SBI PO Mains Memory Based Paper (Held On: 20 July 2019) (Previous Year Paper)

The logic followed here is:

Step I: The second and third digits of each number are interchanged with the fifth and sixth digits respectively.

Step II: The numbers are arranged in ascending order.

Step III: In each number, the odd digits are arranged in ascending order followed by the arrangement of even digits in ascending order.

Step IV: The first and last digits are added and their sum is placed as the first digit. This operation is performed with every 6-digit input number.

Step V: The 5-digit numbers obtained in step IV are arranged in ascending order.

Input: 214261 130145 333421 715620 312451 123456

Step I: 261214 145130 321433 720615 351412 156423

Step II: 145130 156423 261214 321433 351412 720615

Step III: 113504 135246 112246 133324 113524 157026

Step IV: 51350 73524 71224 53332 51352 75702

Step V: 51350 51352 53332 71224 73524 75702

Step V is the final arrangement.

Step I: 261214 145130 321433 720615 351412 156423

The extreme left number in step I is 261214.

Sum of the digits = 2 + 6 + 1 + 2 + 1 + 4 = 16.

Hence, the required sum is 16.

The logic followed here is:

Step I: The second and third digits of each number are interchanged with the fifth and sixth digits respectively.

Step II: The numbers are arranged in ascending order.

Step III: In each number, the odd digits are arranged in ascending order followed by the arrangement of even digits in ascending order.

Step IV: The first and last digits are added and their sum is placed as the first digit. This operation is performed with every 6-digit input number.

Step V: The 5-digit numbers obtained in step IV are arranged in ascending order.

Input: 214261 130145 333421 715620 312451 123456

Step I: 261214 145130 321433 720615 351412 156423

Step II: 145130 156423 261214 321433 351412 720615

Step III: 113504 135246 112246 133324 113524 157026

Step IV: 51350 73524 71224 53332 51352 75702

Step V: 51350 51352 53332 71224 73524 75702

Step V is the final arrangement.

Hence, the fourth number from the left in step II is 321433.

The logic followed here is:

Step I: The second and third digits of each number are interchanged with the fifth and sixth digits respectively.

Step II: The numbers are arranged in ascending order.

Step III: In each number, the odd digits are arranged in ascending order followed by the arrangement of even digits in ascending order.

Step IV: The first and last digits are added and their sum is placed as the first digit. This operation is performed with every 6-digit input number.

Step V: The 5-digit numbers obtained in step IV are arranged in ascending order.

Input: 214261 130145 333421 715620 312451 123456

Step I: 261214 145130 321433 720615 351412 156423

Step II: 145130 156423 261214 321433 351412 720615

Step III: 113504 135246 112246 133324 113524 157026

Step IV: 51350 73524 71224 53332 51352 75702

Step V: 51350 51352 53332 71224 73524 75702

Step V is the final arrangement.

Clearly, every number in step III is an even number.

Hence, there are six even numbers in step III.

The logic followed here is:

Step I: The second and third digits of each number are interchanged with the fifth and sixth digits respectively.

Step II: The numbers are arranged in ascending order.

Step III: In each number, the odd digits are arranged in ascending order followed by the arrangement of even digits in ascending order.

Step IV: The first and last digits are added and their sum is placed as the first digit. This operation is performed with every 6-digit input number.

Step V: The 5-digit numbers obtained in step IV are arranged in ascending order.

Input: 214261 130145 333421 715620 312451 123456

Step I: 261214 145130 321433 720615 351412 156423

Step II: 145130 156423 261214 321433 351412 720615

Step III: 113504 135246 112246 133324 113524 157026

Step IV: 51350 73524 71224 53332 51352 75702

Step V: 51350 51352 53332 71224 73524 75702

Step V is the final arrangement.

The first number from the left in step IV is 51350.

The fourth number from the left in step IV is 53332.

Their difference = 53332 - 51350 = 1982.

Hence, 1982 is the required answer.

The logic followed here is:

Step I: The second and third digits of each number are interchanged with the fifth and sixth digits respectively.

Step II: The numbers are arranged in ascending order.

Step III: In each number, the odd digits are arranged in ascending order followed by the arrangement of even digits in ascending order.

Step IV: The first and last digits are added and their sum is placed as the first digit. This operation is performed with every 6-digit input number.

Step V: The 5-digit numbers obtained in step IV are arranged in ascending order.

Input: 214261 130145 333421 715620 312451 123456

Step I: 261214 145130 321433 720615 351412 156423

Step II: 145130 156423 261214 321433 351412 720615

Step III: 113504 135246 112246 133324 113524 157026

Step IV: 51350 73524 71224 53332 51352 75702

Step V: 51350 51352 53332 71224 73524 75702

Step V is the final arrangement.

Numbers having repeated digits in step IV:

Step IV: 51350 73524 71224 53332 51352 75702

Hence, there are five numbers with repeated digits in step IV.

The diagram for the given direction problem is as follows:

Hence, W is 10 m far with respect to J.

The diagram for the given direction problem is as follows:

Hence, P is in west direction with respect to M.

The diagram for the given direction problem is as follows:

Hence, V $ U is the true statement.

1. 1^st element is number that represents

Number of letters in a word – 2 (if odd no. of letters in the word).

Number of letters in a word + 1 (if even no. of letters in the word).

2. 2^nd element is the letter which is at the same position from the end of the English alphabet series as that of the last letter in the word and is written in its capital form.

3. 3^rd element is symbol that represents

@ → if even no. of vowels in the word.

& → if odd no. of vowels in the word.

Ex:

9O@ for “material”

No. of letters in material is 8.

9 → 8 + 1

Last letter in material is “l”.

“l” is 12^th letter from the starting of alphabet.

The 12^th letter from end of the alphabet is “O”.

Four vowels (a, e, i, a) are present in material.

@ → even number of vowels.

3S@ for “beach”

No. of letters in beach is 5.

3 → 5 – 2

Last letter in beach is “h”.

“h” is 8^th letter from the starting.

The 8^th letter from the end is “S”.

Two vowels (a, e) are present in beach.

@ → even number of vowels. 

Following the same logic “independent” can be written as:

No. of letters in independent is 11.

9 → 11 – 2

Last letter in independent is “t”.

“t” is 20^th element from the starting.

20^th element from the end is “G”.

Four vowels (i, e, e, e) are present in independent.

@ → even number of vowels.

Hence, independent can be coded as “9G@”.

1. 1st element is number that represents

Number of letters in a word – 2 (if odd no. of letters in the word).

Number of letters in a word + 1 (if even no. of letters in the word).

2. 2nd element is the letter which is at the same position from the end of the English alphabet series as that of the last letter in the word and is written in its capital form.

3. 3rd element is symbol that represents

@ → if even no. of vowels in the word.

& → if odd no. of vowels in the word.

Ex:

9O@ for “material”

No. of letters in material is 8.

9 → 8 + 1

Last letter in material is “l”.

“l” is 12th letter from the starting of alphabet.

The 12th letter from end of the alphabet is “O”.

Four vowels (a, e, i, a) are present in material.

@ → even number of vowels.

3S@ for “beach”

No. of letters in beach is 5.

3 → 5 – 2

Last letter in beach is “h”.

“h” is 8th letter from the starting.

The 8th letter from the end is “S”.

Two vowels (a, e) are present in beach.

@ → even number of vowels.

“7I@” is code for flower.

Number of letters in flower is 6.

7 → 6 + 1

Last letter in flower is “r”.

“r” is 18^th letter from the starting.

18^th letter from the end is “I”.

Two vowels (o, e) are present in flower.

@ → even no. of vowels.

Hence, “7I@” is code for flower.

1. 1st element is number that represents

Number of letters in a word – 2 (if odd no. of letters in the word).

Number of letters in a word + 1 (if even no. of letters in the word).

2. 2nd element is the letter which is at the same position from the end of the English alphabet series as that of the last letter in the word and is written in its capital form.

3. 3rd element is symbol that represents

@ → if even no. of vowels in the word.

& → if odd no. of vowels in the word.

Ex:

9O@ for “material”

No. of letters in material is 8.

9 → 8 + 1

Last letter in material is “l”.

“l” is 12th letter from the starting of alphabet.

The 12th letter from end of the alphabet is “O”.

Four vowels (a, e, i, a) are present in material.

@ → even number of vowels.

3S@ for “beach”

No. of letters in beach is 5.

3 → 5 – 2

Last letter in beach is “h”.

“h” is 8th letter from the starting.

The 8th letter from the end is “S”.

Two vowels (a, e) are present in beach.

@ → even number of vowels.

In similar way,

she → 1V&

is → 3H&

innocent → 9G&

girl → 5O&

Hence, code for “she is innocent girl” is “1V&, 5O&, 9G&, 3H&”.

1. 1st element is number that represents

Number of letters in a word – 2 (if odd no. of letters in the word).

Number of letters in a word + 1 (if even no. of letters in the word).

2. 2nd element is the letter which is at the same position from the end of the English alphabet series as that of the last letter in the word and is written in its capital form.

3. 3rd element is symbol that represents

@ → if even no. of vowels in the word.

& → if odd no. of vowels in the word.

Ex:

9O@ for “material”

No. of letters in material is 8.

9 → 8 + 1

Last letter in material is “l”.

“l” is 12th letter from the starting of alphabet.

The 12th letter from end of the alphabet is “O”.

Four vowels (a, e, i, a) are present in material.

@ → even number of vowels.

3S@ for “beach”

No. of letters in beach is 5.

3 → 5 – 2

Last letter in beach is “h”.

“h” is 8th letter from the starting.

The 8th letter from the end is “S”.

Two vowels (a, e) are present in beach.

@ → even number of vowels.

In similar way,

job → 1Y&

is → 3H&

important → 7G&

for → 1I&

all → 1O&

Hence, code for “job is important for all” is 1O&, 1Y&, 7G&, 3H&, 1I&”.

1. 1st element is number that represents

Number of letters in a word – 2 (if odd no. of letters in the word).

Number of letters in a word + 1 (if even no. of letters in the word).

2. 2nd element is the letter which is at the same position from the end of the English alphabet series as that of the last letter in the word and is written in its capital form.

3. 3rd element is symbol that represents

@ → if even no. of vowels in the word.

& → if odd no. of vowels in the word.

Ex:

9O@ for “material”

No. of letters in material is 8.

9 → 8 + 1

Last letter in material is “l”.

“l” is 12th letter from the starting of alphabet.

The 12th letter from end of the alphabet is “O”.

Four vowels (a, e, i, a) are present in material.

@ → even number of vowels.

3S@ for “beach”

No. of letters in beach is 5.

3 → 5 – 2

Last letter in beach is “h”.

“h” is 8th letter from the starting.

The 8th letter from the end is “S”.

Two vowels (a, e) are present in beach.

@ → even number of vowels.

No. of letters in fluctuations is 12.

13 → 12 + 1

Last letter in fluctuations is “s”.

“s” is 19^th letter from the starting.

19^th letter from the end is “H”.

Five vowels (u, u, a, i, o) are present in fluctuations.

& → odd number of vowels.

Hence, code for “fluctuations” is “13H&”.

There is no information given on the work experience of the creative artist or that of the manager, therefore, options 1 and 2 do not follow. 

Being the manager of a company, he possibly has some knowledge about the relation between the two companies (XYZ and ABC) working in the industry and hence the response. Therefore, option 3 follows.

There is no reason to comment on the rivalry between the companies XYZ and ABC, the manager could instead talk about the quality of the content of XYZ and hence option 4 does not follow.

The manager's reply nowhere indicates a lack of trust in the artist and hence option 5 does not follow.

The argument states that homeowners are more likely to purchase ice-cream and in larger amounts. It also states that the common belief that teenagers eat more ice-cream than adults must be false, as a result. This is flawed as the study makes no distinction between purchasing and consuming. The demographic age group could be purchasing more ice cream but that does not necessarily imply that they are eating more of it – they could be purchasing it for their children, who could also be teenagers. Thus option 1 is correct.

Improving news coverage and making it succinct, crisp, factual and riveting can encourage people to start reading newspapers again.

Option 3 is the correct option as it does not talk about the long term effects caused due to the bomb blast whereas the other 4 options do. Option 3 tells us that due to this bombing the ground temperature had reached to a high level but it does not talk or give information on how it had affected the humans so it becomes irrelevant.

Other options could be the possible effects of this bombing on humans such as Leukaemia, an increase of cancer rates among survivors, miscarriage and such.

Answer: 3. It is more comfortable for the youth today to interact with less emotional attachment.

Solution: The given statements suggest 80% of youth today prefers online socializing, therefore option 1 is incorrect. The youth prefers online networking because they think it requires less emotional involvement; hence option 2 cannot be correct. Option 4 can also be eliminated because the statements clearly mention the reason behind youth preferring online interaction is based on emotional attachment and not about personal interaction. Option 5 is not mentioned in the statements. The crux of the statements suggest that the youth today prefers socializing online because it does not require much emotional involvement. Hence, option 3 is correct.

Effect of complaints cannot be assumed as nothing is given about time frames or consequences in the statement. So assumption II is not implicit. The words ‘in spite of…employees have not complained’ tells us that this contrary to norm. Thus generally people must complain. Hence assumption I is implicit.

Information is insufficient to assume the performance of the odd even traffic system from the given statement. Thus this assumption does not hold.

On the other hand, since it has already been mentioned that it has received mixed response it means that some have welcomed it while others haven't.

Thus neither assumptions hold.

For students to get admission in there should be a sufficient number of colleges.

A decrease in the number of schools will not solve the problem.

Vocational courses need to be conducted by the schools for the students who can prepare to get employed after completing school education.

Hence, only conclusions I and III follow.

8 persons: Mridul, Zahida, Nikeeta, Ram, Vishnu, Satya, Hiteshri and Rashi

States: West Bengal, Orissa, Assam

Languages: Chinese, Arabic, Marathi, Spanish, Bengali, English, Hindi and French

People having a blood relation reside in the same state.

At least two people reside in the same state and not more than three persons reside in the same state.

1. Nikeeta resides in West Bengal and is the only female child in her family.

2) Nikeeta likes Hindi.

3) The sibling of Nikeeta likes English.

Therefore, the sibling must be Nikeeta's brother.

4) Ram is the brother of Hiteshri, who likes Chinese. Ram belongs to Orissa.

5) Zahida is the daughter of Satya, who likes French.

Therefore, Zahida is a female and cannot belong to West Bengal.

Also, Zahida and her mother belong to the same state according to the given information.

Because not more than three persons reside in the same state, Zahida and her brother must belong to Assam.

6) Vishnu is a married man. Vishnu belongs to Assam.

7) Zahida's father likes Spanish.

Therefore, Vishnu is Zahida's father and he likes Spanish.

8) The sister of Rashi likes Chinese.

As Nikeeta is the only female child, Rashi and her sister must belong to Orissa.

Therefore, Hiteshri is the sister of Rashi.

9) The family members of Hiteshri do not like Bengali.

Therefore, Zahida likes Bengali.

Now, the only person left is Mridul who must be the sibling of Nikeeta.

As Nikeeta is an only female child, Mridul must be her brother.

10) The Arabic language is liked by a female.

As Ram is a male, Rashi must be the female who likes Arabic.

Therefore, Ram likes Marathi.

Hence, Rashi is the sister of Hiteshri.

8 persons: Mridul, Zahida, Nikeeta, Ram, Vishnu, Satya, Hiteshri and Rashi

States: West Bengal, Orissa, Assam

Languages: Chinese, Arabic, Marathi, Spanish, Bengali, English, Hindi and French

People having a blood relation reside in the same state.

At least two people reside in the same state and not more than three persons reside in the same state.

1. Nikeeta resides in West Bengal and is the only female child in her family.

2) Nikeeta likes Hindi.

3) The sibling of Nikeeta likes English.

Therefore, the sibling must be Nikeeta's brother.

4) Ram is the brother of Hiteshri, who likes Chinese. Ram belongs to Orissa.

5) Zahida is the daughter of Satya, who likes French.

Therefore, Zahida is a female and cannot belong to West Bengal.

Also, Zahida and her mother belong to the same state according to the given information.

Because not more than three persons reside in the same state, Zahida and her brother must belong to Assam.

6) Vishnu is a married man. Vishnu belongs to Assam.

7) Zahida's father likes Spanish.

Therefore, Vishnu is Zahida's father and he likes Spanish.

8) The sister of Rashi likes Chinese.

As Nikeeta is the only female child, Rashi and her sister must belong to Orissa.

Therefore, Hiteshri is the sister of Rashi.

9) The family members of Hiteshri do not like Bengali.

Therefore, Zahida likes Bengali.

Now, the only person left is Mridul who must be the sibling of Nikeeta.

As Nikeeta is an only female child, Mridul must be her brother.

10) The Arabic language is liked by a female.

As Ram is a male, Rashi must be the female who likes Arabic.

Therefore, Ram likes Marathi.

Hence, Zahida likes Bengali.

8 persons: Mridul, Zahida, Nikeeta, Ram, Vishnu, Satya, Hiteshri and Rashi

States: West Bengal, Orissa, Assam

Languages: Chinese, Arabic, Marathi, Spanish, Bengali, English, Hindi and French

People having a blood relation reside in the same state.

At least two people reside in the same state and not more than three persons reside in the same state.

1. Nikeeta resides in West Bengal and is the only female child in her family.

2) Nikeeta likes Hindi.

3) The sibling of Nikeeta likes English.

Therefore, the sibling must be Nikeeta's brother.

4) Ram is the brother of Hiteshri, who likes Chinese. Ram belongs to Orissa.

5) Zahida is the daughter of Satya, who likes French.

Therefore, Zahida is a female and cannot belong to West Bengal.

Also, Zahida and her mother belong to the same state according to the given information.

Because not more than three persons reside in the same state, Zahida and her brother must belong to Assam.

6) Vishnu is a married man. Vishnu belongs to Assam.

7) Zahida's father likes Spanish.

Therefore, Vishnu is Zahida's father and he likes Spanish.

8) The sister of Rashi likes Chinese.

As Nikeeta is the only female child, Rashi and her sister must belong to Orissa.

Therefore, Hiteshri is the sister of Rashi.

9) The family members of Hiteshri do not like Bengali.

Therefore, Zahida likes Bengali.

Now, the only person left is Mridul who must be the sibling of Nikeeta.

As Nikeeta is an only female child, Mridul must be her brother.

10) The Arabic language is liked by a female.

As Ram is a male, Rashi must be the female who likes Arabic.

Therefore, Ram likes Marathi.

Hence, Ram resides in Orissa.

8 persons: Mridul, Zahida, Nikeeta, Ram, Vishnu, Satya, Hiteshri and Rashi

States: West Bengal, Orissa, Assam

Languages: Chinese, Arabic, Marathi, Spanish, Bengali, English, Hindi and French

People having a blood relation reside in the same state.

At least two people reside in the same state and not more than three persons reside in the same state.

1. Nikeeta resides in West Bengal and is the only female child in her family.

2) Nikeeta likes Hindi.

3) The sibling of Nikeeta likes English.

Therefore, the sibling must be Nikeeta's brother.

4) Ram is the brother of Hiteshri, who likes Chinese. Ram belongs to Orissa.

5) Zahida is the daughter of Satya, who likes French.

Therefore, Zahida is a female and cannot belong to West Bengal.

Also, Zahida and her mother belong to the same state according to the given information.

Because not more than three persons reside in the same state, Zahida and her brother must belong to Assam.

6) Vishnu is a married man. Vishnu belongs to Assam.

7) Zahida's father likes Spanish.

Therefore, Vishnu is Zahida's father and he likes Spanish.

8) The sister of Rashi likes Chinese.

As Nikeeta is the only female child, Rashi and her sister must belong to Orissa.

Therefore, Hiteshri is the sister of Rashi.

9) The family members of Hiteshri do not like Bengali.

Therefore, Zahida likes Bengali.

Now, the only person left is Mridul who must be the sibling of Nikeeta.

As Nikeeta is an only female child, Mridul must be her brother.

10) The Arabic language is liked by a female.

As Ram is a male, Rashi must be the female who likes Arabic.

Therefore, Ram likes Marathi.

Hence, Mridul is the brother of Nikeeta.

8 persons: Mridul, Zahida, Nikeeta, Ram, Vishnu, Satya, Hiteshri and Rashi

States: West Bengal, Orissa, Assam

Languages: Chinese, Arabic, Marathi, Spanish, Bengali, English, Hindi and French

People having a blood relation reside in the same state.

At least two people reside in the same state and not more than three persons reside in the same state.

1. Nikeeta resides in West Bengal and is the only female child in her family.

2) Nikeeta likes Hindi.

3) The sibling of Nikeeta likes English.

Therefore, the sibling must be Nikeeta's brother.

4) Ram is the brother of Hiteshri, who likes Chinese. Ram belongs to Orissa.

5) Zahida is the daughter of Satya, who likes French.

Therefore, Zahida is a female and cannot belong to West Bengal.

Also, Zahida and her mother belong to the same state according to the given information.

Because not more than three persons reside in the same state, Zahida and her brother must belong to Assam.

6) Vishnu is a married man. Vishnu belongs to Assam.

7) Zahida's father likes Spanish.

Therefore, Vishnu is Zahida's father and he likes Spanish.

8) The sister of Rashi likes Chinese.

As Nikeeta is the only female child, Rashi and her sister must belong to Orissa.

Therefore, Hiteshri is the sister of Rashi.

9) The family members of Hiteshri do not like Bengali.

Therefore, Zahida likes Bengali.

Now, the only person left is Mridul who must be the sibling of Nikeeta.

As Nikeeta is an only female child, Mridul must be her brother.

10) The Arabic language is liked by a female.

As Ram is a male, Rashi must be the female who likes Arabic.

Therefore, Ram likes Marathi.

Hence, Rashi likes Arabic.

8 persons: A, B, C, D, E, F, G and H

Not more than 4 persons stand on a gate implies that 4 persons are standing on each of the gates.

Also, it is given that the persons standing on the outer gate are facing those on the inner gate.

1) G is facing A who is in the outer gate.

2) H, who is not facing B is 2nd to the left of E who is facing outside.

3. C and F are looking at the same side.

Therefore, C and F are sitting on the outer gate.

4. C is one of the neighbours of D who is not a neighbour of A.

5) E is not facing C. D is facing the same side as C.

6) The person facing the immediate right neighbour of E is not D.

Therefore, case 2 gets eliminated.

This is the final arrangement.

Hence, G and B are the neighbours of E.

8 persons: A, B, C, D, E, F, G and H

Not more than 4 persons stand on a gate implies that 4 persons are standing on each of the gates.

Also, it is given that the persons standing on the outer gate are facing those on the inner gate.

1) G is facing A who is in the outer gate.

2) H, who is not facing B is 2nd to the left of E who is facing outside.

3. C and F are looking at the same side.

Therefore, C and F are sitting on the outer gate.

4. C is one of the neighbours of D who is not a neighbour of A.

5) E is not facing C. D is facing the same side as C.

6) The person facing the immediate right neighbour of E is not D.

Therefore, case 2 gets eliminated.

This is the final arrangement.

Hence, A is facing G.

8 persons: A, B, C, D, E, F, G and H

Not more than 4 persons stand on a gate implies that 4 persons are standing on each of the gates.

Also, it is given that the persons standing on the outer gate are facing those on the inner gate.

1) G is facing A who is in the outer gate.

2) H, who is not facing B is 2nd to the left of E who is facing outside.

3. C and F are looking at the same side.

Therefore, C and F are sitting on the outer gate.

4. C is one of the neighbours of D who is not a neighbour of A.

5) E is not facing C. D is facing the same side as C.

6) The person facing the immediate right neighbour of E is not D.

Therefore, case 2 gets eliminated.

This is the final arrangement.

In each of the options 2, 3, 4 and 5, the persons are facing each other. But in option 1, E and H are not facing each other.

Hence, E and H is the odd pair.

8 persons: A, B, C, D, E, F, G and H

Not more than 4 persons stand on a gate implies that 4 persons are standing on each of the gates.

Also, it is given that the persons standing on the outer gate are facing those on the inner gate.

1) G is facing A who is in the outer gate.

2) H, who is not facing B is 2nd to the left of E who is facing outside.

3. C and F are looking at the same side.

Therefore, C and F are sitting on the outer gate.

4. C is one of the neighbours of D who is not a neighbour of A.

5) E is not facing C. D is facing the same side as C.

6) The person facing the immediate right neighbour of E is not D.

Therefore, case 2 gets eliminated.

This is the final arrangement.

A, C, D and F are standing on the outer gate.

Hence, ACDF is the correct answer.

8 persons: A, B, C, D, E, F, G and H

Not more than 4 persons stand on a gate implies that 4 persons are standing on each of the gates.

Also, it is given that the persons standing on the outer gate are facing those on the inner gate.

1) G is facing A who is in the outer gate.

2) H, who is not facing B is 2nd to the left of E who is facing outside.

3. C and F are looking at the same side.

Therefore, C and F are sitting on the outer gate.

4. C is one of the neighbours of D who is not a neighbour of A.

5) E is not facing C. D is facing the same side as C.

6) The person facing the immediate right neighbour of E is not D.

Therefore, case 2 gets eliminated.

This is the final arrangement.

B, H, G and E are sitting on the inner gate.

Hence, BHGE is the correct answer.

Hence, the starting point is in the North-West direction with respect to the endpoint.

7 persons: A, C, F, J, N, R and V.

1. J sits third from the extreme end.

2. V's father sits second to the right of J.

3) Only 3 people are sitting between V’s father and V’s sister.

4) As many people sit to the left of V’s sister as to the right of V’s mother.

5) Only one person sits between N and V’s mother.

6) N is the son of C.

7) A sits third to the right of C.

8. C is the father of R.

9) F sits to the immediate right of R.

10) V’s wife sits second to the left of V.

Therefore, case 2 gets eliminated as C is a male.

Hence, V is sitting adjacent to the father of V who is N.

7 persons: A, C, F, J, N, R and V.

1. J sits third from the extreme end.

2. V's father sits second to the right of J.

3) Only 3 people are sitting between V’s father and V’s sister.

4) As many people sit to the left of V’s sister as to the right of V’s mother.

5) Only one person sits between N and V’s mother.

6) N is the son of C.

7) A sits third to the right of C.

8. C is the father of R.

9) F sits to the immediate right of R.

10) V’s wife sits second to the left of V.

Therefore, case 2 gets eliminated as C is a male.

Hence, A is the wife of V.

7 persons: A, C, F, J, N, R and V.

1. J sits third from the extreme end.

2. V's father sits second to the right of J.

3) Only 3 people are sitting between V’s father and V’s sister.

4) As many people sit to the left of V’s sister as to the right of V’s mother.

5) Only one person sits between N and V’s mother.

6) N is the son of C.

7) A sits third to the right of C.

8. C is the father of R.

9) F sits to the immediate right of R.

10) V’s wife sits second to the left of V.

Therefore, case 2 gets eliminated as C is a male.

Hence, F is sitting exactly between C and V's mother.

7 persons: A, C, F, J, N, R and V.

1. J sits third from the extreme end.

2. V's father sits second to the right of J.

3) Only 3 people are sitting between V’s father and V’s sister.

4) As many people sit to the left of V’s sister as to the right of V’s mother.

5) Only one person sits between N and V’s mother.

6) N is the son of C.

7) A sits third to the right of C.

8. C is the father of R.

9) F sits to the immediate right of R.

10) V’s wife sits second to the left of V.

Therefore, case 2 gets eliminated as C is a male.

Hence, V's father is sitting at the extreme right end.

7 persons: A, C, F, J, N, R and V.

1. J sits third from the extreme end.

2. V's father sits second to the right of J.

3) Only 3 people are sitting between V’s father and V’s sister.

4) As many people sit to the left of V’s sister as to the right of V’s mother.

5) Only one person sits between N and V’s mother.

6) N is the son of C.

7) A sits third to the right of C.

8. C is the father of R.

9) F sits to the immediate right of R.

10) V’s wife sits second to the left of V.

Therefore, case 2 gets eliminated as C is a male.

Hence, R is sitting second to the left of A.

Seven persons: P, Q, R, S, T, U and V facing the centre of the table.

There are 12 seats because the seating arrangement is in the form of a clock.

1) R is to the immediate left of U who is sitting at 5 O’ clock seat.

2) There is a three hours' difference between R and T’s seat.

3) T is sitting diagonally opposite to Q.

4) Q is to the right of R.

Therefore, case 2 gets eliminated.

5) There is a 6 hours gap between the seats of P and R.

6. V is sitting to the immediate right of P.

7. S is sitting on the seat which is 4 hours less than P.

Therefore, S is sitting at the seat which corresponds to P.

This is the final arrangement.

Hence, U is sitting to the immediate right of R.

Seven persons: P, Q, R, S, T, U and V facing the centre of the table.

There are 12 seats because the seating arrangement is in the form of a clock.

1) R is to the immediate left of U who is sitting at 5 O’ clock seat.

2) There is a three hours' difference between R and T’s seat.

3) T is sitting diagonally opposite to Q.

4) Q is to the right of R.

Therefore, case 2 gets eliminated.

5) There is a 6 hours gap between the seats of P and R.

6. V is sitting to the immediate right of P.

7. S is sitting on the seat which is 4 hours less than P.

Therefore, S is sitting at the seat which corresponds to P.

This is the final arrangement.

Hence, the hour-difference between T and U is 4 hours.

Seven persons: P, Q, R, S, T, U and V facing the centre of the table.

There are 12 seats because the seating arrangement is in the form of a clock.

1) R is to the immediate left of U who is sitting at 5 O’ clock seat.

2) There is a three hours' difference between R and T’s seat.

3) T is sitting diagonally opposite to Q.

4) Q is to the right of R.

Therefore, case 2 gets eliminated.

5) There is a 6 hours gap between the seats of P and R.

6. V is sitting to the immediate right of P.

7. S is sitting on the seat which is 4 hours less than P.

Therefore, S is sitting at the seat which corresponds to P.

This is the final arrangement.

The seat of S corresponds to 8 o' clock.

Hence, 8 is the correct answer.

1. Two boxes are placed between A and C. C is kept at the bottom of Stack 2.

2. Only three boxes are kept between B and E.  E is kept at the bottom but not at the Stack 3.

3. G is not kept in the same stack in which B kept.  G is not kept in the same stack in which A kept.

4. Height of G is 36 cm.

5. O is immediately below G and only one box is kept below O. Number of boxes above G is same as below G. (We can conclude only 5 box are there in each stack)

6. Height of box at the top of the Stack 1 is 50.

7. Less than two boxes are kept below F. F is not immediately above C. (This conclude F is either in stack 1 or stack 3 and either one or no box is kept below F)

8)  As many boxes below F as below D.

By statement 7) and 8) we can conclude only one box is placed below D and F. In stack 3 O is placed so we place F in stack 1)

9. Only one box is kept between J and M. M is not at the top.

10. More than one box is kept between X and Z.

11. Number of boxes above X is same as number of boxes above W.

12. N is not placed at the top.

13. Height of Z is 63 cm. 

14. F’s height is an even number and more than G. Height of F is less than 40 cm. (Height of F is 38 cm)

15. Height of box at the top of the Stack 1 is 50.

16. Difference between the height of box C and G is 25 cm. Height of box C is greater than G. (C – G =25

We know height of G so C – 36 = 25 → C = 61)

15. Sum of box immediately below F and immediately below D is 113 cm. (F + C = 113 so F + 61 = 113)

16. Sum of top and bottom box of stack 2 is 153 cm.  ( J + C =153 so J + 61 =153)

17. Height of M is 15 cm less than height of E.

18. Difference of A and D is 8 cm. Height of D is more than that of A. ( D + A = 110 , D – A =8)

19. N height 16 more than D’s height.

20. Height of X is 3 cm more than height of A.

Note : For height of W =  300 – 215 = 85 cm

21. Height of U is same as height of W

Finally, Height of O = 300 – 238 = 62 cm

Sum of height of the box which is immediate above M and the box which is immediate below W

= 51 + 75  = 126 cm

Hence, ‘126’ is the required sum.

1. Two boxes are placed between A and C. C is kept at the bottom of Stack 2.

2. Only three boxes are kept between B and E.  E is kept at the bottom but not at the Stack 3.

3. G is not kept in the same stack in which B kept.  G is not kept in the same stack in which A kept.

4. Height of G is 36 cm.

5. O is immediately below G and only one box is kept below O. Number of boxes above G is same as below G. (We can conclude only 5 box are there in each stack)

6. Height of box at the top of the Stack 1 is 50.

7. Less than two boxes are kept below F. F is not immediately above C. (This conclude F is either in stack 1 or stack 3 and either one or no box is kept below F)

8)  As many boxes below F as below D.

By statement 7) and 8) we can conclude only one box is placed below D and F. In stack 3 O is placed so we place F in stack 1)

9. Only one box is kept between J and M. M is not at the top.

10. More than one box is kept between X and Z.

11. Number of boxes above X is same as number of boxes above W.

12. N is not placed at the top.

13. Height of Z is 63 cm. 

14. F’s height is an even number and more than G. Height of F is less than 40 cm. (Height of F is 38 cm)

15. Height of box at the top of the Stack 1 is 50.

16. Difference between the height of box C and G is 25 cm. Height of box C is greater than G. (C – G =25

We know height of G so C – 36 = 25 → C = 61)

15. Sum of box immediately below F and immediately below D is 113 cm. (F + C = 113 so F + 61 = 113)

16. Sum of top and bottom box of stack 2 is 153 cm.  ( J + C =153 so J + 61 =153)

17. Height of M is 15 cm less than height of E.

18. Difference of A and D is 8 cm. Height of D is more than that of A. ( D + A = 110 , D – A =8)

19. N height 16 more than D’s height.

20. Height of X is 3 cm more than height of A.

Note : For height of W =  300 – 215 = 85 cm

21. Height of U is same as height of W

Finally, Height of O = 300 – 238 = 62 cm

Hence, X is kept immediately below U.

1. Two boxes are placed between A and C. C is kept at the bottom of Stack 2.

2. Only three boxes are kept between B and E.  E is kept at the bottom but not at the Stack 3.

3. G is not kept in the same stack in which B kept.  G is not kept in the same stack in which A kept.

4. Height of G is 36 cm.

5. O is immediately below G and only one box is kept below O. Number of boxes above G is same as below G. (We can conclude only 5 box are there in each stack)

6. Height of box at the top of the Stack 1 is 50.

7. Less than two boxes are kept below F. F is not immediately above C. (This conclude F is either in stack 1 or stack 3 and either one or no box is kept below F)

8)  As many boxes below F as below D.

By statement 7) and 8) we can conclude only one box is placed below D and F. In stack 3 O is placed so we place F in stack 1)

9. Only one box is kept between J and M. M is not at the top.

10. More than one box is kept between X and Z.

11. Number of boxes above X is same as number of boxes above W.

12. N is not placed at the top.

13. Height of Z is 63 cm. 

14. F’s height is an even number and more than G. Height of F is less than 40 cm. (Height of F is 38 cm)

15. Height of box at the top of the Stack 1 is 50.

16. Difference between the height of box C and G is 25 cm. Height of box C is greater than G. (C – G =25

We know height of G so C – 36 = 25 → C = 61)

15. Sum of box immediately below F and immediately below D is 113 cm. (F + C = 113 so F + 61 = 113)

16. Sum of top and bottom box of stack 2 is 153 cm.  ( J + C =153 so J + 61 =153)

17. Height of M is 15 cm less than height of E.

18. Difference of A and D is 8 cm. Height of D is more than that of A. ( D + A = 110 , D – A =8)

19. N height 16 more than D’s height.

20. Height of X is 3 cm more than height of A.

Note : For height of W =  300 – 215 = 85 cm

21. Height of U is same as height of W

Finally, Height of O = 300 – 238 = 62 cm

J and U are kept at the top of the stack.

Hence, both 1) and 3) are kept at the top of the stacks.

1. Two boxes are placed between A and C. C is kept at the bottom of Stack 2.

2. Only three boxes are kept between B and E.  E is kept at the bottom but not at the Stack 3.

3. G is not kept in the same stack in which B kept.  G is not kept in the same stack in which A kept.

4. Height of G is 36 cm.

5. O is immediately below G and only one box is kept below O. Number of boxes above G is same as below G. (We can conclude only 5 box are there in each stack)

6. Height of box at the top of the Stack 1 is 50.

7. Less than two boxes are kept below F. F is not immediately above C. (This conclude F is either in stack 1 or stack 3 and either one or no box is kept below F)

8)  As many boxes below F as below D.

By statement 7) and 8) we can conclude only one box is placed below D and F. In stack 3 O is placed so we place F in stack 1)

9. Only one box is kept between J and M. M is not at the top.

10. More than one box is kept between X and Z.

11. Number of boxes above X is same as number of boxes above W.

12. N is not placed at the top.

13. Height of Z is 63 cm. 

14. F’s height is an even number and more than G. Height of F is less than 40 cm. (Height of F is 38 cm)

15. Height of box at the top of the Stack 1 is 50.

16. Difference between the height of box C and G is 25 cm. Height of box C is greater than G. (C – G =25

We know height of G so C – 36 = 25 → C = 61)

15. Sum of box immediately below F and immediately below D is 113 cm. (F + C = 113 so F + 61 = 113)

16. Sum of top and bottom box of stack 2 is 153 cm.  ( J + C =153 so J + 61 =153)

17. Height of M is 15 cm less than height of E.

18. Difference of A and D is 8 cm. Height of D is more than that of A. ( D + A = 110 , D – A =8)

19. N height 16 more than D’s height.

20. Height of X is 3 cm more than height of A.

Note : For height of W =  300 – 215 = 85 cm

21. Height of U is same as height of W

Finally, Height of O = 300 – 238 = 62 cm

Boxes which are at the middle N = 75 cm , M = 37 cm, G = 36 cm

Sum = 75 + 37 + 36 = 148 cm

Hence, sum of height of all the boxes which are placed in the middle of each stack is 148 cm

1. Two boxes are placed between A and C. C is kept at the bottom of Stack 2.

2. Only three boxes are kept between B and E.  E is kept at the bottom but not at the Stack 3.

3. G is not kept in the same stack in which B kept.  G is not kept in the same stack in which A kept.

4. Height of G is 36 cm.

5. O is immediately below G and only one box is kept below O. Number of boxes above G is same as below G. (We can conclude only 5 box are there in each stack)

6. Height of box at the top of the Stack 1 is 50.

7. Less than two boxes are kept below F. F is not immediately above C. (This conclude F is either in stack 1 or stack 3 and either one or no box is kept below F)

8)  As many boxes below F as below D.

By statement 7) and 8) we can conclude only one box is placed below D and F. In stack 3 O is placed so we place F in stack 1)

9. Only one box is kept between J and M. M is not at the top.

10. More than one box is kept between X and Z.

11. Number of boxes above X is same as number of boxes above W.

12. N is not placed at the top.

13. Height of Z is 63 cm. 

14. F’s height is an even number and more than G. Height of F is less than 40 cm. (Height of F is 38 cm)

15. Height of box at the top of the Stack 1 is 50.

16. Difference between the height of box C and G is 25 cm. Height of box C is greater than G. (C – G =25

We know height of G so C – 36 = 25 → C = 61)

15. Sum of box immediately below F and immediately below D is 113 cm. (F + C = 113 so F + 61 = 113)

16. Sum of top and bottom box of stack 2 is 153 cm.  ( J + C =153 so J + 61 =153)

17. Height of M is 15 cm less than height of E.

18. Difference of A and D is 8 cm. Height of D is more than that of A. ( D + A = 110 , D – A =8)

19. N height 16 more than D’s height.

20. Height of X is 3 cm more than height of A.

Note : For height of W =  300 – 215 = 85 cm

21. Height of U is same as height of W

Finally, Height of O = 300 – 238 = 62 cm

All are correct except E -54 cm

Hence, E -54 cm does not belong to that group.

GER in 2002 = 0.67 = 2/3 

GER in 2003 = 0.67 × 75/100 = 0.5

​$= \frac{{Female;Population}}{{Male;Population}}$​= 5/10 = 1 : 2

Let Area of city B in 2001 and 2002 are x and y respectively.

⇒ 120/100y = 172.8

⇒ y = 144

⇒ 120/100x = 144

⇒ x = 120

Population of city B in 2001 = 120 × 50 = 6000

Population of city B in 2002 = 144 × 125 = 18000

GER of city B in 2001 $= \frac{{Female;Population}}{{Male;Population}}$ = 1/1.5 = 2/3

Number of female of city B in 2001 = 2400

GER of city B in 2002 $= \frac{{Female;Population}}{{Male;Population}}$ = 1.5 = 3/2

Number of male of city B in 2002 = 7200

Desired Ratio = 7200/2400 = 3 : 1

Given: 

Population of city C in 2001 = 8250

Area of city C increased by 10% each year.

Male population of city C in 2002 = 4356

Male population of city C in 2003 = 7986

Calculation:

Area of city C in 2001 = 8250/75 = 110

Area of city C in 2002 = 110/100 × 110  = 121

Population of city C in 2002 = 121 × 120 = 14520

Female population of city C in 2002 = 14520 - 4356 = 10164

Area of city C in 2003 = 110/100 × 121 = 133.1

Population of city C in 2003 = 133.1 × 180 = 23958

Female population of city C in 2003 = 23958 - 7986 = 15972

GER of city C in 2002 ​​= female population/male population = 10164/4356​

GER of city C in 2003 ​​= female population/male population = 15972/7986​

Desired ratio = [(10164/4356​)]/[(15972/7986​)]

⇒ 7 : 6

Male population = female population/GER

⇒ Male population = 2514/0.4 = 6285

⇒ Total population = 6285 + 2514 = 8799

Population density of city D in 2001 = 110

⇒ Area of city D in 2001 = Population/Population Density = 8799/110

= 79.99 ≈ 80 square unit

Since it is not given in the question that in which year's comparison we have to find the rate at which the area of city D expanded so we can not determine the answer.

Sum of population density of city A for all years = 100 + 200 + 300 = 600

Average of population density of city A for all years = 600/3

Sum of population density of city B for all years = 50 + 125 + 200 = 375

Average of population density of city B for all years = 375/3

Sum of Average of population density of city A and B = 600/3 + 375/3

= 975/3

Sum of population density of city C for all years = 75 + 120 + 180 = 375

Average of population density of city C for all years = 375/3

Sum of population density of city D for all years = 110 + 200 + 250 = 560

Average of population density of city D for all years = 560/3

Sum of Average of population density of city C and D = 375/3 + 560/3

= 935/3

Desired ratio = (975/3)/(935/3) = 195/187

Let Number of Red balls be R, Blue balls be B, Green balls be G and Total balls be T

Box P:

Total number of balls = 32 – 10 – 10 = 12

Since the probability of choosing a Blue ball is 1/4 = 3/12

⇒ Number of blue balls = 3

Since the number of red balls is 2 more than the number of blue balls

⇒ R = B + 2 = 5

⇒ G = 12 – 3 – 5 = 4

Box Q:

Total number of balls = 10

Ratio of probability of getting a Blue and a Red ball is 2 : 3.

Suppose the number of Blue and red balls is 2x & 3x

Sum of Blue and Green balls is 4 more than the number of red balls

⇒ 2x + (10 – 5x) = 4 + 3x

⇒ 6x = 6

⇒ x = 1

⇒ R = 3, B = 2 and G = 10 – 3 – 2 = 5

Box R:

Total number of balls = 10

Since total number of blue balls in all the boxes is 9

⇒ Number of blue balls in box R = 9 – 3 – 2 = 4

Since the probability of getting a red is 1/5 less than the probability of getting a blue

⇒ B/10 – R/10 = 1/5

⇒ B – R = 2

Hence, R = 2

⇒ G = 10 – 4 – 2 = 4

⇒ Number of red balls in Boxes P and R = 5 + 2 = 7

⇒ Number of green balls in boxes Q and R = 5 + 4 = 9

∴ Required ratio = 7 : 9

Let Number of Red balls be R, Blue balls be B, Green balls be G and Total balls be T

Box P:

Total number of balls = 32 – 10 – 10 = 12

Since the probability of choosing a Blue ball is 1/4 = 3/12

⇒ Number of blue balls = 3

Since the number of red balls is 2 more than the number of blue balls

⇒ R = B + 2 = 5

⇒ G = 12 – 3 – 5 = 4

Box Q:

Total number of balls = 10

Ratio of probability of getting a Blue and a Red ball is 2 : 3.

Suppose the number of Blue and red balls is 2x & 3x

Sum of Blue and Green balls is 4 more than the number of red balls

⇒ 2x + (10 – 5x) = 4 + 3x

⇒ 6x = 6

⇒ x = 1

⇒ R = 3, B = 2 and G = 10 – 3 – 2 = 5

Box R:

Total number of balls = 10

Since total number of blue balls in all the boxes is 9

⇒ Number of blue balls in box R = 9 – 3 – 2 = 4

Since the probability of getting a red is 1/5 less than the probability of getting a blue

⇒ B/10 – R/10 = 1/5

⇒ B – R = 2

Hence, R = 2

⇒ G = 10 – 4 – 2 = 4

Total number of balls in box R = 10

Number of blue and green balls in box R is 4 and 4 respectively

∴ Probability of getting either a blue or a green ball from box R = (4 + 4) /10 = 8/10 = 4/5

Let Number of Red balls be R, Blue balls be B, Green balls be G and Total balls be T

Box P:

Total number of balls = 32 – 10 – 10 = 12

Since the probability of choosing a Blue ball is 1/4 = 3/12

⇒ Number of blue balls = 3

Since the number of red balls is 2 more than the number of blue balls

⇒ R = B + 2 = 5

⇒ G = 12 – 3 – 5 = 4

Box Q:

Total number of balls = 10

Ratio of probability of getting a Blue and a Red ball is 2 : 3.

Suppose the number of Blue and red balls is 2x & 3x

Sum of Blue and Green balls is 4 more than the number of red balls

⇒ 2x + (10 – 5x) = 4 + 3x

⇒ 6x = 6

⇒ x = 1

⇒ R = 3, B = 2 and G = 10 – 3 – 2 = 5

Box R:

Total number of balls = 10

Since total number of blue balls in all the boxes is 9

⇒ Number of blue balls in box R = 9 – 3 – 2 = 4

Since the probability of getting a red is 1/5 less than the probability of getting a blue

⇒ B/10 – R/10 = 1/5

⇒ B – R = 2

Hence, R = 2

⇒ G = 10 – 4 – 2 = 4

Probability of drawing red ball from box P = 5/12

Probability of drawing red ball from box Q = 3/10

Probability of drawing red ball from box R = 2/10 = 1/5

∴ Probability of getting all the 3 balls red = 5/12 × 3/10 × 1/5 = 1/40

Let Number of Red balls be R, Blue balls be B, Green balls be G and Total balls be T

Box P:

Total number of balls = 32 – 10 – 10 = 12

Since the probability of choosing a Blue ball is 1/4 = 3/12

⇒ Number of blue balls = 3

Since the number of red balls is 2 more than the number of blue balls

⇒ R = B + 2 = 5

⇒ G = 12 – 3 – 5 = 4

Box Q:

Total number of balls = 10

Ratio of probability of getting a Blue and a Red ball is 2 : 3.

Suppose the number of Blue and red balls is 2x & 3x

Sum of Blue and Green balls is 4 more than the number of red balls

⇒ 2x + (10 – 5x) = 4 + 3x

⇒ 6x = 6

⇒ x = 1

⇒ R = 3, B = 2 and G = 10 – 3 – 2 = 5

Box R:

Total number of balls = 10

Since total number of blue balls in all the boxes is 9

⇒ Number of blue balls in box R = 9 – 3 – 2 = 4

Since the probability of getting a red is 1/5 less than the probability of getting a blue

⇒ B/10 – R/10 = 1/5

⇒ B – R = 2

Hence, R = 2

⇒ G = 10 – 4 – 2 = 4

Probability of drawing a green ball from box Q = 5/10 = ½ = 0.5

Probability of drawing a blue ball from box R = 4/10 = 2/5 = 0.4

∴ Required difference = 0.5 – 0.4 = 0.1

Let Number of Red balls be R, Blue balls be B, Green balls be G and Total balls be T

Box P:

Total number of balls = 32 – 10 – 10 = 12

Since the probability of choosing a Blue ball is 1/4 = 3/12

⇒ Number of blue balls = 3

Since the number of red balls is 2 more than the number of blue balls

⇒ R = B + 2 = 5

⇒ G = 12 – 3 – 5 = 4

Box Q:

Total number of balls = 10

Ratio of probability of getting a Blue and a Red ball is 2 : 3.

Suppose the number of Blue and red balls is 2x & 3x

Sum of Blue and Green balls is 4 more than the number of red balls

⇒ 2x + (10 – 5x) = 4 + 3x

⇒ 6x = 6

⇒ x = 1

⇒ R = 3, B = 2 and G = 10 – 3 – 2 = 5

Box R:

Total number of balls = 10

Since total number of blue balls in all the boxes is 9

⇒ Number of blue balls in box R = 9 – 3 – 2 = 4

Since the probability of getting a red is 1/5 less than the probability of getting a blue

⇒ B/10 – R/10 = 1/5

⇒ B – R = 2

Hence, R = 2

⇒ G = 10 – 4 – 2 = 4

Probability of drawing a red ball from box Q = 3/10

Since the probability increases by 7/60

⇒ Increased probability = 3/10 + 7/60 = 25/60 = 5/12

⇒ Number of increased red balls in box Q = 5 – 3 = 2

∴ x = 2/2 = 1

Total number of people who took loan from Bank S in the year 2006 = 29565

Percentage of defaulters in the year 2006 for Bank S = 21%

∴ Total number of defaulters of Bank S = 21% of 29565 = 6208 ≈ 6210

∴ Total number of defaulters of bank S = 6210

Total number of people who took loan from Bank Q in the year 2004 = 26345

Total number of people who took loan from Bank Q in the year 2005 = 27456

Percentage of defaulters in the year 2004 for bank Q = 9 %

∴ Total number of defaulters for bank Q in the year 2004 = 9 % of 26345 = (26345/100) × 9 = 2371

Percentage of defaulters in the year 2005 for Bank Q = 8 %

∴ Total number of defaulters for Bank Q in the year 2005 = 8 % of 27456 = (27456/100) × 8 = 2196

∴ Difference between the number of defaulters of Bank Q in the year 2004 and 2005 = 2371 – 2196 = 175