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Coding based on Conditions

Directions: 

Digits in the number given in each of the following questions are to be coded based on the codes and the condition given below.

Conditions: 

(i) If both the first and last digits of a number are odd numbers. They both should be coded as the code of the first digit.

(ii) If both the first and last digits of a number are even numbers. They both should be coded as the code of the last digit.

(iii) If both the first and third digits of a number are numbers then they should be coded as the code of the first digit.

(iv) If both the third and fourth digits of a number are even numbers then their code should be interchanged.

Questions:

Find the code of these numbers

1. 6 4 7 8 5 3

a) P%CRML                  

b) PCR%ML      

c) PMLCR%

d) PM%RCL                  

e) None of these.

2. 3 4 6 0 8 5

a) M%PDRM                 

b) M%PDRL      

c) L%PDRL

d) LPDR%L                   

e) None of these.

3. 8 4 7 5 1 2

a) R%CM*R                   

b) @%CM*        

c) CM*%@

d) R%CM*@                 

e) None of these.

4. 2 3 6 8 9 1

a) LRP@#*                    

b) @PRL#*        

c) @LRP#*

d) @LPR#*                    

e) None of these.

5. 5 6 7 8 4 2

a) MPMR%@                                        

b) MRMP%@                                                    

c) MMR%@

d) M@RPM%                                        

e) None of these.

6. 1 2 4 6 8 7

a) *@P%RC                                          

b) *@%PR*                                                       

c) *@%P*R

d) *@P%R*                                           

e) None of these.

Solution:

1. 6 4 7 8 5 3

So, coding of ‘6 4 7 8 5 3’ will be ‘P % C R M L’

So, option (a) is correct.

2. 3 4 6 0 8 5 → here first and last digit both are odd. So, condition (i) applies. Therefore coding of 5 will be as per code of 3:

So, coding of ‘3 4 6 0 8 5’ will be ‘L % P D R L’

So, option (c) is correct.

3. 8 4 7 5 1 2 → Here first and last digit both are even. So, conditions (ii) apply.

So, coding of 8 will be as per the code of 2

So, coding of 8 4 7 5 1 2is ‘@ % CM * @’

So, option (b) is correct.

4. 2 3 6 8 9 1 → Here 3rd and 4th digit both are even. So, condition (iii) apply and their code should be interchanged.

So, coding of 2 3 6 8 9 1 will be ‘@ L R P # *’

So, option (c) is correct.

5.  5 6 7 8 4 2 → Here first and third digit both are odd. So, the coding of the first and third digit will be as per the first digit.

So, coding of ‘5 6 7 8 4 2’ will be ‘M P M R % @’

So, option (a) is correct.

6. 1 2 4 6 8 7 → Here, the first and last digit both are odd. So, coding will be as per the first digit. And third and fourth digit both are even. So, coding will be interchanged for them.

So, condition (i) and (ii) both apply

So, option (d) is correct.