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Mark bought shares for a total value of $1000. Next day, the price of

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Bunuel
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Mark bought shares for a total value of $1000. Next day, the price of [#permalink] New post   29 Jan 2015, 08:41
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Mark bought shares for a total value of $1000. Next day, the price of shares is reduced by 10%. With the same $1000, he could now buy 10 more shares. Which of the following is closest to the original price per share?

A. $10
B. $11
C. $12
D. $13
E. $14
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B
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Re: Mark bought shares for a total value of $1000. Next day, the price of [#permalink] New post   29 Jan 2015, 09:13
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PX = 1000...........(a)

(0.9P)(X+10) = 1000......(b)

0.9PX + 9P = PX...
0.1PX = 9P...
X = 90...
Substitute in original...P = 11.1111
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Re: Mark bought shares for a total value of $1000. Next day, the price of [#permalink] New post   02 Feb 2015, 03:57
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Bunuel wrote:
Mark bought shares for a total value of $1000. Next day, the price of shares is reduced by 10%. With the same $1000, he could now buy 10 more shares. Which of the following is closest to the original price per share?

A. $10
B. $11
C. $12
D. $13
E. $14

Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION:

From the given information,we can conclude that Mark bought x number of shares for $1000. With a 10% reduction in price, he could buy the same x number of shares for 1000*.90 = $900 . With the remaining money ($1000 - $900 = $100), he can buy 10 more shares. So, the reduced price per share = $100/10 = $10.

Therefore, if the reduced price of $10 equals the original price * 9/10, then the original price is approximately $11.
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Re: Mark bought shares for a total value of $1000. Next day, the price of [#permalink] New post   27 Mar 2015, 01:06
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Answer = B = 11

Price/share ................. No of shares ................. Total

x ................................. \(\frac{1000}{x}\) .......................... 1000 (Let x = price per share)

\(\frac{9x}{10}\) .................... \(\frac{1000}{x}\) ........................... 900 (Price of shares reduced by 10% will affect price per share)

\(\frac{9x}{10}\) .................. \(\frac{1000}{x} + 10\) ..................... 1000 (Reduced price per share would give 10 more shares in the same amount)

Setting up the equation

\((\frac{9x}{10})(\frac{1000}{x} + 10) = 1000\)

\(x = \frac{1000}{9} - 100 = \frac{100}{9} = 11.111\)

Answer = B
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Re: Mark bought shares for a total value of $1000. Next day, the price of [#permalink] New post   17 Feb 2020, 07:11
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Bunuel wrote:
Mark bought shares for a total value of $1000. Next day, the price of shares is reduced by 10%. With the same $1000, he could now buy 10 more shares. Which of the following is closest to the original price per share?

A. $10
B. $11
C. $12
D. $13
E. $14

Kudos for a correct solution.


Suppose that number of shares is n
Suppose that price per share is p

1st day:
n*p=1000 (1)

2nd day:
Price per share is 0.9p
Number of shares is 10+n
0.9p * (10+n) =1000
9p + 0.9n*p = 1000 (2)

Replace (1) in (2):
9p + 900 = 1000
9p = 100

p = 11
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Re: Mark bought shares for a total value of $1000. Next day, the price of [#permalink] New post   30 Sep 2021, 22:22
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Lets use a strategy to solve this question under 2min.

For two quantities that are inversely proportional,

(i)an increase of x/y in one causes a corresponding decrease of x/(x +y) in the other

and

(ii)a decrease of x/y in one causes an increase of x/(y-x) in the other.

Price and consumption when expenditure is constant are inversely related.

The price of shares is reduced by 10% = 1/10 as a fraction with x =1 and y=10

Thus corresponding increase in quantity purchases is x/(y-x) = 1/9

1/9 is equivalent to 10 shares

=>1 or the original number of shares he bought = 9*10 = 90 shares

90 shares were bought for a total value of $1000

Price of 1 share = 1000/90 = 11(approx)

(option b )

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Re: Mark bought shares for a total value of $1000. Next day, the price of [#permalink] New post   22 Mar 2015, 15:11
Price x Quantity = Amount(constant)
Since price is reduced to 9/10, so quantity will be increased by reversed amount 10/9 to make the amount constant.
Given increase in number of shares,i.e. 1/9 of Q is 10. so Quantity=90.
Price per share = Amount /quantity= 1000/90=11.11

B
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Re: Mark bought shares for a total value of $1000. Next day, the price of [#permalink] New post   23 Mar 2015, 05:05
Bunuel wrote:
Mark bought shares for a total value of $1000. Next day, the price of shares is reduced by 10%. With the same $1000, he could now buy 10 more shares. Which of the following is closest to the original price per share?

A. $10
B. $11
C. $12
D. $13
E. $14

Kudos for a correct solution.


1000/(0.9x) - 1000/x = 10

let 1000/x = k

k/0.9 - k = 10
=> k = 90

so 1000/x = 90
=> x = 11.11 or 11
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Re: Mark bought shares for a total value of $1000. Next day, the price of [#permalink] New post   16 Jul 2015, 08:52
PX = 1000...........(a)

(0.9P)(X+10) = 1000......(b)

0.9PX + 9P = PX...
0.1PX = 9P...
X = 90...
Substitute in original...P = 11.1111
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Re: Mark bought shares for a total value of $1000. Next day, the price of [#permalink] New post   24 Oct 2016, 08:39
The following equation(s) can be derived from the prompt:

P = price
N = number of stock

(1) PN = 1000
(2) (P-0.10P)(N+10) = 1000

Plug (1) into (2) and you'll find P = 11.111....

Thus, B is the correct answer.
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Re: Mark bought shares for a total value of $1000. Next day, the price of [#permalink] New post   11 Dec 2021, 06:48
Bunuel wrote:
Mark bought shares for a total value of $1000. Next day, the price of shares is reduced by 10%. With the same $1000, he could now buy 10 more shares. Which of the following is closest to the original price per share?

A. $10
B. $11
C. $12
D. $13
E. $14

Kudos for a correct solution.

Price = X
Quantity = Y
Hence, 1000 = XY
ATQ,
Price = .9X
Quantity = Y+10
Now,
XY = .9X*(Y+10)
XY = .9XY + 9X
Y = .9Y + 9Y
Y = 90
We supposed, XY = 1000 and we learned, Y =90
X*90 = 1000 => X = Price = 1000/90 = $11.11 = $11 Approx.

Ans. B
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Re: Mark bought shares for a total value of $1000. Next day, the price of [#permalink] New post   09 Dec 2023, 00:30
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