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.

Share price was at its highest at

• A. 1 pm
• B. 11 am
• C. 12 noon
• D. 10 am
• E. cannot be determined
• A. Share price was at its lowest at 2 pm
• B. more than one of the above
• C. Share price at 1 pm was higher than the share price at 2 pm
• D. Share price at 1 pm was higher than the share price at 12 noon
• J. Share price was at its lowest at 11 am

Answer 1

The Correct Option is D

Step 1: Translate the observations.
Let the prices at 10 am, 11 am, 12 noon, 1 pm, 2 pm, 3 pm be \(a,b,c,d,e,f\).
(i) Abdul bought at 10 am and sold at 3 pm and lost: \(f<a\).
(iii) \(f>e\). (iv) \(c<a\).
(ii) Dane buys one unit each at \(a,b,c\) and sells at \(d,e,f\): profit means \(d+e+f>a+b+c\).
Emily's money grows by the factor \(\dfrac{c}{a}\cdot\dfrac{f}{d}\); profit means \(cf>ad\), i.e. \(d<\dfrac{cf}{a}\).

Step 2: Compare each price with the opening price \(a\).

• \(f<a\) (from i), \(c<a\) (from iv).
• \(e<f<a\) (from iii and i).
• \(d<\dfrac{cf}{a}<f<a\), because \(c<a\) makes \(\dfrac{c}{a}<1\).
• For \(b\): Dane's profit gives \(a+b+c<d+e+f<\dfrac{cf}{a}+2f<c+2a\) (using \(f<a\)). So \(b<a\).

Thus \(b,c,d,e,f\) are all less than \(a\): the price was highest at 10 am.

Step 3: Check that the data are consistent.
Take \(a=195,\ b=65,\ c=190,\ d=180,\ e=90,\ f=185\): Abdul lost (\(185<195\)); \(f>e\); \(c<a\); Dane: \(455>450\); Emily: \(cf=35150>ad=35100\). All conditions hold and the maximum is at 10 am.

Final Answer:
The correct option is (4): 10 am

Answer 2

Step 1: Inequalities.
Let the prices at 10 am, 11 am, 12 noon, 1 pm, 2 pm, 3 pm be \(a,b,c,d,e,f\).
(i) Abdul bought at 10 am and sold at 3 pm and lost: \(f<a\).
(iii) \(f>e\). (iv) \(c<a\).
(ii) Dane buys one unit each at \(a,b,c\) and sells at \(d,e,f\): profit means \(d+e+f>a+b+c\).
Emily's money grows by the factor \(\dfrac{c}{a}\cdot\dfrac{f}{d}\); profit means \(cf>ad\), i.e. \(d<\dfrac{cf}{a}\).

Step 2: 1 pm versus 12 noon (option 4).
\(d<\dfrac{cf}{a}\) and \(f<a\) give \(d<c\). So the 1 pm price is always lower than the 12 noon price: option (4) is necessarily false.

Step 3: Lowest at 2 pm (option 1).
Suppose \(e\) is the lowest price. Then \(e<b\). We also know \(d<c\) and \(f<a\). Adding: \(d+e+f<c+b+a\), which contradicts Dane's profit. So the price cannot be lowest at 2 pm: option (1) is also necessarily false.

Step 4: Options 5 and 3 are possible.
Take \(a=195,\ b=65,\ c=190,\ d=180,\ e=90,\ f=185\) (all four observations hold: \(185<195\), \(185>90\), \(190<195\), \(455>450\), \(35150>35100\)). Here the lowest price is at 11 am (option 5 possible) and the 1 pm price 180 exceeds the 2 pm price 90 (option 3 possible).

Options (1) and (4) are both necessarily false, so more than one of the first four options is necessarily false.

Final Answer:
The correct option is (2): more than one of the above