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07 Jul 2025, 09:51
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All of the non‐hat wearers must be among the first ten from the left. Statement (1) says that, looking at the last six spots in the line, exactly one of those six is hatless. That single non‐hatter must come from positions 1-10, so the block of six “rightmost” positions must reach back far enough to include exactly one of those first ten.
1) If the line has 11 people, the rightmost six are positions 6-11, which includes exactly one non‐hatter (somewhere in positions 6-10).
2) If the line has 12 people, the rightmost six are 7-12, again capturing exactly one of the ten non‐hat wearers.
3) And so on, up through 15 people, where positions 10-15 still include exactly one non‐hatter.
Any total from 11 up to 15 yields the same “one non‐hat among the last six” result. Because this single condition is satisfied by five different line‐lengths, it cannot pin down a unique total.
Statement (2) only tells us about the first eight people in line namely, that three of those eight wear hats (and so five of those eight go hatless). But it says absolutely nothing about anyone standing beyond position 8, nor does it link how many people there are in total to that hat count. Since all the remaining non‐hat wearers must lie somewhere in positions 1-10 and we still don’t know where the eighth‐position window sits relative to the tenth non‐hatter, any number of people beyond eight (as long as it’s at least ten) could leave exactly three hat‐wearers among the first eight. In short, learning about hats in the first eight slots doesn’t constrain how long the overall line is, so you still can’t determine the total.
Hence neither is sufficient not together they are sufficient. E.
1) If the line has 11 people, the rightmost six are positions 6-11, which includes exactly one non‐hatter (somewhere in positions 6-10).
2) If the line has 12 people, the rightmost six are 7-12, again capturing exactly one of the ten non‐hat wearers.
3) And so on, up through 15 people, where positions 10-15 still include exactly one non‐hatter.
Any total from 11 up to 15 yields the same “one non‐hat among the last six” result. Because this single condition is satisfied by five different line‐lengths, it cannot pin down a unique total.
Statement (2) only tells us about the first eight people in line namely, that three of those eight wear hats (and so five of those eight go hatless). But it says absolutely nothing about anyone standing beyond position 8, nor does it link how many people there are in total to that hat count. Since all the remaining non‐hat wearers must lie somewhere in positions 1-10 and we still don’t know where the eighth‐position window sits relative to the tenth non‐hatter, any number of people beyond eight (as long as it’s at least ten) could leave exactly three hat‐wearers among the first eight. In short, learning about hats in the first eight slots doesn’t constrain how long the overall line is, so you still can’t determine the total.
Hence neither is sufficient not together they are sufficient. E.
Bunuel
07 Jul 2025, 09:56
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here when i go with:
A alone --> if the 4/10 people with hat are all on extreme left, then the 5 more new people with hat would be added on the right to satisfy 5/6 condition
similarly the positioning of 4/10 people with hats will determine how many new people are on the right
B Alone--> on similar lines B is also not sufficient
even if you combine both A and B. still we could have different placements where Number of total people could be anywhere between 11 to 15.
Hence together not sufficient.
A alone --> if the 4/10 people with hat are all on extreme left, then the 5 more new people with hat would be added on the right to satisfy 5/6 condition
similarly the positioning of 4/10 people with hats will determine how many new people are on the right
B Alone--> on similar lines B is also not sufficient
even if you combine both A and B. still we could have different placements where Number of total people could be anywhere between 11 to 15.
Hence together not sufficient.
Bunuel
07 Jul 2025, 10:07
Kudos
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(1) 5 of the 6 people from the right are wearing hats.
If last 6 people include k people beyond 10 (all hats), then:
k=5 ⇒ n−10 =5 ⇒ n=15
But if the no hat person
is closer to position 10, fewer beyond 10 people are in last 6, so n n
can be 11 , 12 , 13 , 14 , or 15.
n = 11,12,13,14, or 15, depending on
distribution of hats among first 10.
INSUFFICIENT
(2) 3 of the 8 people from the left are wearing hats.
INSUFFICIENT. Does not help.
Both together - Sufficient.
If last 6 people include k people beyond 10 (all hats), then:
k=5 ⇒ n−10 =5 ⇒ n=15
But if the no hat person
is closer to position 10, fewer beyond 10 people are in last 6, so n n
can be 11 , 12 , 13 , 14 , or 15.
n = 11,12,13,14, or 15, depending on
distribution of hats among first 10.
INSUFFICIENT
(2) 3 of the 8 people from the left are wearing hats.
INSUFFICIENT. Does not help.
Both together - Sufficient.
