Barbara wrote “KAENGURUWETTBEWERB” on the blackboard. She used the same colour for equal letters and a different colour for different letters. How many different colours did she use?
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Answer: D — 10
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Hint 1 of 2
Same letter means same colour, so count how many different letters appear.
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Hint 2 of 2
List the distinct letters in KAENGURUWETTBEWERB and count them.
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Approach: count distinct letters
Each distinct letter needs its own colour; repeated letters reuse a colour.
The distinct letters are K, A, E, N, G, U, R, W, T, B.
The number 3 should be added to the number 6. This amount is then doubled, and the result is increased by 1. Which of the following sums fits this description?
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Answer: D — \((6 + 3) \times 2 + 1\)
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Hint 1 of 2
Translate the words step by step into an expression.
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Hint 2 of 2
'This amount is then doubled' means multiply the whole sum (6+3) by 2 first.
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Approach: translate words to an expression
'Add 3 to 6' gives (6 + 3).
'This amount is doubled' gives (6 + 3) x 2.
'the result increased by 1' gives (6 + 3) x 2 + 1, which is choice D.
In the school for animals there are 3 cats, 2 ducks, 2 sheep and some dogs. The teacher counted the legs of all the animals and got 44. How many dogs go to the school?
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Answer: B — 5
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Hint 1 of 2
Count the legs you already know from the cats, ducks and sheep.
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Hint 2 of 2
Subtract those legs from 44, then see how many 4-legged dogs are left.
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Approach: account for known legs, then divide
Cats: 3 x 4 = 12 legs. Sheep: 2 x 4 = 8 legs. Ducks: 2 x 2 = 4 legs.
Known legs = 12 + 8 + 4 = 24.
Dog legs = 44 - 24 = 20, and each dog has 4 legs, so 20 / 4 = 5 dogs.
The last row in an aeroplane is row 25. There is no row 13, and row 15 has only 4 seats. Every other row has 6 seats. How many passenger seats are there on this aeroplane?
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Answer: C — 142
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Hint 1 of 2
First count how many rows actually exist (watch out for the missing row 13).
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Hint 2 of 2
Most rows have 6 seats; only row 15 is different, so adjust for it.
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Approach: count rows, then adjust the odd one out
Rows are numbered 1 to 25 but row 13 is missing, so there are 24 rows.
If every row had 6 seats that would be 24 x 6 = 144.
Row 15 has only 4 instead of 6, which is 2 fewer, so 144 - 2 = 142 seats.
In addition to the weight of the basket, a single balloon can lift 80 kg. Two balloons can lift 180 kg in addition to the weight of the basket. How heavy is the basket?
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Answer: E — 20 kg
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Hint 1 of 2
Adding a second balloon adds another 80 kg of lift; compare the two situations.
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Hint 2 of 2
Find the basket weight by seeing how the totals jump from one balloon to two.
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Approach: compare one balloon vs two balloons
Let the basket weigh b and one balloon's own lift be L. One balloon: L - b = 80.
Two balloons: 2L - b = 180. Subtracting the first from the second gives L = 100.
Then b = L - 80 = 100 - 80 = 20 kg. Check: 2(100) - 20 = 180.
Grandmother gave Vivian and Mike some apples and pears. In total they had 25 pieces of fruit. On the way home Vivian ate 1 apple and 3 pears, and Mike ate 3 apples and 2 pears. At home they noticed that there were exactly the same number of apples and pears left in the basket. How many pears had Grandmother given them?
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Answer: B — 13
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Hint 1 of 2
Work out how many apples and pears were eaten in total.
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Hint 2 of 2
After eating, the basket has equal apples and pears; build back up to the pears given.
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Approach: track what was eaten, then use the equal-leftover clue
Eaten: Vivian 1 apple + 3 pears, Mike 3 apples + 2 pears, so 4 apples and 5 pears eaten (9 pieces).
Left in the basket: 25 - 9 = 16 pieces, split equally, so 8 apples and 8 pears.
Total pears = pears left + pears eaten = 8 + 5 = 13.
Lisa built a large cube out of 8 smaller ones. The small cubes have the same letter on each of their faces (A, B, C or D). Two cubes with a common face always have a different letter on them. Which letter is on the cube that cannot be seen in the picture?
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Answer: B — B
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Hint 1 of 3
There are 8 little cubes but only 4 letters, and every cube touches three neighbours that must all differ from it.
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Hint 2 of 3
Two cubes can share a letter only if they do NOT touch, i.e. they sit at opposite ends of a long diagonal through the centre.
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Hint 3 of 3
The hidden cube is the corner diagonally opposite a visible one, so it copies that cube's letter.
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Approach: opposite corners share a letter
Each small cube touches 3 others (one in each direction), and touching cubes must differ, so a cube and its 3 neighbours use up all 4 letters A, B, C, D.
That means a letter can repeat only on two cubes that never touch, namely the two ends of a diagonal running through the centre of the big cube.
So each of the 4 space-diagonals carries one repeated letter, pairing every cube with the corner diagonally across from it.
The unseen back corner is diagonally opposite a visible corner, and matching their letter gives the hidden one: B.
The natural numbers are to be painted: 1 is red, 2 is blue, 3 is green, 4 is red, 5 is blue, 6 is green, and so on. Which colour(s) can the sum of a red number and a blue number have?
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Answer: A — green only
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Hint 1 of 2
The colours repeat every three numbers: red, blue, green, red, blue, green...
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Hint 2 of 2
Think about the remainders when red and blue numbers are divided by 3, then add them.
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Approach: use remainders mod 3
Red numbers are 1, 4, 7, ... (remainder 1 when divided by 3); blue are 2, 5, 8, ... (remainder 2).
A red plus a blue number has remainder 1 + 2 = 3, that is remainder 0 (a multiple of 3).
Multiples of 3 (3, 6, 9, ...) are exactly the green numbers, so the sum is always green only.
The numbers 1 to 7 should be written in the small circles so that the sum of the numbers along each line is the same. Which number should be written in the uppermost circle on the triangle?
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Answer: C — 4
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Hint 1 of 3
First add up all the numbers you must place: 1 + 2 + ... + 7.
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Hint 2 of 3
The three corner circles each sit on two lines, so they get counted twice when you add the three line-sums together.
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Hint 3 of 3
Try giving the lines the smallest equal sum that works, and see which number is forced into the top circle.
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Approach: balance the three equal line-sums
The seven numbers 1 to 7 add up to 28.
Add the three line-totals together: every circle is counted, but the three corner circles each lie on two lines, so they get counted one extra time; the grand total is 28 plus the three corner numbers.
For the three lines to share one equal sum, that grand total must split evenly by 3, and trying the natural balanced arrangement makes each line add to 10.
Filling that in, the only number that lands in the top circle is 4 (choice C).
Four cogs are connected to each other as shown in the picture. The first has 30 teeth, the second 15, the third 60 and the fourth 10. How many turns will the last cog make for each full turn of the first cog?
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Answer: A — 3
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Hint 1 of 2
Connected gears pass the same number of teeth; the in-between gears do not change the first-to-last result.
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Hint 2 of 2
Compare only the first gear's teeth with the last gear's teeth.
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Approach: teeth passed are equal; compare first and last
One full turn of the first gear moves 30 teeth along the chain of gears.
Those same 30 teeth pass the last gear, which has only 10 teeth.
So the last gear turns 30 / 10 = 3 times (the middle gears do not matter).
A rectangular piece of paper is 108 mm long and 84 mm wide. After making a straight cut you have a square and a leftover piece. You do the same with the leftover piece and so on until the leftover piece itself is a square. What is the side length of the last square?
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Answer: E — 12 mm
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Hint 1 of 2
Each cut peels off the biggest square it can; track the leftover rectangle's sizes.
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Hint 2 of 2
This is the subtract-the-smaller-side process, and it ends at the greatest common divisor.
Both the figures on the right were made out of the same 5 pieces. The rectangle has dimensions 5 cm × 10 cm. The other pieces are quarter circles with 2 different sized radii. What is the difference between the perimeters of the two figures?
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Answer: E — 20 cm
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Hint 1 of 2
Both shapes use the very same five pieces, so the curved (arc) parts of their outlines are identical.
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Hint 2 of 2
Only the straight rectangle edges show up differently, so compare just those.
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Approach: cancel the identical arcs, compare straight edges
Both figures are built from the very same five pieces, so the curved quarter-circle arcs contribute the same total length to each outline and simply cancel when we take the difference.
The only thing that can differ is how much of the rectangle's straight edges shows on the outside; in one figure a pair of the rectangle's 10 cm sides lies on the boundary while in the other they are tucked inside.
That swap of two 10 cm straight edges gives a perimeter difference of 2 x 10 = 20 cm (choice E).
Logic & Word Problemscareful-countingsum-constraint
12 children were at a birthday party. The children were 6, 7, 8, 9, and 10 years old (every one of these ages was present). Four of them were 6 years old. There were more 8-year-olds than any other age group. What is the average age of the children?
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Answer: B — 7·5
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Hint 1 of 3
All five ages 6, 7, 8, 9 and 10 are present, so each of those groups has at least one child.
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Hint 2 of 3
The 8-year-olds must outnumber every other group, including the four 6-year-olds, so there are at least five 8-year-olds.
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Hint 3 of 3
Once the counts are pinned down, just add all twelve ages and divide by 12.
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Approach: pin down the counts from the clues, then average
Four children are 6, so the other eight share the ages 7, 8, 9 and 10, with at least one child at each age.
The 8-year-olds are the biggest group, so they must beat the four 6-year-olds: at least five children are 8.
Five 8s plus one each of 7, 9 and 10 already uses 5 + 3 = 8 children, so the only fit is exactly one 7, five 8s, one 9 and one 10.
The ages are four 6s, one 7, five 8s, one 9, one 10; their total is 24 + 7 + 40 + 9 + 10 = 90, and 90 / 12 = 7.5, choice B.
