Download [PDF] 2018 AMC Juniors Years 7 and 8 Solutions Australian Mathematics Competition

File Information

Filename:	[PDF] 2018 AMC Juniors Years 7 and 8 Solutions Australian Mathematics Competition.pdf
Filesize:	217.26 KB
Uploaded:	18/08/2021 16:12:56
Keywords:	2018 amc juniors years 7 and 8 solutions australian mathematics competition pdf
Description:	Download file or read online AMC past exam paper 2018 Juniors Years 7 and 8 Solutions - Australian Mathematics Competition.
Downloads:	10

File Preview

Download Urls

Short Page Link

https://www.edufilestorage.com/64Y

Full Page Link

https://www.edufilestorage.com/64Y/PDF_2018_AMC_Juniors_Years_7_and_8_Solutions_Australian_Mathematics_Competition.pdf

HTML Code

<a href="https://www.edufilestorage.com/64Y/PDF_2018_AMC_Juniors_Years_7_and_8_Solutions_Australian_Mathematics_Competition.pdf" target="_blank" title="Download from eduFileStorage.com"><img src="https://www.edufilestorage.com/cache/plugins/filepreviewer/3372/pdf/150x190_middle_46f4e7862b1eb5bd4935adbbba5d79e8.jpg"/></a>

Forum Code

[url=https://www.edufilestorage.com/64Y/PDF_2018_AMC_Juniors_Years_7_and_8_Solutions_Australian_Mathematics_Competition.pdf][img]https://www.edufilestorage.com/cache/plugins/filepreviewer/3372/pdf/150x190_middle_46f4e7862b1eb5bd4935adbbba5d79e8.jpg[/img][/url]

[PDF] 2018 AMC Juniors Years 7 and 8 Solutions Australian Mathematics Competition.pdf | Plain Text

2018 AMC Junior Solutions Solutions { Junior Division 1. 2 + 0 + 1 + 8 = 11, hence (C). 2. (Also UP2) She has 47 + 25 = 72 dollars, hence (E). 3. 4 10000 + 3 1000 + 2 10 + 4 1 = 40000 + 3000 + 20 + 4 = 43000 + 24 = 43024, hence (B). 4. (Also MP7, UP4) The back of the necklace will look like the mirror image of the front of the necklace. So each letter will be mirrored, and the order of the letters will be reversed: KATE KA TE front back hence (A). 5. 60 minutes before 5.34 pm is 4.34 pm, so 58 minutes before 5.34 pm is 2 minutes later, which is 4.36 pm, hence (E). 6. (Also I2) The ma jor markings are 36, 37 and 38, so the minor markings are 0.2 units apart. Then the arrow is 0.3 to the right of 37, so it is on 37.3, hence (E). 7. 1000 7 = 142 with remainder 6, so after 142 subtractions of 7, Ishrak counts 6, hence (E). 8. Alternative 1 40  55  z x  The upper triangle's angles add to 180  , so x = 85  . Then x + z = 180  , so that z = 95  , hence (C). c Australian Mathematics Trust www.amt.edu.au 48

2018 AMC Junior Solutions Alternative 2 In the upper triangle, z = 40  + 55  , since an exterior angle of a triangle is equal to the sum of the other two interior angles, hence (C). 9. Who is youngest? It is not Amelia, Billie, David or Emily. So Charlie is youngest. Take out Charlie, who is youngest? It is not Amelia, Billie or David. So the youngest other than Charlie is Emily, hence (E). Note: In age order, the friends are Charlie, Emily, Billie, Amelia and David. 10. Here is the ribbon: 12 cm The rightmost piece is two-thirds of the right half, so one-third of the right half is 6 cm. That is, the right half is 18 cm. Then the left half is also 18 cm, for a total of 36 cm, hence (C). 11. (Also S7) 1000% means `10 times', since 1000% = 10 100%. So 10 times the number is 100, and the number is 10, hence (C). 12. (Also I6) Alternative 1 Tabulate the letters needed. Letter A D E N O R W Nora 1 - - 1 1 1 - Anne 1 - 1 2 - - - Warren 1 - 1 1 - 2 1 Andrew 1 1 1 1 - 1 1 Needed 1 1 1 2 1 2 1 Then we need 9 letters to cover every name, hence (B). c Australian Mathematics Trust www.amt.edu.au 49

2018 AMC Junior Solutions Alternative 2 ANDREW requires letters ADENRW (in alphabetical order). For WARREN, an additional R is needed: ADENRRW. For ANNE, an additional N is needed: ADENNRRW. For NORA, an O is needed: ADENNORRW. In all, 9 letters are needed, hence (B). 13. (Also I11, S8) To feed 4 dogs for 1 day costs $60 3 = $20, and then to feed 1 dog for 1 day costs $20 4 = $5. To feed 7 dogs for 7 days will cost $5 7 7 = $245, hence (C). 14. (Also UP18, I10, S4) There are many subdivisions of the hexagon into equal areas that show that the shaded area is 2 3 of the total: 2 3 4 6 4 6 8 12 hence (B). 15. Alternative 1 Let the edge of a tile be 1 unit. Then the rectangle has perimeter 12 units, so its height plus width is 6 units. Since the rectangle's height is 1 unit, its width is 5 units. Thus Leila has 5 tiles, hence (B). Alternative 2 Suppose Leila has ntiles, each of side s. The rectangle has sides sand sn, so it has perimeter 2 s+ 2sn. This will equal 3 4s = 12 s, so 12s = 2s+ 2sn. Then 10 s= 2 snand n = 10s 2s = 5, hence (B). 16. Alternative 1 If James's Group A choice is Mandarin, he has 3 Group B options. If James's Group A choice is Japanese, Spanish or Indonesian, he has 4 Group B options, giving 12 options. All of these 15 pairs identi ed are di erent, so there are 15 possible pairs, hence (D). c Australian Mathematics Trust www.amt.edu.au 50

2018 AMC Junior Solutions Alternative 2 James will choose one of four electives in group A and one of four electives in group B. There are 16 such choices. Of these, the only forbidden choice is to choose Mandarin from both groups, so there are 15 possible pairs, hence (D). 17. Given areas shown in the diagram, the shaded area is 3 4.5 10 1 20 10 3 1 4:5 = 1 :5 cm2 ; hence (B). 18. (Also I13) The top centre square must be 8. Then the top two corners must add to 10 and the bottom two corners must add to 14. So in any solution, all four corners add to 24. Checking, there are several possible solutions, as the diagrams show. 3 8 7 10 6 2 5 4 9 4 8 6 8 6 4 6 4 8 8 8 2 0 6 12 10 4 4 hence (D). 19. The folded piece consists of two right-angled triangles, with the fold line being a line of symmetry and the hypotenuse of both triangles. This must be a kite. The other piece is a concave pentagon. Let the side of the square be 2 units, so that its area is 4 square units. The triangle folded over has area 1 2  1 2 = 1 square unit, so that the kite's area is 2 square units. Then the pentagon's area is also 2 square units, equal to the kite, hence (A). 20. (Also I15) When the ob jects in diagrams (B), (C), (D) and (E) are viewed from the top, front-left, front-left and front-right, respectively, and rotated appropriately, then each of them has the view shown. However, when we try to do the same with (A) the best we can get is the mirror-image of this view, hence (A). c Australian Mathematics Trust www.amt.edu.au 51

2018 AMC Junior Solutions 21. A month of 30 days has 720 hours = 43200 minutes, so that 1 1000 of this month is ap- proximately 43 minutes. Other months vary from this by no more that 10%, so the best approximation given is 40 minutes, hence (D). 22. Alternative 1 The eight numbers add to 1 +


+ 8 = 36, so the outer 4 numbers add to 16. Let Tbe the sum of the numbers on a triangle. Then 4T = 220 + 1 16 = 56 so T= 14. We can ll in 11 yand 11 xon the square, and then the nal corner of the square is 20 3 (11 x) (11 y) = x+ y 5. Then the other two outer corners are 14 (11 y) (x +y 5) = 8 xand 14 (11 x) (x +y 5) = 8 y. x + y 5 11 x 11 y 3 8 y 8 x xy 20 14 14 14 14 Now, consider which of the numbers is 8. If x= 8, then 8 x= 0, which is not possible, so x6= 8. Similarly y6= 8. Also if 8 x= 8, then x= 0, which is not possible, so 8 x6= 8. Similarly 8 y6= 8. If 11 x= 8, then x= 3, which is already in the diagram. So 11 x6= 8, and similarly 11 y6= 8. Consequently x+ y 5 = 8 and so x+ y= 13, hence (D). Note: Once we have x+ y 5 = 8 and x+ y= 13, either x= 6 and y= 7 or vice versa. In either case the rest of the diagram can be completed using the formulas above. c Australian Mathematics Trust www.amt.edu.au 52

2018 AMC Junior Solutions Alternative 2 In this solution, sets of shaded circles are used to represent various totals. 1 3 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36 2 3 = 36 20 = 16 3 3 = 3 4 3 = 3 = 16 2 = 8 must be f1;2;6;7g 5 3 = 3 = 20 + 8 = 28 6 3 = 3 = 28 2 = 14 7 3 = 3 = 14 3 = 11 must be f4;5;6;7g 8 3 must be = 6 + 7 = 13 hence (D). 23. The hexagon has two rows of 11 triangles, so each row is not a whole number of rhombuses, and at least one rhombus must have one triangle from each row. This rhombus will be oriented vertically. Once there is one vertical rhombus then due to the 60  angles that need to be lled in the remaining space, all remaining rhombuses can only t in one way, with none of them vertical. For example, ! ! So each of the six choices of vertical rhombus lead to exactly one tiling, and there are six tilings, hence (A). c Australian Mathematics Trust www.amt.edu.au 53

2018 AMC Junior Solutions 24. (Also I18) Alternative 1 Multiply all numbers by lcm(3 ;4; 5; 6;7) = 420 to make an integer version of the problem. 140 105 84 70 60 Some of these numbers will be added, and some subtracted, and the result is a positive integer as close to 0 as possible. So the task is to separate the 5 integers into two sets (added versus subtracted), where the totals of the two sets are as close together as possible. To nd these sets, we make the total of each set as close as possible to a target of (140 + 105 + 84 + 70 + 60) 2 = 229 :5. The number 140 will be in one of the sets. The total of the other number(s) in this set will be close to 229 :5 140 = 89 :5. Clearly 84 is the closest single number. Also the smallest possible choice of 2 or more numbers is 60 + 70 = 130, which is worse. So this set is f140; 84g. Therefore in the best solution, the two sets are f140;84gand f105; 70;60g with totals 224 and 235 respectively, giving this solution. 140 + 10584 + 70 + 60 = 235 224 = 11 Transforming this solution back to the original fraction problem gives this solution. 1 3 + 1 4 1 5 + 1 6 + 1 7 = 11 420 This is approximately 1 40 , which is between 1 50 and 1 20 . Checking, 1 20 = 21 420 > 11 420 and 1 50 < 1 42 < 11 420 , so that 1 50 < 11 420 < 1 20 , hence (C). Alternative 2 As decimals, the numbers in the answers are 0, 0:01, 0:02, 0:05, 0:1 and 1. The shortest interval has length 0 :01, so we try to solve to 3 decimal places. To 3 decimal places, the fractions are 0.333, 0.25, 0.2, 0.167, and 0.143. These add to 1:093, so we try to split into two subsets, each with target 1:093 2 0:546. In the subset with 0:333, the remaining number(s) have target 0:546 0:333 = 0 :213. The closest we can get is 0:2. That is, the best total is 0 :333 + 0:2 = 0:533 and this leaves 0:25 + 0:167 + 0 :143 = 0:56 as the total of the other numbers. Thus the smallest possible positive answer is (0 :25 + 0:167 + 0 :143)(0:333 + 0 :2) = 0:56 0:533 = 0 :027. This answer is between 0 :02 =1 50 and 0 :05 = 1 20 , hence (C). Note: There are approximations in this solution, but they do not a ect the answer. Each fraction is within 0:0005 of the decimal approximation used. Hence the true answer is within 5 0:0005 = 0:0025 of 0 :027, and will still be between 0:02 and 0 :05. c Australian Mathematics Trust www.amt.edu.au 54

2018 AMC Junior Solutions 25. (Also I22, S19) Alternative 1 Start with some simple cases: 1 1 1 1 1 1 1 1 2 2 2 2 1 1 1 0 8 8 8 9 sum of digits = 3 1 + 3 8 + 9 = 36 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 1 1 1 1 0 8 8 8 8 9 sum of digits = 4 1 + 4 8 + 9 = 45 Clearly this pattern continues, and we can generalise. Then the sum of all digits is 49  1 + 0 + 49 8 + 9 = 450, hence (D). Alternative 2 111: : :111 | {z } 100 222 : : :222 | {z } 50 =111 : : :111 | {z } 100 111 : : :111 | {z } 50 111 : : :111 | {z } 50 =111 : : :111 | {z } 50 000 : : :000 | {z } 50 111 : : :111 | {z } 50 =111 : : :110 | {z } 50 999 : : :999 | {z } 50 111 : : :110 | {z } 50 =111 : : :110 | {z } 50 888 : : :889 | {z } 50 In the digit sum of this number, we can pair the 1 + 8 terms, giving 50 9 = 450, hence (D). 26. (Also UP27) LetXbe the sum of each pair of opposite faces, with digits abc. Since all digits on the cube are less than 5, the addition has no carry between place values. We just consider the sums of corresponding digits. In the hundreds, a= 1 + = 2 + , where each box is 0, 1 or 2. Only a= 2 and a= 3 are possible. (If a hidden number has fewer than 3 digits, its hundreds digit is taken to be 0.) In the tens, b= 0 + = 2 + , and only b= 2 is possible. In the units, c= 0 + = 1 + , and only c= 1 and c= 2 are possible. So Xis 221, 222, 321 or 322. If X = 322, then the hidden faces are 102, 201 and 121, and there are only 4 di erent numbers on the cube. So X= 322 is not a solution. If X = 321, then the hidden faces are 101, 200 and 120, and there are 6 di erent numbers on the cube. So X= 321 is the largest solution, hence (321). c Australian Mathematics Trust www.amt.edu.au 55

2018 AMC Junior Solutions 27. (Also I26) Suppose the number nhas digits a,band c, so that s= a+ b+ cis the digit sum. s 2 = n s = 100 a+ 10b +c (a +b+ c) = 99 a+ 9 b Then s2 is a multiple of 9, so that sis a multiple of 3. Also s p 99 >9 and s 27. So we check s= 12 ;15; 18; : : : ; 27 to see whether n= s2 + s= s(s + 1) works. s 12 15 18 21 24 27 n = s(s + 1) 156 240 342 462 600 756 X      Here s= 15; : : : ; 27 all fail since the digit sum of nis not equal to s. The only solution is n = 156, hence (156). 28. (Also I27) Consider a sign XjY that uses exactly two di erent digits, and where Xand Yare numbers with X+Y = 999. There are 10 possibilities for a, the units digit ofX. The units digit of Ywill be b= 9 a, which cannot be the same as a, so the two digits on the sign must be aand b. The tens digit of Xis either aor b, a two-way choice that also determines the tens digit of Y. Similarly the hundreds digits of Xand Yare a two-way choice. This gives 10 2 2 = 40 cases, each of which gives exactly one possible sign. Some of these have `leading zeros' like X= 090, Y= 909. However these still give a single valid sign such as 90j909 . So there are 40 signs that only use two digits, hence (40). 29. Alternative 1 The multiplication is (100X+ 10Y +Z) 18 = (1000Z + 100X+ 10Y+Y) 1800X + 180Y + 18Z= 1000Z + 100X+ 11Y 1700X + 169Y = 982Z Then 169Y is even, so that Ymust be even. Since 169 Yand 982Z di er by 1700X , they have the same last two digits. We list possible values of 169Y and 982Z. Y 2 4 6 8 169Y 3 38 676 1014 1352 Z 1 2 3 4 5 6 7 8 9 982Z 982 1964 2946 3928 4910 5892 6874 7856 88 38 Only when Z= 9 and Y= 2 do the last two digits match. Then 1700 X= 8838 338 = 8500 so that X= 5. Consequently the multiplication is 529 18 = 9522 and the three-digit number X Y Zis 529, hence (529). c Australian Mathematics Trust www.amt.edu.au 56

2018 AMC Junior Solutions Alternative 2 The multiplication gives us this equation. 18(100X+ 10Y+Z) = 1000 Z+ 100X + 10Y+Y 1700X + 169Y 982Z = 0 The coecients 1700, 169 and 982 are equal or close to multiples of 170. 170(10X+Y 6Z ) Y + 38Z = 0 Then 38 Z Y is a multiple of 170, say 38 Z Y = 170 n, where n= 6 Z 10X Y. Since Y and Zare non-zero digits, 0 170n. If n= 1, then the smallest multiple of 38 greater than 170 is 5 38 = 190, so Z 5. Then Y = 38 Z 170 20, which isn't possible if Yis a digit. If n= 2, then the smallest multiple of 38 greater than 340 is 9 38 = 342. Thus 38 9 2 = 340 = 170 nis a solution, where Z= 9 and Y= 2. Then 10 X= 6 Z Y n= 54 2 2 = 50 so that X= 5. Therefore the only solution has X= 5, Y= 2 and Z= 9, hence (529). Alternative 3 The units digit Yis the units digit of 8Z . SinceY6= 0, Z6= 5. The tens digit Yis equal to the units digit of 8Y +Z+ N, where Nis the tens digit of 8Z . Tabulate all possible values of Z. Z 1 2 3 4 6 7 8 9 Y , units digit of 8Z 8 6 4 2 8 6 4 2 N , tens digit of 8Z 0 1 2 3 4 5 6 7 units digit of 8Y 4 8 2 6 4 8 2 6 units digit of 8Y +Z+ N 5 1 7 3 4 0 6 2 The only match between Yand the units digit of 8 Y+Z+ N is with Z= 9. To nd X, consider the complete product (100X+ 29)18 = 9000 + 100 X+ 22 1800X + 522 = 9022 + 100 X 1700X = 8500 X = 5 Consequently the number X Y Zis 529, hence (529). c Australian Mathematics Trust www.amt.edu.au 57

2018 AMC Junior Solutions 30. (Also S26) A number is divisible by 9 if and only if the sum of its digits is divisible by 9. Thus, B and Care divisible by 9. Since Ais a 2018-digit number, B