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01 Dec 2021, 09:00
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
59% (02:34) correct
41%
(02:40)
wrong
based on 289
sessions
History
Date
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Not Attempted Yet
Over the course of 5 days, Monday through Friday, Danny collects a total of 76 baseball cards. Each day, he collects a different number of cards. If Danny collects the largest number of cards on Friday and the second largest number of cards on Thursday, did Danny collect more than 8 cards on Thursday?
(1) On Friday, Danny collected 49 cards.
(2) On one of the first 3 days, Danny collected 6 cards.
(1) On Friday, Danny collected 49 cards.
(2) On one of the first 3 days, Danny collected 6 cards.
02 Dec 2021, 06:13
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Based upon both the statements, following combinations are among the various combinations that are possible. In the first case, Danny collects 8 cards on Thursday; in the second case, Danny collects more than 8 cards on Thursday.
Fr Th Mo Tue Wed
49 8 6 6 7
49 10 6 6 5
So, E.
Fr Th Mo Tue Wed
49 8 6 6 7
49 10 6 6 5
So, E.
07 Dec 2021, 11:15
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Question analysis:
Mon Tue Wed Thu Fri
(b-x) (b-y) (b-z) < (b) < (a)
Need to answer: If b >8?
Only can find 1 definite answer when:
(1) b min >8 --> b always >8
or
(2) b max <8 --> b always <8
Constraints:
Given: a>b, x,y,z >=0 and are different from each other. a,b,x,y,z must be integers (I)
(b-x) + (b-y) + (b-z) + (b) + (a) = 76
--> 4b-(x+y+z) +a =76 (II)
St1:
(1) On Friday, Danny collected 49 cards
from (II) --> 4b-(x+y+z) + 49 =76 --> b = (27 + x+y+z)/4
b min <=> x,y,z = 1,2,3
--> b min = 33/4 >8
--> b always >8 when a=49
--> St1 is sufficient
St2:
(2) On one of the first 3 days, Danny collected 6 cards.
From (I) we can infer 2 scenarios:
Mon, Tue, Wed = 1,2,3
or
Mon, Tue, Wed = 0,1,2
We can only know that b>3 but cannot know whether b>8 or 3<b<8
--> St2 is NOT suff.
So answer is A
Mon Tue Wed Thu Fri
(b-x) (b-y) (b-z) < (b) < (a)
Need to answer: If b >8?
Only can find 1 definite answer when:
(1) b min >8 --> b always >8
or
(2) b max <8 --> b always <8
Constraints:
Given: a>b, x,y,z >=0 and are different from each other. a,b,x,y,z must be integers (I)
(b-x) + (b-y) + (b-z) + (b) + (a) = 76
--> 4b-(x+y+z) +a =76 (II)
St1:
(1) On Friday, Danny collected 49 cards
from (II) --> 4b-(x+y+z) + 49 =76 --> b = (27 + x+y+z)/4
b min <=> x,y,z = 1,2,3
--> b min = 33/4 >8
--> b always >8 when a=49
--> St1 is sufficient
St2:
(2) On one of the first 3 days, Danny collected 6 cards.
From (I) we can infer 2 scenarios:
Mon, Tue, Wed = 1,2,3
or
Mon, Tue, Wed = 0,1,2
We can only know that b>3 but cannot know whether b>8 or 3<b<8
--> St2 is NOT suff.
So answer is A
