O lvl E Maths: Properties of Circles

In the diagram, O is the centre of the circle. TAQ, PAB and PQR are straight lines. Given that APO = 14, OBA = 40, calculate(a) ARB,(b) AOP(c) TAB.

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Answer:

(a) BAO = ABO = 40 (isosceles triangle)AOB = 180 - 40 - 40 = 100 (angles in a triangle add up to 180)ARB = 100 / 2 =50(angle at centre = 2 * angle at circumference)=> Hint: Draw a line from R to B to see it clearer

(b) OAP = 180 - 40 = 140 (angles on a straight line)AOP = 180 - 140 - 14= 26(angles on a triangle add up to 180)

(c) OAQ = (180 - 26) / 2 = 77 (isosceles triangle)TAB = 180 - 77 - 40= 63(angles on a straight line)

O lvl E Maths: Properties of Circles

In the digram, O is the centre of the circle and EG and EF are both tangents to the circle at D and A respectively. Given that AOC = 125 and DCO = 27, find(i) ABC,(ii) DAE,(iii) AED.

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Answer:

(i) ADC = 125 / 2 = 62.5 (angle at centre = 2 * angle at circumference)ABC = 180 - ADC = 180 - 62.5ABC = 117.5(opposite angles in a cyclic quad add up to 180)

(ii) reflex angle AOC = 360 - 125 = 235OAD = 360 - 235 - 27 - 62.5(ADC) = 35.5 (angles in a quadrilateral add up to 360)

DAE = OAE - OAD = 90 - 35.5= 54.5

(iii) ADE = DAE = 54.5 (isosceles triangle)

AED = 180 -54.5 -54.5= 71(angles in a triangle add up to 180 )

O lvl E Maths: Properties of Circles

In the diagram, BD is a diameter of the circle centre O, ATB = 30 and ABD = 48.Calculate,(i) ACD(ii) ACB(iii) BDC(iv) BKC

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Answer:

(i) ACD = ABD (angles subtended by the same arc)ACD = 48

(ii) BCD = 90 (angle at semi-circle)ACB = BCD - ACDACB = 90 - 48ACB = 42

(iii) BDC = DBA - BTD (exterior angle = sum of 2 interior angles)BDC = 48 - 30BDC = 18

(iv) CBD = 90 - 18 = 72BKC = 180 - BCK - CBKBKC = 180 - ACB - CBDBKC = 180 - 42 - 72BKC = 66

O lvl E Maths: Properties of Circles

In the digram, O is the centre of the circle and EG and EF are both tangents to the circle at D and A respectively. Given that AOC = 125 and DCO = 27, find(i) ABC,(ii) DAE,(iii) AED.*************************

Answer:

(i) ADC = 125 / 2 = 62.5 (angle at centre = 2 * angle at circumference)ABC = 180 - ADC = 180 - 62.5ABC = 117.5(opposite angles in a cyclic quad add up to 180)

(ii) reflex angle AOC = 360 - 125 = 235OAD = 360 - 235 - 27 - 62.5(ADC) = 35.5 (angles in a quadrilateral add up to 360)

DAE = OAE - OAD = 90 - 35.5= 54.5

(iii) ADE = DAE = 54.5 (isosceles triangle)

AED = 180 -54.5 -54.5= 71(angles in a triangle add up to 180 )

O lvl E Maths: Properties of CirclesRI Sec 3 EOY 1998 Q11

In the diagram above, DE is the tangent at D to the circle with centre O. AC is the diameter, ACD = 30 and BAC = 67. Calculate(a) CAD,(b) CDE,(c) DBO.

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Answer:

(a) CDA = 90 (angle at semi-circle)CAD = 180 - 90 - 30= 60(angles in a triangle)

(b) CDE = CAD= 60(alternate segment)

(c) ABD = ACD = 30ABO = OAB = 67 (angles in an isosceles triangle)DBO = ABO - ABDDBO = 67 - 30= 37

O lvl E Maths: Properties of CirclesRGS P2 Prelim Q6b 1997

In the figure below, PAQ is the tangent at A to the circle ABCD and CDQ is a straight line. The lines BA and CD are parallel, angle DAQ = 42 and angle DQA = 35. Calculate(i) angle ACD(ii) angle ABC(iii) angle PAB

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Answer:

(i) angle ACD= 42(alternate angles for triangle in a circle)

(ii) angle ADC = 42 + 35 = 77 (Sum of external angles)angle ABD = 180 - 77= 103(Opp angles of a cyclic quad adds up to 180)

(iii) angle ADQ = 180 - 42 - 35= 103 (angles in a triangle)angle BAD = 103 (alternate angles)angle PAB = 180 - 103 - 42 =35(angles on a straight line)

O lvl E Maths: 2D VectorsOABC is a parallelogram whose diagonals intersect at E; the diagonal AB produced to D and OD and CD are joined.

Given thatOA=a,OB=bandDB= 2(a-b), express, as simply as possible, in terms ofaandb(i)AB(ii)ED(iii)OD(iv)CD

Given also that X is the point on BD such thatOX= (8b- 5a), find the value of BX/BD.

If, in addition |a| = |b| = 3 and AOB = 60, find the numerical value of |b-a|.*************************

Answer:

(i)AB=AO+OB= -OA+OB= -a+b

(ii)EB= AB(since E is midpoint of AB, properties of a parallelogram)

ED=EB+BD=AB-DB= -a+b-2(a-b)= -5/2a+ 5/2b

(iii)OD=OB+BD=OB-DB=b- 2(a-b)= -2a+ 3b

(iv)OC=OA+AC=OA+OB(AC=OB)=a+b

CD=CO+OD= -OC+OD= -(a+b) + (-2a+ 3b)= -3a+ 2b

BD= -DB= 2(b-a)

BX=BO+OX= -b+ (8b- 5a)= (5b- 5a)= (5/3)(b-a)= (5/3) BD=(5/6)BD

Hence, BX/BD = 5/6

From (i),AB=b-aHence, |b-a| = length AB

From |a| = |b| = 3, OA = OB = 3

AB = OA + OB - 2 (OA)(OB) cos 60AB = 3 + 3 - 2(3)(3)()AB =9AB = 3

Hence,|b-a|= 3

O lvl E Maths: ProbabilityTwo bags each contains 91 balls. In each bag, there are 3 red balls, 1 white and the rest blue.

(i) One ball is drawn from the first bag. Find the probability that it is not blue.

(ii) One ball is drawn from the second bag. Find the probability that it is blue.

(iii) One ball is drawn from each bag. Find the probability that the two balls are(a) both blue,(b) each of a different colour.*************************

Answer:

In each bag, there are 3 red balls, 1 white and 87 blue.

(i) P(ball is not blue) =4/91

(ii) P(ball is blue) =87/91

(iii)(a) P(both balls are blue) = 87/91 * 87/91=7569/8281

(b) P(both balls same colour)= P(both red) + P(both white) + P(both blue)= 3/91 * 3/91 + 1/91 * 1/91 + 7569/91= 7579/8281

P(both balls are different colour) = 1 - P(both balls same colour)= 1 - 7579/8281= 702/8281

Suppose that 45% and 40% of Singaporeans approve and disapprove the idea of building a casino respectively. If three Singaporeans are interviewed at random,find the probability thata) all of them will either approve or disapprove of the idea

b) at least two of them will not approve of the idea

c) at most one of them will approve of the idea.

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Answer:

Let A = Approve, D = disapprove

(a) P(all of them have opinions) = 0.85 * 0.85 * 0.85 =4913/8000

(b) Let X = not disapproveP(X) = 1 - 0.4 = 0.6

P(at least 2 disapprove) = P(DDX) + P(DXD) + P(XDD) + P(DDD)= 3 (0.4 * 0.4 * 0.6) + 0.4 * 0.4 * 0.4=44/125

(c) Let NA = not approveP(NA) = 1 - 0.45 = 0.55

P(at most 1 approve) = P(all NA) + 3 P(A.NA.NA)= 0.55 * 0.55 * 0.55 + 3 (0.45 * 0.55 * 0.55)= 2299/4000

A student attends either a Mathematics class or an Additional mathematics class everyday. The probability of him bringing the wrong textbook is 3/5 if he attends the Mathematics class and 3/10 if he attends the Additional Mathematics class.

(a) If he attends Mathematics class on three consecutive days, find the probability that he will bring the wrong textbook(i) on all the three days(ii) on just one of the three days.

(b) If he is equally likely to attend the Mathematics class or the Additional Mathematics class, find the probability that he will bring the correct textbook to class on any given day.*************************

Answer:

(a)(i) P(wrong textbook on all three days)= P(1st day wrong textbook) * P(2nd day wrong textbook) * P(3rd day wrong textbook)= 3/5 * 3/5 * 3/5= 27/125

(ii) P(just one of three days wrong textbook)= P(1st day wrong, 2nd, 3rd day correct) + P(2nd day wrong, 1st, 3rd day correct) + P(3rd day wrong, 1st, 2nd day correct)= 3/5 * 2/5 * 2/5 + 2/5 * 3/5 * 2/5 + 2/5 * 2/5 * 3/5= 36/125

(b) P(correct textbook for Mathematics or for Additional Mathematics)= P(attend Mathematics and bring correct txtbook for Mathematics) + P(attend Additional Mathematics and bring correct txtbook for Additional Mathematics)= 1/2 * 2/5 + 1/2 * 7/10= 11/20

A certain brand of boxes of matches is advertised as having "average contents 50 matches". The probability that a box chosen at random will contain exactly 50 matches is 5/8.

(a) Calculate the probability that a box of matches chosen at random will not contain exactly 50 matches.

The probability that a box chosen at random will contain more than 50 matches is twice the probability that it will contain less than 50 matches.

(b) Calculate the probability that(i) one box chosen at random will contain at least 50 matches.(ii) of 2 boxes chosen at random, at least one will contain less than 50 matches.*************************

Answer:

(a) P(a box of matches chosen at random will not contain exactly 50 matches)= 1 - 5/8= 3/8

(b)The probability that a box chosen at random will contain more than 50 matches is twice the probability that it will contain less than 50 matches====> P(a box chosen at random will contain more than 50 matches) = 2/8====> P(a box chosen at random will contain less than 50 matches) = 1/8

(i) P(one box chosen at random will contain at least 50 matches)= P(a box chosen at random will contain more than 50 matches) + P((a box chosen at random will contain exactly 50 matches)= 2/8 + 5/8= 7/8

(ii) P(at least one will contain less than 50 matches)= 1 - P(both boxes contain at least 50 matches)= 1 - 7/8 * 7/8= 15/64

(a) An ordinary six-sided die numbered 1 to 6 is thrown. Write down the probability that the number shown on the die is a prime number.

A six-sided die numbered 1 to 6 and an eight sided die numbered 3 to 10 are thrown together. Giving each answer as a fraction in its lowest terms, find the probability that(i) the sum of the numbers is 10,(ii) the two numbers are not equal.

(b) A pupil has difficulty in waking up for school, and so, to wake himself, he sets 3 alarms to go off at the same time, as the noise from at least 2 alarms is necessary to wake him. Each alarm goes off independently.The probability that each alarm goes off is 0.7, 0.8 and 0.9 respectively. Find the probability that(i) all 3 go off,(ii) the pupil is awakened.*************************

Answer:

(a) Prime numbers are 2, 3, 5Hence, probability that it is a prime number = 3/6 =1/2

(i) P(sum of numbers is 10)= P(1,9) + P(2,8) + P(3,7) + P(4,6) + P(5,5) + P(6,4) => Summing all the possible combinations= 1/6 * 1/8 * 6 times= 1/8

(ii) P(two numbers are not equal)= 1 - P(two numbers are equal)= 1 - {P(3,3) + P(4,4) + P(5,5) + P(6,6)}= 1 - {1/6 * 1/8 * 4 times}= 1 - 1/12= 11/12

(b)(i) Probability that all 3 go off= 7/10 * 8/10 * 9/10= 63/125

(ii) P(Student is awakened)= P(at least 2 alarms goes off)= P(1,2, 3 don't go off) + P(2,3, 1 don't go off) + P(1,3, 2 don't go off) + P(all 3 go off) ==> P(which alarm goes off)= 0.7 * 0.8 * 0.1 + 0.8 * 0.9 * 0.3 + 0.7 * 0.9 * 0.2 + 63/125= 0.398 + 63/125= 451/500

A small box containing chocolates of which 3 contains hazelnuts, 7 contains almonds and 2 contain sultanas. Two chocolates are picked at random, one at a time, without replacement. Find the probability that(a) both contain hazelnuts,(b) one contains almonds and one contains hazelnuts,(c) both are without sultanas.*************************

Answer:

(a) Probability that both contain hazelnuts= probability of first chocolate containing hazelnutsandprobability of second chocolate containing hazelnuts= 3/12 * 2/11= 1/22

(b) Probability that one contains almonds and one contains hazelnuts= probability of first chocolate containing almondsandprobability of second chocolate containing hazelnutsORprobability of first chocolate containing hazelnutsandprobability of second chocolate containing almonds= 7/12 * 3/11 + 3/12 * 7/11= 7/44 + 7/44= 7/22

(c) Probability that both are without sultanas= probability of first chocolateNOTcontaining sultanasandprobability of second chocolateNOTcontaining sultanas= 10/12 * 9/11= 15/22

A certain brand of boxes of matches is advertised as having "average contents 50 matches". The probability that a box chosen at random will contain exactly 50 matches is 5/8.

(a) Calculate the probability that a box of matches chosen at random will not contain exactly 50 matches.

The probability that a box chosen at random will contain more than 50 matches is twice the probability that it will contain less than 50 matches.

(b) Calculate the probability that(i) one box chosen at random will contain at least 50 matches.(ii) of 2 boxes chosen at random, at least one will contain less than 50 matches.*************************

Answer:

(a) P(a box of matches chosen at random will not contain exactly 50 matches)= 1 - 5/8= 3/8

(b)The probability that a box chosen at random will contain more than 50 matches is twice the probability that it will contain less than 50 matches====> P(a box chosen at random will contain more than 50 matches) = 2/8====> P(a box chosen at random will contain less than 50 matches) = 1/8

(i) P(one box chosen at random will contain at least 50 matches)= P(a box chosen at random will contain more than 50 matches) + P((a box chosen at random will contain exactly 50 matches)= 2/8 + 5/8= 7/8

(ii) P(at least one will contain less than 50 matches)= 1 - P(both boxes contain at least 50 matches)= 1 - 7/8 * 7/8= 15/64