Year :
2014
Title :
Mathematics (Core)
Exam :
WASSCE/WAEC MAY/JUNE
Paper 1 | Objectives
41 - 49 of 49 Questions
| # | Question | Ans |
|---|---|---|
| 41. |
 In the diagram, PQR is straight line, ( m + n) = 120o and ( n + r) = 100o. Find (m + r). A. 110o B. 120o C. 140o D. 160o |
C |
| 42. |
 In the diagram, PQR is a straight line, (m + n) = 120o and (n + r) = 100o. Find (m + r) A. 110o B. 120o C. 140o D. 160o Detailed Solutionif m + n = 120o and n + r = 100othen m + n = 120o then r = 180o - 120o = 60o also n + r = 100o m = 180o = 80o m + r is = 80o + 60o = 140o |
|
| 43. |
 In the diagram, ST is parallel to UW, < WVT = xo, < VUT = yo, < RSV = 45o and < VTU = 20o. Find the value of x A. 20 B. 45 C. 65 D. 135 Detailed SolutionFrom the diagram SR//UW given that < RSV = 45o and < WVT = xo< WVT = 45o(corresponding angle) xo = 45 |
|
| 44. |
 In the diagram, ST is parallel to UW, < WVT = xo, < VUT = yo, < RSV = 45o and < VTU = 20o. Calculate the value of y A. 20 B. 25 C. 45 D. 65 Detailed Solutionxo + zo = 180o (sum of < on str. line)45o + zo = 180o zo = 180o - 45o z = 135o z + yo + 20o = 180o (sum of < s in \(\bigtriangleup\)) 135o + yo + 20o = 180o yo + 155o = 180< |
|
| 45. |
 The graph given is for the relation y = 2x2 + x - 1.What are the coordinates of the point S? A. (1, 0.2) B. (1, 0.4) C. (1, 2.0) D. (1, 40) |
C |
| 46. |
 The graph given is for the relation y = 2x2 + x - 1. Find the minimum value of y A. 0.00 B. -0.65 C. -1.25 D. -2.10 |
C |
| 47. |
 In the figures, PQ is a tangent to the circle at R and UT is parallel to PQ. if < TRQ = xo, find < URT in terms of x A. 2xo B. (90 - x)o C. (90 + x)o D. (180 - 2x)o Detailed Solution< URT = < TRQ (angle alternate a tangent and a chord equal to angle in the alternate segment)< RUT = xo In \(\bigtriangleup\) URT < RUT + < RUT + < UTR = 180o (sum of int. < s of \(\bigtriangleup\)) < URT + x + x = 180o < URT = 180o - 2x |
|
| 48. |
 Determine the value of m in the diagram A. 80o B. 90o C. 110o D. 150o |
B |
| 49. |
 In the diagram, O is the centre of the circle of the circle, PR is a tangent to the circle at Q < SOQ = 86o. Calculate the value of < SQR. A. 43o B. 47o C. 54o D. 86o Detailed SolutionConstruction: draw a line from Q to point P and another line from S to point P.< SOQ = 2< QPS (< at centre is twice < on the circumference) < QPS \(\frac{86}{2} = 43\) < SQR = < QPS ( < between a chord and tangent = < in the alternate segment) < SQR = 43o |
| 41. |
In the diagram, PQR is straight line, ( m + n) = 120o and ( n + r) = 100o. Find (m + r). A. 110o B. 120o C. 140o D. 160o |
C |
| 42. |
In the diagram, PQR is a straight line, (m + n) = 120o and (n + r) = 100o. Find (m + r) A. 110o B. 120o C. 140o D. 160o Detailed Solutionif m + n = 120o and n + r = 100othen m + n = 120o then r = 180o - 120o = 60o also n + r = 100o m = 180o = 80o m + r is = 80o + 60o = 140o |
|
| 43. |
In the diagram, ST is parallel to UW, < WVT = xo, < VUT = yo, < RSV = 45o and < VTU = 20o. Find the value of x A. 20 B. 45 C. 65 D. 135 Detailed SolutionFrom the diagram SR//UW given that < RSV = 45o and < WVT = xo< WVT = 45o(corresponding angle) xo = 45 |
|
| 44. |
In the diagram, ST is parallel to UW, < WVT = xo, < VUT = yo, < RSV = 45o and < VTU = 20o. Calculate the value of y A. 20 B. 25 C. 45 D. 65 Detailed Solutionxo + zo = 180o (sum of < on str. line)45o + zo = 180o zo = 180o - 45o z = 135o z + yo + 20o = 180o (sum of < s in \(\bigtriangleup\)) 135o + yo + 20o = 180o yo + 155o = 180< |
|
| 45. |
The graph given is for the relation y = 2x2 + x - 1.What are the coordinates of the point S? A. (1, 0.2) B. (1, 0.4) C. (1, 2.0) D. (1, 40) |
C |
| 46. |
The graph given is for the relation y = 2x2 + x - 1. Find the minimum value of y A. 0.00 B. -0.65 C. -1.25 D. -2.10 |
C |
| 47. |
In the figures, PQ is a tangent to the circle at R and UT is parallel to PQ. if < TRQ = xo, find < URT in terms of x A. 2xo B. (90 - x)o C. (90 + x)o D. (180 - 2x)o Detailed Solution< URT = < TRQ (angle alternate a tangent and a chord equal to angle in the alternate segment)< RUT = xo In \(\bigtriangleup\) URT < RUT + < RUT + < UTR = 180o (sum of int. < s of \(\bigtriangleup\)) < URT + x + x = 180o < URT = 180o - 2x |
|
| 48. |
Determine the value of m in the diagram A. 80o B. 90o C. 110o D. 150o |
B |
| 49. |
In the diagram, O is the centre of the circle of the circle, PR is a tangent to the circle at Q < SOQ = 86o. Calculate the value of < SQR. A. 43o B. 47o C. 54o D. 86o Detailed SolutionConstruction: draw a line from Q to point P and another line from S to point P.< SOQ = 2< QPS (< at centre is twice < on the circumference) < QPS \(\frac{86}{2} = 43\) < SQR = < QPS ( < between a chord and tangent = < in the alternate segment) < SQR = 43o |