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01 Oct 2025, 01:37
Kudos
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
65% (01:39) correct
35%
(01:48)
wrong
based on 146
sessions
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An airline company pays $10,000 less for each additional airplane ordered. What would be the total price for 5 airplanes?
(1) The first 2 airplanes cost 50% of the total price.
(2) The last airplane costs half as much as the first.
(1) The first 2 airplanes cost 50% of the total price.
(2) The last airplane costs half as much as the first.
01 Oct 2025, 02:03
Kudos
Bookmarks
Let the cost of the first airplane = P.
Each next airplane costs $10,000 less than the previous one.
So costs are: P, (P – 10,000), (P – 20,000), (P – 30,000), (P – 40,000).
Total for 5 planes = 5P – (10,000 + 20,000 + 30,000 + 40,000) = 5P – 100,000.
We need P to find the total.
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(1) The first 2 airplanes cost 50% of the total price.
Equation: P + (P – 10,000) = 0.5 × (5P – 100,000).
Simplify: 2P – 10,000 = 2.5P – 50,000 → 40,000 = 0.5P → P = 80,000.
Then total = 5(80,000) – 100,000 = 400,000 – 100,000 = 300,000.
Unique value → Sufficient.
(2) The last airplane costs half as much as the first.
So: (P – 40,000) = 0.5P.
Simplify: P – 40,000 = 0.5P → 0.5P = 40,000 → P = 80,000.
Then total = 300,000.
Unique value → Sufficient.
Answer: D
Each next airplane costs $10,000 less than the previous one.
So costs are: P, (P – 10,000), (P – 20,000), (P – 30,000), (P – 40,000).
Total for 5 planes = 5P – (10,000 + 20,000 + 30,000 + 40,000) = 5P – 100,000.
We need P to find the total.
---
(1) The first 2 airplanes cost 50% of the total price.
Equation: P + (P – 10,000) = 0.5 × (5P – 100,000).
Simplify: 2P – 10,000 = 2.5P – 50,000 → 40,000 = 0.5P → P = 80,000.
Then total = 5(80,000) – 100,000 = 400,000 – 100,000 = 300,000.
Unique value → Sufficient.
(2) The last airplane costs half as much as the first.
So: (P – 40,000) = 0.5P.
Simplify: P – 40,000 = 0.5P → 0.5P = 40,000 → P = 80,000.
Then total = 300,000.
Unique value → Sufficient.
Answer: D
01 Oct 2025, 04:26
Kudos
Bookmarks
Bunuel
Let the initial cost of the plane be a.
The second plane costs b = a - 10000.
Third plane c = a - 20000.
Fourth plane d = a - 30000.
Fifth plane e = a - 40000.
Total ( T) = a + b + c + d + e = 5a - 100000.
If we able to calculate a, then T can be easily found out.
Statement 1:
(1) The first 2 airplanes cost 50% of the total price.
a + (a - 10000) = (1/2) * 5a - 100000
solving it we get , a = 80000.
Hence, T can be found out.
Sufficient.
Statement 2:
(2) The last airplane costs half as much as the first.
e = (1/2)* a
a - 40000 = (1/2) * a
solving it we get , a = 80000.
Hence, T can be found out.
Sufficient.
Option D
