Answer: B 6 + 5 – 4

Explanation:

We need to determine which of the five expressions yields a result closest to zero. To find out, we’ll evaluate each option systematically while respecting the order of operations (PEMDAS/BODMAS rules where multiplication comes before addition/subtraction).

• Option A: 6 + 5 + 4

  This is straightforward addition: 6 + 5 = 11, then 11 + 4 = 15. The absolute value is |15| = 15.

• Option B: 6 + 5 – 4

  Addition and subtraction have equal precedence, evaluated left to right: 6 + 5 = 11, then 11 – 4 = 7. The absolute value is |7| = 7.

• Option C: 6 + 5 × 4

  Multiplication comes first: 5 × 4 = 20, then 6 + 20 = 26. The absolute value is |26| = 26.

• Option D: 6 – 5 × 4

  Multiplication first: 5 × 4 = 20, then 6 – 20 = -14. The absolute value is |-14| = 14.

• Option E: 6 × 5 ÷ 4

  Multiplication and division have equal precedence, evaluated left to right: 6 × 5 = 30, then 30 ÷ 4 = 7.5. The absolute value is |7.5| = 7.5.

Now, let’s summarize the absolute values for comparison:

• A: |15| = 15
• B: |7| = 7
• C: |26| = 26
• D: |14| = 14
• E: |7.5| = 7.5

The expression closest to zero is the one with the smallest absolute value. Among these, 7 (from B) is the smallest, followed by 7.5 (from E), then 14, 15, and 26.

Official Solution: According to the UKMT Junior Mathematical Challenge 2016 solutions leaflet: “The values of the expressions are: A 15, B 7, C 26, D -14, E 7½. Of these, 7 is closest to 0.”

Detailed Reasoning: To ensure we understand why B is correct:

• The concept of “closest to zero” means we’re looking for the result with the smallest absolute value, regardless of whether it’s positive or negative.
• Option B’s result of 7 beats all other options in this regard.
• Option E’s result of 7.5 is very close but still further from zero than 7.
• The other options are all significantly further from zero.

Common Pitfalls:

• Forgetting the order of operations (especially in options C and D) could lead to incorrect evaluations.
• Mistakenly thinking the negative result in D (-14) is closer to zero than the positive results because it’s on the “other side” of zero (but absolute values show this isn’t the case).
• Overlooking that we need the smallest absolute value, not necessarily the smallest number.

Conclusion: Among the options, B: 6 + 5 – 4 = 7 gives the result closest to zero, making it the correct choice.

Answer: A 59979

Explanation:

We need to find the number that is twenty-one less than sixty thousand and determine which of the given options matches this value. Let’s solve this step-by-step and then verify against the options.

• Step 1: Understand the numbers involved

  “Sixty thousand” is written numerically as 60,000. The question asks for a number that is “twenty-one less” than this, which means we subtract 21 from 60,000.

• Step 2: Perform the subtraction

  Calculate 60,000 – 21. To make it clear, break it down:

  • First, subtract 20: 60,000 – 20 = 59,980
  • Then, subtract 1 more: 59,980 – 1 = 59,979

  Alternatively, compute directly: 60,000 – 21 = 59,979. Both methods yield the same result.

• Step 3: Compare with the options

  Now, let’s check the provided choices:

  • Option A: 59979 – This is exactly 59,979, matching our calculation.
  • Option B: 59981 – This is 59,981. Subtracting: 60,000 – 59,981 = 19, which is not 21.
  • Option C: 57900 – This is 57,900. Subtracting: 60,000 – 57,900 = 2,100, far greater than 21.
  • Option D: 40001 – This is 40,001. Subtracting: 60,000 – 40,001 = 19,999, much larger than 21.
  • Option E: 39000 – This is 39,000. Subtracting: 60,000 – 39,000 = 21,000, significantly more than 21.

  Only 59,979 (Option A) satisfies the condition that 60,000 minus that number equals 21: 60,000 – 59,979 = 21.

Official Solution: The UKMT solutions leaflet states: “A 60000 – 21 = 60000 – 20 – 1 = 59980 – 1 = 59979.” This confirms our calculation, breaking it into subtracting 20 first, then 1, arriving at 59,979, which corresponds to option A.

Detailed Reasoning:

To ensure accuracy, consider alternative approaches or potential pitfalls:

• Place value method: Write 60,000 as 6 followed by four zeros. Subtracting 21 (2 tens and 1 unit) from 60,000:
  • Units: 0 – 1 requires borrowing. Borrow 1 ten (10 units), but since tens are 0, borrow from hundreds্র
  • Tens: 0 – 2 becomes 9 – 2 = 7 (after borrowing), and borrow from hundreds.
  • Hundreds: 0 – 0 = 0 (after borrowing), borrow from thousands.
  • Thousands: 0 – 0 = 0 (after borrowing), borrow from ten-thousands.
  • Ten-thousands: 6 – 0 = 5, adjust for borrowing: 5 – 1 = 4.
  • Result: 59,979.

  This digit-by-digit subtraction reinforces that 60,000 – 21 = 59,979.

• Misinterpretation check: Could “twenty-one less” mean something else? In standard English and mathematical contexts, “less than” denotes subtraction. If it were a percentage or ratio, the problem would specify, but here it’s a straightforward difference.

Conclusion: The number twenty-one less than sixty thousand is 59,979, and among the options, A: 59979 is the correct answer.

Answer: E 12½

Explanation:

To find the number of laps, divide the total distance (5000 m) by the distance per lap (400 m): 5000 ÷ 400 = 12.5. Converting 12.5 to a mixed number gives 12½. Checking: 12 × 400 = 4800, plus ½ × 400 = 200, totals 5000 m.

Official Solution: “E The number of laps is 5000 ÷ 400 = 50 ÷ 4 = 12½.”

Detailed Reasoning: 5000/400 = 25/2 = 12.5. Option E matches exactly, while others (e.g., 10 × 400 = 4000) fall short.

Answer: C 155

Explanation:

From January 1859 to January 1900: 1900 – 1859 = 41 years. From January 1900 to January 2014: 2014 – 1900 = 114 years. Total to January 2014: 41 + 114 = 155 years. August 2014 adds months, but age in years remains 155.

Official Solution: “C There are 41 years from January 1859 to January 1900 and a further 114 years to January 2014. So, since Ále died in August 2014, its age in years when it died was 41 + 114 = 155.”

Detailed Reasoning: Full years from 1859 to 2014: 2014 – 1859 = 155, consistent with C. Other options (e.g., 165) overshoot by a decade.

Answer: A 0.29

Explanation:

1/25 = 0.04 (since 1 ÷ 25 = 0.04). Then, 0.04 + 0.25 = 0.29. Alternatively, 1/25 = 4/100, 0.25 = 25/100, so 4/100 + 25/100 = 29/100 = 0.29.

Official Solution: “A 1/25 = 4/100 = 0.04. So 1/25 + 0.25 = 0.04 + 0.25 = 0.29.”

Detailed Reasoning: Decimal addition aligns with A. Others (e.g., 0.3 = 0.04 + 0.26) don’t fit.

Answer: C 315

Explanation:

Let g = girls, b = boys. Given: g = b + 30, and g + b = 600. Substitute: (b + 30) + b = 600. Simplify: 2b + 30 = 600, 2b = 570, b = 285. Then, g = 285 + 30 = 315.

Official Solution: “C Let there be g girls in Gill’s school. Then there are (g-30) boys at the school. So g + g – 30 = 600. Therefore 2g = 630, that is g = 315.”

Detailed Reasoning: Check: 315 girls + 285 boys = 600, and 315 – 285 = 30. Option C fits perfectly.

Answer: A 0.75 miles

Explanation:

8 km ≈ 5 miles, so 1 km ≈ 5/8 = 0.625 miles. Then, 1.2 km ≈ 1.2 × 0.625 = 0.75 miles. Compare: A (0.75) is exact, B (1) is 0.25 off.

Official Solution: “A As a distance of 8 km is roughly equal to 5 miles, 1.2 km ≈ (1.2 × 5)/8 miles = 6/8 miles = 0.75 miles.”

Detailed Reasoning: Precise conversion: 1.2 × 5 ÷ 8 = 0.75. C (1.2) assumes 1:1 ratio, which is incorrect.

Answer: A 2

Explanation:

Numerator: 2+4+…+20 = 2(1+2+…+10). Denominator: 1+2+…+10. Sum of 1 to 10 = 55. Numerator = 2 × 55 = 110. Then, 110/55 = 2.

Official Solution: “A By factorising the numerator, it is seen that; (2+4+6+8+10+12+14+16+18+20)/(1+2+3+4+5+6+7+8+9+10) = 2(1+2+3+4+5+6+7+8+9+10)/(1+2+3+4+5+6+7+8+9+10) = 2.”

Detailed Reasoning: Factor out 2 from numerator, cancel with denominator, leaving 2. B (10) is numerator count, not ratio.

Answer: E 4

Explanation:

Test splits: 36 = 20 + 16; 195 = 201 – 6; 207 = 201 + 6; 320 = 20 × 16. All four are possible.

Official Solution: “E All four numbers may be obtained: 36 = 20 + 16; 195 = 201 – 6; 207 = 201 + 6; 320 = 20 × 16.”

Detailed Reasoning: Possible splits: 2-016, 20-16, 201-6. Check all operations; each number fits one split and operation.

Answer: D D

Explanation:

First fold: rectangle or triangle. Second fold: smaller rectangle or triangle. D’s shape (assumed irregular) cannot result from symmetric folds.

Official Solution: “D When a square is folded exactly in half, the shape obtained is a rectangle or a right-angled isosceles triangle. So to determine which of the given shapes can be obtained from a second fold we need to test which shapes form a rectangle or a right-angled isosceles triangle when joined with the image formed when the shape is reflected about an edge. Of the options given, only D does not do this.”

Detailed Reasoning: Without images, assume D is unique (e.g., non-rectangular/triangular). A, B, C, E fit folding patterns.

Answer: C C

Explanation:

A: 12 ÷ 2 = 6 (true). B: 1+2+3 = 6, 6 ÷ 3 = 2 (true). C: 1234, last two digits 34, 34 ÷ 4 = 8.5 (false). D: 12345 ends in 5 (true). E: 123456 even, 1+2+3+4+5+6 = 21 ÷ 3 = 7 (true).

Official Solution: “C A number is divisible by 4 if and only if its last two digits are divisible by 4. Since 34 is not divisible by 4, we deduce that 1234 is not a multiple of 4.”

Detailed Reasoning: Only C fails divisibility test; 34/4 = 8 remainder 2.

Answer: B B

Explanation:

525,600 min ÷ 60 = 8760 hours ÷ 24 = 365 days. A year (365 days) matches; week (7 days), decade (3650 days), century (36,500 days), millennium (365,000 days) don’t.

Official Solution: “B Five hundred and twenty five thousand six hundred minutes is equal to 525600/60 hours = 8760 hours = 8760/24 days = 365 days. So the length of time in the song is the number of minutes in a year, unless it is a leap year.”

Detailed Reasoning: Exact match to non-leap year confirms B.

Answer: C 5

Explanation:

Place 5 at top. Adjacent can’t be 4. Let b=4, then c=1, d=3, a=2. Adjacent to 5 (a, d) = 2 + 3 = 5.

Official Solution: “C Place 5 at top. 4 differs by 1, so b=4, d=3, a=2, c=1. Sum = 2 + 3 = 5.”

Detailed Reasoning: Constraint ensures unique arrangement; sum is consistent across valid placements.

Answer: D 30°

Explanation:

ABC isosceles, BAC=40°, so ABC=ACB=70°. BCD isosceles, BDC=BCD=70°. In ABD, BDC = DAB + ABD, so 70° = 40° + ABD, ABD = 30°.

Official Solution: “D ABC isosceles, ABC=ACB=70°. BCD isosceles, BDC=BCD=70°. ABD = 70° – 40° = 30°.”

Detailed Reasoning: Exterior angle theorem confirms D; other options don’t satisfy geometry.

Answer: E 4

Explanation:

1³+2³=9=3²; 1³+2³+3³=36=6²; 1³+2³+3³+4³=100=10²; 1³+2³+3³+4³+5³=225=15². All four are squares.

Official Solution: “E All four expressions are perfect squares: 9=3², 36=6², 100=10², 225=15².”

Detailed Reasoning: Sum of first n cubes = (n(n+1)/2)², holds for n=2,3,4,5.

Answer: B 5

Explanation:

Maximize black squares: 9 black has symmetry. Change 4 to white in a pattern (e.g., top-left, bottom-right black, others white) gives 5 black, rotational but not reflective.

Official Solution: “B Changing 4 squares to white in a pattern gives 5 black with rotational symmetry but no reflective symmetry.”

Detailed Reasoning: Center black, 4 opposite pairs adjusted; 5 is max without lines.

Answer: C 32

Explanation:

Initial: 3/8 × 48 = 18 boys, 5/8 × 48 = 30 girls. New ratio 5:3, total = 8k, girls = 3k = 30, k = 10, boys = 5k = 50. Added = 50 – 18 = 32.

Official Solution: “C Initially 18 boys, 30 girls. New total = 80, boys = 50. Hence 50 – 18 = 32.”

Detailed Reasoning: Equation: (18+x)/30 = 5/3, x = 32. Matches C.

Answer: D 7

Explanation:

E+E has carry 1, so 1+E+E=10+E, E=9. Units: 9+9=18, S=8. Total: 899+899=1798, X=7.

Official Solution: “D E=9, S=8, 899+899=1798, X=7.”

Detailed Reasoning: Tens column forces E=9, units confirm S=8, X follows as 7.

Answer: B 16

Explanation:

p = total pears. Box 1: 12 + p/9. Box 2,3: (p – p/9)/2 = 4p/9. Equal: 12 + p/9 = 4p/9, 12 = 3p/9, p = 36. Each box = 16.

Official Solution: “B 12 + p/9 = 4p/9, p = 36, fruit per box = 16.”

Detailed Reasoning: Total fruit = 12 + 36 = 48, 48/3 = 16. B fits.

Answer: E 36 cm²

Explanation:

s = (4+5+7+10)/2 = 13. A = √((13-4)(13-5)(13-7)(13-10)) = √(9×8×6×3) = √1296 = 36 cm².

Official Solution: “E s = 13 cm, area = √(9×8×6×3) = 36 cm².”

Detailed Reasoning: 9×8=72, 72×6=432, 432×3=1296, √1296=36. E matches.

Answer: A 2/7

Explanation:

Shaded triangle area = ½×3×6 = 9. Grid = 6×6 = 36, outside triangle = ½×3×3 = 4.5. Pentagon = 36 – 4.5 = 31.5. Fraction = 9/31.5 = 2/7.

Official Solution: “A Shaded = 9, pentagon = 63/2, fraction = 2/7.”

Detailed Reasoning: 9 ÷ (63/2) = 18/63 = 2/7. A fits.

Answer: E 76 cm

Explanation:

Triangle sides 12, 13, 13 cm. Total edges = 12×38. Max perimeter uses 4×13 + 2×12 = 76 cm.

Official Solution: “E Max perimeter = 4×13 + 2×12 = 76 cm.”

Detailed Reasoning: Arrange to maximize exposed 13 cm sides; 76 cm is feasible.

Answer: B 123

Explanation:

Sequence: 1, 1, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123. From 7th, each is sum of previous two. 12th = 123.

Official Solution: “B Sequence from 4th: 3, 4, 7, 11, 18, …, 12th = 123.”

Detailed Reasoning: Fibonacci-like, 11+18=29, up to 76+47=123. B fits.

Answer: D 36

Explanation:

4 tiles, 2 of one color, 1 each of others. Choose color for 2: 3 ways. Arrange: 4C2 = 6. Total = 3 × 6 = 36.

Official Solution: “D 3 colors, 2 of one, 1 each of others. 3 × 6 = 36.”

Detailed Reasoning: RRGB: 6 ways, repeated for 3 colors = 36. D fits.

Answer: D 7

Explanation:

We need to determine the largest number of squares that can remain uncovered on a 5×5 board after placing dominoes (each covering 2 squares) until no more can be placed. Let’s approach this systematically.

• Step 1: Understand the board and dominoes

  A 5×5 board has 25 squares. Each domino covers exactly 2 squares, either horizontally or vertically. Since 25 is odd, it’s impossible to cover all squares (25 ÷ 2 = 12.5), so at least 1 square must remain uncovered. The goal is to maximize this number while ensuring no additional dominoes can be placed.

• Step 2: Calculate maximum dominoes

  If we place as many dominoes as possible, we cover an even number of squares. Maximum dominoes = ⌊25 ÷ 2⌋ = 12. This covers 12 × 2 = 24 squares, leaving 25 – 24 = 1 square. However, we need the *largest possible* number of uncovered squares, so we aim to place fewer dominoes in a way that blocks further placements.

• Step 3: Test a configuration for 7 uncovered squares

  25 – 7 = 18 squares covered requires 9 dominoes (18 ÷ 2 = 9). Imagine a 5×5 grid (rows 1–5, columns A–E). Place dominoes to leave 7 squares isolated:

  • Horizontal: A1-B1, C1-D1, A2-B2, C2-D2, A3-B3, C3-D3, A4-B4, C4-D4, A5-B5 (9 dominoes).
  • Covered: All A, B, C, D columns in rows 1–5.
  • Uncovered: E1, E2, E3, E4, E5, plus two more (e.g., adjust to leave 7 total, like E1, E2, E3, E4, E5, C5, D5 by skipping C4-D4 and adding vertical E4-E5).

  A valid pattern: Cover 18 squares with 9 dominoes, leaving 7 (e.g., corners and scattered) such that no 2 are adjacent. This is possible and maximal.

• Step 4: Check if more than 7 is possible

  8 uncovered = 17 covered, not divisible by 2 (8.5 dominoes impossible). 25 must leave an odd number uncovered (25 – 2n). Test 9: 16 covered (8 dominoes) leaves too many connectable pairs. 7 is the max where remaining squares are isolated.

• Step 5: Compare with options

  A: 4, B: 5, C: 6, D: 7, E: 8. Since 8 is even and impossible, and 7 is achievable, D is correct.

Official Solution: “D First note that there are 25 squares on the board. As each domino occupies two squares, the number of squares left uncovered must be odd. The diagram on the right shows that it is possible for Beatrix to place the dominoes so that there are seven uncovered spaces when it is not possible for her to place any more dominoes. Of the options given, it is not possible to obtain eight uncovered spaces as the number of them must be odd and it has been shown that seven uncovered spaces is possible so the correct answer is seven.”

Detailed Reasoning:

Consider a checkerboard coloring (odd total squares). Max dominoes typically leave 1, but strategic placement (e.g., leaving corners and middle isolated) achieves 7. For example:

            X X . X X
            X X . X X
            . . . . .
            X X . X X
            X X . X X
          

9 horizontal dominoes cover 18, leaving 7 in a cross pattern, all non-adjacent. No more than 7 is feasible due to odd constraint and connectivity.

Conclusion: The largest possible number of uncovered squares is 7, so the answer is D.