In the four squares of a row there always have to be exactly two coins. In the four squares below each other there also always have to be exactly two coins. On which square does one more coin have to be placed?
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Answer: D — square D
Show hints
Hint 1 of 2
Count the coins in each row and in each column - which one still has only one?
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Hint 2 of 2
One row and one column are each short a coin; they meet at a single empty square.
Show solution
Approach: find the row and column that are short a coin
The third row down has only 1 coin, and the third column across also has only 1 coin.
Both still need one more coin to reach two.
The empty square where that row and column meet is square D, so the coin goes there.
There are 12 children in front of a zoo. Susi is the 7th from the front and Kim is the 2nd from the back. How many children are there between Susi and Kim?
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Answer: B — 3
Show hints
Hint 1 of 2
Find Susi's and Kim's spots in the line first, counting from the same end.
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Hint 2 of 2
Once you know both positions, count only the children strictly in between them.
Show solution
Approach: place both children by position, then count the gap
Counting from the front, Susi is in spot 7.
Kim is 2nd from the back of 12, so Kim is in spot 11.
The children between them sit in spots 8, 9 and 10.
There are two kinds of camels: bactrian camels that have 2 humps, and dromedaries that have 1 hump. Exactly 10 camels live in a certain zoo. Together they have 14 humps. How many bactrian camels are there in this zoo?
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Answer: D — 4
Show hints
Hint 1 of 2
If all 10 camels had just 1 hump, how many humps would that be?
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Hint 2 of 2
Every bactrian camel adds one extra hump beyond that; the extra humps tell you the count.
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Approach: start from all one-hump and add the extra humps
If all 10 camels had 1 hump each, that would be 10 humps.
There are 14 humps, so there are 4 extra humps.
Each bactrian camel has one extra hump, so there are 4 bactrian camels.
Every time a coin is put into the machine, the bottom ball falls out of one of the five tubes. How many coins does Barbara have to put in to be sure she gets at least one white ball?
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Answer: D — 11
Show hints
Hint 1 of 3
Imagine Barbara has the unluckiest day — every ball that is not white comes out first.
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Hint 2 of 3
Only the bottom ball of a tube can come out, so count the non-white balls that could fall before any white one.
Still stuck? Show hint 3 →
Hint 3 of 3
Add up all those non-white balls, then put in one more coin to be sure of a white ball.
Show solution
Approach: imagine the unluckiest order, then add one more coin
Each coin drops the bottom ball of some tube, so the worst luck is all the non-white balls coming out first.
Counting from the bottom of every tube up to its first white ball, 10 non-white balls could fall before any white one.
After those 10 unlucky balls, the next ball must be white, so she needs 10 + 1 = 11 coins. The answer is D.
Amir has stickers of ladybugs with 1, 2, 3 or 4 dots on their wings. He fills the grid so that every row and every column has a ladybug with 1, 2, 3 and 4 dots. What should the top row of the grid look like when he is finished?
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Answer: B
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Hint 1 of 3
In every row and every column you must use a 1-dot, 2-dot, 3-dot and 4-dot ladybug, each exactly once.
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Hint 2 of 3
Look down each column at the ladybugs already placed to see which dot-numbers are still missing.
Still stuck? Show hint 3 →
Hint 3 of 3
A top cell must hold the one dot-number its column does not have yet.
Show solution
Approach: fill the grid so every row and column has 1, 2, 3 and 4 once
Every row and every column must show a 1-dot, 2-dot, 3-dot and 4-dot ladybug, each just once.
Look down each column and find the one dot-number it is still missing — that number must go in the top cell.
Filling each top cell with its column's missing number gives the row shown in option B. The answer is B.
Anna, Bella, Che and Dimitry each have three shapes. Each child shares exactly one of their shapes with one other child. Anna has a triangle, a circle and a square; Bella has a heart, a square and a star; Che has a star, a triangle and a diamond. Which three shapes does Dimitry have?
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Answer: E
Show hints
Hint 1 of 3
Write out the shapes the other three children have and see which ones already come in matching pairs.
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Hint 2 of 3
Some shapes already appear for two children (a pair), but a few shapes appear for only one child so far.
Still stuck? Show hint 3 →
Hint 3 of 3
Dimitry must hold exactly the shapes that still need a partner, so that every shape ends up shared by two children.
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Approach: give Dimitry the shapes that still need a matching partner
Among Anna, Bella and Che, the square (Anna & Bella), the star (Bella & Che) and the triangle (Anna & Che) already come in pairs.
That leaves the circle (only Anna), the heart (only Bella) and the diamond (only Che) without a partner.
So Dimitry must have the circle, the heart and the diamond — this pairs every shape with exactly one other child, which is option E.
Zoran builds towers from three different building blocks (a triangle top, a rectangle, and an hourglass). The picture shows the heights of three towers. How high is the fourth tower?
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Answer: A — 12
Show hints
Hint 1 of 3
Each kind of block is always the same height, so the same block is worth the same number every time.
Still stuck? Show hint 2 →
Hint 2 of 3
Compare two towers that are almost the same — the difference in their heights tells you how tall the extra block is.
Still stuck? Show hint 3 →
Hint 3 of 3
Find the rectangle and the triangle heights, since the fourth tower is just those two stacked.
Show solution
Approach: compare towers to find each block's height
The tall tower (triangle + rectangle + hourglass) is 20, and the short one with just (triangle + hourglass) is 13; the only extra block is the rectangle, so the rectangle is 20 − 13 = 7.
The tower with (rectangle + hourglass) is 15, and the rectangle is 7, so the hourglass is 8; then in the tower of 13 the triangle is 13 − 8 = 5.
The fourth tower is just triangle + rectangle = 5 + 7 = 12.
Andrew throws arrows at a target. He starts with 10 arrows. Each time he hits the target, he gets 2 more arrows. In total Andrew throws 20 arrows, and then he has run out of arrows. How many times did Andrew hit the target?
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Answer: B — 5
Show hints
Hint 1 of 3
He only started with 10 arrows, but he threw 20, so where did the extra arrows come from?
Still stuck? Show hint 2 →
Hint 2 of 3
Every hit is like a little gift of 2 more arrows.
Still stuck? Show hint 3 →
Hint 3 of 3
Figure out how many extra arrows he earned, then see how many hits that took.
Show solution
Approach: count the extra arrows the hits gave him
He started with 10 arrows but threw 20 in total, so 10 extra arrows must have come from hitting the target.
Each hit gives 2 extra arrows, so we count by twos: 2, 4, 6, 8, 10 — that takes 5 hits to reach 10 extra arrows.
The two pictures show the same bridge at different times. All the cars are the same length. The numbers give the distances between the cars, and between a car and the end of the bridge. How long is each car?
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Answer: C — 5 metres
Show hints
Hint 1 of 3
It is the same bridge in both pictures, so the bridge is exactly the same total length both times.
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Hint 2 of 3
In each picture add up all the gap numbers, and count how many cars there are.
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Hint 3 of 3
The bottom picture has one fewer car but more gap — that extra gap must be exactly one car long.
Show solution
Approach: the same bridge length both times
Top picture: the gaps add to 1 + 2 + 1 + 2 = 6 metres, and there are 3 cars.
Bottom picture: the gaps add to 4 + 4 + 3 = 11 metres, and there are 2 cars; the bridge is the same length, so the bottom has 11 − 6 = 5 more metres of gap but exactly one fewer car.
That missing car is filling those extra 5 metres of gap, so each car is 5 metres long.
Logic & Word Problemswork-backwardcareful-counting
Each of the children Ali, Lea, Josef, Vittorio and Sophie gets a birthday cake. The number on top of the cake shows how old the child is. Lea is two years older than Josef, but one year younger than Ali. Vittorio is the youngest. Which cake belongs to Sophie?
Show answer
Answer: C
Show hints
Hint 1 of 3
There are five cakes with five ages on them — match a child to each one.
Still stuck? Show hint 2 →
Hint 2 of 3
Use the clues to label the easy children first, and give the smallest cake to Vittorio.
Still stuck? Show hint 3 →
Hint 3 of 3
Sophie gets the one cake that is left over after everyone else is matched.
Show solution
Approach: match the easy children first; Sophie gets the leftover cake
Josef, then Lea (2 older than Josef), then Ali (1 older than Lea) line up in a row of ages.
Vittorio is the youngest, so he takes the smallest-numbered cake.
Four children are now placed, so the one cake left over is Sophie's: the cake showing 6, option C.
Logic & Word Problemswork-backwardcareful-counting
Three frogs live in a pond. Each night only one of the frogs sings a song. After 9 nights the first frog has sung 2 times. The second frog has listened to 5 songs. How many songs did the third frog listen to?
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Answer: B — 6
Show hints
Hint 1 of 3
There are 9 nights and one song each night, so the three frogs together sang 9 songs.
Still stuck? Show hint 2 →
Hint 2 of 3
On any night, a frog is either singing or listening — so a frog listens on every night it did not sing.
Still stuck? Show hint 3 →
Hint 3 of 3
If the second frog listened to 5 songs, that tells you how many nights it sang instead.
Show solution
Approach: 9 songs in all; a frog listens on every night it does not sing
There are 9 nights with one song each, so 9 songs were sung in all. Frog 1 sang 2 of them.
Frog 2 listened to 5 songs, so out of 9 nights it sang on the other 4 nights.
That leaves Frog 3 singing 9 take away 2 take away 4, which is 3 nights, so it listened on the other 6 nights: 6, option B.
Every time the witch has 3 apples she turns them into 1 banana. Every time she has 3 bananas she turns them into 1 apple. What will she finish with if she starts with 4 apples and 5 bananas?
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Answer: A
Show hints
Hint 1 of 3
Start with her 4 apples and 5 bananas, and just keep following her two rules.
Still stuck? Show hint 2 →
Hint 2 of 3
Trade 3 apples for 1 banana, and trade 3 bananas for 1 apple, again and again.
Still stuck? Show hint 3 →
Hint 3 of 3
Keep swapping in groups of 3 until you can no longer make a group of three.
Show solution
Approach: repeatedly trade groups of three
Start with 4 apples and 5 bananas. Trade 3 apples for 1 banana: 1 apple, 6 bananas.
Trade 6 bananas (two groups of three) for 2 apples: 3 apples, 0 bananas.
Trade those 3 apples for 1 banana: 0 apples, 1 banana.
The five cards shown (2, 3, 4, 5, 6) are placed into 2 boxes. The sums of the numbers in each box are the same. Which number must be in the box with the number 4?
Show answer
Answer: D — 6
Show hints
Hint 1 of 3
Add all five card numbers together first.
Still stuck? Show hint 2 →
Hint 2 of 3
Since the two boxes are equal, split that total in half to see what each box must hold.
Still stuck? Show hint 3 →
Hint 3 of 3
Now figure out which card the 4 needs next to it to reach that box total.
Show solution
Approach: find the equal totals and pair up
All five cards add to 2 + 3 + 4 + 5 + 6 = 20, so each box must hold 10.
Putting 2, 3, and 5 together makes 10, which leaves 4 and 6 for the other box.
Two equal trains, each with 31 numbered wagons, travel in opposite directions. When wagon number 7 of one train is side by side with wagon number 12 of the other train, which wagon is side by side with wagon number 11?
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Answer: A — 8
Show hints
Hint 1 of 2
The two trains line up wagon-to-wagon; use the spot where 7 meets 12.
Still stuck? Show hint 2 →
Hint 2 of 2
The two facing wagon numbers always add up to the same total.
Show solution
Approach: pair wagons from the known matching point
Wagon 7 of one train lines up with wagon 12 of the other.
The wagon numbers run in opposite directions, so the two numbers across from each other always add to 19 (that is 7 plus 12).
Across from wagon 11 is the wagon numbered 19 minus 11 = 8.
Tania bought 14 chocolates, 8 of them round and the rest square. Half were white chocolates and half were dark chocolates. Among the square chocolates, only two are not white. How many dark round chocolates did Tania buy?
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Answer: D — 5
Show hints
Hint 1 of 2
Split the 14 into round/square and into white/dark and fill a little table.
Still stuck? Show hint 2 →
Hint 2 of 2
Use that only two square chocolates are dark to find the round ones.
Show solution
Approach: organise with a 2-by-2 table
There are 8 round and 6 square chocolates; 7 are white and 7 are dark.
Among the 6 square ones only 2 are dark, so 4 squares are white.
That leaves 7 minus 4 = 3 white round chocolates, so round dark = 8 minus 3 = 5.
Rita numbered the circles in the figure from 1 to 8, so that the sum of the three numbers on each of the four sides of the square equals 13. What is the sum of the four numbers written on the coloured circles?
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Answer: E — 16
Show hints
Hint 1 of 2
Adding the four side-sums counts each corner circle twice.
Still stuck? Show hint 2 →
Hint 2 of 2
Compare that double-count to the total 1+2+...+8.
Show solution
Approach: count corners twice via the side sums
The four sides each total 13, so all four sides together total 4 times 13 = 52.
This sum counts every number once, but the four corner circles twice.
Since 1+2+...+8 = 36, the extra from double-counting the corners is 52 minus 36 = 16.
The colored circles are the corners, so their sum is 16.
In the figure, an arrow pointing from one person to another means that the first person is shorter than the second. For example, person B is shorter than person A. Which person is the tallest?
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Answer: C — Person C
Show hints
Hint 1 of 2
An arrow goes from the shorter person to the taller one.
Still stuck? Show hint 2 →
Hint 2 of 2
The tallest person has only arrows pointing toward them and none leaving.
Show solution
Approach: the tallest has no arrow leaving it
An arrow leaving a person means someone is taller, so the tallest person has no arrow pointing away from them.
Checking each person, only person C has every nearby arrow pointing in and none leaving.
Alice has 3 white, 2 black and 2 grey pieces of paper. First she cuts every piece of paper that is not black into two pieces. Then she cuts in half every piece of paper that is not white. How many pieces of paper does she have in the end?
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Answer: D — 18
Show hints
Hint 1 of 3
Keep three little piles in your head — white, black, and grey — and watch each pile change.
Still stuck? Show hint 2 →
Hint 2 of 3
Cutting a piece into two means that pile gets twice as many pieces.
Still stuck? Show hint 3 →
Hint 3 of 3
Do step 1 on the not-black piles, then step 2 on the not-white piles, then add the three piles up.
Show solution
Approach: watch each colour pile through the two cuts, then add
Start with 3 white, 2 black, 2 grey.
Step 1 cuts every not-black piece in two: white 3 becomes 6, grey 2 becomes 4, and the 2 black stay the same — that is 6 + 2 + 4 = 12 pieces.
Step 2 cuts every not-white piece in two: black 2 becomes 4, grey 4 becomes 8, and the 6 white stay the same — so 6 + 4 + 8 = 18 pieces.
Each one of the four keys locks exactly one padlock. Every letter on a padlock stands for exactly one digit, and the same letters mean the same digits. Which letters must be written on the fourth padlock?
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Answer: D — GAG
Show hints
Hint 1 of 2
Match each key's number to the padlock it opens, lining up the digits with the letters in the same spots.
Still stuck? Show hint 2 →
Hint 2 of 2
Once you know which digit each letter is, the leftover key tells you the fourth padlock.
Show solution
Approach: decode letters to digits, then read the leftover key
The lock ADA has its first and last letters the same; the only key like that is 141, so A=1 and D=4.
Now DAG starts 4, 1, so its key is 417, giving G=7; then DGA is 4, 7, 1, which is the key 471.
Those three keys are used, so the fourth padlock is opened by the leftover key 717.
Reading 717 back as letters gives 7=G, 1=A, 7=G, so the fourth padlock is GAG (option D).
Ingrid has 4 red, 3 blue, 2 green and 1 yellow cube. She uses them to build the object shown. Cubes with the same colour don't touch each other. Which colour is the cube with the question mark?
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Answer: A — red
Show hints
Hint 1 of 3
There are 10 cubes and 10 colour-tiles: 4 red, 3 blue, 2 green, 1 yellow.
Still stuck? Show hint 2 →
Hint 2 of 3
The rule says two cubes of the same colour may never sit next to each other.
Still stuck? Show hint 3 →
Hint 3 of 3
Red is the most common colour, so the reds have to be spread far apart all over the pile.
Show solution
Approach: there are exactly enough cubes for each colour, so the spread-out rule forces the marked cube's colour
The pile has 10 cubes, and the colours come in just the right amounts: 4 red, 3 blue, 2 green, 1 yellow.
No two cubes of the same colour may touch, so the 4 reds must be pushed far apart from one another.
Filling the pile while keeping every colour from touching its twin leaves only one colour that can go in the marked spot.
17 squirrels are sitting on 4 trees. There are at least 2 squirrels on each tree. The number of squirrels is different on each tree. What is the largest possible number of squirrels on one tree?
Show answer
Answer: B — 8
Show hints
Hint 1 of 3
To put as many squirrels as possible on one tree, leave as few as possible on the others.
Still stuck? Show hint 2 →
Hint 2 of 3
The other three trees still need at least 2 each, and all four numbers must be different.
Still stuck? Show hint 3 →
Hint 3 of 3
The three smallest different numbers, each at least 2, are 2, 3 and 4.
Show solution
Approach: minimise the other three to maximise one
Each tree needs at least 2 squirrels and all four counts are different.
Make three trees as small as possible: 2, 3 and 4 squirrels.
Those three trees hold 2 + 3 + 4 = 9 squirrels, leaving 17 − 9 = 8 on the last tree. The answer is B.
Logic & Word Problemscareful-countingcomplementary-counting
Chen has these 5 baskets, with 4 toys in each (shown as A, B, C, D, E in the picture). Four of the baskets fall down and the toys lie mixed up on the floor. Which basket did he not drop?
Show answer
Answer: B — B
Show hints
Hint 1 of 3
Look at the floor pile and notice which kind of toy is the most common there.
Still stuck? Show hint 2 →
Hint 2 of 3
The basket Chen kept is the only one with none of a toy that the others all had.
Still stuck? Show hint 3 →
Hint 3 of 3
Count the ducks: every duck is on the floor, so the kept basket has no duck.
Show solution
Approach: find the toy that is missing from the basket that stayed up
On the floor we can find ducks, frogs, ladybugs, and hippos all mixed together.
Count the yellow ducks on the floor: there are 6, which is every duck from the five baskets.
So the basket that did NOT fall has no duck in it — and the only basket with no duck is basket B.
In the table, each shape stands for a different number. The number at the end of each row is the sum of that row, and the number under each column is the sum of that column. What number does the star stand for?
Show answer
Answer: C — 3
Show hints
Hint 1 of 3
Each shape always means the same number, and a line of shapes adds up to the number at its end.
Still stuck? Show hint 2 →
Hint 2 of 3
Start with a line that has two of the SAME shape — that makes the shape easy to figure out.
Still stuck? Show hint 3 →
Hint 3 of 3
Once you know the heart, look at the middle column to find the star.
Show solution
Approach: figure out the easy shapes first, then the star
The left column is two smileys adding to 10, so each smiley is 5 (because 5 and 5 make 10).
The right column is two hearts adding to 4, so each heart is 2 (because 2 and 2 make 4).
The middle column is a heart and a star adding to 5; the heart is 2, and 2 plus 3 makes 5, so the star is 3.
There is an animal asleep in each of the five baskets. The koala and the fox sleep in baskets with the same pattern and the same shape. The kangaroo and the rabbit sleep in baskets with the same pattern. In which basket does the mouse sleep?
Show answer
Answer: E — Basket 5
Show hints
Hint 1 of 2
Find the two baskets that look exactly the same in BOTH pattern and shape - those go to the koala and the fox.
Still stuck? Show hint 2 →
Hint 2 of 2
Then find the other two baskets that share a pattern - those go to the kangaroo and the rabbit; the basket left over is the mouse's.
Show solution
Approach: place the four named animals, then the mouse takes the leftover basket
Baskets 2 and 4 are exactly alike (green with black dots, same shape), so they hold the koala and the fox.
Baskets 1 and 3 share the orange-and-blue woven pattern, so they hold the kangaroo and the rabbit.
That leaves only Basket 5 (the tan basket with a lid) for the mouse, so the answer is Basket 5.
Carl writes down a five-digit number. He then places a shape on each of the five digits (see picture). He places different shapes on different digits. He places the same shape on the same digits. Which number did Carl hide?
Show answer
Answer: A — 34426
Show hints
Hint 1 of 2
Same shape means same digit; different shapes mean different digits.
Still stuck? Show hint 2 →
Hint 2 of 2
The two diamonds sit in positions 2 and 3, so those digits must be equal.
Show solution
Approach: match equal shapes to equal digits
The shapes are heart, diamond, diamond, club, spade - only positions 2 and 3 repeat.
So digits 2 and 3 are equal and all the others are different from them and each other.
Only 34426 has its 2nd and 3rd digits equal (4 and 4) with the rest distinct.
The pink tower is taller than the red tower but shorter than the green tower. The silver tower is taller than the green tower. Which tower is the tallest?
Show answer
Answer: D — silver tower
Show hints
Hint 1 of 3
Line the towers up from shortest to tallest using the clues.
Still stuck? Show hint 2 →
Hint 2 of 3
Green is taller than pink and red; now see how silver compares to green.
Still stuck? Show hint 3 →
Hint 3 of 3
Whoever is taller than green must be the tallest of all.
Show solution
Approach: order the towers by height
The first clues say red is shortest, then pink, then green is taller than both.
The last clue says silver is taller than green, so silver beats green and everyone below it.
These children are standing in a line. Some are facing forwards and others are facing backwards. How many children are holding another child's hand with their right hand?
Show answer
Answer: E — 6
Show hints
Hint 1 of 3
Which hand is a child's right hand depends on which way that child is facing.
Still stuck? Show hint 2 →
Hint 2 of 3
For a child facing you, the right hand is on your left side; for one facing away, it is on your right side.
Still stuck? Show hint 3 →
Hint 3 of 3
Go child by child and only count the ones whose right hand is holding a neighbour.
Show solution
Approach: check each child's facing direction
Look at each child and decide which hand is the right hand based on whether they face front or back.
Now count only the children whose right hand is the one holding a neighbour's hand.
Julia has two pots with flowers, as shown. She keeps the flowers exactly where they are. She buys more flowers and puts them in the pots. After that, each pot has the same number of each type of flower. What is the smallest number of flowers she needs to buy?
Show answer
Answer: D — 8
Show hints
Hint 1 of 3
Look at one type of flower at a time and compare the two pots.
Still stuck? Show hint 2 →
Hint 2 of 3
Whichever pot has fewer of that type needs to be filled up to match the other.
Still stuck? Show hint 3 →
Hint 3 of 3
Add up all the missing flowers across both pots.
Show solution
Approach: match each flower type to the larger count
For each kind of flower, both pots must end up with the bigger of the two amounts they already have.
For every type, count how many are missing in the pot that is short, and add those up.
Adding all the missing flowers gives 8 extra flowers to buy.
A village of 12 houses has four straight streets and four circular streets. The map shows 11 houses. Each straight street has three houses, and each circular street also has three houses. Where should the 12th house be placed on this map?
Show answer
Answer: D — On D
Show hints
Hint 1 of 2
Each straight street needs 3 houses and each circular street needs 3 houses.
Still stuck? Show hint 2 →
Hint 2 of 2
Find the position where a street is currently one house short.
Show solution
Approach: check which street is missing a house
Every street, straight or circular, must hold exactly three houses.
Count the houses already on each street and find the one with only two.
Placing the 12th house at spot D completes that street without overfilling any other.
In a classroom there are two chairs at each table. Each boy in the class sits at a table with a girl, but there are four girls who do not sit at a table with a boy. There are 14 little tables in the classroom. How many girls are in that class?
Show answer
Answer: E — 16
Show hints
Hint 1 of 2
Most tables seat one boy and one girl; a few seat two girls.
Still stuck? Show hint 2 →
Hint 2 of 2
Count the all-girl tables, then add the girls sharing with boys.
Show solution
Approach: split tables into boy-girl and girl-girl
Four girls sit without a boy, two to a table, so 2 tables are all-girl.
That leaves 14 minus 2 = 12 tables, each with one boy and one girl.
Total girls = 12 (one per mixed table) plus 4 = 16.
There are white, grey and black squares. Three children use these to make this pattern. First Anni replaces all black squares with white squares. Then Bob replaces all grey squares with black squares. Finally Chris replaces all white squares with grey squares. Which picture have the three children now created?
Show answer
Answer: A
Show hints
Hint 1 of 2
Do the three colour changes one at a time, in the order Anni, then Bob, then Chris.
Still stuck? Show hint 2 →
Hint 2 of 2
Follow just one square of each starting colour all the way through to see what it turns into.
Show solution
Approach: follow each starting colour through the three changes in order
A black square turns white (Anni), then that white turns grey (Chris), so black ends grey.
A grey square turns black (Bob) and stays black, so grey ends black.
A white square is only changed by Chris, turning grey, so white ends grey.
Recolouring every square this way gives the picture in option A.
Together the three squirrels Anni, Asia and Elli have 10 nuts. Each one has a different number of nuts, but at least 2 nuts. Anni has the least number of nuts. Asia has the most nuts. How many nuts does Elli have?
Show answer
Answer: C — 3
Show hints
Hint 1 of 2
Anni has the fewest and each squirrel has at least 2, so start Anni as low as allowed.
Still stuck? Show hint 2 →
Hint 2 of 2
Try the smallest possible numbers that are all different and add to 10, then read Elli's amount.
Show solution
Approach: use the smallest distinct amounts that sum to 10
All three numbers are different, each at least 2, and they add to 10.
Anni has the fewest, so try Anni = 2; then Elli and Asia must add to 8 with Asia largest.
The only way is Elli = 3 and Asia = 5 (all different, Asia most).
Leo and Max are standing in a queue that is made up of 11 people in total. There are 7 people in front of Leo, and Max stands directly behind him in the queue. How many people are behind Max?
Show answer
Answer: B — 2
Show hints
Hint 1 of 2
First work out Leo's place in line from the 7 people ahead of him.
Still stuck? Show hint 2 →
Hint 2 of 2
Max is right behind Leo; the rest of the 11 are behind Max.
Show solution
Approach: locate each person, then count the tail
7 people are in front of Leo, so Leo is 8th.
Max stands directly behind Leo, so Max is 9th.
There are 11 people total, leaving 11 - 9 = 2 behind Max.
Some pirates are climbing onto a ship one after the other using a rope. Their leader is exactly in the middle. He is the eighth pirate to climb onto the ship. How many pirates board the ship?
Show answer
Answer: B — 15
Show hints
Hint 1 of 2
The leader is the 8th to climb and is exactly in the middle of the line.
Still stuck? Show hint 2 →
Hint 2 of 2
If 7 pirates are ahead of him, the same number must be behind him.
Show solution
Approach: balance the pirates on each side of the middle one
The leader is the 8th to climb, so 7 pirates went before him.
Being exactly in the middle, 7 pirates must also come after him.
Stan has five toys: a ball, a set of blocks, a game, a puzzle and a car. He puts each toy on a different shelf of the bookcase (shelf 1 at the bottom up to shelf 5 at the top). The ball is higher than the blocks and lower than the car. The game is directly above the ball. On which shelf can the puzzle not be placed?
Show answer
Answer: C — 3
Show hints
Hint 1 of 3
The ball is the key: many clues are about where the ball sits.
Still stuck? Show hint 2 →
Hint 2 of 3
The game sits right on top of the ball, with the blocks below the ball and the car above it.
Still stuck? Show hint 3 →
Hint 3 of 3
Try each possible shelf for the ball and see where the puzzle is allowed to land.
Show solution
Approach: test the ball's shelf and rule one out
If the ball is on shelf 2: blocks on 1, game on 3, and the car and puzzle take shelves 4 and 5 in some order — puzzle is on 4 or 5.
If the ball is on shelf 3: game on 4, car on 5, and the blocks and puzzle take shelves 1 and 2 — puzzle is on 1 or 2.
(The ball can not be on 1, with nothing below it, or on 4, leaving no room for the car above the game.)
So the puzzle can land on 1, 2, 4 or 5, but never on shelf 3.
