Latest self Aptitude Question SOLUTION: what is the chance that a leap year selected at random will have 53 sundays

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(#M40010440) SELF question Keep an EYE

what is the chance that a leap year selected at random will have 53 sundays

dola (2 years ago)

Unsolved Read Solution (1)

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Madhu  (2 years ago)

the leap year has 52 weeks and 2 days totally 366 days.
52 weeks contain 52 sundays.
2 days which contain sundays are sunday,monday (or)saturday,sunday out of 7 days.
the probability leap yr contain 53 sundays is=2/7

so the answer is 2/7

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(#M40000113) SELF Numerical Ability Question It is 10 o'clock! Keep an EYE

A clock takes 7 seconds to ring bell at 7 o'clock. How long will it take for the same clock to ring bell at 10 o'clock?

Tijo Jose (5 years ago)

Solved   Sanal Jose Read Solution (4)

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Sanal Jose  (5 years ago)

10.5 seconds; in sense that timer starts at first ring and the first ring is at time t=0 seconds and 7th ring is at time t=7 seconds. So, in 7 seconds, only 2nd, 3rd, 4th, 5th, 6th, and 7th are to be considered. So, 6 rings in 7 seconds means 3 rings in 3.5 seconds. so, at 10, only 9 rings add to the timer, each 3 of which takes 3.5 seconds. so total = 3*3.5 = 10.5 seconds. i think i have read it somewhere, uhh... in some puzzle book.

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Aman  (5 years ago)

10*7/6 = 35/3 secs = 10.6666.... secs

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SATYAJIT PATNAIK  (5 years ago)

same 7 sec.

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Eliza   (5 years ago)

7/10

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(#M40000114) SELF Numerical Ability Question Go home early. Keep an EYE

A man goes to work daily. Everyday his wife picks him from railway station at 6 o'clock. One day, he left office early and reached the station at 5 o'clock. So, he started walking home. On the way, he met his wife, and they returned together in the car. They reached home ten minutes earlier than usual time. How much time did the man walk?

Tijo Jose (5 years ago)

Solved   Aman Read Solution (2)

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Aman  (5 years ago)

55 minutes

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m.raviteja  (2 Months ago)

explain clear answer

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(#M40000115) SELF Question Give me your remainder Keep an EYE

Find the smallest number N such that

N/10 leaves remainder 9

N/9 leaves remainder 8

N/8 leaves remainder 7

N/7 leaves remainder 6

N/6 leaves remainder 5

N/5 leaves remainder 4

N/4 leaves remainder 3

N/3 leaves remainder 2

N/2 leaves remainder 1

N/1 leaves remainder 0

Tijo Jose (5 years ago)

Solved   Aman Read Solution (4)

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Sanal Jose  (5 years ago)

It can be inferred from given conditions that N+1 is divisibly by 1, 2, ..., 10. So, N+1 is LCM. LCM(1,2,...,10) = 2520. So, N=2519

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Aman  (5 years ago)

2519

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Aman  (5 years ago)

2519

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Chaitanya Singh   (2 Months ago)

2 man definitely survive at last because the procedure is kill someone and give it to another only 3 man can complete this at last when 2 will remain then the game will over...........

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(#M40000148) SELF Question find the value Keep an EYE

1=5
2=25
3=125
4=625
5=?

chintha koteswararao (5 years ago)

Solved   Rijo Read Solution (11)

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unni  (5 years ago)

ans is 3125
they arethe powers of 5

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Tijo Jose  (5 years ago)

Given, 1=5. So, 5=1.

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Eliza   (5 years ago)

5=1

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sribharadwaj  (5 years ago)

5=1

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Rijo  (5 years ago)

1 ,already given

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Sanal Jose  (5 years ago)

5=1 :)

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appu  (5 years ago)

1=5^1=5
2=5^2=25
3=5^3=125
4=5^4=625
5=5^5=3125

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shrikant  (4 years ago)

1

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Sanchit Joshi  (2 years ago)

given
1=5
therefore
5=1

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Tania  (4 Months ago)

1=5 is given which implies that 5=1

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Priya  (4 Months ago)

1=5
2=5*5=25
3=5*5*5=125
4=5*5*5*5=625
5=5*5*5*5*5=3125

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(#M40000112) SELF Question Don't kill me! Keep an EYE

1000 men are standing in a circle, numbered 1, 2, ..., 1000. 1 takes a sword, kills 2 and hands over sword to 3, who takes it, kills 4 and gives it to 5. This reapeats in the circle. That is, whoever gets the sword kills the one near him and hands over the sword to the next. If this continues, what is the number of the man that survives?

Tijo Jose (5 years ago)

Solved   Dundi Read Solution (7)

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hitesh gakhar  (5 years ago)

1 to 1000 are in a circle...1 takes a sword, kills 2 and hands over sword to 3, who takes it, kills 4 and gives it to 5....like this the sword moves the complete circle...killing all even positions...now we are left with all odd positions...like men no. 1, 3, 5, ... 999...and when 1000 dies, sword comes to 1 again... now it kills3 , and gives the sword to 5, and so on...on completion of round, the dead people were of form 4n-1...so people are left with form 4n+1..likely, 1, 5, 9,...997..and like this the sword keeps on moving and we are left with... are 1,9 , 17, ..993..but sword is with 993..

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Ranjith  (11 Months ago)

977 is the answer.by using josephus algorithm starting from 1 to 20 we get a formula 2(N-2^m)+1 where n is the total numbers m is the least square of two which is equal to or less than total.

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Tijo Jose  (5 years ago)

All your answers are wrong. Though Hitesh got near. For a hint, search for Josephus Algorithm. Or use a program logic.

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Tijo Jose  (5 years ago)

Or, here is another hint for you... Try finding out the survivor, when initial number of people are 1, 2, 3, 4, ... to some number like 20. Try to find out some sequence out of this. This surely will help.

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Dundi  (5 years ago)

Only one can survive....

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Gokul M Nair  (5 years ago)

1

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Harikrishnan  (9 Months ago)

The right answer is 2(488)+1=977
The problem is quite similar to Josephus problem.

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