Olympiad 2019 : Division J Wednesday 22 May 2019 1

APSMO OLYMPIAD 2019 : DIVISION J WEDNESDAY 22 MAY 2019 1

Total Time Allowed: 30 Minutes

1A. Suggested Time: 3 minutes Write your answers in the What is the value of N that makes the sentence true? boxes on the 2019 + 4038 + 6057 + 8076 + N = 2020 + 4040 + 6060 + 8080 back. ← 1B. Suggested Time: 5 minutes Keep your answers The figure shown is a hexagon with six sides of the hidden by same length. B C folding Not all of the interior angles are the same size. backwards on this line. Angles A and D are the same size. A D

Angles B, C, E, and F are the same size. F E How many lines of symmetry does the figure have?

1C. Suggested Time: 5 minutes

How many counting numbers greater than 100 and less than 1000 are palindromes? [A palindrome is a number that reads the same forwards and backwards such as 252 and 707.]

1D. Suggested Time: 6 minutes Sophie bought 4 pens and 2 pencils for $2.40. Lucas bought 2 pens and 4 pencils for $2.70. What is the cost of 1 pen?

1E. Suggested Time: 7 minutes

Jacob runs at 4 kilometres per hour uphill for half an hour. Then, he runs downhill, along the same route, at 5 kilometres per hour. How many minutes does Jacob take to come down the hill?

Copyright © 2019 Australasian Problem Solving Mathematical Olympiads (APSMO) Inc. and Mathematical Olympiads for Elementary and Middle Schools. All rights reserved. APSMO OLYMPIAD 2019 : DIVISION J WEDNESDAY 22 MAY 2019 1

1A. Student Name: Fold here. Keep your answers hidden. 1B.

1C.

1D.

1E.

Solutions and Answers For teacher use only. Not for Distribution. 1A: 10 1B: 2 1C: 90 1D: 35¢ 1E: 24

1A. METHOD 1 Strategy: Subtract the numbers on the left of the equals sign from the corresponding numbers on the right of the equals sign.

2020 + 4040 + 6060 + 8080 The sum of the numerical terms on the left side is 10 less than the sum of - 2019 - 4038 - 6057 - 8076 the terms on the right side. The non-numerical term on the left side is N. 1 + 2 + 3 + 4 So N = 10.

METHOD 2 Strategy: Examine patterns formed by the numbers on each side of the equation. The terms on the left side of the equation are each multiples of 2019: (1 × 2019) + (2 × 2019) + (3 × 2019) + (4 × 2019) = 10 × 2019 = 20 190. The terms on the right side of the equation are each multiples of 2020: (1 × 2020) + (2 × 2020) + (3 × 2020) + (4 × 2020) = 10 × 2020 = 20 200. Subtract the left value from the right value to get N = 20 200 – 20 190 = 10.

Follow-Up: What value of N makes the statement below true? 1001 + 2002 + 3003 + 4004 = N + 999 + 1999 + 2999 + 3999 [14]

1B. Strategy: Use the diagram to draw lines of symmetry. Since angles B and F are the same, and B C Since angles A and D are B C angles C and E are the same, folding the same, folding along along a horizontal line through A and A D a vertical line midway A D D will match two identical isosceles through segments BC and trapeziums. F E FE will match them up. F E However, when you fold the figure B C However, when you B C across line BE, vertex A does not fold the figure midway coincide with any other point. A D through AB, vertex C does A D That means that BE is not a line of not coincide with any symmetry. F E other point. F E Therefore, there are only 2 lines of symmetry.

Follow-Up: How many lines of symmetry does a regular hexagon have? [6]

1C. METHOD 1 Strategy: Use the multiplication principle (Fundamental Counting Principle). A 3-digit palindrome is of the form ABA, where A ≠ 0. There are 9 choices for A and for each of the different values of A there are 10 choices for B. Thus, there are 9 × 10 = 90 3-digit palindromes.

METHOD 2 Strategy: Make a list and look for a pattern. 101, 111, 121, 131, 141, 151, 161, 171, 181, 191 202, 212, 222, 232, 242, 252, 262, 272, 282, 292 Notice that there are ten 3-digit numbers for each different digit in the hundreds column. Since there are 9 possible digits in that column, there are 9 × 10 = 90 3-digit palindromes.

Follow-Up: Roll a 6-sided die four times to create a 4-digit number. How many different 4-digit palindromic numbers can result from this activity? [36]

1D. METHOD 1 Strategy: Use addition to find the cost of 6 pens + 6 pencils. 4 pens + 2 pencils cost $2.40, and 2 pens + 4 pencils cost $2.70. Therefore: 6 pens + 6 pencils will cost $2.40 + $2.70 = $5.10. 1 pen + 1 pencil will cost $5.10 ÷ 6 = $0.85. 2 pens + 2 pencils will cost 2 × $0.85 = $1.70. Since 4 pens + 2 pencils cost $2.40, and 2 pens + 2 pencils cost $1.70, 2 pens will cost a total of $2.40 – $1.70 = $0.70. Therefore 1 pen will cost $0.70 ÷ 2 = $0.35 or 35¢.

METHOD 2 Strategy: Create two equations and solve using algebra. Let the number of pens be p and the number of pencils be q. Write the cost in terms of cents.

The original statements are: 4p + 2q = 240 ... (1) 2p + 4q = 270 ... (2) Multiplying (1) by 2: 8p + 4q = 480 ... (3) Subtracting (2) from (3): 6p = 210 p = 210 ÷ 6 Therefore p = 35 cents.

Follow-Up: The total cost of 1 apple and 1 banana is $1. The total cost of 3 apples and 2 bananas is $2.40. Find the cost of 1 banana. [60¢]

1E. METHOD 1 Strategy: Compute the distance travelled. To travel at 4 kilometres per hour means that every hour, Jacob travels 4 kilometres. If Jacob only travels for half that time (half an hour), he only travels half as far, or 2 kilometres. On the way down the hill, Jacob runs at 5 kilometres per hour, which is 5 kilometres every 60 minutes. To run at 5 kilometres every 60 minutes is to run at 1 kilometre every 60 ÷ 5 = 12 minutes. Because Jacob has to run 2 kilometres, it takes him 2 × 12 = 24 minutes.

METHOD 2 Strategy: Use algebra. Let T = the amount of time to run downhill, in hours. Distance = 5 km/h × T hrs = 4 km/h × 1 hr 2 We know that distance = speed × time. Since the distance uphill is the same as the distance 5T km = 2km downhill, we can set up and solve the following equation: 2 T = hrs 2 5 So Jacob runs downhill for 5 hours. 2 5 × 60 minutes = 24 minutes.

Follow-Up: Jacob runs up a 2-kilometre hill at 5 kilometres per hour. He then runs downhill retracing the route he ran up the hill. Explain why no matter how fast he runs down the hill he cannot average 10 kilometres per hour for the entire route. [ Jacob runs uphill for 24 minutes. The total distance is 4 km. To have an average speed of 10 km/h, Jacob must run the 4 km in (4 ÷ 10) hours, or 24 minutes. Thus, Jacob has 0 minutes remaining to run down the hill.]