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26 Dec 2019, 03:58
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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:
95%
(hard)
Question Stats:
47% (02:56) correct
53%
(03:26)
wrong
based on 91
sessions
History
Date
Time
Result
Not Attempted Yet
Cakes are sold individually for \($25\), in bundles of \(3\) for \($67\), and in bundles of \(5\) for \($97\). \(100\) cakes are sold for \($2,000\) in total. How many cakes are sold individually?
A. \(3\)
B. \(5\)
C. \(6\)
D. \(7\)
E. \(9\)
A. \(3\)
B. \(5\)
C. \(6\)
D. \(7\)
E. \(9\)
26 Dec 2019, 15:41
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a+3b+5c= 100....(1)
Also, 25a+67b+97c=2000.....(2)
From (1) and (2)
2b+7c=125....(3)
Also, we can notice that average cost of 100 cakes is very close to that of bundle of 5. So we'll try to maximize c .
c=17; b=3; and a=100-9-85=6
25*6+67*3+97*17= 150+201+1649=2000
Also, 25a+67b+97c=2000.....(2)
From (1) and (2)
2b+7c=125....(3)
Also, we can notice that average cost of 100 cakes is very close to that of bundle of 5. So we'll try to maximize c .
c=17; b=3; and a=100-9-85=6
25*6+67*3+97*17= 150+201+1649=2000
MathRevolution
29 Dec 2019, 19:02
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=>
Assume \(\)a is the number of cakes sold individually, \(b\) is the number of bundles with three cakes, and \(c\) is the number of bundles with five cakes.
Then we have \(25a + 67b + 97c = 2,000\) and \(a + 3b + 5c = 100.\)
When we subtract the first equation from \(25\) times the second equation, we have \(25(a + 3b + 5c) – (25a + 67b + 97c) = 25(100) - 2000, 25a + 75b + 125c - 25a - 67b - 97c = 2500 - 2000, 8b + 28c = 500,\) or \(2b + 7c = 125.\)
Since we have \(7c = 125 – 2b, c\) is an odd number, and we can put \(c = 2k + 1\) for some integer \(k ≥ 0.\)
Then we have \(2b = 125 – 7c, 2b = 125 – 7(2k + 1), 2b = 125 - 14k - 7, 2b = 118 – 14k\) or \(b = 59 – 7k > 0.\)
We also have \(a = 100 – 3b – 5c, a = 100 – 3(59 – 7k) - 5(2k + 1), a = 100 – 177 + 21k – 10k - 5, a = 11k – 82.\)
Since \(a > 0, b > 0\) and \(c > 0\), we have \(11k – 82 > 0, 59 – 7k > 0\) and \(2k + 1 > 0.\)
When we put those inequalities together, we have \((\frac{82}{11}) < k < (\frac{59}{7}).\)
Since \(\frac{82}{11} = 7.xxx\) and \(\frac{59}{7} = 8.xxxx\), we have\( k = 8.\)
Therefore, we have \(a = 11k – 82 = 88 – 82 = 6.\)
Therefore, C is the answer.
Answer: C
Assume \(\)a is the number of cakes sold individually, \(b\) is the number of bundles with three cakes, and \(c\) is the number of bundles with five cakes.
Then we have \(25a + 67b + 97c = 2,000\) and \(a + 3b + 5c = 100.\)
When we subtract the first equation from \(25\) times the second equation, we have \(25(a + 3b + 5c) – (25a + 67b + 97c) = 25(100) - 2000, 25a + 75b + 125c - 25a - 67b - 97c = 2500 - 2000, 8b + 28c = 500,\) or \(2b + 7c = 125.\)
Since we have \(7c = 125 – 2b, c\) is an odd number, and we can put \(c = 2k + 1\) for some integer \(k ≥ 0.\)
Then we have \(2b = 125 – 7c, 2b = 125 – 7(2k + 1), 2b = 125 - 14k - 7, 2b = 118 – 14k\) or \(b = 59 – 7k > 0.\)
We also have \(a = 100 – 3b – 5c, a = 100 – 3(59 – 7k) - 5(2k + 1), a = 100 – 177 + 21k – 10k - 5, a = 11k – 82.\)
Since \(a > 0, b > 0\) and \(c > 0\), we have \(11k – 82 > 0, 59 – 7k > 0\) and \(2k + 1 > 0.\)
When we put those inequalities together, we have \((\frac{82}{11}) < k < (\frac{59}{7}).\)
Since \(\frac{82}{11} = 7.xxx\) and \(\frac{59}{7} = 8.xxxx\), we have\( k = 8.\)
Therefore, we have \(a = 11k – 82 = 88 – 82 = 6.\)
Therefore, C is the answer.
Answer: C
22 Jul 2024, 21:07
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MathRevolution
MathRevolution Bunuel KarishmaB
Can we expect these kind of questions on the exam? I find this lengthy I tried this for more than 5 minutes and failed to get the answer.
P.S. - MathRevolution - Great approach to the answer, btw!
