Coding Decoding Questions for MAH-CET

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_ Coding Decoding Questions
_ Coding Decoding Questions

Coding Decoding Questions for MAH-CET

Here you can download a free Coding Decoding questions PDF with answers for MAH MBA CET 2022 by Cracku. These are some tricky questions in the MAH MBA CET 2022 exam that you need to find the answers to the given Coding Decoding questions. These questions will help you to practice and solve the Coding-Decoding questions in the MAH MBA CET exam. Utilize this PDF practice set, which is one of the best sources for practising and includes detailed answers. Click on the below link to download the Coding-Decoding MCQ PDF for MAH MBA CET 2022 for free.

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Instructions

Read the following information and answer the given questions.Symbolise the given number and symbol.
Number    1   2   3   4    5   6  7   8    9
Symbol     x    *   ?    &   $  %  +   !     £
Conditions:
I. if any number begins by the odd number then the odd number symbolised by @
II. if any number ends by even number then the even number symbolised as ©

Question 1: Which of the following is symbol of 673258?

a) %+?*$!

b) @+?*$!

c) @?+$*!

d) %+?*$©

e) None of these

1) Answer (D)

Solution:

Number – 673258

The number ends by an even number, thus condition (ii) is applied and the even number is coded as = ©

Thus, 673258 : %+?*$©

=> Ans – (D)

Question 2: Which of the following is the symbol of 236475?

a) ?*%&x;@

b) © ?%&x;@

c) x?%&+@

d) © ?%&+$

e) None of these

2) Answer (E)

Solution:

Number – 236475

None of the conditions are applied, the codes are written as given.

Thus, 236475 : *?%&+$

=> Ans – (E)

Question 3: Which of the following will be the symbol of 178524?

a) @+!$*©

b) @x+!$©

c) @+!$*&

d) *+!$&©

e) None of these

3) Answer (A)

Solution:

Number – 178524

The number begins by odd number and ends with an even number, thus both conditions are applied.

Thus, 178524 : @+!$*©

=> Ans – (A)

Question 4: Which of the following will be the symbol of 178524?

a) @+!$*©

b) @x+!$©

c) @+!$*&

d) *+!$&©

e) None of these

4) Answer (A)

Solution:

Number – 178524

The number begins by odd number and ends with an even number, thus both conditions are applied.

Thus, 178524 : @+!$*©

=> Ans – (A)

Question 5: Which of the following will be the symbol of 25486 ?

a) *$ &!£

b) *$&!©

c) © $&!%

d) *$!&%

e) None of these

5) Answer (B)

Solution:

Number – 25486

The number ends by an even number, thus condition (ii) is applied and the even number is coded as = ©

Thus, 25486 : *$&!©

=> Ans – (B)

Instructions

 

CONDITIONS:

I. if the first as well the last digit is odd ,both are coded as ©
Ii. if the first as well as last digit is even ,thier codes are to be swapped
Iii. if ‘O’ is the last digit ,it is to be code as ★

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Question 6: 468910

a) $HJ@KL

b) LHJ@K$

c) ★HJ@K$

d) $HJ@K★

e) None of these

6) Answer (D)

Solution:

Expression : 468910

Since the last digit is ‘0’, rule (iii) will be applied and 0 is coded as = ★

=> Code for 468910 = $HJ@K★

=> Ans – (D)

Question 7: 592476

a) H@Q$%T

b) ©@Q$%T

c) H@Q$%©

d) ©@Q$%©

e) None of these

7) Answer (E)

Solution:

Expression : 592476

Since, both the first and last digits are neither odd nor even. Also, last digit is not ‘0’. Thus. no rule is applied.

Code for 592476 = T@Q$%H

=> Ans – (E) : none of these

Question 8: 875306

a) J%T#LH

b) H%T#LH

c) H%T#LJ

d) J%T#LJ

e) NONE OF THESE

8) Answer (C)

Solution:

Expression : 875306

Since, the first and the last digits are even and the last digit is not ‘0’, thus rule (ii) will be applied and codes for 8 and 6 are swapped.

=> Code for 875306 = H%T#LJ

=> Ans – (C)

Question 9: 364279

a) ©H$Q%©

b) #H$Q%@

c) @H$Q%#

d) #H$Q%&

e) None of these

9) Answer (A)

Solution:

Expression : 364279

Since, the first and the last digits are both odd, thus rule (i) is applied and both are coded as ©

=> Code for 364279 = ©H$Q%©

=> Ans – (A)

Question 10: 270514

a) ★%LTK

b) $%LTKQ

c) Q%LTK$

d) $%★TKQ

e) None of these

10) Answer (B)

Solution:

Expression : 270514

Since, the first and the last digits are even and the last digit is not ‘0’, thus rule (ii) will be applied and codes for 2 and 4 are swapped.

=> Code for 270514 = $%LTKQ

=> Ans – (B)

Instructions

Study the following information carefully and answer the questions given below:
In a certain code language,
‘few organic farming techniques’ is written as ‘Ii gs da Cr’
‘fertilizer products few available’ Is written ‘fo pz nb gs”
‘organic waste into fertilizer’ is written ‘nb Cr pt mk’
‘disposal of farming waste’ is written as ‘hu mk li yu’
(All codes are two letter codes only)

Question 11: What is the code for ‘available’ in the given code language?

a) either ‘pz‘ or ‘fo’

b) either `nb‘ or `mk‘

c) li

d) hu

e) Other than those given as options

11) Answer (A)

Solution:

From statement 1 & 2, common word is ‘few’ coded as ‘gs’.

Similarly, from 2 & 3, code for ‘fertilizer = nb’

Now, in statement 2, two words are left, i.e. ‘products’ and ‘available’ coded as ‘fo’ or ‘pz’ interchangeably.

Thus, code for available = ‘pz’ or ‘fo’

=> Ans – (A)

Question 12: If ‘waste management techniques’ is coded as ‘ax da mk’ in the given code language, then how will ‘farming fertilizer management’ be coded as?

a) ax nb cr

b) liaxpt

c) gs nb li

d) nb ax li

e) Other than those given as options

12) Answer (D)

Solution:

From statement 1 & 2, common word is ‘few’ coded as ‘gs’.

From 1 & 4, code for ‘farming = li’

From 1 & 3, code for ‘organic = cr’

From above statements, in statement 1, code for ‘techniques = da’

Similarly, from 2 & 3, code for ‘fertilizer = nb’

From 3 & 4, code for ‘waste = mk’

Now, it is given that code for ‘waste management techniques’ = ‘ax da mk’

=> Since, we already know codes for waste and technique, => code for ‘management = ax’

Thus, code for ‘farming fertilizer management’ is any combination of ‘li nb ax’

=> Ans – (D)

Question 13: In the given code language, what does the code ‘yu. stand for?

a) farming

b) techniques

c) either ‘of’ or ‘disposal’

d) waste

e) either ‘into’ or ‘few’

13) Answer (C)

Solution:

From statement 1 & 4, common word is ‘farming’ coded as ‘li’.

Similarly, from 3 & 4, code for ‘waste = mk’

Now, in statement 4, two words are left, i.e. ‘disposal’ and ‘of’ coded as ‘hu’ or ‘yu’ interchangeably.

Thus, the code ‘yu’ stans for = ‘disposal’ or ‘of’

=> Ans – (C)

Question 14: What is the code for ‘organic’ in the given code language?

a) gs

b) cr

c) da

d) pt

e) Other than those given as options

14) Answer (B)

Solution:

From statements 1 and 3, the common word is organic.

Thus, organic is coded as = ‘cr’

=> Ans – (B)

Question 15: What will be the code for ‘few waste’ in the given code language?

a) gs li

b) pt da

c) da mk

d) Other than those given as options

e) mk gs

15) Answer (E)

Solution:

From statement 1 & 2, common word is ‘few’ coded as ‘gs’.

Similarly, from 3 & 4, code for ‘waste = mk’

Thus, code for ‘few waste’ is the combination of = ‘mk gs’

=> Ans – (E)

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