# Section-1: Numerical Logic Solved QuestionLogical Reasoning Discussion

Common Information

Abdul, Bikram and Chetan are three professional traders who trade in shares of a company XYZ Ltd.

Abdul follows the strategy of buying at the opening of the day at 10 am and selling the whole lot at the close of the day at 3 pm.

Bikram follows the strategy of buying at hourly intervals: 10 am , 11 am, 12 noon, 1 pm and 2 pm, and selling the whole lot at the close of the day. Further, he buys an equal number of shares in each purchase.

Chetan follows a similar pattern as Bikram but his strategy is somewhat different. Chetan’s total investment amount is divided equally among his purchases.

The profit or loss made by each investor is the difference between the sale value at the close of the day less the investment in purchase.

The “return” for each investor is defined as the ratio of the profit or loss to the investment amount expressed as a percentage.

 Q. Common Information Question: 1/5 On a “boom” day the price of XYZ Ltd. keeps rising throughout the day and peaks at the close of the day. Which trader got the minimum return on that day?
 ✔ A. Bikram ✖ B. Chetan ✖ C. Abdul ✖ D. Abdul or Chetan ✖ E. Cannot be determined

Solution:
Option(A) is correct

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.

Table below can be scrolled horizontally

Time of the Day Share Price (in Rs.)
10 am (open) 100
11 am 110
12 noon 140
1 pm 150
2 pm 180
3 pm (close) 200

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

$\therefore$ 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 $180 \times 1 = \text{Rs. }180$.

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

At 3 pm, he sells all 5 shares for $200 \times 5 = \text{Rs. }1,000.$

Thus, his profit will be $1,000 - 680 = \text{Rs. }320$

Hence, Bikram's return is,

$=\dfrac{320}{680}\times 100= 47\%$

Let Chetan invest Rs. 415,800 at each interval.

So, at 10 am, he buys $415,800/100 = 4158$ shares;

at 11 am, he buys $415,800/110 = 3780$ shares;

at 12 noon, he buys $415,800/140 = 2970$ shares;

at 1 pm, he buys $415,800/150 = 2772$ shares;

at 2 pm, he buys $415,800/180 = 2310$ shares.

Thus, he buys $4158 + 3780 + 2970 + 2772 + 2310$ $= 15990$ shares for $415800 \times 5 = \text{Rs. }20,79,000$.

He sells these shares for $200 \times 15990 = \text{Rs. }31,98,000$.

His profit will be $3,198,000 - 2,079,000 = \text{Rs. }1,119,000$.

Hence Chetan's returns,

$=\dfrac{1,119,000}{2,079,000}\times 100$

$=\dfrac{373}{693}\times 100$

$\approx 53\%$

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 $x_1, x_2, ..., x_6$ 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 $\text{Rs. }x_1$ and sells them at $\text{Rs. }x_6$.

$\therefore$ Abdul's returns, $=\dfrac{x_6-x_1}{x_1}$

For Bikram:

Let Bikram have bought $n$ shares at each hourly interval.

His investment amount

$= nx_1 + nx_2 + nx_3 + nx_4 + nx_5$

$= n(x_1 + x_2 + x_3 + x_4 + x_5)$

$=n \times \sum\limits_{i=1}^{5}x_i$

At 3 pm, he sells his shares for $(5n \times x_6)$

Hence, his profit/loss,

$=(n \times 5x_6)- n \times \sum\limits_{i=1}^{5}x_i$

$= n \times \left( 5 x_6 - \sum\limits_{i=1}^{5}x_i\right)$

$\therefore$ Bikram’s returns,

$= \dfrac{n \times (5x_6 - \sum_{i=1}^{5}x_i)}{n \times \sum_{i=1}^{5}x_i}$

$=\left(\dfrac{5x_6}{\sum_{i=1}^{5}x_i}\right) -1$

$=\left(\dfrac{x_6}{\dfrac{\sum_{i=1}^{5}x_i}{5}}\right) -1$

Hence, Bikram’s returns,

$=\left(\dfrac{x_6}{\text{Arithmetic mean of }x_1, x_2, ... x_5}\right)-1$

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:

$\dfrac{P}{x_1}$ shares at 10 am; $\dfrac{P}{x_2}$ at 11 am; and so untill 2 pm.

At 3 pm, he sells each share at $x_6$. So, for all his shares, he receives,

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

$Px_6 \times \left(\dfrac{1}{x_1}+\dfrac{1}{x_2}+\dfrac{1}{x_3}+\dfrac{1}{x_4}+\dfrac{1}{x_5}\right)$

Hence, his profit/loss,

$=Px_6 \times \left(\dfrac{1}{x_1}+\dfrac{1}{x_2}+\dfrac{1}{x_3}+\dfrac{1}{x_4}+\dfrac{1}{x_5}\right) - 5P$

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

$\therefore$ Chetan’s returns,

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

$=\left(\dfrac{x_6 \times \left(\dfrac{1}{x_1}+\dfrac{1}{x_2}+\dfrac{1}{x_3}+\dfrac{1}{x_4}+\dfrac{1}{x_5}\right)}{5}\right)-1$

$=\left(\dfrac{x_6}{\dfrac{5}{\left(\dfrac{1}{x_1}+\dfrac{1}{x_2}+\dfrac{1}{x_3}+\dfrac{1}{x_4}+\dfrac{1}{x_5}\right)}}\right)-1$

$\therefore$ Chetan’s returns,

$=\left(\dfrac{x_6}{\text{Harmonic mean of } x_1, x_2, ... x_5}\right)-1$

Now, let’s compare Bikram’s and Chetan’s returns. Since Arithmetic Mean is always greater than or equal to the Harmonic Mean $(\text{A.M.} \geq \text{H.M.})$, Chetan’s returns will be greater than or equal to Bikram’s.