NMMSE 2024 MAT Questions 21–40
Detailed Solutions – Letter Series and Missing Number Problems
NMMSE 2024 Mental Ability Test:
This article contains detailed solutions to Questions 21 to 40
from the NMMSE 2024 Mental Ability Test. Questions 21–30 are based
on letter-series completion, while Questions 31–40 involve
missing numbers and numerical patterns in figures.
Questions 21–30: Letter Series
Question 21
ba__ba__bac__acb__cbac
Solution:
We have to insert four letters. If we insert C C B A, the series becomes:
BAC / BAC / BAC / BAC / BAC
Thus, the four missing letters are: C, C, B, A.
We have to insert four letters. If we insert C C B A, the series becomes:
BAC / BAC / BAC / BAC / BAC
Thus, the four missing letters are: C, C, B, A.
Answer: (C) CCBA
Question 22
__ccbc__accbcc__c__b
Solution:
Try the missing letters A C A C. The completed series can be grouped into the same repeating pattern:
ACCB / ACCB / ACCB / ACCB
Therefore, the missing letters are A, C, A and C.
Try the missing letters A C A C. The completed series can be grouped into the same repeating pattern:
ACCB / ACCB / ACCB / ACCB
Therefore, the missing letters are A, C, A and C.
Answer: (A) ACAC
Question 23
__aaba__bba__bba__abaa__b
Solution:
Insert A A B A B into the blanks. The resulting series can be grouped as:
AAAB / AABB / ABBB / AAAB / AABB
Notice that the number of consecutive B's increases in the first three groups and then the pattern repeats. Therefore, the missing letters are AABAB.
Insert A A B A B into the blanks. The resulting series can be grouped as:
AAAB / AABB / ABBB / AAAB / AABB
Notice that the number of consecutive B's increases in the first three groups and then the pattern repeats. Therefore, the missing letters are AABAB.
Answer: (A) AABAB
Question 24
__bcc__ac__aabb__ab__cc
Solution:
Put B A C A B in the five blanks. The complete series becomes:
BBCCAA / CCAABB / AABBCC
Each block follows the same cyclic arrangement of pairs of B, C and A. Hence, the missing letters are BACAB.
Put B A C A B in the five blanks. The complete series becomes:
BBCCAA / CCAABB / AABBCC
Each block follows the same cyclic arrangement of pairs of B, C and A. Hence, the missing letters are BACAB.
Answer: (C) BACAB
Question 25
__nmmn__mmnn__mnnm__
Solution:
Insert N N M M. The completed series becomes:
NNMM / NNMM / NNMM / NNMM
Thus, the same four-letter pattern NNMM is repeated throughout the series.
Insert N N M M. The completed series becomes:
NNMM / NNMM / NNMM / NNMM
Thus, the same four-letter pattern NNMM is repeated throughout the series.
Answer: (C) NNMM
Question 26
__op__mo__n____pnmop__
Solution:
The missing letters are: M N P M O N After inserting them, the complete sequence becomes:
MOPN / MOPN / MOPN / MOPN
Thus, the four-letter group MOPN is repeated.
The missing letters are: M N P M O N After inserting them, the complete sequence becomes:
MOPN / MOPN / MOPN / MOPN
Thus, the four-letter group MOPN is repeated.
Answer: (A) MNPMON
Question 27
A, D, H, M, ?
Solution:
Convert the letters into their alphabetical positions:
A = 1
D = 4
H = 8
M = 13
Differences are: +3, +4, +5
Therefore, the next difference should be +6.
13 + 6 = 19
The 19th letter is S.
Convert the letters into their alphabetical positions:
A = 1
D = 4
H = 8
M = 13
Differences are: +3, +4, +5
Therefore, the next difference should be +6.
13 + 6 = 19
The 19th letter is S.
Answer: (B) S
Question 28
KZ, LX, MV, NQ, ?
Solution:
Consider the two letters separately.
First letters:
K → L → M → N → O
So the first letter is O.
Second letters:
Z → X → V
These move backward by 2: Z, X, V Then the pattern continues in a larger backward step: V → Q → L
Therefore, the missing pair is OL.
Consider the two letters separately.
First letters:
K → L → M → N → O
So the first letter is O.
Second letters:
Z → X → V
These move backward by 2: Z, X, V Then the pattern continues in a larger backward step: V → Q → L
Therefore, the missing pair is OL.
Answer: (A) OL
Question 29
b__b__bb____bbb__bb__b
Solution:
Insert A B A B A B. The complete sequence becomes:
BABBBB / BABBBB / BABBBB
Therefore, the six missing letters are ABABAB.
Insert A B A B A B. The complete sequence becomes:
BABBBB / BABBBB / BABBBB
Therefore, the six missing letters are ABABAB.
Answer: (C) ABABAB
Question 30
aaa__bb__aab__baaa__bb
Solution:
Put B A B B in the blanks. The complete sequence becomes:
AAABBB / AAABBB / AAABBB
Hence, the repeated pattern is AAABBB.
Put B A B B in the blanks. The complete sequence becomes:
AAABBB / AAABBB / AAABBB
Hence, the repeated pattern is AAABBB.
Answer: (C) BABB
Questions 31–40: Missing Numbers and Figure Patterns
Question 31
Triangle 1: 841, 784, 729 → Centre = 84
Triangle 2: 225, 196, 169 → Centre = ?
Triangle 2: 225, 196, 169 → Centre = ?
Solution:
First identify the square roots in the first triangle:
√841 = 29
√784 = 28
√729 = 27
Their sum is:
29 + 28 + 27 = 84
Therefore, apply the same rule to the second triangle:
√225 = 15
√196 = 14
√169 = 13
So:
15 + 14 + 13 = 42
First identify the square roots in the first triangle:
√841 = 29
√784 = 28
√729 = 27
Their sum is:
29 + 28 + 27 = 84
Therefore, apply the same rule to the second triangle:
√225 = 15
√196 = 14
√169 = 13
So:
15 + 14 + 13 = 42
Answer: (B) 42
Question 32
Cross figures:
Figure 1 → Top 5, Left 6, Right 15, Bottom 3, Centre 93
Figure 2 → Top 7, Left 9, Right 5, Bottom 6, Centre ?
Figure 3 → Top 18, Left 4, Right 1, Bottom 8, Centre 50
Figure 1 → Top 5, Left 6, Right 15, Bottom 3, Centre 93
Figure 2 → Top 7, Left 9, Right 5, Bottom 6, Centre ?
Figure 3 → Top 18, Left 4, Right 1, Bottom 8, Centre 50
Solution:
The first and third figures give a consistent relationship:
Centre = (Top × Right) + (Left × Bottom)
Check Figure 1:
(5 × 15) + (6 × 3)
= 75 + 18
= 93 ✓
Check Figure 3:
(18 × 1) + (4 × 8)
= 18 + 32
= 50 ✓
Therefore, for Figure 2:
(7 × 5) + (9 × 6)
= 35 + 54
= 89
The first and third figures give a consistent relationship:
Centre = (Top × Right) + (Left × Bottom)
Check Figure 1:
(5 × 15) + (6 × 3)
= 75 + 18
= 93 ✓
Check Figure 3:
(18 × 1) + (4 × 8)
= 18 + 32
= 50 ✓
Therefore, for Figure 2:
(7 × 5) + (9 × 6)
= 35 + 54
= 89
Important:
The mathematically consistent answer is 89, but 89 is not included among the four printed alternatives in the uploaded PDF. Therefore, Question 32 appears to contain a printing/option error. It would be incorrect to manufacture a rule merely to obtain 27, 19, 84 or 5.
The mathematically consistent answer is 89, but 89 is not included among the four printed alternatives in the uploaded PDF. Therefore, Question 32 appears to contain a printing/option error. It would be incorrect to manufacture a rule merely to obtain 27, 19, 84 or 5.
Question 33
9 A 12
B 10 7
8 C 11
B 10 7
8 C 11
Solution:
Each row and each column should have the same total. Using option D:
First row:
9 + 9 + 12 = 30
Second row:
13 + 10 + 7 = 30
Third row:
8 + 11 + 11 = 30
The columns also total 30:
9 + 13 + 8 = 30
9 + 10 + 11 = 30
12 + 7 + 11 = 30
Each row and each column should have the same total. Using option D:
First row:
9 + 9 + 12 = 30
Second row:
13 + 10 + 7 = 30
Third row:
8 + 11 + 11 = 30
The columns also total 30:
9 + 13 + 8 = 30
9 + 10 + 11 = 30
12 + 7 + 11 = 30
Answer: (D) A = 9, B = 13, C = 11
Question 34
Clockwise outer numbers:
? , 3, 16, 81, 406
Centre = 5
Centre = 5
Solution:
Observe the clockwise sequence:
3 → 16 → 81 → 406
Each term is obtained by multiplying by 5 and adding 1:
3 × 5 + 1 = 16
16 × 5 + 1 = 81
81 × 5 + 1 = 406
Therefore, the next number is:
406 × 5 + 1
= 2030 + 1
= 2031
Observe the clockwise sequence:
3 → 16 → 81 → 406
Each term is obtained by multiplying by 5 and adding 1:
3 × 5 + 1 = 16
16 × 5 + 1 = 81
81 × 5 + 1 = 406
Therefore, the next number is:
406 × 5 + 1
= 2030 + 1
= 2031
Answer: (A) 2031
Question 35
72 above 9 and 16
91 above 13 and 14
72 above ? and 12
91 above 13 and 14
72 above ? and 12
Solution:
In the first figure:
9 × 16 ÷ 2 = 144 ÷ 2 = 72
In the second figure:
13 × 14 ÷ 2 = 182 ÷ 2 = 91
Therefore, for the third figure:
? × 12 ÷ 2 = 72
? × 6 = 72
? = 12
In the first figure:
9 × 16 ÷ 2 = 144 ÷ 2 = 72
In the second figure:
13 × 14 ÷ 2 = 182 ÷ 2 = 91
Therefore, for the third figure:
? × 12 ÷ 2 = 72
? × 6 = 72
? = 12
Answer: (B) 12
Question 36
Bottom row: 8, 2, 1, 4
Next row: 7, 3, 4
Next row: 6, 7
Top: ?
Next row: 7, 3, 4
Next row: 6, 7
Top: ?
Solution:
Look at the total of each horizontal level.
Bottom row:
8 + 2 + 1 + 4 = 15
Third row:
7 + 3 + 4 = 14
Second row:
6 + 7 = 13
The row totals decrease by 1 each time:
15, 14, 13, 12
Look at the total of each horizontal level.
Bottom row:
8 + 2 + 1 + 4 = 15
Third row:
7 + 3 + 4 = 14
Second row:
6 + 7 = 13
The row totals decrease by 1 each time:
15, 14, 13, 12
Answer: (B) 12
Question 37
Circle 1: 64, 5, 40
Circle 2: 81, 7, 63
Circle 3: ?, 4, 16
Circle 2: 81, 7, 63
Circle 3: ?, 4, 16
Solution:
In the first circle:
√64 × 5 = 8 × 5 = 40
In the second circle:
√81 × 7 = 9 × 7 = 63
Therefore, for the third circle:
√? × 4 = 16
√? = 4
? = 4² = 16
In the first circle:
√64 × 5 = 8 × 5 = 40
In the second circle:
√81 × 7 = 9 × 7 = 63
Therefore, for the third circle:
√? × 4 = 16
√? = 4
? = 4² = 16
Answer: (A) 16
Question 38
18 7 16
8 3 10
10 4 ?
8 3 10
10 4 ?
Solution:
Compare the first two rows with the third row.
First column:
18 − 8 = 10
Second column:
7 − 3 = 4
Therefore, apply the same rule to the third column:
16 − 10 = 6
Compare the first two rows with the third row.
First column:
18 − 8 = 10
Second column:
7 − 3 = 4
Therefore, apply the same rule to the third column:
16 − 10 = 6
Answer: (B) 6
Question 39
3 6 8
5 8 4
4 7 ?
5 8 4
4 7 ?
Solution:
The sum of the first row is:
3 + 6 + 8 = 17
The sum of the second row is:
5 + 8 + 4 = 17
Therefore, the third row should also total 17.
4 + 7 + ? = 17
11 + ? = 17
? = 6
The sum of the first row is:
3 + 6 + 8 = 17
The sum of the second row is:
5 + 8 + 4 = 17
Therefore, the third row should also total 17.
4 + 7 + ? = 17
11 + ? = 17
? = 6
Answer: (D) 6
Question 40
Figure 1: Top 4, Left 5, Right 9, Bottom 8 → Centre 77
Figure 2: Top 3, Left 9, Right 6, Bottom 7 → Centre ?
Figure 2: Top 3, Left 9, Right 6, Bottom 7 → Centre ?
Solution:
In the first figure:
(Left × Right) + (Top × Bottom)
= (5 × 9) + (4 × 8)
= 45 + 32
= 77
Apply the same rule to the second figure:
(9 × 6) + (3 × 7)
= 54 + 21
= 75
In the first figure:
(Left × Right) + (Top × Bottom)
= (5 × 9) + (4 × 8)
= 45 + 32
= 77
Apply the same rule to the second figure:
(9 × 6) + (3 × 7)
= 54 + 21
= 75
Answer: (D) 75
Final Answer Key: Questions 21–40
| Question | Answer | Question | Answer |
|---|---|---|---|
| 21 | C – CCBA | 31 | B – 42 |
| 22 | A – ACAC | 32 | 89* |
| 23 | A – AABAB | 33 | D |
| 24 | C – BACAB | 34 | A – 2031 |
| 25 | C – NNMM | 35 | B – 12 |
| 26 | A – MNPMON | 36 | B – 12 |
| 27 | B – S | 37 | A – 16 |
| 28 | A – OL | 38 | B – 6 |
| 29 | C – ABABAB | 39 | D – 6 |
| 30 | C – BABB | 40 | D – 75 |
* Question 32:
The consistent numerical rule gives 89, but the uploaded paper
does not provide 89 as an option. Therefore, this question should
be treated as a probable printing/option error rather than forcing
one of the four printed choices.
Quick Tips for NMMSE MAT
For Letter Series:
- Try dividing the series into equal groups.
- Look for a repeated group such as BAC, ACCB, NNMM or AAABBB.
- Check whether letters are cycling in alphabetical order.
- For A–Z series, convert letters into numerical positions.
- Check addition and subtraction first.
- Then check multiplication and division.
- Look for square roots and perfect squares.
- Check whether rows or columns have equal totals.
- For cross and triangle problems, test the same formula on at least two completed figures before applying it to the missing number.
Conclusion:
Questions 21–30 mainly test pattern recognition in letter series,
while Questions 31–40 test numerical relationships and missing
numbers in figures. Practising these patterns can significantly
improve speed and accuracy in NMMSE Mental Ability Test questions.
