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30 Aug 2024, 01:30
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
59% (02:23) correct
41%
(02:30)
wrong
based on 197
sessions
History
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The price of a share of a certain company went up by x% on Monday and on the following day it went down by y% of its opening share price on Tuesday. Is the share price at the end of the day on Tuesday higher than the opening price on Monday?
(1) \(x > y\)
(2) \(x - y > \frac{xy}{100}\)
(1) \(x > y\)
(2) \(x - y > \frac{xy}{100}\)
30 Aug 2024, 05:01
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Let the opening price on Monday = \(o\)
First it went up by \(x\)% and then went down by \(y\)%:
\(o[(1+\frac{x}{100})(1-\frac{y}{100})]\)
\(o[1+\frac{x}{100}-\frac{y}{100}-\frac{xy}{10000}]\)
The share price will be higher on Tuesday than the opening price on Monday if \(\frac{x}{100}-\frac{y}{100}\)is greater than \(\frac{xy}{10000}\)
(1) x>y
Let the cost of the share on Monday = \(50\).
[Case 1] x = 100 and y = 10: Closing price on Monday:\(50*(1+\frac{100}{100}) = 100\), closing price on Tuesday: \(100*(1-\frac{10}{100}) = 90\). In this case the price on Tuesday is greater.
[Case 2] x = 100 and y = 50: Closing price on Monday:\(50*(1+\frac{100}{100}) = 100\), closing price on Tuesday: \(100*(1-\frac{50}{100}) = 50\). In this case the price on Tuesday is the same as Monday.
INSUFFICIENT
(2) x−y>xy/100
Taking \(x−y>\frac{xy}{100}\) and multiplying through by \(\frac{1}{100}\) gives: \(\frac{x−y}{100}>\frac{xy}{10000}\)
This is identical to what one finds in \(o[1+\frac{x}{100}-\frac{y}{100}-\frac{xy}{10000}]\).
As \(\frac{x}{100}-\frac{y}{100}\) is greater than \(\frac{xy}{10000}\); subtracting \(\frac{xy}{10000}\) from \(\frac{x}{100}-\frac{y}{100}\), adding \(1\) and then multiplying that through by \(o\) can only result in a value greater than \(o\). Therefore, the price of the share at the end of Tuesday will be greater than the share's opening price on Monday.
SUFFICIENT
ANSWER B
First it went up by \(x\)% and then went down by \(y\)%:
\(o[(1+\frac{x}{100})(1-\frac{y}{100})]\)
\(o[1+\frac{x}{100}-\frac{y}{100}-\frac{xy}{10000}]\)
The share price will be higher on Tuesday than the opening price on Monday if \(\frac{x}{100}-\frac{y}{100}\)is greater than \(\frac{xy}{10000}\)
(1) x>y
Let the cost of the share on Monday = \(50\).
[Case 1] x = 100 and y = 10: Closing price on Monday:\(50*(1+\frac{100}{100}) = 100\), closing price on Tuesday: \(100*(1-\frac{10}{100}) = 90\). In this case the price on Tuesday is greater.
[Case 2] x = 100 and y = 50: Closing price on Monday:\(50*(1+\frac{100}{100}) = 100\), closing price on Tuesday: \(100*(1-\frac{50}{100}) = 50\). In this case the price on Tuesday is the same as Monday.
INSUFFICIENT
(2) x−y>xy/100
Taking \(x−y>\frac{xy}{100}\) and multiplying through by \(\frac{1}{100}\) gives: \(\frac{x−y}{100}>\frac{xy}{10000}\)
This is identical to what one finds in \(o[1+\frac{x}{100}-\frac{y}{100}-\frac{xy}{10000}]\).
As \(\frac{x}{100}-\frac{y}{100}\) is greater than \(\frac{xy}{10000}\); subtracting \(\frac{xy}{10000}\) from \(\frac{x}{100}-\frac{y}{100}\), adding \(1\) and then multiplying that through by \(o\) can only result in a value greater than \(o\). Therefore, the price of the share at the end of Tuesday will be greater than the share's opening price on Monday.
SUFFICIENT
ANSWER B
General Discussion
08 Oct 2025, 13:27
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Hi Nidzo,
How it can be assumed that opening price on Tuesday will be same as opening price on Monday ? In real markets opening price can be different from the closing price on the previous day. So IMO answer should be E as we dont' have any relation between opening price on Tuesday and closing price on Monday.
How it can be assumed that opening price on Tuesday will be same as opening price on Monday ? In real markets opening price can be different from the closing price on the previous day. So IMO answer should be E as we dont' have any relation between opening price on Tuesday and closing price on Monday.
Nidzo
