Select the set in which the numbers are related in the same way as are the…
2025
Select the set in which the numbers are related in the same way as are the numbers of the following sets.
(Note: Operations should be performed on the whole numbers, without breaking down the numbers into their constituent digits. E.g., 13 — Operations on 13 such as adding/subtracting/multiplying to 13 can be performed. Breaking down 13 into 1 and 3 and then performing mathematical operations on 1 and 3 is not allowed.)
(35, 116, 3)
(49, 164, 5)
Answer: A. (39, 182, 21) — Concept: In a number-triplet analogy, every given set (a, b, c) follows one and the same fixed whole-number rule that links its three entries. The method is…
- A.
(39, 182, 21)
- B.
(23, 99, 11)
- C.
(32, 168, 25)
- D.
(11, 42, 4)
Attempted by 21 students.
Show answer & explanation
Correct answer: A
Concept: In a number-triplet analogy, every given set (a, b, c) follows one and the same fixed whole-number rule that links its three entries. The method is always the same — treat the outer numbers a and c as the inputs, find the single arithmetic rule that produces the middle number b for every given set, and then apply that identical rule to each answer set. The bracketed note narrows the search before any trial begins: the numbers must be used whole, so digit-splitting patterns are not admissible here.
Application — deriving the rule from the two given sets:
Write the two given sets in the form (a, b, c): (35, 116, 3) and (49, 164, 5). In each of them a and c are the outer numbers and b is the middle number that the rule must produce.
Judge the sizes first. 116 is a little more than three times 35, and 164 is a little more than three times 49, so the rule is built on multiplication by a small factor rather than on addition alone.
The third entry is small in both sets (3 and 5), which suggests it joins the first entry instead of acting on its own. Form the combined input a + c: 35 + 3 = 38 and 49 + 5 = 54.
Test the factor 3 on that combined input: 3 × 38 = 114, which is 2 short of 116, and 3 × 54 = 162, which is 2 short of 164. The shortfall is the same in both sets, and that repetition is the signal that one single rule covers both.
Instead of stopping at that observation, set it up as an equation and solve it. Let the rule be b = k × (a + c) + m. Then k × 38 + m = 116 and k × 54 + m = 164.
Subtract the first equation from the second so that m cancels: k × (54 − 38) = 164 − 116, that is 16k = 48, so k = 3.
Substitute k = 3 back into k × 38 + m = 116: 114 + m = 116, so m = 2.
The rule obeyed by both given sets is therefore b = 3 × (a + c) + 2.
Cross-check the derived rule on both given sets before using it:
(35, 116, 3): 3 × (35 + 3) + 2 = 3 × 38 + 2 = 114 + 2 = 116, exactly the middle number given.
(49, 164, 5): 3 × (49 + 5) + 2 = 3 × 54 + 2 = 162 + 2 = 164, exactly the middle number given.
Now apply the same rule to each answer set, using only that set's own outer numbers:
Set (a, b, c) | a + c | 3 × (a + c) + 2 | Middle number in the set | Rule satisfied? |
|---|---|---|---|---|
(39, 182, 21) | 60 | 182 | 182 | Yes |
(23, 99, 11) | 34 | 104 | 99 | No — falls 5 short |
(32, 168, 25) | 57 | 173 | 168 | No — falls 5 short |
(11, 42, 4) | 15 | 47 | 42 | No — falls 5 short |
Why the near misses fail: (23, 99, 11), (32, 168, 25) and (11, 42, 4) each land exactly 5 below the value their own outer numbers demand, so none of them repeats the relationship shown by the examples. Two common slips are worth naming — computing 3 × a + c instead of 3 × (a + c), which gives 3 × 35 + 3 = 108 and not 116 on the first given set, and splitting a number into its digits, which the bracketed note expressly disallows.
Hence the set whose numbers are related in the same way as the given sets is (39, 182, 21), because 3 × (39 + 21) + 2 = 3 × 60 + 2 = 182 reproduces its own middle number exactly.