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30 Dec 2023, 03:05
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Difficulty:
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Question Stats:
64% (02:59) correct
36%
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wrong
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Last month Alex spent $91 on video game rentals, spending $5 for each one-day rental, $8 for each three-day rental, and $10 for each seven-day rental. Alex never rented the same game twice, and he rented fewer than 6 games in one-day rentals. How many different games did Alex rent last month?
(1) Last month Alex spent $20 in seven-day video game rentals.
(2) Last month Alex spent $56 in three-day video game rentals.
Alex.png [ 61.18 KiB | Viewed 13881 times ]
(1) Last month Alex spent $20 in seven-day video game rentals.
(2) Last month Alex spent $56 in three-day video game rentals.
Attachment:
Alex.png [ 61.18 KiB | Viewed 13881 times ]
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30 Dec 2023, 03:51
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MclLaurent
Given
- Alex spend $91 on video games rentals
- Alex never rented the same game twice
- He rented fewer than 6 games in one-day rentals
Price Chart
- $5 → one-day rental
- $8 → three - day rental
- $10 → seven - day rental
Method 1 : Logical Analysis
The crux to solving this question is to observe how can we form a total of $91, without violating any conditions. Observe that $91, ends with 1. A multiple of 5, or 8 or 10 never ends with 1. Hence, we must add an even number with an odd number to obtain $91. One possibility is when the unit digit of one number is 5 and the unit digit of another number is 6.
With this pre-analysis, let's dive into the statements.
Statement 1
(1) Last month Alex spend $20 in seven day video game rentals
This statement tells us that Alex rented two video games for 7 days. Hence, he paid 2 (i.e. number of video games) * $10 (i.e. rental for 7 days) = $20.
The rest amount of $91 - 20 = $71, comes from a one-day rental or from a three-day rental.
Let's assume Alex rents two video games for three days. Rental amount = 2 * $8 = $16
Remaining Amount = $71 - 16 = $55. This must come from one-day rental. Number of games Alex must have rented = 55/5 = 11. However, the constraint set in the premise tells us that Alex rented fewer than 6 games in one-day rentals. So we have to discard this case.
The next possibility is when Alex spends $56 (i.e. $16 + $40) in three-day rental. Number of games rented = $56/$8 = 7
Remaining Amount = $71 - 56 = $15. This must come from one-day rental. Number of games Alex must have rented = 15/5 = 3.
Total games rented = 7 + 3 + 2 = 12
There is no other possibility. The statement alone is sufficient to answer the question. Eliminate B, C, and D.
Statement 2
(2) Last month Alex spend $56 in three days video game rentals
The rest amount of $91 - 56 = $35, comes from a one-day rental or from a seven-day rental.
Number of games rented under three-day rental = 56/8 = 7.
$35 can be split into two ways,
- $10 (from seven-day rental) + $25 (from one-day rental) → In this case, Alex rents one game under a seven-day rental and rents five games under a one-day rental.
Total games rented = 7 + 1 + 5 = 13 - $20 (from seven-day rental) + $15 (from one-day rental) → In this case, Alex rents two games under a seven-day rental and rents 3 games under a one-day rental.
Total games rented = 7 + 2 + 3 = 12
As we are getting two separate answers statement 2 alone is not sufficient to answer the question.
Option A
Method 2 : Algebra
Number of Games Rented
- one-day rental → \(x\) (Given \(x < 6\))
- three - day rental → \(y\)
- seven - day rental → \(z\)
Price Paid
- one-day rental → \(5x\)
- three - day rental → \(8y\)
- seven - day rental → \(10z\)
\(5x + 8y + 10z = 91\)
Attachment:
Screenshot 2023-12-30 164700.png [ 59.47 KiB | Viewed 14315 times ]
Question: \(x + y + z = ?\)
Statement 1
(1) Last month Alex spend $20 in seven day video game rentals
\(10z = 20\)
\(z = 2\)
\(5x + 8y + 10z = 91\)
\(5x + 8y + 20 = 91\)
\(5x + 8y = 71\)
\(5x = 70 + 1 - 5y - 3y\)
\(x = 14 - y + \frac{1 - 3y}{5}\)
If y = 2
\(x = 14 - 2 + \frac{1 - 3*2}{5}\)
\(x = 14 - 2 + \frac{-5}{5}\)
\(x = 14 - 2 - 1 = 11\)
Hence x = 11, and y = 2 is one possible solution.
Next possible solution for equation \(5x + 8y = 71\) in which both x, and y are non negative is
(x,y) → (11-8,2+5) = (3,7)
As (x,y) = (11,2) is not a valid solution, the only valid solution is (x,y) = (3,7)
Total video games rented = 7 + 3 + 2 = 12
The statement alone is sufficient, we can eliminate B, C, and E.
Statement 2
(2) Last month Alex spend $56 in three days video game rentals
\(8y = 56\)
\(y = 7\)
\(5x + 8y + 10z = 91\)
\(5x + 56 + 10z = 91\)
\(5x + 10z = 35\)
\(x + 2z = 7\)
One possible solution (x,z) = (3,2), another possible solution of (x,z) = (5,1).
Number of video games rented when (x,z) = (3,2) ⇒ 7 + 3 + 2 = 12
Number of video games rented when (x,z) = (5,1) ⇒ 7 + 5 + 1 = 13
As we are getting two separate answers statement 2 alone is not sufficient to answer the question.
Option A
15 Mar 2024, 05:05
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MclLaurent
Answer: Option A
Please find the video explanation below
