The clues describe what each child has or does (shown in the picture). Anna has one thing, Barbara gave Eva something, Josef has something, and Bob has something. Using the clues, which picture (A–E) shows Barbara?
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Answer: D
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Hint 1 of 3
Read the clues and cross off any picture that the clue gives to someone who is NOT Barbara.
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Hint 2 of 3
The clue about Barbara is the giving clue: she is the one who gave Eva her thing.
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Hint 3 of 3
Whoever is left after you label Anna, Josef and Bob, and who fits the giving clue, is Barbara.
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Approach: label the children you know, then the leftover picture matching the giving clue is Barbara
First label the pictures the clues hand directly to Anna, Josef and Bob.
The remaining clue says Barbara gave Eva her item, which points to one specific picture.
Five children are talking about the number 325. Andreas: “It is a three-digit number.” Boris: “All the digits are different.” Sara: “The digit sum is 10.” Gerda: “The units digit is 5.” Daniela: “All the digits are odd.” Who has made a mistake?
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Answer: E — Daniela
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Hint 1 of 2
Check each child's statement against the actual number 325.
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Hint 2 of 2
One statement is simply false — look at whether every digit is really odd.
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Approach: test each claim about 325
325 is a three-digit number (Andreas correct), its digits 3, 2, 5 are all different (Boris correct).
3+2+5 = 10 (Sara correct) and the units digit is 5 (Gerda correct).
But 2 is even, so 'all the digits are odd' is wrong — that is Daniela.
Anna starts walking in the direction of the arrow. At each crossing she turns either right or left. She turns right, then left, then left again, then right, then left, then left again. What will she find at the next crossing she reaches?
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Answer: A
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Hint 1 of 3
Put your finger on the start and point it the way the arrow points.
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Hint 2 of 3
Make the turns one at a time in order: right, left, left, right, left, left.
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Hint 3 of 3
After the last turn, look at the very next crossing to see which pictured item is there.
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Approach: walk the path one turn at a time and read off the item at the final crossing
Start at the arrow and trace the path, turning right, left, left, right, left, then left.
Keep your finger moving along the streets so you do not skip a crossing.
The item waiting at the next crossing is the one shown in option A.
Nathalie wanted to build a large cube out of lots of small cubes, just like in Picture 1. How many cubes are missing from Picture 2 that would be needed to build the large cube?
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Answer: C — 7
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Hint 1 of 3
A full big cube like Picture 1 is 3 cubes wide, 3 deep and 3 tall, so it needs 27 small cubes.
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Hint 2 of 3
Count how many small cubes are really in Picture 2, layer by layer.
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Hint 3 of 3
The missing number is 27 take away the cubes you counted in Picture 2.
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Approach: subtract the cubes present from a full cube
A complete large cube is 3 × 3 × 3 = 27 small cubes.
Counting the cubes in picture 2 layer by layer gives 20 cubes.
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?
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Answer: D — 30
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Hint 1 of 3
First work out the total number of bread slices Vera's mum has.
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Hint 2 of 3
A half pack has half of 24 slices, which is 12 slices.
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Hint 3 of 3
Each sandwich eats 2 slices, so split the total into pairs to count the sandwiches.
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Approach: count slices, then divide by 2
Two whole packs and one half pack give 24 + 24 + 12 = 60 slices.
Each sandwich uses 2 slices, so 60 ÷ 2 = 30 sandwiches.
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?
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Answer: D — 23 cm
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Hint 1 of 3
Each lie adds 6 cm to the nose; each truth takes 2 cm away.
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Hint 2 of 3
There are three lies and two truths, so work out the total growing and the total shrinking.
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Hint 3 of 3
Start at 9 cm, add all the growing, then take away all the shrinking.
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Approach: add the lies, subtract the truths
Three lies add 3 × 6 = 18 cm; two truths take off 2 × 2 = 4 cm.
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?
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Answer: C — 16
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Hint 1 of 3
Add up all of the USA's medals, then add up all of China's medals.
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Hint 2 of 3
The lighter way is to compare gold to gold, silver to silver, and bronze to bronze.
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Hint 3 of 3
Find each of those three differences and add the three differences together.
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Approach: compare each colour separately so the numbers stay small
Gold: the USA had 46 and China had 38, that is 8 more gold.
Silver: 29 against 27 is 2 more, and bronze: 29 against 23 is 6 more.
Adding the three extras, 8 + 2 + 6 = 16 more medals, which is answer C.
30 children took part in a sports competition, playing football and handball. 15 took part in football and 20 took part in handball. How many children took part in both sports?
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Answer: E — 5
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Hint 1 of 3
If you add the football players and the handball players, the children who played both get counted twice.
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Hint 2 of 3
Add 15 and 20 and compare that total to the real number of children, 30.
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Hint 3 of 3
The extra amount above 30 is exactly the children who were counted twice.
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Approach: the double-counted children are the ones who played both
Add the football players and handball players: 15 + 20 = 35.
But there are only 30 children, so 35 is 5 too many.
Those extra 5 are the children counted in both lists, so 5 played both, which is answer E.
The number 35 has a special property: it can be divided exactly by its units digit, because \(35 \div 5 = 7\). The number 38 does not have this property. How many numbers bigger than 21 but smaller than 30 have this property?
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Answer: B — 3
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Hint 1 of 3
Bigger than 21 and smaller than 30 means the numbers 22, 23, 24, 25, 26, 27, 28, 29.
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Hint 2 of 3
For each one, can you split it into equal groups of its last digit with none left over?
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Hint 3 of 3
Count how many of the eight numbers pass that test.
Andi, Betti, Clara and Dani were born in the same year. Their birthdays are on 20 February, 12 April, 12 May and 25 May, but not necessarily in that order. Betti and Andi were born in the same month. Andi and Clara were born on the same day, in different months. Who is the oldest?
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Answer: D — Dani
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Hint 1 of 2
'Same month' must be the month that has two of the dates.
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Hint 2 of 2
'Same day, different months' must use the day number that appears twice.
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Approach: match the pairs of dates by the clues
Betti and Andi share a month, and only May has two dates (12 May, 25 May).
Andi and Clara share a day in different months, and only the 12th repeats (12 April, 12 May), so Andi = 12 May, Clara = 12 April; then Betti = 25 May.
Dani gets the leftover 20 February, the earliest date, so Dani is the oldest.
If I join the midpoints of the sides of the large triangle in the picture, a small triangle is formed. If I join the midpoints of the sides of this small triangle, a tiny triangle is formed. How many of these tiny triangles can fit into the largest triangle at the same time?
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Answer: D — 16
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Hint 1 of 3
When you join the midpoints of a triangle, it splits into 4 equal little triangles just like it.
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Hint 2 of 3
So the big triangle holds 4 small triangles, and each small triangle holds 4 tiny ones.
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Hint 3 of 3
Count the tiny ones: 4 small triangles, each made of 4 tiny ones.
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Approach: each midpoint-join splits a triangle into 4 copies, so do it twice
Joining the midpoints cuts the big triangle into 4 equal small triangles.
Doing it again cuts each of those small triangles into 4 tiny ones.
That is 4 groups of 4 tiny triangles, so 4 × 4 = 16 fit in the big triangle, which is answer D.
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?
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Answer: C — 3 and 8
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Hint 1 of 3
Ten marbles two at a time make 5 bags, and all five bags weigh the same.
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Hint 2 of 3
Add all the weights 1 + 2 + ... + 10 = 55, then share that equally among 5 bags.
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Hint 3 of 3
Each bag must weigh 11, so find the marble that pairs with 3 to make 11.
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Approach: equal-sum pairing
The total weight is 1+2+···+10 = 55, and 5 bags means each weighs 11 dag.
Pairs making 11: (1,10), (2,9), (3,8), (4,7), (5,6).
Baris has a few dominoes, shown in the picture. He wants to lay them in a line following the rules of dominoes: two dominoes can be placed next to each other only if the touching squares have the same number of dots. What is the greatest number of these dominoes that he can lay in a single line?
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Answer: C — 5
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Hint 1 of 3
Two dominoes may touch only when the halves that meet show the same number of dots.
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Hint 2 of 3
Treat it like a chain: the right end of one domino must equal the left end of the next.
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Hint 3 of 3
Try building the longest single chain you can, joining matching ends, and you may flip a domino around.
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Approach: build the longest chain where touching halves match
Pick a starting domino, then add a domino whose end matches its end.
Keep linking matching ends, flipping a domino around when that helps the numbers meet.
The longest single line you can make uses 5 dominoes, which is answer C.
Peter has bought a rug that is 36 dm wide and 60 dm long. The rug is covered in squares that each contain either a sun or a moon, as shown in the picture. There are exactly nine squares along the width of the rug. The full length of the rug cannot be seen. How many moons would you see if you could see the entire rug?
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Answer: B — 67
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Hint 1 of 3
If 9 squares fit across the 36 dm width, work out how wide one square is.
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Hint 2 of 3
Use that square size to find how many squares fit along the 60 dm length.
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Hint 3 of 3
Multiply rows by columns for the total squares, then the sun/moon pattern splits them almost in half.
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Approach: find the grid size, then count one colour
36 dm ÷ 9 = 4 dm per square, so the length 60 dm holds 60 ÷ 4 = 15 squares.
The rug is 9 × 15 = 135 squares in a checkerboard of suns and moons.
The moons fall on the smaller of the two colour groups, giving 67 moons.
Beatrice has several grey tiles that all look exactly like the one pictured. At least how many of these tiles does she need in order to make a complete square?
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Answer: B — 4
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Hint 1 of 3
Try fitting copies of the tile together so they make a big square with no gaps and no overlaps.
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Hint 2 of 3
Turn and rotate the copies so their stair-step edges lock into each other.
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Hint 3 of 3
Start small: see whether 2 or 3 copies can ever make a square before trying 4.
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Approach: rotate copies of the tile so they lock into a square, using as few as possible
Slide and rotate copies of the tile so their jagged edges fit together with no gaps.
Two or three copies cannot close up into a full square, but four copies do fit together into one.
So the fewest she needs is 4 tiles, which is answer B.
