🦘 Math Kangaroo Grade All Felix 1-2 Ecolier 3-4 Benjamin 5-6 Kadett 7-8 Junior 9-10 Student 11-12 ⇄ switch contest
Topic

Arithmetic & Operations

Order of operations, adding smartly, structure in a calculation.

70 problems 📖 Read the lesson
Practice
Problem 1 · 2025 Math Kangaroo Easy
Arithmetic & Operations place-value

Nico and his little sister play with shells and marbles. Each shell is worth 6 and each marble is worth 1 (shell = 6, marble = 1). Which picture shows the value 16?

Figure for Math Kangaroo 2025 Problem 1
Show answer
Answer: E
Show hints
Hint 1 of 2
Each shell counts as 6 and each marble counts as 1 - add up each picture's total.
Still stuck? Show hint 2 →
Hint 2 of 2
You need a total of exactly 16, so look for two shells plus four marbles.
Show solution
Approach: add the shell and marble values in each option
  1. A shell is worth 6 and a marble is worth 1.
  2. Two shells give 12, and you need 4 more to reach 16, so 4 marbles.
  3. The picture with two shells and four marbles totals 12 + 4 = 16.
  4. That is option E.
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Problem 4 · 2025 Math Kangaroo Easy
Arithmetic & Operations order-of-operationssum-constraint

Simona writes the numbers 2, 0, 2 and 5 in the boxes, one number per box (see picture). In what order can she write them so that the calculation gives the biggest result?

Figure for Math Kangaroo 2025 Problem 4
Show answer
Answer: E — 5, 2, 0, 2
Show hints
Hint 1 of 2
The third box is the one that gets subtracted, so put the smallest number there.
Still stuck? Show hint 2 →
Hint 2 of 2
Put the 0 in the subtracted (third) box and add everything else.
Show solution
Approach: minimise what is subtracted, maximise what is added
  1. The calculation is first + second minus third + fourth.
  2. To make it biggest, subtract the smallest number, which is 0.
  3. Then the other three (5, 2, 2) are all added: 5 + 2 - 0 + 2 = 9.
  4. The order 5, 2, 0, 2 does this, which is option E.
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Problem 2 · 2023 Math Kangaroo Easy
Arithmetic & Operations sum-constraint

The two markers with a question mark have the same value: \(20 + 10 + 10 + ? + ? + 1 = 51\). Which value do you have to use instead of the question mark so that the calculation is correct?

Figure for Math Kangaroo 2023 Problem 2
Show answer
Answer: C — 5
Show hints
Hint 1 of 2
Add up the numbers you can already read.
Still stuck? Show hint 2 →
Hint 2 of 2
Subtract that running total from 51, then split the leftover between the two equal markers.
Show solution
Approach: fill the known values, then split the remainder
  1. The known markers add to 20 + 10 + 10 + 1 = 41.
  2. The two equal question marks must make up 51 − 41 = 10.
  3. Since the two are equal, each is 10 ÷ 2 = 5.
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Problem 2 · 2022 Math Kangaroo Easy
Arithmetic & Operations arithmetic-series

On every birthday Maria gets as many teddies as the age she turns: 1 teddy on her first birthday, 2 teddies on her second birthday, and so on. How many teddies does Maria have in total the day after her sixth birthday?

Show answer
Answer: C — 21
Show hints
Hint 1 of 2
On each birthday she gets a number of teddies equal to her age that day.
Still stuck? Show hint 2 →
Hint 2 of 2
Add up the teddies from birthday 1 through birthday 6.
Show solution
Approach: add the gifts from each birthday
  1. Birthdays 1 to 6 give 1, 2, 3, 4, 5 and 6 teddies.
  2. Add them up: 1 + 2 + 3 + 4 + 5 + 6 = 21.
  3. So the day after her sixth birthday she has 21 teddies (choice C).
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Problem 3 · 2021 Math Kangaroo Easy
Arithmetic & Operations order-of-operations

When you put the 4 puzzle pieces together correctly, they form a rectangle with a calculation on it. What is the result of this calculation?

Figure for Math Kangaroo 2021 Problem 3
Show answer
Answer: B — 15
Show hints
Hint 1 of 2
The four jigsaw pieces fit into one row, so the symbols line up left to right into a calculation.
Still stuck? Show hint 2 →
Hint 2 of 2
Arrange 3, 2, 1 and + so the tabs match, then read the calculation and work it out.
Show solution
Approach: assemble the pieces into a valid calculation and evaluate
  1. The pieces 3, 2, 1 and + interlock into a single row.
  2. Matching the tabs, the 1 and the 3 sit together to make 13, then the + and the 2 follow: 13 + 2.
  3. 13 + 2 = 15.
  4. So the result is 15 (B).
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Problem 7 · 2021 Math Kangaroo Easy
Arithmetic & Operations work-backwardfair-share

Denise fired a silver and a gold rocket at the same time. The rockets exploded into 20 stars in total. The gold rocket exploded into 6 more stars than the silver one. How many stars did the gold rocket explode into?

Show answer
Answer: D — 13
Show hints
Hint 1 of 2
The two rockets together make 20 stars, and the gold makes 6 more than the silver.
Still stuck? Show hint 2 →
Hint 2 of 2
Take the 6 extra off the 20 first, split what is left evenly, then give the gold its extra back.
Show solution
Approach: split the total after removing the difference
  1. Remove the 6 extra gold stars: 20 - 6 = 14 shared equally.
  2. Each rocket's base share is 14 / 2 = 7, so the silver has 7.
  3. The gold has 7 + 6 = 13.
  4. So the gold rocket made 13 stars (D).
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Problem 3 · 2020 Math Kangaroo Easy
Arithmetic & Operations order-of-operations
Figure for Math Kangaroo 2020 Problem 3
Show answer
Answer: C
Show hints
Hint 1 of 2
Work out each little calculation and check which ones equal 24.
Still stuck? Show hint 2 →
Hint 2 of 2
Paint a cell only when its value is exactly 24.
Show solution
Approach: evaluate each cell and paint the ones equal to 24
  1. Top row: 28−4 = 24, 4×6 = 24, 18+6 = 24 — all three are painted.
  2. Bottom row: 19+6 = 25, 8×3 = 24, 29−6 = 23 — only the middle one is painted.
  3. So the whole top row is shaded and only the middle of the bottom row is, which is picture C.
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Problem 2 · 2019 Math Kangaroo Easy
Arithmetic & Operations place-value

The diagram shows the number 8. A dot stands for the number 1 and a line for the number 5. Which diagram represents the number 12?

Figure for Math Kangaroo 2019 Problem 2
Show answer
Answer: C
Show hints
Hint 1 of 2
A dot is worth 1 and a line is worth 5; the example 8 uses 3 dots and 1 line.
Still stuck? Show hint 2 →
Hint 2 of 2
Make 12 using as many 5s (lines) as possible, then add dots for the rest.
Show solution
Approach: build 12 from fives and ones
  1. A line = 5 and a dot = 1.
  2. Two lines give 10, and two dots give 2, for a total of 12.
  3. The diagram with 2 lines and 2 dots is C.
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Problem 8 · 2019 Math Kangaroo Easy
Arithmetic & Operations work-backward

Which number goes into the field with the question mark, if all calculations are solved correctly?

Figure for Math Kangaroo 2019 Problem 8
Show answer
Answer: B — 5
Show hints
Hint 1 of 2
Start at START and follow each small calculation step by step into the next box.
Still stuck? Show hint 2 →
Hint 2 of 2
Fill the boxes in order; the value flowing into the question-mark box is the answer.
Show solution
Approach: follow the calculation chain
  1. Begin from START and apply each operation along the path.
  2. Work box by box, carrying each result into the next calculation.
  3. The value that lands in the question-mark box is 5 (B).
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Problem 2 · 2018 Math Kangaroo Easy
Arithmetic & Operations addition

Susanne is 6 years old. Her sister Lisa is 2 years younger. Her brother Max is 2 years older than Susanne. How old are the three siblings altogether?

Show answer
Answer: D — 18
Show hints
Hint 1 of 2
Work out each sibling's age from Susanne's.
Still stuck? Show hint 2 →
Hint 2 of 2
Lisa is younger, Max is older — then just add the three ages.
Show solution
Approach: find each age, then add
  1. Susanne is 6.
  2. Lisa is 2 younger: 6 − 2 = 4.
  3. Max is 2 older: 6 + 2 = 8.
  4. Total: 6 + 4 + 8 = 18.
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Problem 1 · 2017 Math Kangaroo Easy
Arithmetic & Operations place-value

Which one of the domino pieces A to E has to be placed in between the shown pieces, so that both calculations are correct?

Figure for Math Kangaroo 2017 Problem 1
Show answer
Answer: D
Show hints
Hint 1 of 2
Work out 16 - 3 first, then see what the second domino half must equal.
Still stuck? Show hint 2 →
Hint 2 of 2
One half of the new piece must match 16 - 3; the other half must produce 2017.
Show solution
Approach: match each half of the inserted domino to the two required calculations
  1. The left calculation must equal 16 - 3 = 13, so the touching half must read = 13.
  2. The right calculation must equal 2017, so the other half must be 2000 + 17.
  3. Only the piece with = 13 and 2000 + 17 fits both, which is D.
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Problem 5 · 2017 Math Kangaroo Easy
Arithmetic & Operations total-then-divide

How many white squares need to be coloured in black, so that there are exactly twice as many white squares as there are black squares?

Figure for Math Kangaroo 2017 Problem 5
Show answer
Answer: B — 3
Show hints
Hint 1 of 2
Count the squares in the grid, then count how many are already black.
Still stuck? Show hint 2 →
Hint 2 of 2
If the whites are to be twice the blacks, then out of every 3 squares exactly 1 is black.
Show solution
Approach: split the grid into thirds: one third must end up black
  1. The grid has 24 squares in all.
  2. Twice as many white as black means 1 of every 3 squares is black, so 24 ÷ 3 = 8 squares must be black.
  3. There are already 5 black squares, so colour 8 − 5 = 3 more.
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Problem 6 · 2017 Math Kangaroo Easy
Arithmetic & Operations order-of-operations

Which number is hidden behind the panda?

Figure for Math Kangaroo 2017 Problem 6
Show answer
Answer: A — 16
Show hints
Hint 1 of 2
Fill in each box one at a time, following the arrows from the start.
Still stuck? Show hint 2 →
Hint 2 of 2
Do exactly what each arrow says (add or take away) before moving to the next box.
Show solution
Approach: follow the chain step by step
  1. First box: 10 + 6 = 16.
  2. Add 8: 16 + 8 = 24; then 24 - 6 = 18.
  3. Add 8: 18 + 8 = 26; then 26 - 10 = 16.
  4. So the panda hides 16.
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Problem 1 · 2016 Math Kangaroo Easy
Arithmetic & Operations careful-counting

Amy, Bert, Carl, Doris and Ernst each throw two dice (see picture). Who has got the biggest total altogether?

Figure for Math Kangaroo 2016 Problem 1
Show answer
Answer: E — Ernst
Show hints
Hint 1 of 2
For each child, add the dots showing on their two dice.
Still stuck? Show hint 2 →
Hint 2 of 2
You only need to find the single biggest sum, so spot the pair with the most dots.
Show solution
Approach: add the pips on each pair of dice and compare
  1. Count the dots on both dice for every child and add them.
  2. Ernst's two dice show the most dots of anyone.
  3. So Ernst has the biggest total.
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Problem 2 · 2016 Math Kangaroo Easy
Arithmetic & Operations

A kangaroo is 7 weeks and 2 days old. In how many days is it 8 weeks old?

Show answer
Answer: E — 5
Show hints
Hint 1 of 2
One week is 7 days, so think about how many days are left until the next full week.
Still stuck? Show hint 2 →
Hint 2 of 2
From 7 weeks 2 days, count the days needed to reach exactly 8 weeks.
Show solution
Approach: count days to the next whole week
  1. Going from 7 weeks up to 8 weeks is one more full week, which is 7 days.
  2. The kangaroo has already lived 2 of those days.
  3. So 7 − 2 = 5 more days until it is 8 weeks old.
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Problem 3 · 2016 Math Kangaroo Easy
Arithmetic & Operations order-of-operations

Follow the arrows through the boxes (see diagram). What is the final result?

Figure for Math Kangaroo 2016 Problem 3
Show answer
Answer: A — 24
Show hints
Hint 1 of 2
Work the two top boxes first, then feed both results into the bottom + box.
Still stuck? Show hint 2 →
Hint 2 of 2
The left box is 17 + 3 and the right box is 20 − 16.
Show solution
Approach: follow the flow chart from top to bottom
  1. Left box: 17 + 3 = 20.
  2. Right box: 20 − 16 = 4.
  3. Bottom box: 20 + 4 = 24.
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Problem 3 · 2015 Math Kangaroo Easy
Arithmetic & Operations total-then-divide
Figure for Math Kangaroo 2015 Problem 3
Show answer
Answer: B
Show hints
Hint 1 of 2
First add up everything Lucy had in her purse.
Still stuck? Show hint 2 →
Hint 2 of 2
Take away the 7 Kangas she paid; match what is left to a picture.
Show solution
Approach: total the money, then subtract the price
  1. The purse holds a 10, two 2-coins and a 1-coin: 10 + 2 + 2 + 1 = 15 Kangas.
  2. Paying 7 for the ball leaves 15 − 7 = 8 Kangas.
  3. The purse that adds up to 8 is the 5 + 2 + 1 picture.
  4. That is option B.
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Problem 1 · 2013 Math Kangaroo Easy
Arithmetic & Operations careful-countingarithmetic-series

How many more bricks does the right-hand pyramid have than the left-hand pyramid?

Figure for Math Kangaroo 2013 Problem 1
Show answer
Answer: B — 5
Show hints
Hint 1 of 3
Count the bricks in each pyramid, going one row at a time from the top.
Still stuck? Show hint 2 →
Hint 2 of 3
Notice the right pyramid is just the left one with one extra row added at the bottom.
Still stuck? Show hint 3 →
Hint 3 of 3
So you only need to count the bricks in that one extra bottom row.
Show solution
Approach: the right pyramid is the left pyramid plus one extra bottom row
  1. Both pyramids look the same at the top; the right one has one more row at the bottom.
  2. Count that extra bottom row: it has 5 bricks.
  3. So the right pyramid has 5 more bricks than the left one, which is answer B.
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Problem 5 · 2012 Math Kangaroo Easy
Arithmetic & Operations division

13 children play hide and seek. One of them is the seeker. After a little while 9 children have been found. How many are still hiding?

Show answer
Answer: A — 3
Show hints
Hint 1 of 2
One of the 13 is the seeker, so how many are actually hiding?
Still stuck? Show hint 2 →
Hint 2 of 2
Subtract the children already found from the number hiding.
Show solution
Approach: subtract in two steps
  1. One child seeks, so 13 - 1 = 12 children are hiding.
  2. Of those, 9 have been found, leaving 12 - 9 = 3.
  3. So 3 children are still hiding.
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Problem 6 · 2012 Math Kangaroo Easy
Arithmetic & Operations careful-counting

Mike and Jake play darts. Each of them throws three darts (see picture). Who won, and by how many points?

Figure for Math Kangaroo 2012 Problem 6
Show answer
Answer: E — Mike won. He had 4 points more.
Show hints
Hint 1 of 2
Read off the value of each ring the dart lands in, for both boys.
Still stuck? Show hint 2 →
Hint 2 of 2
Add each boy's three scores, then compare the totals.
Show solution
Approach: add each player's three darts and compare
  1. Mike's darts land on 25, 35 and 7, giving 25 + 35 + 7 = 67.
  2. Jake's darts land on 15, 45 and 3, giving 15 + 45 + 3 = 63.
  3. Mike scores 67 and Jake 63, so Mike wins by 67 - 63 = 4.
  4. The answer is Mike won by 4 (E).
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Problem 8 · 2012 Math Kangaroo Easy
Arithmetic & Operations careful-countingoff-by-one

On the 24th of February 2012 Grandfather's chicks hatched. There are 29 days in February in 2012. How old are the chicks today, on the 15th of March 2012?

Show answer
Answer: D — 20 days
Show hints
Hint 1 of 2
Count the days from 24 February up to 15 March, using that February has 29 days.
Still stuck? Show hint 2 →
Hint 2 of 2
Find how many days from the 24th to the end of February, then add the March days.
Show solution
Approach: count days across the month boundary
  1. From 24 February to 29 February is 5 days.
  2. From 1 March to 15 March is another 15 days.
  3. Altogether that is 5 + 15 = 20 days.
  4. The chicks are 20 days old.
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Problem 1 · 2011 Math Kangaroo Easy
Arithmetic & Operations careful-countingoff-by-one

Bernd wants to paint the word KANGAROO. He begins on a Wednesday and paints one letter each day. On which day will he paint the last letter?

Show answer
Answer: C — Wednesday
Show hints
Hint 1 of 2
Write the days out one letter at a time, starting on Wednesday.
Still stuck? Show hint 2 →
Hint 2 of 2
KANGAROO has eight letters — land on the right weekday for the eighth one.
Show solution
Approach: count days letter by letter
  1. KANGAROO has 8 letters, painted on 8 days in a row starting Wednesday.
  2. Wed, Thu, Fri, Sat, Sun, Mon, Tue are the first seven days; the eighth day is the next Wednesday, answer C.
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Problem 7 · 2011 Math Kangaroo Easy
Arithmetic & Operations careful-counting

The bell of a clocktower rings every full hour (8:00, 9:00, 10:00 etc.) and rings as many times as the number of hours. It also rings once on every half hour (8:30, 9:30, 10:30 etc.). How often will it ring between 7:55 and 10:45?

Show answer
Answer: D — 30 times
Show hints
Hint 1 of 2
List every full hour and every half hour strictly between 7:55 and 10:45.
Still stuck? Show hint 2 →
Hint 2 of 2
A full hour rings as many times as the hour number; each half hour rings once.
Show solution
Approach: count rings by occasion
  1. Full hours in the window: 8:00 rings 8, 9:00 rings 9, 10:00 rings 10, total 27 rings.
  2. Half hours 8:30, 9:30, 10:30 each ring once, adding 3.
  3. Total = 27 + 3 = 30 times, answer 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?

Show answer
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 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 1 · 2009 Math Kangaroo Easy
Arithmetic & Operations order-of-operations

\(2 \times 9 + 200 + 9 = {?}\)

Show answer
Answer: E — 227
Show hints
Hint 1 of 2
Do the multiplication before the additions.
Still stuck? Show hint 2 →
Hint 2 of 2
Then just add the three numbers in any order.
Show solution
Approach: evaluate the expression with order of operations
  1. First multiply: 2 × 9 = 18.
  2. Now add: 18 + 200 + 9.
  3. 18 + 9 = 27, and 27 + 200 = 227.
  4. So the value is 227.
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Problem 2 · 2009 Math Kangaroo Easy
Arithmetic & Operations proportion

Four sticks have 8 ends. How many ends do 7 sticks have?

Show answer
Answer: E — 14
Show hints
Hint 1 of 2
Each stick has exactly 2 ends.
Still stuck? Show hint 2 →
Hint 2 of 2
So the number of ends is just twice the number of sticks.
Show solution
Approach: 2 ends per stick
  1. Every single stick has 2 ends, no matter how many sticks there are.
  2. 7 sticks × 2 ends each = 14 ends.
  3. So 7 sticks have 14 ends.
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Problem 14 · 2025 Math Kangaroo Hard
Arithmetic & Operations total-then-divide

Rudi feeds six sheep in the petting zoo. The six sheep get a total of 210 g of food. Each of the five large sheep gets the same amount, and the small sheep gets twice as much as a large sheep. How much food does the small sheep get?

Show answer
Answer: C — 60 g
Show hints
Hint 1 of 2
The small sheep eats as much as two large sheep, so count it as two large sheep.
Still stuck? Show hint 2 →
Hint 2 of 2
Then the 210 g is shared into 7 equal large-sheep portions.
Show solution
Approach: count everything in equal large-sheep portions
  1. The small sheep eats as much as two large sheep, so pretend it is two large sheep.
  2. Now there are 5 + 2 = 7 equal large-sheep portions sharing 210 g.
  3. Each portion is 210 ÷ 7 = 30 g.
  4. The small sheep gets two portions: 30 + 30 = 60 g, option C.
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Problem 18 · 2024 Math Kangaroo Stretch
Arithmetic & Operations groupingtotal-then-divide

Lucy weighs building blocks two at a time and reads these scale values: 200 g, 100 g and 240 g (see picture). How much do the three different building blocks weigh all together?

Figure for Math Kangaroo 2024 Problem 18
Show answer
Answer: A — 270 g
Show hints
Hint 1 of 2
Each reading is the weight of two of the three blocks together.
Still stuck? Show hint 2 →
Hint 2 of 2
Add the three known readings — that counts every block twice.
Show solution
Approach: add the pair-weights and halve
  1. The three readings 200, 100, 240 each weigh two blocks, so together they count all three blocks twice.
  2. Sum: 200 + 100 + 240 = 540 grams.
  3. Half of that is the weight of the three blocks once: 540 ÷ 2 = 270.
  4. Answer: 270 g (A).
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Problem 13 · 2023 Math Kangaroo Hard
Arithmetic & Operations sum-constraint

In a queue in front of a ferry there are 8 cars with 19 people in total. There are either 2 or 3 people in each car. How many cars are there with exactly 2 people?

Show answer
Answer: D — 5
Show hints
Hint 1 of 2
If every car held 2 people there would be only 16; you need 3 more.
Still stuck? Show hint 2 →
Hint 2 of 2
Each car upgraded from 2 to 3 people adds exactly one person.
Show solution
Approach: start from all-twos and account for the extra people
  1. Eight cars with 2 people each would carry 16 people, but there are 19.
  2. The 3 extra people come from 3 cars holding 3 instead of 2.
  3. So 8 − 3 = 5 cars hold exactly 2 people.
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Problem 22 · 2021 Math Kangaroo Stretch
Arithmetic & Operations part-of-totalcareful-counting

Each of the 5 boxes contains either apples or bananas, but not both. The total weight of all the bananas is 3 times the weight of all the apples. Which boxes contain apples?

Figure for Math Kangaroo 2021 Problem 22
Show answer
Answer: E — 1 and 4
Show hints
Hint 1 of 2
If the bananas weigh 3 times the apples, then for every 1 kg of apples there are 3 kg of bananas — so apples are 1 part out of 4 equal parts of the whole.
Still stuck? Show hint 2 →
Hint 2 of 2
Add all five box weights, take a quarter of that for the apples, then find which boxes add up to it.
Show solution
Approach: apples are one quarter of the total weight
  1. The five boxes weigh 7, 5, 6, 2, 16, totalling 36 kg.
  2. Bananas are 3 times the apples, so apples make up 36 / 4 = 9 kg.
  3. Boxes summing to 9 kg are the 7 kg and 2 kg ones, i.e. boxes 1 and 4.
  4. So the apple boxes are 1 and 4 (E).
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Problem 20 · 2017 Math Kangaroo Stretch
Arithmetic & Operations gridcareful-counting

A number is written into every square of a 4 × 4 table. Mary is looking for the 2 × 2 table where the sum of the four numbers is greatest. How big is this sum?

1213
4112
1732
2131
Show answer
Answer: D — 14
Show hints
Hint 1 of 2
Slide a 2 x 2 window over the table and add its four numbers each time.
Still stuck? Show hint 2 →
Hint 2 of 2
The biggest entry, 7, should sit inside the best window.
Show solution
Approach: check the 2 x 2 block containing the largest numbers
  1. Try 2 x 2 blocks, focusing on the area around the 7.
  2. The block with 7, 3 (its right neighbour) and the 1, 3 below them gives 7+3+1+3.
  3. That total is 14, larger than any other 2 x 2 block.
  4. So the greatest sum is 14.
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Problem 21 · 2017 Math Kangaroo Stretch
Arithmetic & Operations groupingtotal-then-divide

Five boys share 10 bags of marbles between themselves. Everyone gets exactly two bags (see picture). Alex gets 5 marbles, Bob 7, Charles 9 and Dennis 15. Eric gets the two bags that are left over. How many marbles does he get?

Figure for Math Kangaroo 2017 Problem 21
Show answer
Answer: E — 19
Show hints
Hint 1 of 2
The ten bags hold 1, 2, 3, ... up to 10 marbles; add them all first.
Still stuck? Show hint 2 →
Hint 2 of 2
Subtract the four known boys' totals from the grand total to get Eric's two bags.
Show solution
Approach: total all bags, then subtract the known amounts
  1. The ten bags hold 1 through 10 marbles, a total of 55.
  2. Alex, Bob, Charlie and Dennis took 5 + 7 + 9 + 15 = 36 marbles.
  3. Eric gets the two leftover bags: 55 - 36 = 19.
  4. So Eric gets 19 marbles.
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Problem 23 · 2017 Math Kangaroo Stretch
Arithmetic & Operations careful-countingtotal-then-divide

Kate has four flowers, which have 6, 7, 8 and 11 petals respectively. She now tears off one petal from each of three different flowers. She repeats this until it is no longer possible to tear off one petal from each of three different flowers. What is the minimum number of petals left over?

Show answer
Answer: B — 2
Show hints
Hint 1 of 2
Every single round takes away exactly 3 petals (one from each of three flowers).
Still stuck? Show hint 2 →
Hint 2 of 2
Since 3 petals leave each round, the petals removed always count up by threes; think about what is left from 32.
Show solution
Approach: petals leave 3 at a time, so the leftover is what 32 has past a count-by-three
  1. There are 6 + 7 + 8 + 11 = 32 petals to start.
  2. Each round takes one petal from three different flowers, so exactly 3 petals leave every round.
  3. Counting by threes (3, 6, 9, ..., 30), the most she can remove is 30, which still keeps three flowers alive long enough to do every round.
  4. That leaves 32 - 30 = 2 petals.
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Problem 18 · 2016 Math Kangaroo Stretch
Arithmetic & Operations sum-constraintages

Tick, Trick and Track are triplets. Their brother Franz is exactly 3 years older. All four children are having their birthdays today. How old can the four brothers be altogether?

Show answer
Answer: B — 27
Show hints
Hint 1 of 2
The three triplets are all the same age, and Franz is just 3 more than that.
Still stuck? Show hint 2 →
Hint 2 of 2
If you take Franz's extra 3 years off the total, what is left should split into four equal ages.
Show solution
Approach: take off Franz's extra 3, then split the rest into four equal ages
  1. All four boys would be the same age except Franz, who has 3 extra years, so take those 3 away from the total first.
  2. What is left must split evenly into four equal ages (one for each boy).
  3. Take 3 off each choice and see which splits into 4 equal whole numbers: 27 − 3 = 24, and 24 shared by 4 is 6 each, so the ages are 6, 6, 6 and 9.
  4. So the four brothers can be 27 years old altogether, choice B.
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Problem 19 · 2016 Math Kangaroo Stretch
Arithmetic & Operations sum-constraint

In a magic garden there are magic trees. On each tree there are either 6 pears and 3 apples, or 8 pears and 4 apples. In total there are 25 apples on the magic trees. How many pears in total are hanging on the magic trees altogether?

Show answer
Answer: D — 50
Show hints
Hint 1 of 2
Use the apple counts (3 per first kind of tree, 4 per second) to total 25 apples.
Still stuck? Show hint 2 →
Hint 2 of 2
Notice each apple comes with a fixed bundle of pears, so the pear total may not depend on the exact mix.
Show solution
Approach: find the tree counts from apples, then total the pears
  1. A '6-pears/3-apples' tree carries 2 pears per apple, and an '8-pears/4-apples' tree also carries 2 pears per apple.
  2. Since every apple is matched by exactly 2 pears on either kind of tree, the 25 apples come with 2 × 25 = 50 pears.
  3. So there are 50 pears in total.
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Problem 20 · 2016 Math Kangaroo Stretch
Arithmetic & Operations sum-constraint

Lisa's dogs have 18 more legs than noses. How many dogs does Lisa have?

Show answer
Answer: C — 6
Show hints
Hint 1 of 2
Each dog has 4 legs but only 1 nose, so each dog adds 3 extra legs over noses.
Still stuck? Show hint 2 →
Hint 2 of 2
Divide the 18 extra legs by 3 per dog.
Show solution
Approach: count the leg-minus-nose difference per dog
  1. Every dog has 4 legs and 1 nose, so it contributes 4 − 1 = 3 more legs than noses.
  2. All dogs together give 18 extra legs, so there are 18 ÷ 3 = 6 dogs.
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Problem 22 · 2016 Math Kangaroo Stretch
Arithmetic & Operations

Eva writes seven numbers on a piece of paper, one of which is 201. She adds up these seven numbers and gets 2016. Now she replaces the 201 with the number 102 and again adds up the seven numbers. Which result does she get now?

Show answer
Answer: C — 1917
Show hints
Hint 1 of 2
Only one number changes, so only the change in that number changes the total.
Still stuck? Show hint 2 →
Hint 2 of 2
Find how much smaller 102 is than 201, and subtract that from 2016.
Show solution
Approach: adjust the sum by the change in the single number
  1. Replacing 201 with 102 lowers that number by 201 − 102 = 99.
  2. The total drops by the same 99: 2016 − 99 = 1917.
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Problem 21 · 2014 Math Kangaroo Stretch
Arithmetic & Operations work-backward

Heinzi the kangaroo has bought some toys. For them he gave 150 kangoo-coins (KC) and received 20 kangoo-coins back. Just before leaving the shop he changed his mind and exchanged one of the toys he had bought for another one. Because of this he received a further 5 kangoo-coins back from the shopkeeper. Which of the toys in the picture has Heinzi taken home with him? (The price of each toy is shown on its tag.)

Figure for Math Kangaroo 2014 Problem 21
Show answer
Answer: A — Carriage and Aeroplane
Show hints
Hint 1 of 3
Work out how many coins Heinzi really spent in the end, after all the change he got back.
Still stuck? Show hint 2 →
Hint 2 of 3
He handed over 150, then got 20 back, then 5 more back, so take both amounts away.
Still stuck? Show hint 3 →
Hint 3 of 3
Now look at the price tags and find the toys that add up to exactly that many coins.
Show solution
Approach: net spending, then match a price pair
  1. He paid 150 and got 20 back, then 5 more back after the exchange: net 150 − 20 − 5 = 125.
  2. The toys he kept must cost 125 together.
  3. Carriage (73) + Aeroplane (52) = 125; no other pair fits.
  4. Answer: Carriage and Aeroplane.
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Problem 18 · 2013 Math Kangaroo Stretch
Arithmetic & Operations total-then-divide

In February 2013, Schnurrli the tomcat slept for exactly three weeks. For how many hours during this month was he awake?

Show answer
Answer: A — 168
Show hints
Hint 1 of 3
February 2013 has 28 days, and three weeks of sleeping is 3 × 7 = 21 days.
Still stuck? Show hint 2 →
Hint 2 of 3
Take the sleeping days away from the whole month to find the awake days.
Still stuck? Show hint 3 →
Hint 3 of 3
Each awake day is 24 hours, so turn the awake days into hours.
Show solution
Approach: awake days times 24
  1. February 2013 has 28 days; he slept 21 days, so he was awake 28 − 21 = 7 days.
  2. 7 × 24 = 168 hours awake.
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Problem 21 · 2013 Math Kangaroo Stretch
Arithmetic & Operations sum-constrainttotal-then-divide

Chrissi wants to sell 10 glass marbles that each have a different weight. Their weights are 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10 dag. They are to be packed into bags, two marbles at a time, so that each bag has the same weight. Which two marbles will be put into the same bag?

Show answer
Answer: C — 3 and 8
Show hints
Hint 1 of 3
Ten marbles two at a time make 5 bags, and all five bags weigh the same.
Still stuck? Show hint 2 →
Hint 2 of 3
Add all the weights 1 + 2 + ... + 10 = 55, then share that equally among 5 bags.
Still stuck? Show hint 3 →
Hint 3 of 3
Each bag must weigh 11, so find the marble that pairs with 3 to make 11.
Show solution
Approach: equal-sum pairing
  1. The total weight is 1+2+···+10 = 55, and 5 bags means each weighs 11 dag.
  2. Pairs making 11: (1,10), (2,9), (3,8), (4,7), (5,6).
  3. So 3 goes with 8.
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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 11 · 2025 Math Kangaroo Medium
Arithmetic & Operations total-then-divide

Anna, Bonnie and Caspar have some kangaroo cookies on their plates (see picture). There are 15 more cookies left over. They share these out so that each child ends up with the same number of cookies on their plate. How many cookies are added to Anna's plate?

Figure for Math Kangaroo 2025 Problem 11
Show answer
Answer: D — 6
Show hints
Hint 1 of 2
First find how many cookies there are altogether, then share them equally among the three children.
Still stuck? Show hint 2 →
Hint 2 of 2
Subtract the cookies Anna already has from her fair share.
Show solution
Approach: find the equal share, then see how many Anna still needs
  1. The plates already hold 3 (Anna) + 4 (Bonnie) + 5 (Caspar) = 12 cookies.
  2. With the 15 extra there are 12 + 15 = 27 cookies in all.
  3. Shared equally, each child should have 27 ÷ 3 = 9 cookies.
  4. Anna has 3, so she gets 9 − 3 = 6 more, option D.
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Problem 12 · 2024 Math Kangaroo Medium
Arithmetic & Operations divisioncareful-counting

Penguin Peter goes fishing every day and brings home 9 fish for his two children. Each day he gives 5 fish to the first child he sees, and the other child gets the remaining 4 fish. Over the last few days, one child has received 26 fish in total. How many fish did the other child get?

Show answer
Answer: D — 28
Show hints
Hint 1 of 3
Every single day the two children together get 5 + 4 = 9 fish.
Still stuck? Show hint 2 →
Hint 2 of 3
Each day a child gets either 5 or 4, so try how many days it takes one child to reach 26.
Still stuck? Show hint 3 →
Hint 3 of 3
Once you know the number of days, the other child's total is 9 per day minus 26.
Show solution
Approach: find the number of days, then take the rest of the daily 9s
  1. Each day one child gets 5 fish and the other gets 4, so together they get 9 fish a day.
  2. For one child to reach 26 (made of 5s and 4s), it takes 6 days: 5 + 5 + 4 + 4 + 4 + 4 = 26.
  3. In all, the children got 9 × 6 = 54 fish over those 6 days.
  4. So the other child got 54 − 26 = 28 (D).
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Problem 10 · 2022 Math Kangaroo Medium
Arithmetic & Operations sum-constraint

The kangaroos in a family are 2, 4, 5, 6, 8 and 10 years old. Four of them add up to 22 years. How old are the other two kangaroos?

Show answer
Answer: C — 5 and 8
Show hints
Hint 1 of 3
Add up all six ages first to get the grand total.
Still stuck? Show hint 2 →
Hint 2 of 3
If four of them add to 22, the two left over must make up the rest of the total.
Still stuck? Show hint 3 →
Hint 3 of 3
Take 22 away from the grand total, then find which two ages on the list add to that.
Show solution
Approach: add all the ages, then take away 22
  1. All six ages add up to 2 + 4 + 5 + 6 + 8 + 10 = 35.
  2. If four of them make 22, the other two must make 35 − 22 = 13.
  3. On the list, the pair that adds to 13 is 5 and 8.
  4. So the other two kangaroos are 5 and 8 (choice C).
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Problem 12 · 2021 Math Kangaroo Medium
Arithmetic & Operations work-backwardcareful-counting

The 5 balls shown begin to move at the same time in the directions of their arrows. When two balls going in opposite directions collide, the bigger ball swallows the smaller one and grows by the smaller ball's value, then keeps moving in its own direction (see the worked example). What is the final result of the collisions of the 5 balls?

Figure for Math Kangaroo 2021 Problem 12
Show answer
Answer: C
Show hints
Hint 1 of 2
When two balls bump head-on, the bigger one eats the smaller one and grows by its number, still going the same way.
Still stuck? Show hint 2 →
Hint 2 of 2
No ball ever disappears, so the final ball's number is just every ball's number added together — you only need its direction too.
Show solution
Approach: all the balls merge into one, so add the values and find the direction
  1. Nothing is ever lost in a collision, so the one ball left at the end is worth 10 + 9 + 3 + 7 + 20 = 49.
  2. Following the bumps shows the big left-going 20 keeps winning the head-on hits, so the surviving ball moves left.
  3. Matching '49 moving left' to the pictures gives option C.
  4. So the result is a ball worth 49 moving left (C).
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Problem 13 · 2021 Math Kangaroo Medium
Arithmetic & Operations unit-ratework-backward

In an ice cream shop there is some money in a drawer. After 6 ice creams were sold, there were 70 euros left in the drawer. After a total of 16 ice creams were sold, there were 120 euros left in the drawer. How many euros were there in the drawer at the start?

Show answer
Answer: C — 40
Show hints
Hint 1 of 2
Between the two moments, 10 more ice creams were sold and the drawer grew by 50 euros.
Still stuck? Show hint 2 →
Hint 2 of 2
Find the price of one ice cream, then step back to before any were sold.
Show solution
Approach: find the price per ice cream, then work back to the start
  1. From 6 to 16 sold is 10 more ice creams, and the money rose from 70 to 120, i.e. +50 euros.
  2. So each ice cream costs 50 / 10 = 5 euros.
  3. The first 6 brought in 6 x 5 = 30 euros, so the start amount was 70 - 30 = 40 euros.
  4. So there were 40 euros at the start (C).
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Problem 15 · 2021 Math Kangaroo Medium
Arithmetic & Operations differenceswork-backward

On a tall building there are 4 fire escape ladders, as shown. The heights of 3 of the ladders are marked at their tops (32, 48 and 36). What is the height of the shortest ladder?

Figure for Math Kangaroo 2021 Problem 15
Show answer
Answer: D — 20
Show hints
Hint 1 of 2
Each ladder stands on a ledge, and a taller ledge means a taller ladder top — the ledges step up by equal amounts.
Still stuck? Show hint 2 →
Hint 2 of 2
Find the step between two ladders whose tops you know, then step the same amount down to the shortest ladder.
Show solution
Approach: find the equal step between ladder tops and step down to the shortest
  1. The ladders sit on ledges that rise by the same step each time, so their tops rise by that same step too.
  2. Two of the marked tops are 48 and 36, a step of 48 - 36 = 12.
  3. Stepping down 12 from the 32 ladder gives the shortest ladder: 32 - 12 = 20.
  4. So the shortest ladder is 20 (D).
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Problem 13 · 2019 Math Kangaroo Medium
Arithmetic & Operations divisionwork-backward

Sara has 16 blue marbles. She can swap her marbles in the following way: for 3 blue marbles she gets 1 red marble, and for 2 red marbles she gets 5 green marbles. What is the maximum number of green marbles she can get?

Show answer
Answer: B — 10
Show hints
Hint 1 of 2
First turn as many blue marbles as possible into red ones, then trade reds for greens.
Still stuck? Show hint 2 →
Hint 2 of 2
Watch the leftovers: trades only happen in fixed bundles (3 blue, 2 red).
Show solution
Approach: trade in bundles and track leftovers
  1. 16 blue ÷ 3 gives 5 red marbles (1 blue left over).
  2. 5 red ÷ 2 gives 2 trades = 10 green marbles (1 red left over).
  3. The maximum number of green marbles is 10 (B).
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Problem 9 · 2017 Math Kangaroo Medium
Arithmetic & Operations work-backward

Every box shows the result of the addition of the numbers on the very left and on the very top (for example: 6 + 2 = 8). Which number is written behind the question mark?

Figure for Math Kangaroo 2017 Problem 9
Show answer
Answer: E — 15
Show hints
Hint 1 of 2
Every box is its left number added to its top number; find a box where you already know both to fill in a missing edge number.
Still stuck? Show hint 2 →
Hint 2 of 2
The box showing 10 sits under the top number 2, so its left number must make 10.
Show solution
Approach: use a known box to find the missing left number, then add
  1. Each box is the number at the left of its row plus the number at the top of its column.
  2. The box showing 10 is under the top number 2, so its left number is 8 (because 8 + 2 = 10).
  3. The question-mark box is in that same row, under the top number 7.
  4. So it is 8 + 7 = 15.
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Problem 11 · 2017 Math Kangaroo Medium
Arithmetic & Operations careful-counting

Balloons are sold in packages of 5, 10 or 25 pieces each. Marius buys exactly 70 balloons. What is the minimum number of packages he has to buy?

Show answer
Answer: B — 4
Show hints
Hint 1 of 2
Use the biggest packages first to cut down the number of packages.
Still stuck? Show hint 2 →
Hint 2 of 2
Try two 25-packs, then fill the remaining 20 with the fewest packs.
Show solution
Approach: use the largest packages first
  1. Two packs of 25 give 50, leaving 20 balloons to reach 70.
  2. Make 20 with two packs of 10.
  3. That is 2 + 2 = 4 packages, and three packages cannot total exactly 70.
  4. So the minimum is 4.
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Problem 16 · 2017 Math Kangaroo Medium
Arithmetic & Operations work-backwardcustom-operation

Which number must be written into the circle with the question mark so that the calculation is correct?

Figure for Math Kangaroo 2017 Problem 16
Show answer
Answer: B — 9
Show hints
Hint 1 of 2
Going all the way around the loop must bring you back to the same number.
Still stuck? Show hint 2 →
Hint 2 of 2
The x0 step forces one circle to be 0; start there and go around.
Show solution
Approach: use that the loop closes; the x0 step pins a value
  1. Following the arrows must return to the starting circle.
  2. One step is 'x0', whose result is 0, so the circle it leads to holds 0.
  3. From there: 0, then +6 gives 6, then x4 gives 24, then -15 gives the question mark.
  4. So the question mark is 24 - 15 = 9.
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Problem 16 · 2015 Math Kangaroo Medium
Arithmetic & Operations grouping

There are 10 balls, numbered 0 to 9 in a basket. John and George play a game. Each person is allowed to take three balls from the basket and calculate the total of the numbers on the balls. What is the biggest possible difference between the John and George's totals?

Figure for Math Kangaroo 2015 Problem 16
Show answer
Answer: E — 21
Show hints
Hint 1 of 2
To make the gap as large as possible, one player grabs the biggest numbers and the other the smallest.
Still stuck? Show hint 2 →
Hint 2 of 2
They draw from the same basket, so the two sets of three balls cannot overlap.
Show solution
Approach: maximise one total and minimise the other
  1. One player can take the three largest balls: 9 + 8 + 7 = 24.
  2. The other is then left to take the three smallest: 0 + 1 + 2 = 3.
  3. The biggest possible difference is 24 − 3 = 21.
  4. The answer is 21.
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Problem 11 · 2014 Math Kangaroo Medium
Arithmetic & Operations work-backward

Leo writes numbers in the multiplication pyramid. In a multiplication pyramid, you multiply two numbers that are next to each other to get the number directly above them (in the middle). Which number must Leo write in the grey field?

Figure for Math Kangaroo 2014 Problem 11
Show answer
Answer: E — 8
Show hints
Hint 1 of 3
Each block is found by multiplying the two blocks right under it.
Still stuck? Show hint 2 →
Hint 2 of 3
Start at the bottom row, which you know, and build one row up at a time.
Still stuck? Show hint 3 →
Hint 3 of 3
Keep multiplying neighbouring blocks until you reach the grey one.
Show solution
Approach: build the pyramid upward, multiplying each pair of neighbours
  1. Start with the bottom row 1, 2, 2, 1.
  2. Multiply each neighbouring pair to get the next row up: 1×2 = 2, 2×2 = 4, 2×1 = 2, so that row is 2, 4, 2.
  3. Multiply neighbours again: 2×4 = 8 and 4×2 = 8, so the grey block is 8.
  4. Answer: 8.
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Problem 15 · 2014 Math Kangaroo Medium
Arithmetic & Operations casework

Hubert the rabbit loves cabbages and carrots. In one day he eats either 9 carrots, or 2 cabbages, or one cabbage and 4 carrots. In one week Hubert ate 30 carrots. How many cabbages did he eat during this week?

Show answer
Answer: B — 7
Show hints
Hint 1 of 3
There are 7 days in a week, so Hubert eats one of his three menus on each of the 7 days.
Still stuck? Show hint 2 →
Hint 2 of 3
Only two of the menus give carrots: the 9-carrot day and the 1-cabbage-and-4-carrots day.
Still stuck? Show hint 3 →
Hint 3 of 3
Try a few of each carrot-day until the carrots add up to 30, then count cabbages on every day.
Show solution
Approach: find which days give the 30 carrots, then count cabbages
  1. Only two kinds of day give carrots: a 9-carrot day, or a day of 1 cabbage and 4 carrots.
  2. Two 9-carrot days give 18 carrots, and three of the ‘1 cabbage + 4 carrots’ days give 12 more — that is 18 + 12 = 30 carrots, using 5 days.
  3. Those three mixed days give 3 cabbages, and the 2 days left over are 2-cabbage days, giving 2 × 2 = 4 more.
  4. Total cabbages = 3 + 4 = 7.
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Problem 10 · 2013 Math Kangaroo Medium
Arithmetic & Operations total-then-divide

Vera’s mum has made sandwiches, each using two slices of bread. There are 24 slices in a pack. How many sandwiches can she make with two whole packs and one half pack of bread?

Show answer
Answer: D — 30
Show hints
Hint 1 of 3
First work out the total number of bread slices Vera's mum has.
Still stuck? Show hint 2 →
Hint 2 of 3
A half pack has half of 24 slices, which is 12 slices.
Still stuck? Show hint 3 →
Hint 3 of 3
Each sandwich eats 2 slices, so split the total into pairs to count the sandwiches.
Show solution
Approach: count slices, then divide by 2
  1. Two whole packs and one half pack give 24 + 24 + 12 = 60 slices.
  2. Each sandwich uses 2 slices, so 60 ÷ 2 = 30 sandwiches.
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Problem 12 · 2013 Math Kangaroo Medium
Arithmetic & Operations careful-counting

Pinocchio’s nose is 9 cm long. Each time he lies, his nose grows by 6 cm. Each time he tells the truth, it shrinks by 2 cm. He tells three lies and twice tells the truth. How long is Pinocchio’s nose now?

Show answer
Answer: D — 23 cm
Show hints
Hint 1 of 3
Each lie adds 6 cm to the nose; each truth takes 2 cm away.
Still stuck? Show hint 2 →
Hint 2 of 3
There are three lies and two truths, so work out the total growing and the total shrinking.
Still stuck? Show hint 3 →
Hint 3 of 3
Start at 9 cm, add all the growing, then take away all the shrinking.
Show solution
Approach: add the lies, subtract the truths
  1. Three lies add 3 × 6 = 18 cm; two truths take off 2 × 2 = 4 cm.
  2. 9 + 18 − 4 = 23 cm.
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Problem 14 · 2013 Math Kangaroo Medium
Arithmetic & Operations careful-counting

In a shop you can buy oranges in bags of 4 or bags of 10. Pedro wants to buy exactly 48 oranges. What is the smallest number of bags he must buy?

Show answer
Answer: C — 6
Show hints
Hint 1 of 3
Fewer bags means use the big bags of 10 as much as you can.
Still stuck? Show hint 2 →
Hint 2 of 3
Try four bags of 10, which is 40 oranges, and see how many oranges are still needed.
Still stuck? Show hint 3 →
Hint 3 of 3
Fill the leftover with bags of 4, then count how many bags you used altogether.
Show solution
Approach: use the largest bags first
  1. Four bags of 10 give 40, leaving 8, which is two bags of 4.
  2. That is 4 + 2 = 6 bags, and no smaller number of bags reaches exactly 48.
  3. So he needs 6 bags.
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Problem 15 · 2013 Math Kangaroo Medium
Arithmetic & Operations careful-counting

At the London 2012 Olympic Games the USA won the most medals: 46 gold, 29 silver and 29 bronze. China was second with 38 gold, 27 silver and 23 bronze. How many more medals did the USA win than China?

Show answer
Answer: C — 16
Show hints
Hint 1 of 3
Add up all of the USA's medals, then add up all of China's medals.
Still stuck? Show hint 2 →
Hint 2 of 3
The lighter way is to compare gold to gold, silver to silver, and bronze to bronze.
Still stuck? Show hint 3 →
Hint 3 of 3
Find each of those three differences and add the three differences together.
Show solution
Approach: compare each colour separately so the numbers stay small
  1. Gold: the USA had 46 and China had 38, that is 8 more gold.
  2. Silver: 29 against 27 is 2 more, and bronze: 29 against 23 is 6 more.
  3. Adding the three extras, 8 + 2 + 6 = 16 more medals, which is answer C.
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Problem 14 · 2012 Math Kangaroo Medium
Arithmetic & Operations total-then-divide

15 tables were set for a party. 5 plates were laid on 6 of the tables. 3 plates were laid on the rest of the tables. How many plates were needed in total?

Show answer
Answer: C — 57
Show hints
Hint 1 of 2
Find how many tables get 3 plates after 6 tables get 5 plates.
Still stuck? Show hint 2 →
Hint 2 of 2
Work out each group's plates and add them.
Show solution
Approach: split into two groups of tables
  1. 6 tables get 5 plates each: 6 x 5 = 30 plates.
  2. The other 15 - 6 = 9 tables get 3 plates each: 9 x 3 = 27 plates.
  3. Altogether 30 + 27 = 57 plates are needed.
  4. The answer is 57.
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Problem 16 · 2011 Math Kangaroo Medium
Arithmetic & Operations careful-counting

During a party, two identical cakes were each cut into four identical pieces. Each of these pieces was then cut into three identical pieces. Each person at the party got a piece of cake, and there were three pieces left over. How many people were at the party?

Show answer
Answer: B — 21
Show hints
Hint 1 of 2
Work out how many small pieces the two cakes are cut into altogether.
Still stuck? Show hint 2 →
Hint 2 of 2
Each person ate one piece, and three pieces were left.
Show solution
Approach: count pieces, then subtract leftovers
  1. Two cakes × 4 = 8 quarters, and each quarter × 3 = 24 small pieces in all.
  2. Three pieces were left over, so the number of people = 24 − 3 = 21, answer B.
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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 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 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?

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Answer: E — 30
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Hint 1 of 2
Each of the 5 full trips carries either 10 cars or 6 lorries.
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Hint 2 of 2
Start by pretending every trip was lorries, then see how far short of 42 you are.
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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?

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Answer: C — 31
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Hint 1 of 2
Each round doubles the number of new people: 1, 2, 4, then 8, 16.
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Hint 2 of 2
Add up the new people from all five rounds.
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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 20 · 2025 Math Kangaroo Stretch
Arithmetic & Operations sum-constrainttotal-then-divide

Jana writes down how much her toys weigh: 22 g, 23 g, 25 g, 34 g and 36 g. She wants to share all her toys into two boxes so that both boxes weigh the same. Which two toys do not go in the same box?

Figure for Math Kangaroo 2025 Problem 20
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Answer: C
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Hint 1 of 2
Add all five weights and split into two equal boxes.
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Hint 2 of 2
Find which toys sum to half the total; the two that land in different boxes are your answer.
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Approach: split the weights into two equal halves
  1. The weights 22, 23, 25, 34, 36 add to 140 g, so each box holds 70 g.
  2. One box is 22 + 23 + 25 = 70 (balloon, car, boat); the other is 34 + 36 = 70 (helicopter, plane).
  3. So the balloon and the plane end up in different boxes.
  4. That pairing is option C.
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Problem 21 · 2023 Math Kangaroo Stretch
Arithmetic & Operations sum-constraint

Adam has 9 marbles and Brenda also has 9 marbles. Together they have 8 white and 10 black marbles. Brenda has twice as many black marbles as white marbles. How many black marbles does Adam have?

Figure for Math Kangaroo 2023 Problem 21
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Answer: B — 4
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Hint 1 of 2
Brenda has 9 marbles, and her black pile is twice as big as her white pile.
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Hint 2 of 2
Once you know how many black marbles Brenda has, the rest of the 10 black ones must be Adam's.
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Approach: split Brenda's 9 into equal groups, then give the rest of the black marbles to Adam
  1. Brenda's black pile is twice her white pile, so think of 1 white group and 2 matching black groups: that is 3 equal groups making 9, so each group is 3.
  2. Brenda then has 3 white and 6 black marbles.
  3. There are 10 black marbles altogether, so Adam has the leftover 10 − 6 = 4 black marbles.
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Problem 21 · 2020 Math Kangaroo Stretch
Arithmetic & Operations sum-constraint

The teacher wrote the numbers 1 to 8 on the board. Then he covered the numbers with triangles, squares and one circle (see picture). The sum of the numbers covered by the triangles equals the sum of the numbers covered by the squares, and the number covered by the circle is a quarter of that sum. What is the sum of the numbers covered by the triangles and the circle?

Figure for Math Kangaroo 2020 Problem 21
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Answer: C — 20
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Hint 1 of 3
First add up all the hidden numbers: 1 + 2 + 3 + ... + 8.
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Hint 2 of 3
The triangle pile and the square pile weigh the same, and the circle is just a small extra equal to a quarter of one of those piles.
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Hint 3 of 3
Try to split the total into two equal big piles plus a small piece that is a quarter of one big pile.
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Approach: split the total 36 into two equal piles plus a quarter-size circle
  1. The hidden numbers are 1 through 8, and 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36.
  2. The triangles and the squares make two equal piles, and the circle adds a quarter of one of those piles, so 36 splits as one pile + one equal pile + a quarter-pile.
  3. That is the same as four-and-a-quarter quarter-piles making 36, so each quarter-pile is 4; one full pile (the triangles) is four of them, which is 16, and the circle is one quarter-pile, which is 4.
  4. The triangles cover 16 and the circle covers 4, so together they cover 16 + 4 = 20, choice C.
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