# Mathematics 2017 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.

**11. **Simplify 3 \(^{n − 1}\) × \(\frac{27^{n + 1}}{81^n}\)

**A.**3\(^{2n}\)**B.**9**C.**3^{n}**D.**3 \(^{n + 1}\)

**Correct Option: Answer is B**

3\(^{n – 1}\) × \(\frac{27^{n + 1}}{81^n}\)

= 3\(^{n – 1}\) × \(\frac{3^{3(n + 1)}}{3^{4n}}\)

= 3\(^{n – 1 + 3n + 3 − 4n}\)

= 3\(^{4n − 4n − 1 + 3}\)

= \(3^{2}\)

= 9

**12. **The locus of a point which is equidistant from the line PQ forms a

**A.**circle centre P**B.**pair of parallel lines each opposite to PQ**C.**circle centre Q**D.**perpendicular line to PQ

**Correct Option: Answer is D**

**13. **Given T = {even numbers from 1 to 12}

N = {common factors of 6, 8 and 12} Find T n N

**A.**{2, 3}**B.**{2, 3, 4}**C.**{3, 4, 6}**D.**{2}

**Correct Option: Answer is D**

Given T = {even numbers from 1 to 12}

= { 2, 4, 6, 8,10, 12}

N = {common factors of 6, 8 and 12}

= {2} Find T n N = {2}

**14**.

^{o}

**A.**100^{o}**B.**140^{o}**C.**120^{o}**D.**10^{o}

**Correct Option: Answer is C**

**SOLUTION**

If RST = 60^{o}

RXT = 2 × RST

(angle at the centre twice angle at the circumference)

RXT = 2 × 60

= 120^{o}

**15. **Find the sum of the range and the mode of the set of numbers 10, 9, 10, 9, 8, 7, 7, 10, 8, 10, 8, 4, 6, 9, 10, 9, 7, 10, 6, 5

**A.**16**B.**14**C.**12**D.**10

**Correct Option: Answer is A**

SOLUTION

Range = Highest Number – Lowest Number

Mode is the number with highest occurrence

10, 9, 10, 9, 8, 7, 7, 10, 8, 4, 6,, 9, 10, 9, 7, 10, 6, 5

Range = 10 − 4 = 6

Mode = 10

Sum of range and mode = range + mode = 6 + 10

= 16

**16. **Find the sum to infinity of the series \(\frac{1}{4}\), \(\frac{1}{8}\), \(\frac{1}{16}\),……….

**A.**\(\frac{1}{2}\)**B.**\(\frac{3}{5}\)**C.**\(\frac{-1}{5}\)**D.**\(\frac{73}{12}\)

**Correct Option: Answer is A**

Sum to infinity

∑ = arn − 1

= \(\frac{a}{1}\) − r

a = \(\frac{1}{4}\)

r = \(\frac{1}{8}\) ÷ \(\frac{1}{4}\)

r = \(\frac{1}{s}\) × \(\frac{4}{1}\)

= \(\frac{1}{2}\)

S = \(\frac{1 \div 4}{1}\) − \(\frac{1}{2}\)

= \(\frac{1}{4}\) ÷ \(\frac{1}{2}\)

= \(\frac{1}{4}\) × \(\frac{2}{1}\)

= \(\frac{1}{2}\)

**17. **

**A.**6**B.**2**C.**4**D.**5

**Correct Option: Answer is B**

**18. **The value of x + x ( xx) when x = 2 is

**A.**16**B.**10**C.**18**D.**24

**Correct Option: Answer is B**

when x=2, we have

2+2(2)²=

2+8=10

**19. **In a regular polygon, each interior angle doubles its corresponding exterior angle. Find the number of sides of the polygon

**A.**8**B.**6**C.**4**D.**3

**Correct Option: Answer is B**

2x + x = 180^{o}

3x = 180^{o}

x = 60^{o} (exterior angle of the polygon)

angle = \(\frac{\text{total angle}}{\text{number of sides}}\)

60 = \(\frac{360}{n}\)

n = \(\frac{360}{60}\)

n = 6 sides

**20. **A cylindrical tank has a capacity of 3080m^{3}. What is the depth of the tank if the diameter of its base is 14m? Take pi = 22/7.

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

**Correct Option: Answer is C**

Capacity = Volume = 3080m^{3}

base diameter = 14m

radius = \(\frac{\text{diameter}}{2}\)

= 7m

Volume of Cylidner = Capacity of cylinder

πr^{2}h = 3080

\(\frac{22}{7}\) × 7 × 7 × h = 3080

h = \(\frac{3080}{22 \times 7}\)

h = 20m