LR - Mathematical Reasoning - Previous Year CAT/MBA Questions

Answer: 2

Text Explanation :

As per the data given in the question the gold coins were equally distributed amongst the 3 children. We are also told that the remaining assets (apart from the gold coins) are also equally distributed and that no flat or house is split or divided between 2 or more people. Only the bank deposit of Rs.70 lakhs can be split. Apart from the gold coins, the total value of the assets is 70 + 50 + 30 + 30 + 30 = 210 lakhs

So 210 lakhs split equally would means that each of the 3 daughters would get Rs.70 lakhs. This would mean that the person receiving the Rs.50 lakh house would get Rs.20 lakhs from the bank deposit. Now there are 3 flats of Rs.30 lakhs each and a balance of 50 lakhs in the bank deposit. The only way these assets can be split equally is if one of the 2 remaining people gets one of the flats of Rs.30 lakhs and receives 40 lakhs from the bank deposit. The second person would receive 70-40-20 = 10 Lakhs as bank deposit and 2 flats of 30 lakhs each, the sum of which adds upto 70 lakhs.
As Neeta receives the least amount in bank deposits, she will receive 10 lakhs. Also, as Geeta receives the highest amount in bank deposits she wil receive 40 lakhs.

Since Neeta is the person who receives the least amount in bank deposits i.e., 10 lakhs, he will receive 2 flats of 30 lakhs each.

Answer: 2

Answer: Option B

Text Explanation :

As per the data given in the question the gold coins were equally distributed amongst the 3 children. We are also told that the remaining assets (apart from the gold coins) are also equally distributed and that no flat or house is split or divided between 2 or more people. Only the bank deposit of Rs.70 lakhs can be split. Apart from the gold coins, the total value of the assets is 70 + 50 + 30 + 30 + 30 = 210 lakhs

So 210 lakhs split equally would means that each of the 3 daughters would get Rs.70 lakhs. This would mean that the person receiving the Rs.50 lakh house would get Rs.20 lakhs from the bank deposit. Now there are 3 flats of Rs.30 lakhs each and a balance of 50 lakhs in the bank deposit. The only way these assets can be split equally is if one of the 2 remaining people gets one of the flats of Rs.30 lakhs and receives 40 lakhs from the bank deposit. The second person would receive 70-40-20 = 10 Lakhs as bank deposit and 2 flats of 30 lakhs each, the sum of which adds upto 70 lakhs.
As Neeta receives the least amount in bank deposits, she will receive 10 lakhs. Also, as Geeta receives the highest amount in bank deposits she wil receive 40 lakhs.

Let the total value of assets held by Neeta, Geeta and Seeta be x, 2x and 3x and the total number of gold coins held by each of them be 2y, 3y and 4y. So the total assets (apart from the gold coins) held by Neeta, Geeta and Seeta will be x – 2y, 2x – 3y and 3x – 4y. 

Now total assets apart from gold coin = 210 lakhs 

∴ (x – 2y) + (2x -3y) + (3x – 4y) = 210 

⇒ 6x – 9y = 210

On simplifying we get 2x – 3y = 70

This means that the total assets held by Geeta apart from the gold coins = 70 lakhs We are further told that one child got all 3 flats and one person other than Geeta got 30 lakhs in bank deposits. So then the only way Geeta could have got assets (other than gold coins) worth 70 lakhs is by getting the house of 50 lakhs and 20 lakhs from the bank deposit. This is also because all the 3 flats of Rs. 30 lakh each have gone to one person and obviously that person cannot be Geeta. This means that the 3rd sister has got 20 lakhs from the bank deposit (since one more sister has got 30 lakhs). Since Geeta cannot receive 30 lakhs from the bank deposit, it would imply that Neeta will receive 30 lakhs and Geeta will receive 20 lakhs from the bank deposit.
Now, all 3 flats go to either Neeta or Geeta.

If all 3 flats are received by Neeta, then the ratio of value of assets of Neeta and Geeta

(in terms of y) will be (30 + 90 + 2y): (20 + 4y)

 ⇒ (120 + 2y) : (20 + 4y) = $\frac{120+2y}{20+4y}$

$=\frac{1}{3}$

⇒ 360 + 6y = 20 + 4y ⇒ 2y = -340 or y = -170 

However ‘y’ cannot be a negative value. So all 3 flats will go to Geeta 

∴ Ratio of value of assets of Neeta and Geeta (in terms of y) will be (30 + 2y) : (20 + 90 + 4y) 

$\Rightarrow \frac{30+2y}{110+4y}=\frac{1}{3}$

⇒ 90 + 6y = 110 + 4y ⇒ y = 10 

Total number of gold coins = 2y + 3y + 4y = 9y = 90. 

Hence, option (b).

Answer: 20

Text Explanation :

As per the data given in the question the gold coins were equally distributed amongst the 3 children. We are also told that the remaining assets (apart from the gold coins) are also equally distributed and that no flat or house is split or divided between 2 or more people. Only the bank deposit of Rs.70 lakhs can be split. Apart from the gold coins, the total value of the assets is 70 + 50 + 30 + 30 + 30 = 210 lakhs

So 210 lakhs split equally would means that each of the 3 daughters would get Rs.70 lakhs. This would mean that the person receiving the Rs.50 lakh house would get Rs.20 lakhs from the bank deposit. Now there are 3 flats of Rs.30 lakhs each and a balance of 50 lakhs in the bank deposit. The only way these assets can be split equally is if one of the 2 remaining people gets one of the flats of Rs.30 lakhs and receives 40 lakhs from the bank deposit. The second person would receive 70-40-20 = 10 Lakhs as bank deposit and 2 flats of 30 lakhs each, the sum of which adds upto 70 lakhs.
As Neeta receives the least amount in bank deposits, she will receive 10 lakhs. Also, as Geeta receives the highest amount in bank deposits she wil receive 40 lakhs.

Geeta gets 20 lakhs from bank deposits.

Answer: 20

Answer: Option A

Text Explanation :

Using the information given in the questions let us tabulate the rates of the seats in the different rows.

​​​​​​​

Following condition (4) as Jayanta, Ajit and Byomkesh sit on seats marked by the same letter in increasing order of row in numbers and paid different amounts, the only way this would be possible is if they sat in rows 10, 11 and 12 respectively, either on the window seat or the aisle seat. So this would lead to 2 possible cases.
Case 1: Jayanta, Ajit and Byomkesh sit on the window seat in consecutive rows. In this case, as Manik sits adjacent to Jayanta, it would imply Manik sits on the middle seat next to Jayanta in row 10. This further implies that Manik did not pay any additional amount for his seat. Now Jayanta, Ajit and Byomkesh would pay an amount of 300 + 200 + 1000 = Rs. 1500. So the balance amount of 4600 – 1500 = 3100 must have been paid by Gargi, Kikira, Pradosh and Tapesh and all of them have to be occupying aisle seats. However, no contribution of amounts changeable for aisle seats adds upto 3100. So case is ruled out.
Case 2. Jayanta, Ajit and Byomkesh sit on the aisle seat in rows 10, 11 and 12 respectively. Here Manik can occupy either the middle seat next to Jayanta or the aisle seat next to Jayanta. Let us assume that Manik occupies the middle seat in row 10 next to Jayanta. In that case, cost of 3 aisle seats adds upto 4600 – (1000 + 500 + 400) = 2900. However, no contribution of amounts chargeable for the above mentioned seats adds upto 2900. So then Manik can only sit on the aisle seat adjacent to Jayanta in row 10. The total amount of seats chargeable to Jayanta, Ajit, Byomkesh and Manik adds upto 1000 + 500 + 400 + 500 = 2400. Now the balance amount of 4600 – 2400
= 2200 has to be accounted for by 4 people i.e., Gagi, Kikira, Pradosh and Tapesh. 3 of them sit on 3 window seats and one on the middle seat. Now out of Gargi and Kikira, one of them will sit on the middle seat and other on the window seat as they sit adjacent to each other. Which means that both Pradosh and Tapesh sit on the window seat. Since the amounts chargeable for the window/middle seats can only the amongst 1000, 300, 200 or 0 and the total we need to account for is 2200, at least one of the seats has to be a 1000 Rs. Seat. If we assume one of seats is Rs. 1000, then the balance amount of Rs 1200, cannot be accounted for. So there have to be at least 2 seats of Rs. 1000 each. Which means 2 out of Gargi, Kikira, Pradosh and Tapesh have to sit in the 13th row. Now these 2 people will have to be Gargi and Kikirsa as per condition (5). So each of Gargi and Kikira pay Rs. 1000 for their seat and these 2 seats will be a window and middle seat (both adjacent to each other)in the 13th row. The balance amount of Rs. 200 will be paid by Pradosh who sits on the window seat of the 20th row. This also implies that Tapesh will sit just behind Pradosh in the window seat of the 21st row. Let us represent all this information in the table given below

​​​​​​​

J – Jayanta, M – Manik, A – Ajit, B – Byomkesh, G – Gargi, K – Kikira, P – Pradosh, T – Tapesh

Using this data let us answer the question.

As can be seen from the table, Manik was sitting in row 10.

Hence, option (a).

Answer: Option C

Text Explanation :

Using the information given in the questions let us tabulate the rates of the seats in the different rows.

​​​​​​​

Following condition (4) as Jayanta, Ajit and Byomkesh sit on seats marked by the same letter in increasing order of row in numbers and paid different amounts, the only way this would be possible is if they sat in rows 10, 11 and 12 respectively, either on the window seat or the aisle seat. So this would lead to 2 possible cases.
Case 1: Jayanta, Ajit and Byomkesh sit on the window seat in consecutive rows. In this case, as Manik sits adjacent to Jayanta, it would imply Manik sits on the middle seat next to Jayanta in row 10. This further implies that Manik did not pay any additional amount for his seat. Now Jayanta, Ajit and Byomkesh would pay an amount of 300 + 200 + 1000 = Rs. 1500. So the balance amount of 4600 – 1500 = 3100 must have been paid by Gargi, Kikira, Pradosh and Tapesh and all of them have to be occupying aisle seats. However, no contribution of amounts changeable for aisle seats adds upto 3100. So case is ruled out.
Case 2. Jayanta, Ajit and Byomkesh sit on the aisle seat in rows 10, 11 and 12 respectively. Here Manik can occupy either the middle seat next to Jayanta or the aisle seat next to Jayanta. Let us assume that Manik occupies the middle seat in row 10 next to Jayanta. In that case, cost of 3 aisle seats adds upto 4600 – (1000 + 500 + 400) = 2900. However, no contribution of amounts chargeable for the above mentioned seats adds upto 2900. So then Manik can only sit on the aisle seat adjacent to Jayanta in row 10. The total amount of seats chargeable to Jayanta, Ajit, Byomkesh and Manik adds upto 1000 + 500 + 400 + 500 = 2400. Now the balance amount of 4600 – 2400
= 2200 has to be accounted for by 4 people i.e., Gagi, Kikira, Pradosh and Tapesh. 3 of them sit on 3 window seats and one on the middle seat. Now out of Gargi and Kikira, one of them will sit on the middle seat and other on the window seat as they sit adjacent to each other. Which means that both Pradosh and Tapesh sit on the window seat. Since the amounts chargeable for the window/middle seats can only the amongst 1000, 300, 200 or 0 and the total we need to account for is 2200, at least one of the seats has to be a 1000 Rs. Seat. If we assume one of seats is Rs. 1000, then the balance amount of Rs 1200, cannot be accounted for. So there have to be at least 2 seats of Rs. 1000 each. Which means 2 out of Gargi, Kikira, Pradosh and Tapesh have to sit in the 13th row. Now these 2 people will have to be Gargi and Kikirsa as per condition (5). So each of Gargi and Kikira pay Rs. 1000 for their seat and these 2 seats will be a window and middle seat (both adjacent to each other)in the 13th row. The balance amount of Rs. 200 will be paid by Pradosh who sits on the window seat of the 20th row. This also implies that Tapesh will sit just behind Pradosh in the window seat of the 21st row. Let us represent all this information in the table given below

​​​​​​​

J – Jayanta, M – Manik, A – Ajit, B – Byomkesh, G – Gargi, K – Kikira, P – Pradosh, T – Tapesh

Using this data let us answer the question.

Jayanta paid Rs.500 for his choice of seat.

Hence, option (c).

Answer: Option D

Text Explanation :

Using the information given in the questions let us tabulate the rates of the seats in the different rows.

​​​​​​​

Following condition (4) as Jayanta, Ajit and Byomkesh sit on seats marked by the same letter in increasing order of row in numbers and paid different amounts, the only way this would be possible is if they sat in rows 10, 11 and 12 respectively, either on the window seat or the aisle seat. So this would lead to 2 possible cases.
Case 1: Jayanta, Ajit and Byomkesh sit on the window seat in consecutive rows. In this case, as Manik sits adjacent to Jayanta, it would imply Manik sits on the middle seat next to Jayanta in row 10. This further implies that Manik did not pay any additional amount for his seat. Now Jayanta, Ajit and Byomkesh would pay an amount of 300 + 200 + 1000 = Rs. 1500. So the balance amount of 4600 – 1500 = 3100 must have been paid by Gargi, Kikira, Pradosh and Tapesh and all of them have to be occupying aisle seats. However, no contribution of amounts changeable for aisle seats adds upto 3100. So case is ruled out.
Case 2. Jayanta, Ajit and Byomkesh sit on the aisle seat in rows 10, 11 and 12 respectively. Here Manik can occupy either the middle seat next to Jayanta or the aisle seat next to Jayanta. Let us assume that Manik occupies the middle seat in row 10 next to Jayanta. In that case, cost of 3 aisle seats adds upto 4600 – (1000 + 500 + 400) = 2900. However, no contribution of amounts chargeable for the above mentioned seats adds upto 2900. So then Manik can only sit on the aisle seat adjacent to Jayanta in row 10. The total amount of seats chargeable to Jayanta, Ajit, Byomkesh and Manik adds upto 1000 + 500 + 400 + 500 = 2400. Now the balance amount of 4600 – 2400
= 2200 has to be accounted for by 4 people i.e., Gagi, Kikira, Pradosh and Tapesh. 3 of them sit on 3 window seats and one on the middle seat. Now out of Gargi and Kikira, one of them will sit on the middle seat and other on the window seat as they sit adjacent to each other. Which means that both Pradosh and Tapesh sit on the window seat. Since the amounts chargeable for the window/middle seats can only the amongst 1000, 300, 200 or 0 and the total we need to account for is 2200, at least one of the seats has to be a 1000 Rs. Seat. If we assume one of seats is Rs. 1000, then the balance amount of Rs 1200, cannot be accounted for. So there have to be at least 2 seats of Rs. 1000 each. Which means 2 out of Gargi, Kikira, Pradosh and Tapesh have to sit in the 13th row. Now these 2 people will have to be Gargi and Kikirsa as per condition (5). So each of Gargi and Kikira pay Rs. 1000 for their seat and these 2 seats will be a window and middle seat (both adjacent to each other)in the 13th row. The balance amount of Rs. 200 will be paid by Pradosh who sits on the window seat of the 20th row. This also implies that Tapesh will sit just behind Pradosh in the window seat of the 21st row. Let us represent all this information in the table given below

​​​​​​​

J – Jayanta, M – Manik, A – Ajit, B – Byomkesh, G – Gargi, K – Kikira, P – Pradosh, T – Tapesh

Using this data let us answer the question.

Gargi paid Rs.1000 extra for her choice of seat.

Hence, option (d).

Answer: Option D

Text Explanation :

Using the information given in the questions let us tabulate the rates of the seats in the different rows.

​​​​​​​

Following condition (4) as Jayanta, Ajit and Byomkesh sit on seats marked by the same letter in increasing order of row in numbers and paid different amounts, the only way this would be possible is if they sat in rows 10, 11 and 12 respectively, either on the window seat or the aisle seat. So this would lead to 2 possible cases.
Case 1: Jayanta, Ajit and Byomkesh sit on the window seat in consecutive rows. In this case, as Manik sits adjacent to Jayanta, it would imply Manik sits on the middle seat next to Jayanta in row 10. This further implies that Manik did not pay any additional amount for his seat. Now Jayanta, Ajit and Byomkesh would pay an amount of 300 + 200 + 1000 = Rs. 1500. So the balance amount of 4600 – 1500 = 3100 must have been paid by Gargi, Kikira, Pradosh and Tapesh and all of them have to be occupying aisle seats. However, no contribution of amounts changeable for aisle seats adds upto 3100. So case is ruled out.
Case 2. Jayanta, Ajit and Byomkesh sit on the aisle seat in rows 10, 11 and 12 respectively. Here Manik can occupy either the middle seat next to Jayanta or the aisle seat next to Jayanta. Let us assume that Manik occupies the middle seat in row 10 next to Jayanta. In that case, cost of 3 aisle seats adds upto 4600 – (1000 + 500 + 400) = 2900. However, no contribution of amounts chargeable for the above mentioned seats adds upto 2900. So then Manik can only sit on the aisle seat adjacent to Jayanta in row 10. The total amount of seats chargeable to Jayanta, Ajit, Byomkesh and Manik adds upto 1000 + 500 + 400 + 500 = 2400. Now the balance amount of 4600 – 2400
= 2200 has to be accounted for by 4 people i.e., Gagi, Kikira, Pradosh and Tapesh. 3 of them sit on 3 window seats and one on the middle seat. Now out of Gargi and Kikira, one of them will sit on the middle seat and other on the window seat as they sit adjacent to each other. Which means that both Pradosh and Tapesh sit on the window seat. Since the amounts chargeable for the window/middle seats can only the amongst 1000, 300, 200 or 0 and the total we need to account for is 2200, at least one of the seats has to be a 1000 Rs. Seat. If we assume one of seats is Rs. 1000, then the balance amount of Rs 1200, cannot be accounted for. So there have to be at least 2 seats of Rs. 1000 each. Which means 2 out of Gargi, Kikira, Pradosh and Tapesh have to sit in the 13th row. Now these 2 people will have to be Gargi and Kikirsa as per condition (5). So each of Gargi and Kikira pay Rs. 1000 for their seat and these 2 seats will be a window and middle seat (both adjacent to each other)in the 13th row. The balance amount of Rs. 200 will be paid by Pradosh who sits on the window seat of the 20th row. This also implies that Tapesh will sit just behind Pradosh in the window seat of the 21st row. Let us represent all this information in the table given below

​​​​​​​

J – Jayanta, M – Manik, A – Ajit, B – Byomkesh, G – Gargi, K – Kikira, P – Pradosh, T – Tapesh

Using this data let us answer the question.

As Tapesh was sitting in row 21, he did not pay any extra amount for his seat.

Hence, option (d).

Answer: Option C

Text Explanation :

Since Aditya didn’t get a call from any of the colleges, so for each college, he either didn’t clear one of the sectional cut-offs or he didn’t clear the aggregate cut-off or both.

If he didn’t clear one of the sectional cut-offs, then for that section he scored less marks than the least cut-off among the given cut-offs of all the colleges.

For example, for section A, it is given that the cut-offs for colleges 1, 4 and 5 are 42, 43 and 45 respectively. The least cut-off among them is 42.

So, in order to not clear the sectional cut-off of section A for colleges 1, 4 and 5, he should have scored less than 42.

Similarly,

For colleges 1, 2 and 6, Aditya’s Section B marks < 41

For colleges 1, 2, 3 and 5, Aditya’s Section C marks < 42

For colleges 4 and 6, Aditya’s Section D marks < 44

If he scores less in Section C and D, he would not get calls for any colleges. Also in order to maximise the score we would assume that he got just one less than the cut-off in section C and D and he scored maximum marks (50) in other sections.

∴ Maximum marks obtained by Aditya such that he doesn’t get any calls = 41 + 43 + 50 + 50 = 184

Hence, option (c).

Answer: Option B

Text Explanation :

Since Bhama got calls from all colleges, she must have cleared each of the 4 sections. This means that for a particular section she scored more marks than the greatest cut-off for that section across the six colleges.

For example, for section A, it is given that the cut-offs for colleges 1, 4 and 5 are 42, 43 and 45 respectively. The greatest cut-off among them is 45.

So, in order to clear the sectional cut-off of section A for all the colleges, she should have scored at least 45.

Since we wish to minimise her marks, we should take her score in section A as 45.

Similarly, in sections B, C and D, she scored 45, 46, and 45 marks respectively.

∴ Bhama’s minimum marks such that she gets calls from all the colleges = 45 + 45 + 46 + 45 = 181

Hence, option (b).

Note: This is already greater than the highest aggregate cut-off of all colleges (which is 180 for college 5). So, she will get calls from all 6 colleges.

Answer: Option C

Text Explanation :

The aggregate cut-off for each college is given in the common data. In order for Charlie to get minimum marks in one of the sections, he should have got maximum marks (i.e. 50) in the other three sections.

For example, the aggregate cut-off in college 1 is 176. Since, we want minimum marks in a section he should have gotten an aggregate of exactly 176. To minimise one of the sections, assume that he got 50 marks in the 3 sections whose cut-off is given in the common data. Then, Charlie will get a call from college 1 if he gets at least 176 – (50 × 3) = 26 marks in section D, provided that the cut-off for this section is also 26.

Now, there is at least one unknown sectional cut-off for each of the colleges, so we can use the same logic as used above for each of the remaining colleges.

For college 2, the minimum marks that Charlie needs to get a call = 175 – 150 = 25

For college 3, the minimum marks that Charlie needs to get a call = 171 – 150 = 21

For college 4, the minimum marks that Charlie needs to get a call = 178 – 150 = 28

For college 5, the minimum marks that Charlie needs to get a call = 180 – 150 = 30

For college 6, the minimum marks that Charlie needs to get a call = 176 – 150 = 26

The question states that Charlie only gets a call from 2 of the colleges. So, Charlie got 25 marks.

Hence, option (c).

Answer: Option C

Text Explanation :

Let the age of the employee being transferred from the

1. Marketing department to the Finance department be x.

2. Finance department to the Marketing department be y.

3. Marketing department to the HR department be z.

Now,

The sum of the ages of all employees in Finance originally was 30 × 20 = 600. Later, an employee of x years of age joined the department and one of y years of age left it.

So, the new average age for the Finance department = $\frac{600+x-y}{20}$ = 31 (given)

∴ 600 + x – y = 620

∴ x – y = 20                                            … (i)

The sum of the ages of all employees in Marketing originally was 35 × 30 = 1050.

Later, two employees of x years and z years of age left the department and one of y years of age joined it. Since 2 employees left and 1 joined, hence the number of employees currently in this department is 29

So, the new average age for the Marketing department = .$\frac{1050-x+y-z}{29}$ = 35 (given)

∴ 1050 – x + y – z = 1015

∴ x – y + z = 35                                       … (ii)

From equations (i) and (ii), we get,

20 + z = 35

∴ z = 15

The sum of the ages of all employees in HR originally was 45 × 5 = 225.

Later, one employee of z years of age joined the department. Also, the number of employees increases by one to 6.

So, the new average age for the HR department = $\frac{225+z}{6}=\frac{225+15}{6}=\frac{240}{6}=40$

Hence, option (c).

Answer: Option C

Text Explanation :

The average age of the Marketing department is 35 years and that of the HR department is 45 years. So, the employee is being transferred from a department with a lower average age to one with a higher average age, which means that he gets an additional allowance of 10% of basic pay over his current allowance.

His current allowance = 80% of 8000 = 6400

Therefore, his new allowance = 6400 + 10% of 8000 = 6400 + 800 = 7200

After the transfer, his gross pay = 8000 + 7200 = 15200

Initially, the average gross pay of the HR department = 5000 + 70% of 5000 = 8500

The new average gross pay of the HR department (i.e. after the transfer of the 40-yr old)

$=\frac{8500\times 5+15200}{6}=\frac{57700}{6}\approx 9617$

∴ The percentage change in the average gross pay of the HR department

$=\frac{9617-8500}{8500}$ × 100 ≈ 13.13%

Hence, option (c).

Answer: Option B

Text Explanation :

Note that in this question, the percentage change in basic pay is asked. According to the common data, only the allowances (and hence the gross pay) is affected when a person is transferred. The basic pay of a person remains unaltered.

∴ The average basic pay after the transfers have taken place

$$\begin{gathered} =\frac{(5000\times 5)+(6000\times 2)+\left(8000\right)}{8} \\ =\frac{45000}{8}=5625 \end{gathered}$$

∴ The percentage change in the average basic pay of the HR department

$=\frac{5625-5000}{5000}$ × 100 = 12.5%

Hence, option (b).

Answer: Option A

Text Explanation :

Firstly, let us try to understand the way the investments of the three traders behave.

Abdul buys shares at 10 am everyday and sells them at a particular price at 3 pm. So his return is determined by the difference in the share price at these two times.

Bikram and Chetan buy shares at equal intervals. But since Chetan buys them in equal amount he would end up buying more when the price is less and less when the price is more.

Whether the prices are continuously rising or continuously falling down or in a fluctuating market, Chetan always has a higher proportion of lower priced shares as compared to Bikram. This increases his profit in a rising market and reduces his loss in a falling market. Therefore Chetan never has return lower than that of Bikram.

We have explained this concept by taking examples. For more depth we have also provided the theoretical explanation. The theoretical explanation is only for better understanding and may not be suitable in a test environment.

Consider the scenario when the share price keeps rising throughout the day.

Let the share price at 10 am be Rs. 100, 11 am be Rs. 110, 12 noon be Rs. 140, 1 pm be Rs. 150, 2 pm be Rs. 180, and finally at 3 pm be Rs. 200.

​​​​​​​

Abdul buys shares at Rs. 100 at 10 am and sells them at Rs. 200 at 3 pm.

∴ Abdul’s return is 100%.

Let Bikram buy one share at each interval. So, at 10 am, he buys a share for Rs. 100; at 11 am, he buys a share for Rs. 110; at 12 noon, he buys a share for Rs. 140; at 1 pm, he buys a share for Rs. 150; and at 2 pm, he buys a share for 180 × 1 = Rs. 180.

Thus, he buys a total of 5 shares for 100 + 110 + 140 + 150 + 180 = Rs. 680

At 3 pm, he sells all 5 shares for 200 × 5 = Rs. 1,000. Thus, his profit will be 1,000 − 680 = Rs. 320

Hence, Bikram's return is $\frac{320}{800}$ × 100 ≈ 47%

Let Chetan invest Rs. 415,800 at each interval. So, at 10 am, he buys 415800/100 = 4158 shares; at 11 am, he buys 415800/110 = 3780 shares; at 12 noon, he buys 415800/140 = 2970 shares; at 1 pm, he buys 415800/150 = 2772 shares; at 2 pm, he buys 415800/180 = 2310 shares.

Thus, he buys 4158 + 3780 + 2970 + 2772 + 2310 = 15990 shares for 415800 × 5 = Rs. 20,79,000. He sells these shares for 200 × 15990 = Rs. 31,98,000. His profit will be 3198000 − 2079000 = Rs. 11,19,000.

Hence Chetan's returns = $\frac{1119000}{2079000}\times 100=\frac{373}{693}$ × 100 ≈ 53%

From the above example, we see that in case of continuously rising share prices,

Abdul’s return > Chetan’s return > Bikram’s return

Thus Bikram gets the minimum return on a “boom” day.

Hence, option (a).

Note: Theoretical Explanation:

Let x1, x2, … , x6 be the share prices at 10 am, 11 am, 12 noon, 1 pm, 2 pm and 3 pm respectively.

For Abdul:

Abdul buys shares at Rs. x1 and sells them at Rs. x6.

∴ Abdul's returns = $\frac{x_6-x_1}{x_1}$

For Bikram:

Let Bikram have bought n shares at each hourly interval.

His investment amount = nx₁ + nx₂ + nx₃ + nx₄ + nx₅

= n(x₁ + x₂ + x₃ + x₄ + x₅)

$=n\times \underset{i=1}{\overset{5}{\sum x_t}}$

At 3 pm, he sells his shares for (5n × x₆)

Hence, his profit/loss = (n × 5x₆) - n × $\underset{i=1}{\overset{5}{\sum x_t}}$

= n × $\left(5x_6-\underset{i=1}{\overset{5}{\sum x_t}}\right)$

∴ Bikram's returns = $\frac{n\times (5x_6-{\displaystyle \underset{i=1}{\overset{5}{\sum x_t}}}}{n\times \underset{i=1}{\overset{5}{\sum x_t}}}=\frac{5x_6}{\underset{i=1}{\overset{5}{\sum x_t}}}-1=\frac{x_6}{{\displaystyle \frac{\underset{i=1}{\overset{5}{\sum x_t}}}{5}}}-1$

Hence, Bikram's returns = $\frac{x_6}{(\mathrm{Arithmetic} \mathrm{mean} \mathrm{of} x_1,x_2,....,x_5)}-1$

For Chetan:

Let Chetan invest Rs. P at each hourly interval.

His investment amount = 5P

Since he invests Rs. P at each interval, he buys $\frac{P}{x_1}$ shares at 10 am; $\frac{P}{x_2}$ at 11 am; and so on until 2 pm.

At 3 pm, he sells each share at x6. So, for all his shares, he receives,

Rs. $\left(\frac{P}{x_1}+\frac{P}{x_2}+\frac{P}{x_3}+\frac{P}{x_4}+\frac{P}{x_5}\right)\times x_6$

= Px₆ × $\left(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}+\frac{1}{x_5}\right)$

Hence, his profit/loss = Px₆ × $\left(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}+\frac{1}{x_5}\right)$ - 5P

= P × $\left[x_6\left(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}+\frac{1}{x_5}\right)-5\right]$

∴ Chetan's returns = $\frac{P\left[x_6\left({\displaystyle \frac{1}{x_1}}+{\displaystyle \frac{1}{x_2}}+{\displaystyle \frac{1}{x_3}}+{\displaystyle \frac{1}{x_4}}+{\displaystyle \frac{1}{x_5}}\right)-5\right]}{5P}$

$$\begin{gathered} =\frac{x_6\left({\displaystyle \frac{1}{x_1}}+{\displaystyle \frac{1}{x_2}}+{\displaystyle \frac{1}{x_3}}+{\displaystyle \frac{1}{x_4}}+{\displaystyle \frac{1}{x_5}}\right)-5}{5} \\ =\frac{x_6\left(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}+\frac{1}{x_5}\right)}{5}-1 \\ =\frac{x_6}{\left[{\displaystyle \frac{5}{\left(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}+\frac{1}{x_5}\right)}}\right]}-1 \end{gathered}$$

∴ Chetan's returns = $\frac{x_6}{(\mathrm{Harmonic} \mathrm{Mean} \mathrm{of} {\mathrm{x}}_1,{\mathrm{x}}_2,....,{\mathrm{x}}_5)}-1$

Now, let’s compare Bikram’s and Chetan’s returns. Since Arithmetic Mean is always greater than or equal to the Harmonic Mean, Chetan’s returns will be greater than or equal to Bikram’s.

Answer: Option E

Text Explanation :

Since Chetan’s return is always higher than or equal to that of Bikram, the trader with the maximum return would be either Abdul or Chetan.

If it is a continuously rising market then Abdul would end up having the highest gain as seen in the example above.

But there might be a scenario when the share price of XYZ would go down after 10 AM and rise in the end at 3 PM to a higher value.

In such a case, if Chetan gets the shares at lower prices than what the price was at 10 AM he would end up making more profit and hence higher return.

​​​​​​​

Here, Abdul’s returns remain unaltered as 100%.

Let Chetan always buy shares worth Rs. 100.

So he would end up buying 1 + 10 + 10 + 10 + 10 = 41 shares.

When he sells the same at Rs. 200 he gets Rs. 8,200 for the same.

∴ Chetan’s profit = 8200 − 500 = 7700

∴ Chetan's returns = $\frac{7700}{500}>100\%$

∴ We cannot say for sure who would have higher returns. 

Hence, option (e).

Answer: Option E

Text Explanation :

From the explanation seen till now we can rule out options 1, 3 and 4.

Now option 2 is only partially correct. We have seen that Chetan’s return would be higher than or equal to that of Bikram. It would be equal to Bikram’s return in the scenario when the share price remains at a constant value throughout the day.

∴ Option 2 is not always true.

Hence, option (e).

Answer: Option A

Text Explanation :

Let x₁, x₂, … , x₆ be the share prices at 10 am, 11 am, 12 noon, 1 pm, 2 pm and 3 pm respectively.

Now, since Abdul lost money in the transaction,

x₁ > x₆

Combining the above, we have,

x₁ > x₆ > x₅

and x₁ > x₃,

Also, let the money Emily invests at 10 am be Rs. P. Then,

Her investment = Rs. P

And the number of shares she buys = P/x₁

So, after selling these shares at 12 noon, she will get Rs. (P/x₁) × x₃

Now, she invests the money at 1 pm, and the number of shares she buys = $\frac{P{x}_3}{x_1{x}_4}$

So, after selling these shares at 3 pm, she gets Rs. $\frac{P{x}_3}{x_1{x}_4}\times x_6$

So, her returns = $\frac{{\displaystyle \frac{P{x}_3{x}_6}{x_1{x}_4}}-P}{P}=\frac{x_3{x}_6}{x_1{x}_4}-1$

Since she made profit, her returns > 0;

i.e. $\frac{x_3{x}_6}{x_1{x}_4}-1>$ or $\frac{x_3{x}_6}{x_1{x}_4}>1$

Now, we know that x₁ > x₆; so $\frac{x_6}{x_1}$ cannot be > 1.

$\therefore \frac{x_3}{x_4}$ has to be > 1; i.e. x₃ > x₄

∴ The share price at 12 noon is greater than that at 1 pm. 

Hence, option (d) is definitely false.

Also, since in the first half, Emily invests at 10 am and sells at 12 noon, and we know that the share price at 10 am was greater than at 12 noon; hence she must have suffered a loss during this transaction. However, she makes a net profit in the end. So, she must have made profit during the second part of the transaction; i.e. the share price at 1 pm must have been less than that at 3 pm.

i.e. x₄ < x₆,

Also, let Dane buy n shares at 10 am, 11 am and 12 noon.

Hence, her investment = n(x₁ + x₂ + x₃)

And she sells these at 1 pm, 2 pm and 3 pm for n(x₄ + x₅ + x₆)

∴ her returns = $\frac{n(x_4+x_5+x_6)-n(x_1+x_2+x_3)}{n(x_1+x_2+x_3)}=\frac{(x_4+x_5+x_6)}{(x_1+x_2+x_3)}-1$

Since she made profit, her returns are greater than 0;

i.e. $\frac{(x_4+x_5+x_6)}{(x_1+x_2+x_3)}-1>$ or $\frac{(x_4+x_5+x_6)}{(x_1+x_2+x_3)}>1$

Hence, (x₄+ x₅+x₆ )> (x₁+ x₂+x₃ )

Since, x₁ > x₆ and x₃ > x₄, hence x₅ > x₂

So far, we have,

x₁ > x₆ > x₅ > x_2, x₄ < x₆ and x₁ > x₃ > x₄

Now from Dane’s investment, we know that,

(x₄+ x₅+ x₆) - (x₁+ x₂+x₃ ) > 0                         … (i)

Keeping in mind the relationships between the share prices, we have

x₆ = x₁ – b

x₄ = x₁ – b – c

x₃ = x₁ – b – c + a

x₅ = x₁ – d, where a, b, c and d are all positive.

Substituting the above in equation (i), we have,

(x₁ – b – c + x₁ – d + x₁ – b) – (x₁ + x₂ + x₁ – b – c + a) > 0

∴ x₁ - x₂ > b + d + a (which is > 0, since all the variables are positive)

i.e. x₁ > x₂

∴ x₂ < x₁ – b – a – d

∴ x₂ is definitely less than x₆ and x₅.

∴ Although we don’t know when the share price is at its lowest, we do know that x₅ > x₂.

∴ x₅, i.e. the share price at 2 pm is not the lowest.

Hence, option (a) is also definitely false.

Thus there are two options which are correct for this question. This is an ambiguity and therefore, we are not indicating any option as correct.

Answer: Option A

Text Explanation :

From the solution of the last question, we can see that,

x₁ > x₆ > x₅ > x₂, x₄ < x₆ and x₁ > x₃ > x₄

∴ x₁,  i.e. the share price at 10 is the highest.

Hence, option (a).

Answer: Option B

Text Explanation :

Let F and E have Erdös numbers f and e respectively at the beginning of the conference. 

On the third day, A’s and C’s Erdös numbers become (f + 1)

The sum of Erdös numbers changed to 8 × 3 = 24

At the end of the third day, five members had identical Erdös numbers while the other three had distinct ones.

On the fifth day, E’s Erdös numbers became f + 1 and this reduced the group’s average by 0.5. This means that E’s Erdös numbers was not f + 1 on the third day.

Thus we have, 
At the end of the third day, 5(f + 1) + f + e + y = 24
Hence 6f + 5 + e + y = 24
Hence 6f + e + y = 19

At the end of the fifth day,
6(f + 1) + f + y = 2.5 × 8 = 20
Hence 7f + y = 14

Among the eight mathematicians, F has the smallest Erdös number.
Let f = 2
∴ y = 0

However, only Paul Erdös himself has an Erdös number of 0. So f cannot be equal to 2. Any other value greater than 2, would render y as a negative number, which is also not possible.
So, f = 1
∴ y = 7
∴ e = 6

Now, we can solve all the questions.

From the above explanation, the largest Erdös number at the end of the conference would be 7.

Hence, option (b). 

Answer: Option D

Text Explanation :

Let F and E have Erdös numbers f and e respectively at the beginning of the conference. 

On the third day, A’s and C’s Erdös numbers become (f + 1)

The sum of Erdös numbers changed to 8 × 3 = 24

At the end of the third day, five members had identical Erdös numbers while the other three had distinct ones.

On the fifth day, E’s Erdös numbers became f + 1 and this reduced the group’s average by 0.5. This means that E’s Erdös numbers was not f + 1 on the third day.

Thus we have, 
At the end of the third day, 5(f + 1) + f + e + y = 24
Hence 6f + 5 + e + y = 24
Hence 6f + e + y = 19

At the end of the fifth day,
6(f + 1) + f + y = 2.5 × 8 = 20
Hence 7f + y = 14

Among the eight mathematicians, F has the smallest Erdös number.
Let f = 2
∴ y = 0

However, only Paul Erdös himself has an Erdös number of 0. So f cannot be equal to 2. Any other value greater than 2, would render y as a negative number, which is also not possible.
So, f = 1
∴ y = 7
∴ e = 6

Now, we can solve all the questions.

From the above explanation, the Erdös numbers of B, D, G, H and F did not change during the conference.

Hence, option (d).

Answer: Option B

Text Explanation :

Let F and E have Erdös numbers f and e respectively at the beginning of the conference. 

On the third day, A’s and C’s Erdös numbers become (f + 1)

The sum of Erdös numbers changed to 8 × 3 = 24

At the end of the third day, five members had identical Erdös numbers while the other three had distinct ones.

On the fifth day, E’s Erdös numbers became f + 1 and this reduced the group’s average by 0.5. This means that E’s Erdös numbers was not f + 1 on the third day.

Thus we have, 
At the end of the third day, 5(f + 1) + f + e + y = 24
Hence 6f + 5 + e + y = 24
Hence 6f + e + y = 19

At the end of the fifth day,
6(f + 1) + f + y = 2.5 × 8 = 20
Hence 7f + y = 14

Among the eight mathematicians, F has the smallest Erdös number.
Let f = 2
∴ y = 0

However, only Paul Erdös himself has an Erdös number of 0. So f cannot be equal to 2. Any other value greater than 2, would render y as a negative number, which is also not possible.
So, f = 1
∴ y = 7
∴ e = 6

Now, we can solve all the questions.

From the above explanation, C’s Erdös number was f + 1 = 2 on the third day and thereafter.

Hence, option (b).

Answer: Option C

Text Explanation :

Let F and E have Erdös numbers f and e respectively at the beginning of the conference. 

On the third day, A’s and C’s Erdös numbers become (f + 1)

The sum of Erdös numbers changed to 8 × 3 = 24

At the end of the third day, five members had identical Erdös numbers while the other three had distinct ones.

On the fifth day, E’s Erdös numbers became f + 1 and this reduced the group’s average by 0.5. This means that E’s Erdös numbers was not f + 1 on the third day.

Thus we have, 
At the end of the third day, 5(f + 1) + f + e + y = 24
Hence 6f + 5 + e + y = 24
Hence 6f + e + y = 19

At the end of the fifth day,
6(f + 1) + f + y = 2.5 × 8 = 20
Hence 7f + y = 14

Among the eight mathematicians, F has the smallest Erdös number.
Let f = 2
∴ y = 0

However, only Paul Erdös himself has an Erdös number of 0. So f cannot be equal to 2. Any other value greater than 2, would render y as a negative number, which is also not possible.
So, f = 1
∴ y = 7
∴ e = 6

Now, we can solve all the questions.

From the above explanation, E’s Erdös number was 6.

Hence, option (c).

Answer: Option B

Text Explanation :

Since 5 participants had identical Erdös numbers at the end of day three and two of these were A and C whose Erdös numbers had changed on the same day, three had the same Erdös numbers at the beginning of the conference.

Hence, option (b).

Answer: Option C

Text Explanation :

Let the price of the share rise on x days and fall on y days. As the price increases by Rs. 10 in the five days, we have:
x + y = 5 and 10x – 10y = 10

Solving the above two equations simultaneously, we have
x = 3 and y = 2

The price of the share goes up on 3 days and falls on 2 days.

The three days on which the price rises can be selected in ⁵C₃ = 10 ways

The following are the 10 cases:

​​​​​​​

Consider Case 5:

Chetan sells on Days 1, 2 and 4 and buys on days 3 and 5.

Change in the number of shares he has = –30 + 20 = –10

Change in his cash = 10 × (110 + 120 + 120) – 10 × (110 + 110) = Rs. 1300

Michael sells on days 2 and 4, but never buys as the share price does not go below Rs. 90.

Change in the number of shares he has = –20

Change in his cash = 10 × (120 + 120) = Rs. 2400

The other cases are evaluated in a similar manner and the data is tabulated as shown above.

Chetan sold on three consecutive days: Cases 1, 2 and 3.

Michael sold only once: Case 3.

∴ The price of the share at the end of day 3 = Rs. 110

Hence, option (c).

Answer: Option E

Text Explanation :

Let the price of the share rise on x days and fall on y days. As the price increases by Rs. 10 in the five days, we have:
x + y = 5 and 10x – 10y = 10

Solving the above two equations simultaneously, we have
x = 3 and y = 2

The price of the share goes up on 3 days and falls on 2 days.

The three days on which the price rises can be selected in ⁵C₃ = 10 ways

The following are the 10 cases:

​​​​​​​

Consider Case 5:

Chetan sells on Days 1, 2 and 4 and buys on days 3 and 5.

Change in the number of shares he has = –30 + 20 = –10

Change in his cash = 10 × (110 + 120 + 120) – 10 × (110 + 110) = Rs. 1300

Michael sells on days 2 and 4, but never buys as the share price does not go below Rs. 90.

Change in the number of shares he has = –20

Change in his cash = 10 × (120 + 120) = Rs. 2400

The other cases are evaluated in a similar manner and the data is tabulated as shown above.

Chetan sold on three consecutive days: Cases 1, 2 and 3.

Michael sold only once: Case 3.

Michael ends up with Rs. 100 less cash than Chetan in cases 3, 6 and 8. In each of these cases, both of them hold the same number of shares at the end of day 5.

Hence, option (e).