UK I NTERMEDIATE MATHEMATICAL OLYMPIAD

Size: px
Start display at page:

Download "UK I NTERMEDIATE MATHEMATICAL OLYMPIAD"

Transcription

1 UK I NTERMEITE MTHEMTIL OLYMPI ayley Question Papers and Solutions 2008 to 2010 Organised by the United Kingdom Mathematics Trust

2 i UKMT UKMT UKMT UK Intermediate Mathematical Olympiad 2008 to 2010 ayley Question Papers and Solutions Organised by the United Kingdom Mathematics Trust ontents ackground ii Rules and Guidelines paper paper paper solutions solutions solutions 18 UKMT 2011

3 ii ackground The Intermediate Mathematical Olympiad and Kangaroo (IMOK) are the follow-up competitions for pupils who do extremely well in the UKMT Intermediate Mathematical hallenge (about 1 in 200 are invited to take part). The IMOK was established in There are three written papers (ayley, Hamilton, Maclaurin) and two multiple-choice papers (the Pink and Grey Kangaroo). The written papers each take two hours and contain six questions. oth Kangaroo papers are one hour long and contain 25 questions. The ayley paper is for pupils in: Y9 or below (England and Wales); S2 or below (Scotland); School Year 10 or below (Northern Ireland).

4 1 The United Kingdom Mathematics Trust Intermediate Mathematical Olympiad and Kangaroo (IMOK) Olympiad ayley Paper ll candidates must be in School Year 9 or below (England and Wales), S2 or below (Scotland), or School Year 10 or below (Northern Ireland). RE THESE INSTRUTIONS REFULLY EFORE STRTING 1. Time allowed: 2 hours. 2. The use of calculators, protractors and squared paper is forbidden. Rulers and compasses may be used. 3. Solutions must be written neatly on 4 paper. Sheets must be STPLE together in the top left corner with the over Sheet on top. 4. Start each question on a fresh 4 sheet. You may wish to work in rough first, then set out your final solution with clear explanations and proofs. o not hand in rough work. 5. nswers must be FULLY SIMPLIFIE, and EXT. They may contain symbols such as π, fractions, or square roots, if appropriate, but NOT decimal approximations. 6. Give full written solutions, including mathematical reasons as to why your method is correct. Just stating an answer, even a correct one, will earn you very few marks; also, incomplete or poorly presented solutions will not receive full marks. 7. These problems are meant to be challenging! The earlier questions tend to be easier; the last two questions are the most demanding. o not hurry, but spend time working carefully on one question before attempting another. Try to finish whole questions even if you cannot do many: you will have done well if you hand in full solutions to two or more questions. O NOT OPEN THE PPER UNTIL INSTRUTE Y THE INVIGILTOR TO O SO! The United Kingdom Mathematics Trust is a Registered harity. Enquiries should be sent to: Maths hallenges Office, School of Maths Satellite, University of Leeds, Leeds, LS2 9JT. (Tel )

5 2 dvice to candidates o not hurry, but spend time working carefully on one question before attempting another. Try to finish whole questions even if you cannot do many. You will have done well if you hand in full solutions to two or more questions. nswers must be FULLY SIMPLIFIE, and EXT. They may contain symbols such as π, fractions, or square roots, if appropriate, but NOT decimal approximations. Give full written solutions, including mathematical reasons as to why your method is correct. Just stating an answer, even a correct one, will earn you very few marks. Incomplete or poorly presented solutions will not receive full marks. o not hand in rough work.

6 How many four-digit multiples of 9 consist of four different odd digits? 2. hexagon is made by cutting a small equilateral triangle from each corner of a larger equilateral triangle. The sides of the smaller triangles have lengths 1, 2 and 3 units. The lengths of the perimeters of the hexagon and the original triangle are in the ratio 5 : 7. What fraction of the area of the original triangle remains? 3. In the rectangle the midpoint of is M and : = 2 : 1. The point X is such that triangle MX is equilateral, with X and lying on opposite sides of the line M. Find the value of X. 4. The number N is the product of the first 99 positive integers. The number M is the product of the first 99 positive integers after each has been reversed. That is, for example, the reverse of 8 is 8; of 17 is 71; and of 20 is 02. Find the exact value of N M. 5. kite has sides and of length 25 cm and sides and of length 39 cm. The perpendicular distance from to is 24 cm. The perpendicular distance from to is h cm. Find the value of h. 39 cm 39 cm h cm 24 cm 25cm 25cm 6. regular tetrahedron has edges of length 2 units. The midpoint of the edge is M and the midpoint of the edge is N. Find the exact length of the segment MN.

7 n aquarium contains 280 tropical fish of various kinds. If 60 more clownfish were added to the aquarium, the proportion of clownfish would be doubled. How many clownfish are in the aquarium? 2. The boundary of the shaded figure consists of four semicircular arcs whose radii are all different. The centre of each arc lies on the line, which is 10 cm long. What is the length of the perimeter of the figure? 3. Two different rectangles are placed together, edge-to-edge, to form a large rectangle. 2 The length of the perimeter of the large rectangle is 3 of the total perimeter of the original two rectangles. Prove that the final rectangle is in fact a square. 4. In the rectangle, the side has length 2 and the side has length 1. Let the circle with centre and passing through meet at X. Find X (in degrees). 5. Two candles are the same height. The first takes 10 hours to burn completely whilst the second takes 8 hours to burn completely. oth candles are lit at midday. t what time is the height of the first candle twice the height of the second candle? 6. Teams,, and competed against each other once. The results table was as follows: Team Win raw Loss Goals for Goals against (a) Find (with proof) which team won in each of the six matches. (b) Find (with proof) the scores in each of the six matches.

8 The sum of three positive integers is 11 and the sum of the cubes of these numbers is 251. Find all such triples of numbers. 2. The diagram shows a square and an equilateral triangle E. The point F lies on so that E = EF. alculate the angle EF. E F 3. Find all possible solutions to the word sum on the right. Each letter stands for one of the digits 0 9 and has the same meaning each time it occurs. ifferent letters stand for different digits. No number starts with a zero. O + O E V E N 4. Walking at constant speeds, Eoin and his sister ngharad take 40 minutes and 60 minutes respectively to walk to the nearest town. Yesterday, Eoin left home 12 minutes after ngharad. How long was it before he caught up with her? 5. square sheet of paper is folded along FG, as shown, so that the corner is folded onto the midpoint M of. Prove that the sides of triangle GM have lengths in the ratio 3 : 4 : 5. M G F 6. qprime number is a positive integer which is the product of exactly two different primes, that is, one of the form q p, where q and p are prime and q p. What is the length of the longest possible sequence of consecutive integers all of which are qprime numbers?

9 Solutions 1. How many four-digit multiples of 9 consist of four different odd digits? First solution There are five odd digits: 1, 3, 5, 7 and 9. The sum of the four smallest odd digits is 16 and the sum of the four largest is 24. Hence the digit sum of any four-digit number with different odd digits lies between 16 and 24, inclusive. However, the sum of the digits of a multiple of 9 is also a multiple of 9, and the only multiple of 9 between 16 and 24 is 18. Hence the sum of the four digits is 18. Now = 18, so that the four digits can be 1, 3, 5 and 9. If 7 is one of the four digits then the sum of the other three is 11, which is impossible. So 7 cannot be one of the digits and therefore the four digits can only be 1, 3, 5 and 9. The number of arrangements of these four digits is = 24. Hence there are 24 four-digit multiples of 9 that consist of four different odd digits. Second solution The sum of all five odd digits is = 25. Subtracting 1, 3, 5, 7 and 9 in turn we get 24, 22, 20, 18 and 16, only one of which is a multiple of 9, namely 18 = Since the sum of the digits of a multiple of 9 is also a multiple of 9, it follows that the four digits can only be 1, 3, 5 and 9. The number of arrangements of these four digits is = 24. Hence there are 24 four-digit multiples of 9 that consist of four different odd digits. 2. hexagon is made by cutting a small equilateral triangle from each corner of a larger equilateral triangle. The sides of the smaller triangles have lengths 1, 2 and 3 units. The lengths of the perimeters of the hexagon and the original triangle are in the ratio 5 : 7. What fraction of the area of the original triangle remains? First solution Let the side length of the large equilateral triangle be x units; this triangle therefore has a perimeter of length 3x units. Now consider the hexagon, which has sides of lengths 1, x 3, 2, x 5, 3 and x 4 units. Hence the hexagon has perimeter length 3x 6 units. Since the ratio of the perimeter lengths of the hexagon and the large triangle is 5 : 7, we have 3x 6 3x Rearranging and solving for x we obtain x = 7. = 5 7. ( )

10 7 Now, in order to find the area of the large equilateral triangle, we determine the height units using Pythagoras' theorem: h 2 = 7 2 ( 7 2) 2 = 49 (1 1 4) = Hence h = = Therefore the area of the large equilateral triangle is = We may find the areas of the three small equilateral triangles in a similar way. These areas are 3 4, and The area of the hexagon is the area of the large equilateral triangle minus the areas of the three small equilateral triangles, that is, ( ) = Finally, the fraction of the original equilateral triangle remaining is = 5 7. Second solution Having established that the large triangle has sides of length 7 (equation (*) in the solution above), we may proceed as follows: The four equilateral triangles in the problem are similar. Now the ratio of the areas of similar figures is equal to the ratio of the squares of their sides. Hence the four triangles have areas in the ratio 1 2 : 2 2 : 3 2 : 7 2 = 1 : 4 : 9 : 49. Hence the ratio of the areas of the hexagon and the large triangle is 49 ( ) : 49 = 35 : 49 = 5 : 7. This may be illustrated by dividing the large triangle into 49 small triangles, as shown. h Note: The observant reader will have noticed that the answer to this problem is surprising: the ratio of the areas is the same as the ratio of the perimeters. There is no reason to expect this to happen.

11 8 3. In the rectangle the midpoint of is M and : = 2 : 1. The point X is such that triangle MX is equilateral, with X and lying on opposite sides of the line M. Find the value of X. Solution The key to this solution is to draw M and consider triangle MX. We are given that is a rectangle, so that = and M = 90 = M. We are also given that = 2 and that M is the midpoint of. Therefore = M = M =. X It follows that triangles M and M are congruent (SS) and we deduce that M = M. ut triangle MX is equilateral, so MX = M and hence MX = M. In other words, triangle MX is isosceles. Now consider the angles at M. 1. Triangle M is right-angled with M = 90. It is also isosceles, so M = M = 45, since the angle sum is Similarly, from triangle M, M = Finally, because triangle MX is equilateral, MX = 60. Hence M MX = = 30 since angles on a straight line add up to 180. Lastly, we consider the angles at. We know that triangle XM is isosceles and that MX = 30. Hence each base angle 1 is 2 ( ) = 75 ; in particular, XM = 75. lso, is a rectangle, so = 90, and triangle M is right-angled and isosceles, so M = 45. Therefore M = M = = 45. We can now calculate the value of X. We have X = XM M = = 30.

12 9 4. The number N is the product of the first 99 positive integers. The number M is the product of the first 99 positive integers after each has been reversed. That is, for example, the reverse of 8 is 8; of 17 is 71; and of 20 is 02. Find the exact value of N M. First solution From the given definition we have N = (1 2 9) 10 ( ) (91 99), which rearranges to lso N = (1 2 9) ( ) (91 99) ( ). M = (1 2 9) 01 ( ) (19 99) which rearranges to M = (1 2 9) ( ) (91 99) ( ) = (1 2 9) ( ) (91 99) (1 2 9). omparing these arrangements for M and N, we see that M has the same terms as N except that the product is replaced by the product Thus when we divide N by M all the common terms cancel and we are left with N M = = Second solution We may place the numbers from 1 to 99 into three categories, determined by how they are transformed when they are reversed: 1. single digit numbers a are unchanged; 2. a two-digit number ab, where neither a nor b is zero, is transformed to the twodigit number ba ; and 3. a multiple of 10 such as a0 is transformed to 0a = a, a single-digit number. Thus there is a correspondence between the factors in N and M, as shown in the table: N M a a aa aa ab and ba ba and ab a0 a Single-digit numbers are unchanged; two-digit numbers with a repeated digit are unchanged; pairs of two-digit numbers, with different digits and neither digit zero, are unchanged as a pair; the multiples of 10 in N are replaced by single-digit numbers in M. Thus when we divide N by M all the identical factors cancel and we are left with N M = = 10 9.

13 10 5. kite has sides and of length 25 cm and sides and of length 39 cm. The perpendicular distance from to is 24 cm. The perpendicular distance from to is h cm. Find the value of h. 39 cm 39 cm h cm 24cm 25 cm 25 cm First solution s shown in the figure below, let the perpendicular from to the line meet the line at the point X; let the perpendicular from to the line meet the line at the point Y and let the distance Y be y cm. 39 y 39 Y h y X 25 onsidering triangle X and using Pythagoras' Theorem we obtain X = = 7 cm. Similarly, from triangle X we have = cm (25 7) 2 cm = 30 cm. ( ) Now from triangles Y and Y, again by Pythagoras' theorem, we deduce that h 2 + (39 y) 2 = 39 2 and h 2 + y 2 = (1) Subtract to get (39 y) 2 y 2 = , which simplifies to 78y = 900, 150 so that y =. 13 Finally, by substituting in equation (1), we find h = Second solution nother solution uses the length of obtained in (*) above to find the area of isosceles triangle. Once the area is known the value of the height h may be found from area = h. an you see how to find the area of triangle and so complete the solution? Note: Triangle Y is a 5, 12, 13 triangle.

14 6. regular tetrahedron has edges of length 2 units. The midpoint of the edge is M and the midpoint of the edge is N. Find the exact length of the segment MN. M 11 First solution We make use of the following result. Theorem (Median of isosceles triangle): The line joining the apex to the midpoint of the base of an isosceles triangle is perpendicular to the base. That is, in the following figure, PSR = 90. N R pplying the theorem to triangle, we find that M = 90. Similarly, in triangle, M = 90. Now applying Pythagoras' theorem to the triangles M and M we get M 2 = 2 M 2 = and M 2 = 2 M 2 = Hence M = 3 and M = 3, so triangle M is isosceles. Now apply the theorem to triangle M to obtain NM = 90. Then by Pythagoras' theorem in triangle NM Therefore MN = 2. MN 2 = M 2 N 2 = 3 1. Second solution tetrahedron may be formed by joining face diagonals of a cube, as shown below. Since the faces of the cube are congruent squares the face diagonals have equal length and so the tetrahedron is regular. Now M and N are midpoints of opposite edges of the tetrahedron. Therefore they are midpoints of opposite face diagonals of the cube, that is, centres of opposite faces of the cube. Hence MN = R. Letting the sides of the cube have length a, from Pythagoras' theorem in triangle R we get P so that 2 = R 2 + R = a 2 + a 2 = 2a 2. Hence a = 2 and therefore MN = 2. S M P N S Q R Q

15 Solutions 1. n aquarium contains 280 tropical fish of various kinds. If 60 more clownfish were added to the aquarium, the proportion of clownfish would be doubled. How many clownfish are in the aquarium? Solution Let there be x clownfish in the aquarium. If 60 clownfish are added there are x + 60 clownfish and 340 tropical fish in total. Since the proportion of clownfish is then doubled, we have 2 x 280 = x Multiplying both sides by 20 we get and hence It follows that x 7 = x x = 7 (x + 60). x = 42 and thus there are 42 clownfish in the aquarium. 2. The boundary of the shaded figure consists of four semicircular arcs whose radii are all different. The centre of each arc lies on the line, which is 10 cm long. What is the length of the perimeter of the figure? Solution The centre of the large semicircular arc lies on, so we know that is a diameter of the large semicircle. ut is 10 cm long, so the radius of the large semicircle is 5 cm. Let the radii of the other three semicircles be r 1 cm, r 2 cm and r 3 cm. The centres of these arcs also lie on, so the sum of their diameters is equal to the length of. It follows that 2r 1 + 2r 2 + 2r 3 = 10 and hence r 1 + r 2 + r 3 = 5. Now the lengths, in cm, of the semicircular arcs are 5π, πr 1, πr 2 and πr 3. Therefore the perimeter of the figure has length, in cm, 5π + πr 1 + πr 2 + πr 3 = π (5 + r 1 + r 2 + r 3 ) Hence the perimeter of the figure has length = π (5 + 5) = 10π. 10π cm.

16 3. Two different rectangles are placed together, edge-to-edge, to form a large rectangle. 2 The length of the perimeter of the large rectangle is 3 of the total perimeter of the original two rectangles. 13 Prove that the final rectangle is in fact a square. First solution Since the smaller rectangles are placed together edge-to-edge, they have a side length in common. Let this side have length y and let the other sides have lengths x 1 and x 2 as shown. y x 1 x 2 The perimeters of the smaller rectangles are 2x 1 + 2y and 2x 2 + 2y, so the total perimeter of the two smaller rectangles is 2x 1 + 2x 2 + 4y. The perimeter of the large rectangle is 2 (x 1 + x 2 ) + 2y = 2x 1 + 2x 2 + 2y. 2 We are given that the length of the perimeter of the large rectangle is 3 of the total perimeter of the two original rectangles. Hence we may form the equation 2x 1 + 2x 2 + 2y = 2 3 (2x 1 + 2x 2 + 4y). We may simplify this equation by multiplying both sides by 3 and expanding the brackets, to obtain which simplifies to 6x 1 + 6x 2 + 6y = 4x 1 + 4x 2 + 8y, x 1 + x 2 = y. This means that the length and width of the large rectangle are the same. In other words, the rectangle is actually a square. Second solution The total perimeter length P of the original two rectangles is equal to the perimeter length of the large rectangle added to the lengths of the two edges which are joined together. 2 ut the perimeter length of the large rectangle is 3 P and hence the two edges which are 1 joined together have total length 3P. However, the two edges which are joined together are parallel to two sides of the large rectangle and have the same length as them. Hence these two sides of the large rectangle 1 have total length 3 P. Since the perimeter length of the large rectangle is 2 3 P, the other two sides of the large 1 rectangle also have total length 3 P. It follows that all the sides of the rectangle are equal in length, in other words, the rectangle is a square.

17 14 4. In the rectangle, the side has length 2 and the side has length 1. Let the circle with centre and passing through meet at X. Find X (in degrees). Solution We begin with a diagram showing the information given in the question. We have used the fact that is a rectangle, so that = = 1 and = = 2, and angles, and are right angles. 2 1 X Since X and are both radii of the circle, X also has length 1. This means that triangle X is isosceles and so X = X. Furthermore, since is a right angle, X and X are both equal to 45. From the fact that is a right angle, it follows that X = = 45. We may use Pythagoras' theorem in triangle X to obtain X 2 = X = = 2 and so X = 2. We are given that also has length 2 and so triangle X is isosceles. This means that X and X are equal, and so each is equal to ( ) 2 = Lastly, we use the fact that is a right angle to conclude that X = =

18 15 5. Two candles are the same height. The first takes 10 hours to burn completely whilst the second takes 8 hours to burn completely. oth candles are lit at midday. t what time is the height of the first candle twice the height of the second candle? Solution h Let the initial height of each candle be h cm. In one hour the first candle will burn 10 cm h and the second candle will burn cm. Thus in hours, the candles will burn 8 t ht 10 cm and ht 8 cm, respectively. If both candles are lit at midday, then t hours after midday the heights of the first and second candles will be ( h 10) ht ( cm h ht 8 ) and cm, respectively. We are asked to find the time at which the height of the first candle is twice the height of the second candle. We therefore need to find the value of t such that h ht 10 ( = 2 h ht 8 ). We may divide every term by h, since we know that h is not zero, and expand the brackets to obtain the equation 1 t 10 = 2 t 4. Multiplying both sides by 20, we get and so 20 2t = 40 5t, t = 20 3 = Hence the height of the first candle is twice that of the second after 6 hours and 40 minutes, in other words, this happens at 18:40.

19 16 6. Teams,, and competed against each other once. The results table was as follows: Team Win raw Loss Goals for Goals against (a) Find (with proof) which team won in each of the six matches. (b) Find (with proof) the scores in each of the six matches. Solution (a) Team won all three games and so beat teams, and. Of the three games that team played, the one that was lost can only have been against team. Therefore team drew against teams and. If we consider the three games that team played, the game against team was lost, the game against team was a draw and so the remaining game, that team won, was against team. In summary: beat, beat, beat ; drew with, beat ; and drew with. (b) onsider the following table in which the rows give the number of goals scored for each team and the columns give the number of goals against each team. Goals against ll z Goals x t 2 for z x y 5 y 3 ll We have let the number of goals scored by team against team be x, so that the number of goals scored by team against team is also x, since their match was a draw. Similarly, we have let the number of goals scored by team against team be y, so that this is also the number scored by team against team. Furthermore, we have let the number of goals scored by team against team be z, so that the number of goals scored by team against team is z + 1 since the difference between the number of goals scored and conceded by team is 1. Finally, we have let the number of goals scored by team against team be t. Then t is at least 1 since team beat team.

20 17 We observe that the row for now means that x + y + z = 5 (which agrees with the column for ). From the column for we see that z is at most 1, since the total in that column is 1. Similarly, from the row for, we see that y is at most 3, and from the row for we see that x is at most 1 since t is at least 1. ut we have x + y + z = 5, so that the only possibilities are x = 1, y = 3 and z = 1. It follows that t = 1. Therefore the table is: Goals against ll 2 5 Goals for ll We may now complete the table by, for example, first noting that all other entries in the column for are 0, and then filling in the rows from the bottom. Goals against ll Goals for ll In summary, the scores in each match were as follows: beat 1 0 beat 2 1 beat 2 0 drew with 1 1 beat 1 0 drew with 3 3

21 Solutions 1 The sum of three positive integers is 11 and the sum of the cubes of these numbers is 251. Find all such triples of numbers. Solution Let us calculate the first few cubes in order to see what the possibilities are: 1 3 = 1, 2 3 = 8, 3 3 = 27, 4 3 = 64, 5 3 = 125, 6 3 = 216 and 7 3 = 343. ( ) The sum of the cubes of the positive integers is 251, which is less than 343, hence none of the integers is greater than 6. Now 3 = > 64 = 4 3, therefore at least one of the integers is 5 or more. 251 If one of the integers is 6, then the other two cubes add up to = = 35. From (*) above, = = 35 is the only possibility. lso, = 11 so that 6, 3 and 2 is a possible triple of numbers. If one of the integers is 5, then the other two cubes add up to = = 126. From (*) above = = 126 is the only possibility. lso, = 11 so that 5, 5 and 1 is a possible triple of numbers. Hence 2, 3, 6 and 1, 5, 5 are the triples of numbers satisfying the given conditions. 2 The diagram shows a square and an equilateral triangle E. The point F lies on so that E = EF. alculate the angle EF. E F Solution The diagram seems to include several isosceles triangles, and we solve the problem by proving this is the case. For example, since the square and the equilateral triangle E share the side, all their sides are the same length. That means the triangle E is isosceles. E 75 F Now, angle E is = 30 (since it is the difference between the interior angle of a square and the 30 interior angle of an equilateral triangle). Hence angles 1 E and E are each 2 ( ) = 75, because 60 they are the base angles of an isosceles triangle. We are also given that triangle EF is isosceles. Since we have worked out that angle FE = 75, we deduce that angle EF = 180 (2 75 ) = 30. Finally, we find that angle EF = E EF = = 45.

22 3 Find all possible solutions to the word sum on the right. Each letter stands for one of the digits 0 9 and has the same meaning each time it occurs. ifferent letters stand for different digits. No number starts with a zero. O + O E V E N Solution Firstly, it is clear that the three-digit number O lies between 100 and 999. Therefore, since EVEN = 2 O, we have 200 < EVEN < Hence the first digit E of EVEN is 1 since it is a four-digit number. 19 We are left with the following problem: O + O 1 V 1 N Now the same numbers are added in the tens and units columns, but N 1, otherwise N and E would be equal. The only way for different totals to occur in these columns is for there to be a carry to the tens column, and the greatest possible carry is 1, so that N = 0. There are two possible digits that give N = 0, namely 0 and 5. ut 0 is already taken as the value of N, so that = 5. The problem is thus: O O V 1 0 Now, the digit O has to be big enough to produce a carry, but cannot be 5, which is already taken as the value of. So the possibilities are but the second and fourth of these are not allowed since V repeats a digit used for another letter. We are left with the two possibilities and it is clear that both of these work Walking at constant speeds, Eoin and his sister ngharad take 40 minutes and 60 minutes respectively to walk to the nearest town. Yesterday, Eoin left home 12 minutes after ngharad. How long was it before he caught up with her? Solution Let the distance from home to town be km. Now in every minute Eoin travels onefortieth of the way to town: that is, a distance of. So after minutes, he has 40 km t travelled a distance t 40 km.

23 20 Similarly, in every minute ngharad travels one-sixtieth of the way to town: that is, a distance of. ut she has had 12 minutes extra walking time. So after Eoin has 60 km been walking for t minutes, she has been walking for t + 12 minutes and so has travelled a distance (t + 12) km. 60 We are asked how long Eoin has been walking when they meet. They meet when they have travelled equal distances, which is when t (t + 12) = We cancel the from each side and multiply both sides by 120 to obtain Simplifying, we get 120 t 40 = 120 t t = 2 (t + 12), which we solve to give t = 24. Thus Eoin catches up with ngharad after he has walked for 24 minutes. 5 square sheet of paper is folded along FG, as shown, so that the corner is folded onto the midpoint M of. Prove that the sides of triangle GM have lengths in the ratio 3 : 4 : 5. M G F Solution This problem does not give us units, and so we choose them so that the side length of the square is 2s. Since M is the midpoint of, we have M = s. Then we define x = G. Since = 2s, G = 2s x. ut, as GM is the image of G after folding, GM = 2s x too. s M s 2s x x G 2s x F Now Pythagoras' theorem for triangle MG gives us s 2 + x 2 = (2s x) 2. We multiply out to get s 2 + x 2 = 4s 2 4sx + x 2.

24 Eliminating the which has the solution x = 3 4s. x 2 terms and dividing by s (which is not zero), we obtain s = 4s 4x, 21 Thus the triangle GM has sides of length x = 3 4 s, s and 2s x = 5 4 s. Multiplying all the sides by 4, we get 3s, 4s and 5s, so the side lengths are in the ratio 3 : 4 : 5, as required. 6 qprime number is a positive integer which is the product of exactly two different primes, that is, one of the form q p, where q and p are prime and q p. What is the length of the longest possible sequence of consecutive integers all of which are qprime numbers? Solution To help to understand this problem, it is natural to test the first few numbers to see which small numbers are qprime, and which are not: 1 is not a qprime since it has no prime factors. 2 and 3 are not qprimes since they are prime. 4 is not qprime since it is is not qprime, since it is prime. 6 = 2 3 is the first qprime number. 7 is not qprime. 8 is not qprime, since it is is not, since it is = 2 5 is another qprime. 11 is not. 12 is not, since it is Of course, we cannot prove a general result just by continuing the list, but it can guide us to a proof, such as the one that follows. We note that no multiple of 4 is ever qprime, since a multiple of 4 is a multiple of 2 2. This means that a string of consecutive qprime numbers can be of length at most three, because any sequence of four or more consecutive integers includes a multiple of 4. We are therefore led to ask whether any strings of three consecutive qprime numbers exist. We have looked as far as 12 and not found any, but we will continue searching, using the fact that none of the numbers is a multiple of 4: For (13, 14, 15), the number 13 is prime and so not qprime. For (17, 18, 19), 17 is not qprime (nor are the others). For (21, 22, 23), 23 is not qprime. For (25, 26, 27), 25 is not qprime (nor is 27). For (29, 30, 31), 29 is not qprime (nor are the others). For (33, 34, 35), all three are qprime (being 3 11, 2 17 and 5 7). So we have found a sequence of three consecutive qprimes, and have also proved that no sequence of four (or more) consecutive qprimes exists. Thus the longest possible sequence of consecutive integers all of which are qprime numbers has length 3.

SENIOR KANGAROO MATHEMATICAL CHALLENGE. Friday 2nd December Organised by the United Kingdom Mathematics Trust

SENIOR KANGAROO MATHEMATICAL CHALLENGE. Friday 2nd December Organised by the United Kingdom Mathematics Trust UKMT UKMT UKMT SENIOR KNGROO MTHEMTIL HLLENGE Friday 2nd December 2011 Organised by the United Kingdom Mathematics Trust The Senior Kangaroo paper allows students in the UK to test themselves on questions

More information

Junior Mathematical Olympiad

Junior Mathematical Olympiad UKMT UKMT UKMT United Kingdom Mathematics Trust Junior Mathematical Olympiad Organised by the United Kingdom Mathematics Trust s These are polished solutions and do not illustrate the process of exploration

More information

SENIOR KANGAROO MATHEMATICAL CHALLENGE. Friday 29th November Organised by the United Kingdom Mathematics Trust

SENIOR KANGAROO MATHEMATICAL CHALLENGE. Friday 29th November Organised by the United Kingdom Mathematics Trust SENIOR KNGROO MTHEMTIL HLLENGE Friday 29th November 203 Organised by the United Kingdom Mathematics Trust The Senior Kangaroo paper allows students in the UK to test themselves on questions set for the

More information

UKMT UKMT UKMT. IMOK Olympiad. Thursday 16th March Organised by the United Kingdom Mathematics Trust. Solutions

UKMT UKMT UKMT. IMOK Olympiad. Thursday 16th March Organised by the United Kingdom Mathematics Trust. Solutions UKMT UKMT UKMT IMOK Olympiad Thursday 16th March 2017 Organised by the United Kingdom Mathematics Trust s These are polished solutions and do not illustrate the process of failed ideas and rough work by

More information

UK I NTERMEDIATE MATHEMATICAL OLYMPIAD

UK I NTERMEDIATE MATHEMATICAL OLYMPIAD UK I NTERMEDIATE MATHEMATICAL OLYMPIAD Maclaurin Question Papers and Solutions 008 to 00 Organised by the United Kingdom Mathematics Trust i UKMT UKMT UKMT UK Intermediate Mathematical Olympiad 008 to

More information

Mathematical Olympiad for Girls

Mathematical Olympiad for Girls UKMT UKMT UKMT Mathematical Olympiad for Girls Organised by the United Kingdom Mathematics Trust These are polished solutions and do not illustrate the process of failed ideas and rough work by which candidates

More information

2009 Math Olympics Level II

2009 Math Olympics Level II Saginaw Valley State University 009 Math Olympics Level II 1. f x) is a degree three monic polynomial leading coefficient is 1) such that f 0) = 3, f 1) = 5 and f ) = 11. What is f 5)? a) 7 b) 113 c) 16

More information

2016 Pascal Contest (Grade 9)

2016 Pascal Contest (Grade 9) The CENTRE for EDUCATION in MATHEMATICS and COMPUTING cemc.uwaterloo.ca 06 Pascal Contest (Grade 9) Wednesday, February, 06 (in North America and South America) Thursday, February, 06 (outside of North

More information

2003 Solutions Pascal Contest (Grade 9)

2003 Solutions Pascal Contest (Grade 9) Canadian Mathematics Competition An activity of The Centre for Education in Mathematics and Computing, University of Waterloo, Waterloo, Ontario 00 Solutions Pascal Contest (Grade 9) for The CENTRE for

More information

Mathematical Olympiad for Girls

Mathematical Olympiad for Girls UKMT UKMT UKMT United Kingdom Mathematics Trust Mathematical Olympiad for Girls Organised by the United Kingdom Mathematics Trust These are polished solutions and do not illustrate the process of failed

More information

48th AHSME If a is 50% larger than c, and b is 25% larger than c, then a is what percent larger than b?

48th AHSME If a is 50% larger than c, and b is 25% larger than c, then a is what percent larger than b? 48th HSME 1997 2 1 If a and b are digits for which then a + b = 2 a b 6 9 9 2 9 8 9 () () 4 () 7 () 9 (E) 12 2 The adjacent sides of the decagon shown meet at right angles What is its perimeter? () 22

More information

UK Junior Mathematical Olympiad 2010

UK Junior Mathematical Olympiad 2010 UK Junior Mathematical Olympiad 010 Organised by The United Kingdom Mathematics Trust Tuesday 15th June 010 RULES AND GUIDELINES : READ THESE INSTRUCTIONS CAREFULLY BEFORE STARTING 1. Time allowed: hours..

More information

number. However, unlike , three of the digits of N are 3, 4 and 5, and N is a multiple of 6.

number. However, unlike , three of the digits of N are 3, 4 and 5, and N is a multiple of 6. C1. The positive integer N has six digits in increasing order. For example, 124 689 is such a number. However, unlike 124 689, three of the digits of N are 3, 4 and 5, and N is a multiple of 6. How many

More information

3. A square has 4 sides, so S = 4. A pentagon has 5 vertices, so P = 5. Hence, S + P = 9. = = 5 3.

3. A square has 4 sides, so S = 4. A pentagon has 5 vertices, so P = 5. Hence, S + P = 9. = = 5 3. JHMMC 01 Grade Solutions October 1, 01 1. By counting, there are 7 words in this question.. (, 1, ) = 1 + 1 + = 9 + 1 + =.. A square has sides, so S =. A pentagon has vertices, so P =. Hence, S + P = 9..

More information

2005 Cayley Contest. Solutions

2005 Cayley Contest. Solutions anadian Mathematics ompetition n activity of the entre for Education in Mathematics and omputing, University of Waterloo, Waterloo, Ontario 005 ayley ontest (Grade 10) Wednesday, February 3, 005 Solutions

More information

Preliminary chapter: Review of previous coursework. Objectives

Preliminary chapter: Review of previous coursework. Objectives Preliminary chapter: Review of previous coursework Objectives By the end of this chapter the student should be able to recall, from Books 1 and 2 of New General Mathematics, the facts and methods that

More information

2008 Cayley Contest. Solutions

2008 Cayley Contest. Solutions Canadian Mathematics Competition An activity of the Centre for Education in Mathematics and Computing, University of Waterloo, Waterloo, Ontario 008 Cayley Contest (Grade 10) Tuesday, February 19, 008

More information

2018 Best Student Exam Solutions Texas A&M High School Students Contest October 20, 2018

2018 Best Student Exam Solutions Texas A&M High School Students Contest October 20, 2018 08 Best Student Exam Solutions Texas A&M High School Students Contest October 0, 08. You purchase a stock and later sell it for $44 per share. When you do, you notice that the percent increase was the

More information

2009 Math Olympics Level II Solutions

2009 Math Olympics Level II Solutions Saginaw Valley State University 009 Math Olympics Level II Solutions 1. f (x) is a degree three monic polynomial (leading coefficient is 1) such that f (0) 3, f (1) 5 and f () 11. What is f (5)? (a) 7

More information

GCSE Mathematics Non Calculator Higher Tier Free Practice Set 6 1 hour 45 minutes ANSWERS. Marks shown in brackets for each question (2) A* A B C D E

GCSE Mathematics Non Calculator Higher Tier Free Practice Set 6 1 hour 45 minutes ANSWERS. Marks shown in brackets for each question (2) A* A B C D E MathsMadeEasy GCSE Mathematics Non Calculator Higher Tier Free Practice Set 6 1 hour 45 minutes ANSWERS Marks shown in brackets for each question A* A B C D E 88 75 60 45 25 15 3 Legend used in answers

More information

Mathematics Higher Tier, June /2H (Paper 2, calculator)

Mathematics Higher Tier, June /2H (Paper 2, calculator) Link to past paper on AQA website: www.aqa.org.uk The associated question paper is available to download freely from the AQA website. To navigate around the website, choose QUALIFICATIONS, GCSE, MATHS,

More information

Candidate Name Centre Number Candidate Number MATHEMATICS UNIT 1: NON-CALCULATOR INTERMEDIATE TIER SPECIMEN PAPER SUMMER 2017

Candidate Name Centre Number Candidate Number MATHEMATICS UNIT 1: NON-CALCULATOR INTERMEDIATE TIER SPECIMEN PAPER SUMMER 2017 GCSE MATHEMATICS Specimen Assessment Materials 27 Candidate Name Centre Number Candidate Number 0 GCSE MATHEMATICS UNIT 1: NON-CALCULATOR INTERMEDIATE TIER SPECIMEN PAPER SUMMER 2017 1 HOUR 45 MINUTES

More information

(b) [1] (c) [1]

(b) [1] (c) [1] GCSE MATHEMATICS Specimen Assessment Materials 29 1. Calculate the following. (a) 5 2 2 3 [2] (b) 0 3 0 6 (c) 8 7 5 25 (d) 7 1 8 4 [2] GCSE MATHEMATICS Specimen Assessment Materials 30 2. (a) Write down

More information

Please read these instructions carefully, but do not open the question paper until you are told that you may do so.

Please read these instructions carefully, but do not open the question paper until you are told that you may do so. TST OF MTHMTIS FOR UNIVRSITY MISSION PPR SPIMN Time: 75 minutes dditional Materials: nswer sheet INSTRUTIONS TO NITS Please read these instructions carefully, but do not open the question paper until you

More information

Q 1 Find the square root of 729. 6. Squares and Square Roots Q 2 Fill in the blank using the given pattern. 7 2 = 49 67 2 = 4489 667 2 = 444889 6667 2 = Q 3 Without adding find the sum of 1 + 3 + 5 + 7

More information

2017 OHMIO Individual Competition

2017 OHMIO Individual Competition 2017 OHMIO Individual Competition 1. On a winter hike with friends (all of whom were wearing either a scarlet or gray hat), I saw twice as many scarlet hats as gray. That s silly, said a friend. I see

More information

Plane geometry Circles: Problems with some Solutions

Plane geometry Circles: Problems with some Solutions The University of Western ustralia SHL F MTHMTIS & STTISTIS UW MY FR YUNG MTHMTIINS Plane geometry ircles: Problems with some Solutions 1. Prove that for any triangle, the perpendicular bisectors of the

More information

1999 Solutions Fermat Contest (Grade 11)

1999 Solutions Fermat Contest (Grade 11) Canadian Mathematics Competition n activity of The Centre for Education in Mathematics and Computing, University of Waterloo, Waterloo, Ontario 999 s Fermat Contest (Grade ) for the wards 999 Waterloo

More information

State Math Contest (Senior)

State Math Contest (Senior) Name: Student I: State Math ontest (Senior) Instructions: o not turn this page until your proctor tells you. nter your name, grade, and school information following the instructions given by your proctor.

More information

221 MATH REFRESHER session 3 SAT2015_P06.indd 221 4/23/14 11:39 AM

221 MATH REFRESHER session 3 SAT2015_P06.indd 221 4/23/14 11:39 AM Math Refresher Session 3 1 Area, Perimeter, and Volume Problems Area, Perimeter, and Volume 301. Formula Problems. Here, you are given certain data about one or more geometric figures, and you are asked

More information

Save My Exams! The Home of Revision For more awesome GCSE and A level resources, visit us at 4H June 2017.

Save My Exams! The Home of Revision For more awesome GCSE and A level resources, visit us at   4H June 2017. 4H June 2017 Model Answers Level Subject Exam Board Paper Booklet IGCSE Maths Edexcel June 2017 4H Model Answers Time Allowed: Score: Percentage: 120 minutes / 100 /100 Grade Boundaries: 9 8 7 6 5 4 3

More information

BC Exam Solutions Texas A&M High School Math Contest October 24, p(1) = b + 2 = 3 = b = 5.

BC Exam Solutions Texas A&M High School Math Contest October 24, p(1) = b + 2 = 3 = b = 5. C Exam Solutions Texas &M High School Math Contest October 4, 01 ll answers must be simplified, and If units are involved, be sure to include them. 1. p(x) = x + ax + bx + c has three roots, λ i, with

More information

2. In the diagram, PQ and TS are parallel. Prove that a + b + c = 360. We describe just two of the many different methods that are possible. Method 1

2. In the diagram, PQ and TS are parallel. Prove that a + b + c = 360. We describe just two of the many different methods that are possible. Method 1 s to the Olympiad Cayley Paper 1. The digits p, q, r, s and t are all different. What is the smallest five-digit integer pqrst that is divisible by 1, 2, 3, 4 and 5? Note that all five-digit integers are

More information

(A) 20% (B) 25% (C) 30% (D) % (E) 50%

(A) 20% (B) 25% (C) 30% (D) % (E) 50% ACT 2017 Name Date 1. The population of Green Valley, the largest suburb of Happyville, is 50% of the rest of the population of Happyville. The population of Green Valley is what percent of the entire

More information

BRITISH COLUMBIA SECONDARY SCHOOL MATHEMATICS CONTEST,

BRITISH COLUMBIA SECONDARY SCHOOL MATHEMATICS CONTEST, BRITISH COLUMBIA SECONDARY SCHOOL MATHEMATICS CONTEST, 014 Solutions Junior Preliminary 1. Rearrange the sum as (014 + 01 + 010 + + ) (013 + 011 + 009 + + 1) = (014 013) + (01 011) + + ( 1) = 1 + 1 + +

More information

8 LEVELS 5 7 PAPER. Paper 2. Year 8 mathematics test. Calculator allowed. First name. Last name. Class. Date YEAR

8 LEVELS 5 7 PAPER. Paper 2. Year 8 mathematics test. Calculator allowed. First name. Last name. Class. Date YEAR Ma YEAR 8 LEVELS 5 7 PAPER 2 Year 8 mathematics test Paper 2 Calculator allowed Please read this page, but do not open your booklet until your teacher tells you to start. Write your details in the spaces

More information

MATH CIRCLE Session # 2, 9/29/2018

MATH CIRCLE Session # 2, 9/29/2018 MATH CIRCLE Session # 2, 9/29/2018 SOLUTIONS 1. The n-queens Problem. You do NOT need to know how to play chess to work this problem! This is a classical problem; to look it up today on the internet would

More information

1997 Solutions Cayley Contest(Grade 10)

1997 Solutions Cayley Contest(Grade 10) Canadian Mathematics Competition n activity of The Centre for Education in Mathematics and Computing, University of Waterloo, Waterloo, Ontario 199 Solutions Cayley Contest(Grade 10) for the wards 199

More information

THE KENNESAW STATE UNIVERSITY HIGH SCHOOL MATHEMATICS COMPETITION PART I MULTIPLE CHOICE (A) 0 (B) 1 (C) 2 (D) 3 (E) 4

THE KENNESAW STATE UNIVERSITY HIGH SCHOOL MATHEMATICS COMPETITION PART I MULTIPLE CHOICE (A) 0 (B) 1 (C) 2 (D) 3 (E) 4 THE 007 008 KENNESAW STATE UNIVERSITY HIGH SCHOOL MATHEMATICS COMPETITION PART I MULTIPLE CHOICE For each of the following questions, carefully blacken the appropriate box on the answer sheet with a #

More information

Math Wrangle Practice Problems

Math Wrangle Practice Problems Math Wrangle Practice Problems American Mathematics Competitions December 22, 2011 ((3!)!)! 1. Given that, = k. n!, where k and n are positive integers and n 3. is as large as possible, find k + n. 2.

More information

The CENTRE for EDUCATION in MATHEMATICS and COMPUTING cemc.uwaterloo.ca Cayley Contest. (Grade 10) Tuesday, February 28, 2017

The CENTRE for EDUCATION in MATHEMATICS and COMPUTING cemc.uwaterloo.ca Cayley Contest. (Grade 10) Tuesday, February 28, 2017 The CENTRE for EDUCATION in MATHEMATICS and COMPUTING cemc.uwaterloo.ca 2017 Cayley Contest (Grade 10) Tuesday, February 28, 2017 (in North America and South America) Wednesday, March 1, 2017 (outside

More information

2015 Canadian Intermediate Mathematics Contest

2015 Canadian Intermediate Mathematics Contest The CENTRE for EDUCATION in MATHEMATICS and COMPUTING cemc.uwaterloo.ca 015 Canadian Intermediate Mathematics Contest Wednesday, November 5, 015 (in North America and South America) Thursday, November

More information

Math is Cool Masters

Math is Cool Masters Sponsored by: GENIE Industries 7 th Grade November 19, 2005 Individual Contest Express all answers as reduced fractions unless stated otherwise. Leave answers in terms of π where applicable. Do not round

More information

Math Contest, Fall 2017 BC EXAM , z =

Math Contest, Fall 2017 BC EXAM , z = Math Contest, Fall 017 BC EXAM 1. List x, y, z in order from smallest to largest fraction: x = 111110 111111, y = 1 3, z = 333331 333334 Consider 1 x = 1 111111, 1 y = thus 1 x > 1 z > 1 y, and so x

More information

Instructions. Do not open your test until instructed to do so!

Instructions. Do not open your test until instructed to do so! st Annual King s College Math Competition King s College welcomes you to this year s mathematics competition and to our campus. We wish you success in this competition and in your future studies. Instructions

More information

Western Australian Junior Mathematics Olympiad 2011

Western Australian Junior Mathematics Olympiad 2011 Western Australian Junior Mathematics Olympiad 2011 Individual Questions 100 minutes General instructions: Each solution in this part is a positive integer less than 100. No working is needed for Questions

More information

2010 Fermat Contest (Grade 11)

2010 Fermat Contest (Grade 11) Canadian Mathematics Competition An activity of the Centre for Education in Mathematics and Computing, University of Waterloo, Waterloo, Ontario 010 Fermat Contest (Grade 11) Thursday, February 5, 010

More information

GCSE Mathematics Non-Calculator Higher Tier Free Practice Set 1 1 hour 45 minutes ANSWERS. Grade Boundaries A* A B C D E.

GCSE Mathematics Non-Calculator Higher Tier Free Practice Set 1 1 hour 45 minutes ANSWERS. Grade Boundaries A* A B C D E. MathsMadeEasy GCSE Mathematics Non-Calculator Higher Tier Free Practice Set 1 1 hour 45 minutes ANSWERS Grade Boundaries A* A B C D E 88 71 57 43 22 13 3 Authors Note Every possible effort has been made

More information

Solutions to the Olympiad Maclaurin Paper

Solutions to the Olympiad Maclaurin Paper s to the Olympiad Maclaurin Paper M1. The positive integer N has five digits. The six-digit integer Pis formed by appending the digit 2 to the front of N. The six-digit integer Q is formed by appending

More information

Yavapai County Math Contest College Bowl Competition. January 28, 2010

Yavapai County Math Contest College Bowl Competition. January 28, 2010 Yavapai County Math Contest College Bowl Competition January 28, 2010 Is your adrenalin engaged? What is 1 2 + 3 4? 82 Solve for x in: 2x + 7 = 1 3x. x=-6/5 (or x=-1.2) If a fair die is rolled once, what

More information

Warm-Up Let K = = 151,200. How many positive integer divisors does K have? 104.

Warm-Up Let K = = 151,200. How many positive integer divisors does K have? 104. Warm-Up 8 101. Pamela Wickham writes a sequence of four consecutive integers on a sheet of paper. The sum of three of these integers is 206. What is the other integer? 102. seconds enjamin starts walking

More information

2005 Euclid Contest. Solutions

2005 Euclid Contest. Solutions Canadian Mathematics Competition An activity of the Centre for Education in Mathematics and Computing, University of Waterloo, Waterloo, Ontario 2005 Euclid Contest Tuesday, April 19, 2005 Solutions c

More information

GCSE EDEXCEL MATHS. Year 10 Revision REVISION BOOKLET. Foundation. Name:

GCSE EDEXCEL MATHS. Year 10 Revision REVISION BOOKLET. Foundation. Name: GCSE EDEXCEL MATHS Year 10 Revision REVISION BOOKLET Foundation Name: 1 Contents Page: Number: Types of number 3 Place value 6 Directed numbers 8 Algebra: Coordinates 12 Patterns and sequences 15 Collecting

More information

Mathematics Enhancement Programme

Mathematics Enhancement Programme 1A 1B UNIT 3 Theorem Lesson Plan 1 Introduction T: We looked at angles between 0 and 360 two weeks ago. Can you list the different types of angles? (Acute, right, reflex, obtuse angles; angles on straight

More information

NUMERACY TOOLKIT TOOLKIT NUMERACY

NUMERACY TOOLKIT TOOLKIT NUMERACY NUMERACY TOOLKIT TOOLKIT NUMERACY Addition Calculating methods Example 534 + 2678 Place the digits in the correct place value columns with the numbers under each other. Th H T U Begin adding in the units

More information

Solutions Parabola Volume 49, Issue 2 (2013)

Solutions Parabola Volume 49, Issue 2 (2013) Parabola Volume 49, Issue (013) Solutions 1411 140 Q1411 How many three digit numbers are there which do not contain any digit more than once? What do you get if you add them all up? SOLUTION There are

More information

Mathematics A *P43380A0132* Pearson Edexcel GCSE P43380A. Paper 2 (Calculator) Foundation Tier. Friday 13 June 2014 Morning Time: 1 hour 45 minutes

Mathematics A *P43380A0132* Pearson Edexcel GCSE P43380A. Paper 2 (Calculator) Foundation Tier. Friday 13 June 2014 Morning Time: 1 hour 45 minutes Write your name here Surname Other names Pearson Edexcel GCSE Centre Number Mathematics A Paper 2 (Calculator) Friday 13 June 2014 Morning Time: 1 hour 45 minutes Candidate Number Foundation Tier Paper

More information

State Math Contest Senior Exam SOLUTIONS

State Math Contest Senior Exam SOLUTIONS State Math Contest Senior Exam SOLUTIONS 1. The following pictures show two views of a non standard die (however the numbers 1-6 are represented on the die). How many dots are on the bottom face of figure?

More information

2015 Canadian Team Mathematics Contest

2015 Canadian Team Mathematics Contest The CENTRE for EDUCATION in MATHEMATICS and COMPUTING cemc.uwaterloo.ca 205 Canadian Team Mathematics Contest April 205 Solutions 205 University of Waterloo 205 CTMC Solutions Page 2 Individual Problems.

More information

Australian Intermediate Mathematics Olympiad 2016

Australian Intermediate Mathematics Olympiad 2016 A u s t r a l i a n M at h e m at i c a l O ly m p i a d C o m m i t t e e a d e p a r t m e n t o f t h e a u s t r a l i a n m at h e m at i c s t r u s t Australian Intermediate Mathematics Olympiad

More information

Math is Cool Championships

Math is Cool Championships Individual Contest Express all answers as reduced fractions unless stated otherwise. Leave answers in terms of π where applicable. Do not round any answers unless stated otherwise. Record all answers on

More information

2018 LEHIGH UNIVERSITY HIGH SCHOOL MATH CONTEST

2018 LEHIGH UNIVERSITY HIGH SCHOOL MATH CONTEST 08 LEHIGH UNIVERSITY HIGH SCHOOL MATH CONTEST. A right triangle has hypotenuse 9 and one leg. What is the length of the other leg?. Don is /3 of the way through his run. After running another / mile, he

More information

The CENTRE for EDUCATION in MATHEMATICS and COMPUTING cemc.uwaterloo.ca Euclid Contest. Tuesday, April 12, 2016

The CENTRE for EDUCATION in MATHEMATICS and COMPUTING cemc.uwaterloo.ca Euclid Contest. Tuesday, April 12, 2016 The CENTRE for EDUCATION in MATHEMATICS and COMPUTING cemc.uwaterloo.ca 016 Euclid Contest Tuesday, April 1, 016 (in North America and South America) Wednesday, April 13, 016 (outside of North America

More information

2009 Cayley Contest. Solutions

2009 Cayley Contest. Solutions Canadian Mathematics Competition An activity of the Centre for Education in Mathematics and Computing, University of Waterloo, Waterloo, Ontario 2009 Cayley Contest (Grade 10) Wednesday, February 18, 2009

More information

BRITISH COLUMBIA COLLEGES High School Mathematics Contest 2003 Solutions

BRITISH COLUMBIA COLLEGES High School Mathematics Contest 2003 Solutions BRITISH COLUMBIA COLLEGES High School Mathematics Contest 3 Solutions Junior Preliminary. Solve the equation x = Simplify the complex fraction by multiplying numerator and denominator by x. x = = x = (x

More information

Ciphering MU ALPHA THETA STATE 2008 ROUND

Ciphering MU ALPHA THETA STATE 2008 ROUND Ciphering MU ALPHA THETA STATE 2008 ROUND SCHOOL NAME ID CODE Circle one of the following Mu Alpha Theta Euclidean Round 1 What is the distance between the points (1, -6) and (5, -3)? Simplify: 5 + 5 5

More information

= 126 possible 5-tuples.

= 126 possible 5-tuples. 19th Philippine Mathematical Olympiad 1 January, 017 JUDGES COPY EASY 15 seconds, points 1. If g (x) = x x 5 Answer: 14 Solution: Note that x x and f ( g ( x)) = x, find f (). x 6 = = x = 1. Hence f ()

More information

MATHEMATICS UNIT 2: CALCULATOR-ALLOWED INTERMEDIATE TIER

MATHEMATICS UNIT 2: CALCULATOR-ALLOWED INTERMEDIATE TIER Surname Centre Number Candidate Number Other Names 0 GCSE NEW 3300U40-1 A16-3300U40-1 MATHEMATICS UNIT 2: CALCULATOR-ALLOWED INTERMEDIATE TIER THURSDAY, 10 NOVEMBER 2016 MORNING 1 hour 45 minutes For s

More information

BRITISH COLUMBIA COLLEGES High School Mathematics Contest 2004 Solutions

BRITISH COLUMBIA COLLEGES High School Mathematics Contest 2004 Solutions BRITISH COLUMBI COLLEGES High School Mathematics Contest 004 Solutions Junior Preliminary 1. Let U, H, and F denote, respectively, the set of 004 students, the subset of those wearing Hip jeans, and the

More information

High School Math Contest

High School Math Contest High School Math Contest University of South Carolina February th, 017 Problem 1. If (x y) = 11 and (x + y) = 169, what is xy? (a) 11 (b) 1 (c) 1 (d) (e) 8 Solution: Note that xy = (x + y) (x y) = 169

More information

Gauss School and Gauss Math Circle 2017 Gauss Math Tournament Grade 7-8 (Sprint Round 50 minutes)

Gauss School and Gauss Math Circle 2017 Gauss Math Tournament Grade 7-8 (Sprint Round 50 minutes) Gauss School and Gauss Math Circle 2017 Gauss Math Tournament Grade 7-8 (Sprint Round 50 minutes) 1. Compute. 2. Solve for x: 3. What is the sum of the negative integers that satisfy the inequality 2x

More information

Investigation Find the area of the triangle. (See student text.)

Investigation Find the area of the triangle. (See student text.) Selected ACE: Looking For Pythagoras Investigation 1: #20, #32. Investigation 2: #18, #38, #42. Investigation 3: #8, #14, #18. Investigation 4: #12, #15, #23. ACE Problem Investigation 1 20. Find the area

More information

GCSE style questions arranged by topic

GCSE style questions arranged by topic Write your name here Surname Other names In the style of: Pearson Edecel GCSE Centre Number Candidate Number Mathematics A* type questions Model Answers GCSE style questions arranged by topic Higher Tier

More information

Nozha Directorate of Education Form : 2 nd Prep. Nozha Language Schools Ismailia Road Branch

Nozha Directorate of Education Form : 2 nd Prep. Nozha Language Schools Ismailia Road Branch Cairo Governorate Department : Maths Nozha Directorate of Education Form : 2 nd Prep. Nozha Language Schools Sheet Ismailia Road Branch Sheet ( 1) 1-Complete 1. in the parallelogram, each two opposite

More information

MOEMS What Every Young Mathlete Should Know

MOEMS What Every Young Mathlete Should Know MOEMS What Every Young Mathlete Should Know 2018-2019 I. VOCABULARY AND LANGUAGE The following explains, defines, or lists some of the words that may be used in Olympiad problems. To be accepted, an answer

More information

Chapter 1. Some Basic Theorems. 1.1 The Pythagorean Theorem

Chapter 1. Some Basic Theorems. 1.1 The Pythagorean Theorem hapter 1 Some asic Theorems 1.1 The ythagorean Theorem Theorem 1.1 (ythagoras). The lengths a b < c of the sides of a right triangle satisfy the relation a 2 + b 2 = c 2. roof. b a a 3 2 b 2 b 4 b a b

More information

Candidate Name Centre Number Candidate Number MATHEMATICS UNIT 2: CALCULATOR-ALLOWED FOUNDATION TIER SPECIMEN PAPER SUMMER 2017

Candidate Name Centre Number Candidate Number MATHEMATICS UNIT 2: CALCULATOR-ALLOWED FOUNDATION TIER SPECIMEN PAPER SUMMER 2017 GCSE MATHEMATICS Specimen Assessment Materials 105 Candidate Name Centre Number Candidate Number 0 GCSE MATHEMATICS UNIT 2: CALCULATOR-ALLOWED FOUNDATION TIER SPECIMEN PAPER SUMMER 2017 1 HOUR 30 MINUTES

More information

MATHCOUNTS State Competition Countdown Round Problems This section contains problems to be used in the Countdown Round.

MATHCOUNTS State Competition Countdown Round Problems This section contains problems to be used in the Countdown Round. MATHCOUNTS 2011 State Competition Countdown Round Problems 1 80 This section contains problems to be used in the Countdown Round. National Sponsors Raytheon Company * National Defense Education Program

More information

Australian Intermediate Mathematics Olympiad 2016

Australian Intermediate Mathematics Olympiad 2016 Australian Intermediate Mathematics Olympiad 06 Questions. Find the smallest positive integer x such that x = 5y, where y is a positive integer. [ marks]. A 3-digit number in base 7 is also a 3-digit number

More information

2018 Pascal Contest (Grade 9)

2018 Pascal Contest (Grade 9) The CENTRE for EDUCATION in MATHEMATICS and COMPUTING cemc.uwaterloo.ca 018 Pascal Contest (Grade 9) Tuesday, February 7, 018 (in North America and South America) Wednesday, February 8, 018 (outside of

More information

Unofficial Solutions

Unofficial Solutions Canadian Open Mathematics Challenge 2016 Unofficial Solutions COMC exams from other years, with or without the solutions included, are free to download online. Please visit http://comc.math.ca/2016/practice.html

More information

Written test, 25 problems / 90 minutes

Written test, 25 problems / 90 minutes Sponsored by: UGA Math Department and UGA Math Club Written test, 5 problems / 90 minutes October, 06 WITH SOLUTIONS Problem. Let a represent a digit from to 9. Which a gives a! aa + a = 06? Here aa indicates

More information

Solutions Math is Cool HS Championships Mental Math

Solutions Math is Cool HS Championships Mental Math Mental Math 9/11 Answer Solution 1 30 There are 5 such even numbers and the formula is n(n+1)=5(6)=30. 2 3 [ways] HHT, HTH, THH. 3 6 1x60, 2x30, 3x20, 4x15, 5x12, 6x10. 4 9 37 = 3x + 10, 27 = 3x, x = 9.

More information

Mathematics Paper 2 (Calculator)

Mathematics Paper 2 (Calculator) www.themathsprofessor.com Write your name here Surname Other names Pearson Edexcel Level 1/Level GCSE (9-1) Centre Number Candidate Number Mathematics Paper (Calculator) Specimen Papers Set Time: 1 hour

More information

Diagnostic Test. Month Balance Change February $ March $ $13.10 April $1, $ May $ $ June $ $163.

Diagnostic Test. Month Balance Change February $ March $ $13.10 April $1, $ May $ $ June $ $163. Diagnostic Test Select the best answer for questions 1 60. Fill in the correct bubble on your answer sheet. 1. The chart shows the balance in Neil s savings account and the change from the previous month.

More information

Theorem 1.2 (Converse of Pythagoras theorem). If the lengths of the sides of ABC satisfy a 2 + b 2 = c 2, then the triangle has a right angle at C.

Theorem 1.2 (Converse of Pythagoras theorem). If the lengths of the sides of ABC satisfy a 2 + b 2 = c 2, then the triangle has a right angle at C. hapter 1 Some asic Theorems 1.1 The ythagorean Theorem Theorem 1.1 (ythagoras). The lengths a b < c of the sides of a right triangle satisfy the relation a + b = c. roof. b a a 3 b b 4 b a b 4 1 a a 3

More information

Mathematics Higher Tier, November /2H (Paper 2, calculator)

Mathematics Higher Tier, November /2H (Paper 2, calculator) Link to past paper on AQA website: www.aqa.org.uk This question paper is available to download freely from the AQA website. To navigate around the website, you want QUALIFICATIONS, GCSE, MATHS, MATHEMATICS,

More information

Odd numbers 4 2 = 4 X 4 = 16

Odd numbers 4 2 = 4 X 4 = 16 Even numbers Square numbers 2, 4, 6, 8, 10, 12, 1 2 = 1 x 1 = 1 2 divides exactly into every even number. 2 2 = 2 x 2 = 4 3 2 = 3 x 3 = 9 Odd numbers 4 2 = 4 X 4 = 16 5 2 = 5 X 5 = 25 1, 3, 5, 7, 11, 6

More information

CONTENTS NUMBER SYSTEMS. Number Systems

CONTENTS NUMBER SYSTEMS. Number Systems NUMBER SYSTEMS CONTENTS Introduction Classification of Numbers Natural Numbers Whole Numbers Integers Rational Numbers Decimal expansion of rational numbers Terminating decimal Terminating and recurring

More information

The Alberta High School Mathematics Competition Solution to Part I, 2014.

The Alberta High School Mathematics Competition Solution to Part I, 2014. The Alberta High School Mathematics Competition Solution to Part I, 2014. Question 1. When the repeating decimal 0.6 is divided by the repeating decimal 0.3, the quotient is (a) 0.2 (b) 2 (c) 0.5 (d) 0.5

More information

Key Stage 3 Subject: Maths Foundation Year: Year 7 Year 8 Year 9 Topic/Module: Geometry

Key Stage 3 Subject: Maths Foundation Year: Year 7 Year 8 Year 9 Topic/Module: Geometry Subject: Foundation Topic/Module: Geometry Time Geometry 1 234 Metric Units Angles/Polygons Bearings Transformations watch Clips N21 N7 N7 N7 N7 N7, R2, 112 112 G10, 46 G16, 122 G14 G13, G17, 45, 121 G23

More information

The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION COURSE II. Friday, January 26, :15 a.m. to 12:15 p.m.

The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION COURSE II. Friday, January 26, :15 a.m. to 12:15 p.m. The University of the State of New York REGENTS HIGH SCHOOL EXMINTION THREE-YER SEQUENCE FOR HIGH SCHOOL MTHEMTICS COURSE II Friday, January 26, 2001 9:15 a.m. to 12:15 p.m., only Notice... Scientific

More information

Answers. Investigation 4. ACE Assignment Choices. Applications. The number under the square root sign increases by 1 for every new triangle.

Answers. Investigation 4. ACE Assignment Choices. Applications. The number under the square root sign increases by 1 for every new triangle. Answers Investigation 4 ACE Assignment Choices Problem 4. Core, Other Connections 6 Problem 4. Core, 4, Other Applications 6 ; Connections 7, 6, 7; Extensions 8 46; unassigned choices from earlier problems

More information

Instructions. Information. Advice

Instructions. Information. Advice Instructions Use black ink 7C or ball-point pen. Fill in the boxes at the top of this page with your name, centre number and candidate number. Answer all questions. Answer the questions in the spaces provided

More information

UNIVERSITY OF NORTH CAROLINA CHARLOTTE 1995 HIGH SCHOOL MATHEMATICS CONTEST March 13, 1995 (C) 10 3 (D) = 1011 (10 1) 9

UNIVERSITY OF NORTH CAROLINA CHARLOTTE 1995 HIGH SCHOOL MATHEMATICS CONTEST March 13, 1995 (C) 10 3 (D) = 1011 (10 1) 9 UNIVERSITY OF NORTH CAROLINA CHARLOTTE 5 HIGH SCHOOL MATHEMATICS CONTEST March, 5. 0 2 0 = (A) (B) 0 (C) 0 (D) 0 (E) 0 (E) 0 2 0 = 0 (0 ) = 0 2. If z = x, what are all the values of y for which (x + y)

More information

UNCC 2001 Comprehensive, Solutions

UNCC 2001 Comprehensive, Solutions UNCC 2001 Comprehensive, Solutions March 5, 2001 1 Compute the sum of the roots of x 2 5x + 6 = 0 (A) (B) 7/2 (C) 4 (D) 9/2 (E) 5 (E) The sum of the roots of the quadratic ax 2 + bx + c = 0 is b/a which,

More information

GCSE Mathematics Non-Calculator Higher Tier Free Practice Set 6 1 hour 45 minutes. Answers at:

GCSE Mathematics Non-Calculator Higher Tier Free Practice Set 6 1 hour 45 minutes. Answers at: First Name Last Name Date Total Marks / 100 marks MathsMadeEasy 3 GCSE Mathematics Non-Calculator Higher Tier Free Practice Set 6 1 hour 45 minutes Answers at: http://www.mathsmadeeasy.co.uk/gcsemathspapers-free.htm

More information

Marquette University

Marquette University Marquette University 2 0 7 C O M P E T I T I V E S C H O L A R S H I P E X A M I N A T I O N I N M A T H E M A T I C S Do not open this booklet until you are directed to do so.. Fill out completely the

More information

NMC Sample Problems: Grade 7

NMC Sample Problems: Grade 7 NMC Sample Problems: Grade 7. If Amy runs 4 4 mph miles in every 8 4. mph hour, what is her unit speed per hour? mph. mph 6 mph. At a stationary store in a state, a dozen of pencils originally sold for

More information

Math is Cool Masters

Math is Cool Masters 8th Grade November 19, 2005 Individual Contest Express all answers as reduced fractions unless stated otherwise. Leave answers in terms of π where applicable. Do not round any answers unless stated otherwise.

More information