MCQ Test: Alphanumeric Series - 1 - Bank Exams MCQ

The given numbers:
428    548    943    297    567    459
Arranging all the numbers in ascending order, we get:
297    428    459    548    567    943
The smallest and the second highest numbers are 297 and 567.
Now, required sum = 297 + 567 = 864
Multiplication of all the digits of 864 = 8 X 6 X 4 = 192
Hence, the correct answer is option B.

The given numbers:
428    548    943    297    567    459
Arranging all the numbers in ascending order, we get:
297    428    459    548    567    943
The third lowest number = 459, the last digit of 459 is 9.
The second highest number = 567, the last digit of 567 is 7.
Square of 9 = 81 
Square of 7 = 49 
Difference of 81 and 49 = 32
Hence, the correct answer is option B.

The given numbers:
428    548    943    297    567    459
After applying above condition, we get:
482     584   934    279   576    495
Now, arranging all the numbers in ascending order, we get:
279   482   495   576   584   934 
The second lowest and the highest number of the sequence are 482 and 934 respectively 
The middle digit of 482 = 8
The middle digit of 934 = 3
Now, difference between the middle digits of these two numbers = 8 – 3 = 5
Hence, the correct answer is option C.

The given numbers:
428    548    943    297    567    459
Applying the above condition in the given sequence, we get:
429    549   944    298   568   460
Here, we can see the 944, 568 and 460 are multiple of 4.
Hence, the correct answer is option E.

The given numbers:
428    548    943    297    567    459
Applying the above condition in the sequence, we get:
248    458    493    927    657    549
Now, arranging all the numbers in ascending order, we get:
248    458    493   549    657    927
Third highest number will be 549.
Hence, the correct answer is option C.

Given series : 528 ,  496 ,  328 ,  629 ,  963
New series : 636 ,  274 ,  436 ,  437 ,  744
Clearly 436 has two different composite digits, which is represented by 328.
Hence option C is correct.

Given series : 528 ,  496 ,  328 ,  629 ,  963
New series : 428 , 486 , 228 , 628 ,  862
In such a case, 3 numbers i.e. 528, 328 and 629 will be fully divisible by 4.
Hence option B is correct.

Given series : 528 ,  496 ,  328 ,  629 ,  963
First digit of the second highest number is 6.
First digit of the smallest number is 3. 
Numbers that come between 3 and 6 is 4 and 5 i.e. 2 numbers.
Hence option B is correct.

Given series : 528 ,  496 ,  328 ,  629 ,  963
New series : 258, 469 , 238 , 269 , 369
Thus the second smallest number is 528.
Hence option A is correct.

Given series : 528 ,  496 ,  328 ,  629 ,  963
Second digit of the highest number is 6.
Third digit of the second smallest number is 6.
Required sum is 12.
Hence option C is correct.

Given series – B @ C 7 N R % 5 $ G 6 K M & 4 S # P U 5 
After operation – B C @ 7 N R % 5 G $ 6 K M & 4 S P # U 5
So, third element from the right end = #

Given series – B @ C 7 N R % 5 $ G 6 K M & 4 S # P U 5 
After dropping symbols – B C 7 N R 5 G 6 K M 4 S P U 5
So, 12th from the right end – N
Hence, second to the left of N is C.

Given series – B @ C 7 N R % 5 $ G 6 K M & 4 S # P U 5 
So, there are two letters (@C7, $G6). Therefore option c is the correct answer.

Given series – B @ C 7 N R % 5 $ G 6 K M & 4 S # P U 5 
So, SPU is the correct answer.

Given series – B @ C 7 N R % 5 $ G 6 K M & 4 S # P U 5 
After operation – B @ 7 C N R % 5 $ 6 G K M & 4 S # P 5 U
After changing – B @ 7 D N R % 5 $ 6 H K M & 4 S # P 5 V
So, fourth from the left end and eleventh from the right end – D and 6
Therefore, option d is the correct answer.

Given Series – * E 7 O & 6 ^ 4 Z 5 * 3 9 T ! D # 2 $ U
After dropping digits – * E O & ^ Z * T ! D # $ U
So, eleventh from the right end – O
Therefore, third to the right of O – Z

Given Series – * E 7 O & 6 ^ 4 Z 5 * 3 9 T ! D # 2 $ U
So, there is one digit (4). Hence, option (b) is the correct answer.

Given Series – * E 7 O & 6 ^ 4 Z 5 * 3 9 T ! D # 2 $ U
After operation – * E 9 O & 7 ^ 6 Z 5 * 4 3 T ! D # 2 $ U
So, 2 digits that is 5 and 2 has not changed. Hence, option (c) is the correct answer.

Given Series – * E 7 O & 6 ^ 4 Z 5 * 3 9 T ! D # 2 $ U
9th from the right end – 3
So, the 7th element to the left of 3 – &
Hence, option a) is the correct answer.

Given Series – * E 7 O & 6 ^ 4 Z 5 * 3 9 T ! D # 2 $ U
After dropping symbols – E 7 O 6 4 Z 5 3 9 T D 2 U
So, 9th from the left end – 9
Hence, option (d) is the correct answer.