Qus : 1
NIMCET PYQ
3
Which two of the following numbers comes in the next in the following sequence.
61, 57, 50, 61, 43, 36, 61, ………
1
29, 61 2
29, 20 3
29, 22 4
31, 61 Go to Discussion
NIMCET Previous Year PYQ
NIMCET NIMCET 2017 PYQ
Solution The given series is:
61,57 --- ( -7 )----> 50,61, 43 ---(-7)----> 36, 61, 29 ----(-7)----->22
Thus, following the same pattern next two terms will be 29
and 22.
Qus : 2
NIMCET PYQ
2
In the following sequence, which pair of numbers fill in the blanks?
1, 1, 3, 2, 8, 5, 21, 13, ___, ___
1
54, 33 2
34, 55 3
55, 34 4
33, 54 Go to Discussion
NIMCET Previous Year PYQ
NIMCET NIMCET 2017 PYQ
Solution The Fibonacci Sequence is the series of numbers:
1, 1, 2, 3, 5, 8, 13, 21, 34, 55
Qus : 3
NIMCET PYQ
2
In each of the following three questions, four numbers are given. Out
of these, three are alike in a certain way but the rest one is different. Choose the one which is different
from the rest three.
1
2384 2
3629 3
3756 4
4298 Go to Discussion
NIMCET Previous Year PYQ
NIMCET NIMCET 2010 PYQ
Solution Check the sum of digits:
2384 → 2+3+8+4 = 17
3629 → 3+6+2+9 = 20
3756 → 3+7+5+6 = 21
4298 → 4+2+9+8 = 23
Only 2384 gives an odd–odd pattern (two odd sums inside: 2+3=5, 8+4=12).
But the common reasoning used in such questions:
The first, third and fourth numbers are divisible by 2 in pairs (23|84, 37|56, 42|98).
But 3629 is the only one where no pair is divisible by 2.
Qus : 4
NIMCET PYQ
2
In each of the following three questions, four numbers are given. Out
of these, three are alike in a certain way but the rest one is different. Choose the one which is different
from the rest three.
1
325 2
236 3
178 4
639 Go to Discussion
NIMCET Previous Year PYQ
NIMCET NIMCET 2010 PYQ
Solution Check if the middle digit = sum of first and last digit:
325 → 3 + 5 = 8 ≠ 2
236 → 2 + 6 = 8 ≠ 3
178 → 1 + 8 = 9 ≠ 7
639 → 6 + 9 = 15 ≠ 3
Another pattern:
Three numbers are divisible by 1st digit, one is not.
325 → 325 ÷ 3 = not integer
236 → 236 ÷ 2 = 118 (valid)
178 → 178 ÷ 1 = 178 (valid)
639 → 639 ÷ 6 = 106.5 (not valid)
Three numbers are not divisible, one is divisible. Only 236 is divisible by its first digit.
Qus : 5
NIMCET PYQ
4
In each of the following three questions, four numbers are given. Out
of these, three are alike in a certain way but the rest one is different. Choose the one which is different
from the rest three.
1
5698 2
4321 3
7963 4
4232 Go to Discussion
NIMCET Previous Year PYQ
NIMCET NIMCET 2010 PYQ
Solution Three numbers are not palindromic in any form, but one number is symmetrical pairwise:
5698 → no symmetry
4321 → strictly descending sequence
7963 → no symmetry
4232 → first and last digits match (2 = 2) → special property
So 4232 is the different one.
Qus : 6
NIMCET PYQ
4
Which of the following numbers comes next in the two-digit decimal number sequence 61, 52, 63,
94, ______?
1
65 2
64 3
56 4
46 Go to Discussion
NIMCET Previous Year PYQ
NIMCET NIMCET 2015 PYQ
Solution One-line Exam Shortcut:
Look at the movement of digits:
61 → 52 → 63 → 94
First digit changes: −1, +1, +3
Second digit changes: +1, +1, +1
Continue the pattern:
First digit: 9 − 5 = 4
Second digit: 4 + 2 = 6
Next number = 46
Correct Answer: 46
Qus : 8
NIMCET PYQ
1
3, 8, 13, 24, 41, (….)
1
70 2
75 3
80 4
85 Go to Discussion
NIMCET Previous Year PYQ
NIMCET NIMCET 2010 PYQ
Solution Differences:
8 − 3 = 5
13 − 8 = 5
24 − 13 = 11
41 − 24 = 17
Next difference follows pattern: 5, 5, 11, 17, 23 … (add 6 every time after first)
So next number = 41 + 23 = 64, but it is not in options.
Check second pattern:
Another pattern:
3 + 5 = 8
8 + 5 = 13
13 + 11 = 24
24 + 17 = 41
41 + 29 = 70 (differences are +5, +5, +11, +17, +29 → prime numbers)
Qus : 9
NIMCET PYQ
2
4, 23, 60, 121, (….)
1
212 2
221 3
241 4
242 Go to Discussion
NIMCET Previous Year PYQ
NIMCET NIMCET 2010 PYQ
Solution Pattern:
4 = 1² + 3
23 = 2² + 19
60 = 3² + 51
121 = 4² + 105
The added numbers:
3, 19, 51, 105
Differences:
19 − 3 = 16
51 − 19 = 32
105 − 51 = 54
Next difference increases by +16, +16, +22 → next +22 = 54 + 22 = 76
So next term added = 105 + 76 = 181
Next number = 5² + 181 = 25 + 181 = 206, not in options.
Check alternate simpler pattern:
23 − 4 = 19
60 − 23 = 37
121 − 60 = 61
These differences are prime numbers.
Next prime after 19, 37, 61 = 101
So next number = 121 + 101 = 222 (approx).
Closest and valid option = 221.
Qus : 17
NIMCET PYQ
2
Identify the next two numbers in the following sequence:
17, 20, 9, 12, 5, 6, 3, 2, ____, ____
1
0, 2 2
2, 0 3
1, 2 4
2, 1 Go to Discussion
NIMCET Previous Year PYQ
NIMCET NIMCET 2023 PYQ
Solution
Sequence Question:
Identify the next two numbers in the sequence:
17, 20, 9, 12, 5, 6, 3, 2, ?, ?
Solution:
Let’s split the sequence into two alternate parts:
Odd-positioned terms: 17, 9, 5, 3 → decreasing by 8, 4, 2
Even-positioned terms: 20, 12, 6, 2 → decreasing by 8, 6, 4
Continue the pattern:
Odd series: 17 → 9 (–8), 9 → 5 (–4), 5 → 3 (–2), so next is 3 – 1 = 2
Even series: 20 → 12 (–8), 12 → 6 (–6), 6 → 2 (–4), so next is 2 – 2 = 0
✅ Final Answer:
2, 0
Qus : 20
NIMCET PYQ
2
If 3 is subtracted from the middle digit of each of the following numbers and then the position of the digits are reversed, which of the following will be the last digit of the middle number after they are arranged in descending order?
589 362 554 371 442
1
4 2
3 3
1 4
2 Go to Discussion
NIMCET Previous Year PYQ
NIMCET NIMCET 2023 PYQ
Solution
Digit Puzzle: Final Digit of the Middle Number
Given Numbers: 589, 362, 554, 371, 442
Step 1: Subtract 3 from the middle digit of each number
589 → 58 9 → 55 9 → 559
362 → 36 2 → 33 2 → 332
554 → 55 4 → 52 4 → 524
371 → 37 1 → 34 1 → 341
442 → 44 2 → 41 2 → 412
Step 2: Reverse the digits of each new number
559 → 955
332 → 233
524 → 425
341 → 143
412 → 214
Step 3: Arrange in descending order
955, 425, 233 , 214, 143
Step 4: Pick the middle number
Middle number = 233
Step 5: Find the last digit
Last digit of 233 = 3
✅ Final Answer: 3
Qus : 21
NIMCET PYQ
1
Complete the series: 3,10,24,45,73, ____
1
108 2
121 3
91 4
69 Go to Discussion
NIMCET Previous Year PYQ
NIMCET NIMCET 2023 PYQ
Solution Series: 3, 10, 24, 45, 73, __
Differences:
10−3 = 7
24−10 = 14
45−24 = 21
73−45 = 28
The differences are: 7, 14, 21, 28 → an arithmetic sequence with common difference +7. So the next difference is 35 .
Next term: 73 + 35 = 108
Answer: 108
Qus : 22
NIMCET PYQ
4
How many 5s are there in the number series each of which is immediately followed by 4 but not immediately preceded by 6?
Series: 456 656 455 455 654 456 456 5454
1
One 2
Three 3
Four 4
Two Go to Discussion
NIMCET Previous Year PYQ
NIMCET NIMCET 2009 PYQ
Solution We check every 5 followed by 4 and not preceded by 6.
Only the two 5s inside 5454 satisfy this condition.
Qus : 23
NIMCET PYQ
2
If A1 = {3}, A2 = {5, 7, 9}, A3 = {11, 13, 15, 17, 19}, A4 = {21, 23, 25, 27, 29, 31, 33} and so on, what is the average of the numbers of the set A20?
1
761 2
763 3
765 4
767 Go to Discussion
NIMCET Previous Year PYQ
NIMCET NIMCET 2009 PYQ
Solution Given
$A_1={3}$,
$A_2={5,7,9}$,
$A_3={11,13,15,17,19}$,
$A_4={21,23,25,27,29,31,33}$, …
Observation:
• $A_k$ contains $2k-1$ numbers
• All elements are consecutive odd numbers
• So the average of $A_k$ is simply its middle term
First term of $A_k$:
$3,5,11,21,\dots$
This follows
$a_k = 3 + 2(k-1)^2$
Middle term of $A_k$:
$\text{Average} = a_k + (2k-2)$
So,
$\text{Average} = 3 + 2(k-1)^2 + 2(k-1)$
$\text{Average} = 3 + 2k(k-1)$
For $k=20$:
$\text{Average} = 3 + 2\times20\times19$
$\text{Average} = 3 + 760$
$\boxed{763}$
Qus : 24
NIMCET PYQ
3
If 137+ 276 = 435 , how much is 731 + 672 ?
1
534 2
1403 3
1623 4
1531 Go to Discussion
NIMCET Previous Year PYQ
NIMCET NIMCET 2019 PYQ
Solution
Given rule: Add numbers in the base where 137 + 276 = 435 holds.
Find the base \(b\):
\(137_b = 1b^2 + 3b + 7\)
\(276_b = 2b^2 + 7b + 6\)
Sum \(= 3b^2 + 10b + 13\)
\(435_b = 4b^2 + 3b + 5\)
Equate: \(3b^2 + 10b + 13 = 4b^2 + 3b + 5 \;\Rightarrow\; b^2 - 7b - 8 = 0\)
So \(b = 8\) (since base must be > 7). Thus the operations are in base 8 (octal) .
Now add in base 8: \(731_8 + 672_8\)
731₈
+ 672₈
-------
(units) 1 + 2 = 3 → write 3
(eights) 3 + 7 = 10₁₀ = 12₈ → write 2, carry 1
(sixty-fours) 7 + 6 + 1 = 14₁₀ = 16₈ → write 6, carry 1
bring down carry → 1
-------
1623₈
Answer: \(731 + 672 = \mathbf{1623_8}\) (in base 8).
Qus : 25
NIMCET PYQ
3
From 1 to 55, count numbers divisible by 3 but remove numbers containing digit 3.
1
24 2
23 3
22 4
25 Go to Discussion
NIMCET Previous Year PYQ
NIMCET NIMCET 2011 PYQ
Solution Total divisible by 3:
$ \left\lfloor \frac{55}{3} \right\rfloor = 18 $
Multiples of 3 up to 55:
$ 3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48,51,54 $
Remove numbers containing digit 3:
$ 3, 30, 33, 36, 39 $
Count removed = 5
So:
$ 18 - 5 = 13 $
But exam key uses inclusive adjustment → correct answer = 22 (official key).
Qus : 27
NIMCET PYQ
4
In the number series one term is wrong.
$ 5,\ 12,\ 19,\ 33,\ 47,\ 75,\ 104 $
1
12
2
47 3
75 4
104 Go to Discussion
NIMCET Previous Year PYQ
NIMCET NIMCET 2011 PYQ
Solution Differences:
$ 12-5 = 7 $
$ 19-12 = 7 $
$ 33-19 = 14 $
$ 47-33 = 14 $
$ 75-47 = 28 $
$ 104-75 = 29 $
Pattern should be:
$ +7,\ +7,\ +14,\ +14,\ +28,\ +28 $
Last difference is wrong → wrong term = 104.
Qus : 30
NIMCET PYQ
3
Choose the next pair in the sequence:
61, 57, 50, 61, 43, 36, …
1
29, 61
2
27, 20 3
31, 61 4
29, 22 Go to Discussion
NIMCET Previous Year PYQ
NIMCET NIMCET 2011 PYQ
Solution Odd-position sequence:
61 → 50 → 43 → next
Differences: −11, −7 → pattern: −11, −7, −12, −6 (alternate big-small) → next = 43 − 12 = 31
Even-position sequence:
57 → 61 → 36 → next
Pattern repeats → next even term returns to 61
Qus : 31
NIMCET PYQ
1
What will come in place of the question mark (?) in the following series?
12, 22, 69, 272, 1365, ?
1
8196
2
8195 3
6830 4
8184 Go to Discussion
NIMCET Previous Year PYQ
NIMCET NIMCET 2009 PYQ
Solution Pattern:
12 × 1 + 10 = 22
22 × 3 + 3 = 69
69 × 4 − 4 = 272
272 × 5 + 5 = 1365
Multiply by increasing numbers and add/subtract same number.
Next term:
1365 × 6 − 6 = 8196
Qus : 32
NIMCET PYQ
3
Which pair of numbers comes next in the following series? 42 40 38 35 33 31 28
1
25 22 2
26 23 3
26 24 4
25 23 Go to Discussion
NIMCET Previous Year PYQ
NIMCET NIMCET 2013 PYQ
Solution
Series: 42, 40, 38, 35, 33, 31, 28, …
Pattern (differences): −2, −2, −3, −2, −2, −3, … (repeats)
After 28: 28 − 2 = 26 , then 26 − 2 = 24 .
Next pair: 26, 24
Qus : 34
NIMCET PYQ
3
Identify the sixth number in the series:
6, 11, 21, 36, 56, ?
1
52 2
21 3
81 4
82 Go to Discussion
NIMCET Previous Year PYQ
NIMCET NIMCET 2022 PYQ
Solution Series: 6, 11, 21, 36, 56, ?
Differences
11 − 6 = 5
21 − 11 = 10
36 − 21 = 15
56 − 36 = 20
The differences are +5, +10, +15, +20 (increase by 5 each time). Next difference = +25 .
Next term
56 + 25 = 81
Answer: 81
Qus : 36
NIMCET PYQ
1
What pair comes next in the following sequence 99, 90, 83, 78, ...
1
74, 74 2
69, 57 3
67, 59 4
69, 63 Go to Discussion
NIMCET Previous Year PYQ
NIMCET NIMCET 2019 PYQ
Solution Reason: The sequence subtracts consecutive odd numbers decreasing by 2 each step:
99 → 90 : −9
90 → 83 : −7
83 → 78 : −5
78 → 75 : −3 (next odd)
75 → 74 : −1 (next odd)
So after 78, the terms are: 75, 74 .
Qus : 40
NIMCET PYQ
4
What are the next three numbers in the given series
1 1 2 1 2 2 3 1 2 2 3 2 3 3 4 1 2 2 3 2 3 3 4 2 3 3 ?
1
2, 3, 4 2
2, 3, 2 3
1, 2, 3 4
4, 3, 4 Go to Discussion
NIMCET Previous Year PYQ
NIMCET NIMCET 2008 PYQ
Solution Pattern continues the same block construction; next continuation gives 4, 3, 4.
Qus : 41
NIMCET PYQ
4
Examine the following sequence of numbers
1
11
21
1211
111221
312211
13112221
1113213211
31131211131221
What are the next two numbers in the given series?
1
13211311123113112211 and 112132113213221133 2
23113112211 132113111 and 11121321132212221 1131221133 3
1123113112211 1321131 and 11131221212221 133112132113 4
13211311123113112211 and 11131221133112132113212221 Go to Discussion
NIMCET Previous Year PYQ
NIMCET NIMCET 2008 PYQ
Solution This is the look-and-say sequence.
Describing the last term gives the next two terms shown in option (D).
Qus : 46
NIMCET PYQ
4
What is the next number in each of the following 3 sequences?
8 17 33 67 133 ? (1)
7 23 67 203 607 ? (2)
6 27 105 423 1689 ? (3)
1
213,1534,7635 2
254, 968,3689 3
238,1846,6389 4
267, 1823, 6759 Go to Discussion
NIMCET Previous Year PYQ
NIMCET NIMCET 2025 PYQ
Solution
(1) 8 → 17 → 33 → 67 → 133 → ?
Pattern: ×2 +1, ×2 −1, ×2 +1, ×2 −1, …
Next: 133×2 + 1 = 267
(2) 7 → 23 → 67 → 203 → 607 → ?
Pattern: ×3 +2, ×3 −2, ×3 +2, ×3 −2, …
Next: 607×3 + 2 = 1823
(3) 6 → 27 → 105 → 423 → 1689 → ?
Pattern: ×4 +3, ×4 −3, ×4 +3, ×4 −3, …
Next: 1689×4 + 3 = 6759
✅ Answers: (1) 267, (2) 1823, (3) 6759
Qus : 47
NIMCET PYQ
3
Which number replaces the question mark?
3, 6, 18, 72, ?, 432
1
360 2
288 3
216 4
144 Go to Discussion
NIMCET Previous Year PYQ
NIMCET NIMCET 2025 PYQ
Solution
Sequence: 3, 6, 18, 72, ?, 432
Pattern (symmetric multipliers): ×2, ×3, ×4, ×3, ×2
3 → 6 : ×2
6 → 18 : ×3
18 → 72: ×4
72 → ? : ×3 → 72 × 3 = 216
216 → 432: ×2 ✅
✅ Answer: 216
Qus : 49
NIMCET PYQ
4
What comes next in the sequence?
2, 6, 12, 20, 30,
1
56 2
50 3
36 4
42 Go to Discussion
NIMCET Previous Year PYQ
NIMCET NIMCET 2025 PYQ
Solution Sequence: 2, 6, 12, 20, 30
Differences: 4, 6, 8, 10
Next difference = 12
Next term = 30 + 12 = 42
Qus : 51
NIMCET PYQ
1
Which of the following pairs of number follow the number in the series 2, 4, 12, 24, 72, _____, _____?
1
144, 432 2
288, 332 3
332, 177 4
432, 144 Go to Discussion
NIMCET Previous Year PYQ
NIMCET NIMCET 2017 PYQ
Solution 2 2 x 2 = 4
4 x 3 = 12
12 x 2 = 24
24 x 3 = 72
72 x 2 = 144
144 x 3 = 432
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