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Math Kangaroo — Ecolier

2010 Math Kangaroo — Ecolier

24 problems — read each, give it a real try, then peek at the hints.

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Problem 1 · 2010 Math Kangaroo Easy
Spatial & Visual Reasoning path-tracing
Figure for Math Kangaroo 2010 Problem 1
Show answer
Answer: E
Show hints
Hint 1 of 2
The road piece has to join the cat to the milk and the mouse to the cheese, yet keep those two routes from ever touching.
Still stuck? Show hint 2 →
Hint 2 of 2
Look at which sides of the missing square each road must enter and leave, then find the piece whose roads connect exactly those sides without crossing.
Show solution
Approach: match the road piece to the required connections
  1. The cat must reach the milk, and the mouse the cheese, but the two animals' paths must stay separate.
  2. So the missing piece needs two roads that link the correct opposite sides while never meeting in the middle.
  3. Only the curved piece E carries the two routes past each other without letting them join.
  4. So the piece is E.
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Problem 2 · 2010 Math Kangaroo Easy
Logic & Word Problems off-by-one

A 40 minute long lesson began at 11:50. Exactly in the middle of the lesson a bird flew into the classroom. At what time did this happen?

Show answer
Answer: C — 12:10
Show hints
Hint 1 of 2
"Exactly in the middle" of a 40-minute lesson is 20 minutes after it starts.
Still stuck? Show hint 2 →
Hint 2 of 2
Add half the length to the start time.
Show solution
Approach: add half the lesson length to the start
  1. Half of 40 minutes is 20 minutes.
  2. Start 11:50, then 20 minutes later is 12:10.
  3. So the bird flew in at 12:10.
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Problem 3 · 2010 Math Kangaroo Easy
Spatial & Visual Reasoning careful-counting
Figure for Math Kangaroo 2010 Problem 3
Show answer
Answer: D
Show hints
Hint 1 of 2
Count one type of shape at a time in each square.
Still stuck? Show hint 2 →
Hint 2 of 2
You need the square with exactly 3 four-sided shapes, 3 circles and 4 hearts.
Show solution
Approach: count each shape type and match the target
  1. Go square by square and tally the squares, circles and hearts separately.
  2. Look for the one with exactly 3 squares, 3 circles and 4 hearts.
  3. That square is D.
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Problem 4 · 2010 Math Kangaroo Easy
Arithmetic & Operations grouping

In a cafe the soup costs €4, the main course €9 and the dessert €5. The three courses ordered together cost €15. How many euros cheaper is this than ordering the same three courses separately?

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Answer: A — €3
Show hints
Hint 1 of 2
First add up the three separate prices.
Still stuck? Show hint 2 →
Hint 2 of 2
Then compare that total with the combined price of 15.
Show solution
Approach: compare separate total with bundle price
  1. Separately the meal costs 4 + 9 + 5 = 18 Euro.
  2. Together it costs 15 Euro.
  3. So the saving is 18 − 15 = 3 Euro.
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Problem 5 · 2010 Math Kangaroo Easy
Spatial & Visual Reasoning spatial-reasoning

Six coins make a triangle (see the picture). What is the smallest number of coins that must be moved to make the circle?

Figure for Math Kangaroo 2010 Problem 5
Show answer
Answer: B — 2
Show hints
Hint 1 of 2
Picture the 6-coin ring on top of the triangle and see which coins already sit in the right spots.
Still stuck? Show hint 2 →
Hint 2 of 2
Count how many coins are NOT already where the ring needs them.
Show solution
Approach: keep the coins already in place, move only the rest
  1. Lay the target ring of 6 coins over the triangle.
  2. Four of the coins already sit where the ring needs them; only two are out of place.
  3. So the smallest number of coins to move is 2.
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Problem 6 · 2010 Math Kangaroo Easy
Logic & Word Problems work-backward

Four friends each eat some ice cream. Mike eats more than Franz, Jaroslav eats more than Veit, and Jaroslav eats less than Franz. Put the friends in order by how much ice cream they ate, starting with the largest amount.

Show answer
Answer: C — Mike, Franz, Jaroslav, Veit
Show hints
Hint 1 of 2
Turn each clue into a simple "more than" comparison and chain them together.
Still stuck? Show hint 2 →
Hint 2 of 2
Jaroslav eats less than Franz but more than Veit, and Mike eats more than Franz.
Show solution
Approach: chain the inequalities into one order
  1. Mike > Franz, Jaroslav > Veit, and Jaroslav < Franz.
  2. Combine: Mike is biggest, then Franz, then Jaroslav, then Veit.
  3. Largest first: Mike, Franz, Jaroslav, Veit — option C.
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Problem 7 · 2010 Math Kangaroo Easy
Spatial & Visual Reasoning tiling-tessellation
Figure for Math Kangaroo 2010 Problem 7
Show answer
Answer: B
Show hints
Hint 1 of 2
Each tile is a square split by one diagonal (or a half-shaded diamond); the shaded part is always a triangle.
Still stuck? Show hint 2 →
Hint 2 of 2
Try to build each pattern from those triangle halves — one pattern needs a piece the tiles can't make.
Show solution
Approach: try to assemble each pattern from the triangular tiles
  1. The tiles only give you right-triangle halves of a square.
  2. Four of the patterns can be tiled with these halves.
  3. Pattern B requires a shape the given tiles cannot form, so it cannot be made.
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Problem 8 · 2010 Math Kangaroo Easy
Arithmetic & Operations work-backward

Eva is a centipede with exactly 100 feet. Yesterday she bought 16 pairs of shoes and put them on right away. Even so, she still had 14 feet with no shoes. On how many feet was she already wearing shoes before she went shopping yesterday?

Show answer
Answer: C — 54
Show hints
Hint 1 of 2
Each pair of shoes covers 2 feet; first find how many feet have shoes now.
Still stuck? Show hint 2 →
Hint 2 of 2
Subtract the feet she put new shoes on today from the total now wearing shoes.
Show solution
Approach: count shod feet, then remove today's new shoes
  1. She has 100 feet and 14 are bare, so 100 − 14 = 86 feet wear shoes now.
  2. Today she put on 16 pairs = 32 shoes, covering 32 feet.
  3. So before shopping she already wore shoes on 86 − 32 = 54 feet.
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Problem 9 · 2010 Math Kangaroo Medium
Arithmetic & Operations matching

Suppose  +  + 6 =  +  +  + . Which number should replace ?

Show answer
Answer: B — 3
Show hints
Hint 1 of 2
Both sides have two triangles, so cover those up with your finger on each side.
Still stuck? Show hint 2 →
Hint 2 of 2
Whatever is left over on the two sides must still be equal.
Show solution
Approach: match the same triangles on both sides and see what is left
  1. The left side is two triangles and a 6; the right side is two triangles and two more triangles.
  2. Cover the two matching triangles on each side, and the 6 is left on the left while two triangles are left on the right.
  3. So two triangles make 6, which means one triangle is 6 split into 2 equal parts, or 3.
  4. The triangle is 3, choice B.
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Problem 10 · 2010 Math Kangaroo Medium
Spatial & Visual Reasoning foldingsymmetry

Maria folds a square piece of paper so that the two kangaroos land exactly on top of each other. Along how many of the lines shown is this possible?

Figure for Math Kangaroo 2010 Problem 10
Show answer
Answer: C — 2
Show hints
Hint 1 of 2
A fold line works only if the two halves are mirror images across it.
Still stuck? Show hint 2 →
Hint 2 of 2
Check each drawn line and count how many are true lines of symmetry for the four kangaroos.
Show solution
Approach: count the lines of symmetry of the figure
  1. A fold makes the kangaroos overlap exactly only along a line of symmetry.
  2. Test the drawn lines: only two of them reflect the figure onto itself.
  3. So the answer is 2 lines.
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Problem 11 · 2010 Math Kangaroo Medium
Arithmetic & Operations off-by-one

Matthias and Klara live in a tower block. Klara lives 12 floors above Matthias. One day Matthias climbs the stairs to visit Klara. When he is halfway there he is on the 8th floor. On which floor does Klara live?

Show answer
Answer: B — 14th
Show hints
Hint 1 of 2
Halfway up the climb, Matthias has gone up half of the 12 floors.
Still stuck? Show hint 2 →
Hint 2 of 2
Find Matthias's floor first, then add 12 for Klara.
Show solution
Approach: use the halfway floor to find the start
  1. Half of the 12-floor climb is 6 floors, and that point is the 8th floor.
  2. So Matthias starts on the 8 − 6 = 2nd floor.
  3. Klara lives 12 floors higher: 2 + 12 = 14th floor.
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Problem 12 · 2010 Math Kangaroo Medium
Spatial & Visual Reasoning cube-views

A large cube is made from 64 small cubes. The 5 visible faces of the large cube are green and the bottom face is red. How many of the small cubes have 3 green faces?

Figure for Math Kangaroo 2010 Problem 12
Show answer
Answer: A — 4
Show hints
Hint 1 of 2
A small cube shows 3 green faces only if it is a corner of the big cube with all three faces on green sides.
Still stuck? Show hint 2 →
Hint 2 of 2
The bottom face is red, so bottom corners can't have 3 green faces — only the top corners can.
Show solution
Approach: check which corner cubes touch three green faces
  1. Only corner cubes can show three faces.
  2. The four bottom corners each touch the red bottom, so they have at most 2 green faces.
  3. The four top corners each touch the green top and two green sides — 3 green faces.
  4. So 4 small cubes have three green faces.
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Problem 13 · 2010 Math Kangaroo Medium
Counting & Probability careful-counting

Kangi walks straight from the zoo to the school (Schule) and counts the flowers along the way. Which of these numbers can he not get this way?

Figure for Math Kangaroo 2010 Problem 13
Show answer
Answer: C — 11
Show hints
Hint 1 of 2
He can take either branch of the first loop and either branch of the second loop, always crossing the middle.
Still stuck? Show hint 2 →
Hint 2 of 2
Add up the flowers for each route choice and see which of the listed totals never comes out.
Show solution
Approach: add the flowers for every route and find the missing total
  1. The walk has a choice in the first loop, the fixed middle path, and a choice in the second loop.
  2. Adding the flowers for the four possible routes gives several totals.
  3. The number that none of the routes produces is 11.
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Problem 14 · 2010 Math Kangaroo Medium
Arithmetic & Operations guess-and-check

A ferry boat can carry, in one trip, either 10 cars or 6 lorries. Yesterday the boat crossed the river 5 times. It was always full and carried 42 vehicles in all. How many of these were cars?

Show answer
Answer: E — 30
Show hints
Hint 1 of 2
Each of the 5 full trips carries either 10 cars or 6 lorries.
Still stuck? Show hint 2 →
Hint 2 of 2
Start by pretending every trip was lorries, then see how far short of 42 you are.
Show solution
Approach: start from all-lorry trips and swap until the total is right
  1. If all 5 trips were lorry trips, that would be 6 + 6 + 6 + 6 + 6 = 30 vehicles, which is 12 short of 42.
  2. Changing one lorry trip (6) into a car trip (10) adds 4 vehicles, and 12 needs three such changes.
  3. So 3 trips were car trips: 10 + 10 + 10 = 30 cars, choice E.
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Problem 15 · 2010 Math Kangaroo Medium
Arithmetic & Operations patterndoubling

Hans started a chain e-mail. He sent an e-mail to his friend Peter, who sent it on to 2 more people. Each person who gets the e-mail sends it on to 2 more people. After 3 rounds, 1 + 2 + 4 = 7 people have received the e-mail. How many people have received the e-mail after 5 rounds?

Show answer
Answer: C — 31
Show hints
Hint 1 of 2
Each round doubles the number of new people: 1, 2, 4, then 8, 16.
Still stuck? Show hint 2 →
Hint 2 of 2
Add up the new people from all five rounds.
Show solution
Approach: sum the doubling rounds
  1. Each round the number of new people doubles: 1, 2, 4, then 8, then 16.
  2. Add up all five rounds: 1 + 2 + 4 + 8 + 16 = 31.
  3. So 31 people have the e-mail, choice C.
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Problem 16 · 2010 Math Kangaroo Medium
Logic & Word Problems proportion

On the playground some children measure the length of the playground in strides. Anni takes 15 strides, Betty 17, Denis 12 and Ivo 14. Who has the longest stride?

Show answer
Answer: C — Denis
Show hints
Hint 1 of 2
They all cross the same length, so fewer strides means each stride is longer.
Still stuck? Show hint 2 →
Hint 2 of 2
Find who used the fewest strides.
Show solution
Approach: fewest strides means the longest stride
  1. The playground length is fixed, so the longest stride belongs to whoever takes the fewest steps.
  2. Denis takes only 12 strides, the fewest of all.
  3. So Denis has the longest stride.
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Problem 17 · 2010 Math Kangaroo Stretch
Arithmetic & Operations arithmetic-seriessum-constraint

Which number must replace the question mark if the total of the numbers in each row is the same?

12345678910199
11121314151617181920?
Show answer
Answer: A — 99
Show hints
Hint 1 of 2
Add up the top row, then make the bottom row reach the same total.
Still stuck? Show hint 2 →
Hint 2 of 2
The top row is 1+2+…+10 plus 199.
Show solution
Approach: match the two row sums
  1. Top row: 1 + 2 + … + 10 = 55, plus 199 gives 254.
  2. Bottom row: 11 + 12 + … + 20 = 155.
  3. So the missing number is 254 − 155 = 99.
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Problem 18 · 2010 Math Kangaroo Stretch
Arithmetic & Operations number-systems

The number \(60 \times 60 \times 24 \times 7\) is the same as

Show answer
Answer: D — the number of seconds in one week
Show hints
Hint 1 of 2
Read the factors as time conversions: 60 seconds, 60 minutes, 24 hours, 7 days.
Still stuck? Show hint 2 →
Hint 2 of 2
Multiplying them turns seconds all the way up to one week.
Show solution
Approach: interpret the product as a chain of time units
  1. 60 × 60 turns seconds into hours, × 24 turns hours into days, × 7 turns days into a week.
  2. So 60 × 60 × 24 × 7 is the number of seconds in one week.
  3. That matches option D.
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Problem 19 · 2010 Math Kangaroo Stretch
Arithmetic & Operations ages

Two years ago the cats Tim and Tom were 15 years old together. Now Tom is 13 years old. In how many years will Tim be 9 years old?

Show answer
Answer: C — 3
Show hints
Hint 1 of 2
Two years pass for both cats, so their combined age grows by 4.
Still stuck? Show hint 2 →
Hint 2 of 2
Find Tim's age now, then count up to 9.
Show solution
Approach: track the ages forward
  1. Two years ago Tim + Tom = 15, so now Tim + Tom = 15 + 4 = 19.
  2. Tom is now 13, so Tim is 19 − 13 = 6.
  3. Tim reaches 9 in 9 − 6 = 3 years.
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Problem 20 · 2010 Math Kangaroo Stretch
Spatial & Visual Reasoning gridsequence-of-figures
Figure for Math Kangaroo 2010 Problem 20
Show answer
Answer: C
Show hints
Hint 1 of 2
In a 5-column table, the number just below another is always 5 more, and the one to the right is 1 more.
Still stuck? Show hint 2 →
Hint 2 of 2
Check each piece: do the shown numbers line up with those +5 and +1 steps for their positions?
Show solution
Approach: test each piece against the table's +5 (down) and +1 (right) rule
  1. With 5 columns, moving down adds 5 and moving right adds 1.
  2. Check the relative positions of the two given numbers in each piece against that rule.
  3. Only piece C has its numbers in positions consistent with the table.
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Problem 21 · 2010 Math Kangaroo Stretch
Number Theory casework

The teacher said, “In our school library there are roughly 2010 books.” The pupils then guessed exactly how many there are. Artur guessed 2010, Beate guessed 1998 and Carlos guessed 2015. Their guesses are off by 12, 7 and 5, but not in that order. How many books are in the library?

Show answer
Answer: A — 2003
Show hints
Hint 1 of 2
The real number differs from the three guesses by 12, 7 and 5 in some order.
Still stuck? Show hint 2 →
Hint 2 of 2
Try a value near 2010 and check that its distances to 2010, 1998 and 2015 are exactly 12, 7 and 5.
Show solution
Approach: find the value whose distances to the guesses are 12, 7, 5
  1. Test 2003: |2003 − 2010| = 7, |2003 − 1998| = 5, |2003 − 2015| = 12.
  2. Those are exactly 7, 5 and 12 — the required errors.
  3. So the library has 2003 books.
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Problem 22 · 2010 Math Kangaroo Stretch
Spatial & Visual Reasoning paper-cuttingnet-folding

Lines are drawn on a piece of paper and some of the lines are numbered. The paper is cut along some of these lines and then folded as shown in the picture. What is the total of the numbers on the lines that were cut?

Figure for Math Kangaroo 2010 Problem 22
Show answer
Answer: D — 20
Show hints
Hint 1 of 2
A line is cut only if the fold could not bring its two sides together; uncut lines are the fold creases.
Still stuck? Show hint 2 →
Hint 2 of 2
Figure out which numbered lines stayed as folds, and add up the rest.
Show solution
Approach: separate fold creases from cut lines, then add the cut numbers
  1. Match the folded result to the flat sheet to see which lines were creases and which were cut.
  2. Add the numbers on the lines that were cut.
  3. That total is 20.
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Problem 23 · 2010 Math Kangaroo Stretch
Logic & Word Problems casework

Andrew, Stefan, Robert and Marko meet at a concert in Zagreb. They come from different cities — Paris, Dubrovnik, Rome and Berlin (not necessarily in this order).

• Andrew and the friend from Berlin arrive first in Zagreb. Neither of these two has ever been to Paris or Rome.
• Robert is not from Berlin, but he arrives together with the friend from Paris.
• Marko and the friend from Paris enjoyed the concert very much.

Which city does Marko come from?

Show answer
Answer: D — Berlin
Show hints
Hint 1 of 2
Andrew is not from Paris, Rome, or Berlin, so his city is forced.
Still stuck? Show hint 2 →
Hint 2 of 2
Place each person's city one clue at a time until only Marko's is left.
Show solution
Approach: eliminate cities person by person
  1. Andrew and the Berlin friend are two people, and Andrew hasn't been to Paris or Rome, so Andrew is from Dubrovnik.
  2. Robert isn't from Berlin and arrives with the Paris friend, so Robert is from Rome.
  3. That leaves Paris and Berlin for Stefan and Marko; since Marko isn't the Paris friend, Marko is from Berlin.
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Problem 24 · 2010 Math Kangaroo Stretch
Number Theory caseworksum-constraint

Berti’s friends each add together the day and the month of their birthday. They all get the answer 35, but no two of them have the same birthday. What is the largest number of friends Berti can have?

Show answer
Answer: B — 8
Show hints
Hint 1 of 2
You need months and days with month + day = 35, and each birthday must be a real date.
Still stuck? Show hint 2 →
Hint 2 of 2
Start from December and step down through the months, checking the day fits that month.
Show solution
Approach: count valid (month, day) pairs summing to 35
  1. List month + day = 35 with a valid day: Dec 23, Nov 24, Oct 25, Sep 26, Aug 27, Jul 28, Jun 29, May 30.
  2. April would need day 31, which doesn't exist, and earlier months need impossible days.
  3. That gives 8 different birthdays, so at most 8 friends.
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