125 =  \(1111101_{two}\)

97.

For what value of x is \(\frac{4 - 2x}{x + 1}\) undefined.

A.

2

B.

-1

C.

1

D.

-2

Correct answer is B

A rational expression is undefined when the denominator is equal to zero.

when x = -1

The denominator in this equation : x + 1

--> -1 + 1 = 0

This expression is undefined when x = -1

98.

The equation of a line is given as 3 x - 5y = 7. Find its gradient (slope)

A.

\(\frac{5}{3}\).

B.

\(\frac{3}{5}\).

C.

\(\frac{-3}{5}\).

D.

\(\frac{-5}{3}\).

Correct answer is B

the form y=mx+c

where m is the gradient and c is the y-intercept.

the equation to gives 5y=3x - 7.

divide both side by 5 the coefficient of y → \(\frac{3}{5}\)x - \(\frac{7}{5}\) 

comparing this with the general equation y=mx+c,

you can see that m= the gradient= \(\frac{3}{5}\).

99.

A man will be (x+10)years old in 8years time. If 2years ago he was 63 years., find the value of x

A.

55

B.

63

C.

57

D.

67

Correct answer is B

A man will be (x+10) years old in 8years time.

 As at today, he is x + 2 years of age.
 

The man was 63 years old 2 years ago, so he is 63+2=65 now.


8 years from now, he will be 65+8=73.

He will be (x+10) years old when he is 73. So

x+10=73
x=73-10=63

100.

A fair die is tossed twice what is the probability of get a sum of at least 10.

A.

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

B.

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

C.

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

D.

\(\frac{1}{6}\)

Correct answer is D

\(\begin{array}{c|c}
& 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
1 & 1,1 & 1,2 & 1,3 & 1,4 & 1,5 & 1,6 \\ \hline 2 & 2,1 & 2,2 & 2,3 & 2,4 & 2,5 & 2,6 \\ \hline 3 & 3,1 & 3,2 & 3,3 & 3,4 & 3,5 & 3,6 \\ \hline 4 & 4,1 & 4,2 & 4,3 & 4,4 & 4,5 & 4,6 \\ 5 & 5,1 & 5,2 & 5,3 & 5,4 & 5,5 & 5,6 \\ \hline 6 & 6,1 & 6,2 & 6,3 & 6,4 & 6,5 & 6,6\end{array}\)

From the table above, event space, n(E) = 6

sample space, n(S) = 36

Hence, probability sum of scores is at least 10, is;

\(\frac{n(E)}{n(S)}\)

= \(\frac{6}{36}\)

= \(\frac{1}{6}\)