Year : 
1984
Title : 
Mathematics (Core)
Exam : 
JAMB Exam

Paper 1 | Objectives

11 - 20 of 45 Questions

# Question Ans
11.

A trader in country where their currency 'MONI'(M) is in base five bought 1035 oranges at M145 each. If he sold the oranges at M245 each, what will be his gain?

A. 1035

B. 10305

C. 1025

D. 20025

E. 30325

Detailed Solution

Total cost of 1035 oranges at N145 each

= 1035 x 145

= 20025

Total selling price at N245 each

= (103)5 x 245

= 30325

Hence his gain = 30325 - 20025

= 10305
12.

Rationalize \(\frac{5\sqrt{7} - 7\sqrt{5}}{\sqrt{7} - \sqrt{5}}\)

A. -2 \sqrt{35}\)

B. 4\sqrt{7}\) - 6\(\sqrt{5}\)

C. -\(\sqrt{35}\)

D. 4\sqrt{7}\) - 8\(\sqrt{5}\)

E. \(\sqrt{35}\)

Detailed Solution

\(\frac{5\sqrt{7} - 7\sqrt{5}}{\sqrt{7} - \sqrt{5}}\) = \(\frac{5\sqrt{7} - 7\sqrt{5}}{\sqrt{7} - \sqrt{5}}\) x \(\frac{\sqrt{7} + \sqrt{5}}{\sqrt{7} + \sqrt{5}}\)

= \(\frac{(5 \times 7) + (5 \sqrt{7} \times 5) - (7 \times \sqrt{5} \times 7) (-7 \times 5)}{(\sqrt{7})^2}\)

= \(\frac{5 \sqrt{35} - 7\sqrt{35}}{2}\)

= \(\frac{-2\sqrt{35}}{2}\)

= - \(\sqrt{35}\)
13.

Simplify \(\frac{3^n - 3^{n - 1}}{3^3 \times 3^n - 27 \times 3^{n - 1}}\)

A. 1

B. 6

C. \(\frac{1}{27}\)

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

Detailed Solution

\(\frac{3^n - 3^{n - 1}}{3^3 \times 3^n - 27 \times 3^{n - 1}}\) = \(\frac{3^n - 3^{n - 1}}{3^3(3^n - 3^{n - 1})}\)

= \(\frac{3^n - 3^{n - 1}}{27(3^n - 3^{n - 1})}\)

= \(\frac{1}{27}\)
14.

P varies directly as the square of q and inversely as r. If p = 36, when q = 36, when q = 3 and r = 4, find p when q = 5 and r = 2

A. 72

B. 100

C. 90

D. 200

E. 125

Detailed Solution

P \(\alpha\) \(\frac{q^2}{r}\)

P = \(\frac{kq^2}{r}\)

k = \(\frac{pr}{q^2}\)

= \(\frac{36 x 4}{(3)^2}\)

p = \(\frac{16q^2}{r}\)

= \(\frac{16 \times 25}{2}\)

= 200
15.

Factorize 6x2 - 14x - 12

A. 2(x + 3)(3x - 2)

B. 6(x - 2)(x + 1)

C. 2(x - 3)(3x + 2)

D. 6(x + 2)(x - 1)

E. (3x - 4)(2x + 3)

Detailed Solution

6x2 - 14x - 12 = 6x2 - 18x + 4x - 12

(3x + 2)(2x - 6)

= 3x(2x - 6) + 2(2x - 6)

= (3x + 2) 2(x - 3)
16.

A straight line y = mx meets the curve y = x2 - 12x + 40 in two distinct points. If one of them is (5, 5) find the other

A. (5, 6)

B. (8, 8)

C. (8, 5)

D. (7,7)

E. (7, 5)

Detailed Solution

When y = 5, y = x2 - 12x + 40, becomes

x2 - 12x + 40 = 5

x2 - 12x + 40 - 5 = 0

x2 + 12x + 35 = 0

x2 - 7x - 5x + 35 = 0

x(x - 7) - 5(x - 7) = 0

= (x - 5)(x - 7)

x = 5 or 7
17.

The table below is drawn for a graph y = x3 - 3x + 1
\(\begin{array}{c|c} x & -3 & -2 & -1 & 0 & 1 & 2 & 3 \\\hline y = x^3 - 3x + 1 & 1 & -1 & 3 & 1 & -1 & 3 & 1\end{array}\)
From x = -2 to x = 1, the graph crosses the x-axis in the range(s)

A. -1 < x < 0 and 0 < x < 1

B. -2 < x < -1 and 0 < x < 1

C. -2 < x < -1 and -1 < x < 0

D. 0 < x < 1

Detailed Solution

If the graph of y = x3 - 3x + 1 is plotted,the graph crosses the x-axis in the ranges -2 < x < -1 and 0 < x < 1
18.

In a racing competition, Musa covered a distance 5x km in the first hour and (x + 10)km in the next hour. He was second to Nzozi who covered a total distance of 118km in the two hours. Which of the following inequalities is correct?

A. -1 < x < x < 0

B. -3 < x < 3

C. 0 \(\leq\) x < 18

D. 15 < x < 18

Detailed Solution

Total distance covered by Musa in 2 hrs

= x + 10 + 5x

= 6x + 10

Ngozi = 118 km

If they are equal, 6x + 10 = 118

but 6x + 10 < 118

6x < 108

= x < 18

0 < x < 18 = 0 \(\leq\) x < 18
19.

If 2x + 3y = 1 and x - 2y = 11, find (x + y)

A. 5

B. -3

C. 8

D. 2

E. -2

Detailed Solution

2x + 3y = 1 x 2.......(i)

x - 2y = 11 x 3.......(ii)

4x + 6y = 2........(iii)

3x - 6Y = 33........(4)

7x = 35

x = 5

Subt. for x

10 + 3y = 1

3y = -9

y = -3

x + y = 5 + -3

5 - 3 = 2
20.

Tunde and Shola can do a piece of work in 18 days. Tunde can do it alone in x days, whilst Shola takes 15 days longer to do it alone. Which of the following equations is satisfied by x?

A. x2 - 5x - 10 = 0

B. x2 - 20x + 360 = 0

C. x2 - 21x - 270 = 0

D. 3x2 - 65x + 362 = 0

Detailed Solution

Tunde and Shola can do the work in 18 days.
Both will do the work in \(\frac{1}{18}\) days.
But Tunde can do the whole work in x days; Hence he does \(\frac{1}{x}\) of the work in 1 day.
Shola does the work in (x + 15) days; hence, he does \(\frac{1}{x + 15}\) of the work in 1 day.
\(\frac{1}{x} + \frac{1}{x + 15} = \frac{1}{18}\)
\(\frac{2x + 15}{x^{2} + 15x} = \frac{1}{18}\)
\(x^{2} + 15x = 36x + 270\)
\(x^{2} + 15x - 36x - 270 = 0\)
\(x^{2} - 21x - 270 = 0\)
11.

A trader in country where their currency 'MONI'(M) is in base five bought 1035 oranges at M145 each. If he sold the oranges at M245 each, what will be his gain?

A. 1035

B. 10305

C. 1025

D. 20025

E. 30325

Detailed Solution

Total cost of 1035 oranges at N145 each

= 1035 x 145

= 20025

Total selling price at N245 each

= (103)5 x 245

= 30325

Hence his gain = 30325 - 20025

= 10305
12.

Rationalize \(\frac{5\sqrt{7} - 7\sqrt{5}}{\sqrt{7} - \sqrt{5}}\)

A. -2 \sqrt{35}\)

B. 4\sqrt{7}\) - 6\(\sqrt{5}\)

C. -\(\sqrt{35}\)

D. 4\sqrt{7}\) - 8\(\sqrt{5}\)

E. \(\sqrt{35}\)

Detailed Solution

\(\frac{5\sqrt{7} - 7\sqrt{5}}{\sqrt{7} - \sqrt{5}}\) = \(\frac{5\sqrt{7} - 7\sqrt{5}}{\sqrt{7} - \sqrt{5}}\) x \(\frac{\sqrt{7} + \sqrt{5}}{\sqrt{7} + \sqrt{5}}\)

= \(\frac{(5 \times 7) + (5 \sqrt{7} \times 5) - (7 \times \sqrt{5} \times 7) (-7 \times 5)}{(\sqrt{7})^2}\)

= \(\frac{5 \sqrt{35} - 7\sqrt{35}}{2}\)

= \(\frac{-2\sqrt{35}}{2}\)

= - \(\sqrt{35}\)
13.

Simplify \(\frac{3^n - 3^{n - 1}}{3^3 \times 3^n - 27 \times 3^{n - 1}}\)

A. 1

B. 6

C. \(\frac{1}{27}\)

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

Detailed Solution

\(\frac{3^n - 3^{n - 1}}{3^3 \times 3^n - 27 \times 3^{n - 1}}\) = \(\frac{3^n - 3^{n - 1}}{3^3(3^n - 3^{n - 1})}\)

= \(\frac{3^n - 3^{n - 1}}{27(3^n - 3^{n - 1})}\)

= \(\frac{1}{27}\)
14.

P varies directly as the square of q and inversely as r. If p = 36, when q = 36, when q = 3 and r = 4, find p when q = 5 and r = 2

A. 72

B. 100

C. 90

D. 200

E. 125

Detailed Solution

P \(\alpha\) \(\frac{q^2}{r}\)

P = \(\frac{kq^2}{r}\)

k = \(\frac{pr}{q^2}\)

= \(\frac{36 x 4}{(3)^2}\)

p = \(\frac{16q^2}{r}\)

= \(\frac{16 \times 25}{2}\)

= 200
15.

Factorize 6x2 - 14x - 12

A. 2(x + 3)(3x - 2)

B. 6(x - 2)(x + 1)

C. 2(x - 3)(3x + 2)

D. 6(x + 2)(x - 1)

E. (3x - 4)(2x + 3)

Detailed Solution

6x2 - 14x - 12 = 6x2 - 18x + 4x - 12

(3x + 2)(2x - 6)

= 3x(2x - 6) + 2(2x - 6)

= (3x + 2) 2(x - 3)
16.

A straight line y = mx meets the curve y = x2 - 12x + 40 in two distinct points. If one of them is (5, 5) find the other

A. (5, 6)

B. (8, 8)

C. (8, 5)

D. (7,7)

E. (7, 5)

Detailed Solution

When y = 5, y = x2 - 12x + 40, becomes

x2 - 12x + 40 = 5

x2 - 12x + 40 - 5 = 0

x2 + 12x + 35 = 0

x2 - 7x - 5x + 35 = 0

x(x - 7) - 5(x - 7) = 0

= (x - 5)(x - 7)

x = 5 or 7
17.

The table below is drawn for a graph y = x3 - 3x + 1
\(\begin{array}{c|c} x & -3 & -2 & -1 & 0 & 1 & 2 & 3 \\\hline y = x^3 - 3x + 1 & 1 & -1 & 3 & 1 & -1 & 3 & 1\end{array}\)
From x = -2 to x = 1, the graph crosses the x-axis in the range(s)

A. -1 < x < 0 and 0 < x < 1

B. -2 < x < -1 and 0 < x < 1

C. -2 < x < -1 and -1 < x < 0

D. 0 < x < 1

Detailed Solution

If the graph of y = x3 - 3x + 1 is plotted,the graph crosses the x-axis in the ranges -2 < x < -1 and 0 < x < 1
18.

In a racing competition, Musa covered a distance 5x km in the first hour and (x + 10)km in the next hour. He was second to Nzozi who covered a total distance of 118km in the two hours. Which of the following inequalities is correct?

A. -1 < x < x < 0

B. -3 < x < 3

C. 0 \(\leq\) x < 18

D. 15 < x < 18

Detailed Solution

Total distance covered by Musa in 2 hrs

= x + 10 + 5x

= 6x + 10

Ngozi = 118 km

If they are equal, 6x + 10 = 118

but 6x + 10 < 118

6x < 108

= x < 18

0 < x < 18 = 0 \(\leq\) x < 18
19.

If 2x + 3y = 1 and x - 2y = 11, find (x + y)

A. 5

B. -3

C. 8

D. 2

E. -2

Detailed Solution

2x + 3y = 1 x 2.......(i)

x - 2y = 11 x 3.......(ii)

4x + 6y = 2........(iii)

3x - 6Y = 33........(4)

7x = 35

x = 5

Subt. for x

10 + 3y = 1

3y = -9

y = -3

x + y = 5 + -3

5 - 3 = 2
20.

Tunde and Shola can do a piece of work in 18 days. Tunde can do it alone in x days, whilst Shola takes 15 days longer to do it alone. Which of the following equations is satisfied by x?

A. x2 - 5x - 10 = 0

B. x2 - 20x + 360 = 0

C. x2 - 21x - 270 = 0

D. 3x2 - 65x + 362 = 0

Detailed Solution

Tunde and Shola can do the work in 18 days.
Both will do the work in \(\frac{1}{18}\) days.
But Tunde can do the whole work in x days; Hence he does \(\frac{1}{x}\) of the work in 1 day.
Shola does the work in (x + 15) days; hence, he does \(\frac{1}{x + 15}\) of the work in 1 day.
\(\frac{1}{x} + \frac{1}{x + 15} = \frac{1}{18}\)
\(\frac{2x + 15}{x^{2} + 15x} = \frac{1}{18}\)
\(x^{2} + 15x = 36x + 270\)
\(x^{2} + 15x - 36x - 270 = 0\)
\(x^{2} - 21x - 270 = 0\)