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Mrs. Stark paid $6,000 for 100 shares of stock A and $4,000 for 100 sh

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Bunuel
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Mrs. Stark paid $6,000 for 100 shares of stock A and $4,000 for 100 sh [#permalink] New post   07 Feb 2020, 07:51
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Question Stats:

61% (02:33) correct 39% (02:39) wrong based on 127 sessions
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Mrs. Stark paid $6,000 for 100 shares of stock A and $4,000 for 100 shares of stock B. She later sold the same shares of both stocks, gaining 3/8 the amount she paid for stock B. If the prices she had paid for the stocks had been reversed and everything else remained the same, then the net result would have been:

(A) The same.

(B) A loss 1 1/2 times as much.

(C) A loss 2 times as much.

(D) A loss 9 times as much.

(E) A gain \(1 \frac{1}{2}\) times as much.


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E
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Re: Mrs. Stark paid $6,000 for 100 shares of stock A and $4,000 for 100 sh [#permalink] New post   07 Feb 2020, 17:09
Bunuel wrote:
Mrs. Stark paid $6,000 for 100 shares of stock A and $4,000 for 100 shares of stock B. She later sold the same shares of both stocks, gaining 3/8 the amount she paid for stock B. If the prices she had paid for the stocks had been reversed and everything else remained the same, then the net result would have been:

(A) The same.

(B) A loss 1 1/2 times as much.

(C) A loss 2 times as much.

(D) A loss 9 times as much.

(E) A gain \(1 \frac{1}{2}\) times as much.


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Given:
1. Mrs. Stark paid $6,000 for 100 shares of stock A and $4,000 for 100 shares of stock B.
2. She later sold the same shares of both stocks, gaining 3/8 the amount she paid for stock B.

Asked: If the prices she had paid for the stocks had been reversed and everything else remained the same, then the net result would have been:

Stock A:
Number of shares = 100
Total amount invested = $6000
Purchase price = $6000/100 = $600

Stock B:
Number of shares = 100
Total amount invested = $4000
Purchase price = $4000/100 = $400

Total gain by selling all shares = 3/8 * $4000 = $1500

If price for shares had been reversed

Stock A:
Number of shares = 100
Total amount invested = $4000
Purchase price = $4000/100 = $400

Stock B:
Number of shares = 100
Total amount invested = $6000
Purchase price = $6000/100 = $600

Total gain by selling all shares = 3/8*$6000 = $2250

Net gain = $2250/$1500 = 3/2 = 1.5

IMO E
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Re: Mrs. Stark paid $6,000 for 100 shares of stock A and $4,000 for 100 sh [#permalink] New post   07 Feb 2020, 20:11
Quick solution (time taken :40 secs)

net gain in the 1st transaction = 3/8 of 4000 = 1500 Rs.
now the prices are reversed. Since the number of shares is the same for Stock A and B, shares in Stock B were purchased at 6000 and stock A at 4000.
Net gain now = 3/8 * 6000 = 2250
the CP is same in both the cases. Since the profit increased, only one option depicts that IMO E
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Re: Mrs. Stark paid $6,000 for 100 shares of stock A and $4,000 for 100 sh [#permalink] New post   10 Feb 2020, 02:49
Net price per share of A = 6000/100 ; 60 and B = 4000/100 = 40
net gain of B ; 40*3/8 ; 15$
and when price are reversed net gain of B now becomes; 60*3/8 ; 22.5
so net result of gain ; 22.5/15 ; 1.5 $
IMO E

Bunuel wrote:
Mrs. Stark paid $6,000 for 100 shares of stock A and $4,000 for 100 shares of stock B. She later sold the same shares of both stocks, gaining 3/8 the amount she paid for stock B. If the prices she had paid for the stocks had been reversed and everything else remained the same, then the net result would have been:

(A) The same.

(B) A loss 1 1/2 times as much.

(C) A loss 2 times as much.

(D) A loss 9 times as much.

(E) A gain \(1 \frac{1}{2}\) times as much.


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Re: Mrs. Stark paid $6,000 for 100 shares of stock A and $4,000 for 100 sh [#permalink] New post   16 Aug 2021, 01:42
Expert Reply
Method I:

Cost price for share A = 100 * 6,000 = 60,000
Cost price for share B = 100 * 4,000 = 40,000

Gain happens: \(\frac{3}{ 8}\) * CP of share B = \(\frac{3}{ 8} * 40,000 = 15,000\)

NOTE :TRAP: Common Error: Now, if the price is reversed, the number of each share is still the same i.e. 100 so, it won't affect the CP. Hence, the net result will be the same..

But

The gain amount will change as it is \(\frac{3}{ 8}\) of CP of share B.

New CP of share B = Cost price for share B = 100 * 6,000 = 60,000

Gain happens: \(\frac{3}{ 8}\) * CP of share B = \(\frac{3}{ 8} * 60,000 = 22,500\)

Net result: \(\frac{22,500}{ 15,000} = \frac{225}{150} = \frac{3}{2} \)

Answer E


----------------------------------------------------------------------

Method II:

The number of shares remains the same even if the price is reversed.

Earlier A = 6 and B = 4

Later A= 4 and B = 6

\(\frac{3}{ 8}\) of CP of B.

Net result : \(\frac{3}{ 8}\) of 6 divided by \(\frac{3}{ 8}\) of 4 :

=> \(\frac{6}{4} = \frac{3}{2} \)

Answer E
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Re: Mrs. Stark paid $6,000 for 100 shares of stock A and $4,000 for 100 sh [#permalink]
16 Aug 2021, 01:42
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