# Mathematics 2017 Past Questions | WAEC

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. **Given that t = \(2 ^{-x}\), find \(2 ^{x + 1}\) in terms of t.

**A.**\(\frac{2}{t}\)**B.**\(\frac{t}{2}\)**C.**\(\frac{1}{2t}\)**D.**t

**Correct Option: Answer is A**

**32. **Two bottles are drawn with replacement from a crate containing 8 coke, 12 and 4 sprite bottles. What is the probability that the first is coke and the second is not coke?

**A.**\(\frac{1}{12}\)**B.**\(\frac{1}{6}\)**C.**\(\frac{2}{9}\)**D.**\(\frac{3}{8}\)

**Correct Option: Answer is C**

**33. **If the simple interest on a certain amount of money saved in a bank for 5 years at 212$\frac{1}{2}$% annum is N500.00, calculate the total amount due after 6 years at the same rate

**A.**N2,500.00**B.**N2,600.00**C.**N4,500.00**D.**N4,600.00

**Correct Option: Answer is D**

**34**. Calculate the variance of 2, 3, 3, 4, 5, 5, 5, 7, 7 and 9

**A.**2.2**B.**3.4**D.**4.0**D.**4.2

**Correct Option: Answer is D**

**35. **A circular pond of radius 4m has a path of width 2.5m round it. Find, correct to two decimal places, the area of the path. [Take22/7]

**A.**7.83m2${m}^{2}$**B.**32.29m2${m}^{2}$**C.**50.29m2${m}^{2}$**D.**82.50m2

**Correct Option: Answer is D**

**36.**

Fig. 1 and Fig. 2 are the addition and multiplication tables respectively in modulo 5. Use these tables to solve the equation (n ⊕4$\oplus 4$)

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

**Correct Option: Answer is C**

(n $\oplus $ 4) $\oplus $ 3 = 0 (mod 5)

(3 $\oplus $ 4) $\oplus $ 3

12 $\oplus $ 3 = 15 (mod 5)

(5 x 3 + 0) = 0 (mod 5)

**37.**

The diagram shows a circle centre O. if <STR = 29 and <RST = 450, calculate the value of <STO

**A.**12º**B.**15º**C.**29º**D.**34º

**Correct Option: Answer is A**

SRT = 180 – (46 + 29) sum of < s in a

= 180 – 75

= 105

SOT = 2 x 46 < at the centre is twice all the circle = 92

RTO = 180 – (96 + 43)

= 41

STO = 41 – 29

= 12º

**38.**

In the diagram, XY is a straight line. <POX = <POQ and <ROY = <QOR. Find the value of <POQ + <ROY.

**A.**60º**B.**90º**C.**100º**D.**120º

**Correct Option: Answer is C**

<POX = <POQ; <ROY = QOR

2 <POQ + 2 <ROY = 180

2(<POQ = <ROY) = 180

<POQ + <ROY = 90

**39**

The diagram shows a circle O. If < ZYW = 33º

, find < ZWX

**A.**33º**B.**57º**C.**90º**D.**100º

**Correct Option: Answer is C**

In ZY = 90º < subtends In a semi O

ZWY = 180 – (90 + 33)

= 57

ZWX = 57 + 33 = 90º

**40. **

In the diagram, PQ and PS are tangents to the circle O. If PSQ = m, <SPQ = n and <SQR = 33º, find the value of (m + n)

**A.**103º**B.**123º**C.**133º**D.**143º

**Correct Option: Answer is B**

< SQP = 180 – (90 + 33) < on a straight line

= 180 – (123)

= 57º

Therefore, (m + n) = 123º