There are two holes in the cover of a book. The book lies on the table opened up (see diagram). After closing up the book, which vehicles can Olaf see through the two holes?
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Answer: D
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Hint 1 of 2
When the cover folds over, the holes line up above the page on the right.
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Hint 2 of 2
The folded cover flips left-to-right, so the vehicles appear in mirror order through the two windows.
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Approach: fold the cover and look through the holes
Folding the cover onto the page flips it like a mirror.
The two holes then sit over two groups of vehicles, but in reversed left-right order.
Matching the windows to the line of vehicles gives the set in option D.
Linda fixes 3 photos on a pin board next to each other. She uses 8 pins to do so (see picture). Peter wants to fix 7 photos in the same way. How many pins does he need for that?
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Answer: B — 16
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Hint 1 of 2
Look at the picture: the first photo needs 4 pins (its 4 corners), but each new photo touches the one before it.
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Hint 2 of 2
Because neighbours share their pins, every photo after the first adds the same small number of pins.
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Approach: see the repeating add-on for each photo
The top row of pins has one more pin than the number of photos, and so does the bottom row.
For 3 photos that is 4 pins on top and 4 pins on the bottom, which makes 8 pins — this matches the picture.
For 7 photos it is 8 pins on top and 8 pins on the bottom.
Mother halves the birthday cake. One half she then halves again. Of that she again halves one of the smaller pieces. Of these smaller pieces she once more halves one of them (see diagram). One of the two smallest pieces weighs 100 g. How much does the entire cake weigh?
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Answer: D — 1600 g
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Hint 1 of 2
Each halving makes a piece half as big; the smallest piece is a fraction of the whole cake.
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Hint 2 of 2
Halving four times gives a sixteenth, so the 100 g piece is 1/16 of the cake.
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Approach: track the fraction of the whole
Halving the cake repeatedly gives pieces of 1/2, 1/4, 1/8, and finally 1/16.
The smallest piece is 1/16 of the cake and weighs 100 g.
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?
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Answer: B — 10
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Hint 1 of 2
First turn as many blue marbles as possible into red ones, then trade reds for greens.
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Hint 2 of 2
Watch the leftovers: trades only happen in fixed bundles (3 blue, 2 red).
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Approach: trade in bundles and track leftovers
16 blue ÷ 3 gives 5 red marbles (1 blue left over).
5 red ÷ 2 gives 2 trades = 10 green marbles (1 red left over).
Steven wants to write each of the digits 2, 0, 1 and 9 into the boxes of this addition (a three-digit number plus a single-digit number). He wants to obtain the biggest result possible. Which digit does he have to use for the single-digit number?
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Answer: A — either 0 or 1
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Hint 1 of 2
The biggest sum comes from putting the largest digit, 9, in the hundreds place of the three-digit number.
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Hint 2 of 2
The single digit and the units of the three-digit number both count once, so swapping them does not change the total.
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Approach: maximise by place value
Put 9 in the hundreds place; the sum becomes 900 + (tens, units, single from 2,1,0).
Put 2 in the tens place to add the most: 920 + (remaining 1 and 0).
The leftover 1 and 0 fill the units and the single-digit number, each adding the same amount.
So the single digit can be either 0 or 1: answer A.
The pictures show how much 2 pieces of fruit cost altogether. The first three show pairs costing 5, 7 and 10 Taler. How much do the three fruits in the last picture cost altogether?
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Answer: D — 11 Taler
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Hint 1 of 2
Add the three given pair-prices together; each fruit then appears exactly twice.
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Hint 2 of 2
The last picture shows all three fruits, whose total is half of that combined sum.
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Approach: add the pairs, then halve
The three given totals are 5, 7 and 10 Taler.
Adding them gives 22, which counts each fruit twice.
All three fruits together cost 22 ÷ 2 = 11 Taler (D).
Each shape represents exactly one digit. The sum of the digits in each row is stated on the right hand-side of each row. Which digit does the star stand for?
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Answer: E — 6
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Hint 1 of 2
A row of three identical circles tells you one circle's value right away.
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Hint 2 of 2
Substitute the circle into the other rows to peel out the star and heart.
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Approach: solve the shape values one at a time
Row of three circles: 3×circle = 12, so circle = 4.
Top row: circle + star + heart = 15, so star + heart = 11.
Bottom row: star + heart + heart = 16, so subtracting gives heart = 5 and star = 6.
Anna uses 32 small grey squares to frame a 7 cm by 7 cm big picture. How many small grey squares does she have to use to frame a 10 cm by 10 cm big picture?
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Answer: C — 44
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Hint 1 of 2
The grey squares make a ring one square thick all the way around the picture.
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Hint 2 of 2
Picture the four sides of the ring, and be careful not to count the four corner squares twice.
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Approach: count the four sides of the grey ring
Around the 7×7 picture, each side of the grey ring is 9 squares long (the 7 picture squares plus one corner at each end), and 4 sides of 9 with the 4 corners counted once give 32 — matching the picture.
Around the 10×10 picture, each side of the ring is 12 squares long.
Four sides of 12 is 48, but the 4 corners were each counted twice, so take 4 away: 48 − 4 = 44.
The pages of a book are numbered with 1, 2, 3, 4, 5 and so on. The digit 5 appears exactly 16 times. What is the maximum number of pages the book can have?
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Answer: B — 64
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Hint 1 of 2
Count where the digit 5 shows up as you list page numbers 1, 2, 3, ...
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Hint 2 of 2
The block 50-59 alone contributes ten 5s (the tens digit); add those to the single 5s like 5, 15, 25, ...
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Approach: count occurrences of the digit 5
Up to 49 the 5s appear at 5, 15, 25, 35, 45: five of them.
The block 50-59 has a 5 in every tens digit: ten more, plus the extra units-5 in 55, reaching the 16th by page 59.
Pages 60-64 add no new 5s, but page 65 would add a 17th.
So the most pages with exactly sixteen 5s is 64 (B).
Logic & Word Problemssum-constraintcomplementary-counting
There live exactly 15 animals on a farm: cows, cats and kangaroos. We know that exactly 10 animals are not cows and exactly 8 animals are not cats. How many kangaroos live on the farm?
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Answer: B — 3
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Hint 1 of 2
'Not cows' counts cats and kangaroos; 'not cats' counts cows and kangaroos.
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Hint 2 of 2
Find the cows and cats first, then subtract from 15 to get the kangaroos.
One of the 5 children Alex, Bartek, Cora, Dani and Emil has eaten a cake. Alex says: “I did not eat a cake.” Bartek says: “I ate a cake.” Cora says: “Emil has not eaten a cake.” Dani says: “I did not eat a cake.” Emil says: “Alex has eaten a cake.” One of the children lies. Which child has eaten a cake?
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Answer: B — Bartek
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Hint 1 of 2
Exactly one statement is false; test who the eater could be and count the lies.
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Hint 2 of 2
Suppose Bartek is the eater and check whether only one child ends up lying.
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Approach: test the eater so that exactly one lies
Assume Bartek ate the cake.
Then Alex, Cora and Dani all speak truthfully, and Bartek's 'I ate' is true.
Only Emil's 'Alex ate' is false — exactly one liar, as required.
From above, the corridor of a school looks like in the diagram. A cat walks along the dotted line drawn in the middle of the room. How many meters does the cat walk?
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Answer: E — 83 m
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Hint 1 of 2
The cat follows the dashed centre line, so use the middle of each corridor section, not the outer walls.
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Hint 2 of 2
Break the path into the three straight middle-line pieces and add their lengths.
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Approach: add the centre-line segments
The corridor has three arms; the cat's dashed path runs along the middle of each.
Bottom arm: the vertical part is 40 − 36 = 4 m wide, so its middle sits 2 m in from the right wall; the centre line runs 36 + 2 = 38 m across.
Vertical arm: it is 20 m up to the start of the top arm, which is 6 m tall, so the centre line climbs to the middle of the top arm; together this part of the path is 19 m.
Top arm: from the middle of the vertical arm out to the far end is 26 m.
