Logical Reasoning based on Rankings

LOGICAL REASONING BASED ON ARRANGEMENTS

(Matching Puzzle):
Mr Bankatlal acted as a judge for the beauty contest. There were four participants, viz. Ms Andhra Pradesh, Ms Uttar Pradesh, Ms West Bangal and Ms Maharashtra. Mrs Bankatlal, who was very anxious about the result asked about it as soon as he was back home. Mr Bankatlal told her just that the one who was wearing the yellow saree won the contest. When Mrs Bankatlal pressed for further details, he elaborated as follows:
i. All of them were sitting in a row.
ii. All of them wore sarees of different colours, viz. green, yellow, white, red.
iii. There was only one runner-up and she was sitting beside Ms Maharashtra.
iv. The runner-up was wearing the green saree.
v. Ms West Bengal was not sitting at the end and was not the runner-up.
vi. The winner and the runner up are not sitting adjacent to each other.
vii. Ms Maharashtra was wearing a white saree.
viii. Ms Andhra Pradesh was wearing a green saree.
ix. Participants wearing yellow saree and white saree were at the ends.

Q.1

Who wore the red saree?

Ms Andhra Pradesh

Ms West Bangal

Ms Uttar Pradesh

Ms Maharashtra

Ans .

Explanation :
Based on the information given, we can come up with the following alternatives:
Possibility 1:

Runner Up
Miss UP Miss WB Miss AP Miss Maharashtra

Yellow Red Green White

Possibility 2:

Runner Up
Miss Maharashtra Miss AP Miss WB Miss UP

White Green Red Yellow

Miss West Bengal wore red. Option (b) is correct.

Q.2

Ms West Bengal was sitting adjacent to

Ms Andhra Pradesh and Ms Maharashtra

Ms Uttar Pradesh and Ms Maharashtra

Ms Andhra Pradesh and Ms Uttar Pradesh

Ms Uttar Pradesh only

Ans .

Explanation :
Based on the information given, we can come up with the following alternatives:
Possibility 1:

Runner Up
Miss UP Miss WB Miss AP Miss Maharashtra

Yellow Red Green White

Possibility 2:

Runner Up
Miss Maharashtra Miss AP Miss WB Miss UP

White Green Red Yellow

Miss West Bengal was adjacent to Miss Andhra Pradesh and Miss Uttar Pradesh. Option (c) is correct.

Q.3

Which saree was worn by Ms Andhra Pradesh?

Yellow

Red

Green

White

Ans .

Explanation :
Based on the information given, we can come up with the following alternatives:
Possibility 1:

Runner Up
Miss UP Miss WB Miss AP Miss Maharashtra

Yellow Red Green White

Possibility 2:

Runner Up
Miss Maharashtra Miss AP Miss WB Miss UP

White Green Red Yellow

Miss Andhra Pradesh wore the green saree. Option (c) is correct.

Q.4

Who was the runner-up?

Ms Andhra Pradesh

Ms West Bengal

Ms Uttar Pradesh

Ms Maharashtra

Ans .

Explanation :
Based on the information given, we can come up with the following alternatives:
Possibility 1:

Runner Up
Miss UP Miss WB Miss AP Miss Maharashtra

Yellow Red Green White

Possibility 2:

Runner Up
Miss Maharashtra Miss AP Miss WB Miss UP

White Green Red Yellow

Miss Andhra Pradesh was the runner up as she wears green and sits next to Miss Maharashtra in both cases. Option (a) is correct.

Q.5

Who was the winner of the beauty contest?

Ms Andhra Pradesh

Ms West Bengal

Ms Uttar Pradesh

Ms Maharashtra

Ans .

Explanation :
Based on the information given, we can come up with the following alternatives:
Possibility 1:

Runner Up
Miss UP Miss WB Miss AP Miss Maharashtra

Yellow Red Green White

Possibility 2:

Runner Up
Miss Maharashtra Miss AP Miss WB Miss UP

White Green Red Yellow

Miss Uttar Pradesh was the winner. Option (c) is correct.

Directions for Questions 6 to 10: Study the following information carefully and answer the questions given below:
Five geeks (enthusiasts) entered a comic book character costume contest held during the comic con festival. The contestants dressed up and were given two awards—one for being best in a particular category (i.e. strongest, smartest, most efficient, scariest and powerful) and one for being ranked (i.e.1st, 2nd, 3rd, 4th and 5th). Determine who wore what costume and what two awards they received.
(i) The winner of the most efficient character’s costume was ranked just above Sacha’s character’s costume (which wasn’t the superman).
(ii) The winner of the strongest category (which wasn’t the Wonderwoman) was not worn by Billy.
(iii) The joker’s costume was ranked just above Sacha’s costume (which wasn’t the strongest costume winner).
(iv) The Wonderwoman costume ranked just above the scariest costume and just below Vladimirs’s costume.
(v) The Batman costume was placed just higher than the winner of the strongest costume and just lower than Catherine’s costume.
(vi) The Lex Luthor costume ranked just above Jelena’s (which wasn’t the Superman) and just below the smartest.

Q.6

Who among the following ranked fifth?

Billy

Jelena

Vladimir

Can’t be determined

Ans .

Explanation :
This question seems to be quite confusing due to the reason that all the clues seem to be the same/similar in nature. While that might be true on the surface, there are a few distinctions in the ways the clues can be used—especially when we look at the clues in the correct order.
On combining clues (i) and clues (iii) we would get the following Table 1:

Costume Person Category best
Joker Most efficient

Sacha

Note: We combine Clues (i) and (iii) because Clue (ii) is not usable at this stage, as it is what can be described as “a checking” clue, i.e. once we have multiple final figure possibilities in place, we use clues of the nature of clue (ii) and also clue (i) (Sacha was not the superman), to eliminate one of the possibilities. Clue (iv) will give us Table 2:

Costume Person Category best
Vladimir

Wonderwoman

Scariest

Clue (v) would give us Table 3:

Costume Person Category best
Catherine

Batman

Strongest

Clue (vi) would give us Table 4:

Costume Person Category best
Smartest

Lex Luthor

Jelena

From this point we need to focus on Sacha’s costume. We would realise that Sacha cannot wear Superman (given clue (i)); Cannot wear joker (as joker is above Sacha according to Table 1) and cannot wear Lex Luthor because we cannot superimpose Table 4 on Table 1 by making Sacha correspond to Lex Luthor as it leads to the joker being the smartest (but Table 1 has joker as most efficient). This leaves us with 2 possibilities for Sacha’s costume—Wonderwoman or Batman. Let us take a look at how possibility 1 rolls out: For Sacha to be Wonderwoman we would need to merge Tables 1 and 2 to get Table 5 below:

Costume Person Category best
Joker Vladimir Most Efficient

Wonderwoman Sacha

Scariest

To this we need to merge Table 3, which would give us Table 6 below:

Costume Person Category best
Joker Vladimir Most Efficient

Wonderwoman Sacha

Catherine Scariest

Batman

Strongest

There is only one way from this point that Table 4 can get merged into Table 6. That would give us:

Costume Person Category best
Joker Vladimir Most Efficient

Wonderwoman Sacha

Catherine Scariest

Batman Jelena

Strongest

That leaves us with a final solution as below.

Costume Person Category best
Joker Vladimir Most Efficient

Wonderwoman Sacha

Catherine Scariest

Batman Jelena Most powerful

Superman Billy Strongest

We can clearly see that this solution table contradicts clue (ii) – i.e. Billy was not the strongest. Thus, we reject Possibility 1 and move to a possibility which would give us that Sacha must be Batman. In this case the thinking pattern would go as follows: Combining Table 3 with Table 1:

Costume Person Category best
Joker Catherine Most Efficient

Batman Sacha

Strongest

To this we merge Table 3 and get:

Costume Person Category best
Joker Catherine Most Efficient

Batman Sacha

Vladimir Strongest

Wonderwoman

Scariest

To this we merge Table 4 and get:
Lex Luthor

Costume Person Category best
Joker Catherine Most Efficient

Batman Sacha Smartest

Vladimir Strongest

Wonderwoman Jelena

Scariest

This leaves us with the task of placing the remaining values in each column. When we do this we get:
Lex Luthor

Costume Person Category best
Joker Catherine Most Efficient

Batman Sacha Smartest

Vladimir Strongest

Wonderwoman Jelena Powerful

Superman Billy Scariest

Billy ranked fifth. Option (a)

Q.7

Who among the following got the award of the scariest costume?

Billy

Catharine

Sacha

Vladimir

Ans .

Explanation :
This question seems to be quite confusing due to the reason that all the clues seem to be the same/similar in nature. While that might be true on the surface, there are a few distinctions in the ways the clues can be used—especially when we look at the clues in the correct order.
On combining clues (i) and clues (iii) we would get the following Table 1:

Costume Person Category best
Joker Most efficient

Sacha

Note: We combine Clues (i) and (iii) because Clue (ii) is not usable at this stage, as it is what can be described as “a checking” clue, i.e. once we have multiple final figure possibilities in place, we use clues of the nature of clue (ii) and also clue (i) (Sacha was not the superman), to eliminate one of the possibilities. Clue (iv) will give us Table 2:

Costume Person Category best
Vladimir

Wonderwoman

Scariest

Clue (v) would give us Table 3:

Costume Person Category best
Catherine

Batman

Strongest

Clue (vi) would give us Table 4:

Costume Person Category best
Smartest

Lex Luthor

Jelena

From this point we need to focus on Sacha’s costume. We would realise that Sacha cannot wear Superman (given clue (i)); Cannot wear joker (as joker is above Sacha according to Table 1) and cannot wear Lex Luthor because we cannot superimpose Table 4 on Table 1 by making Sacha correspond to Lex Luthor as it leads to the joker being the smartest (but Table 1 has joker as most efficient). This leaves us with 2 possibilities for Sacha’s costume—Wonderwoman or Batman. Let us take a look at how possibility 1 rolls out: For Sacha to be Wonderwoman we would need to merge Tables 1 and 2 to get Table 5 below:

Costume Person Category best
Joker Vladimir Most Efficient

Wonderwoman Sacha

Scariest

To this we need to merge Table 3, which would give us Table 6 below:

Costume Person Category best
Joker Vladimir Most Efficient

Wonderwoman Sacha

Catherine Scariest

Batman

Strongest

There is only one way from this point that Table 4 can get merged into Table 6. That would give us:

Costume Person Category best
Joker Vladimir Most Efficient

Wonderwoman Sacha

Catherine Scariest

Batman Jelena

Strongest

That leaves us with a final solution as below.

Costume Person Category best
Joker Vladimir Most Efficient

Wonderwoman Sacha

Catherine Scariest

Batman Jelena Most powerful

Superman Billy Strongest

We can clearly see that this solution table contradicts clue (ii) – i.e. Billy was not the strongest. Thus, we reject Possibility 1 and move to a possibility which would give us that Sacha must be Batman. In this case the thinking pattern would go as follows: Combining Table 3 with Table 1:

Costume Person Category best
Joker Catherine Most Efficient

Batman Sacha

Strongest

To this we merge Table 3 and get:

Costume Person Category best
Joker Catherine Most Efficient

Batman Sacha

Vladimir Strongest

Wonderwoman

Scariest

To this we merge Table 4 and get:
Lex Luthor

Costume Person Category best
Joker Catherine Most Efficient

Batman Sacha Smartest

Vladimir Strongest

Wonderwoman Jelena

Scariest

This leaves us with the task of placing the remaining values in each column. When we do this we get:
Lex Luthor

Costume Person Category best
Joker Catherine Most Efficient

Batman Sacha Smartest

Vladimir Strongest

Wonderwoman Jelena Powerful

Superman Billy Scariest

Billy got the scariest costume. Option (a)

Q.8

Which of the following is the costume of Sacha?

Joker

Batman

Lex Luthor

Wonderwoman

Ans .

Explanation :
This question seems to be quite confusing due to the reason that all the clues seem to be the same/similar in nature. While that might be true on the surface, there are a few distinctions in the ways the clues can be used—especially when we look at the clues in the correct order.
On combining clues (i) and clues (iii) we would get the following Table 1:

Costume Person Category best
Joker Most efficient

Sacha

Note: We combine Clues (i) and (iii) because Clue (ii) is not usable at this stage, as it is what can be described as “a checking” clue, i.e. once we have multiple final figure possibilities in place, we use clues of the nature of clue (ii) and also clue (i) (Sacha was not the superman), to eliminate one of the possibilities. Clue (iv) will give us Table 2:

Costume Person Category best
Vladimir

Wonderwoman

Scariest

Clue (v) would give us Table 3:

Costume Person Category best
Catherine

Batman

Strongest

Clue (vi) would give us Table 4:

Costume Person Category best
Smartest

Lex Luthor

Jelena

From this point we need to focus on Sacha’s costume. We would realise that Sacha cannot wear Superman (given clue (i)); Cannot wear joker (as joker is above Sacha according to Table 1) and cannot wear Lex Luthor because we cannot superimpose Table 4 on Table 1 by making Sacha correspond to Lex Luthor as it leads to the joker being the smartest (but Table 1 has joker as most efficient). This leaves us with 2 possibilities for Sacha’s costume—Wonderwoman or Batman. Let us take a look at how possibility 1 rolls out: For Sacha to be Wonderwoman we would need to merge Tables 1 and 2 to get Table 5 below:

Costume Person Category best
Joker Vladimir Most Efficient

Wonderwoman Sacha

Scariest

To this we need to merge Table 3, which would give us Table 6 below:

Costume Person Category best
Joker Vladimir Most Efficient

Wonderwoman Sacha

Catherine Scariest

Batman

Strongest

There is only one way from this point that Table 4 can get merged into Table 6. That would give us:

Costume Person Category best
Joker Vladimir Most Efficient

Wonderwoman Sacha

Catherine Scariest

Batman Jelena

Strongest

That leaves us with a final solution as below.

Costume Person Category best
Joker Vladimir Most Efficient

Wonderwoman Sacha

Catherine Scariest

Batman Jelena Most powerful

Superman Billy Strongest

We can clearly see that this solution table contradicts clue (ii) – i.e. Billy was not the strongest. Thus, we reject Possibility 1 and move to a possibility which would give us that Sacha must be Batman. In this case the thinking pattern would go as follows: Combining Table 3 with Table 1:

Costume Person Category best
Joker Catherine Most Efficient

Batman Sacha

Strongest

To this we merge Table 3 and get:

Costume Person Category best
Joker Catherine Most Efficient

Batman Sacha

Vladimir Strongest

Wonderwoman

Scariest

To this we merge Table 4 and get:
Lex Luthor

Costume Person Category best
Joker Catherine Most Efficient

Batman Sacha Smartest

Vladimir Strongest

Wonderwoman Jelena

Scariest

This leaves us with the task of placing the remaining values in each column. When we do this we get:
Lex Luthor

Costume Person Category best
Joker Catherine Most Efficient

Batman Sacha Smartest

Vladimir Strongest

Wonderwoman Jelena Powerful

Superman Billy Scariest

Sacha’s costume is Batman. Option (b) is correct.

Q.9

Which of the following shows the correct order of ranks 1st, 2nd, 3rd, 4th and 5th respectively?

Billy, Jelena, Vladimir, Sacha, Catherine

Vladimir, Sacha, Catherine, Billy, Jelena

Catherine, Sacha, Vladimir, Jelena, Billy

Vladimir, Billy, Catherine, Sacha, Jelena

Ans .

Explanation :
This question seems to be quite confusing due to the reason that all the clues seem to be the same/similar in nature. While that might be true on the surface, there are a few distinctions in the ways the clues can be used—especially when we look at the clues in the correct order.
On combining clues (i) and clues (iii) we would get the following Table 1:

Costume Person Category best
Joker Most efficient

Sacha

Note: We combine Clues (i) and (iii) because Clue (ii) is not usable at this stage, as it is what can be described as “a checking” clue, i.e. once we have multiple final figure possibilities in place, we use clues of the nature of clue (ii) and also clue (i) (Sacha was not the superman), to eliminate one of the possibilities. Clue (iv) will give us Table 2:

Costume Person Category best
Vladimir

Wonderwoman

Scariest

Clue (v) would give us Table 3:

Costume Person Category best
Catherine

Batman

Strongest

Clue (vi) would give us Table 4:

Costume Person Category best
Smartest

Lex Luthor

Jelena

From this point we need to focus on Sacha’s costume. We would realise that Sacha cannot wear Superman (given clue (i)); Cannot wear joker (as joker is above Sacha according to Table 1) and cannot wear Lex Luthor because we cannot superimpose Table 4 on Table 1 by making Sacha correspond to Lex Luthor as it leads to the joker being the smartest (but Table 1 has joker as most efficient). This leaves us with 2 possibilities for Sacha’s costume—Wonderwoman or Batman. Let us take a look at how possibility 1 rolls out: For Sacha to be Wonderwoman we would need to merge Tables 1 and 2 to get Table 5 below:

Costume Person Category best
Joker Vladimir Most Efficient

Wonderwoman Sacha

Scariest

To this we need to merge Table 3, which would give us Table 6 below:

Costume Person Category best
Joker Vladimir Most Efficient

Wonderwoman Sacha

Catherine Scariest

Batman

Strongest

There is only one way from this point that Table 4 can get merged into Table 6. That would give us:

Costume Person Category best
Joker Vladimir Most Efficient

Wonderwoman Sacha

Catherine Scariest

Batman Jelena

Strongest

That leaves us with a final solution as below.

Costume Person Category best
Joker Vladimir Most Efficient

Wonderwoman Sacha

Catherine Scariest

Batman Jelena Most powerful

Superman Billy Strongest

We can clearly see that this solution table contradicts clue (ii) – i.e. Billy was not the strongest. Thus, we reject Possibility 1 and move to a possibility which would give us that Sacha must be Batman. In this case the thinking pattern would go as follows: Combining Table 3 with Table 1:

Costume Person Category best
Joker Catherine Most Efficient

Batman Sacha

Strongest

To this we merge Table 3 and get:

Costume Person Category best
Joker Catherine Most Efficient

Batman Sacha

Vladimir Strongest

Wonderwoman

Scariest

To this we merge Table 4 and get:
Lex Luthor

Costume Person Category best
Joker Catherine Most Efficient

Batman Sacha Smartest

Vladimir Strongest

Wonderwoman Jelena

Scariest

This leaves us with the task of placing the remaining values in each column. When we do this we get:
Lex Luthor

Costume Person Category best
Joker Catherine Most Efficient

Batman Sacha Smartest

Vladimir Strongest

Wonderwoman Jelena Powerful

Superman Billy Scariest

Option (c) is the correct order.

Q.10

Which of the following are not correctly matched?

	Costume	Category
I. Joker	Most Efficient
II. Batman	Smartest
Ill. Lex Luthor	Strongest
IV. Wonderwoman	Scariest
V. Superman	Powerful

Only IV and V

Only III and IV

Only ll and III

Only I and II

Ans .

Explanation :
This question seems to be quite confusing due to the reason that all the clues seem to be the same/similar in nature. While that might be true on the surface, there are a few distinctions in the ways the clues can be used—especially when we look at the clues in the correct order.
On combining clues (i) and clues (iii) we would get the following Table 1:

Costume Person Category best
Joker Most efficient

Sacha

Note: We combine Clues (i) and (iii) because Clue (ii) is not usable at this stage, as it is what can be described as “a checking” clue, i.e. once we have multiple final figure possibilities in place, we use clues of the nature of clue (ii) and also clue (i) (Sacha was not the superman), to eliminate one of the possibilities. Clue (iv) will give us Table 2:

Costume Person Category best
Vladimir

Wonderwoman

Scariest

Clue (v) would give us Table 3:

Costume Person Category best
Catherine

Batman

Strongest

Clue (vi) would give us Table 4:

Costume Person Category best
Smartest

Lex Luthor

Jelena

From this point we need to focus on Sacha’s costume. We would realise that Sacha cannot wear Superman (given clue (i)); Cannot wear joker (as joker is above Sacha according to Table 1) and cannot wear Lex Luthor because we cannot superimpose Table 4 on Table 1 by making Sacha correspond to Lex Luthor as it leads to the joker being the smartest (but Table 1 has joker as most efficient). This leaves us with 2 possibilities for Sacha’s costume—Wonderwoman or Batman. Let us take a look at how possibility 1 rolls out: For Sacha to be Wonderwoman we would need to merge Tables 1 and 2 to get Table 5 below:

Costume Person Category best
Joker Vladimir Most Efficient

Wonderwoman Sacha

Scariest

To this we need to merge Table 3, which would give us Table 6 below:

Costume Person Category best
Joker Vladimir Most Efficient

Wonderwoman Sacha

Catherine Scariest

Batman

Strongest

There is only one way from this point that Table 4 can get merged into Table 6. That would give us:

Costume Person Category best
Joker Vladimir Most Efficient

Wonderwoman Sacha

Catherine Scariest

Batman Jelena

Strongest

That leaves us with a final solution as below.

Costume Person Category best
Joker Vladimir Most Efficient

Wonderwoman Sacha

Catherine Scariest

Batman Jelena Most powerful

Superman Billy Strongest

We can clearly see that this solution table contradicts clue (ii) – i.e. Billy was not the strongest. Thus, we reject Possibility 1 and move to a possibility which would give us that Sacha must be Batman. In this case the thinking pattern would go as follows: Combining Table 3 with Table 1:

Costume Person Category best
Joker Catherine Most Efficient

Batman Sacha

Strongest

To this we merge Table 3 and get:

Costume Person Category best
Joker Catherine Most Efficient

Batman Sacha

Vladimir Strongest

Wonderwoman

Scariest

To this we merge Table 4 and get:
Lex Luthor

Costume Person Category best
Joker Catherine Most Efficient

Batman Sacha Smartest

Vladimir Strongest

Wonderwoman Jelena

Scariest

This leaves us with the task of placing the remaining values in each column. When we do this we get:
Lex Luthor

Costume Person Category best
Joker Catherine Most Efficient

Batman Sacha Smartest

Vladimir Strongest

Wonderwoman Jelena Powerful

Superman Billy Scariest

Only (IV) and (V) are not properly matched. Hence, Option (a) is correct.

Directions for Questions 11 to 15: Sprinters in Olympics running on the track are carrying numbers from 1 to 7 at the back of their vests. They complete the race when their feet touches the finish line at 9.10, 9.20, 9.30 seconds till 9.7 seconds i.e. at an interval of one mili second each. Each athlete is ranked from 1 (for the fastest) to 7 (for the slowest). The following additional information is known:
(i) The square of the highest total of the vest number and rank is 169 which occurs only once.
(ii) The square of the lowest total of vest number and rank is 16 and occurs only once.
(iii) The square of the total of vest number and rank = 81, occurs thrice.
(iv) The winners’ vest no. exceeds that of the first runner up.

Q. 11

The vest number of the sprinter coming first is:

Ans .

Explanation :
There are essentially two logical forks in this question: The first one comes from Clue (i), according to which the highest total of vest number and time must be 13 (as the square is 169). This could mean two scenarios: Vest number 7 with time=6; or vest number 6 with time=7. The two scenarios can be shown as:

Possibility 1A
Vest number Rank Vest number Rank

1 1

2 2

3 3

4 4

5 5

6 6 7

7 6 7

Inside these two scenarios we need to merge the other three way fork: The square of the least total of vest number and rank is 16 and occurs only once. It gives us 3 scenarios on the surface: However in this scenario we can reject Vest 3 ranking 1 because then the second ranker’s vest number would exceed the winner’s vest number (Clue (iv))—as we cannot put the 2 nd rank for either 1 or 2 as the total of 4 is the least sum and occurs only once. This leaves us with 4 principal situations 1A, 1B, 2A and 2B. Let us evaluate each of these separately:

Possibility 1A (starting thought)
Vest number Rank

1 3

2 4/5/7

3 2/4/5/7

4 1/2/4/5/7

5 1/2/4/5/7

6 1/2/4/5

7 6

In the above situation, the total of vest number + rank has to occur thrice. There are only 3 places where this can actually occur given the possibilities listed above. These are 2-7; 4-5 and 5-4. Hence, the following conclusions can be made:

Possibility 1A (starting thought)
Vest number Rank

1 3

2 7

3 2

4 5

5 4

6 1

7 6

3-2 and 6-1 get automatically selected. As for the vest number 3, only rank possibility left is 2 nd rank and after than rank 1 must go to vest number 6. If we go for a similar analysis with the other possibilities you can see that they do not work out. For instance in Possibility 1B, if we put down what is possible to be placed against vest numbers 1 and 3 respectively we get the following figure.

Possibility 1B
Vest number Rank

1 4/5/7

2 2

3 3/4/5/7

4

5

6

7 6

From this point we know that there have to be exactly 3 totals of 9, which can only be achieved by matching 4-5; 5-4 and 6-3. Then, 3 rd vest must be 7 th rank but that leaves us with no rank possibility for vest 1 and also rank 1 not getting allotted anywhere. Thus, this possibility does not exist. Similarly we can reject possibilities 2A and 2B:

Possibility 2A
Vest number Rank

1 3

2 4/5/6

3 2/4/5/6

4 1/2/4/5/6

5 1/2/4/5/6

6 7

7 1/2/4/5/6

In the above table we need three 9s. There are 4 possibilities for the sum of vest number and rank to be 9. These are 3-6, 4-5, 5-4 and 7-2. Only 3 of these 4 have to make up 9, hence we will have to leave out one of the 4. If we leave out 3-6, 4-5 or 5-4 then vest number 7 must take Position 2 (runner up). This will contradict Clue (iv) as the winner’s vest number > the runner-ups vest number. If we leave out 7-2 and place 3-6, 4-5 and 5-4 there will be no place to put rank 2. Thus, this possibility gets rejected. Possibility 2B also gets rejected as follows:

Possibility 2A
Vest number Rank

1 4/5/6

2 2

3 3/4/5/6

4 1/3/4/5/6

5 1/3/4/5/6

6 7

7 1/3/4/5/6

In the table above, there are only 3 cases of total of vest number + rank = 9. Thus, the table should become:

Possibility 2A
Vest number Rank

1 4/5/6 no place for 3rd rank??

2 2

3 6

4 5

5 4

6 7

7 1

Vest number 6. Option (b)

Q.12

The sprinter number finishing second is:

Ans .

Explanation :
There are essentially two logical forks in this question: The first one comes from Clue (i), according to which the highest total of vest number and time must be 13 (as the square is 169). This could mean two scenarios: Vest number 7 with time=6; or vest number 6 with time=7. The two scenarios can be shown as:

Possibility 1A
Vest number Rank Vest number Rank

1 1

2 2

3 3

4 4

5 5

6 6 7

7 6 7

Inside these two scenarios we need to merge the other three way fork: The square of the least total of vest number and rank is 16 and occurs only once. It gives us 3 scenarios on the surface: However in this scenario we can reject Vest 3 ranking 1 because then the second ranker’s vest number would exceed the winner’s vest number (Clue (iv))—as we cannot put the 2 nd rank for either 1 or 2 as the total of 4 is the least sum and occurs only once. This leaves us with 4 principal situations 1A, 1B, 2A and 2B. Let us evaluate each of these separately:

Possibility 1A (starting thought)
Vest number Rank

1 3

2 4/5/7

3 2/4/5/7

4 1/2/4/5/7

5 1/2/4/5/7

6 1/2/4/5

7 6

In the above situation, the total of vest number + rank has to occur thrice. There are only 3 places where this can actually occur given the possibilities listed above. These are 2-7; 4-5 and 5-4. Hence, the following conclusions can be made:

Possibility 1A (starting thought)
Vest number Rank

1 3

2 7

3 2

4 5

5 4

6 1

7 6

3-2 and 6-1 get automatically selected. As for the vest number 3, only rank possibility left is 2 nd rank and after than rank 1 must go to vest number 6. If we go for a similar analysis with the other possibilities you can see that they do not work out. For instance in Possibility 1B, if we put down what is possible to be placed against vest numbers 1 and 3 respectively we get the following figure.

Possibility 1B
Vest number Rank

1 4/5/7

2 2

3 3/4/5/7

4

5

6

7 6

From this point we know that there have to be exactly 3 totals of 9, which can only be achieved by matching 4-5; 5-4 and 6-3. Then, 3 rd vest must be 7 th rank but that leaves us with no rank possibility for vest 1 and also rank 1 not getting allotted anywhere. Thus, this possibility does not exist. Similarly we can reject possibilities 2A and 2B:

Possibility 2A
Vest number Rank

1 3

2 4/5/6

3 2/4/5/6

4 1/2/4/5/6

5 1/2/4/5/6

6 7

7 1/2/4/5/6

In the above table we need three 9s. There are 4 possibilities for the sum of vest number and rank to be 9. These are 3-6, 4-5, 5-4 and 7-2. Only 3 of these 4 have to make up 9, hence we will have to leave out one of the 4. If we leave out 3-6, 4-5 or 5-4 then vest number 7 must take Position 2 (runner up). This will contradict Clue (iv) as the winner’s vest number > the runner-ups vest number. If we leave out 7-2 and place 3-6, 4-5 and 5-4 there will be no place to put rank 2. Thus, this possibility gets rejected. Possibility 2B also gets rejected as follows:

Possibility 2A
Vest number Rank

1 4/5/6

2 2

3 3/4/5/6

4 1/3/4/5/6

5 1/3/4/5/6

6 7

7 1/3/4/5/6

In the table above, there are only 3 cases of total of vest number + rank = 9. Thus, the table should become:

Possibility 2A
Vest number Rank

1 4/5/6 no place for 3rd rank??

2 2

3 6

4 5

5 4

6 7

7 1

Vest number 3. Option (d)

Q.13

The sprinter number finishing third is:

Ans .

Explanation :
There are essentially two logical forks in this question: The first one comes from Clue (i), according to which the highest total of vest number and time must be 13 (as the square is 169). This could mean two scenarios: Vest number 7 with time=6; or vest number 6 with time=7. The two scenarios can be shown as:

Possibility 1A
Vest number Rank Vest number Rank

1 1

2 2

3 3

4 4

5 5

6 6 7

7 6 7

Inside these two scenarios we need to merge the other three way fork: The square of the least total of vest number and rank is 16 and occurs only once. It gives us 3 scenarios on the surface: However in this scenario we can reject Vest 3 ranking 1 because then the second ranker’s vest number would exceed the winner’s vest number (Clue (iv))—as we cannot put the 2 nd rank for either 1 or 2 as the total of 4 is the least sum and occurs only once. This leaves us with 4 principal situations 1A, 1B, 2A and 2B. Let us evaluate each of these separately:

Possibility 1A (starting thought)
Vest number Rank

1 3

2 4/5/7

3 2/4/5/7

4 1/2/4/5/7

5 1/2/4/5/7

6 1/2/4/5

7 6

In the above situation, the total of vest number + rank has to occur thrice. There are only 3 places where this can actually occur given the possibilities listed above. These are 2-7; 4-5 and 5-4. Hence, the following conclusions can be made:

Possibility 1A (starting thought)
Vest number Rank

1 3

2 7

3 2

4 5

5 4

6 1

7 6

3-2 and 6-1 get automatically selected. As for the vest number 3, only rank possibility left is 2 nd rank and after than rank 1 must go to vest number 6. If we go for a similar analysis with the other possibilities you can see that they do not work out. For instance in Possibility 1B, if we put down what is possible to be placed against vest numbers 1 and 3 respectively we get the following figure.

Possibility 1B
Vest number Rank

1 4/5/7

2 2

3 3/4/5/7

4

5

6

7 6

From this point we know that there have to be exactly 3 totals of 9, which can only be achieved by matching 4-5; 5-4 and 6-3. Then, 3 rd vest must be 7 th rank but that leaves us with no rank possibility for vest 1 and also rank 1 not getting allotted anywhere. Thus, this possibility does not exist. Similarly we can reject possibilities 2A and 2B:

Possibility 2A
Vest number Rank

1 3

2 4/5/6

3 2/4/5/6

4 1/2/4/5/6

5 1/2/4/5/6

6 7

7 1/2/4/5/6

In the above table we need three 9s. There are 4 possibilities for the sum of vest number and rank to be 9. These are 3-6, 4-5, 5-4 and 7-2. Only 3 of these 4 have to make up 9, hence we will have to leave out one of the 4. If we leave out 3-6, 4-5 or 5-4 then vest number 7 must take Position 2 (runner up). This will contradict Clue (iv) as the winner’s vest number > the runner-ups vest number. If we leave out 7-2 and place 3-6, 4-5 and 5-4 there will be no place to put rank 2. Thus, this possibility gets rejected. Possibility 2B also gets rejected as follows:

Possibility 2A
Vest number Rank

1 4/5/6

2 2

3 3/4/5/6

4 1/3/4/5/6

5 1/3/4/5/6

6 7

7 1/3/4/5/6

In the table above, there are only 3 cases of total of vest number + rank = 9. Thus, the table should become:

Possibility 2A
Vest number Rank

1 4/5/6 no place for 3rd rank??

2 2

3 6

4 5

5 4

6 7

7 1

Vest number 2. Option (d)

Q.14

The sprinter number finishing last is:

Ans .

Explanation :
There are essentially two logical forks in this question: The first one comes from Clue (i), according to which the highest total of vest number and time must be 13 (as the square is 169). This could mean two scenarios: Vest number 7 with time=6; or vest number 6 with time=7. The two scenarios can be shown as:

Possibility 1A
Vest number Rank Vest number Rank

1 1

2 2

3 3

4 4

5 5

6 6 7

7 6 7

Inside these two scenarios we need to merge the other three way fork: The square of the least total of vest number and rank is 16 and occurs only once. It gives us 3 scenarios on the surface: However in this scenario we can reject Vest 3 ranking 1 because then the second ranker’s vest number would exceed the winner’s vest number (Clue (iv))—as we cannot put the 2 nd rank for either 1 or 2 as the total of 4 is the least sum and occurs only once. This leaves us with 4 principal situations 1A, 1B, 2A and 2B. Let us evaluate each of these separately:

Possibility 1A (starting thought)
Vest number Rank

1 3

2 4/5/7

3 2/4/5/7

4 1/2/4/5/7

5 1/2/4/5/7

6 1/2/4/5

7 6

In the above situation, the total of vest number + rank has to occur thrice. There are only 3 places where this can actually occur given the possibilities listed above. These are 2-7; 4-5 and 5-4. Hence, the following conclusions can be made:

Possibility 1A (starting thought)
Vest number Rank

1 3

2 7

3 2

4 5

5 4

6 1

7 6

3-2 and 6-1 get automatically selected. As for the vest number 3, only rank possibility left is 2 nd rank and after than rank 1 must go to vest number 6. If we go for a similar analysis with the other possibilities you can see that they do not work out. For instance in Possibility 1B, if we put down what is possible to be placed against vest numbers 1 and 3 respectively we get the following figure.

Possibility 1B
Vest number Rank

1 4/5/7

2 2

3 3/4/5/7

4

5

6

7 6

From this point we know that there have to be exactly 3 totals of 9, which can only be achieved by matching 4-5; 5-4 and 6-3. Then, 3 rd vest must be 7 th rank but that leaves us with no rank possibility for vest 1 and also rank 1 not getting allotted anywhere. Thus, this possibility does not exist. Similarly we can reject possibilities 2A and 2B:

Possibility 2A
Vest number Rank

1 3

2 4/5/6

3 2/4/5/6

4 1/2/4/5/6

5 1/2/4/5/6

6 7

7 1/2/4/5/6

In the above table we need three 9s. There are 4 possibilities for the sum of vest number and rank to be 9. These are 3-6, 4-5, 5-4 and 7-2. Only 3 of these 4 have to make up 9, hence we will have to leave out one of the 4. If we leave out 3-6, 4-5 or 5-4 then vest number 7 must take Position 2 (runner up). This will contradict Clue (iv) as the winner’s vest number > the runner-ups vest number. If we leave out 7-2 and place 3-6, 4-5 and 5-4 there will be no place to put rank 2. Thus, this possibility gets rejected. Possibility 2B also gets rejected as follows:

Possibility 2A
Vest number Rank

1 4/5/6

2 2

3 3/4/5/6

4 1/3/4/5/6

5 1/3/4/5/6

6 7

7 1/3/4/5/6

In the table above, there are only 3 cases of total of vest number + rank = 9. Thus, the table should become:

Possibility 2A
Vest number Rank

1 4/5/6 no place for 3rd rank??

2 2

3 6

4 5

5 4

6 7

7 1

Vest number 2. Option (b)

Q.15

The sprinter number finishing last but one is:

Ans .

Explanation :
There are essentially two logical forks in this question: The first one comes from Clue (i), according to which the highest total of vest number and time must be 13 (as the square is 169). This could mean two scenarios: Vest number 7 with time=6; or vest number 6 with time=7. The two scenarios can be shown as:

Possibility 1A
Vest number Rank Vest number Rank

1 1

2 2

3 3

4 4

5 5

6 6 7

7 6 7

Inside these two scenarios we need to merge the other three way fork: The square of the least total of vest number and rank is 16 and occurs only once. It gives us 3 scenarios on the surface: However in this scenario we can reject Vest 3 ranking 1 because then the second ranker’s vest number would exceed the winner’s vest number (Clue (iv))—as we cannot put the 2 nd rank for either 1 or 2 as the total of 4 is the least sum and occurs only once. This leaves us with 4 principal situations 1A, 1B, 2A and 2B. Let us evaluate each of these separately:

Possibility 1A (starting thought)
Vest number Rank

1 3

2 4/5/7

3 2/4/5/7

4 1/2/4/5/7

5 1/2/4/5/7

6 1/2/4/5

7 6

In the above situation, the total of vest number + rank has to occur thrice. There are only 3 places where this can actually occur given the possibilities listed above. These are 2-7; 4-5 and 5-4. Hence, the following conclusions can be made:

Possibility 1A (starting thought)
Vest number Rank

1 3

2 7

3 2

4 5

5 4

6 1

7 6

3-2 and 6-1 get automatically selected. As for the vest number 3, only rank possibility left is 2 nd rank and after than rank 1 must go to vest number 6. If we go for a similar analysis with the other possibilities you can see that they do not work out. For instance in Possibility 1B, if we put down what is possible to be placed against vest numbers 1 and 3 respectively we get the following figure.

Possibility 1B
Vest number Rank

1 4/5/7

2 2

3 3/4/5/7

4

5

6

7 6

From this point we know that there have to be exactly 3 totals of 9, which can only be achieved by matching 4-5; 5-4 and 6-3. Then, 3 rd vest must be 7 th rank but that leaves us with no rank possibility for vest 1 and also rank 1 not getting allotted anywhere. Thus, this possibility does not exist. Similarly we can reject possibilities 2A and 2B:

Possibility 2A
Vest number Rank

1 3

2 4/5/6

3 2/4/5/6

4 1/2/4/5/6

5 1/2/4/5/6

6 7

7 1/2/4/5/6

In the above table we need three 9s. There are 4 possibilities for the sum of vest number and rank to be 9. These are 3-6, 4-5, 5-4 and 7-2. Only 3 of these 4 have to make up 9, hence we will have to leave out one of the 4. If we leave out 3-6, 4-5 or 5-4 then vest number 7 must take Position 2 (runner up). This will contradict Clue (iv) as the winner’s vest number > the runner-ups vest number. If we leave out 7-2 and place 3-6, 4-5 and 5-4 there will be no place to put rank 2. Thus, this possibility gets rejected. Possibility 2B also gets rejected as follows:

Possibility 2A
Vest number Rank

1 4/5/6

2 2

3 3/4/5/6

4 1/3/4/5/6

5 1/3/4/5/6

6 7

7 1/3/4/5/6

In the table above, there are only 3 cases of total of vest number + rank = 9. Thus, the table should become:

Possibility 2A
Vest number Rank

1 4/5/6 no place for 3rd rank??

2 2

3 6

4 5

5 4

6 7

7 1

Vest number 7. Option (a)

Directions for Questions 16 to 20: In a bike racing competition, ten selected bikers from a city were to compete in various stunts. The structure of the competition requires all bikers to compete in five events before the elimination round begins so as to choose the top five bikers who will participate from that city at the national level. The ranks attained by each biker in the five events from first to tenth are:

	Names	Event A	Event B	Event C	Event D	Event E
Alex	2	5	8	5	1
Ben	8	1	7	1	2
Charlie	1	10	2	10	3
Devon
Ethan	9	4	9	2	7
Frank	6	7	6	6	8
Garry	3	2	3	7	4
Harry	10	8	5	3	10
Ian	7	9	10	8	5
John	4	3	4	9	6

It is also stated that:
i. The biker with the first rank in most events gets to eliminate the biker with the last rank in most of the events.
ii. The first rank holder of event E is not to be eliminated.
iii. A biker whose highest rank and the lowest rank difference is the second highest among all bikers is not to be eliminated.
iv. A biker who does not rank in the top five positions in any of the five events gets eliminated.
v. The biker with the same rank in most of the events gets to eliminate that event E biker with an immediately lower rank than his same ranks in all events.
vi. A biker who got the second highest rank in event D gets to eliminate the biker who got the same rank in event B.
vii. Any biker who gets the first rank in two or more events will not be eliminated.
viii. Among the bikers who got the lowest rank in event B and the one who got the ninth rank in event D, the one who got a higher rank in the event E gets to eliminate the other.
ix. The biker whose sum of all the ranks in the events is the highest and does not get eliminated by any other condition will not be eliminated.
It is also known that if more than 5 people are getting eliminated by one rule or the other, then it is allowed. However, if less than 5 people are getting eliminated, then there is a toss up between the remaining ones and the results are determined by the toss of a coin.

Q.16

Who among the following is a biker who does not get a chance to eliminate another biker?

Ben

Charlie

Devon

Frank

Ans .

Explanation :
There are essentially two logical forks in this question: The first one comes from Clue (i), according to which the highest total of vest number and time must be 13 (as the square is 169). This could mean two scenarios: Vest number 7 with time=6; or vest number 6 with time=7. The two scenarios can be shown as:

Possibility 1A
Vest number Rank Vest number Rank

1 1

2 2

3 3

4 4

5 5

6 6 7

7 6 7

Inside these two scenarios we need to merge the other three way fork: The square of the least total of vest number and rank is 16 and occurs only once. It gives us 3 scenarios on the surface: However in this scenario we can reject Vest 3 ranking 1 because then the second ranker’s vest number would exceed the winner’s vest number (Clue (iv))—as we cannot put the 2 nd rank for either 1 or 2 as the total of 4 is the least sum and occurs only once. This leaves us with 4 principal situations 1A, 1B, 2A and 2B. Let us evaluate each of these separately:

Possibility 1A (starting thought)
Vest number Rank

1 3

2 4/5/7

3 2/4/5/7

4 1/2/4/5/7

5 1/2/4/5/7

6 1/2/4/5

7 6

In the above situation, the total of vest number + rank has to occur thrice. There are only 3 places where this can actually occur given the possibilities listed above. These are 2-7; 4-5 and 5-4. Hence, the following conclusions can be made:

Possibility 1A (starting thought)
Vest number Rank

1 3

2 7

3 2

4 5

5 4

6 1

7 6

3-2 and 6-1 get automatically selected. As for the vest number 3, only rank possibility left is 2 nd rank and after than rank 1 must go to vest number 6. If we go for a similar analysis with the other possibilities you can see that they do not work out. For instance in Possibility 1B, if we put down what is possible to be placed against vest numbers 1 and 3 respectively we get the following figure.

Possibility 1B
Vest number Rank

1 4/5/7

2 2

3 3/4/5/7

4

5

6

7 6

From this point we know that there have to be exactly 3 totals of 9, which can only be achieved by matching 4-5; 5-4 and 6-3. Then, 3 rd vest must be 7 th rank but that leaves us with no rank possibility for vest 1 and also rank 1 not getting allotted anywhere. Thus, this possibility does not exist. Similarly we can reject possibilities 2A and 2B:

Possibility 2A
Vest number Rank

1 3

2 4/5/6

3 2/4/5/6

4 1/2/4/5/6

5 1/2/4/5/6

6 7

7 1/2/4/5/6

In the above table we need three 9s. There are 4 possibilities for the sum of vest number and rank to be 9. These are 3-6, 4-5, 5-4 and 7-2. Only 3 of these 4 have to make up 9, hence we will have to leave out one of the 4. If we leave out 3-6, 4-5 or 5-4 then vest number 7 must take Position 2 (runner up). This will contradict Clue (iv) as the winner’s vest number > the runner-ups vest number. If we leave out 7-2 and place 3-6, 4-5 and 5-4 there will be no place to put rank 2. Thus, this possibility gets rejected. Possibility 2B also gets rejected as follows:

Possibility 2A
Vest number Rank

1 4/5/6

2 2

3 3/4/5/6

4 1/3/4/5/6

5 1/3/4/5/6

6 7

7 1/3/4/5/6

In the table above, there are only 3 cases of total of vest number + rank = 9. Thus, the table should become:

Possibility 2A
Vest number Rank

1 4/5/6 no place for 3rd rank??

2 2

3 6

4 5

5 4

6 7

7 1

Ben gets a chance to eliminate (Clue a); Frank gets a chance to eliminate (Clue e); Devon gets a chance to eliminate (Clue h); Only Charlie does not get a chance to eliminate someone. Option (b) is correct.

Q.17

Who among the following is a biker who does not rank in the top four positions in any of the five events but does not get eliminated?

Ethan

Frank

Harry

Ian

Ans .

Explanation :
There are essentially two logical forks in this question: The first one comes from Clue (i), according to which the highest total of vest number and time must be 13 (as the square is 169). This could mean two scenarios: Vest number 7 with time=6; or vest number 6 with time=7. The two scenarios can be shown as:

Possibility 1A
Vest number Rank Vest number Rank

1 1

2 2

3 3

4 4

5 5

6 6 7

7 6 7

Inside these two scenarios we need to merge the other three way fork: The square of the least total of vest number and rank is 16 and occurs only once. It gives us 3 scenarios on the surface: However in this scenario we can reject Vest 3 ranking 1 because then the second ranker’s vest number would exceed the winner’s vest number (Clue (iv))—as we cannot put the 2 nd rank for either 1 or 2 as the total of 4 is the least sum and occurs only once. This leaves us with 4 principal situations 1A, 1B, 2A and 2B. Let us evaluate each of these separately:

Possibility 1A (starting thought)
Vest number Rank

1 3

2 4/5/7

3 2/4/5/7

4 1/2/4/5/7

5 1/2/4/5/7

6 1/2/4/5

7 6

In the above situation, the total of vest number + rank has to occur thrice. There are only 3 places where this can actually occur given the possibilities listed above. These are 2-7; 4-5 and 5-4. Hence, the following conclusions can be made:

Possibility 1A (starting thought)
Vest number Rank

1 3

2 7

3 2

4 5

5 4

6 1

7 6

3-2 and 6-1 get automatically selected. As for the vest number 3, only rank possibility left is 2 nd rank and after than rank 1 must go to vest number 6. If we go for a similar analysis with the other possibilities you can see that they do not work out. For instance in Possibility 1B, if we put down what is possible to be placed against vest numbers 1 and 3 respectively we get the following figure.

Possibility 1B
Vest number Rank

1 4/5/7

2 2

3 3/4/5/7

4

5

6

7 6

From this point we know that there have to be exactly 3 totals of 9, which can only be achieved by matching 4-5; 5-4 and 6-3. Then, 3 rd vest must be 7 th rank but that leaves us with no rank possibility for vest 1 and also rank 1 not getting allotted anywhere. Thus, this possibility does not exist. Similarly we can reject possibilities 2A and 2B:

Possibility 2A
Vest number Rank

1 3

2 4/5/6

3 2/4/5/6

4 1/2/4/5/6

5 1/2/4/5/6

6 7

7 1/2/4/5/6

In the above table we need three 9s. There are 4 possibilities for the sum of vest number and rank to be 9. These are 3-6, 4-5, 5-4 and 7-2. Only 3 of these 4 have to make up 9, hence we will have to leave out one of the 4. If we leave out 3-6, 4-5 or 5-4 then vest number 7 must take Position 2 (runner up). This will contradict Clue (iv) as the winner’s vest number > the runner-ups vest number. If we leave out 7-2 and place 3-6, 4-5 and 5-4 there will be no place to put rank 2. Thus, this possibility gets rejected. Possibility 2B also gets rejected as follows:

Possibility 2A
Vest number Rank

1 4/5/6

2 2

3 3/4/5/6

4 1/3/4/5/6

5 1/3/4/5/6

6 7

7 1/3/4/5/6

In the table above, there are only 3 cases of total of vest number + rank = 9. Thus, the table should become:

Possibility 2A
Vest number Rank

1 4/5/6 no place for 3rd rank??

2 2

3 6

4 5

5 4

6 7

7 1

Frank and Ian satisfy the condition of not being in the top four in any event. Of them Ian would not get eliminated (Clue i); Frank would get eliminated (Clue d). Option (d)

Q.18

If a new condition is included in the problem stating that, ‘any biker who gets second rank in any event will not be eliminated’, then who among the following will not get eliminated?

Garry

Harry

Ian

John

Ans .

Explanation :
There are essentially two logical forks in this question: The first one comes from Clue (i), according to which the highest total of vest number and time must be 13 (as the square is 169). This could mean two scenarios: Vest number 7 with time=6; or vest number 6 with time=7. The two scenarios can be shown as:

Possibility 1A
Vest number Rank Vest number Rank

1 1

2 2

3 3

4 4

5 5

6 6 7

7 6 7

Inside these two scenarios we need to merge the other three way fork: The square of the least total of vest number and rank is 16 and occurs only once. It gives us 3 scenarios on the surface: However in this scenario we can reject Vest 3 ranking 1 because then the second ranker’s vest number would exceed the winner’s vest number (Clue (iv))—as we cannot put the 2 nd rank for either 1 or 2 as the total of 4 is the least sum and occurs only once. This leaves us with 4 principal situations 1A, 1B, 2A and 2B. Let us evaluate each of these separately:

Possibility 1A (starting thought)
Vest number Rank

1 3

2 4/5/7

3 2/4/5/7

4 1/2/4/5/7

5 1/2/4/5/7

6 1/2/4/5

7 6

In the above situation, the total of vest number + rank has to occur thrice. There are only 3 places where this can actually occur given the possibilities listed above. These are 2-7; 4-5 and 5-4. Hence, the following conclusions can be made:

Possibility 1A (starting thought)
Vest number Rank

1 3

2 7

3 2

4 5

5 4

6 1

7 6

3-2 and 6-1 get automatically selected. As for the vest number 3, only rank possibility left is 2 nd rank and after than rank 1 must go to vest number 6. If we go for a similar analysis with the other possibilities you can see that they do not work out. For instance in Possibility 1B, if we put down what is possible to be placed against vest numbers 1 and 3 respectively we get the following figure.

Possibility 1B
Vest number Rank

1 4/5/7

2 2

3 3/4/5/7

4

5

6

7 6

From this point we know that there have to be exactly 3 totals of 9, which can only be achieved by matching 4-5; 5-4 and 6-3. Then, 3 rd vest must be 7 th rank but that leaves us with no rank possibility for vest 1 and also rank 1 not getting allotted anywhere. Thus, this possibility does not exist. Similarly we can reject possibilities 2A and 2B:

Possibility 2A
Vest number Rank

1 3

2 4/5/6

3 2/4/5/6

4 1/2/4/5/6

5 1/2/4/5/6

6 7

7 1/2/4/5/6

In the above table we need three 9s. There are 4 possibilities for the sum of vest number and rank to be 9. These are 3-6, 4-5, 5-4 and 7-2. Only 3 of these 4 have to make up 9, hence we will have to leave out one of the 4. If we leave out 3-6, 4-5 or 5-4 then vest number 7 must take Position 2 (runner up). This will contradict Clue (iv) as the winner’s vest number > the runner-ups vest number. If we leave out 7-2 and place 3-6, 4-5 and 5-4 there will be no place to put rank 2. Thus, this possibility gets rejected. Possibility 2B also gets rejected as follows:

Possibility 2A
Vest number Rank

1 4/5/6

2 2

3 3/4/5/6

4 1/3/4/5/6

5 1/3/4/5/6

6 7

7 1/3/4/5/6

In the table above, there are only 3 cases of total of vest number + rank = 9. Thus, the table should become:

Possibility 2A
Vest number Rank

1 4/5/6 no place for 3rd rank??

2 2

3 6

4 5

5 4

6 7

7 1

Garry who is getting eliminated by Clue f would not be eliminated due to this Clue.

Q.19

Which of the following conditions if not included in the problem do not change the final solution?

viii

Ans .

Explanation :
There are essentially two logical forks in this question: The first one comes from Clue (i), according to which the highest total of vest number and time must be 13 (as the square is 169). This could mean two scenarios: Vest number 7 with time=6; or vest number 6 with time=7. The two scenarios can be shown as:

Possibility 1A
Vest number Rank Vest number Rank

1 1

2 2

3 3

4 4

5 5

6 6 7

7 6 7

Inside these two scenarios we need to merge the other three way fork: The square of the least total of vest number and rank is 16 and occurs only once. It gives us 3 scenarios on the surface: However in this scenario we can reject Vest 3 ranking 1 because then the second ranker’s vest number would exceed the winner’s vest number (Clue (iv))—as we cannot put the 2 nd rank for either 1 or 2 as the total of 4 is the least sum and occurs only once. This leaves us with 4 principal situations 1A, 1B, 2A and 2B. Let us evaluate each of these separately:

Possibility 1A (starting thought)
Vest number Rank

1 3

2 4/5/7

3 2/4/5/7

4 1/2/4/5/7

5 1/2/4/5/7

6 1/2/4/5

7 6

In the above situation, the total of vest number + rank has to occur thrice. There are only 3 places where this can actually occur given the possibilities listed above. These are 2-7; 4-5 and 5-4. Hence, the following conclusions can be made:

Possibility 1A (starting thought)
Vest number Rank

1 3

2 7

3 2

4 5

5 4

6 1

7 6

3-2 and 6-1 get automatically selected. As for the vest number 3, only rank possibility left is 2 nd rank and after than rank 1 must go to vest number 6. If we go for a similar analysis with the other possibilities you can see that they do not work out. For instance in Possibility 1B, if we put down what is possible to be placed against vest numbers 1 and 3 respectively we get the following figure.

Possibility 1B
Vest number Rank

1 4/5/7

2 2

3 3/4/5/7

4

5

6

7 6

From this point we know that there have to be exactly 3 totals of 9, which can only be achieved by matching 4-5; 5-4 and 6-3. Then, 3 rd vest must be 7 th rank but that leaves us with no rank possibility for vest 1 and also rank 1 not getting allotted anywhere. Thus, this possibility does not exist. Similarly we can reject possibilities 2A and 2B:

Possibility 2A
Vest number Rank

1 3

2 4/5/6

3 2/4/5/6

4 1/2/4/5/6

5 1/2/4/5/6

6 7

7 1/2/4/5/6

In the above table we need three 9s. There are 4 possibilities for the sum of vest number and rank to be 9. These are 3-6, 4-5, 5-4 and 7-2. Only 3 of these 4 have to make up 9, hence we will have to leave out one of the 4. If we leave out 3-6, 4-5 or 5-4 then vest number 7 must take Position 2 (runner up). This will contradict Clue (iv) as the winner’s vest number > the runner-ups vest number. If we leave out 7-2 and place 3-6, 4-5 and 5-4 there will be no place to put rank 2. Thus, this possibility gets rejected. Possibility 2B also gets rejected as follows:

Possibility 2A
Vest number Rank

1 4/5/6

2 2

3 3/4/5/6

4 1/3/4/5/6

5 1/3/4/5/6

6 7

7 1/3/4/5/6

In the table above, there are only 3 cases of total of vest number + rank = 9. Thus, the table should become:

Possibility 2A
Vest number Rank

1 4/5/6 no place for 3rd rank??

2 2

3 6

4 5

5 4

6 7

7 1

Clue b does not have any usage in this question.

Q. 20

It is found that the results of event D are wrong and the last two ranks have been mistakenly interchanged, then who among the following bikers cannot be definitely eliminated in the elimination round?

Ethan & Devon

Garry & John

Harry & Devon

John & Devon

Ans .

Explanation :
There are essentially two logical forks in this question: The first one comes from Clue (i), according to which the highest total of vest number and time must be 13 (as the square is 169). This could mean two scenarios: Vest number 7 with time=6; or vest number 6 with time=7. The two scenarios can be shown as:

Possibility 1A
Vest number Rank Vest number Rank

1 1

2 2

3 3

4 4

5 5

6 6 7

7 6 7

Inside these two scenarios we need to merge the other three way fork: The square of the least total of vest number and rank is 16 and occurs only once. It gives us 3 scenarios on the surface: However in this scenario we can reject Vest 3 ranking 1 because then the second ranker’s vest number would exceed the winner’s vest number (Clue (iv))—as we cannot put the 2 nd rank for either 1 or 2 as the total of 4 is the least sum and occurs only once. This leaves us with 4 principal situations 1A, 1B, 2A and 2B. Let us evaluate each of these separately:

Possibility 1A (starting thought)
Vest number Rank

1 3

2 4/5/7

3 2/4/5/7

4 1/2/4/5/7

5 1/2/4/5/7

6 1/2/4/5

7 6

In the above situation, the total of vest number + rank has to occur thrice. There are only 3 places where this can actually occur given the possibilities listed above. These are 2-7; 4-5 and 5-4. Hence, the following conclusions can be made:

Possibility 1A (starting thought)
Vest number Rank

1 3

2 7

3 2

4 5

5 4

6 1

7 6

3-2 and 6-1 get automatically selected. As for the vest number 3, only rank possibility left is 2 nd rank and after than rank 1 must go to vest number 6. If we go for a similar analysis with the other possibilities you can see that they do not work out. For instance in Possibility 1B, if we put down what is possible to be placed against vest numbers 1 and 3 respectively we get the following figure.

Possibility 1B
Vest number Rank

1 4/5/7

2 2

3 3/4/5/7

4

5

6

7 6

From this point we know that there have to be exactly 3 totals of 9, which can only be achieved by matching 4-5; 5-4 and 6-3. Then, 3 rd vest must be 7 th rank but that leaves us with no rank possibility for vest 1 and also rank 1 not getting allotted anywhere. Thus, this possibility does not exist. Similarly we can reject possibilities 2A and 2B:

Possibility 2A
Vest number Rank

1 3

2 4/5/6

3 2/4/5/6

4 1/2/4/5/6

5 1/2/4/5/6

6 7

7 1/2/4/5/6

In the above table we need three 9s. There are 4 possibilities for the sum of vest number and rank to be 9. These are 3-6, 4-5, 5-4 and 7-2. Only 3 of these 4 have to make up 9, hence we will have to leave out one of the 4. If we leave out 3-6, 4-5 or 5-4 then vest number 7 must take Position 2 (runner up). This will contradict Clue (iv) as the winner’s vest number > the runner-ups vest number. If we leave out 7-2 and place 3-6, 4-5 and 5-4 there will be no place to put rank 2. Thus, this possibility gets rejected. Possibility 2B also gets rejected as follows:

Possibility 2A
Vest number Rank

1 4/5/6

2 2

3 3/4/5/6

4 1/3/4/5/6

5 1/3/4/5/6

6 7

7 1/3/4/5/6

In the table above, there are only 3 cases of total of vest number + rank = 9. Thus, the table should become:

Possibility 2A
Vest number Rank

1 4/5/6 no place for 3rd rank??

2 2

3 6

4 5

5 4

6 7

7 1

John is getting eliminated due to Clue h, which pertains to 9th rank in the event D. Also, Devon is getting eliminated due to maximum number of last ranks, which would no longer be true. Thus, Option (d) is correct.

Directions for Questions 21 to 23: Refer to the following information to answer the questions that follow.
A. Seven students A, B, C, D, E, F and G take a series of tests.
B. No two students get similar marks.
C. G always scores more than A.
D. A always scores more than B.
E. Each time either C scores the highest and E gets the least, or alternatively D scores the highest and F or B scores the least.

Q.21

If G is ranked fifth, which of these must be true?

D scores the highest.

C is ranked second.

E is ranked third.

B is ranked fourth.

Ans .

Explanation :
There are essentially two logical forks in this question: The first one comes from Clue (i), according to which the highest total of vest number and time must be 13 (as the square is 169). This could mean two scenarios: Vest number 7 with time=6; or vest number 6 with time=7. The two scenarios can be shown as:

Possibility 1A
Vest number Rank Vest number Rank

1 1

2 2

3 3

4 4

5 5

6 6 7

7 6 7

Inside these two scenarios we need to merge the other three way fork: The square of the least total of vest number and rank is 16 and occurs only once. It gives us 3 scenarios on the surface: However in this scenario we can reject Vest 3 ranking 1 because then the second ranker’s vest number would exceed the winner’s vest number (Clue (iv))—as we cannot put the 2 nd rank for either 1 or 2 as the total of 4 is the least sum and occurs only once. This leaves us with 4 principal situations 1A, 1B, 2A and 2B. Let us evaluate each of these separately:

Possibility 1A (starting thought)
Vest number Rank

1 3

2 4/5/7

3 2/4/5/7

4 1/2/4/5/7

5 1/2/4/5/7

6 1/2/4/5

7 6

In the above situation, the total of vest number + rank has to occur thrice. There are only 3 places where this can actually occur given the possibilities listed above. These are 2-7; 4-5 and 5-4. Hence, the following conclusions can be made:

Possibility 1A (starting thought)
Vest number Rank

1 3

2 7

3 2

4 5

5 4

6 1

7 6

3-2 and 6-1 get automatically selected. As for the vest number 3, only rank possibility left is 2 nd rank and after than rank 1 must go to vest number 6. If we go for a similar analysis with the other possibilities you can see that they do not work out. For instance in Possibility 1B, if we put down what is possible to be placed against vest numbers 1 and 3 respectively we get the following figure.

Possibility 1B
Vest number Rank

1 4/5/7

2 2

3 3/4/5/7

4

5

6

7 6

From this point we know that there have to be exactly 3 totals of 9, which can only be achieved by matching 4-5; 5-4 and 6-3. Then, 3 rd vest must be 7 th rank but that leaves us with no rank possibility for vest 1 and also rank 1 not getting allotted anywhere. Thus, this possibility does not exist. Similarly we can reject possibilities 2A and 2B:

Possibility 2A
Vest number Rank

1 3

2 4/5/6

3 2/4/5/6

4 1/2/4/5/6

5 1/2/4/5/6

6 7

7 1/2/4/5/6

In the above table we need three 9s. There are 4 possibilities for the sum of vest number and rank to be 9. These are 3-6, 4-5, 5-4 and 7-2. Only 3 of these 4 have to make up 9, hence we will have to leave out one of the 4. If we leave out 3-6, 4-5 or 5-4 then vest number 7 must take Position 2 (runner up). This will contradict Clue (iv) as the winner’s vest number > the runner-ups vest number. If we leave out 7-2 and place 3-6, 4-5 and 5-4 there will be no place to put rank 2. Thus, this possibility gets rejected. Possibility 2B also gets rejected as follows:

Possibility 2A
Vest number Rank

1 4/5/6

2 2

3 3/4/5/6

4 1/3/4/5/6

5 1/3/4/5/6

6 7

7 1/3/4/5/6

In the table above, there are only 3 cases of total of vest number + rank = 9. Thus, the table should become:

Possibility 2A
Vest number Rank

1 4/5/6 no place for 3rd rank??

2 2

3 6

4 5

5 4

6 7

7 1

If G is fifth, then A and B must be ranked 6 th and 7th respectively. This can only happen if D scores the highest (as, if C is highest, then E must be 7th. Option (a)

Q. 22

If D is ranked second, which of the following can be true?

A gets more than C.

G gets more than D.

A gets more than G.

F gets more than G.

Ans .

Explanation :
There are essentially two logical forks in this question: The first one comes from Clue (i), according to which the highest total of vest number and time must be 13 (as the square is 169). This could mean two scenarios: Vest number 7 with time=6; or vest number 6 with time=7. The two scenarios can be shown as:

Possibility 1A
Vest number Rank Vest number Rank

1 1

2 2

3 3

4 4

5 5

6 6 7

7 6 7

Inside these two scenarios we need to merge the other three way fork: The square of the least total of vest number and rank is 16 and occurs only once. It gives us 3 scenarios on the surface: However in this scenario we can reject Vest 3 ranking 1 because then the second ranker’s vest number would exceed the winner’s vest number (Clue (iv))—as we cannot put the 2 nd rank for either 1 or 2 as the total of 4 is the least sum and occurs only once. This leaves us with 4 principal situations 1A, 1B, 2A and 2B. Let us evaluate each of these separately:

Possibility 1A (starting thought)
Vest number Rank

1 3

2 4/5/7

3 2/4/5/7

4 1/2/4/5/7

5 1/2/4/5/7

6 1/2/4/5

7 6

In the above situation, the total of vest number + rank has to occur thrice. There are only 3 places where this can actually occur given the possibilities listed above. These are 2-7; 4-5 and 5-4. Hence, the following conclusions can be made:

Possibility 1A (starting thought)
Vest number Rank

1 3

2 7

3 2

4 5

5 4

6 1

7 6

3-2 and 6-1 get automatically selected. As for the vest number 3, only rank possibility left is 2 nd rank and after than rank 1 must go to vest number 6. If we go for a similar analysis with the other possibilities you can see that they do not work out. For instance in Possibility 1B, if we put down what is possible to be placed against vest numbers 1 and 3 respectively we get the following figure.

Possibility 1B
Vest number Rank

1 4/5/7

2 2

3 3/4/5/7

4

5

6

7 6

From this point we know that there have to be exactly 3 totals of 9, which can only be achieved by matching 4-5; 5-4 and 6-3. Then, 3 rd vest must be 7 th rank but that leaves us with no rank possibility for vest 1 and also rank 1 not getting allotted anywhere. Thus, this possibility does not exist. Similarly we can reject possibilities 2A and 2B:

Possibility 2A
Vest number Rank

1 3

2 4/5/6

3 2/4/5/6

4 1/2/4/5/6

5 1/2/4/5/6

6 7

7 1/2/4/5/6

In the above table we need three 9s. There are 4 possibilities for the sum of vest number and rank to be 9. These are 3-6, 4-5, 5-4 and 7-2. Only 3 of these 4 have to make up 9, hence we will have to leave out one of the 4. If we leave out 3-6, 4-5 or 5-4 then vest number 7 must take Position 2 (runner up). This will contradict Clue (iv) as the winner’s vest number > the runner-ups vest number. If we leave out 7-2 and place 3-6, 4-5 and 5-4 there will be no place to put rank 2. Thus, this possibility gets rejected. Possibility 2B also gets rejected as follows:

Possibility 2A
Vest number Rank

1 4/5/6

2 2

3 3/4/5/6

4 1/3/4/5/6

5 1/3/4/5/6

6 7

7 1/3/4/5/6

In the table above, there are only 3 cases of total of vest number + rank = 9. Thus, the table should become:

Possibility 2A
Vest number Rank

1 4/5/6 no place for 3rd rank??

2 2

3 6

4 5

5 4

6 7

7 1

The first three options are not feasible. Option (d) is the only thing that can be true.

Q.23

If D is ranked sixth and B is ranked fifth, which of these can be true?

G is ranked first or fourth.

C is ranked second or third.

A is ranked second or fifth.

F is ranked third or fourth.

Ans .

Explanation :
There are essentially two logical forks in this question: The first one comes from Clue (i), according to which the highest total of vest number and time must be 13 (as the square is 169). This could mean two scenarios: Vest number 7 with time=6; or vest number 6 with time=7. The two scenarios can be shown as:

Possibility 1A
Vest number Rank Vest number Rank

1 1

2 2

3 3

4 4

5 5

6 6 7

7 6 7

Inside these two scenarios we need to merge the other three way fork: The square of the least total of vest number and rank is 16 and occurs only once. It gives us 3 scenarios on the surface: However in this scenario we can reject Vest 3 ranking 1 because then the second ranker’s vest number would exceed the winner’s vest number (Clue (iv))—as we cannot put the 2 nd rank for either 1 or 2 as the total of 4 is the least sum and occurs only once. This leaves us with 4 principal situations 1A, 1B, 2A and 2B. Let us evaluate each of these separately:

Possibility 1A (starting thought)
Vest number Rank

1 3

2 4/5/7

3 2/4/5/7

4 1/2/4/5/7

5 1/2/4/5/7

6 1/2/4/5

7 6

In the above situation, the total of vest number + rank has to occur thrice. There are only 3 places where this can actually occur given the possibilities listed above. These are 2-7; 4-5 and 5-4. Hence, the following conclusions can be made:

Possibility 1A (starting thought)
Vest number Rank

1 3

2 7

3 2

4 5

5 4

6 1

7 6

3-2 and 6-1 get automatically selected. As for the vest number 3, only rank possibility left is 2 nd rank and after than rank 1 must go to vest number 6. If we go for a similar analysis with the other possibilities you can see that they do not work out. For instance in Possibility 1B, if we put down what is possible to be placed against vest numbers 1 and 3 respectively we get the following figure.

Possibility 1B
Vest number Rank

1 4/5/7

2 2

3 3/4/5/7

4

5

6

7 6

From this point we know that there have to be exactly 3 totals of 9, which can only be achieved by matching 4-5; 5-4 and 6-3. Then, 3 rd vest must be 7 th rank but that leaves us with no rank possibility for vest 1 and also rank 1 not getting allotted anywhere. Thus, this possibility does not exist. Similarly we can reject possibilities 2A and 2B:

Possibility 2A
Vest number Rank

1 3

2 4/5/6

3 2/4/5/6

4 1/2/4/5/6

5 1/2/4/5/6

6 7

7 1/2/4/5/6

In the above table we need three 9s. There are 4 possibilities for the sum of vest number and rank to be 9. These are 3-6, 4-5, 5-4 and 7-2. Only 3 of these 4 have to make up 9, hence we will have to leave out one of the 4. If we leave out 3-6, 4-5 or 5-4 then vest number 7 must take Position 2 (runner up). This will contradict Clue (iv) as the winner’s vest number > the runner-ups vest number. If we leave out 7-2 and place 3-6, 4-5 and 5-4 there will be no place to put rank 2. Thus, this possibility gets rejected. Possibility 2B also gets rejected as follows:

Possibility 2A
Vest number Rank

1 4/5/6

2 2

3 3/4/5/6

4 1/3/4/5/6

5 1/3/4/5/6

6 7

7 1/3/4/5/6

In the table above, there are only 3 cases of total of vest number + rank = 9. Thus, the table should become:

Possibility 2A
Vest number Rank

1 4/5/6 no place for 3rd rank??

2 2

3 6

4 5

5 4

6 7

7 1

Option (d) is possible.

Directions for Questions 24 to 28: Each of the following questions gives information about an individual attribute through statements 1, 2, and 3. For each question if the first two statement are true, what conclusion can be drawn about the third statement?

Q.24

(1) Gulabjamuns are sweeter than rosagullas but laddoos are sweeter than gulabjamuns.
(2) Rosagullas are sweeter than chamchams but pethas are sweeter than rosagullas.
(3) Of the five kinds of sweets, chamchams are the least in terms of sweetness.
If the first two statements are true, the third statement is

True

False

Uncertain

Ans .

Explanation :
There are essentially two logical forks in this question: The first one comes from Clue (i), according to which the highest total of vest number and time must be 13 (as the square is 169). This could mean two scenarios: Vest number 7 with time=6; or vest number 6 with time=7. The two scenarios can be shown as:

Possibility 1A
Vest number Rank Vest number Rank

1 1

2 2

3 3

4 4

5 5

6 6 7

7 6 7

Inside these two scenarios we need to merge the other three way fork: The square of the least total of vest number and rank is 16 and occurs only once. It gives us 3 scenarios on the surface: However in this scenario we can reject Vest 3 ranking 1 because then the second ranker’s vest number would exceed the winner’s vest number (Clue (iv))—as we cannot put the 2 nd rank for either 1 or 2 as the total of 4 is the least sum and occurs only once. This leaves us with 4 principal situations 1A, 1B, 2A and 2B. Let us evaluate each of these separately:

Possibility 1A (starting thought)
Vest number Rank

1 3

2 4/5/7

3 2/4/5/7

4 1/2/4/5/7

5 1/2/4/5/7

6 1/2/4/5

7 6

In the above situation, the total of vest number + rank has to occur thrice. There are only 3 places where this can actually occur given the possibilities listed above. These are 2-7; 4-5 and 5-4. Hence, the following conclusions can be made:

Possibility 1A (starting thought)
Vest number Rank

1 3

2 7

3 2

4 5

5 4

6 1

7 6

3-2 and 6-1 get automatically selected. As for the vest number 3, only rank possibility left is 2 nd rank and after than rank 1 must go to vest number 6. If we go for a similar analysis with the other possibilities you can see that they do not work out. For instance in Possibility 1B, if we put down what is possible to be placed against vest numbers 1 and 3 respectively we get the following figure.

Possibility 1B
Vest number Rank

1 4/5/7

2 2

3 3/4/5/7

4

5

6

7 6

From this point we know that there have to be exactly 3 totals of 9, which can only be achieved by matching 4-5; 5-4 and 6-3. Then, 3 rd vest must be 7 th rank but that leaves us with no rank possibility for vest 1 and also rank 1 not getting allotted anywhere. Thus, this possibility does not exist. Similarly we can reject possibilities 2A and 2B:

Possibility 2A
Vest number Rank

1 3

2 4/5/6

3 2/4/5/6

4 1/2/4/5/6

5 1/2/4/5/6

6 7

7 1/2/4/5/6

In the above table we need three 9s. There are 4 possibilities for the sum of vest number and rank to be 9. These are 3-6, 4-5, 5-4 and 7-2. Only 3 of these 4 have to make up 9, hence we will have to leave out one of the 4. If we leave out 3-6, 4-5 or 5-4 then vest number 7 must take Position 2 (runner up). This will contradict Clue (iv) as the winner’s vest number > the runner-ups vest number. If we leave out 7-2 and place 3-6, 4-5 and 5-4 there will be no place to put rank 2. Thus, this possibility gets rejected. Possibility 2B also gets rejected as follows:

Possibility 2A
Vest number Rank

1 4/5/6

2 2

3 3/4/5/6

4 1/3/4/5/6

5 1/3/4/5/6

6 7

7 1/3/4/5/6

In the table above, there are only 3 cases of total of vest number + rank = 9. Thus, the table should become:

Possibility 2A
Vest number Rank

1 4/5/6 no place for 3rd rank??

2 2

3 6

4 5

5 4

6 7

7 1

The ordering in terms of sweetness would be L>G>R>C andpethas (P) have to be greater than chamchams (C). Thus, clearly the third statement is true.

Q.25

On the day the triplets were born to Moti the female dog,
(1) A weighs more than B.
(2) B weighs less than C.
(3) Of the three puppies, C weighs the most.
If the first two statements are true, the third statement is

True

False

Uncertain

Ans .

Explanation :
There are essentially two logical forks in this question: The first one comes from Clue (i), according to which the highest total of vest number and time must be 13 (as the square is 169). This could mean two scenarios: Vest number 7 with time=6; or vest number 6 with time=7. The two scenarios can be shown as:

Possibility 1A
Vest number Rank Vest number Rank

1 1

2 2

3 3

4 4

5 5

6 6 7

7 6 7

Inside these two scenarios we need to merge the other three way fork: The square of the least total of vest number and rank is 16 and occurs only once. It gives us 3 scenarios on the surface: However in this scenario we can reject Vest 3 ranking 1 because then the second ranker’s vest number would exceed the winner’s vest number (Clue (iv))—as we cannot put the 2 nd rank for either 1 or 2 as the total of 4 is the least sum and occurs only once. This leaves us with 4 principal situations 1A, 1B, 2A and 2B. Let us evaluate each of these separately:

Possibility 1A (starting thought)
Vest number Rank

1 3

2 4/5/7

3 2/4/5/7

4 1/2/4/5/7

5 1/2/4/5/7

6 1/2/4/5

7 6

In the above situation, the total of vest number + rank has to occur thrice. There are only 3 places where this can actually occur given the possibilities listed above. These are 2-7; 4-5 and 5-4. Hence, the following conclusions can be made:

Possibility 1A (starting thought)
Vest number Rank

1 3

2 7

3 2

4 5

5 4

6 1

7 6

3-2 and 6-1 get automatically selected. As for the vest number 3, only rank possibility left is 2 nd rank and after than rank 1 must go to vest number 6. If we go for a similar analysis with the other possibilities you can see that they do not work out. For instance in Possibility 1B, if we put down what is possible to be placed against vest numbers 1 and 3 respectively we get the following figure.

Possibility 1B
Vest number Rank

1 4/5/7

2 2

3 3/4/5/7

4

5

6

7 6

From this point we know that there have to be exactly 3 totals of 9, which can only be achieved by matching 4-5; 5-4 and 6-3. Then, 3 rd vest must be 7 th rank but that leaves us with no rank possibility for vest 1 and also rank 1 not getting allotted anywhere. Thus, this possibility does not exist. Similarly we can reject possibilities 2A and 2B:

Possibility 2A
Vest number Rank

1 3

2 4/5/6

3 2/4/5/6

4 1/2/4/5/6

5 1/2/4/5/6

6 7

7 1/2/4/5/6

In the above table we need three 9s. There are 4 possibilities for the sum of vest number and rank to be 9. These are 3-6, 4-5, 5-4 and 7-2. Only 3 of these 4 have to make up 9, hence we will have to leave out one of the 4. If we leave out 3-6, 4-5 or 5-4 then vest number 7 must take Position 2 (runner up). This will contradict Clue (iv) as the winner’s vest number > the runner-ups vest number. If we leave out 7-2 and place 3-6, 4-5 and 5-4 there will be no place to put rank 2. Thus, this possibility gets rejected. Possibility 2B also gets rejected as follows:

Possibility 2A
Vest number Rank

1 4/5/6

2 2

3 3/4/5/6

4 1/3/4/5/6

5 1/3/4/5/6

6 7

7 1/3/4/5/6

In the table above, there are only 3 cases of total of vest number + rank = 9. Thus, the table should become:

Possibility 2A
Vest number Rank

1 4/5/6 no place for 3rd rank??

2 2

3 6

4 5

5 4

6 7

7 1

Both A and C weigh more than B. However, it is not certain whether C weighs the most. Thus, we choose Option (c).

Q.26

At a car showroom,
(1) the price of Car A was lower than that of Car B.
(2) the price of Car C was lower than that of Car B.
(3) the price of Car A was higher than that of Car C.
If the first two statements are true, the third statement is

True

False

Uncertain

Ans .

Explanation :
There are essentially two logical forks in this question: The first one comes from Clue (i), according to which the highest total of vest number and time must be 13 (as the square is 169). This could mean two scenarios: Vest number 7 with time=6; or vest number 6 with time=7. The two scenarios can be shown as:

Possibility 1A
Vest number Rank Vest number Rank

1 1

2 2

3 3

4 4

5 5

6 6 7

7 6 7

Inside these two scenarios we need to merge the other three way fork: The square of the least total of vest number and rank is 16 and occurs only once. It gives us 3 scenarios on the surface: However in this scenario we can reject Vest 3 ranking 1 because then the second ranker’s vest number would exceed the winner’s vest number (Clue (iv))—as we cannot put the 2 nd rank for either 1 or 2 as the total of 4 is the least sum and occurs only once. This leaves us with 4 principal situations 1A, 1B, 2A and 2B. Let us evaluate each of these separately:

Possibility 1A (starting thought)
Vest number Rank

1 3

2 4/5/7

3 2/4/5/7

4 1/2/4/5/7

5 1/2/4/5/7

6 1/2/4/5

7 6

In the above situation, the total of vest number + rank has to occur thrice. There are only 3 places where this can actually occur given the possibilities listed above. These are 2-7; 4-5 and 5-4. Hence, the following conclusions can be made:

Possibility 1A (starting thought)
Vest number Rank

1 3

2 7

3 2

4 5

5 4

6 1

7 6

3-2 and 6-1 get automatically selected. As for the vest number 3, only rank possibility left is 2 nd rank and after than rank 1 must go to vest number 6. If we go for a similar analysis with the other possibilities you can see that they do not work out. For instance in Possibility 1B, if we put down what is possible to be placed against vest numbers 1 and 3 respectively we get the following figure.

Possibility 1B
Vest number Rank

1 4/5/7

2 2

3 3/4/5/7

4

5

6

7 6

From this point we know that there have to be exactly 3 totals of 9, which can only be achieved by matching 4-5; 5-4 and 6-3. Then, 3 rd vest must be 7 th rank but that leaves us with no rank possibility for vest 1 and also rank 1 not getting allotted anywhere. Thus, this possibility does not exist. Similarly we can reject possibilities 2A and 2B:

Possibility 2A
Vest number Rank

1 3

2 4/5/6

3 2/4/5/6

4 1/2/4/5/6

5 1/2/4/5/6

6 7

7 1/2/4/5/6

In the above table we need three 9s. There are 4 possibilities for the sum of vest number and rank to be 9. These are 3-6, 4-5, 5-4 and 7-2. Only 3 of these 4 have to make up 9, hence we will have to leave out one of the 4. If we leave out 3-6, 4-5 or 5-4 then vest number 7 must take Position 2 (runner up). This will contradict Clue (iv) as the winner’s vest number > the runner-ups vest number. If we leave out 7-2 and place 3-6, 4-5 and 5-4 there will be no place to put rank 2. Thus, this possibility gets rejected. Possibility 2B also gets rejected as follows:

Possibility 2A
Vest number Rank

1 4/5/6

2 2

3 3/4/5/6

4 1/3/4/5/6

5 1/3/4/5/6

6 7

7 1/3/4/5/6

In the table above, there are only 3 cases of total of vest number + rank = 9. Thus, the table should become:

Possibility 2A
Vest number Rank

1 4/5/6 no place for 3rd rank??

2 2

3 6

4 5

5 4

6 7

7 1

Uncertain. Option (c).

Q. 27

In the country of Mohenjo Daro,
(1) Cricket is bigger than Football and smaller than Tennis.
(2) Hockey is smaller than Tennis and bigger than Cricket.
(3) Football is bigger than Hockey.
If the first two statements are true, the third statement is

True

False

Uncertain

Ans .

Explanation :
There are essentially two logical forks in this question: The first one comes from Clue (i), according to which the highest total of vest number and time must be 13 (as the square is 169). This could mean two scenarios: Vest number 7 with time=6; or vest number 6 with time=7. The two scenarios can be shown as:

Possibility 1A
Vest number Rank Vest number Rank

1 1

2 2

3 3

4 4

5 5

6 6 7

7 6 7

Inside these two scenarios we need to merge the other three way fork: The square of the least total of vest number and rank is 16 and occurs only once. It gives us 3 scenarios on the surface: However in this scenario we can reject Vest 3 ranking 1 because then the second ranker’s vest number would exceed the winner’s vest number (Clue (iv))—as we cannot put the 2 nd rank for either 1 or 2 as the total of 4 is the least sum and occurs only once. This leaves us with 4 principal situations 1A, 1B, 2A and 2B. Let us evaluate each of these separately:

Possibility 1A (starting thought)
Vest number Rank

1 3

2 4/5/7

3 2/4/5/7

4 1/2/4/5/7

5 1/2/4/5/7

6 1/2/4/5

7 6

In the above situation, the total of vest number + rank has to occur thrice. There are only 3 places where this can actually occur given the possibilities listed above. These are 2-7; 4-5 and 5-4. Hence, the following conclusions can be made:

Possibility 1A (starting thought)
Vest number Rank

1 3

2 7

3 2

4 5

5 4

6 1

7 6

3-2 and 6-1 get automatically selected. As for the vest number 3, only rank possibility left is 2 nd rank and after than rank 1 must go to vest number 6. If we go for a similar analysis with the other possibilities you can see that they do not work out. For instance in Possibility 1B, if we put down what is possible to be placed against vest numbers 1 and 3 respectively we get the following figure.

Possibility 1B
Vest number Rank

1 4/5/7

2 2

3 3/4/5/7

4

5

6

7 6

From this point we know that there have to be exactly 3 totals of 9, which can only be achieved by matching 4-5; 5-4 and 6-3. Then, 3 rd vest must be 7 th rank but that leaves us with no rank possibility for vest 1 and also rank 1 not getting allotted anywhere. Thus, this possibility does not exist. Similarly we can reject possibilities 2A and 2B:

Possibility 2A
Vest number Rank

1 3

2 4/5/6

3 2/4/5/6

4 1/2/4/5/6

5 1/2/4/5/6

6 7

7 1/2/4/5/6

In the above table we need three 9s. There are 4 possibilities for the sum of vest number and rank to be 9. These are 3-6, 4-5, 5-4 and 7-2. Only 3 of these 4 have to make up 9, hence we will have to leave out one of the 4. If we leave out 3-6, 4-5 or 5-4 then vest number 7 must take Position 2 (runner up). This will contradict Clue (iv) as the winner’s vest number > the runner-ups vest number. If we leave out 7-2 and place 3-6, 4-5 and 5-4 there will be no place to put rank 2. Thus, this possibility gets rejected. Possibility 2B also gets rejected as follows:

Possibility 2A
Vest number Rank

1 4/5/6

2 2

3 3/4/5/6

4 1/3/4/5/6

5 1/3/4/5/6

6 7

7 1/3/4/5/6

In the table above, there are only 3 cases of total of vest number + rank = 9. Thus, the table should become:

Possibility 2A
Vest number Rank

1 4/5/6 no place for 3rd rank??

2 2

3 6

4 5

5 4

6 7

7 1

We will get from the first two statements T>H>C>F. Thus Statement 3 is false if the first two statements are true.

Q. 28

Ramu has 3 marble boxes. (1) Box A contains more marbles than Box B. (2) Box B contains more marbles than Box C. (3) Box C contains more marbles than Box A. If the first two statements are true, the third statement is

True

False

Uncertain

Ans .

Explanation :
There are essentially two logical forks in this question: The first one comes from Clue (i), according to which the highest total of vest number and time must be 13 (as the square is 169). This could mean two scenarios: Vest number 7 with time=6; or vest number 6 with time=7. The two scenarios can be shown as:

Possibility 1A
Vest number Rank Vest number Rank

1 1

2 2

3 3

4 4

5 5

6 6 7

7 6 7

Inside these two scenarios we need to merge the other three way fork: The square of the least total of vest number and rank is 16 and occurs only once. It gives us 3 scenarios on the surface: However in this scenario we can reject Vest 3 ranking 1 because then the second ranker’s vest number would exceed the winner’s vest number (Clue (iv))—as we cannot put the 2 nd rank for either 1 or 2 as the total of 4 is the least sum and occurs only once. This leaves us with 4 principal situations 1A, 1B, 2A and 2B. Let us evaluate each of these separately:

Possibility 1A (starting thought)
Vest number Rank

1 3

2 4/5/7

3 2/4/5/7

4 1/2/4/5/7

5 1/2/4/5/7

6 1/2/4/5

7 6

In the above situation, the total of vest number + rank has to occur thrice. There are only 3 places where this can actually occur given the possibilities listed above. These are 2-7; 4-5 and 5-4. Hence, the following conclusions can be made:

Possibility 1A (starting thought)
Vest number Rank

1 3

2 7

3 2

4 5

5 4

6 1

7 6

3-2 and 6-1 get automatically selected. As for the vest number 3, only rank possibility left is 2 nd rank and after than rank 1 must go to vest number 6. If we go for a similar analysis with the other possibilities you can see that they do not work out. For instance in Possibility 1B, if we put down what is possible to be placed against vest numbers 1 and 3 respectively we get the following figure.

Possibility 1B
Vest number Rank

1 4/5/7

2 2

3 3/4/5/7

4

5

6

7 6

From this point we know that there have to be exactly 3 totals of 9, which can only be achieved by matching 4-5; 5-4 and 6-3. Then, 3 rd vest must be 7 th rank but that leaves us with no rank possibility for vest 1 and also rank 1 not getting allotted anywhere. Thus, this possibility does not exist. Similarly we can reject possibilities 2A and 2B:

Possibility 2A
Vest number Rank

1 3

2 4/5/6

3 2/4/5/6

4 1/2/4/5/6

5 1/2/4/5/6

6 7

7 1/2/4/5/6

In the above table we need three 9s. There are 4 possibilities for the sum of vest number and rank to be 9. These are 3-6, 4-5, 5-4 and 7-2. Only 3 of these 4 have to make up 9, hence we will have to leave out one of the 4. If we leave out 3-6, 4-5 or 5-4 then vest number 7 must take Position 2 (runner up). This will contradict Clue (iv) as the winner’s vest number > the runner-ups vest number. If we leave out 7-2 and place 3-6, 4-5 and 5-4 there will be no place to put rank 2. Thus, this possibility gets rejected. Possibility 2B also gets rejected as follows:

Possibility 2A
Vest number Rank

1 4/5/6

2 2

3 3/4/5/6

4 1/3/4/5/6

5 1/3/4/5/6

6 7

7 1/3/4/5/6

In the table above, there are only 3 cases of total of vest number + rank = 9. Thus, the table should become:

Possibility 2A
Vest number Rank

1 4/5/6 no place for 3rd rank??

2 2

3 6

4 5

5 4

6 7

7 1

False, as we get A>B>C and hence C>A is definitely false. Option (b)

Directions for Questions 29 to 32: Five cities all got more rain than usual this year. The five cities are: Aurangabad, Ahmednagar, Pune, Mumbai and Nagpur. The cities are located in five different areas of the country: the mountains, the forest, the coast, the desert, and in a valley. The rainfall amounts were: 24 inches, 54 inches, 64 inches, 88 inches and 130 inches.
■ The city in the desert got the least rain; the city in the forest got the most rain.
■ Pune is in the mountains.
■ Aurangabad got more rain than Mumbai.
■ Ahmednagar got more rain than Nagpur, but less rain than Pune.
■ Mumbai got 88 inches of rain.
■ The city in the mountains got 64 inches of rain; the city on the coast got 54 inches of rain.

Q.29

Which city is in the desert?

Aurangabad

Ahmednagar

Pune

Nagpur

Ans .

Explanation :

Rainfall (From Max to Min) Location City
130 Forest Aurangabad

88 Valley Mumbai

64 Mountains Pune

54 Coast Ahmednagar

24 Desert Nagpur

Nagpur (Option d)

Q. 30

Which city got the most rain?

Aurangabad

Ahmednagar

Pune

Mumbai

Ans .

Explanation :

Rainfall (From Max to Min) Location City
130 Forest Aurangabad

88 Valley Mumbai

64 Mountains Pune

54 Coast Ahmednagar

24 Desert Nagpur

Aurangabad (Option a)

Q.31

How much rain did Ahmednagar get?

24 inches

54 inches

64 inches

88 inches

Ans .

Explanation :

Rainfall (From Max to Min) Location City
130 Forest Aurangabad

88 Valley Mumbai

64 Mountains Pune

54 Coast Ahmednagar

24 Desert Nagpur

Ahmednagar ranked fourth and got 54 inches of rain. (Option b).

Q.32

Where is Mumbai located?

The mountains

The coast

In a valley

The desert

Ans .

Explanation :

Rainfall (From Max to Min) Location City
130 Forest Aurangabad

88 Valley Mumbai

64 Mountains Pune

54 Coast Ahmednagar

24 Desert Nagpur

Mumbai is located in a valley (Option c).

upscfever.com

Journey of a Thousand Miles begins with One Click!!!

Logical Reasoning based on Rankings

UPSCFEVER - POPULAR PAGES

Join The Movement

Discussions and Comments

	Runner Up
Miss UP	Miss WB	Miss AP	Miss Maharashtra
Yellow	Red	Green	White

	Costume	Person	Category best
Joker	Vladimir	Most Efficient
Wonderwoman	Sacha
		Scariest

	Possibility 1A
Vest number	Rank	Vest number	Rank
1		1
2		2
3		3
4		4
5		5
6		6	7
7	6	7

	Possibility 1A (starting thought)
Vest number	Rank
1	3
2	4/5/7
3	2/4/5/7
4	1/2/4/5/7
5	1/2/4/5/7
6	1/2/4/5
7	6

	Possibility 2A
Vest number	Rank
1	3
2	4/5/6
3	2/4/5/6
4	1/2/4/5/6
5	1/2/4/5/6
6	7
7	1/2/4/5/6

	Rainfall (From Max to Min)	Location	City
130	Forest	Aurangabad
88	Valley	Mumbai
64	Mountains	Pune
54	Coast	Ahmednagar
24	Desert	Nagpur