Year : 
1994
Title : 
Mathematics (Core)
Exam : 
JAMB Exam

Paper 1 | Objectives

21 - 30 of 48 Questions

# Question Ans
21.

\(\begin{array}{c|c} \oplus mod 10 & 2 & 4 & 6 & 8 \\ \hline 2 & 4 & 8 & 2 & 6 \\4 & 8 & 6 & 4 & 2\\ 4 & 8 & 6 & 4 & 2\\ 6 & 2 & 4 & 6 & 8\\ 8 & 6 & 2 & 8 & 4\end{array}\)
The multiplication table above has modulo 10 on the set S = (2, 4, 6, 8). Find the inverse of 2

A. 2

B. 4

C. 6

D. 8

Detailed Solution

The inverse of 2 is 6 since 2 x 6 = 12; under mod 10

12 = 2 which is also the value required
22.

Solve for x and y \(\begin{pmatrix} 1 & 1 \\ 3 & y \end{pmatrix}\)\(\begin{pmatrix} x \\ 1 \end{pmatrix}\) = \(\begin{pmatrix} 4 \\ 1\end{pmatrix}\)

A. x = 3, y = 8

B. x = 3, y = -8

C. x = -8, y = 3

D. x = 8, y = -3

Detailed Solution

\(\begin{pmatrix} 1 & 1 \\ 3 & y \end{pmatrix}\)\(\begin{pmatrix} x \\ 1 \end{pmatrix}\) = \(\begin{pmatrix} 4 \\ 1\end{pmatrix}\) = x + 1 = 4

x = 4 - 1

= 3

3x + y =1

3(3) = y = 1

= 9 + y = 1

y = 1 - 9

= -8
23.

The determination of the matrix \(\begin{pmatrix} 1 & 3 & 3 \\ 4 & 5 & 6\\ 2 & 0 & 1 \end{pmatrix}\) is

A. -67

B. -57

C. -3

D. 3

Detailed Solution

4 x 2 = -8 upward arrows = +ve

2 x 5 x 3 = 30

0 x 6 = \(\frac{0}{22}\) downward arrows = -ve

-1(1 x 5x - 1) = 5 - (2 x 6 x 2) = -24

= -(4 x 0 x 1)

= \(\frac{0}{-19}\)

therefore 22 - 19 = 3
24.

\(\begin{array}{c|c} \text{Age in years} & 10 & 11 & 12 \\ \hline \text{Number of pupils} & 6 & 27 & 7\end{array}\)
The table above shows the number of pupils in each age group in a class. What is the probability that a pupil chosen at random is at least 11 years old?

A. \(\frac{27}{40}\)

B. \(\frac{17}{20}\)

C. \(\frac{33}{40}\)

D. \(\frac{3}{20}\)

Detailed Solution

Total number of pupils = 7 + 27 + 6 = 40
n(at least 11 years old) = 27 + 7 = 34.
P(picking a pupil who is at least 11 years old) = \(\frac{34}{40} = \frac{17}{20}\).
25.

If three angles of a quadrilateral are (3y - x - z)o, 3xo, (2z - 2y - x)o find the fourth angle in terms of x, y and z

A. (360 - x- y - z)o

B. (360 + x + y - z)o

C. (180 - x + y + z)o

D. (180 - x + y + z)o

Detailed Solution

The sum of angles of a quadrilateral is 360o

∴ (3y - x - z)o + 3xo + (2z - 2y - x)o + po = 360o

Where P is the fourth angle

3y - x - z + 3x + 2z - 2y - x + p = 360o

p = 360 - (x + y + z)

∴ p = (360 - x - y - z)o
26.

An open rectangular box is made of wood 2cm thick. If the internal dimensions of the box are 50cm long, 36cm wide and 20cm deep, the box volume of wood in the box is

A. 11 520cm3

B. 36 000cm3

C. 38 200cm3

D. 47 520cm3

Detailed Solution

Internal dimension are 50cm, 36cm and 20cm

internal volume = 50 x 36 x 20cm3

1000 x 36cm3

= 36000cm3

External dimension are 54cm x 40cm x 22cm

= 2160cm2 x 22cm = 47520cm3

Volume of wood = Ext. volume - Int. volume

= 47,520cm3 - 36,000cm3

= 11,520cm3
27.

Calculate the perimeter, in cm, of a sector of a circle of radius 8cm and angle 45o

A. 2\(\pi\)

B. 8 + 2\(\pi\)

C. 16 + 2\(\pi\)

D. 16 + 16\(\pi\)

Detailed Solution

Perimeter = OP + OQ + PQ

= 8 + 8 + PQ

length PQ = \(\frac{\theta}{360 \times 2\pi r}\)

= \(\frac{45}{360}\) x 2 x \(\pi\) x 8

= 2\(\pi\)

Perimeter of sector 2r + L

Where l = length of arc and r = radius

∴ P = 2(8) + 2\(\pi\)

= 16 + 2\(\pi\)
28.

What is the locus of a point P which moves on one side of a straight line XY, so that the angle XPY is always equal to 90o?

A. the perpendicular bisector of XY

B. a right-angled triangle

C. a circle

D. a semi circle

Detailed Solution

Since XY is a fixed line and

XPY = 90o P is on one side of XY

P1P2P3......Pn are all possible cases where

XPY = 90o the only possible tendency is a semicircle because angles in semicircle equals 90o
29.

If M(4, q) is the mid-point of the line joining L(p, -2) and N(q, p). Find the values of p and q

A. P = 2, q = 4

B. p = 3, q = 1

C. p = 5, q = 3

D. p = 6, q = 2

Detailed Solution

\(\frac{p + q}{2}\) = 4

p + q = 8 ....(i)

\(\frac{p - 2}{2}\) = q

p - 2q 2 ....(ii)

q = 2, p = 6

3q = 6
30.

The angle of depression of a boat from the top of a cliff 10m high is 30. How far is the boat from the foot of the cliff?

A. \(\frac{5√3m}{3}\)

B. 5√3m

C. 10√3m

D. \(\frac{10√3m}{3}\)

Detailed Solution

RQP = 90o

PQS = 30o but

RQS = 90o = \(\frac{RS}{10}\)

RS = 10 tan 60o

RS = 10 tan 60o

RS = 10√3m
21.

\(\begin{array}{c|c} \oplus mod 10 & 2 & 4 & 6 & 8 \\ \hline 2 & 4 & 8 & 2 & 6 \\4 & 8 & 6 & 4 & 2\\ 4 & 8 & 6 & 4 & 2\\ 6 & 2 & 4 & 6 & 8\\ 8 & 6 & 2 & 8 & 4\end{array}\)
The multiplication table above has modulo 10 on the set S = (2, 4, 6, 8). Find the inverse of 2

A. 2

B. 4

C. 6

D. 8

Detailed Solution

The inverse of 2 is 6 since 2 x 6 = 12; under mod 10

12 = 2 which is also the value required
22.

Solve for x and y \(\begin{pmatrix} 1 & 1 \\ 3 & y \end{pmatrix}\)\(\begin{pmatrix} x \\ 1 \end{pmatrix}\) = \(\begin{pmatrix} 4 \\ 1\end{pmatrix}\)

A. x = 3, y = 8

B. x = 3, y = -8

C. x = -8, y = 3

D. x = 8, y = -3

Detailed Solution

\(\begin{pmatrix} 1 & 1 \\ 3 & y \end{pmatrix}\)\(\begin{pmatrix} x \\ 1 \end{pmatrix}\) = \(\begin{pmatrix} 4 \\ 1\end{pmatrix}\) = x + 1 = 4

x = 4 - 1

= 3

3x + y =1

3(3) = y = 1

= 9 + y = 1

y = 1 - 9

= -8
23.

The determination of the matrix \(\begin{pmatrix} 1 & 3 & 3 \\ 4 & 5 & 6\\ 2 & 0 & 1 \end{pmatrix}\) is

A. -67

B. -57

C. -3

D. 3

Detailed Solution

4 x 2 = -8 upward arrows = +ve

2 x 5 x 3 = 30

0 x 6 = \(\frac{0}{22}\) downward arrows = -ve

-1(1 x 5x - 1) = 5 - (2 x 6 x 2) = -24

= -(4 x 0 x 1)

= \(\frac{0}{-19}\)

therefore 22 - 19 = 3
24.

\(\begin{array}{c|c} \text{Age in years} & 10 & 11 & 12 \\ \hline \text{Number of pupils} & 6 & 27 & 7\end{array}\)
The table above shows the number of pupils in each age group in a class. What is the probability that a pupil chosen at random is at least 11 years old?

A. \(\frac{27}{40}\)

B. \(\frac{17}{20}\)

C. \(\frac{33}{40}\)

D. \(\frac{3}{20}\)

Detailed Solution

Total number of pupils = 7 + 27 + 6 = 40
n(at least 11 years old) = 27 + 7 = 34.
P(picking a pupil who is at least 11 years old) = \(\frac{34}{40} = \frac{17}{20}\).
25.

If three angles of a quadrilateral are (3y - x - z)o, 3xo, (2z - 2y - x)o find the fourth angle in terms of x, y and z

A. (360 - x- y - z)o

B. (360 + x + y - z)o

C. (180 - x + y + z)o

D. (180 - x + y + z)o

Detailed Solution

The sum of angles of a quadrilateral is 360o

∴ (3y - x - z)o + 3xo + (2z - 2y - x)o + po = 360o

Where P is the fourth angle

3y - x - z + 3x + 2z - 2y - x + p = 360o

p = 360 - (x + y + z)

∴ p = (360 - x - y - z)o
26.

An open rectangular box is made of wood 2cm thick. If the internal dimensions of the box are 50cm long, 36cm wide and 20cm deep, the box volume of wood in the box is

A. 11 520cm3

B. 36 000cm3

C. 38 200cm3

D. 47 520cm3

Detailed Solution

Internal dimension are 50cm, 36cm and 20cm

internal volume = 50 x 36 x 20cm3

1000 x 36cm3

= 36000cm3

External dimension are 54cm x 40cm x 22cm

= 2160cm2 x 22cm = 47520cm3

Volume of wood = Ext. volume - Int. volume

= 47,520cm3 - 36,000cm3

= 11,520cm3
27.

Calculate the perimeter, in cm, of a sector of a circle of radius 8cm and angle 45o

A. 2\(\pi\)

B. 8 + 2\(\pi\)

C. 16 + 2\(\pi\)

D. 16 + 16\(\pi\)

Detailed Solution

Perimeter = OP + OQ + PQ

= 8 + 8 + PQ

length PQ = \(\frac{\theta}{360 \times 2\pi r}\)

= \(\frac{45}{360}\) x 2 x \(\pi\) x 8

= 2\(\pi\)

Perimeter of sector 2r + L

Where l = length of arc and r = radius

∴ P = 2(8) + 2\(\pi\)

= 16 + 2\(\pi\)
28.

What is the locus of a point P which moves on one side of a straight line XY, so that the angle XPY is always equal to 90o?

A. the perpendicular bisector of XY

B. a right-angled triangle

C. a circle

D. a semi circle

Detailed Solution

Since XY is a fixed line and

XPY = 90o P is on one side of XY

P1P2P3......Pn are all possible cases where

XPY = 90o the only possible tendency is a semicircle because angles in semicircle equals 90o
29.

If M(4, q) is the mid-point of the line joining L(p, -2) and N(q, p). Find the values of p and q

A. P = 2, q = 4

B. p = 3, q = 1

C. p = 5, q = 3

D. p = 6, q = 2

Detailed Solution

\(\frac{p + q}{2}\) = 4

p + q = 8 ....(i)

\(\frac{p - 2}{2}\) = q

p - 2q 2 ....(ii)

q = 2, p = 6

3q = 6
30.

The angle of depression of a boat from the top of a cliff 10m high is 30. How far is the boat from the foot of the cliff?

A. \(\frac{5√3m}{3}\)

B. 5√3m

C. 10√3m

D. \(\frac{10√3m}{3}\)

Detailed Solution

RQP = 90o

PQS = 30o but

RQS = 90o = \(\frac{RS}{10}\)

RS = 10 tan 60o

RS = 10 tan 60o

RS = 10√3m