Year :
2021
Title :
Mathematics (Core)
Exam :
WASSCE/WAEC MAY/JUNE
Paper 1 | Objectives
31 - 40 of 49 Questions
| # | Question | Ans |
|---|---|---|
| 31. |
Find, correct to two decimal, the mean of 1\(\frac{1}{2}\), 2\(\frac{2}{3}\), 3\(\frac{3}{4}\), 4\(\frac{4}{5}\), and 5\(\frac{5}{6}\). A. 3.71 B. 3.70 C. 3.69 D. 3.72 Detailed SolutionThe mean = sum of 1\(\frac{1}{2}\), 2\(\frac{2}{3}\), 3\(\frac{3}{4}\), 4\(\frac{4}{5}\), and 5\(\frac{5}{6}\) divided by their quantity.Mean = ( 1\(\frac{1}{2}\) + 2\(\frac{2}{3}\) + 3\(\frac{3}{4}\) + 4\(\frac{4}{5}\) + and 5\(\frac{5}{6}\) ) / 5 = 3.708 or 3.71( 2 d.p ) |
|
| 32. |
A cyclist moved at a speed of Xkm/h for 2 hours. He then increased his speed by 2 km/h for the next 3 hours.If the total distance covered is 36 km, calculate his initials speed. A. 12km/h B. 3km/h C. 4km/h D. 6km/h |
|
| 33. |
 Find the value of (x+y) A. 215° B. 70° C. 135° D. 145° Detailed Solutionsum of angle in a straight line = 180°35 + m = 180 m = 180 - 35 = 145° Corresponding angles are congruent i.e equal to each other Angle X and M are congruent = 145 sum of angle in a straight line = 180° 110 + n = 180 m = 180 - 110 = 70° Angle Y and N are congruent = 70° \(\therefore\) ( x + y) = ( 145 + 70)° (x + y) = 215° |
|
| 34. |
 In the diagram, \(\overline{MP}\) is a tangent to the circle NQR, ∠NQR, ∠PNQ = 64 and | \(\overline{RQ}\) | = | \(\overline{RN}\) |. Find the angle market t. A. 130° B. 115° C. 58° D. 68° Detailed SolutionAlternate segments are equal : R° = 64°An isosceles triangle has two angles ( Q° & N° ) equal as A° R° + A° + A° = 180° 2A°= 180 - 64 2A° = 116° A = \(\frac{116}{2}\) A = 58. : Q° = 58, N° = 58 and T° = 58 As Alternate segments are equal: T = Q |
|
| 35. |
Find the first quartile of 7,8,7,9,11,8,7,9,6 and 8. A. 8.5 B. 7.0 C. 7.5 D. 8.0 Detailed SolutionRearrange data in increasing order: 6,7,7,7,8,8,8,9,9 and 11First quartile (ungrouped data) = \(\frac{n}{4}\)th value = \(\frac{10}{4}\) = 2.5th value = \(\frac{7 + 7}{2}\) = 7.0 |
|
| 36. |
 In the diagram, PQRS is a circle. find the value of x. A. 50° B. 30° C. 80° D. 100° Detailed SolutionOpp. angles in a cyclic quadrilateral always add up to 180°∠P + ∠R & ∠Q + ∠S = 180 x + x+y = 180 2x + y = 180... i 2y - 30 + x = 180 2y + x = 180 + 30 x + 2y = 210 ... ii Elimination method: (2x + y = 180) * 1 --> 2x + y = 180 (x + 2y = 210) * 2 --> 2x + 4y = 420 Subtracting both equations - 3y = - 240 y = 80° using eqn i 2x + y = 180 2x + 80 = 180 2x = 100 x = 50° |
|
| 37. |
A cone has a base radius of 8cm and height 11cm. calculate , correct to 2d.p, the curved surface area A. 341.98cm\(^2\) B. 276.57cm\(^2\) C. 201.14cm\(^2\) D. 477.71cm\(^2\) Detailed SolutionWhere l\(^2\) = h\(^2\) + r\(^2\)l\(^2\) = 11\(^2\) + 8\(^2\) l = √185 l = 13.60cm The formula of CSA of Cone is πrl \(\frac{22}{7}\) * 8 * 13.60 = 341.979 or 341.98 (2d.p) |
|
| 38. |
Given that sin x = 3/5, 0 ≤ x ≤ 90, evaluate (tanx + 2cosx) A. 2\(\frac{11}{20}\) B. \(\frac{11}{20}\) C. 2\(\frac{7}{20}\) D. \(\frac{1}{20}\) Detailed SolutionSin x = \(\frac{opp}{hyp}\)sinx = \(\frac{3}{5}\) using Pythagoras' theorem hyp\(^2\) = opp\(^2\) + adj\(^2\) adj\(^2\) = 5\(^2\) - 3\(^2\) = 25 - 9 adj\(^2\) = 16 adj = √ 16 adj = 4. tanx = \(\frac{opp}{adj}\) = \(\frac{3}{4}\) cosx = \(\frac{adj}{hyp}\) = \(\frac{4}{5}\) (tanx + 2cosx) = \(\frac{3}{4}\) + 2(\(\frac{4}{5}\)) = \(\frac{15 + 32}{20}\) = \(\frac{47}{20}\) or 2 \(\frac{7}{20}\) |
|
| 39. |
In the diagram, line \(\overline{EC}\) is a diameter of the circle ABCDE.If angle ABC equals 158°, find ∠ADE A. 112 B. 90 C. 68 D. 22 |
C |
| 40. |
 Table: The table shows the height of 37 players of a basketball team calculates correct to one decimal place the mean height of the players. A. 163.0 B. 162.0 C. 160.0 D. 165.0 Detailed Solution∑fx = (160 * 4) + (161 * 6) + (162 * 3) + (163 * 7) + (164 * 8) + (165 * 9)=640 + 966 + 486 + 1,141 + 1,312 + 1,485 = 6,030 ∑f = 4 + 6 + 3 + 7 + 8 + 9 = 37 = \(\frac{∑fx}{∑f}\) = \(\frac{6030}{37}\) = 162.97 or 163 |
| 31. |
Find, correct to two decimal, the mean of 1\(\frac{1}{2}\), 2\(\frac{2}{3}\), 3\(\frac{3}{4}\), 4\(\frac{4}{5}\), and 5\(\frac{5}{6}\). A. 3.71 B. 3.70 C. 3.69 D. 3.72 Detailed SolutionThe mean = sum of 1\(\frac{1}{2}\), 2\(\frac{2}{3}\), 3\(\frac{3}{4}\), 4\(\frac{4}{5}\), and 5\(\frac{5}{6}\) divided by their quantity.Mean = ( 1\(\frac{1}{2}\) + 2\(\frac{2}{3}\) + 3\(\frac{3}{4}\) + 4\(\frac{4}{5}\) + and 5\(\frac{5}{6}\) ) / 5 = 3.708 or 3.71( 2 d.p ) |
|
| 32. |
A cyclist moved at a speed of Xkm/h for 2 hours. He then increased his speed by 2 km/h for the next 3 hours.If the total distance covered is 36 km, calculate his initials speed. A. 12km/h B. 3km/h C. 4km/h D. 6km/h |
|
| 33. |
Find the value of (x+y) A. 215° B. 70° C. 135° D. 145° Detailed Solutionsum of angle in a straight line = 180°35 + m = 180 m = 180 - 35 = 145° Corresponding angles are congruent i.e equal to each other Angle X and M are congruent = 145 sum of angle in a straight line = 180° 110 + n = 180 m = 180 - 110 = 70° Angle Y and N are congruent = 70° \(\therefore\) ( x + y) = ( 145 + 70)° (x + y) = 215° |
|
| 34. |
In the diagram, \(\overline{MP}\) is a tangent to the circle NQR, ∠NQR, ∠PNQ = 64 and | \(\overline{RQ}\) | = | \(\overline{RN}\) |. Find the angle market t. A. 130° B. 115° C. 58° D. 68° Detailed SolutionAlternate segments are equal : R° = 64°An isosceles triangle has two angles ( Q° & N° ) equal as A° R° + A° + A° = 180° 2A°= 180 - 64 2A° = 116° A = \(\frac{116}{2}\) A = 58. : Q° = 58, N° = 58 and T° = 58 As Alternate segments are equal: T = Q |
|
| 35. |
Find the first quartile of 7,8,7,9,11,8,7,9,6 and 8. A. 8.5 B. 7.0 C. 7.5 D. 8.0 Detailed SolutionRearrange data in increasing order: 6,7,7,7,8,8,8,9,9 and 11First quartile (ungrouped data) = \(\frac{n}{4}\)th value = \(\frac{10}{4}\) = 2.5th value = \(\frac{7 + 7}{2}\) = 7.0 |
| 36. |
In the diagram, PQRS is a circle. find the value of x. A. 50° B. 30° C. 80° D. 100° Detailed SolutionOpp. angles in a cyclic quadrilateral always add up to 180°∠P + ∠R & ∠Q + ∠S = 180 x + x+y = 180 2x + y = 180... i 2y - 30 + x = 180 2y + x = 180 + 30 x + 2y = 210 ... ii Elimination method: (2x + y = 180) * 1 --> 2x + y = 180 (x + 2y = 210) * 2 --> 2x + 4y = 420 Subtracting both equations - 3y = - 240 y = 80° using eqn i 2x + y = 180 2x + 80 = 180 2x = 100 x = 50° |
|
| 37. |
A cone has a base radius of 8cm and height 11cm. calculate , correct to 2d.p, the curved surface area A. 341.98cm\(^2\) B. 276.57cm\(^2\) C. 201.14cm\(^2\) D. 477.71cm\(^2\) Detailed SolutionWhere l\(^2\) = h\(^2\) + r\(^2\)l\(^2\) = 11\(^2\) + 8\(^2\) l = √185 l = 13.60cm The formula of CSA of Cone is πrl \(\frac{22}{7}\) * 8 * 13.60 = 341.979 or 341.98 (2d.p) |
|
| 38. |
Given that sin x = 3/5, 0 ≤ x ≤ 90, evaluate (tanx + 2cosx) A. 2\(\frac{11}{20}\) B. \(\frac{11}{20}\) C. 2\(\frac{7}{20}\) D. \(\frac{1}{20}\) Detailed SolutionSin x = \(\frac{opp}{hyp}\)sinx = \(\frac{3}{5}\) using Pythagoras' theorem hyp\(^2\) = opp\(^2\) + adj\(^2\) adj\(^2\) = 5\(^2\) - 3\(^2\) = 25 - 9 adj\(^2\) = 16 adj = √ 16 adj = 4. tanx = \(\frac{opp}{adj}\) = \(\frac{3}{4}\) cosx = \(\frac{adj}{hyp}\) = \(\frac{4}{5}\) (tanx + 2cosx) = \(\frac{3}{4}\) + 2(\(\frac{4}{5}\)) = \(\frac{15 + 32}{20}\) = \(\frac{47}{20}\) or 2 \(\frac{7}{20}\) |
|
| 39. |
In the diagram, line \(\overline{EC}\) is a diameter of the circle ABCDE.If angle ABC equals 158°, find ∠ADE A. 112 B. 90 C. 68 D. 22 |
C |
| 40. |
Table: The table shows the height of 37 players of a basketball team calculates correct to one decimal place the mean height of the players. A. 163.0 B. 162.0 C. 160.0 D. 165.0 Detailed Solution∑fx = (160 * 4) + (161 * 6) + (162 * 3) + (163 * 7) + (164 * 8) + (165 * 9)=640 + 966 + 486 + 1,141 + 1,312 + 1,485 = 6,030 ∑f = 4 + 6 + 3 + 7 + 8 + 9 = 37 = \(\frac{∑fx}{∑f}\) = \(\frac{6030}{37}\) = 162.97 or 163 |