Marie wants to put the digit 3 somewhere into the number 2014. Where must she put the 3 so that the new number (with all 5 digits) is as small as possible?
Show answer
Answer: D — between 1 and 4
Show hints
Hint 1 of 3
A number is smaller when its first (left-most) digits are smaller.
Still stuck? Show hint 2 →
Hint 2 of 3
Putting the big digit 3 near the front pushes the number up, so push the 3 as far right as you can.
Still stuck? Show hint 3 →
Hint 3 of 3
Write out each new number and read them like words to see which comes first.
Show solution
Approach: keep the small left-hand digits and push the 3 to the right
Try the 3 in each gap: 32014, 23014, 20314, 20134, 20143.
Reading them like a race, the one that stays smallest the longest at the front is 20134.
Whenever Koko the koala bear is awake, he always eats 50 grams of leaves in one hour. Yesterday Koko slept for 20 hours. How many grams of leaves did he eat yesterday?
Show answer
Answer: D — 200 grams
Show hints
Hint 1 of 3
Koko only eats when he is awake, so first work out how many hours he was awake.
Still stuck? Show hint 2 →
Hint 2 of 3
A whole day is 24 hours; take away the hours he slept.
Still stuck? Show hint 3 →
Hint 3 of 3
For each awake hour add another 50 grams of leaves.
Show solution
Approach: find the awake hours, then add 50 grams for each one
A day is 24 hours and Koko slept 20 of them, so he was awake 24 − 20 = 4 hours.
Each awake hour he eats 50 grams, so count by fifties for the 4 hours: 50, 100, 150, 200.
Christopher worked out the sums written next to the dots and got the answers 0, 1, 2, 3, 4 and 5. He joined the dots in order, starting at the dot with answer 0 and finishing at the dot with answer 5. Which shape was he left with? (Choose the matching picture.)
Show answer
Answer: A
Show hints
Hint 1 of 3
First work out the little sum next to each dot and write its answer on the dot.
Still stuck? Show hint 2 →
Hint 2 of 3
Now you have dots labelled 0, 1, 2, 3, 4 and 5.
Still stuck? Show hint 3 →
Hint 3 of 3
Draw a line from 0 to 1 to 2 and on to 5, then see which picture your line looks like.
Show solution
Approach: label each dot with its answer, then join them in order
Work out each sum and write the answer on its dot, so the dots are now numbered 0, 1, 2, 3, 4 and 5.
Start your pencil at the 0 dot and draw to the 1 dot, then the 2, and so on up to the 5.
The line zig-zags from side to side and makes a clear shape.
Anita has built fewer sandcastles than Hans but more than Stefan. Fabian has built more sandcastles than Anita and more than Hans. Bruno has built more sandcastles than Hans but fewer than Fabian. Who has built the most sandcastles?
Show answer
Answer: E — Fabian
Show hints
Hint 1 of 3
Picture the children standing in a line, tallest pile of sandcastles at the top.
Still stuck? Show hint 2 →
Hint 2 of 3
Each sentence tells you who stands above whom, so place them one at a time.
Still stuck? Show hint 3 →
Hint 3 of 3
Look for the one child who never has anybody above them.
Show solution
Approach: stand the children in order, most sandcastles at the top
Hans is above Anita, and Anita is above Stefan, so far Hans, then Anita, then Stefan.
Fabian is above both Anita and Hans, so Fabian goes near the top.
Bruno is above Hans but below Fabian, so the full line is Fabian, Bruno, Hans, Anita, Stefan.
Nobody is above Fabian, so the most sandcastles belong to Fabian.
Mr Hofer drew a picture of flowers on the inside of a shop window (the large picture). What do these flowers look like when you walk outside and look at the picture through the glass? (Choose the matching picture.)
Show answer
Answer: E
Show hints
Hint 1 of 3
Looking through the glass from the other side is just like looking in a mirror.
Still stuck? Show hint 2 →
Hint 2 of 3
A mirror swaps left and right, but keeps top and bottom the same.
Still stuck? Show hint 3 →
Hint 3 of 3
Hold the picture up to a mirror in your mind: what is on the left jumps to the right.
Show solution
Approach: flip the picture left-to-right like a mirror
Seeing the drawing from outside the glass is the same as seeing it in a mirror.
A mirror keeps each flower the right way up but swaps the left side with the right side.
So flowers on the left of the drawing should now be on the right.
With which square do you have to swap the question-mark square so that the white area and the black area become the same size? (Choose the matching picture.)
Show answer
Answer: B
Show hints
Hint 1 of 3
Count how many small black parts and how many small white parts the picture has right now.
Still stuck? Show hint 2 →
Hint 2 of 3
If there is more black than white, the new square must add some white to even them out (or the other way round).
Still stuck? Show hint 3 →
Hint 3 of 3
Pick the square that swaps in just the right amount to make black and white match.
Show solution
Approach: count the black and white parts and find the square that balances them
Count up all the black little pieces and all the white little pieces as the picture stands.
One colour is ahead, so the question-mark square needs to be replaced by one that gives back exactly that difference in the other colour.
Try each choice and see which one makes the black total equal the white total.
A bowl was full of sweets. Raphael took half of them out. Afterwards Emanuel took out half of the remaining sweets. Now there are only 12 sweets left in the bowl. How many sweets were in the bowl to begin with?
Show answer
Answer: E — 48
Show hints
Hint 1 of 3
Start at the end with the 12 sweets that are left and walk backwards.
Still stuck? Show hint 2 →
Hint 2 of 3
Each boy took half, so the 12 left is half of what was there just before him.
Still stuck? Show hint 3 →
Hint 3 of 3
Going back one step means doubling, so double, then double again.
Show solution
Approach: walk backwards, doubling at each step
The 12 left over is half of what was in the bowl before Emanuel reached in, so before Emanuel there were 12 + 12 = 24.
Those 24 are half of what was there before Raphael, so the bowl started with 24 + 24 = 48.
The solid in the diagram is built from 8 identical cubes. What does the solid look like when you look straight down at it from above? (Choose the matching picture.)
Show answer
Answer: C
Show hints
Hint 1 of 3
Imagine you are a bird flying right over the top, looking straight down.
Still stuck? Show hint 2 →
Hint 2 of 3
From up there you cannot tell how tall a stack is, only which floor squares are covered.
Still stuck? Show hint 3 →
Hint 3 of 3
Shade in every square that has at least one cube under it and match that shape.
Show solution
Approach: draw the shape of the floor squares the cubes cover
Looking from above, tall and short stacks look the same; all that matters is which floor squares are filled.
Mark each square that has a cube sitting on it, ignoring how high the pile goes.
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?
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
Start with the bottom row 1, 2, 2, 1.
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.
Multiply neighbours again: 2×4 = 8 and 4×2 = 8, so the grey block is 8.
Katja throws darts at the target shown on the right. If she does not hit the target she gets no points. She throws twice and adds her points. What can her total not be?
Show answer
Answer: D — 90
Show hints
Hint 1 of 3
Each single throw can only score one of the numbers on the target (or 0 for a miss).
Still stuck? Show hint 2 →
Hint 2 of 3
Try adding two of those numbers together in every way you can.
Still stuck? Show hint 3 →
Hint 3 of 3
Check each answer choice to find the one total you can never build.
Show solution
Approach: add the two throws in every way and find the missing total
One throw scores 0, 30, 50 or 70, and the total is two throws added.
Erwin has the four paper pieces shown. He has to cover a special shape exactly with these four pieces. In which drawing can he do this, when the one piece is placed as shown? (Choose the matching picture.)
Show answer
Answer: C
Show hints
Hint 1 of 3
The piece that is already placed covers part of the shape, so look at the empty gap that is left.
Still stuck? Show hint 2 →
Hint 2 of 3
Ask whether the other three pieces can fill that gap with no holes and no sticking out.
Still stuck? Show hint 3 →
Hint 3 of 3
Try each drawing and keep the only one that the pieces fit perfectly.
Show solution
Approach: see which outline the leftover three pieces fill exactly
Once the shown piece is set down, the empty space that remains has a fixed shape.
Imagine sliding the other three pieces in like a little jigsaw, covering every square with no gaps and no overlaps.
Only one of the drawings lets all four pieces fit exactly.
Gerhard has the same number of white, grey and black counters. He has thrown some of these round pieces together onto a pile. All the pieces he used can be seen in the picture. He has, however, got 5 counters left that will not stay on the pile. How many black counters did he have to begin with?
Show answer
Answer: B — 6
Show hints
Hint 1 of 3
He began with the same number of white, grey and black, so think of them in equal groups.
Still stuck? Show hint 2 →
Hint 2 of 3
Count the counters on the pile, colour by colour, from the picture.
Still stuck? Show hint 3 →
Hint 3 of 3
The 5 left over are the extras that did not fit, so add them back to find each starting group.
Show solution
Approach: count the pile by colour, then add back the leftovers to make equal groups
Count how many white, grey and black counters are actually on the pile in the picture.
He started with the same number of each colour, and 5 counters were left over that did not stay on.
Sharing everything back into three equal colour groups, each group had 6 counters.
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
Only two kinds of day give carrots: a 9-carrot day, or a day of 1 cabbage and 4 carrots.
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.
Those three mixed days give 3 cabbages, and the 2 days left over are 2-cabbage days, giving 2 × 2 = 4 more.
On the Kangaroo planet each kangoo-year has 20 kangoo-months. Each kangoo-month has 6 kangoo-weeks. How many kangoo-weeks are in a quarter of a kangoo-year?
Show answer
Answer: B — 30
Show hints
Hint 1 of 3
First find how many kangoo-weeks fill a whole kangoo-year.
Still stuck? Show hint 2 →
Hint 2 of 3
Every one of the 20 months holds 6 weeks, so count by sixes (or multiply).
Still stuck? Show hint 3 →
Hint 3 of 3
A quarter means splitting that whole year into 4 equal parts and taking one.
Show solution
Approach: find the weeks in a whole year, then split into four equal parts
A year has 20 months and each month has 6 weeks, so a year is 20 × 6 = 120 kangoo-weeks.
A quarter is one of four equal pieces, so share 120 into 4 parts: 120 ÷ 4 = 30.
Seven children stand in a circle. Nowhere are two boys standing next to each other. Nowhere are three girls standing next to each other. What is possible for the number of girls? The number of girls can…
Show answer
Answer: C — …only be 4.
Show hints
Hint 1 of 3
Try drawing 7 dots in a ring and colouring some as boys (B) and some as girls (G).
Still stuck? Show hint 2 →
Hint 2 of 3
No two B's may touch, and no run of three G's is allowed, so the boys must spread out to break up the girls.
Still stuck? Show hint 3 →
Hint 3 of 3
See how many boys you must have, then the rest are girls.
Show solution
Approach: place boys to keep girls in short runs, and count what is left
Since no two boys may stand together, the boys must be spaced apart around the ring of 7.
If there were only 2 boys, the other 5 girls would have to bunch up and three girls would end up together, which is not allowed.
So we need 3 boys spread out, breaking the 7 children into girl-runs of at most two; that leaves exactly 4 girls (like B G G B G G B around the ring).
Elisabeth sorts the cards shown above. With each move she is allowed to swap any two cards with each other. What is the smallest number of moves she needs in order to get the word KANGAROO?
Show answer
Answer: B — 3
Show hints
Hint 1 of 3
Write KANGAROO under the cards and mark every letter that is already in the right place.
Still stuck? Show hint 2 →
Hint 2 of 3
Only the wrong letters need to move, so look at just those.
Still stuck? Show hint 3 →
Hint 3 of 3
One swap trades two cards, so it can drop two wrong letters into the right spots at once.
Show solution
Approach: count the swaps that fix two misplaced letters at a time
O A R G O N K A must become K A N G A R O O; the A and the G are already in place.
The six remaining letters split into three pairs that each swap into place: (K↔O), (N↔R), (A↔O).
The black diamonds ◆ and white diamonds ◇ follow a fixed pattern. The first 3 levels are shown on the right. Each level (from the 2nd level on) has one more row than the level before. In every level, the two outermost diamonds of the last row are white, and all the other diamonds are black. How many black diamonds are there in level 6?
Show answer
Answer: C — 26
Show hints
Hint 1 of 3
Each level is a little triangle of rows: 1 diamond, then 2, then 3, and so on down.
Still stuck? Show hint 2 →
Hint 2 of 3
Count ALL the diamonds in level 6 first, then take away the white ones.
Still stuck? Show hint 3 →
Hint 3 of 3
In every level only the two ends of the very bottom row are white.
Show solution
Approach: count all the diamonds, then take away the two white ones
Level 6 has 6 + 1 = 7 rows, with 1, 2, 3, 4, 5, 6, 7 diamonds in them.
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.)
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
He paid 150 and got 20 back, then 5 more back after the exchange: net 150 − 20 − 5 = 125.
The toys he kept must cost 125 together.
Carriage (73) + Aeroplane (52) = 125; no other pair fits.
In each box exactly one of the digits 0, 1, 2, 3, 4, 5 and 6 is to be written. Each digit is used only once. The picture on the right shows two 2-digit numbers being added to give a 3-digit number. Which digit has to be written in the grey box so that the sum is correct?
Show answer
Answer: D — 5
Show hints
Hint 1 of 3
The answer has 3 digits, so the two 2-digit numbers must add up to at least 100.
Still stuck? Show hint 2 →
Hint 2 of 3
You only have the digits 0 to 6 once each, so the hundreds digit of the answer can only be 1.
Still stuck? Show hint 3 →
Hint 3 of 3
Once you know the answer starts with 1, fit the remaining digits and read the grey (ones) box.
Show solution
Approach: find the only valid sum, then read the grey (units) box
Two 2-digit numbers add to a 3-digit number using 0..6 once each.
The only working sum is 105 (e.g. 42 + 63), using digits 0,1,2,3,4,5,6.
The grey box is the units digit of the result, which is 5.
In the figure on the right a few of the small squares will be painted grey. While doing this, no 2×2 block made of four small grey squares is allowed to appear. At most how many of the squares in the figure can be painted grey?
Show answer
Answer: D — 21
Show hints
Hint 1 of 3
The rule is broken the moment four grey squares make a full 2×2 block, so every 2×2 block needs at least one white square.
Still stuck? Show hint 2 →
Hint 2 of 3
To colour the MOST squares grey, leave as few white squares as you can while still breaking every 2×2 block.
Still stuck? Show hint 3 →
Hint 3 of 3
Spread your white squares out cleverly so each one spoils several 2×2 blocks at once.
Show solution
Approach: leave the fewest white squares that still break every 2x2 block
Every little 2×2 group of squares must have at least one square left white, or it would be a forbidden block.
To keep the most grey, place the white squares far apart so each white square breaks as many 2×2 blocks as possible.
Doing this for the whole figure leaves just a few white squares, and the rest, 21 of them, can be grey.
Albin has put each of the digits from 1 to 9 in the fields of the table. In the diagram only 4 of these digits are shown. For the field containing the number 5, Albin noticed that the sum of the numbers in the neighbouring fields is 13 (neighbouring fields are fields which share a side). He noticed exactly the same for the field containing the digit 6. Which digit had Albin written in the grey field?
Show answer
Answer: D — 8
Show hints
Hint 1 of 3
The grey centre square touches all four edge squares, and the four corners 1, 2, 3, 4 are already filled in.
Still stuck? Show hint 2 →
Hint 2 of 3
The missing numbers are 5, 6, 7, 8 and 9, and they go in the centre and the four edge squares.
Still stuck? Show hint 3 →
Hint 3 of 3
Try placing them so that the neighbours of 5 add to 13 and the neighbours of 6 also add to 13.
Show solution
Approach: place 5–9 so both neighbour-sum clues hold
The four corners are 1, 2, 4 and 3; the digits 5, 6, 7, 8, 9 go in the centre and the four edge cells.
The cell holding 5 and the cell holding 6 must each have neighbour-sum 13.
The only arrangement that satisfies both forces 8 into the grey centre.
