# Mathematics 2020 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.

**31**. The chord ST of a circle is equal to the radius r of the circle. Find the length of arc ST

**A.**πr/3**B.**πr/2**C.**πr/12**D.**πr/6

**Correct Option: Answer is A**

Solution:
\(\frac{ \frac{r}{2}}{r}\) Sin \(\theta\) = \(\frac{1}{2}\)

\(\theta\) = sin\(^{-1}\) (\(\frac{1}{2}\)) = 30\(^o\) = 60\(^o\)

Length of arc (minor)

ST = \(\frac{\theta}{360}\) x 2\(\pi r\)

\(\frac{60}{360} \times 2 \pi \times r = \frac{\pi}{3}\)

**32. **A sector of circle of radius 7.2cm which subtends an angle of 300^{o}at the centre is used to form a cone. What s the radius of the base of the cone?

**A.**8cm**B.**9cm**C.**6cm**D.**7cm

**Correct Option: Answer is C**

Solution:

The length of the arc subtended by the sector of angle 300° = circumference of the base of the cone

\(\frac{300}{360} \times 2 \pi \times 7.2 = 2{\pi}r\)

5/6 x 2π x 7.2 = 2πr

5 x π x 2.4 = 2πr

5 x 2.4 = 2r

12 = 2r => r =6cm

**33. **A cylindrical tank has a capacity of 3080m3. What is the depth of the tank if the diameter of its base is 14m?

**A.**25m**B.**23m**C.**22m**D.**20m

**Correct Option: Answer is D**

Solution:

V = 2080cm\(^3\), h = ?

r = 7cm

V = V\(\pi r^2h\)

h = \(\frac{V}{\pi r^2} = \frac{3080}{\frac{22}{7} \times 49}\)

\(\frac{3080}{54}\) = 20cm

**34. **The acres for rice, pineapple, cassava, cocoa and palm oil in a certain district are given respectively as 2, 5, 3, 11 and 9. What is the angle of the sector of cassava in a pie chart?

**A.**180^{o}**B.**36^{o}**C.**60^{o}**D.**108^{o}

**Correct Option: Answer is B**

Total number of acres = 2 + 5 + 3 + 11 + 9 = 30

The angle of acres = 2 + 5 + 3 + 11 + 9 = 30

The angle of the sector for cassava in a pie chart = 3/30×360^{o}=36^{o}

**35. **Three consecutive terms of a geometric progression are give as n – 2, n and n + 3. Find the common ratio

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

**Correct Option: Answer is A**

Solution:

\(\frac{h}{n – 2} = \frac{n + 3}{n}\)

n\(^2\) = (n + 3) (n – 2)

n\(^2\) = n\(^2\) + n – 6

n\(^2\) + n – 6 – n\(^2\) = 0

n – 6 = 0

n = 6

Common ratio: \(\frac{n}{n – 2} = \frac{6}{6 – 2} = \frac{6}{4}\) = \(\frac{3}{2}\)

**36. **In a class of 40 students, 32 offer mathematics, 24 offer physics and 4 offer neither mathematics nor physics. How many offer both mathematics and physics?

**A.**4**B.**8**C.**16**D.**20

**Correct Option: Answer is A**

Solution:

40 = 32 – x + x + 24 + 4

40 = 60 – x

x = 60 – 40

x = 20

**37. **The sum of the interior angle of pentagon is 6x + 6y. Find y in terms of x.

**A.**y = 6 – x**B.**y = 90 – x**C.**y = 120 – x**D.**y = 150 – x

**Correct Option: Answer is B**

Solution:

Sum of interior angles = (2n – 4) 90o

For pentagon, n = 5

Sum of interior angles = 6 x 90o = 540o

6x + 6y = 540

6(x + y) = 540

x + y = 540/6 = 90

y = 90

y = 90 – x

**38. **The mean age group of some students is 15years. When the age of a teacher, 45 years old, is added to the ages of the students, the mean of their ages become 18 years. Find the number of students in the group.

**A.**7**B.**9**C.**15**D.**42

**Correct Option: Answer is A**

Solution:

**39. **A surveyor walks 500m up a hill which slopes at an angle of 30o. Calculate the vertical height through which he rises

**A.**252m**B.**500m**C.**250m**D.**255m

**Correct Option: Answer is C**

h/500 = sin 30o

= 500 sin 30o

= 500 x 1/2

= 250m

**40. **Find the non-zero positive value of x which satisfies the equation

**A.**2**B.**√3**C.**√2**D.**1

**Correct Option: Answer is B**

S.I = PRT/100

I = 1,240 – 1000 240

P = 1000; R = ? T = 3

∴ 240 = 240 x 100/ 1000 x 3

= 8%