Discussion

Question:

One day, two other traders, Dane and Emily joined Abdul, Bikram and Chetan for trading in the shares of XYZ Ltd. Dane followed a strategy of buying equal numbers of shares at 10 am, 11 am and 12 noon, and selling the same numbers at 1 pm, 2 pm and 3 pm. Emily, on the other hand, followed the strategy of buying shares using all her money at 10 am and selling all of them at 12 noon and again buying the shares for all the money at 1 pm and again selling all of them at the close of the day at 3 pm. At the close of the day the following was observed:

i. Abdul lost money in the transactions.
ii. Both Dane and Emily made profits.
iii. There was an increase in share price during the closing hour compared to the price at 2 pm.
iv. Share price at 12 noon was lower than the opening price.

Which of the following is necessarily false?

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.

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