# 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.

**41.**

Marks | 0 | 1 | 2 | 3 | 4 | 5 |

Frequency | 7 | 4 | 18 | 12 | 8 | 11 |

The table gives the distribution of marks obtained by a number of pupils in a class test. Using this information, Find the median of the distribution

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

**Correct Option: Answer is B**

Median is n/2=6/2

= 3

Median = 3

**42.**

Marks | 0 | 1 | 2 | 3 | 4 | 5 |

Frequency | 7 | 4 | 18 | 12 | 8 | 11 |

The table gives the distribution of marks obtained by a number of pupils in a class test. Using this information, find the first quartile

**A.**1.0**B.**1.5**C.**2.0**D.**2.5

**Correct Option: Answer is C**

First quartile = n/4=60/4

=15

The 15th value is 2

**43. **In a class of 45 students, 28 offer chemistry and 25 offer Biology. If each student offers at least one of the two subjects, calculate the probability that a student selected at random from the class the class offers chemistry only.

**A.**2/9**B.**4/9**C.**5/9**D.**7/9

**Correct Option: Answer is B**

28 – x + x + 25 – x = 45

53 – x = 45

x = 53 – 45

x = 8

chemistry only = 28 – 8

= 20

Probability = 20/45

= 4/9

44. ** **

In the diagram, NQ//TS, <RTS = 50o${}^{o}$ and <PRT = 100º. Find the value of <NPR

**A.**110${o}^{}$**B.**130${o}^{}$**C.**140${o}^{}$**D.**150º

**Correct Option: Answer is B**

< TSR = 180 – (80 + 50)

= 180 – (130)

= 50o${}^{o}$

< QPR = < TSR corresponding < s

< NPR + QPR = < NPR

180º – < QPR = < NPR

180º – 50 = < NPR

< NPR = 130º

45. A stationary boat is observed from a height of 100m. If the horizontal distance between the observer and the boat is 80m, calculate, correct to two decimal places, the angles of depression of the boat from point of observation

**A.**36.87º**B.**39.70º**C.**51.34º**D.**53.13o

**Correct Option: Answer is C**

Tan \(x^o = \frac{100m}{80}\)

Tan \(x^o = Tan^{-1} 1.25\)

x = 51.34\(^o\)

**46. **The diagonal of a square is 60 cm. Calculate its peremeter

**A.**20\(\sqrt{2}\)**B.**40\(\sqrt{2}\)**C.**90\(\sqrt{2}\)**D.**120\(\sqrt{2}\)

**Correct Option: Answer is D**

60²+x²+x²

360²=2x²

x² = 1800

x = √$\sqrt{1800}$

x = 42.4264

x = 42.4264

perimeter = 4x

= 4 x 42.4264

= 169.7056

= 120√$\sqrt{2}$

= 120√2

**47. **Find the value of m in the diagram

**A.**72º**B.**68º**C.**44º**D.**34º

**Correct Option: Answer is C**

2x + m = 180

x + m = 112

x = 122 – m

2(112 – m) + m = 180

224 – 2m + m = 180

224 – m = 180

224 – 180 = m

m = 44º

**48. **The graph of y = $a{\mathrm{x\xb2}}^{}+bx+c$is shown on the diagram. Find the minimum value of y

**A.**-2, 0**B.**-2, 1**C.**-2, 3**D.**-2, 5

**Correct Option: Answer is B**

**49. **In the diagram, PR is a diameter of the circle RSP, RP is produced to T and TS is a tangent to the circle at S. If < PRS = 24º, calculate the value of < STR

**A.**24º**B.**42º**C.**48º**D.**66º

**Correct Option: Answer is A**

RSP = 90 < substance in semi a circle

RPS = 180 – (90 + 24)

= 180 – (114)

= 66

TPS = 180 – 66

= 114

RST = 24

< STR = 180 – (114 + 24)

= 180 – 138

= 42º

**50. **

The table shows the distribution of goals scored by 25 teams in a football competition. Calculate the probability that a team selected at random scored at most 3 goals.

**A.**$\frac{\mathrm{3/}}{25}$**B.**$\frac{\mathrm{1/}}{5}$**C.**$\frac{\mathrm{6/}}{25}$**D.**2/5