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01 Aug 2022, 07:39
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
55% (01:50) correct
45%
(03:41)
wrong
based on 11
sessions
History
Date
Time
Result
Not Attempted Yet
On the occasion of his daughter’s birthday, Charles planned to buy ‘x’ number of chocolates for $ y (where, x and y are integers greater than 1). If Charles had bought 15 more chocolates, the total amount paid by him will be $ 3 and thus he would have saved $ 0.40 per dozen of chocolates. What is the value of x?
A. 8
B. 12
C. 15
D. 18
E. 20
(adapted from beatthegmat)
A. 8
B. 12
C. 15
D. 18
E. 20
(adapted from beatthegmat)
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01 Aug 2022, 08:19
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pintukr
On the occasion of his daughter’s birthday, Charles planned to buy ‘x’ number of chocolates for $ y (where, x and y are integers greater than 1). If Charles had bought 15 more chocolates, the total amount paid by him will be $ 3 and thus he would have saved $ 0.40 per dozen of chocolates. What is the value of x?
A. 8
B. 12
C. 15
D. 18
E. 20
(adapted from beatthegmat)
A. 8
B. 12
C. 15
D. 18
E. 20
(adapted from beatthegmat)
Two ways:
1. Algebra
y is an integer greater than 1. It is also less than 3 since we get to a total price of $3 only by buying 15 more chocolates. Therefore, y = 2.
The original cost for a dozen chocolates is \(12*\frac{2}{x}\)
The final cost for a dozen chocolates is \(12*\frac{3}{x+15}\)
Final cost for a dozen chocolates is $0.4 less than the original cost, so
\(12*\frac{2}{x}-0.4 = 12*\frac{3}{x+15}\)
\(\frac{24}{x}-\frac{0.4x}{x} = \frac{36}{x+15}\)
\(\frac{24-0.4x}{x} = \frac{36}{x+15}\)
\(36x = (24-0.4x)(x+15)\)
\(36x = 24x-0.4x^2+360-6x\)
\(0 = -0.4x^2-18x+360\)
\(0 = 4x^2-180x+3600\)
\(0 = (2x-30)(2x+12)\)
2x-30 = 0 ... or ... 2x+12 = 0
x=15 ... or ... x=-6
x has to be positive, so x=15.
Answer choice C.
2. Plug In The Answers (PITA)
y is an integer greater than 1. It is also less than 3 since we get to a total price of $3 only by buying 15 more chocolates. Therefore, y = 2.
The original cost for a dozen chocolates is \(12*\frac{2}{x}\)
The final cost for a dozen chocolates is \(12*\frac{3}{x+15}\)
Let's try B:
Original cost = 24/12 = 2
Final cost = 36/27 = 1.25
That's a difference of more than $0.40.
Let's try D:
Original cost = 24/18 = 1.333
Final cost = 36/33 = 12/11 = 1.09
That's a difference of less than $0.40.
B was MORE, D was LESS. We need something between those.
Answer choice C.
ThatDudeKnowsPITA
01 Aug 2022, 10:57
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Given: On the occasion of his daughter’s birthday, Charles planned to buy ‘x’ number of chocolates for $ y (where, x and y are integers greater than 1). If Charles had bought 15 more chocolates, the total amount paid by him will be $ 3 and thus he would have saved $ 0.40 per dozen of chocolates.
Asked: What is the value of x?
1 < y < 3
y = 2
Original cost of a dozen chocolates = 12*2/x = 24/x
Final cost of a dozen chocolates = 12*3/(x+15) = 36/(x+15)
24/x - 36/(x+15) = .4
24(x+15) - 36x = .4x(x+15)
60(x+15) - 90x = x^2 + 15x
900 - 30x = x^2 + 15x
x^2 + 45x - 900 = 0
x^2 + 60x - 15x - 900 =0
(x+60)(x-15)=0
x = 15 since x >1
IMO C
Asked: What is the value of x?
1 < y < 3
y = 2
Original cost of a dozen chocolates = 12*2/x = 24/x
Final cost of a dozen chocolates = 12*3/(x+15) = 36/(x+15)
24/x - 36/(x+15) = .4
24(x+15) - 36x = .4x(x+15)
60(x+15) - 90x = x^2 + 15x
900 - 30x = x^2 + 15x
x^2 + 45x - 900 = 0
x^2 + 60x - 15x - 900 =0
(x+60)(x-15)=0
x = 15 since x >1
IMO C
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
