SOLUTIONS TO PRISM PROBLEMS Juior Level 04. (B) Sice 50% of 50 is 50 5 ad 50% of 40 is the secod by 5 0 5. 40 0, the first exceeds. (A) Oe way of comparig the magitudes of the umbers,,, 5 ad 0.7 is 4 5 7 to express them all as decimals. Sice 0.75, 0.6. 0.667, 5 0.74, the largest of the five umbers is. 4 5 7 4 Alteratively but far more tediously i the preset problem, we could express all umbers as fractios with a commo deomiator. I the order give, the umbers are 5, 5, 80, 00 94 ad. ad we agai see that the largest is the 40 40 40 40 40 first-appearig fractio which is 5 40 4. (C) The umbers betwee ad 40 that are multiples of 7 are 7,4,,8 ad 5. Theumbersbetweead40thatedia7are7,7,7 ad 7. Note that the umber 7 appears o both lists above. The umbers betwee ad 40 that are either a multiple of 7 or ed i a 7 are the 7,4,,8,5,7,7, ad 7. There are thus 8 such umbers. 4. (C) Halfway betwee 8:04 ad 8:8 is.. 5. (D) Let L,E ad C be the price or value of a loaf of bread, a egg ad a chocolate bar, respectively. We are give that L 4E ad C E. We use these two equatios to get a relatioship betwee L ad C. If we multiply the equatio C E across by, we get C 4E (that is chocolate bars cost the same as 4 eggs). But the first equatio says that a loaf costs the same as four eggs. Hece we have that chocolate bars are equal to oe loaf, that is C L. Multiplyig across by we see that 6C L, i.e. 6 chocolate bars are worth loaves. 6. (C) If Peter ate oly, say 6 sweets, there would be 8 6 sweets to share amog the other two boys so oe or both of them could receive 6, which would cotradict the fact that Peter ate more tha each of the other two boys. This cotradictios meas that Peter did ot eat 6 sweets. A similar argumet shows that Peter caot have ate less tha 6 sweets. It is however possible for Peter to eat 7 sweets, as the there would be left for the other two boys ad the most each of these could receive (while esurig that Peter gets more tha each of them) is 6 ad 5. Thus 7 is the miimum umber of sweets that Peter could have received.
7. (D) The first positio i the row could be occupied by ay oe of 4 studets, the secod positio the by ay of three, the third positio by oe of the remaiig two, ad fially we the have just oe choice of perso for the fourth positio. The total umber of ways of arragig the four people is the 4 4. Studets who are uclear o why we multiple the four umbers 4,,, adare requested to read the followig ote which will prove very useful i their Leavig Certificate course! Note [About the fudametal coutig rule ad the -factorial formula.]. Studets will fid it helpful to recall the fudametal theorem of coutig. Oe versio of this says that if we have a ways of doig oe experimet, the b ways of doig a secod experimet, the c ways of doig a third experimet, ad so o, there are a b c... ways of performig the combied experimet. I the above problem, there were 4 ways of occupyig the first positio, the ways of choosig a perso for the secod positio, the choices for the third positio ad fially choice for the last positio. By the above rule, the total umber of ways of seatig the four people is the 4 4. Note that the fudametal coutig rule shows that there are... ways of arragig distict objects i a row, because ay oe of objects ca be placed i the first positio, the ay oe of the remaiig objects ca be put i the secod positio, ad so o to way of placig the last object. This umber... is deoted! [read -factorial ] 8. (A) For each team to play oe game, 5 games will be played (because each team will play some other team). Similarly, for each team to play their secod game, aother 5 games will be played. Fially whe each team has played its third game, aother 5 games will have bee played. The total umber of games played is the 5 5 5 5. 9. (C) Oe way of proceedig is to write each fractio as a decimal. We fid that 0.4, 4 5 0.4, 0.44, 0.46 ad 0.4667. We thus see that the 5 0 5 50 75 largest is 5. 75 Alteratively (ad perhaps easier), studets might compare the fractios by represetig them all with the same commo deomiator. Clearly the deomiators of the five give umbers all divide ito 50 (which i fact is the least commo multiple of the deomiators). We ca write 0 60, 4 60, 66, 69 5 ad 70. 5 5 0 50 0 50 5 50 50 50 75 50 The largest umerator is ow 70 so 5 70 is the greatest of the five fractios. 75 50
0. (D) Oe way of proceedig is by log divisio, from which we see that is exactly 6500 equal to 0. 00006, so there are six digits after the decimal poit. Slightly shorter would be to factorize 6500 as 6500 65 00 5 5 0 0. The sice 0.04, we have that 4 5 00 6500 5 5 0 0 that follow the decimal poit. 0.04 0.04 0. 0. 0.00006, so agai there are 6 digits. (B) We solve the simultaeous equatios 4x y 4 ad 4x y 4. The coefficiet of x is the same i both equatios so we ca immediately elimiate x by subtractig the secod equatio from he first. This immediately gives y 0, uiquely. The correspodig value of x is got by isertig y 0 ito oe of the equatios, say the first. We get 4x 0 y 4 so 4x 4 ad x. The equatios thus have exactly oe solutio x,y 0.. (D) Let x be the price of a sweater ad let y be the price of a shirt. We are give that x 8y 8x y 0. Hece 4x 4y 0, so x y 5. That is, a sweater costs 5 more tha a shirt.. (A) I the Ve diagram above, the etire rectagle represets all 70 studets. The smaller egg represets the set of 40 male studets ad the larger egg is the set of 50 Galway studets. The itersectio of these two eggs is the set of male Galway studets ad we are give that the umber of studets i this set is 0. Hece the umber of studets who are male ad ot from Galway is 40 0 0 ad the umber of studets who are from Galway but ot male is 50 0 0. Fially the the umber of studets who are either male or from Galway is the set of studets withi the rectagle who do ot belog to either egg, ad this umber is 70 0 0 0 70 70 0.
4. (D) Let s be the distace from the bottom to the top of the hill so s is also the distace from the top to the bottom. Also let t be the time take to get from the bottom to the top of the hill, ad let t be the time take to go from the top to the bottom. Sice speed is distace divided by time, we have 0 s/t ad 45 s/t. Hece s 0t 45t. Accordigly, t 45 t 0 t. Now the total distace is the distace to top distace to the bottom 0t 45t. The average speed is the total distace divided by the total time, i.e. twice the distace to the top divided by the total time, i.e. fid that the average speed is 90t 90 90 t t 5 45t t t. Substitutig t t /, we 6. Note: A study of the argumet above shows that i computig average speed we do ot use the arithmetic average of the speeds up ad dow the moutais, but rather the harmoic mea, speed up speed dow speed up speed dow of the two speeds! which ca be writte as, i.e. the reciprocal of the arithmetic mea of the reciprocals 5.(A) The total amout, T, placed o the board is T... 6 64. It would be very tedious to add these umbers term-wise. However there is a well-kow trick! We write the sum dow with the umbers reversed, that is, we write T... 6 64 as T 64 6... We ow add these last two equatios to get T T 64 6 6... 6 6 64. That is, T 65 65 65... 65 65 65 64 terms Thus T 64 65 ad so T 64 65 65 080. Note to studets: The above method is oly oe way of addig ay umber of cosecutive itegers ad was developed by a youg precocious child amed Gauss. If you google the word Gauss you will get a great may hits about this famous cetury mathematicia who was bor i 777 ad is sometimes referred to as the Price of Mathematicias ad the greatest mathematicia sice atiquity! The geeral formula for the sum of the first itegers,,,..., is... ad this formula which is well worth memorizig is a special case of the formula for the sum of a arithmetic progressio. 4
6. (C) Whe Adrew crosses the fiish lie, Bria will have covered 800 metres. Thus the ratio of Bria s speed to Adrew s speed is 800 0.8. That is, Bria travelled at 000 90% of Adrew s speed. Similarly, Charlie travelled at 80% of Bria s speed, so Charlie travelled at 80% of 80% 64% of Adrew s speed. Accordigly, whe Adrew has covered the 000 metres, Charlie will have covered 0.64 000 640 metres. Hece Adrew beat Charlie by 000 640 60 metres. 7. (B) We are give that ab 6,ac 8 ad bc 7. We require abc. May studets will proceed by rial ad error. A rigorous approach would ote that sice ab 6, we have abc 6c, so we could easily solve the problem if we ca evaluate c. From ab 6,ac 8, we have a 6 ad a 8 b c. Thus 6 8 b c so c b. But bc 7 so b 7 c. Substitutig b 7 c ito c b gives c 7 c, or c 8 ad (sice c 0 we have c 9. Thus abc ab c 6 9 54 8. (D) First ote that a square with sides of legth 6 has perimeter 5 5 5 5 0. Next by Pythagoras Theorem if the legth of the side of the square is x the x x 50. Thus x 50, or x 5 or x 5. Thus the square i B) has agai perimeter 4 5 0. For C) ote that a square with area 5 cm has side legth 5 5, so agai the perimeter of this square is 5. The circle i D) has radius r.5 so its perimeter is r.5 7.5 ad sice is about.4, this perimeter is somewhat greater tha 7.5.5 Fially, if r deotes the radius of the circle i E), we are give that the area is r 9. Hece r 9. so r. The perimeter of this circle is 6 which is certaily less tha all of the aswers i A), B), C) ad D). Puttig everythig together, we see that the figure with the largest perimeter is the circle i D). 5
9. (A) Let x be the legth of a side of the square. By Pythagoras Theorem, the square of the diagoal is x x x. We are give that x 4 4x. Thus x 8.Accordigly the area of the square is 8 8 64 cm 0. (B) Let the marks obtaied by the studets be x,x,...,x ad let x x x... x be their average mark. Note that this average is obtaied by addig up the marks of all of the studets ad dividig the result by the umber of studets. Whe m of the studets have their marks raised by ad the remaiig m have their marks raised by, the ew average will be x x... x...... x NEW m times m times Notice that this ca be writte as x NEW x x... x m m x x... x m x x... x m x m. The differece x NEW x is the m ad we are give that this equals.5 Writig m.5 as m.5, we have m 0.5 so m 4 6