Year : 
1980
Title : 
Mathematics (Core)
Exam : 
JAMB Exam

Paper 1 | Objectives

21 - 30 of 49 Questions

# Question Ans
21.

In triangle FGB, < G = 90o, < H = 60o, while in triangle XYZ, < X = 60o and < Y = 30o. Form XYZ write down the ratio equal to \(\frac{FG}{FH}\)

A. \(\frac{YX}{ZX}\)

B. \(\frac{YX}{YZ}\)

C. \(\frac{ZX}{YZ}\)

D. \(\frac{YZ}{XY}\)

Detailed Solution

The two triangles are similar

\(\frac{FG}{YZ}\) = \(\frac{FH}{XY}\) = \(\frac{GH}{YZ}\) = \(\frac{FG}{FH}\) = \(\frac{YZ}{XZ}\)
22.

A pentagon has four of its angles equal. If the size of the fifth angle is 60o, Find the size of each of the four equal angles.

A. 60o

B. 108o

C. 120o

D. 150o

E. None of the above

Detailed Solution

The sum of interior angle of pentagon

(5 sides) = (2n - 4) x 90o

n = 5

(2 x 5 - 4) x 90o = 6 x 90o

= 540o

540o - 60o = 480o \(\div\) 4

= 120o
23.

Find the product of (2\(\sqrt{y - 3y}\)) and (3y = 2y)

A. 4y = y2

B. 4y + 9y2

C. 4y - 9y2

D. -4y - 9y2

E. 2y + y2

Detailed Solution

(2\(\sqrt{y - 3y}\))(3y = 2y)

= 6y y + 4y - 9y2 - 6y\(\sqrt{y}\)

= 4y - 9y2
24.

(\(\frac{x^a}{x^b}\))a + b by (\(\frac{x^{a + b}}{x^{a - b}}\))\(\frac{a^2}{b}\)

A. x-a2

B. xb2

C. xa2 - b2

D. \(\frac{1}{x^{a2 + b2}}\)

E. xb2 - a2

Detailed Solution

(\(\frac{x^a}{x^b}\))a + b = (\(\frac{x^{a2 + ab}}{x^{b2 + ab}}\))

= xa2 - b2

{\(\frac{xa + b}{xa - b}\)} = xa + b - a + b

= x2b

= x2a

= xa2 - b2

= xb2 \(\div \) a2 = \(\frac{1}{x^\text a2 + b2}\)
25.

What will be the value of k so that the quadratic equation kx2 - 4x + 1 = 0 has two equal roots?

A. 2

B. 3

C. 4

D. 8

E. \(\frac{1}{4}\)

Detailed Solution

kx2 - 4x + 1 = 0, comparing with ax2 + bx + c = 0

a = k, b = -4, c = 1 for equal root b2 = 4ac

(-4)2 = 4k

k = \(\frac{16}{4}\)

= 4
26.

Find the solution of the equation x + 2\(\sqrt{x - 8}\) = 0

A. (4, 16)

B. (2, 4)

C. (4, 1)

D. (1, 16)

E. (16, 16)

Detailed Solution

x + 2\(\sqrt{x - 8}\) = 0, Let x = y

x = y2

y2 + 2y - 8 = 0

(y + 4)(x - 2) = 0

y = -4 or 2

x = 16 or 4
27.

If it is given that 5x + 1 + 5x = 150, then the value of x is equal to

A. 3

B. 4

C. 1

D. 2

E. \(\frac{1}{2}\)

Detailed Solution

5x + 1 + 5x = 150

5(5x) + 5x = 150

6(5x) = 150

5x = \(\frac{150}{6}\)

= 25

= 52

= 2
28.

Solve the system of equation 2x + y = 32, 33y - x = 27

A. (3, 2)

B. (-3, 2)

C. (3, -2)

D. (-3, -2)

E. (2, 2)

Detailed Solution

2x + y = 32, 33y - x = 27

2x + y = 25

33y + x = 33

x + y = 5

\(\frac{3y - x = 3}{4y = 8}\)

y = 2
29.

Simplify the given expression \(\sqrt{\frac{1 - cos x}{1 + cos x}}\)

A. \(\frac{1 - cos x}{sin x}\)

B. 1 - cos x

C. sin x

D. 1 + cos x

E. \(\frac{1 + cos x}{sin x}\)

Detailed Solution

\(\sqrt{\frac{1 - cos x}{1 + cos x}}\) = a

a2 = \(\frac{1 - cosx}{1 + cosx}\)

\(\frac{1 - cosx}{1 + cosx}\) = \(\frac{1 - cosx}{1 - cosx}\)

= \(\frac{(1 - cosx)^2}{cos^2 x}\)

a2 = \(\frac{(1 - cos x)^2}{sin^2 x}\)

a = \(\frac{1 - cos x}{sin x}\)
30.

Find the area of the curved surface of a cone whose base radius is 6cm and whose height is 8cm. (take \(\pi\) = \(\frac{22}{7}\))

A. 188.57cm2

B. 1320cm2

C. 188cm2

D. 188.08cm2

E. 10cm2

Detailed Solution

S = curved surface area = \(\pi\)rL

= \(\frac{22}{7}\) x 6 x 10

= 188.57cm2
21.

In triangle FGB, < G = 90o, < H = 60o, while in triangle XYZ, < X = 60o and < Y = 30o. Form XYZ write down the ratio equal to \(\frac{FG}{FH}\)

A. \(\frac{YX}{ZX}\)

B. \(\frac{YX}{YZ}\)

C. \(\frac{ZX}{YZ}\)

D. \(\frac{YZ}{XY}\)

Detailed Solution

The two triangles are similar

\(\frac{FG}{YZ}\) = \(\frac{FH}{XY}\) = \(\frac{GH}{YZ}\) = \(\frac{FG}{FH}\) = \(\frac{YZ}{XZ}\)
22.

A pentagon has four of its angles equal. If the size of the fifth angle is 60o, Find the size of each of the four equal angles.

A. 60o

B. 108o

C. 120o

D. 150o

E. None of the above

Detailed Solution

The sum of interior angle of pentagon

(5 sides) = (2n - 4) x 90o

n = 5

(2 x 5 - 4) x 90o = 6 x 90o

= 540o

540o - 60o = 480o \(\div\) 4

= 120o
23.

Find the product of (2\(\sqrt{y - 3y}\)) and (3y = 2y)

A. 4y = y2

B. 4y + 9y2

C. 4y - 9y2

D. -4y - 9y2

E. 2y + y2

Detailed Solution

(2\(\sqrt{y - 3y}\))(3y = 2y)

= 6y y + 4y - 9y2 - 6y\(\sqrt{y}\)

= 4y - 9y2
24.

(\(\frac{x^a}{x^b}\))a + b by (\(\frac{x^{a + b}}{x^{a - b}}\))\(\frac{a^2}{b}\)

A. x-a2

B. xb2

C. xa2 - b2

D. \(\frac{1}{x^{a2 + b2}}\)

E. xb2 - a2

Detailed Solution

(\(\frac{x^a}{x^b}\))a + b = (\(\frac{x^{a2 + ab}}{x^{b2 + ab}}\))

= xa2 - b2

{\(\frac{xa + b}{xa - b}\)} = xa + b - a + b

= x2b

= x2a

= xa2 - b2

= xb2 \(\div \) a2 = \(\frac{1}{x^\text a2 + b2}\)
25.

What will be the value of k so that the quadratic equation kx2 - 4x + 1 = 0 has two equal roots?

A. 2

B. 3

C. 4

D. 8

E. \(\frac{1}{4}\)

Detailed Solution

kx2 - 4x + 1 = 0, comparing with ax2 + bx + c = 0

a = k, b = -4, c = 1 for equal root b2 = 4ac

(-4)2 = 4k

k = \(\frac{16}{4}\)

= 4
26.

Find the solution of the equation x + 2\(\sqrt{x - 8}\) = 0

A. (4, 16)

B. (2, 4)

C. (4, 1)

D. (1, 16)

E. (16, 16)

Detailed Solution

x + 2\(\sqrt{x - 8}\) = 0, Let x = y

x = y2

y2 + 2y - 8 = 0

(y + 4)(x - 2) = 0

y = -4 or 2

x = 16 or 4
27.

If it is given that 5x + 1 + 5x = 150, then the value of x is equal to

A. 3

B. 4

C. 1

D. 2

E. \(\frac{1}{2}\)

Detailed Solution

5x + 1 + 5x = 150

5(5x) + 5x = 150

6(5x) = 150

5x = \(\frac{150}{6}\)

= 25

= 52

= 2
28.

Solve the system of equation 2x + y = 32, 33y - x = 27

A. (3, 2)

B. (-3, 2)

C. (3, -2)

D. (-3, -2)

E. (2, 2)

Detailed Solution

2x + y = 32, 33y - x = 27

2x + y = 25

33y + x = 33

x + y = 5

\(\frac{3y - x = 3}{4y = 8}\)

y = 2
29.

Simplify the given expression \(\sqrt{\frac{1 - cos x}{1 + cos x}}\)

A. \(\frac{1 - cos x}{sin x}\)

B. 1 - cos x

C. sin x

D. 1 + cos x

E. \(\frac{1 + cos x}{sin x}\)

Detailed Solution

\(\sqrt{\frac{1 - cos x}{1 + cos x}}\) = a

a2 = \(\frac{1 - cosx}{1 + cosx}\)

\(\frac{1 - cosx}{1 + cosx}\) = \(\frac{1 - cosx}{1 - cosx}\)

= \(\frac{(1 - cosx)^2}{cos^2 x}\)

a2 = \(\frac{(1 - cos x)^2}{sin^2 x}\)

a = \(\frac{1 - cos x}{sin x}\)
30.

Find the area of the curved surface of a cone whose base radius is 6cm and whose height is 8cm. (take \(\pi\) = \(\frac{22}{7}\))

A. 188.57cm2

B. 1320cm2

C. 188cm2

D. 188.08cm2

E. 10cm2

Detailed Solution

S = curved surface area = \(\pi\)rL

= \(\frac{22}{7}\) x 6 x 10

= 188.57cm2