# Mathematics 2021 Past Questions | JAMB

Study the following Mathematics past questions and answers for JAMB, WAEC NECO and Post-JAMB. Get prepared with official past questions and answers for upcoming examinations.

**16**. P(-6, 1) and Q(6, 6) are the two ends of the diameter of a given circle. Calculate the radius.

**A.**6.5 units**B.**13.0 units**C.**3.5 units**D.**7.0 units

**Correct Option: Answer is B**

Solution:

**17. **The angle of a sector of s circle, radius 10.5cm, is 48°, Calculate the perimeter of the sector

**A.**8.8cm**B.**25.4cm**C.**25.6cm**D.**29.8cm

**Correct Option: Answer is D**

Solution:

**18. **Find the length of a side of a rhombus whose diagonals are 6cm and 8cm

**A.**8cm**B.**5cm**C.**4cm**D.**3cm

**Correct Option: Answer is B**

Solution:

A rhombus has two diagonals that bisect each other at right angles.

i.e this splits 6cm into 3cm each AND 8cm to 4cm

Using Hyp2 = adj2 + opp2

Hyp^{2} = 3^{2} + 4^{2}

Hyp^{2} = 25

Hyp = 5

**19. **Each of the interior angles of a regular polygon is 140°. How many sides has the polygon?

**A.**9**B.**8**C.**7**D.**5

**Correct Option: Answer is A**

Each interior angle = (n-2)180 /n

140 = (n-2)180 /n

Cross multiply:

140n = 180n – 360

40n = 360

n = 9 sides

**20. **A cylinder pipe, made of metal is 3cm thick. If the internal radius of the pope is 10cm. Find the volume of metal used in making 3m of the pipe.

**A.**153πcm^{3}**B.**207πcm^{3}**C.**15300πcm^{3}**D.**20700πcm^{3}

**Correct Option: Answer is D**

Solution:

Volume of a cylinder = πr^{2}h

First convert 3m to cm by multiplying by 100

Volume of External cylinder = π x times 13^{2} x 300

Volume of Internal cylinder = π x 10^{2} x 300

Hence; Volume of External cylinder – Volume of Internal cylinder

Total volume (v) = π (169 – 100) x 300

V = π x 69 x 300

V = 20700πcm^{3}

**21. **The locus of a point which moves so that it is equidistant from two intersecting straight lines is the?

**A.**perpendicular bisector of the two lines**B.**angle bisector of the two lines**C.**bisector of the two lines**D.**line parallel to the two lines

**Correct Option: Answer is B**

Solution:

Angle bisector of the two lines.

What is an Angle Bisector? Angle bisector of two lines i.e. the line which bisects the angle between the two lines is **the locus of a point which is equidistant from the two lines**. In other words, an angle bisector has equal perpendicular distance from the two lines

**22. **4, 16, 30, 20, 10, 14 and 26 are represented on a pie chart. Find the sum of the angles of the bisectors representing all numbers equals to or greater than 16

**A.**48^{o}**B.**84^{o}**C.**92^{o}**D.**276^{o}

**Correct Option: Answer is D**

Solution:

Given that 4, 16, 30, 20, 10, 14 and 26

total = 120

numbers equals to or greater than 16 are:

16 + 30 + 20 + 26 = 92

The requires sum of angles = 92/120 x 360 = 276^{o}

**23. **The mean of ten positive numbers is 16. When another number is added, the mean becomes 18. Find the eleventh number

**A.**3**B.**16**C.**38**D.**30

**Correct Option: Answer is C**

Solution:

Mean of 10 numbers = 16

The total sum of numbers = 16 x 10 = 160

Mean of 11 numbers = 18

Total sum of numbers = 11 x 18 = 198

The 11th number = 198 – 160 = 38

**24. **Two numbers are removed at random from the numbers 1, 2, 3 and 4. What is the probability that the sum of the numbers removed is even?

**A.**2/3**B.**2/15**C.**1/2**D.**1/4

**Correct Option: Answer is C**

Solution:

sample space = 16

sum of numbers removed are (2), 3, (4), 5

3, (4), 5, (6)

(4), 5, (6), 7

(5), 6, 7, (8)

Even number = 8

Pr(even sum) = 8/16 = 1/22

**25. **Find the probability that a number selected at random from 41 to 56 is a multiply of 9

**A.**1/8**B.**2/15**C.**3/16**D.**7/8

**Correct Option: Answer is A**

- Given from 41 to 56

41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56

The nos multiple of 9 are: 45, 54

P(multiple of 9) = 2/16 = 1/8

**26. **Musa borrows N10.00 at 2% per month simple interest and repays N8.00 after 4 months. How much does he still owe?

**A.**N10.80**B.**N10.67**C.**N2.80**D.**N2.67

**Correct Option: Answer is C**

S.I = PRT/100

=> 10 x 2 x 4 / 100

=> 4/5 = 0.8

Total amount = N10.80

He pays N8.00

Remainder = 10.80 – 8.00 = N2.80

**27. **Simplify 2log2/5 – log72/125 + log 9

**A.**1 – 4 log3**B.**-1 + 2 log 3**C.**-1 + 5 log2**D.**1 – 2log 2

**Correct Option: Answer is D**

**28. **A car travels from calabar to Enugu, a distance of P km with an average speed of U km per hour and continues to benin, a distance of Q km, with an average speed of Wkm per hour. Find its average speed from Calabar to Benin

**A.**(\frac{(p + q)}{pw + qu}\)**B.**\(\frac{uw(p + q)}{pw + qu}\)**C.**\(\frac{uw(p + q)}{pw}\)**D.**\(\frac{uw}{pw + qu}\)

**Correct Option: Answer is D**

Average speed = Total Distance / Total Time

Calabar to Enugu in time t1, hence

t1 = P/U also from Enugu to Benin

t2 = q/w

Averagw speed = p + q / t1 + t2

= \(\frac{p + q}{\frac{p}{u} + \frac{q}{w}\)

= p + q x \(\frac{uw}{pw + qu}\)

= \(\frac{uw(p + q)}{pw + qu}\)

**29. **If w varies inversely as uv / v+u and is equal to 8 when

u = 2 and v = 6, find a relationship between u, v, w.

**A.**uvw = 16(u + v)**B.**16uv = 3w(u + v)**C.**uvw = 12(u + v)**D.**12uvw = u + v

**Correct Option: Answer is D**

**30. **If g(x) = x^{2}+ 3x find g(x + 1) – g(x)

**A.**(x + 2)**B.**2(x + 2)**C.**(2x + 1)**D.**(x^{2}+ 4)

**Correct Option: Answer is B**

g(x) = x^{2} + 3x

When g(x + 1) = (x + 1)^{2} + 3(x + 1)

= x^{2} + 1 + 2x + 3x + 3

= x^{2} + 5x + 4

g(x + 1) – g(x) = x2 + 5x + 8 – (x^{2} + 3x)

= x^{2} + 5x + 4 – x^{2} -3x

= 2x + 4 or 2(x + 4)

= 2(x + 2)