JAMB Mathematics Past Questions & Answers - Page 182

907.

Simplify f\(\frac{1}{2}\)g2h\(\frac{1}{2}\) \(\div\) f\(\frac{5}{2}\)goh\(\frac{7}{3}\)

A.

(\(\frac{g}{fh}\))2

B.

f2g2h2

C.

\(\frac{5}{4}\)goh\(\frac{7}{9}\)

D.

\(\frac{g^2}{f^5h^7}\)

E.

\(\frac{1}{f^2h^2}\)

Correct answer is A

f\(\frac{1}{2}\)g2h\(\frac{1}{2}\) \(\div\) f\(\frac{5}{2}\)goh\(\frac{7}{3}\) = f\(\frac{1}{2}\) - \(\frac{5}{2}\) g2 - 0 h\(\frac{1}{2}\) - \(\frac{7}{3}\)

f-2 g2 h-2

= \(\frac{g^2}{f^2h^2}\)

= (\(\frac{g}{fh}\))2

908.

The square base of a pyramid of side 3cm has height 8cm. If the pyramid is cut into two parts by a plane parallel to the base midway between the base and the vertex, the volumes of the two sections are

A.

(21cm3, 3cm3)

B.

(24cm3, 3cm3)

C.

(24cm3, 21cm3)

D.

(72cm3, 9cm3)

E.

(63cm3, 9cm3)

Correct answer is B

Vol. of the 1st section is side x height

vol. = 3 x 8

= 24cm3

vol. of the second section is 3 x 1 = 3

= 24cm3, 3cm3

909.

The area of the curved surface of the cone generated by the sector of a circle radius 6cm and arc length 22cm is (\(\pi\) = \(\frac{22}{7}\))

A.

58 sq.cm

B.

34 sq.cm

C.

132 sq.cm

D.

77 sq.cm

E.

66 sq.cm

Correct answer is E

Given: length of the arc AOB = 22cm ; L = 6cm

Curved surface area of cone = \(\pi\)rl

Length of an arc = \(\frac{\theta}{360}\) x 2 \(\pi\)L

= 22cm

but length of an arc = circumference of the cone = 2\(\pi\)

where r is the radius of the cone circle

2\(\pi\)r = 22, r = \(\frac{22}{2\pi}\)r

= 11 x \(\frac{7}{22}\)

= \(\frac{7}{2}\)

curved surface area = \(\pi\)rl

= \(\frac{22}{7}\) x \(\frac{7}{2}\) x 6

= 66 sq.cm

910.

The solution to the simultaneous equations 3x + 5y = 4, 4x + 3y = 5 is

A.

(\(\frac{-13}{11}, \frac{1}{11}\))

B.

(\(\frac{13}{11}, \frac{1}{11}\))

C.

(\(\frac{13}{11}, \frac{-1}{11}\))

D.

(\(\frac{11}{13}, \frac{1}{11}\))

E.

(13, 11)

Correct answer is B

3x + 5y = 4, 4x + 3y = 5

3x + 5y = 4 x 4

4x + 3y = 5 x 3

12x + 20y = 16.....(i)

12x + 9y = 15.......(ii)

subtract eqn.(ii) from eqn.(i)

11y = 1

y = \(\frac{1}{11}\)

12x + 20 x \(\frac{1}{11}\) = 16

12x = \(\frac{156}{11}\)

x = \(\frac{13}{11}\)

= \(\frac{13}{11}, \frac{1}{11}\)