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Given statement:

'for the victory sign' is coded as 'si vi th fo', 

‘empty the apartment now’ is coded as ‘no ap th em’

‘now for the finale’ is coded as ‘fi th fo no’

‘for empty sign victory’ is coded as ‘vi si em fo’

We get:

for - fo

the - th

empty - em

now – no

The codes for ‘apartment’ and ‘finale’ is determined by elimination, i.e. all the other codes in the sentence are determined:

apartment - ap

finale – fi

We can not clearly determine the codes for ‘victory’ and ‘sign’:

victory – vi/si

sign – si/vi

The code for 'Victory' is either vi or si.

Given statement:

'for the victory sign' is coded as 'si vi th fo', 

‘empty the apartment now’ is coded as ‘no ap th em’

‘now for the finale’ is coded as ‘fi th fo no’

‘for empty sign victory’ is coded as ‘vi si em fo’

We get:

for - fo

the - th

empty - em

now – no

The codes for ‘apartment’ and ‘finale’ is determined by elimination, i.e. all the other codes in the sentence are determined:

apartment - ap

finale – fi

We can not clearly determine the codes for ‘victory’ and ‘sign’:

victory – vi/si

sign – si/vi

The code for 'Finale' is fi.

Given statement:

'for the victory sign' is coded as 'si vi th fo', 

‘empty the apartment now’ is coded as ‘no ap th em’

‘now for the finale’ is coded as ‘fi th fo no’

‘for empty sign victory’ is coded as ‘vi si em fo’

We get:

for - fo

the - th

empty - em

now – no

The codes for ‘apartment’ and ‘finale’ is determined by elimination, i.e. all the other codes in the sentence are determined:

apartment - ap

finale – fi

We can not clearly determine the codes for ‘victory’ and ‘sign’:

victory – vi/si

sign – si/vi

The code for 'the empty sign' is 'si em th'.

Given statement:

'for the victory sign' is coded as 'si vi th fo', 

‘empty the apartment now’ is coded as ‘no ap th em’

‘now for the finale’ is coded as ‘fi th fo no’

‘for empty sign victory’ is coded as ‘vi si em fo’

We get:

for - fo

the - th

empty - em

now – no

The codes for ‘apartment’ and ‘finale’ is determined by elimination, i.e. all the other codes in the sentence are determined:

apartment - ap

finale – fi

We can not clearly determine the codes for ‘victory’ and ‘sign’:

victory – vi/si

sign – si/vi

The code for 'apartment' is 'ap'.

Given words: PROPELLER

1^st, 3^rd, 6^th, and 8^th letter of the word PROPELLER is: P, O, L, and E

Combining and un-jumbling the letters we get: POLE and LOPE

Since 2 words are possible from the letters.

The answer is Y.

The pattern for the code is as follows,

As per the English Alphabetical Series,

Z - 2 = X; X - 2 = V

D - 2 = B; B - 2 = Z

G + 2 = I; I + 2 = K

F - 2 = D; D - 2 = B

K - 2 = I; I - 2 = G

'GIK' is the odd one out.

Given:

5 designations - CEO, CFO, Manager, Clerk, and Peon

CEO is the senior most and Peon is the juniormost person. 

They have a different number of pens such that no two persons have the same number of pens.

Now,

1. Two persons are junior to Z. It means Z is a Manager.

2. The one who is just senior to the one who is Manager has 33 pens. It means CFO has 33 pens.

3. One designation lies between the one who has 33 pens and A. It means A is Clerk.

 

4. A is just senior to the one who has 45 pens. It means Peon has 45 pens.

5. There is one designation between the Manager and R. It means R can be either CEO or Peon.

Case I

 

Case II

 

6. A has 11 pens more and 9 pens more than W and Q respectively. It means A = W+11 and A = Q+9. W+11 = Q+9. Q-W = 2. Q has more pens than W. In case I, Q has 45 pens and W has 33 pens. But 45-33 is not equal to 2. So, the case I get eliminated. In case II, again there are two possibilities:

Possibility I - W have 33 pens. So Q will have 35 pens and A will have 44 pens.

Possibility II - Q has 33 pens so W will have 31 pens and A will have 42 pens.

Case II (a)

 

Case II (b)

 

7. The difference in the number of pens between R and Z is 14 and Z has the least number of pens. So, R – 14 = Z. Z has 45-14 =31 pens. But no two persons can have the same number of pens. So, case II (b) gets eliminated as W also has 31 pens in it. 

The final arrangement is -

 

The total number of pens by the persons who are on the post of CEO and CFO is 35 + 33 = 68.

Hence, The correct answer is 68.

Given:

5 designations - CEO, CFO, Manager, Clerk, and Peon

CEO is the senior most and Peon is the juniormost person. 

They have a different number of pens such that no two persons have the same number of pens.

Now,

1. Two persons are junior to Z. It means Z is a Manager.

2. The one who is just senior to the one who is Manager has 33 pens. It means CFO has 33 pens.

3. One designation lies between the one who has 33 pens and A. It means A is Clerk.

 

4. A is just senior to the one who has 45 pens. It means Peon has 45 pens.

5. There is one designation between the Manager and R. It means R can be either CEO or Peon.

Case I

 

Case II

 

6. A has 11 pens more and 9 pens more than W and Q respectively. It means A = W+11 and A = Q+9. W+11 = Q+9. Q-W = 2. Q has more pens than W. In case I, Q has 45 pens and W has 33 pens. But 45-33 is not equal to 2. So, the case I get eliminated. In case II, again there are two possibilities:

Possibility I - W have 33 pens. So Q will have 35 pens and A will have 44 pens.

Possibility II - Q has 33 pens so W will have 31 pens and A will have 42 pens.

Case II (a)

 

Case II (b)

 

7. The difference in the number of pens between R and Z is 14 and Z has the least number of pens. So, R – 14 = Z. Z has 45-14 =31 pens. But no two persons can have the same number of pens. So, case II (b) gets eliminated as W also has 31 pens in it. 

The final arrangement is -

 

Hence, The correct answer is CEO-Q-35.

Given:

5 designations - CEO, CFO, Manager, Clerk, and Peon

CEO is the senior most and Peon is the juniormost person. 

They have a different number of pens such that no two persons have the same number of pens.

Now,

1. Two persons are junior to Z. It means Z is a Manager.

2. The one who is just senior to the one who is Manager has 33 pens. It means CFO has 33 pens.

3. One designation lies between the one who has 33 pens and A. It means A is Clerk.

 

4. A is just senior to the one who has 45 pens. It means Peon has 45 pens.

5. There is one designation between the Manager and R. It means R can be either CEO or Peon.

Case I

 

Case II

 

6. A has 11 pens more and 9 pens more than W and Q respectively. It means A = W+11 and A = Q+9. W+11 = Q+9. Q-W = 2. Q has more pens than W. In case I, Q has 45 pens and W has 33 pens. But 45-33 is not equal to 2. So, the case I get eliminated. In case II, again there are two possibilities:

Possibility I - W have 33 pens. So Q will have 35 pens and A will have 44 pens.

Possibility II - Q has 33 pens so W will have 31 pens and A will have 42 pens.

Case II (a)

 

Case II (b)

 

7. The difference in the number of pens between R and Z is 14 and Z has the least number of pens. So, R – 14 = Z. Z has 45-14 =31 pens. But no two persons can have the same number of pens. So, case II (b) gets eliminated as W also has 31 pens in it. 

The final arrangement is -

 

Hence, The correct answer is Three.

Given:

5 designations - CEO, CFO, Manager, Clerk, and Peon

CEO is the senior most and Peon is the juniormost person. 

They have a different number of pens such that no two persons have the same number of pens.

Now,

1. Two persons are junior to Z. It means Z is a Manager.

2. The one who is just senior to the one who is Manager has 33 pens. It means CFO has 33 pens.

3. One designation lies between the one who has 33 pens and A. It means A is Clerk.

 

4. A is just senior to the one who has 45 pens. It means Peon has 45 pens.

5. There is one designation between the Manager and R. It means R can be either CEO or Peon.

Case I

 

Case II

 

6. A has 11 pens more and 9 pens more than W and Q respectively. It means A = W+11 and A = Q+9. W+11 = Q+9. Q-W = 2. Q has more pens than W. In case I, Q has 45 pens and W has 33 pens. But 45-33 is not equal to 2. So, the case I get eliminated. In case II, again there are two possibilities:

Possibility I - W have 33 pens. So Q will have 35 pens and A will have 44 pens.

Possibility II - Q has 33 pens so W will have 31 pens and A will have 42 pens.

Case II (a)

 

Case II (b)

 

7. The difference in the number of pens between R and Z is 14 and Z has the least number of pens. So, R – 14 = Z. Z has 45-14 =31 pens. But no two persons can have the same number of pens. So, case II (b) gets eliminated as W also has 31 pens in it. 

The final arrangement is -

 

Hence, The correct answer is W.