JAMB Mathematics Past Questions & Answers - Page 152

756.

In a geometric progression, the first term is 153 and the sixth term is \(\frac{17}{27}\). The sum of the first four terms is

A.

\(\frac{860}{3}\)

B.

\(\frac{680}{3}\)

C.

\(\frac{608}{3}\)

D.

\(\frac{680}{3}\)

Correct answer is B

a = 153 - 1st term, 6th term = \(\frac{17}{27}\)

nth term = arn

Sn = a(1 - rn) where r < 1

6th term = 153\(\frac{1 - 0.4^4}{1 - 0.4}\)

= \(\frac{680}{3}\)

757.

A pyramid is constructed on a cuboid. The figure has

A.

Twelve faces

B.

Thirteen vertices

C.

Fourteen edges

D.

Fifteen edges

E.

Sixteen edges

Correct answer is E

The pyramid has 8 edges in itself while the cuboid has 12 edges. When merging the two shapes together, the edge of the base of the pyramid becomes same as the edges of the top of the cuboid.

Hence, the new structure will have (12 + 8) - 4 = 16 edges.

758.

A man runs a distance of 9km at a constant speed for the first 4 km and then 2 km\h faster for the rest of the distance. The whole run takes him one hour. His average speed for the first 4 km is

A.

6 km/h

B.

8 km/h

C.

9 km\h

D.

11 km\h

E.

13 km\h

Correct answer is B

Let the average speed for the first 4 km = x km/h.

Hence, the last 5 km, speed = (x + 2) km/h

Recall: \(Time = \frac{Distance}{Speed}\)

Total time = 1 hour.

\(\therefore \frac{4}{x} + \frac{5}{x + 2} = 1\)

\(\frac{4(x + 2) + 5x}{x(x + 2)} = 1\)

\(9x + 8 = x^{2} + 2x\)

\(x^{2} + 2x - 9x - 8 = 0 \implies x^{2} - 7x - 8 = 0\)

\(x^{2} - 8x + x - 8 = 0\)

\(x(x - 8) + 1(x - 8) = 0\)

\(x = -1; x = 8\)

Since speed cannot be negative, x = 8km/h.

759.

A canal has rectangular cross section of width 10cm and breadth 1m. If water of uniform density 1 gm cm-3 flows through it at a constant speed of 1000mm per minute, the adjacent sea is

A.

100000

B.

1000000

C.

120000

D.

30000

E.

350000

Correct answer is A

The canal's width = 10cm = 100mm (given) The speed of water = 1000mm 10mm = 1cm 1000mm = 100cm The adjacent sea must give speed x width = 1000 x 100 = 100,000

760.

If \(3x - \frac{1}{4})^{\frac{1}{2}} > \frac{1}{4} - x \), then the interval of values of x is

A.

x > \(\frac{1}{3}\)

B.

x < \(\frac{1}{3}\)

C.

x < \(\frac{1}{4}\)

D.

x < \(\frac{9}{16}\)

E.

x > \(\frac{9}{16}\)

Correct answer is E

\(3x - (\frac{1}{4})^{-\frac{1}{2}} > \frac{1}{4} - x\)

= \(3x - 4^{\frac{1}{2}} > \frac{1}{4} - x\)

= \(3x - 2 > \frac{1}{4} - x\)

= \(3x + x > \frac{1}{4} + 2 \implies 4x > \frac{9}{4}\)

\(x > \frac{9}{16}\)