JAMB Mathematics Past Questions & Answers - Page 175

871.

The difference between the length and width of a rectangle is 6cm and the area is 135cm². What is the length?

A.

25cm

B.

18cm

C.

15cm

D.

24cm

E.

27cm

Correct answer is C

Area = L x B

L - B = 6

L = 6 + B

area = 135

B(6 + B) = B2 + 6B - 135 = 0

B = 9

L = 6 + 9

= 15

872.

The positive root of t in the following equation, 4t2 + 7t - 1 = 0, correct to 4 places of decimal, is

A.

1.0622

B.

10.6225

C.

0.1328

D.

0.3218

E.

2.0132

Correct answer is C

\(4t^{2} + 7t - 1 = 0\)

Using a = 4, b = 7, c = -1.

\(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\)

= \(\frac{-7 \pm \sqrt{7^{2} - 4(4)(-1)}}{2(4)}\)

= \(\frac{-7 \pm \sqrt{49 + 16}}{8}\)

= \(\frac{-7 \pm \sqrt{65}}{8}\)

= \(\frac{-7 \pm 8.0623}{8}\)

The positive answer = \(\frac{-7 + 8.0623}{8} = \frac{1.0623}{8}\)

\(\approxeq 0.1328\) (4 decimal place)

873.

Evaluate \(\frac{6.3 \times 10^5}{8.1 \times 10^3}\) to 3 significant fiqures

A.

77.80

B.

778.0

C.

7.870

D.

8.770

E.

88.70

Correct answer is A

\(\frac{6.3 \times 10^5}{8.1 \times 10^3}\)

\(\frac{7}{9} \times 10^{2}\)

= \(0.77778 \times 10^{2}\)

= \(77.778 \approxeq 77.80\)

874.

The number 25 when converted from the tens and units base to the binary base (base two) is one of the following

A.

10011

B.

111011

C.

111000

D.

11001

E.

110011

Correct answer is D

\(\begin{array}{c|c} 2 & 25\\2 & 12 R 1\\2 & 6 R 0\\2 & 3 R 0\\2 & 1 R 1\end{array}\)

= (11001)

875.

Simplify \(\frac{6^{2n + 1} \times 9^n \times 4^{2n}}{18^n \times 2^n \times 12^{2n}}\)

A.

32n

B.

3 x 23n - 1

C.

2n

D.

6

E.

1

Correct answer is D

\(\frac{6^{2n + 1} \times 9^n \times 4^{2n}}{18^n \times 2^n \times 12^{2n}}\) = \(\frac{(2 \times 3^{2n + 1} \times 3^{2n}) \times 2^{4n}}{(2 \times 9)^n \times 2^n \times (6 \times 2^{2n})}\)

= \(\frac{(2^{2n + 1} \times 3^{2n + 1}) \times 3^{2n} \times 2^{4n}}{2^n \times 3^{2n} \times 2^n \times 2^{4n} \times 3^{2n}}\)

= \(\frac{2^{2n} + 1 + 4^n \times 3^{2n} + 1 + 2^n}{2^{n + n + 4n} \times 3^{2n + 4n} \times 3^{2n + 2n}}\)

= \(\frac{2^{6n + 1} \times 3^{4n + 1}}{2^{6n} x 3^{4n}}\)

= 26n + 1 - 6n x 34n + 1 - 4n

2 x 3 = 6