d "AMERICAN MATHEMATICS COMPETITIONS AJHSME SOLUTIONS eee FOR STUDENTS AND TEA 9th ANNUAL AMERICAN JUNIOR HIGH SCHOOL MATHEMATICS EXAMINATION (AJHSME) THURSDAY, Pree 18, 1993 Siena dlicatrnee cancpbianiaand Society of Actuaries Mu Alpha Theta isa Prof Walter B Mientka, AMC Executive Director of Mathematics and Statistics Daieraity of Nebaneks Lincoln, 68588-0658 Copyright © 1998, Commities on the Ametican Mathematics Competitions2 SOLUTIONS 1993 AJHSME 1. (€) 5x (-72) = -30, Rahs oe Toke vie 2) a= 738 = 12" The sum of the numerator and the denominator is 7 +12 = 19. 3. (B) Factoring each number into prime factors yields 39=3x13, 51=3x17, 77=7x11, 91=7x13 and 121=11 x11. The largest of these prime factors is 17, which is a factor of 51. 4. (E) 1000 x 1993 x 0.1993 x 10 = ((1000 x 10) x 0.1998) x 1993 = (10,000 x 0.1993) x 1993 = 1993 x 1993 = (1993)?. 5. (C) The unshaded area is half the total, and each of the shaded areas is one fourth of the total. This is represented in bar graph (C). 6. (B) Three cans of soup are needed for 15 children, so the remaining 2 cans of soup will feed 2x 3=6 adults. 7. (A) 3° +39 + 39 = 3(85) = (3 x3 x 3) =3x3x3x3=34. OR 3° +39 + 3° — 27 +27 +27=81=9x9=3x3x3x3=34. 8. (D) Since she takes one half of a pill every other day, one pill will last 4 days. Hence 60 pills will last 60 x 4 = 240 days, or about 8 months. 9, (D) Substituting the values from the table yields (244)*(1*3)=343=4. Query. Would evaluating the products ((2*4)*1)*3, (2*(4*1))*3, 2*((441))*3) and 2+(44(1+3)), yield the same result?SOLUTIONS 1993 AJHSME 3 10. (B) The graph shows the following changes in the price of the card: Jan: $2.50 to $2.00 drop of $0.50 Feb: $2.00 to $4.00 rise of $2.00 Mar: $4.00 to $1.50 drop of $2.50 Apr: $1.50 to $4.50 rise of $3.00 May: $4.50 to $3.00 drop of $1.50 Jun: $3.00 to $1.00 drop of $2.00 The greatest drop occurred during March. 12. (C) Since 81 took the test, the median (middle) score is the 41*t. The test interval containing the 41% score is labeled 70. 12. (B) The six permutations of +,—and x yield these results: 5x44+6-3=204+6-3=23 5x4—-643=20-6+3=17 54+4x6-3=5+424—-3=26 5—-4x6+3=5—-244+3=-16 54+4-6x3=5+4-18=-9 5-4+6x3=5—-4+18=19. The only result listed is 19. 13. (D) The white portion can be partitioned into rectangles as shown, The sum of the areas of the white rectangles is 4(2) + 3(5) +8+1+4— 36. OR Compute the area of the black letters and subtract it from 5 x 15=75, the total area of the sign: Hi: 2(1 x 5)+1x1=11. B: 1x543(2x 1)=11. EL: 1x5+1x2=7. P: 1x54+2(1x1)+1x3=10. The area of the white portion is 75 — (11 + 11+7+10) = 36.4 SOLUTIONS 1993 AJHSME OR Superimpose a 1 x 1 grid on the sign and count the 36 white squares: 14. (C) Only 3’s can complete the 2 by 2 square whose diagonal is given. If two ape, a fon os column are, ope tte lands eteerued Use this to 15. (A) The sum of the four numbers is 4 x 85 = 340, so the sum of the three numbers is 340 — 97 = 243. Thus the mean of these three numbers is 243/3 = 81. 16. (C) Using the common denominators and simplifying yieldwe CB TS aaa Oe oes eee ete near 8 OPOIGGR OSES sheet of cardboard after the corners have been removed. The area of the sheet is 30 x 20 = 600 and the area of each of ra the square corners removed is 5 x 5 = 25, so the answer is 600 — (4 x 25) = 500. 18. (A) Rectangle AC'DE has area 32 x 20 = 640. Triangle BCD has area (16 x 20)/2 = 160, and triangle DEF has area (10 x 32)/2 = 160. The remaining area, ABDF, is 640 — (160 + 160) = 320. OR Insert diagonal AD. The areas of triangles ABD and BOD pee ABIL) ane (BC)(CD)}/2 which are equal since 4 oC AB = BC. Hence half the area in the rectangle above AD is, in ABDF. Similarly, triangles ADF and DEF have equal eee ge ee ena a ot ABDPF. Thus, the area of ABDF is (32 x 20)/2 = 320. OR Draw a perpendicular from point B to BD to show that the area of ABCD is one fourth of the area of ACDE. Similarly, draw a perpendicular from point F to OD to show that the area of ADEF is one fourth of the area of ACDE. ‘Thus the area of ABDF is one half of the area of ACDE, or (32 x 20)/2 = 320.6 ; SOLUTIONS 1993 AJHSME 19. (A) Each number in the first set of numbers is 1800 more than the corre- sponding number in the second set: 1901, 1902, 1903,..., 199% 101, 102, =103, ..., =193 1800, 1800, 1800, ..., 1800 Thus the sum of the first set of numbers is 93 x 1800 = 167,400 more than the sum of the second set. 93 zeros. 20. (D) Since 10°? = 190---00, we have 100---000 me a and the sum of the digits is (91 x 9) + 7 = 826. 91 nines OR Look for a pattern using simpler cases: 10? - 93 = 100-93 = 07 19°-93= 1,000-93 = 907 10*-93= 10,000-93 = 9907 10°-93= 100,000-93 = 99907 10" — 93 = = 999...907 91 nines Thus the sum of the digits is (91 x 9) + 7 = 826. 21. (D) When a problem indicates a general result, then it must hold for any specific case. ‘Therefore, suppose the original rectangle is 10 by 10 with area 100. The new length is 104-2 = 12 and the new width is 1045 = 15. Hence the new area is 12 x 15 = 180 for an increase of 80%. OR ‘The length is changed to 120%, or 1.2 times its original value, and the width is changed to 150%, or 1.5 times its original value. Since area is length times width, the new areais 1.2x 1.5= 1.8 times the original area. Thus the area is increased by 80%.SOLUTIONS 1993 AJHSME 7 22, (D) Ten 2’s are needed in the unit’s place in counting to 100 and ten more 2’s are used in the ten’s place. With the remaining two 2’s he can number 102 and 112 and continue all the way to 119 before needing another 2, 23. (C) Since P, T and @ must finish in front of S, S cannot be third. Since P is the winner, P cannot be third. Thus the only possible orders are PRTQS, PTRQS, PTQRS and PTQSR, which show that anyone except P and S could finish third. OR Since PTQS must finish in that order and R can finish anyplace except ahead of P, it follows that the only possible orders are PRTQS, PTRQS, PTQRS, PTQSR. Thus T, Rand Q might have finished third, but P and S could not have finished third. 24. (C) After completing some more rows, the pattern of squares as the last entry in each row becomes apparent. Thus, the line containing 142 ends in 144, and the line above it ends in 121. Therefore 121 is directly above 143, and 120 is above 142. Bl 2 3 fal Ouch aT eetoee) 10 11 12 13 14 15 fie] ; + 119 120 [121] ve 1d 142143, [4a] 25. (E) Using the Pythagorean rem, the of the diagonal of the card is perce = w 21 This is longer than 2, the length of two adjacent squares. The figure shows 12 squares being touched. Query. Is 12 the maximum number of squares that can be touched?