plz HELP...Balu has र216 in two rupee denomination. He wanted to allocate them into a number of bags in such a way that he could give any amount (even number of rupes) between र2 and र216 by handing out one or more bags without opening them. What is the minimum number of bags he would require
7
Amounts would be 2, 4, 8, 16, 32, 64, 90.
Funda being he wants to give any number of the form 2k from k = 1 to k = 108. So we can write in powers of two and then hand out the binary equivalent (someone else can explain this better perhaps, I am not finding the words offhand 😞 ) regards scrabbler
One year, India plays 30 one-day matches. Having played 40% of the scheduled matches, India achieves a success rate of 75%. If by the end of the year, India wants to achieve 80% success rate, then what percentage of the remaining matches should it win??
HELP!!One year, India plays 30 one-day matches. Having played 40% of the scheduled matches, India achieves a success rate of 75%. If by the end of the year, India wants to achieve 80% success rate, then what percentage of the remaining matches should it win??
Small numbers, so simplest would be to take actual values...(we could also use weighted averages etc, but that would make it longer in this case!)
30 total, out of which we need to win 80% i.e. 24..
Out of 12 already won 75% i.e. 9.
So in remaining 30 - 12 = 18 we need to win 24 - 9 = 15.
So required %age is 15/18 = 83.33% regards scrabbler
HELP!!One year, India plays 30 one-day matches. Having played 40% of the scheduled matches, India achieves a success rate of 75%. If by the end of the year, India wants to achieve 80% success rate, then what percentage of the remaining matches should it win??
total matches to be played in one year=30
40% matches already played=12
winning percentage till now=75%=9 matches won
so,number of matches to be played=30-12=18
to achieve 80% success,total matches to be won=80/100*30=24
so,out of last 18 matches ,India need to win=24-9=15
so,percentage of remaining matches to be won=15/18*100=83.33%...Ans.
One year, India plays 30 one-day matches. Having played 40% of the scheduled matches, India achieves a success rate of 75%. If by the end of the year, India wants to achieve 80% success rate, then what percentage of the remaining matches should it win??
u can ignore the 30 matches here...
just consider 100 total
played 40. won 30.
total matches required to be won =80
=> out of 60 remaining matches 50 have to be won.
=> 50/60 = 83.33%
P:S in such types directly ignore the initial total value that is given.
_ P1 _ P2 _ P3 _ g = gap between any two person g1 + g2 + g3 + g4 = 9 (where g2 and g3 should be greater than 0) g2 = g2' + 1 and g3 = g3' + 1 g1 + g2' + g3' + g4 = 7 => number of integral solution = 10C3 = 120
HELP!!One year, India plays 30 one-day matches. Having played 40% of the scheduled matches, India achieves a success rate of 75%. If by the end of the year, India wants to achieve 80% success rate, then what percentage of the remaining matches should it win??
83.33 % ?
matches won = 30*.4*.75 = 9
let x matches more india must win to achieve a success rate of 80%
Absolutely! Medians are good fun.Try this:Consider a triangle ABC such that AB = 11, AC = 12 and BC = 13. D, E and F are points on AC such that BD is an altitude, BE an angle bisector and BF a median. Find the lengths DF and EF.(Also try generalising this to any triangle with sides k-1, k and k+1 with the three lines drawn to the line of length k. Something curious I noticed on a long journey when I ran out of better things to do )regardsscrabbler
No...be careful....the 3 lines are being drawn from B to AC (the side of length 12). The final answers will be reasonably nice numbers. No very complicated forms :)
@scrabbler A very silly doubt!!How did you do calculation sfor this line...any shortcuts??? 2^90 divided by 13 => (2^12)^7 * 2^6 => 1^7 * 12 => 12 hence 13m+12
No...be careful....the 3 lines are being drawn from B to AC (the side of length 12). The final answers will be reasonably nice numbers. No very complicated forms regardsscrabbler
i am getting the length of the height and the median same.. so basically even the angle bisector is also the same.. but this is the property of an equilateral triangle, right?
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