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22 Mar 2018, 11:01
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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:
75%
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Question Stats:
61% (02:33) correct
39%
(02:35)
wrong
based on 343
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Josh has a big drawer full of exactly 40 packets, each containing a marker which is either black or blue or red in colour. All 40 packets are sealed and are completely identical from outside. It is not possible to know which colour marker is inside unless the packet is fully opened. Out of 40, 23 packets contain a black marker each. How many packets contain a blue marker?
(1) Drawer has less packets containing blue markers than those containing red markers.
(2) If Josh withdraws packets without looking at their contents, he needs to draw minimum 20 packets to ensure that he has exactly 8 markers of any single colour out of red, blue, black.
(1) Drawer has less packets containing blue markers than those containing red markers.
(2) If Josh withdraws packets without looking at their contents, he needs to draw minimum 20 packets to ensure that he has exactly 8 markers of any single colour out of red, blue, black.
22 Mar 2018, 13:56
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pushpitkc
2. If he needs to withdraw 20 markers to have 8 of a color,
2 cases are possible such that 8 markers of either color can be taken out.
Case 1: Red = 9, Blue = 8
Case 2: Red = 8, Blue = 9
We cannot come to a unique value for the number of packets containing Blue markers(Insufficient)
2 cases are possible such that 8 markers of either color can be taken out.
Case 1: Red = 9, Blue = 8
Case 2: Red = 8, Blue = 9
We cannot come to a unique value for the number of packets containing Blue markers(Insufficient)
What if Josh selects 7 black, 7 blue, 6 red? In that case, he will not have 8 markers of any one color.
Here's what I did...
40 packets
23 of them have black markers
(# of packets with red markers) + (# of packets with blue markers) = 17
We're asked to find the # of packets with blue markers.
Statement 1
(# of packets with blue markers) < (# of packets with red markers)
We can have (7 blue, 10 red), or (6 blue, 11 red).
Insufficient
Statement 2
This means that when Josh selects 19 packets, he is not guaranteed to have 8 packets of any one color. In the worst case scenario, Josh will select 7 packets of one color, 7 packets of another, and 5 packets of another color (total selected = 19 packets). The 20th packet he selects MUST give him 8 packets of one color, thus one of the colors must have 5 packets. So we either have (5 red, 12 blue) or (12 red, 5 blue).
Insufficient
Combine Statements 1 & 2
Per Stmt 1: red < blue
Per Stmt 2: (5 red, 12 blue) or (12 red, 5 blue)
Combined, we get 5 red, 12 blue
Answer: C
02 May 2021, 22:01
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Anush56
Hello VeritasKarishma
Can you please help me out with this question ?
I'm not understanding how from the second condition we can assume the below
Case 1: Red = 12, Blue = 5
Case 2: Red = 5, Blue = 12
I saw a similar question in Veritas question bank.
Can you please help me out with this question ?
I'm not understanding how from the second condition we can assume the below
Case 1: Red = 12, Blue = 5
Case 2: Red = 5, Blue = 12
I saw a similar question in Veritas question bank.
Think of it this way -
Stmnt 2 tells you that you need to withdraw 20 markers to ensure that you have at least 8 of the same colour.
So this means that you could pick 19 markers and still not have 8 of the same colour. But when you pick the 20th, you MUST have 8 of a colour.
So then, worst case scenario, how many of what colour markers could this 19 be such that the 20th has to ensure that you have 8 markers of 1 colour, no matter which marker you pick next?
We would assume that this happens in case we have 7 markers of all colours so that when we pick the next marker, it gives us 8 of one of the colours but that is not possible since we have only 19 markers. This means we must have total only 5 markers of one colour and 7 of the other two to make up 19. Now the leftover markers would only be of the other two colours. So whichever marker we pick next, it will give us 8 of one of the two other colours.
Since we know there are 23 black markers, it means we have only 5 markers of either red or blue.
If we have 5 red markers, then we have total 12 blue markers.
If we have 5 blue markers, then we have total 12 red markers.
Either case is possible.
