Year : 
1988
Title : 
Mathematics (Core)
Exam : 
JAMB Exam

Paper 1 | Objectives

41 - 49 of 49 Questions

# Question Ans
41.

If x and y represent the mean and the median respectively of the following set of numbers 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, find the \(\frac{x}{y}\) correct to one decimal place

A. 18

B. 15

C. 1.0

D. 5

Detailed Solution

Mean \(\bar{x}\) = \(\frac{156}{10}\) = 15.6

Median = \(\bar{y}\) = \(\frac{15 + 16}{2}\)

\(\frac{31}{2}\) = 15.5

\(\frac{x}{y}\) = \(\frac{15.6}{15.5}\) = 1.0065

1.0(1 d.p)
42.

If two dice are thrown together, what is the probability of obtaining at least a score of 10?

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

B. \(\frac{1}{12}\)

C. \(\frac{5}{6}\)

D. \(\frac{11}{12}\)

Detailed Solution

The total sample space when two dice are thrown together is 6 x 6 = 36

\(\begin{array}{c|c} & 1 & 2 & 3 & 4 & 5 & 6\\ 1. & 1.1 & 1.2 & 1.3 & 1.4 & 1.5 & 1.6\\ 2 & 2.1 & 2.2 & 2.3 & 2.4 & 2.5 & 2.6\\3 & 3.1 & 3.2 & 3.3 & 3.4 & 3.5 & 3.6\\ 4 & 4.1 & 4.2 & 4.3 & 4.4 & 4.5 & 4.6\\ 5 & 5.1 & 5.2 & 5.3 & 5.4 & 5.5 & 5.6\\6 & 6.1 & 6.2 & 6.3 & 6.4 & 6.5 & 6.6 \end{array}\)

At least 10 means 10 and above

P(at least 10) = \(\frac{6}{36}\)

= \(\frac{1}{6}\)
43.

In the figure, PQRS is a circle. If chords QR and RS are equal, calculate the value of x

A. 80o

B. 60o

C. 45o

D. 40o

Detailed Solution

SRT is a straight line, where QRT = 120

SRQ = 180o - 120o = 60o - (angle on a straight line)

also angle QRS = 180o - 100o (angle on a straight line) . In angles where QR = SR and angle SRQ = 60o

x = 100 - 60 = 40o
44.

In the figure, PQ is a parallel to ST and QRS = 40o. Find the value of x

A. 55o

B. 60o

C. 65o

D. 75o

Detailed Solution

From the figure, 3x + x - 40o = 180o

4x = 180o + 40o

4x = 220o

x = \(\frac{220}{4}\)

= 55o
45.

In the figure, PS = 7cm and RY = 9cm. IF the area of parallelogram PQRS is 56cm2. Find the area of trapezium PQTS

A. 56cm2

B. 112cm2

C. 120cm2

D. 176cm2

Detailed Solution

From the figure, PS = QR = YT = 7cm

Area of parallelogram PQRS = 56cm

56 = base x height, where base = 7

7 x h = 56cm,

h = \(\frac{56}{7}\)

= 8cm

Area of trapezium \(\frac{1}{2}\) (sum of two sides)x height where two sides are QT and PS but QT = QR + RY + YT = 7 +9 + 7 = 23cm

Area of trapezium PQTS = \(\frac{1}{2}\)(23 + 7) x 8

\(\frac{1}{2}\) x 30 x 8 = 120cmsq
46.

In the figure, STQ = SRP, PT = TQ = 6cm and QS = 5cm. Find SR

A. \(\frac{47}{5}\)

B. 5

C. \(\frac{32}{5}\)

D. \(\frac{22}{5}\)

Detailed Solution

From similar triangle, \(\frac{QS}{QP} = \frac{TQ}{QR} = \frac{5}{12} = \frac{6}{QR}\)

QT = \(\frac{6 \times 12}{5} = \frac{72}{5} = SR = QR - QS\)

= \(\frac{72}{5} - 5 = \frac{72 - 25}{5}\)

= \(\frac{47}{5}\)
47.

In the figure, PS = RS = QS and QRS = 50°. Find QPR

A. 25O

B. 40O

C. 50O

D. 65O

Detailed Solution

In the figure PS = RS = QS, they will have equal base QR = RP

In angle SQR, angle S = 50O

In angle QRP, 65 + 65 = 130O

Since RQP = angle RPQ = \(\frac{180 - 130}{2}\)

= \(\frac{50}{2} = 25^o\)

QPR = 25°
48.

In the figure, XR and YQ are tangents to the circle YZXP if ZXR = 45° and YZX = 55°, Find ZYQ

A. 135O

B. 125O

C. 100O

D. 90O

Detailed Solution

< RXZ = < ZYX = 45O(Alternate segment)

< ZYQ = 90 + 45

= 135°
49.

In the figure, a solid consists of a hemisphere surmounted by a right circular cone, with radius 3.0cm and height 6.0cm. Find the volume of the solid

A. 36\(\pi\)cm 3

B. 54\(\pi\)cm 2

C. 18\(\pi\)cm 2

D. 108\(\pi\)cm 2

Detailed Solution

The volume of the solid = vol. of cone + vol. of hemisphere

volume of cone = \(\frac{1}{3} \pi r^2 h\)

= \(\frac{1 \pi}{3} \times (3)^2 x 6 = 18 \pi cm^2\)

vol. of hemisphere = \(\frac{4 \pi r^3}{6} = \frac{2 \pi r^3}{3}\)

= \(\frac{2 \pi}{3} \times (3)^3 = 18\pi cm^3\)

vol. of solid = 18\(\pi\) + 18\(\pi\)

= 36\(\pi\)cm3
41.

If x and y represent the mean and the median respectively of the following set of numbers 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, find the \(\frac{x}{y}\) correct to one decimal place

A. 18

B. 15

C. 1.0

D. 5

Detailed Solution

Mean \(\bar{x}\) = \(\frac{156}{10}\) = 15.6

Median = \(\bar{y}\) = \(\frac{15 + 16}{2}\)

\(\frac{31}{2}\) = 15.5

\(\frac{x}{y}\) = \(\frac{15.6}{15.5}\) = 1.0065

1.0(1 d.p)
42.

If two dice are thrown together, what is the probability of obtaining at least a score of 10?

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

B. \(\frac{1}{12}\)

C. \(\frac{5}{6}\)

D. \(\frac{11}{12}\)

Detailed Solution

The total sample space when two dice are thrown together is 6 x 6 = 36

\(\begin{array}{c|c} & 1 & 2 & 3 & 4 & 5 & 6\\ 1. & 1.1 & 1.2 & 1.3 & 1.4 & 1.5 & 1.6\\ 2 & 2.1 & 2.2 & 2.3 & 2.4 & 2.5 & 2.6\\3 & 3.1 & 3.2 & 3.3 & 3.4 & 3.5 & 3.6\\ 4 & 4.1 & 4.2 & 4.3 & 4.4 & 4.5 & 4.6\\ 5 & 5.1 & 5.2 & 5.3 & 5.4 & 5.5 & 5.6\\6 & 6.1 & 6.2 & 6.3 & 6.4 & 6.5 & 6.6 \end{array}\)

At least 10 means 10 and above

P(at least 10) = \(\frac{6}{36}\)

= \(\frac{1}{6}\)
43.

In the figure, PQRS is a circle. If chords QR and RS are equal, calculate the value of x

A. 80o

B. 60o

C. 45o

D. 40o

Detailed Solution

SRT is a straight line, where QRT = 120

SRQ = 180o - 120o = 60o - (angle on a straight line)

also angle QRS = 180o - 100o (angle on a straight line) . In angles where QR = SR and angle SRQ = 60o

x = 100 - 60 = 40o
44.

In the figure, PQ is a parallel to ST and QRS = 40o. Find the value of x

A. 55o

B. 60o

C. 65o

D. 75o

Detailed Solution

From the figure, 3x + x - 40o = 180o

4x = 180o + 40o

4x = 220o

x = \(\frac{220}{4}\)

= 55o
45.

In the figure, PS = 7cm and RY = 9cm. IF the area of parallelogram PQRS is 56cm2. Find the area of trapezium PQTS

A. 56cm2

B. 112cm2

C. 120cm2

D. 176cm2

Detailed Solution

From the figure, PS = QR = YT = 7cm

Area of parallelogram PQRS = 56cm

56 = base x height, where base = 7

7 x h = 56cm,

h = \(\frac{56}{7}\)

= 8cm

Area of trapezium \(\frac{1}{2}\) (sum of two sides)x height where two sides are QT and PS but QT = QR + RY + YT = 7 +9 + 7 = 23cm

Area of trapezium PQTS = \(\frac{1}{2}\)(23 + 7) x 8

\(\frac{1}{2}\) x 30 x 8 = 120cmsq
46.

In the figure, STQ = SRP, PT = TQ = 6cm and QS = 5cm. Find SR

A. \(\frac{47}{5}\)

B. 5

C. \(\frac{32}{5}\)

D. \(\frac{22}{5}\)

Detailed Solution

From similar triangle, \(\frac{QS}{QP} = \frac{TQ}{QR} = \frac{5}{12} = \frac{6}{QR}\)

QT = \(\frac{6 \times 12}{5} = \frac{72}{5} = SR = QR - QS\)

= \(\frac{72}{5} - 5 = \frac{72 - 25}{5}\)

= \(\frac{47}{5}\)
47.

In the figure, PS = RS = QS and QRS = 50°. Find QPR

A. 25O

B. 40O

C. 50O

D. 65O

Detailed Solution

In the figure PS = RS = QS, they will have equal base QR = RP

In angle SQR, angle S = 50O

In angle QRP, 65 + 65 = 130O

Since RQP = angle RPQ = \(\frac{180 - 130}{2}\)

= \(\frac{50}{2} = 25^o\)

QPR = 25°
48.

In the figure, XR and YQ are tangents to the circle YZXP if ZXR = 45° and YZX = 55°, Find ZYQ

A. 135O

B. 125O

C. 100O

D. 90O

Detailed Solution

< RXZ = < ZYX = 45O(Alternate segment)

< ZYQ = 90 + 45

= 135°
49.

In the figure, a solid consists of a hemisphere surmounted by a right circular cone, with radius 3.0cm and height 6.0cm. Find the volume of the solid

A. 36\(\pi\)cm 3

B. 54\(\pi\)cm 2

C. 18\(\pi\)cm 2

D. 108\(\pi\)cm 2

Detailed Solution

The volume of the solid = vol. of cone + vol. of hemisphere

volume of cone = \(\frac{1}{3} \pi r^2 h\)

= \(\frac{1 \pi}{3} \times (3)^2 x 6 = 18 \pi cm^2\)

vol. of hemisphere = \(\frac{4 \pi r^3}{6} = \frac{2 \pi r^3}{3}\)

= \(\frac{2 \pi}{3} \times (3)^3 = 18\pi cm^3\)

vol. of solid = 18\(\pi\) + 18\(\pi\)

= 36\(\pi\)cm3