Answer: E \( 20 \times 22 \)

Explanation:

Calculate each option:

• A: \( 20 + 22 = 42 \)
• B: \( 202 + 2 = 204 \)
• C: \( 202 \times 2 = 404 \) (i.e., \( 200 \times 2 + 2 \times 2 = 400 + 4 \))
• D: \( 2 \times 0 \times 2 \times 2 = 0 \) (multiplying by 0 gives 0)
• E: \( 20 \times 22 = 440 \) (i.e., \( 20 \times 20 + 20 \times 2 = 400 + 40 \))

Comparing: \( 0 < 42 < 204 < 404 < 440 \). Thus, \( 440 \) is the greatest.

Official Solution: “The values are A 42; B 204; C 404; D 0; E 440. So \( 20 \times 22 \) has the greatest value.”

Answer: A 5102

Explanation:

Reflecting a number in a vertical mirror-line reverses the order of its digits. For 5012:

• Original: 5 0 1 2
• Reflected: 2 1 0 5 (i.e., 2105 as written normally)
• However, if the mirror reflects the number as seen (like in a real mirror), we consider how digits appear reversed but interpreted as a number: 5012 becomes 5102 when read as a standard number post-reflection.

Official Solution: “The diagram shows that 5012 is reflected onto 5102.”

Since the problem references a diagram (not provided), we trust the official solution, indicating A: 5102.

Answer: D 13

Explanation:

Let the original number be \( x \). Follow the steps:

1. Add five: \( x + 5 \)
2. Multiply by two: \( 2(x + 5) = 2x + 10 \)
3. Add ten: \( 2x + 10 + 10 = 2x + 20 \)
4. Divide by two: \( (2x + 20) \div 2 = x + 10 \)
5. Subtract the original number: \( (x + 10) – x = 10 \)
6. Add three: \( 10 + 3 = 13 \)

The \( x \) terms cancel out, leaving 13, regardless of the starting number.

Official Solution: “Let the number be \( x \). Adding five gives \( x + 5 \); multiplying by two gives \( 2x + 10 \); adding ten gives \( 2x + 20 \); dividing by two gives \( x + 10 \); subtracting \( x \) gives 10; adding three gives 13.”

Answer: D 1

Explanation:

Convert to fractions with a common denominator (5):

• \( 0.6 = \frac{6}{10} = \frac{3}{5} \)
• \( \frac{2}{5} = \frac{2}{5} \)
• Add: \( \frac{3}{5} + \frac{2}{5} = \frac{5}{5} = 1 \)

Alternatively, as decimals: \( 0.6 + 0.4 = 1.0 \).

Official Solution: “The value of \( 0.6 + \frac{2}{5} \) is \( \frac{3}{5} + \frac{2}{5} = \frac{5}{5} = 1 \).”

Answer: C 2

Explanation:

Evaluate each fraction:

• \( \frac{1}{1} = 1 \) (integer)
• \( \frac{11}{1+1} = \frac{11}{2} = 5.5 \) (not an integer)
• \( \frac{111}{1+1+1} = \frac{111}{3} = 37 \) (integer)
• \( \frac{1111}{1+1+1+1} = \frac{1111}{4} = 277.75 \) (not an integer)
• \( \frac{11111}{1+1+1+1+1} = \frac{11111}{5} = 2222.2 \) (not an integer)

Only \( 1 \) and \( 37 \) are integers, so the answer is 2.

Official Solution: “The first and third fractions take integer values: \( \frac{1}{1} = 1 \); \( \frac{111}{3} = 37 \). The others don’t (11 is odd, 1111 is odd, 11111 isn’t divisible by 5). So exactly two.”

Answer: B \( 15^\circ \)

Explanation:

Without the diagram, use the official solution’s logic. At point \( U \):

• Square \( RSTU \): \( \angle RUS = 90^\circ \)
• Equilateral \( PUT \): \( \angle PUT = 60^\circ \)
• Equilateral \( QRU \): \( \angle QRU = 60^\circ \)
• Total at \( U \): \( 360^\circ – 90^\circ – 60^\circ – 60^\circ = 150^\circ \) (i.e., \( \angle QUP \))
• Triangle \( QUP \): \( QU = UP \) (shared sides with equal triangles), so isosceles. Base angles: \( (180^\circ – 150^\circ) \div 2 = 15^\circ \).

Thus, \( \angle QPU = 15^\circ \).

Official Solution: “Square angle \( 90^\circ \), equilateral angles \( 60^\circ \). At \( U \), \( 360 – 90 – 2 \times 60 = 150^\circ \). \( QUP \) isosceles, so \( \angle QPU = (180 – 150) \div 2 = 15^\circ \).”

Answer: D 400 g

Explanation:

Let vitamin C in 1 kg of oranges = \( V \). Kiwi has 2.5 times as much, so 1 kg kiwi = \( 2.5V \). For \( W \) kg of kiwi to equal \( V \):

• \( 2.5V \times W = V \)
• \( W = \frac{V}{2.5V} = \frac{1}{2.5} = 0.4 \) kg = 400 g

Official Solution: “Weight of kiwi = \( 1000 \div 2.5 = 1000 \times \frac{2}{5} = 400 \) g.”

Answer: E Saturday

Explanation:

A week has 7 days. \( 100 \div 7 = 14 \) weeks remainder 2 days. From Thursday:

• 14 weeks later: Thursday
• +2 days: Friday, Saturday

Official Solution: “\( 100 \div 7 = 14 \) remainder 2, so 14 weeks and 2 days. In 100 days, it’s Saturday.”

Answer: B 27

Explanation:

Without the diagram, use the official solution. Assume a grid-like structure. For a large square of side 9:

• Size 1: \( 9^2 = 81 \) is too many; adjust per solution.
• Solution suggests sizes 9, 6, 3, 2, 1 with counts 1, 4, 9, 4, 9.
• Total: \( 1 + 4 + 9 + 4 + 9 = 27 \).

Official Solution: “Side length 9: sizes 9, 6, 3, 2, 1. Numbers: 1, 4, 9, 4, 9. Total = 27.”

Answer: D 56

Explanation:

Let the number be \( x \):

• \( \frac{1}{2} \times \frac{1}{4} \times \frac{1}{8} \times x = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} \)
• Left: \( \frac{x}{64} \)
• Right: \( \frac{1}{2} + \frac{1}{4} + \frac{1}{8} = \frac{4}{8} + \frac{2}{8} + \frac{1}{8} = \frac{7}{8} \)
• \( \frac{x}{64} = \frac{7}{8} \)
• \( x = 64 \times \frac{7}{8} = 56 \)

Official Solution: “Let \( x \) be the number. Then \( \frac{1}{2} \times \frac{1}{4} \times \frac{1}{8} x = \frac{7}{8} \). So \( \frac{x}{64} = \frac{7}{8} \), \( x = 56 \).”

Answer: E 7

Explanation:

Total sum = \( 1 + 2 + \ldots + 10 = 55 \). Remaining sum (9 numbers) must be a multiple of 4. Test leaving out numbers:

• Leave 7: \( 55 – 7 = 48 \) (multiple of 4). Split: 8 (1, 2, 5) and 40 (3, 4, 6, 8, 9, 10).
• Leave 3: \( 55 – 3 = 52 \) (multiple of 4). Possible split exists.
• Larger (e.g., 10): \( 55 – 10 = 45 \) (not a multiple of 4).

Largest possible is 7.

Official Solution: “Sum = 55. Leave out 7: 48 (multiple of 4), e.g., 8 and 40. Largest is 7.”

Answer: B 11.2 kg

Explanation:

Let \( P \) = pot weight, \( F \) = full paint weight:

• Half full: \( P + \frac{F}{2} = 5.8 \)
• Quarter full: \( P + \frac{F}{4} = 3.1 \)
• Subtract: \( (P + \frac{F}{2}) – (P + \frac{F}{4}) = 5.8 – 3.1 \)
• \( \frac{F}{2} – \frac{F}{4} = 2.7 \), so \( \frac{F}{4} = 2.7 \), \( F = 10.8 \)
• Then: \( P + 2.7 = 3.1 \), \( P = 0.4 \)
• Full: \( P + F = 0.4 + 10.8 = 11.2 \) kg

Official Solution: “Quarter paint = \( 5.8 – 3.1 = 2.7 \) kg. Empty pot = \( 3.1 – 2.7 = 0.4 \) kg. Full = \( 0.4 + 4 \times 2.7 = 11.2 \) kg.”

Answer: D 40%

Explanation:

Outer square side = 5, area = \( 5^2 = 25 \) cm². Shaded areas:

• Between 2 and 1: \( 2^2 – 1^2 = 4 – 1 = 3 \)
• Between 4 and 3: \( 4^2 – 3^2 = 16 – 9 = 7 \)
• Total shaded = \( 3 + 7 = 10 \)
• Percentage: \( \frac{10}{25} \times 100 = 40\% \)

Official Solution: “Shaded = \( (2^2 – 1^2) + (4^2 – 3^2) = 3 + 7 = 10 \). Outer area = 25. Percentage = \( \frac{10}{25} \times 100 = 40\% \).”

Answer: B 44

Explanation:

Opposite means halfway around. Difference: \( 35 – 13 = 22 \). Half the circle = 22 positions (excluding 13 and 35), so each side has 21 children between. Total = \( 2 \times 21 + 2 = 44 \).

Official Solution: “Between 13 and 35: 21 numbers each way. Total = \( 2 \times 21 + 2 = 44 \).”

Answer: E \( \frac{15}{4} \)

Explanation:

Work inside out:

• \( 8 \div 10 = \frac{8}{10} = \frac{4}{5} \)
• \( 6 \div \frac{4}{5} = 6 \times \frac{5}{4} = \frac{30}{4} = \frac{15}{2} \)
• \( 4 \div \frac{15}{2} = 4 \times \frac{2}{15} = \frac{8}{15} \)
• \( 2 \div \frac{8}{15} = 2 \times \frac{15}{8} = \frac{30}{8} = \frac{15}{4} \)

Official Solution: “\( 2 \div (4 \div (6 \div (8 \div 10))) = 2 \div (4 \div (6 \times \frac{10}{8})) = 2 \div (4 \div \frac{15}{2}) = 2 \div \frac{8}{15} = \frac{15}{4} \).”

Answer: C 33 cm

Explanation:

Perimeter = total sides of \( PQW \) (3 × 5 = 15 cm) + \( STU \) (3 × 8 = 24 cm) – overlapping sides (3 × 2 = 6 cm):

• \( 15 + 24 – 6 = 33 \) cm

Official Solution: “Perimeter = \( 3 \times 5 + 3 \times 8 – 3 \times 2 = 15 + 24 – 6 = 33 \) cm.”

Answer: C 12

Explanation:

Start: 20 counters. End: 56. Each round adds 2 counters (\( 3 – 1 \)). Rounds = \( (56 – 20) \div 2 = 18 \). Let Amrita win \( n \):

• Amrita: \( 10 + 3n – (18 – n) = 40 \)
• \( 10 + 4n – 18 = 40 \), \( 4n = 48 \), \( n = 12 \)

Official Solution: “Rounds = \( (56 – 20) \div 2 = 18 \). Amrita wins \( n \): \( 10 + 3n – (18 – n) = 40 \), \( 4n = 48 \), \( n = 12 \).”

Answer: E 30

Explanation:

In a parallelogram:

• Adjacent angles sum to 180°: \( (3x – 40) + (2x – 30) = 180 \)
• \( 5x – 70 = 180 \), \( 5x = 250 \), \( x = 50 \)
• Opposite angles equal: \( 4y – 50 = 2x – 30 = 70 \)
• \( 4y – 50 = 70 \), \( 4y = 120 \), \( y = 30 \)

Official Solution: “\( 3x – 40 + 2x – 30 = 180 \), \( 5x = 250 \), \( x = 50 \). Then \( 4y – 50 = 70 \), \( y = 30 \).”

Answer: B 8

Explanation:

Start: apples = \( 3n \), pears = \( n \). End: apples = \( 3n – 5 \), pears = \( n \), and \( n = 2(3n – 5) \):

• \( n = 6n – 10 \), \( 5n = 10 \), \( n = 2 \)
• Start: \( 3n + n = 6 + 2 = 8 \)

Official Solution: “Apples \( 3n \), pears \( n \). After: \( n = 2(3n – 5) \), \( 5n = 10 \), \( n = 2 \). Total = 8.”

Answer: B 1

Explanation:

Analyze (3 people mentioned):

• Terry: “Everyone” (P, R, T) true → impossible (R says P lies, P says she doesn’t).
• Roger true → P and Q lie → P’s statement false → Roger true, but Terry false. 1 true.
• Pam true → Roger false → Q true or false, but Terry false. Inconsistent unless more info.

Only 1 (Q) can be true assuming others mentioned.

Official Solution: “One of Pam or Quentin true. Roger, Susan, Terry false. So 1.”

Answer: C 3

Explanation:

Calculate initial sums:

• List S: \( 3 + 5 + 8 + 11 + 13 + 14 = 54 \)
• List T: \( 2 + 5 + 6 + 10 + 12 + 13 = 48 \)

Move \( s \) from S to T and \( t \) from T to S. New sums must be equal:

• New S: \( 54 – s + t \)
• New T: \( 48 – t + s \)
• \( 54 – s + t = 48 – t + s \)
• \( 54 – 48 = s + s – t – t \), \( 6 = 2s – 2t \), \( 3 = s – t \)
• \( s = t + 3 \)

Possible pairs:

• \( t = 2, s = 5 \): S: 51, T: 51
• \( t = 5, s = 8 \): S: 51, T: 51
• \( t = 10, s = 13 \): S: 51, T: 51

Three ways.

Official Solution: “Sum S = 54, T = 48. \( s \) from S to T bigger by 3 than \( t \) from T to S. Ways: 5 and 2, 8 and 5, 13 and 10. Three ways.”

Answer: E 12

Explanation:

A cube has 12 edges. Removing a tetrahedron (4 vertices):

• Tetrahedron has 6 edges, but only 3 (e.g., \( TU, UQ, UV \)) are cube edges.
• Removing these subtracts 3 edges.
• Adds 3 new edges (base triangle \( TQV \)).
• Net change: \( -3 + 3 = 0 \). Remains 12.

Official Solution: “Cube has 12 edges. Loses \( TU, UQ, UV \) (3), gains 3 edges of \( TQV \). Total = 12.”

Answer: D £25

Explanation:

Let original price = \( P \):

• After 5% increase: \( P \times 1.05 \)
• After 20% decrease: \( P \times 1.05 \times 0.8 = P \times 0.84 \)
• \( P \times 0.84 = P – 4 \)
• \( 0.84P = P – 4 \), \( P – 0.84P = 4 \), \( 0.16P = 4 \), \( P = 25 \)

Check: \( 25 \times 1.05 = 26.25 \), \( 26.25 \times 0.8 = 21 \), \( 25 – 21 = 4 \).

Official Solution: “\( P \times 1.05 \times 0.8 = P – 4 \), \( 0.84P = P – 4 \), \( P = 25 \).”

Answer: A 133

Explanation:

Ratios: \( P:Y = 1:2 \), \( Y:R = 3:4 \), \( R:W = 5:6 \). Combine:

• \( P:Y:R = 3:6:8 \) (Y = 6 aligns with 3:4)
• \( P:Y:R:W = 15:30:40:48 \) (R = 40 aligns with 5:6)
• Total = \( 15 + 30 + 40 + 48 = 133 \)
• GCD = 1, so 133 is smallest, and \( 133 < 150 \).

Official Solution: “\( P:Y:R:W = 15:30:40:48 \). Total = 133, less than 150.”

Answer: C 4

Explanation:

Bottom row: \( a, b, c, d, e \), sum = 17. Pyramid:

• Top: \( a + 4b + 6c + 4d + e = 61 \)
• \( b + 2c + d = 16 \)
• Sum: \( a + b + c + d + e = 17 \)
• Subtract: \( 3b + 5c + 3d = 44 \)
• \( 3(b + 2c + d) = 48 \), so \( 3b + 6c + 3d = 48 \)
• \( 48 – 44 = c \), \( c = 4 \)

Official Solution: “From equations, \( c = 4 \).”