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16 Oct 2016, 11:29
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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:
85%
(hard)
Question Stats:
51% (02:25) correct
49%
(02:28)
wrong
based on 484
sessions
History
Date
Time
Result
Not Attempted Yet
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
I am not 100% convinced with OA and explanation provided, please add some comment to help me,
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
I am not 100% convinced with OA and explanation provided, please add some comment to help me,
16 Oct 2016, 11:46
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At my first attempt I got it wrong as I thought they were asking the value of Chipwiches at 9:00 AM instead of 5:00 PM, but here is the explanation from my second attempt. HTH
Let chipwiches as C and mud pies as M
No of Chipwiches sold during the day = Z
As per question
At 9:00 AM --> C= 2M
At 5:00 PM --> 2( C - Z) = M
Substituting C with 2M
We get M = 2 * (Z/3) or Z = 3 *(M/2)
I. The number of chipwiches sold must be a multiple of 3 --- Z is the number of Chipwiches sold which as calculated above is a multiple of 3.
II. The number of mud pies must be even -- No of Mud pies is M as calculated above and is a multiple of 2 so it will be even. (Since Z will be an integer)
III. The number of chipwiches the cart has at 5:00 PM must be even -- No of Chipwiches at 5:00 PM
= C - Z
= 2M - 3* (M/2)
= M/2 (We don't know the value of M so we can't tell if M/2 will be even or odd)
So only 1 and 2 are correct, so option C.
Let chipwiches as C and mud pies as M
No of Chipwiches sold during the day = Z
As per question
At 9:00 AM --> C= 2M
At 5:00 PM --> 2( C - Z) = M
Substituting C with 2M
We get M = 2 * (Z/3) or Z = 3 *(M/2)
I. The number of chipwiches sold must be a multiple of 3 --- Z is the number of Chipwiches sold which as calculated above is a multiple of 3.
II. The number of mud pies must be even -- No of Mud pies is M as calculated above and is a multiple of 2 so it will be even. (Since Z will be an integer)
III. The number of chipwiches the cart has at 5:00 PM must be even -- No of Chipwiches at 5:00 PM
= C - Z
= 2M - 3* (M/2)
= M/2 (We don't know the value of M so we can't tell if M/2 will be even or odd)
So only 1 and 2 are correct, so option C.
General Discussion
23 Nov 2016, 14:59
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I think the best way to solve this problem is by correctly interpreting the number of chipwiches relative to mudpies as follows:
9am:
2x = chipwiches
x = mudpies
5pm:
1/2x = chipwiches
x=mudpies
This calculation is done as such to allow mudpies to remain as x throughout the day based on the specifications indicated in the prompt
Test #1:
9am: 20 = 2x = # chipwiches
10=x=# mudpies
5pm: 10 mudpies remain
5 chipwiches remain (i.e. 15 sold)
Test #2:
9am: 12 chipwiches
6 mudpies
5pm: 6 mudpies remain
3 chipwiches remain (i.e. 9 sold)
So, from these two different examples, we can see that I and II are true. III can be ruled out by looking at Test #2 (i.e. number of chipwiches remaining is an odd number and therefore violating that chipwiches remaining must be even).
9am:
2x = chipwiches
x = mudpies
5pm:
1/2x = chipwiches
x=mudpies
This calculation is done as such to allow mudpies to remain as x throughout the day based on the specifications indicated in the prompt
Test #1:
9am: 20 = 2x = # chipwiches
10=x=# mudpies
5pm: 10 mudpies remain
5 chipwiches remain (i.e. 15 sold)
Test #2:
9am: 12 chipwiches
6 mudpies
5pm: 6 mudpies remain
3 chipwiches remain (i.e. 9 sold)
So, from these two different examples, we can see that I and II are true. III can be ruled out by looking at Test #2 (i.e. number of chipwiches remaining is an odd number and therefore violating that chipwiches remaining must be even).
