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17 Mar 2023, 01:30
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Difficulty:
45%
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Question Stats:
65% (01:54) correct
35%
(01:45)
wrong
based on 109
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The average (arithmetic mean) price of three items sold by a store to a particular customer, before discounts, was $90. If a discount of 20% was applied at the point of purchase to all items of at least $100, what was the total discount saved by the customer?
(1) The pre-discount price of the most expensive item that the customer purchased from the store was $180.
(2) The pre-discount price of the least expensive item that the customer purchased from the store was $10.
(1) The pre-discount price of the most expensive item that the customer purchased from the store was $180.
(2) The pre-discount price of the least expensive item that the customer purchased from the store was $10.
17 Mar 2023, 03:38
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The average (arithmetic mean) price of three items sold by a store to a particular customer, before discounts, was $90. If a discount of 20% was applied at the point of purchase to all items of at least $100, what was the total discount saved by the customer?
(1) The pre-discount price of the most expensive item that the customer purchased from the store was $180.
(2) The pre-discount price of the least expensive item that the customer purchased from the store was $10.
Solution:
average price of three items = 90
Let the three items be = x,y,z
\(\frac{x+y+z}{3}=90\)
\(x+y+z = 270\)
We are also given that 20% was applied to all items of at least $100. So, we need to know the exact number of items with at least a price of $100 in order to find the total discount saved by the customer.
(1) The pre-discount price of the most expensive item that the customer purchased from the store was $180.
x+y+180=270
x+y=90
This means the price of other two items must be less than $100. So, there is only one item on which the 20% discount is applicable. Sufficient. Eliminate B,C, and E
(2) The pre-discount price of the least expensive item that the customer purchased from the store was $10.
10+y+z=270
y+z=260
From this we do not know the exact price of the other two items. The price of both the remaining items could be more than $100 or it could more than $100 for one and less than $100 for the other. Hence, we cannot answer the question. Eliminate B. Hence, A IMO
(1) The pre-discount price of the most expensive item that the customer purchased from the store was $180.
(2) The pre-discount price of the least expensive item that the customer purchased from the store was $10.
Solution:
average price of three items = 90
Let the three items be = x,y,z
\(\frac{x+y+z}{3}=90\)
\(x+y+z = 270\)
We are also given that 20% was applied to all items of at least $100. So, we need to know the exact number of items with at least a price of $100 in order to find the total discount saved by the customer.
(1) The pre-discount price of the most expensive item that the customer purchased from the store was $180.
x+y+180=270
x+y=90
This means the price of other two items must be less than $100. So, there is only one item on which the 20% discount is applicable. Sufficient. Eliminate B,C, and E
(2) The pre-discount price of the least expensive item that the customer purchased from the store was $10.
10+y+z=270
y+z=260
From this we do not know the exact price of the other two items. The price of both the remaining items could be more than $100 or it could more than $100 for one and less than $100 for the other. Hence, we cannot answer the question. Eliminate B. Hence, A IMO
19 Mar 2023, 07:03
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Let price of three items be x,y,z respectively
Average price of three items = \( \frac{x+y+z}{3} \) = 90
=> x+y+z = 270
(1) The pre-discount price of the most expensive item that the customer purchased from the store was $180.
Let x = $180
=> 180+y+z = 270
=> y+z = 90
=> price of other two items is less than $100. Therefore, 20% discount is not applicable on y and z.
=> Total discount saved by customer = 20% * 180 = $36
(1) is sufficient.
(2) The pre-discount price of the least expensive item that the customer purchased from the store was $10.
Let x = $10
=> 10+y+z = 270
=> y+z = 260
=> Values of y and z could be $170, $90 or $130, $130.
Therefore, exact value of discounts could not be answered.
(2) is insufficient
Answer A
Average price of three items = \( \frac{x+y+z}{3} \) = 90
=> x+y+z = 270
(1) The pre-discount price of the most expensive item that the customer purchased from the store was $180.
Let x = $180
=> 180+y+z = 270
=> y+z = 90
=> price of other two items is less than $100. Therefore, 20% discount is not applicable on y and z.
=> Total discount saved by customer = 20% * 180 = $36
(1) is sufficient.
(2) The pre-discount price of the least expensive item that the customer purchased from the store was $10.
Let x = $10
=> 10+y+z = 270
=> y+z = 260
=> Values of y and z could be $170, $90 or $130, $130.
Therefore, exact value of discounts could not be answered.
(2) is insufficient
Answer A
