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D
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Difficulty:
95%
(hard)
Question Stats:
59% (03:00) correct
41%
(03:02)
wrong
based on 583
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At a craft fair, Ruth's total revenue from selling 2 types of floral arrangements was $832. She sold x arrangements of Type A for y dollars each and (131 - x) arrangements of Type B for (y + 3) dollars each. If x and y are both integers, did Ruth sell more Type A arrangements than Type B arrangements?
(1) If Ruth had sold 2 more Type A arrangements and 5 more Type B arrangements, her total revenue would have been $882.
(2) If Ruth had sold (131 - x) Type A arrangements and x Type B arrangements, her total revenue would have been more than $832.
Q.png [ 82.47 KiB | Viewed 10069 times ]
(1) If Ruth had sold 2 more Type A arrangements and 5 more Type B arrangements, her total revenue would have been $882.
(2) If Ruth had sold (131 - x) Type A arrangements and x Type B arrangements, her total revenue would have been more than $832.
Attachment:
Q.png [ 82.47 KiB | Viewed 10069 times ]
09 Nov 2023, 03:41
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At a craft fair, Ruth's total revenue from selling 2 types of floral arrangements was $832. She sold x arrangements of Type A for y dollars each and (131 - x) arrangements of Type B for (y + 3) dollars each. If x and y are both integers, did Ruth sell more Type A arrangements than Type B arrangements?
Any of the two info
(a) Total are x+131-x or 131 for $832, that is $6.3 each. So we should know the ratio of per piece price or value of x
(b) The equation can be written as x*y+(131-x)(y+3)=832. Two variables, so we require another equation.
(1) If Ruth had sold 2 more Type A arrangements and 5 more Type B arrangements, her total revenue would have been $882.
2y+5(y+3)=882-832
We will get y and can find x.
Sufficient
(2) If Ruth had sold (131 - x) Type A arrangements and x Type B arrangements, her total revenue would have been more than $832.
The numbers have been interchanged. Revenue will increase if we are purchasing more of higher value.
As y+3 is of higher value, and we are buying x of them, x>131-x
Sufficient
D
Any of the two info
(a) Total are x+131-x or 131 for $832, that is $6.3 each. So we should know the ratio of per piece price or value of x
(b) The equation can be written as x*y+(131-x)(y+3)=832. Two variables, so we require another equation.
(1) If Ruth had sold 2 more Type A arrangements and 5 more Type B arrangements, her total revenue would have been $882.
2y+5(y+3)=882-832
We will get y and can find x.
Sufficient
(2) If Ruth had sold (131 - x) Type A arrangements and x Type B arrangements, her total revenue would have been more than $832.
The numbers have been interchanged. Revenue will increase if we are purchasing more of higher value.
As y+3 is of higher value, and we are buying x of them, x>131-x
Sufficient
D
06 Nov 2024, 00:21
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Blackcrow1972
The question asks whether x > 131 - x, which translates to asking if x > 65.5. So, essentially, the question is asking whether Ruth sold more than half of the total 131 arrangements of Type A.
From the stem, we have xy + (131 - x)(y + 3) = 832.
From (1), we have 2y + 5(y + 3) = 50. Together with the equation above, we find that x = 72 and y = 5 (though solving isn’t strictly necessary here; the main point is that we can determine x to answer the question). So, the answer is YES.
From (2), we know that by interchanging the quantities of Type A and Type B arrangements sold, Ruth would have paid more than with the original quantities ((131 - x)y + x(y + 3) > 832). Since the price of Type A, y, is less than the price of Type B, y + 3, this implies that x must be greater than 131 - x (because assigning x to the higher price, y + 3, results in a greater total than assigning it to y). So, the answer is again YES. Note that x = 72 and y = 5 still satisfy (131 - x)y + x(y + 3) > 832, so there’s no contradiction between the statements.
Answer: D.
Hope it's clear.
