Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Apr 3001:30 AM EDT
-02:30 AM EDT
At one point, she believed GMAT wasn’t for her. After scoring 595, self-doubt crept in and she questioned her potential. But instead of quitting, she made the right strategic changes. The result? A remarkable comeback to 695. Check out how Saakshi did it.
May 0111:00 AM PDT
-12:00 PM PDT
Free- full-length test + 15 concept videos + 150+ short videos + study plan!
May 0306:30 AM PDT
-08:30 AM PDT
Verbal trouble on GMAT? Fix it NOW! Join Sunita Singhvi for a focused webinar on actionable strategies to boost your Verbal score and take your performance to the next level.
26 Dec 2019, 03:58
Kudos
Bookmarks
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% (02:58) correct
53%
(03:26)
wrong
based on 90
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
Kudos
Bookmarks
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
Kudos
Bookmarks
=>
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
