sameerspice wrote:
At 9:00 AM, a certain ice cream cart has twice as many chipwiches as it does mud pies. During the day, only chipwiches are sold, and at 5:00 PM, the cart has the exact same number of mud pies as it did at 9:00 AM, but now the number of mud pies is twice the number of chipwiches. Assuming no items were otherwise lost or gained, how many of the following must be true?
I. The number of chipwiches sold must be a multiple of 3
II. The number of mud pies must be even
III. The number of chipwiches the cart has at 5:00 PM must be even
A. I only
B. II only
C. I and II only
D. II and III only
E. I, II, and III
We can let the number of mud pies be 10 at 9 AM. Thus, the number of mud pies at 5 PM is still 10. The number of chipwiches at 9 AM is 10 x 2 = 20; however, the number of chipwiches at 5 PM is 10/2 = 5. We see that Roman numeral I could be true since 20 - 5 = 15 is a multiple of 3. We also see that II could be true since we assume there are 10 mudpies. However, III is definitely not true since we only have 5 chipwiches at 5 PM.
To prove I and II must be true, we can let x = the number of mud pies at 9 AM or 5 PM. Thus there are 2x chipwiches at 9 AM and x/2 chipwiches at 5 PM. We see that x must be even; otherwise, x/2 wouldn’t be a whole number. So II must be true. Finally, the number of chipwiches sold during the day is 2x - x/2 = 4x/2 - x/2 = 3x/2 = 3(x/2). Since x/2 is a whole number, 3(x/2) must be a multiple of 3. So I must be true also.
Answer: C
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