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Williams has x eggs. He sells y of them at a profit of 10 pe

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Williams has x eggs. He sells y of them at a profit of 10 pe [#permalink] New post   29 Apr 2019, 02:25
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Williams has x eggs. He sells y of them at a profit of 10 percent and the remaining eggs at a loss of 10 percent. He made a profit overall.

Quantity A
Quantity B
x-y
y


A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
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B
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grewhiz
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Re: Williams has x eggs. He sells y of them at a profit of 10 pe [#permalink] New post   16 May 2019, 04:54
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Hi,

The answer is (D).

First, mentally, assume the cost price of 1 egg to be 'k'. Williams first sells 'y' eggs at 10% profit, which means yk*1.1.
Second, he sells the remaining eggs at a loss of 10%, which means k(x-y)*0.9.

Now, it is mentioned that he made a profit, and this is where you need to catch that without the values of 'x' and 'y', it is impossible to determine the greater value between '(x-y)' and 'y'.

For instance, if (x-y)=y, then it is still a profit of 20%, as mentioned in the question. However, if (x-y)<y, even then, there is a profit.

Take an assumption to understand this.

x=10
y=6
Cost Price per egg = $100

Therefore, the Selling Price for 6 eggs is 100*1.1*6 = 660.

x-y=4, therefore selling price for 4 eggs is 100*0.9*4 = 360. Therefore, the overall profit is 660+360-200=820.

But if x-y=y, then Selling Price 1 for 5 eggs is 100*1.1*5=550, and Selling Price 2 is 100*0.9*5=450. Therefore, overall profit is 1000-200=800.

Therefore, in both cases, we find that Williams makes an overall profit. The moment there are two equally plausible cases, the answer becomes unverifiable, and is therefore (D).
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Re: Williams has x eggs. He sells y of them at a profit of 10 pe [#permalink] New post   30 Jul 2019, 07:18
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Answer is B. if x and y were same then there would be no profit or loss. But here is get profit so Y>x. Quantity A something negative. Answer is B
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Re: Williams has x eggs. He sells y of them at a profit of 10 pe [#permalink] New post   30 Jul 2019, 17:05
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Answer is B.

X = # of eggs
Y = # of eggs sold
X - Y = # of eggs NOT sold

Since he made a profit: Y > X - Y, therefore B must be greater than A
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Re: Williams has x eggs. He sells y of them at a profit of 10 pe [#permalink] New post   31 Jul 2019, 07:23
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grewhiz wrote:
Hi,

The answer is (D).

First, mentally, assume the cost price of 1 egg to be 'k'. Williams first sells 'y' eggs at 10% profit, which means yk*1.1.
Second, he sells the remaining eggs at a loss of 10%, which means k(x-y)*0.9.

Now, it is mentioned that he made a profit, and this is where you need to catch that without the values of 'x' and 'y', it is impossible to determine the greater value between '(x-y)' and 'y'.

For instance, if (x-y)=y, then it is still a profit of 20%, as mentioned in the question. However, if (x-y)<y, even then, there is a profit.

Take an assumption to understand this.

x=10
y=6
Cost Price per egg = $100

Therefore, the Selling Price for 6 eggs is 100*1.1*6 = 660.

x-y=4, therefore selling price for 4 eggs is 100*0.9*4 = 360. Therefore, the overall profit is 660+360-200=820.

But if x-y=y, then Selling Price 1 for 5 eggs is 100*1.1*5=550, and Selling Price 2 is 100*0.9*5=450. Therefore, overall profit is 1000-200=800.

Therefore, in both cases, we find that Williams makes an overall profit. The moment there are two equally plausible cases, the answer becomes unverifiable, and is therefore (D).

Please review the calculations, Answer is (B)
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Re: Williams has x eggs. He sells y of them at a profit of 10 pe [#permalink] New post   01 Aug 2019, 13:56
As he is always making profit y will be more than 50% of x so the answer will be always b
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Re: Williams has x eggs. He sells y of them at a profit of 10 pe [#permalink] New post   02 Aug 2019, 11:05
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Y eggs value = p(y) x 1.1
Rest Eggs value = P(x-y) x 0.9

p(y) x 1.1 + P(x-y) x 0.9 > p(x)

above equation defined due to profit mentioned

leads to 0.2y > x-0.9x
0.2y > 0.1x
2y > x
y > x/2

using this condition answer will be "B"
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Re: Williams has x eggs. He sells y of them at a profit of 10 pe [#permalink] New post   05 Aug 2020, 15:23
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Carcass wrote:
Williams has x eggs. He sells y of them at a profit of 10 percent and the remaining eggs at a loss of 10 percent. He made a profit overall.

Quantity A
Quantity B
x-y
y


A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.



Hi the Answer is D

By using Fill in, here are the situations:

1) X = 100
- 50 * 1.1 = 55 (y)
- 50 * 0.9 - 45
profit = 10

A: 100 - 50 = 50
B: 50

2) X = 100
- 47 * 1.1 = 51.7
- 53 * 0.9 = 47.7
profit = 4

A: 100 - 47 = 53
B: 47

3) X = 100
- 53 * 1.1 = 58.3
- 47 * 0.9 = 42.3
profit = 16

A: 47
B: 53
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Williams has x eggs. He sells y of them at a profit of 10 pe [#permalink] New post   Updated on: 02 Jul 2021, 08:29
Roro22 wrote:
Carcass wrote:
Williams has x eggs. He sells y of them at a profit of 10 percent and the remaining eggs at a loss of 10 percent. He made a profit overall.

Quantity A
Quantity B
x-y
y


A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.



Hi the Answer is D

By using Fill in, here are the situations:

1) X = 100
- 50 * 1.1 = 55 (y)
- 50 * 0.9 - 45
profit = 10

A: 100 - 50 = 50
B: 50

2) X = 100
- 47 * 1.1 = 51.7
- 53 * 0.9 = 47.7
profit = 4

A: 100 - 47 = 53
B: 47

3) X = 100
- 53 * 1.1 = 58.3
- 47 * 0.9 = 42.3
profit = 16

A: 47
B: 53


I'm afraid your answer does not take into account the cost price of each egg.

For instance, consider the first situation in your solution.
x=100, y=50, x-y=50.
Let c be the cost price per egg.
Then, C.P. = 100c
S.P. = (1.1)(50c) + (0.9)(50c) = 55c + 45c = 100c
So, Profit = S.P. - C.P. =100c -100c = 0, which is a contradiction to the assumption that he makes a profit overall.
So, x=100, y=50 does not work in this case.


Now, let's consider the second situation.
x=100, y=47, x-y=53.
Let c be the cost price per egg.
Then, C.P. = 100c
S.P. = (1.1)(47c) + (0.9)(53c) = (51.7)c + (47.7)c = (99.4)c
So, Profit = S.P. - C.P. =(99.4)c -100c = -(0.6)c, which again leads to a contradiction since profit is negative in this case (which, in turn, implies that he made a loss over all).
So, x=100, y=47 does not work as well.

Originally posted by BM137 on 02 Jul 2021, 07:51.
Last edited by BM137 on 02 Jul 2021, 08:29, edited 1 time in total.
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Williams has x eggs. He sells y of them at a profit of 10 pe [#permalink] New post   02 Jul 2021, 08:06
Let c be the cost price per egg. Then, total C.P. = cx.
From the question, it is easy to see that the selling price of y eggs = (1.1c)y = (1.1)cy.
Similarly, selling price of the remaining (x-y) eggs = (0.9c)(x-y) = (0.9)xc - (0.9)cy.
So, total S.P. = (1.1)cy + [(0.9)xc - (0.9)cy] = (0.2)cy + (0.9)cx.
Hence, profit = total S.P. - total C.P. = [(0.2)cy + (0.9)cx] - cx = (0.2)cy - (0.1)cx.
Since Williams made a profit over all, profit > 0 ---> (0.2)cy - (0.1)cx > 0 ---> (0.2)cy > (0.1)cx ---> 2y > x ---> y > x-y.
So, Quantity B is greater and therefore the answer is B.
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Re: Williams has x eggs. He sells y of them at a profit of 10 pe [#permalink] New post   02 Jul 2021, 10:24
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Carcass wrote:
Williams has x eggs. He sells y of them at a profit of 10 percent and the remaining eggs at a loss of 10 percent. He made a profit overall.

Quantity A
Quantity B
x-y
y


A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.


Col. A: x - y
Col. B: y

Adding y to both sides,

Col. A: x
Col. B: 2y

Now let us make cases;

For example:
Let the original price of each egg be 10
Therefore, at 10% profit; the price would be 11 and,
at 10% loss; the price would be 9

Case I: x = 2y - Not possible
Let x = 100, y = 50
remaining = 50

Cost Price = 100(10) = 1000
Sale Price = 50(11) + 50(9) = 550 + 450 = 1000
Overall profit = 0

Case II: x > 2y - Not Possible
Let x = 100, y = 49
remaining = 51

Cost Price = 100(10) = 1000
Sale Price = 49(11) + 51(9) = 539 + 459 = 998
Overall Loss = 2

Case III: x < 2y - Only Possible case
Let x = 100, y = 51
remaining = 49

Cost Price = 100(10) = 1000
Sale Price = 51(11) + 49(9) = 561 + 441 = 1002
Overall profit = 2

Since, x can only be less than 2y

Hence, option B
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Re: Williams has x eggs. He sells y of them at a profit of 10 pe [#permalink] New post   15 Jul 2024, 03:40
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Hi Answer is D.

We know only the percent changes here not the quantities of eggs as it is.

Loss or profit doesn't provide a logical explanation with the number of eggs at the first place.
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Re: Williams has x eggs. He sells y of them at a profit of 10 pe [#permalink]
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