JAMB Mathematics Past Questions & Answers - Page 302

1,506.

Evaluate [\(\frac{1}{0.03}\) \(\div\) \(\frac{1}{0.024}\)]-1 correct to 2 decimal places

A.

3.76

B.

1.25

C.

0.94

D.

0.75

Correct answer is B

[\(\frac{1}{0.03}\) + \(\frac{1}{0.024}\)]

= [\(\frac{1}{0.03 \times 0.024}\)]-1

= [\(\frac{0.024}{0.003}\)]-1

= \(\frac{0.03}{0.024}\)

= \(\frac{30}{24}\) = 1.25

1,507.

If 10112 + x7 = 2510, solve for X.

A.

207

B.

14

C.

20

D.

24

Correct answer is A

10112 + x7 = 2510 = 10112 = 1 x 23 + 0 x 22 + 1 x 21 + 1 x 2o

= 8 + 0 + 2 + 1

= 1110

x7 = 2510 - 1110

= 1410

\(\begin{array}{c|c}
7 & 14 \\ 7 & 2 R 0 \\ & 0 R 2
\end{array}\)

X = 207

1,508.

Find the roots of x\(^3\) - 2x\(^2\) - 5x + 6 = 0

A.

1, -2, 3

B.

1, 2, -3,

C.

-1, -2, 3

D.

-1, 2, -3

Correct answer is A

Equation: x\(^3\) - 2x\(^2\) - 5x + 6 = 0.

First, bring out a\(_n\) which is the coefficient of x\(^3\) = 1.

Then, a\(_0\) which is the coefficient void of x = 6.

The factors of a\(_n\) = 1; The factors of a\(_0\) = 1, 2, 3 and 6.

The numbers to test for the roots are \(\pm (\frac{a_0}{a_n})\).

= \(\pm (1, 2, 3, 6)\).

Test for +1: 1\(^3\) - 2(1\(^2\)) - 5(1) + 6 = 1 - 2 - 5 + 6 = 0.

Therefore x = 1 is a root of the equation.

Using long division method, \(\frac{x^3 - 2x^2 - 5x + 6}{x - 1}\) = x\(^2\) - x - 6.

x\(^2\) - x - 6 = (x - 3)(x + 2).

x = -2, 3.

\(\therefore\) The roots of the equation = 1, -2 and 3.

1,509.

If x = \(\begin{pmatrix} 1 & 0 & 1 \\ 2 & -1 & 0 \\ -1 & 0 & 1\end{pmatrix}\) and y = \(\begin{pmatrix} -1 & 1 & 2 \\ 0 & -1 & -1 \\ 2 & -1 & 1\end{pmatrix}\)
find 2x - y

A.

\(\begin{pmatrix} 3 & -1 & 0 \\ 4 & -3 & -1 \\ -4 & 1 & 1\end{pmatrix}\)

B.

\(\begin{pmatrix} 3 & -1 & 0 \\ 4 & -1 & 1 \\ -4 & 1 & 1\end{pmatrix}\)

C.

\(\begin{pmatrix} 3 & -1 & 0 \\ 4 & -3 & 1 \\ -4 & 1 & 1\end{pmatrix}\)

D.

\(\begin{pmatrix} 3 & -1 & 0 \\ 4 & 1 & 1 \\ -4 & -1 & 1\end{pmatrix}\)

Correct answer is B

2(1) - (-1) = 3      2(2) - (0) = 4          2(-1) - (2) = -4

2(0) - (1) = -1      2(-1) - (-1) = -1      2(0) - (-1) = 1

2(1) - 2 = 0         2(0) - (-1) = 1         2(1) - (1) = 1

\(\begin{pmatrix} 3 & -1 & 0 \\ 4 & -1 & 1 \\ -4 & 1 & 1\end{pmatrix}\)

1,510.

Find p, q for which \(\begin{pmatrix} 2p & 8 \\ 3 & -5q \end{pmatrix}\)\(\begin{pmatrix} 1 \\ 2\end{pmatrix}\) = \(\begin{pmatrix}24 \\ -17\end{pmatrix}\)

A.

-4

B.

45

C.

4,2

D.

2

Correct answer is C

2p x 1 + 8 x 2 = 24

\(\to\) 4p = 24 - 8 = 16,

p = 4

3 x 1 + -5q x 2 = -17

\(\to\) -10q = -17 - 3

-10q = -20

q = 2