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

Paper 1 | Objectives

11 - 20 of 49 Questions

# Question Ans
11.

The ratio of the areas of similar triangles is necessarily equal to

A. the ratio of the corresponding sides

B. the ratio of the squares of corresponding sides

C. the ratio of the corresponding heights of the triangles

D. half the ratio of the corresponding heights of the triangles

E. the ratio of the corresponding bases to the heights of the triangles

Detailed Solution

The ratio of the areas of two similar triangles is equal to the square of ratio of their corresponding sides.
12.

P and Q are fixed points and X is a variable point which moves so that angle PXQ = 45o. What is the locus of x?

A. A pair of straight lines parallel to PQ

B. The perpendicular bisector of PQ

C. An arc of a circle passing through P and Q

D. A circle with diameter PQ

E. The bisector of angle PXQ

B

13.

A man and his wife went to buy an article costing N400. The woman had 10% of the cost and the man 40% of the remainder. How much did they have altogether?

A. N216

B. N200

C. N184

D. N144

E. N100

Detailed Solution

10% of 400 = N40, Remainder = N360

40% of 360 = N144

Altogether they 144 + 40

= N184
14.

Simplify \(\frac{log_{10}8}{log_{10}4}\)

A. log102

B. log104

C. \(\frac{3}{2}\)

D. 2

E. None of the above

Detailed Solution

\(\frac{log_{10}8}{log_{10}4}\) = \(\frac{log_{10}2^3}{log_{10}2^2}\)

= \(\frac{3 log_{10}2}{2log_{10}2}\)

= \(\frac{3}{2}\)
15.

Make c the subject of formula v = 1 - \(\frac{a}{5}\)(b + \(\frac{3c}{7}\))

A. [\(\frac{7}{3}\) + \(\frac{5}{a}\)(v - 1)] + b

B. [\(\frac{7b}{3}\) + \(\frac{5}{a}\)(v - 1)]

C. \(\frac{7}{3}\)[b + \(\frac{5}{a}\)(v - 1)]

D. \(\frac{7}{3}\)[b + \(\frac{5v - 1}{a}\)]

A

16.

What factor is common to all the expressions x2 - x, 2x2 + x - 1 and x2 - 1?

A. x

B. x + 1

C. x - 1

D. No common factor

E. (2x - 1)

Detailed Solution

x2 - x = x(x - 1)

= 2x2 + x - 1

= (2x - 1)(x + 10

x2 - 1 = (x + 1)(x - 1)

No factor
17.

Which of the formula below represents the general terms of the following set of numbers(-1, \(\frac{2}{3}\), -\(\frac{1}{2}\), \(\frac{2}{5}\)......) for n = 1, 2, 3, 4.......?

A. \(\frac{2}{n + 1}\)

B. (-1)n + 1 \(\frac{2}{n + 1}\)

C. (-1)n \(\frac{2}{n + 1}\)

D. \(\frac{n}{2n - 1}\)

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

Detailed Solution

(-1, \(\frac{2}{3}\), -\(\frac{1}{2}\), \(\frac{2}{5}\)......)

nthterm is (-1)n\(\frac{2}{n + 1}\)

Since (-1)n will always be negative for n = odd
18.

Three numbers, x, y and z are connected by the relationships y = \(\frac{4x}{9}\) + 1. If x = 99, Find z = \(\frac{4y}{9}\) + 1

A. 6\(\frac{1}{3}\)

B. 20

C. 21

D. 176\(\frac{4}{9}\)

E. None of the above

Detailed Solution

y = \(\frac{4x}{9}\) + 1, z = \(\frac{4y}{9}\) + 1, z = \(\frac{4}{9}\)(\(\frac{4}{9}\) x + 1) + 1

= [\(\frac{4}{9}\)(\(\frac{4}{9}\)(99) + 1]

= \(\frac{4}{9}\)(45) + 1

= 21
19.

In a school, there are 35 students in class 2A and 40 in class 2B. The mean score for class 2 in an English Literature examination is 60.0 and that for 2B in the same paper is 52.5. Find, to one place of decimal, the mean for the combined classes

A. 2.5

B. 56.0

C. 56.2

D. 56.3

E. 6.5

Detailed Solution

Total score for 2A & 2B = 4200

mean score = \(\frac{4200}{40 + 35}\)

= \(\frac{4200}{75}\)

= 56
20.

A set of data contains a total of 130 items which are divided into six groups for display on a pie chart. If one group contains 26 items, then the sector representing this group on the pie chart contains an angle xo at the centre of the circle where x is

A. 36o

B. 60o

C. 70o

D. 72o

E. 75o

Detailed Solution

\(\frac{26}{130}\) = \(\frac{x}{360}\)

130x = 26 x 360

= 9360

x = \(\frac{9360}{130}\)

= 72o
11.

The ratio of the areas of similar triangles is necessarily equal to

A. the ratio of the corresponding sides

B. the ratio of the squares of corresponding sides

C. the ratio of the corresponding heights of the triangles

D. half the ratio of the corresponding heights of the triangles

E. the ratio of the corresponding bases to the heights of the triangles

Detailed Solution

The ratio of the areas of two similar triangles is equal to the square of ratio of their corresponding sides.
12.

P and Q are fixed points and X is a variable point which moves so that angle PXQ = 45o. What is the locus of x?

A. A pair of straight lines parallel to PQ

B. The perpendicular bisector of PQ

C. An arc of a circle passing through P and Q

D. A circle with diameter PQ

E. The bisector of angle PXQ

B

13.

A man and his wife went to buy an article costing N400. The woman had 10% of the cost and the man 40% of the remainder. How much did they have altogether?

A. N216

B. N200

C. N184

D. N144

E. N100

Detailed Solution

10% of 400 = N40, Remainder = N360

40% of 360 = N144

Altogether they 144 + 40

= N184
14.

Simplify \(\frac{log_{10}8}{log_{10}4}\)

A. log102

B. log104

C. \(\frac{3}{2}\)

D. 2

E. None of the above

Detailed Solution

\(\frac{log_{10}8}{log_{10}4}\) = \(\frac{log_{10}2^3}{log_{10}2^2}\)

= \(\frac{3 log_{10}2}{2log_{10}2}\)

= \(\frac{3}{2}\)
15.

Make c the subject of formula v = 1 - \(\frac{a}{5}\)(b + \(\frac{3c}{7}\))

A. [\(\frac{7}{3}\) + \(\frac{5}{a}\)(v - 1)] + b

B. [\(\frac{7b}{3}\) + \(\frac{5}{a}\)(v - 1)]

C. \(\frac{7}{3}\)[b + \(\frac{5}{a}\)(v - 1)]

D. \(\frac{7}{3}\)[b + \(\frac{5v - 1}{a}\)]

A

16.

What factor is common to all the expressions x2 - x, 2x2 + x - 1 and x2 - 1?

A. x

B. x + 1

C. x - 1

D. No common factor

E. (2x - 1)

Detailed Solution

x2 - x = x(x - 1)

= 2x2 + x - 1

= (2x - 1)(x + 10

x2 - 1 = (x + 1)(x - 1)

No factor
17.

Which of the formula below represents the general terms of the following set of numbers(-1, \(\frac{2}{3}\), -\(\frac{1}{2}\), \(\frac{2}{5}\)......) for n = 1, 2, 3, 4.......?

A. \(\frac{2}{n + 1}\)

B. (-1)n + 1 \(\frac{2}{n + 1}\)

C. (-1)n \(\frac{2}{n + 1}\)

D. \(\frac{n}{2n - 1}\)

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

Detailed Solution

(-1, \(\frac{2}{3}\), -\(\frac{1}{2}\), \(\frac{2}{5}\)......)

nthterm is (-1)n\(\frac{2}{n + 1}\)

Since (-1)n will always be negative for n = odd
18.

Three numbers, x, y and z are connected by the relationships y = \(\frac{4x}{9}\) + 1. If x = 99, Find z = \(\frac{4y}{9}\) + 1

A. 6\(\frac{1}{3}\)

B. 20

C. 21

D. 176\(\frac{4}{9}\)

E. None of the above

Detailed Solution

y = \(\frac{4x}{9}\) + 1, z = \(\frac{4y}{9}\) + 1, z = \(\frac{4}{9}\)(\(\frac{4}{9}\) x + 1) + 1

= [\(\frac{4}{9}\)(\(\frac{4}{9}\)(99) + 1]

= \(\frac{4}{9}\)(45) + 1

= 21
19.

In a school, there are 35 students in class 2A and 40 in class 2B. The mean score for class 2 in an English Literature examination is 60.0 and that for 2B in the same paper is 52.5. Find, to one place of decimal, the mean for the combined classes

A. 2.5

B. 56.0

C. 56.2

D. 56.3

E. 6.5

Detailed Solution

Total score for 2A & 2B = 4200

mean score = \(\frac{4200}{40 + 35}\)

= \(\frac{4200}{75}\)

= 56
20.

A set of data contains a total of 130 items which are divided into six groups for display on a pie chart. If one group contains 26 items, then the sector representing this group on the pie chart contains an angle xo at the centre of the circle where x is

A. 36o

B. 60o

C. 70o

D. 72o

E. 75o

Detailed Solution

\(\frac{26}{130}\) = \(\frac{x}{360}\)

130x = 26 x 360

= 9360

x = \(\frac{9360}{130}\)

= 72o