Odd Man Out and Series Questions

Practice Odd Man Out and Series MCQs with answers and explanations. Page 9 of 34.

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Aptitude
Topic
Odd Man Out and Series
Page
9 / 34
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Questions

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Number series – missing term with “×k then +5” rule: 14, 33, 104, ?, 2110
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Number series – insert the missing term by successive division: 840, ?, 420, 140, 35, 7
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Number series – find the missing term by decreasing differences: 125, 80, 45, ?, 5
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Number series – missing term with doubling increments: 4, 19, 49, ?, 229
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Number series – fill the missing term in the “×2 + 2” sequence: 2, 6, 14, 30, …, 126, ?
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Odd-man-out (faulty term) – Fibonacci-like addition pattern: 7, 9, 16, 25, 41, 68, 107, 173, ? — Identify the term that breaks the rule among the options.
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Odd-man-out – identify the inconsistent term: 16, 4, 2, 1.5, 1.75, 1.875
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Number series – next term via halving decrements: 656, 432, 320, 264, 236, ?
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Number series – insert the missing term (growing differences): 6, 13, 32, ?, 130, 221
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Compound multipliers – next value with rising percentage factors: 50, 60, 75, 97.5, ?, 184.275, 267.19875
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Recursive construction – multiply by n and add n: 1, 2, 6, 21, 88, 445, ? — Find the next term.
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Mixed pattern series – find the missing term: 600, 125, 30, ?, 7.2, 6.44, 6.288
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Number series — alternating multipliers ×2 and ×3: Complete the series: 12, 24, 72, 144, 432, ?, ...
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Number series — differences increase by a constant amount: Complete the series: 39, 52, 78, 117, 169, ?
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Number series — subtract consecutive perfect squares: Complete the series: 41, 40, 36, ?, 11
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Split 124 into four terms in AP with a product condition: Divide 124 into four parts in arithmetic progression such that the product of the 1st and 4th is 128 less than the product of the 2nd and 3rd. Identify the four parts.
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Split 20 into four terms in AP with a product ratio 2:3: Divide 20 into four parts in arithmetic progression such that (1st × 4th) : (2nd × 3rd) = 2 : 3. Identify the four parts.
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Savings in an arithmetic progression over 10 years: A man saves ₹ 1,45,000 in ten years. Each year after the first, he saves ₹ 2,000 more than in the previous year. How much did he save in the first year?
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Common terms count in two APs: How many terms are common to the two sequences: AP1 = 1, 3, 5, ... (120 terms) and AP2 = 3, 6, 9, ... (80 terms)?
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Summation of product-pattern series: Find S(n) = 1·2·4 + 2·3·5 + 3·4·6 + ... up to n terms.
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