Number Series – MBA CET questions with solutions

Approach: Number series with ×2 + 2 rule (differences doubling).

Solution:
10 × 2 + 2 = 22
22 × 2 + 2 = 46
46 × 2 + 2 = 94
94 × 2 + 2 = 190
Check: differences 12, 24, 48 double → next 96; 94 + 96 = 190.

Cetking shortcut: Differences 12, 24, 48 double each time — add 96.

Answer: C

Approach: Number series in a circle — ×2 plus an increasing number.

Solution:
Clockwise from the top-right sector: 4, 9, 20, 43, 90, ?
4 × 2 + 1 = 9
9 × 2 + 2 = 20
20 × 2 + 3 = 43
43 × 2 + 4 = 90
90 × 2 + 5 = 185

Why the other options are wrong:
B: 126 assumes n³ + 1, which fails (2³ + 1 = 9 but 3³ + 1 = 28 ≠ 20).
C, D, E: break the ×2 + n pattern.

Cetking shortcut: Each term is a little more than double the previous — 2 × 90 = 180, so the answer is just above 180: 185.

Answer: A

Approach: Missing number in figure — compare corresponding cells.

Solution:
First circle: 8, 59, 25, 42. Second circle: 12, 63, 29, ?
8 → 12 (+4), 59 → 63 (+4), 25 → 29 (+4).
So 42 + 4 = 46.
Check within circles: 59 − 8 = 51 = 63 − 12, and 42 − 25 = 17 = 46 − 29 ✓.

Cetking shortcut: Every cell of the second circle is +4 over the same cell of the first.

Answer: A

Approach: Missing number in grid — n² + 1 sequence read row-wise.

Solution:
Row-wise: 2, 5, 10, 17, ?, 37, 50, 65, 82.
These are n² + 1: 1²+1, 2²+1, 3²+1, 4²+1, …
Missing term = 5² + 1 = 26.
Check: 6² + 1 = 37 ✓, 9² + 1 = 82 ✓.

Cetking shortcut: Differences 3, 5, 7, … (odd numbers): 17 + 9 = 26, 26 + 11 = 37 ✓.

Answer: B

Approach: Number series: each term appears twice, then 13 is subtracted.

Solution:
53, 53 → 53 − 13 = 40, 40 → 40 − 13 = 27, 27.
27 has already appeared twice, so the next term is 27 − 13 = 14.

Why the other options are wrong:
B (27): 27 is already repeated twice.
C, D, E: do not follow the −13 step.

Cetking shortcut: Repeated-pair series: drop the duplicates (53, 40, 27, ?) and the rule is −13.

Answer: A

Approach: Number series with two alternating operations.

Solution:
4 + 2 = 6; 6 × 2 = 12; 12 + 2 = 14; 14 × 2 = 28; 28 + 2 = 30.
Next operation is ×2: 30 × 2 = 60.

Why the other options are wrong:
A (32): applies +2 twice in a row.
C (62): applies ×2 then +2.
D, E: fit no step of the pattern.

Cetking shortcut: Look at the jumps: +2, ×2 alternate; the last step was +2, so now ×2.

Answer: B

Approach: Alternate-end digit removal.

Solution:
589654237 → drop first digit (5) → 89654237
89654237 → drop last digit (7) → 8965423
8965423 → drop first digit (8) → 965423
965423 → next drop is the last digit (3) → 96542

Why the other options are wrong:
A (58965), C (89654), E (38965): contain digits (5/8) already removed from the front.
B (65423): drops the first digit again; the pattern alternates, so the last digit goes now.

Cetking shortcut: Compare lengths: each term is one digit shorter, so the answer must be 5 digits starting with 9 — only D.

Answer: D

Approach: Difference series with alternating signs.

Solution:
125 → 126 (+1) → 124 (−2) → 127 (+3) → 123 (−4) → next should be 123 + 5 = 128.
The series shows 129, so 129 is the wrong term.

Why the other options are wrong:
A, B, C, E: 126, 124, 123, 125 all fit +1, −2, +3, −4.

Cetking shortcut: Write the differences first: 1, −2, 3, −4, 6 — the 6 breaks the 1-2-3-4-5 run.

Answer: D

Approach: Mark each number O (odd) or E (even), then scan for the pattern.

Solution:
Sequence: 5 1 4 7 3 9 8 5 7 2 6 3 1 5 8 6 3 8 5 2 2 4 3 4 9 6.
Parity: O O E O O O E O O E E O O O E E O E O E E E O E O E.
Odd → odd pairs: (5,1), (7,3), (3,9), (5,7), (3,1), (1,5).
Count = 6.

Why the other options are wrong:
E 5: misses one overlapping pair such as (3,9) after (7,3).
A 8, B 2, C 3: miscount the pairs.

Cetking shortcut: Each run of k odd numbers gives k − 1 pairs: runs 5 1 (1), 7 3 9 (2), 5 7 (1), 3 1 5 (2) → 6.

Answer: D

Approach: Mark each number O (odd) or E (even), then scan for the pattern.

Solution:
Sequence: 5 1 4 7 3 9 8 5 7 2 6 3 1 5 8 6 3 8 5 2 2 4 3 4 9 6.
Pattern needed: odd, even, even.
7 2 6 ✓; 5 8 6 ✓; 5 2 2 ✓. (9 8 5, 3 8 5, 3 4 9, 1 4 7 fail.)
Count = 3.

Why the other options are wrong:
A 1, B 2: miss a match.
D 4, E more than 4: include cases like 3 8 5 or 3 4 9.

Cetking shortcut: Check only evens that start an even pair: 2 6, 8 6, 2 2, 2 4 — then see which has an odd before it.

Answer: C

Approach: Mark each number O (odd) or E (even), then scan for the pattern.

Solution:
Sequence: 5 1 4 7 3 9 8 5 7 2 6 3 1 5 8 6 3 8 5 2 2 4 3 4 9 6.
Pattern needed: even, odd, even.
6 3 8 ✓; 8 5 2 ✓; 4 3 4 ✓; 4 9 6 ✓.
Count = 4.

Why the other options are wrong:
A 1, B 2, C 3: miss a match near the end.
E more than 4: includes cases like 4 7 3 or 8 5 7.

Cetking shortcut: Isolated odd numbers (odd with evens on both sides) are easy to spot on the parity string: E O E.

Answer: D

Approach: Alternating add/subtract with consecutive odd numbers.

Solution:
353 + 1 = 354
354 − 3 = 351
351 + 5 = 356
356 − 7 = 349
Next: 349 + 9 = 358

Cetking shortcut: When terms zig-zag up and down, write the differences first: +1, −3, +5, −7 → the next is +9.

Answer: E

Approach: Differences are consecutive odd numbers.

Solution:
6 + 3 = 9
9 + 5 = 14
14 + 7 = 21
21 + 9 = 30
30 + 11 = 41

Cetking shortcut: Terms are n² + 5 (1+5, 4+5, 9+5, 16+5, 25+5) → 36 + 5 = 41.

Answer: E

Approach: Multiplier series ×0.5, ×1.5, ×2.5, ×3.5, ×4.5 — find the term that breaks it.

Solution:
4 × 0.5 = 2 ✓
2 × 1.5 = 3 (series shows 3.5 ✗)
3 × 2.5 = 7.5 ✓ (the next term fits 3, not 3.5)
7.5 × 3.5 = 26.25 ✓
26.25 × 4.5 = 118.125 ✓
So 3.5 is the wrong term; it should be 3.

Cetking shortcut: Check ratios from the end: 118.125/26.25 = 4.5, 26.25/7.5 = 3.5 — the multiplier drops by 1 each step, so 7.5/2.5 = 3 must be the third term.

Answer: B

Approach: Fibonacci series — each term is the sum of the previous two.

Solution:
1 + 1 = 2, 1 + 2 = 3, 2 + 3 = 5, 3 + 5 = 8, 5 + 8 = 13
Next: 8 + 13 = 21

Answer: A

Approach: Repeating block — count full cycles and the leftover terms.

Solution:
Block = 1, 2, 3, 4 with sum 10 and length 4.
2007 ÷ 4 = 501 full blocks, remainder 3.
Sum of full blocks = 501 × 10 = 5010
Leftover terms = 1 + 2 + 3 = 6
Total = 5010 + 6 = 5016

Cetking shortcut: 2008 terms would be 502 × 10 = 5020; drop the last term (4) → 5016.

Answer: A

Approach: Differences are consecutive odd numbers.

Solution:
5 + 5 = 10
10 + 7 = 17
17 + 9 = 26
26 + 11 = 37
Check: 37 + 13 = 50 ✓

Cetking shortcut: Terms are n² + 1: 4+1, 9+1, 16+1, 25+1, 36+1, 49+1.

Answer: B

Approach: Multiplicative series: next term = previous × 2 + 2.

Solution:
2 × 2 + 2 = 6
6 × 2 + 2 = 14
14 × 2 + 2 = 30
30 × 2 + 2 = 62
62 × 2 + 2 = 126

Cetking shortcut: Differences 4, 8, 16, 32 double each time, so the next difference is 64: 62 + 64 = 126. Also every term = 2ⁿ⁺¹ − 2.

Answer: B

Approach: Multiplicative series: next term = previous × 3 − 1.

Solution:
5 × 3 − 1 = 14
14 × 3 − 1 = 41
41 × 3 − 1 = 122
122 × 3 − 1 = 365

Cetking shortcut: Differences 9, 27, 81 are powers of 3, so the next difference is 243: 122 + 243 = 365.

Answer: B

Approach: Difference table — second differences in AP.

Solution:
Terms: 7, 20, 47, 94, 167
1st differences: 13, 27, 47, 73
2nd differences: 14, 20, 26 (rising by 6)
Next 2nd difference = 26 + 6 = 32
Next 1st difference = 73 + 32 = 105
Next term = 167 + 105 = 272

Answer: A

Approach: Power-based series: term = n³ − n² = n²(n − 1), n = 2, 3, 4…

Solution:
2³ − 2² = 4
3³ − 3² = 18
4³ − 4² = 48
5³ − 5² = 100
6³ − 6² = 216 − 36 = 180
Check: 7³ − 7² = 343 − 49 = 294 ✓

Cetking shortcut: Factor each term: 4 = 2²×1, 18 = 3²×2, 48 = 4²×3, 100 = 5²×4 → 6²×5 = 180.

Answer: C

Approach: Power-based series: term = n³ − n = (n − 1) × n × (n + 1).

Solution:
6 = 1×2×3, 24 = 2×3×4, 60 = 3×4×5
120 = 4×5×6, 210 = 5×6×7
Next = 6×7×8 = 336

Cetking shortcut: Second differences 18, 24, 30 rise by 6 → next 36; differences 18, 36, 60, 90 → next 126; 210 + 126 = 336.

Answer: A

Approach: Centre number = sum of the four outer numbers.

Solution: Figure 1: 11 + 44 + 22 + 33 = 110 ✔.
Figure 2: 16 + 40 + 24 + 32 = 112 ✔.
Figure 3: ? + 12 + 23 + 34 = 114 → ? = 114 − 69 = 45.

Cetking shortcut: When the centre is much larger than the arms and close across figures, test the plain sum first.

Answer: D

Approach: Column rule: (row 1 × row 2) ÷ row 3 = row 4.

Solution: Column 1: 18 × 12 = 216; 216 ÷ 3 = 72 ✔.
Column 3: 32 × 16 = 512; 512 ÷ 4 = 128 ✔.
Column 2: 24 × 14 = 336; 336 ÷ ? = 112 → ? = 3.

Cetking shortcut: The large bottom row usually comes from a product; divide the product by the bottom value to find the link.

Answer: C

Approach: Interleaved series — a +3 series with two odd terms (35, 40) inserted after every two terms.

Solution:
Mark the terms that clearly break the flow: 35 and 40.
Remaining terms: 7, 10, 13, __, 19, 22 → each term is +3.
13 + 3 = 16, and 16 + 3 = 19 confirms it.
Full series: 7, 10, [35], 13, 16, [40], 19, 22 — the inserted terms themselves rise by 5 (35 → 40).

Why the other options are wrong: 18, 23, 31, 28 do not keep the 13 → ? → 19 gap of +3, +3.

Cetking shortcut: When one or two terms are far out of line, strike them out first and read the remaining series.

Answer: A

Approach: Interleaved series — main terms with alternating differences, and a fixed 5 after every two terms.

Solution:
Main terms: 28 → 25 (–3) → 21 (–4) → 18 (–3) → 14 (–4) → next –3 = 11.
Pattern of the series: two main terms, then 5 → 28, 25, 5, 21, 18, 5, 14, 11, 5.

Why the other options are wrong: 10 uses –4 twice in a row; (5, 10) and (11, 8) break the 'two terms then 5' rhythm.

Answer: E

Approach: Differences are consecutive squares.

Solution:
107 – 71 = 36 (6²), 71 – 46 = 25 (5²), 46 – 30 = 16 (4²), 30 – 21 = 9 (3²).
So ? → 21 is 2² = 4 → ? = 17, and 16 → 17 is 1² = 1 ✓.

Answer: D

Approach: Sort the numbers, pick the required digits, then divide.

Solution:
Ascending: 253, 467, 535, 738, 883.
Second lowest = 467 → first digit = 4.
Highest = 883 → second digit = 8.
'4 divides 8' → 8 ÷ 4 = 2.

Answer: E

Approach: The centre number 5 drives the chain around the outer circles: next = previous × 5 + 1.

Solution: Going clockwise from the right:
3 × 5 + 1 = 16
16 × 5 + 1 = 81
81 × 5 + 1 = 406
406 × 5 + 1 = 2031 → the top circle.

Why the other options are wrong: 1625 and 2120 do not fit ×5 + 1 from 406; 1 and 835 break the growth.

Cetking shortcut: Ratio of consecutive outer numbers ≈ centre number → try ×centre ± small constant.

Answer: A

Approach: Multiplicative series with multipliers rising by 1 each step.

Solution:
Multipliers: ×0.5, ×1.5, ×2.5, ×3.5, ×4.5.
4 × 0.5 = 2
2 × 1.5 = 3
3 × 2.5 = 7.5 ✓
7.5 × 3.5 = 26.25 ✓
26.25 × 4.5 = 118.125 ✓

Cetking shortcut: Find a ratio from the known end: 118.125 ÷ 26.25 = 4.5, 26.25 ÷ 7.5 = 3.5 ⇒ multipliers fall by 1 going back ⇒ ?: 7.5 ÷ 2.5 = 3.

Answer: E

Approach: Alternating operations series: ×2, −8.

Solution:
14 × 2 = 28
28 − 8 = 20
20 × 2 = 40
40 − 8 = 32
32 × 2 = 64
64 − 8 = 56

Cetking shortcut: Every second term doubles (14, 20, 32 → 28, 40, 64); the next step is the −8 one.

Answer: B

Approach: Prime-number series: every third prime.

Solution:
Primes in order: 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71.
Series picks every 3rd prime: 11, (13, 17) 19, (23, 29) 31, (37, 41) 43, (47, 53) 59, (61, 67) 71.
So the missing term is 59.

Cetking shortcut: Notice all terms are prime; count how many primes are skipped between 11 and 19 (two) – keep skipping two.

Answer: C

Approach: Factorial series: each term = previous term × next natural number.

Solution:
1 ×2 = 2, 2 ×3 = 6, 6 ×4 = 24, 24 ×5 = 120, 120 ×6 = 720.
(Terms are 1!, 2!, 3!, 4!, 5!, 6!.)

Cetking shortcut: 1, 2, 6, 24, 120 is the factorial sequence – recognise it on sight.

Answer: A

Approach: Number series with symmetric differences.

Solution:
Differences: 99 − 94 = 5, 108 − 99 = 9, 119 − 108 = 11, 130 − 119 = 11.
The gaps mirror: 5, 9, 11, 11, 9, 5.
130 + 9 = 139, and 139 + 5 = 144 ✓

Cetking shortcut: When gaps rise then repeat (11, 11), expect them to fall back in reverse – verify with the last term 144.

Answer: C

Approach: Break the series into repeating blocks that end in 9.

Solution:
Group the terms at every 9:
0 | 9
0 1 | 9
0 1 2 | 9
0 1 2 3 | 9
0 1 2 3 4 | 9
0 1 2 3 4 5 | 9
0 1 2 3 4 5 | ?
Each block counts up from 0 one step further than the previous block before the 9.
The last block must be 0 1 2 3 4 5 6 (then 9 would follow).

Why the other options are wrong:
0 – a block restarts with 0 only after a 9; here 5 is followed by the next count.
9 – the block 0–5 already appeared; this block must go one step further.
4, 5 – would repeat or go backwards within a counting block.

Cetking shortcut: Use 9 as a separator — the length of each block before 9 grows by one.

Answer: B

Approach: Ratio of about 2 between terms → test ×2 + constant.

Solution:
5 × 2 + 3 = 13
13 × 2 + 3 = 29
29 × 2 + 3 = 61
61 × 2 + 3 = 125
125 × 2 + 3 = 253 ✓

Cetking shortcut: Differences are 8, 16, ?, ?, 128 — doubling: 8, 16, 32, 64, 128 → 29 + 32 = 61.

Answer: D

Approach: Multipliers grow as even numbers (2, 4, 6, 8, 10) with a square added each step.

Solution:
Rule: next = previous × 2n + n².
1.5 × 2 + 1² = 4
4 × 4 + 2² = 20
20 × 6 + 3² = 129 (printed as 126)
126 × 8 + 4² = 1008 + 16 = 1024
1024 × 10 + 5² = 10240 + 25 = 10265 ✓

Cetking shortcut: Work backwards from the last term: 10265 − 25 = 10240 = 1024 × 10. Only 1024 among the options gives 10265 with ×10; e.g. 1052 × 10 = 10520 already exceeds it.

Answer: A

Approach: Check successive ratios.

Solution:
89250 ÷ 5 = 17850
17850 ÷ 2.5 = 7140
7140 ÷ 5 = 1428
1428 ÷ 2.5 = 571.2
571.2 ÷ 5 = 114.24

Cetking shortcut: Divisors alternate 5, 2.5 — the next divisor is 5, and 571.2 ÷ 5 = 114.24 (÷10 × 2).

Answer: A

Approach: Differences are cubes of decreasing even numbers.

Solution:
3 + 12³ (1728) = 1731
1731 + 10³ (1000) = 2731
2731 + 8³ (512) = 3243
3243 + 6³ (216) = 3459
3459 + 4³ (64) = 3523 ✓

Cetking shortcut: Read the known differences first: 3459 − 3243 = 216 = 6³ and 3523 − 3459 = 64 = 4³ → cubes of 12, 10, 8, 6, 4; then 1731 + 1000 = 2731.

Answer: E

Approach: Check differences.

Solution:
Known differences: 123 − 102 = 21, 146 − 123 = 23 → differences rise by 2.
Sequence of differences: 15, 17, 19, 21, 23.
66 + 17 = 83; check 83 + 19 = 102 ✓.

Answer: D

Approach: Find one rule that turns the first two numbers of a row into the third.

Solution:
Rule: third = 2 × first − second − 8.
Row 1: 2 × 67 − 25 − 8 = 134 − 33 = 101 ✓
Row 2: 2 × 61 − 17 − 8 = 122 − 25 = 97 ✓
Row 3: 2 × 45 − 19 − 8 = 90 − 27 = 63.

Why the other options are wrong: 71 = 2 × 45 − 19 forgets the −8 that rows 1 and 2 need; 92, 81, 49 fit no rule consistent with both rows.

Cetking shortcut: Compute 2 × first − second for the known rows (109, 105): both exceed the third number by exactly 8 — apply the same to row 3.

Answer: C

Approach: Find the constant ratio between terms.

Solution:
21 × 5 = 105, 105 × 5 = 525, 525 × 5 = 2625, 2625 × 5 = 13125, 13125 × 5 = 65625.
The term given as 2615 should be 2625 → 2615 is wrong.

Cetking shortcut: All correct terms of a ×5 series from 525 onward end in 25; 2615 does not.

Answer: D

Approach: Check the differences between consecutive terms.

Solution:
110 − 20 = 90, 210 − 110 = 100, 320 − 210 = 110.
Differences increase by 10 → next difference 120.
320 + 120 = 440.

Answer: C

Approach: Split into interleaved series of period 3.

Solution:
Positions 1, 4, 7: 12, 22, 32 → +10 each.
Positions 2, 5, 8: 16, 26, 36 → +10 each.
Positions 3, 6, 9: 78, 78, 78 → constant.
Next pair (8th, 9th) = 36, 78.

Cetking shortcut: A repeating number (78) every third term signals a period-3 interleave.

Answer: A

Approach: Interleaved series — the terms run in groups of three, each position following its own rule.

Solution: Group the terms in threes: (4, 7, 26), (10, 13, 20), (16, ?, ?).
1st terms: 4 → 10 → 16 (+6).
2nd terms: 7 → 13 → 19 (+6).
3rd terms: 26 → 20 → 14 (−6).
So the next pair is 19, 14.

Why the other options are wrong: A (14 4), B (14 17), C (18 14) break the +6 rule for the 2nd position; D (19 13) gives −7 instead of −6 for the 3rd position.

Cetking shortcut: When a series jumps up and down, split it into groups of 2 or 3 and check each position separately.

Answer: E

Approach: Differences form a series.

Solution: 7−4 = 3, 12−7 = 5, 19−12 = 7, 28−19 = 9 → differences are consecutive odd numbers.
Next difference = 11 → 28 + 11 = 39.

Cetking shortcut: Terms are n² + 3 (1+3, 4+3, 9+3, 16+3, 25+3) → 36 + 3 = 39.

Answer: C

Approach: Missing number in figure: relate the three outer numbers to the inside number.

Solution: Figure 1: (5 × 6 × 4) ÷ 10 = 120 ÷ 10 = 12 ✓
Figure 2: (6 × 7 × 5) ÷ 10 = 210 ÷ 10 = 21 ✓
Figure 3: (4 × 8 × 10) ÷ 10 = 320 ÷ 10 = 32.

Cetking shortcut: Inside numbers 12 and 21 are small compared with the sides' product (120, 210) – try 'product ÷ 10'.

Answer: B

Approach: Interleaved series — a repeating constant plus a main series with alternating differences.

Solution:
28, 25, 5, 21, 18, 5, 14, ...
The number 5 appears after every two terms.
Remove the 5s: 28, 25, 21, 18, 14 → differences −3, −4, −3, −4.
Next difference is −3: 14 − 3 = 11.
After 14 and 11 comes the pair-breaker 5 again.
So the next two terms are 11, 5.

Why the other options are wrong:
B (10,7), C (11,8): 7 and 8 break the repeating 5.
D (5,10): 5 cannot come before 11; 10 also breaks −3/−4.
E (10,5): 14 − 4 = 10 uses the wrong difference (−3 is due).

Cetking shortcut: When a number repeats at fixed gaps, strike it out first — the rest is usually a simple difference series.

Answer: A

Approach: Two alternating series.

Solution:
Odd positions: 13, 15, 17, 19 → +2 each time.
Even positions: 29, 26, 23 → −3 each time.
The last given term 19 is an odd-position term, so the next term belongs to the even series: 23 − 3 = 20.
Then the odd series continues: 19 + 2 = 21.
Next pair = 20, 21.

Why the other options are wrong:
A (21,23): puts the odd-series term first and repeats 23.
C (20,17): 17 already used; odd series is increasing.
D, E: neither number fits either series in order.

Cetking shortcut: Note which series the last given term belongs to — the very next term comes from the other series.

Answer: B

Approach: Two alternating arithmetic series.

Solution:
Odd positions: 8, 11, __, 17 → +3 each time, so blank = 11 + 3 = 14 (check: 14 + 3 = 17).
Even positions: 43, 41, 39 → −2 each time.
Missing number = 14.

Why the other options are wrong:
A (13), B (15), D (16): do not keep the +3 step from 11 to 17.
E (40): belongs to the pattern of the even series, but the blank is an odd-position term.

Cetking shortcut: The blank sits between 11 and 17 in its own series — just take the midpoint: (11 + 17)/2 = 14.

Answer: C