An (infinitely small) ball starting out in the middle of a 5 pointed star table (outer 5 points - 10m radius..... inner 5 points - 5m radius) has a starting angle of a random value from 0 to 360 degrees. The ball is now set loose and travels around the table.On average, how many sides will have been hit once the ball has travelled 1000m ?
My approach:00:00 to 23:59Here, Total conditions= 24(00-23) * 60(00-59)Favourable ways:Minutes section:01_______1 way10-19____10 ways21,31,41 and 51___1 way eachSo total= 15 waysHour section:01___1 way10-19____10 ways21__1 wayTotal= 12 waysSo, fraction = 12*15/(24*60) = 1/8??
But...you have taken min AND hour both to have a 1....it should be OR.
So 12 hours full, and in the remaining 12 hours, 15 min each = 3 more hours. So total should be 15/24 or 5/8 (my previous answer was a calculation typo...took 11 instead of 12 😞 ) regards scrabbler
sir dekho maine 12 hrs vali ghadi li hai therefore hours 1, 10, 11,12 and the minutes 01, 10, 11, 12, ...19, 21, 31, 41, 51......minutes ho gaye 15*16=240=4hrs......therefore 8+4/24=1/2....atleast one 1 bola hai plz correct me if i m wrong
24 hours makes more sense I feel...it says the dial shows "only hours and minutes" if it were a 12 hour system would also have to show "am or pm". If 12 hours then I guess you're right!
sir dekho maine 12 hrs vali ghadi li hai therefore hours 1, 10, 11,12 and the minutes 01, 10, 11, 12, ...19, 21, 31, 41, 51......minutes ho gaye 15*16=240=4hrs......therefore 8+4/24=1/2....atleast one 1 bola hai plz correct me if i m wrong
since you said hours and minutes are the "only things" that are displayed i took it as 24 hours watch. i assumed AM PM wont be displayed.!!
When a trader buys goods at a price which is 10.5% less than the usual cost price and sells them at a price which is 20% less than the usual sale price, his profit decreases by an amount equal to 12.5% of the usual sale price. What is the usual percentage of profit made by the trader?
When a trader buys goods at a price which is 10.5% less than the usual cost price and sells them at a price which is 20% less than the usual sale price, his profit decreases by an amount equal to 12.5% of the usual sale price. What is the usual percentage of profit made by the trader?
When a trader buys goods at a price which is 10.5% less than the usual cost price and sells them at a price which is 20% less than the usual sale price, his profit decreases by an amount equal to 12.5% of the usual sale price. What is the usual percentage of profit made by the trader?
When a trader buys goods at a price which is 10.5% less than the usual cost price and sells them at a price which is 20% less than the usual sale price, his profit decreases by an amount equal to 12.5% of the usual sale price. What is the usual percentage of profit made by the trader?
When a trader buys goods at a price which is 10.5% less than the usual cost price and sells them at a price which is 20% less than the usual sale price, his profit decreases by an amount equal to 12.5% of the usual sale price. What is the usual percentage of profit made by the trader?
I think you are talking about this problem:-The 150 contestants of Miss India 2010 are given individual numbers from 1 to 150. Several rounds happen before the final winner is selected. The elimination in each round follows an interesting pattern. In the 1st round starting from first contestant, every 3rd contestant is eliminated i.e. 1st, 4th, 7th, .... This repeats again from the first numbered (among the remaining) contestant in the next round (leaving 3, 5, 8, 9, ... ). This process is carried out repeatedly until there is only the winner left. What is the number of Miss India 2010?The logic that you mentioned goes like this:-Convert 150 in base 3150 in base 3 is 12120Now remove first two digits and append them to the end of the remainder number.So, new number becomes 12012It is 140 is base 10So, contestant number 140 will be the winner.(I don't know the reason why it works)But somehow I'm not sure if its the correct way to go about it.For example has the number been 159, then in base 3 its 12220After removing first digits and appending them at the end it will become 22012 which is 221 in base 10 (which doesn't make any sense)So, I'm not sure about this method.Alternatively:-Backtracking will also help.If any contestants position in a round is n, then in the previous round its position will be 3n/2 or [3n/2] + 1So, positions of winner are 1, 2, 3, 5, 8, 12, 18, 27, 41, 62, 93, 140, 210, ...Means for any n (no of contestants) between 62 and 92 winner will be contestant no 62 and so onSince here we have 150 contestants winner will be 140But here the contestants are not standing around the circle (for that you can read about Josephus problem)
Why have you taken (3n/2 or [3n/2+1])? What is logic behind it?
An automobile car assembly system consists of two independent subsystems: chassis assembly (C) and engine assembly (E). Based on the prior maintenance data, the following probabilities are available. The probability that C fails is 0.2. The probability that both C and E fail together is 0.15. The probability that E fails alone is 0.15. The event €œC fails alone €? implies that C fails when E is working and vice versa. 1)What is the probability that C fails when E has failed? 2)What is the probability that C fails alone?
When a trader buys goods at a price which is 10.5% less than the usual cost price and sells them at a price which is 20% less than the usual sale price, his profit decreases by an amount equal to 12.5% of the usual sale price. What is the usual percentage of profit made by the trader?
@gnehagarg You could work the pattern backwards in this way In the first round the no.s are 1 2 3 4 5 6 7 8 9 10 11 12 13 14 .... in the second round it becomes _ 2 3 _ 5 6 _ 8 9 _ 11 12 _ 14 ....... - Space (_) represents cancelled no.
Lets look at the second round list and try to estimate the position of the no. 12 in the fisrt round. In the second round 12 is placed at the 8th position.
Looking at the second round list we can see that there is a a space ( _ ) after very two no.s. So if the no. is in 8th position there are three couples before it (1st and 2nd, 3rd and fourth, 5th and 6th) Since the first space (_) appears before the first couple therefore the total no. of cancelled no.s (of previous round) is equal to = [the no. of couples before the 8th no.] + 1 = 3 + 1 = 4.
So, effectively, if the no. of cancelled no.s = No. of couples including the couple 11 and 12. So, No. of couples = 8/2 = 4
So previous position of the 8th no. (12 in this case) = 8 + (8/2) general case = n + (n/2)
This works if the no. is at even position. If it is odd position like 7th position, then for the formation of couples we would have to find the couples for 7 + 1. So the general case for odd position would be n + (n+1)/2.
For eg. To find the previous position of a no. that is at 18th in this round 18 + (18/2) = 27
An automobile car assembly system consists of two independent subsystems: chassis assembly (C) and engine assembly (E). Based on the prior maintenance data, the following probabilities are available. The probability that C fails is 0.2. The probability that both C and E fail together is 0.15. The probability that E fails alone is 0.15. The event €œC fails alone €? implies that C fails when E is working and vice versa.1)What is the probability that C fails when E has failed? 2)What is the probability that C fails alone?
I am not too sure about these but it looks like this : A) 1/2 -> 0.15/0.30 B) 1/20 -> 0.20 - 0.15
When a trader buys goods at a price which is 10.5% less than the usual cost price and sells them at a price which is 20% less than the usual sale price, his profit decreases by an amount equal to 12.5% of the usual sale price. What is the usual percentage of profit made by the trader?
An automobile car assembly system consists of two independent subsystems: chassis assembly (C) and engine assembly (E). Based on the prior maintenance data, the following probabilities are available. The probability that C fails is 0.2. The probability that both C and E fail together is 0.15. The probability that E fails alone is 0.15. The event €œC fails alone €? implies that C fails when E is working and vice versa.1)What is the probability that C fails when E has failed? 2)What is the probability that C fails alone?
When a trader buys goods at a price which is 10.5% less than the usual cost price and sells them at a price which is 20% less than the usual sale price, his profit decreases by an amount equal to 12.5% of the usual sale price. What is the usual percentage of profit made by the trader?
When a trader buys goods at a price which is 10.5% less than the usual cost price and sells them at a price which is 20% less than the usual sale price, his profit decreases by an amount equal to 12.5% of the usual sale price. What is the usual percentage of profit made by the trader?
CP's = 200 >>> 179 SP's = 100x >>> 80x
100x - 200 - 80x + 179 = 12.5x => x = 2.8
so , usual profit will be = 200*z% = 80 which is 40 %
