JAMB Mathematics Past Questions & Answers - Page 159

791.

In \(\bigtriangleup\)PQR, PQ = 10cm, QR = 8cm and RP = 6cm, the perpendicular RS is drawn from R to PQ. Find the length of RS

A.

4cm

B.

32cm

C.

\(\frac{30}{7}\)

D.

\(\frac{40}{7}\)

E.

4.8cm

Correct answer is E

Cos Q = \(\frac{r^2 + p^2 - q^2}{2rp}\)

= \(\frac{10^2 + 8^2 - 6^2}{2(10)(8)}\)

= \(\frac{164 - 36}{160}\)

= \(\frac{128}{160}\)

= 0.8

Q = Cos-1 o.8

= 37o x rep. from rt< RSQ, Let RS = x

\(\frac{x}{sin 37^o}\) = \(\frac{8}{sin 90^o}\)

but sin 90o = 1

x = 8 sin 37o

x = 4.8cm

792.

(3.2)2 - (1.8)2 equals

A.

7.0

B.

2.56

C.

13.48

D.

2.0

E.

0.07

Correct answer is A

(3.2)2 - (1.8)2 = (3.2 + 1.8)(3.2 - 1.8)

= 5 x 1.4

= 7

793.

If the value of \(\pi\) is taken to be \(\frac{22}{7}\), the area of a semi-circle of diameter 42m is

A.

5544m2

B.

1386m2

C.

132m2

D.

264m2

E.

693m2

Correct answer is E

\(\pi\) = \(\frac{22}{7}\), Diameter D = 42m

Area of a circle = \(\frac{\pi D^2}{4}\) when a diameter is given

area of a semicircle \(\frac{\pi D^2}{4}\) x \(\frac{1}{2}\)

= \(\frac{22}{7}\) x \(\frac{4262}{4}\) x \(\frac{1}{2}\)

= \(\frac{11}{7} \times 441\)

= 693m\(^{2}\)

794.

\((1.28 \times 10^{4}) \div (6.4 \times 10^{2})\) equals

A.

2 x 10-5

B.

2 x 10-1

C.

2 x 101

D.

2 x 10-4

Correct answer is C

\((1.28 \times 10^{4}) \div (6.4 \times 10^{2})\)

= \(\frac{1.28 \times 10^{4}}{6.4 \times 10^{2}}\)

\(\equiv \frac{12.8 \times 10^{3}}{6.4 \times 10^{2}}\)

= \(2 \times 10^{1}\)

795.

The mean of the numbers 1.2, 1.0, 0.9, 1.4, 0.8, 0.8, 1.2 and 1.1 is

A.

1.5

B.

0.8

C.

1.0

D.

1.02

E.

1.05

Correct answer is E

\(\frac{1.2 + 1.0 + 0.9 + 1.4 + 0.8 + 0.8 + 1.2}{8}\) = \(\frac{8.4}{8}\)

= 1.05