@Enceladus said:€Ž4 dies are thrown. find the probability that the sum of them will be 18?
is the ans 80 by any chance.....?
doing calculations after a while!!!
doing calculations after a while!!!
@Enceladus said:€Ž4 dies are thrown. find the probability that the sum of them will be 18?
welcome back sir 😃 😃 seeing you after a long time 😃
case 1 : 6 6 5 1 and 6 6 4 2 : 2*4!/2! = 12
case 2 : 6 6 3 3 = 4!/2!2! = 3
case 3 : 6 5 5 2 = 12...
case 4 : 6 5 4 3 = 24
case 5 : 5 5 5 3 = 4
case 6 : 5 5 4 4 = 3
case 7 : 4 4 6 4 = 3
total = 12+3+12+24+4+3+3 =61??
jaroor galat hai.. and there s a better approach for sure 
@shadowwarrior said:is the ans 80 by any chance.....?doing calculations after a while!!!
@chandrakant.k said:welcome back sir seeing you after a long time case 1 : 6 6 5 1 and 6 6 4 2 : 2*4!/2! = 12 case 2 : 6 6 3 3 = 4!/2!2! = 3case 3 : 6 5 5 2 = 12... case 4 : 6 5 4 3 = 24 case 5 : 5 5 5 3 = 4case 6 : 5 5 4 4 = 3case 7 : 4 4 6 4 = 3total = 12+3+12+24+4+3+3 =61?? jaroor galat hai.. and there s a better approach for sure
OA - 80/6^4.
@Enceladus said:OA - 80/6^4.
@shadowwarrior : approach batado bhai.. konsa cases miss kiya hai maine..
@Enceladus said:€Ž4 dies are thrown. find the probability that the sum of them will be 18?
it simply the coeff of x^18 in (x+x^2+x^3+x^4+x^5+x^6)^4
take x common x^4(1+x+x^2+x^3+x^4+x^5)^4
now we have to find coeff of x^14 in (1-x^6)^4(1-x)^-4
x^14 in (1-4x^6+6x^12)(1-x)^-4
C(17,3) - 4*C(11,3) +6*C(5,3) = 80
so prob is 80/6^4
@chandrakant.k said:@shadowwarrior : approach batado bhai.. konsa cases miss kiya hai maine..
when 4 dices are rolled. we have the following sums possible:
4,5,6,7,8,9,10,11,12,13,[14],15,16,17,18,19,20,21,22,23,24.
Now 14 is the mid point.
This implies the probability of getting a sum on each side of the mid point is similar as these both are mirror images.
Hence, ways(of getting 18) = ways(of getting 10) = C(9,6) = 84.
but we need to remove those 4 cases jispe max sum aya hai.
4,5,6,7,8,9,10,11,12,13,[14],15,16,17,18,19,20,21,22,23,24.
Now 14 is the mid point.
This implies the probability of getting a sum on each side of the mid point is similar as these both are mirror images.
Hence, ways(of getting 18) = ways(of getting 10) = C(9,6) = 84.
but we need to remove those 4 cases jispe max sum aya hai.
Hence, prob = 84-4/6^4.
^^ ye acha shortcut hai for ppl who do not know binomial. (mere jaise). :D
^^ ye acha shortcut hai for ppl who do not know binomial. (mere jaise). :D
@chandrakant.k said:welcome back sir seeing you after a long time case 1 : 6 6 5 1 and 6 6 4 2 : 2*4!/2! = 12 case 2 : 6 6 3 3 = 4!/2!2! = 3case 3 : 6 5 5 2 = 12... case 4 : 6 5 4 3 = 24 case 5 : 5 5 5 3 = 4case 6 : 5 5 4 4 = 3case 7 : 4 4 6 4 = 3total = 12+3+12+24+4+3+3 =61?? jaroor galat hai.. and there s a better approach for sure
I think u hav taken all cases except some calc mistake
case 1) 2*4!/2!=24
case2) 4!/2!.2!=6
case 3) 12...........
case 4)24...........
case 5) 4.............
case 6) 6
case 7) 4
tot=80
case 1) 2*4!/2!=24
case2) 4!/2!.2!=6
case 3) 12...........
case 4)24...........
case 5) 4.............
case 6) 6
case 7) 4
tot=80
A certain number is first increased by x% and then decreased by x% and the value of that number decreased by 2000. Next again the new number is decreased by x/2 % and increased by x/2% and the ultimate value obtained is 47,520. What was the original number ?
@19rsb said:I think u hav taken all cases except some calc mistakecase 1) 2*4!/2!=24case2) 4!/2!.2!=6case 3) 12...........case 4)24...........case 5) 4.............case 6) 6case 7) 4tot=80
horrible ... effect of no pencil and paper . thanku bhai now i am assured myself that i should not trust my brain processed result 😛 😛 completely rely on pencil and paer :D@Enceladus said:A certain number is first increased by x% and then decreased by x% and the value of that number decreased by 2000. Next again the new number is decreased by x/2 % and increased by x/2% and the ultimate value obtained is 47,520. What was the original number ?
bhai calculation bahut ho raha hai.. i guess it will be easier to solve using option IMO
"A person travels from P to Q at the speed of 40 kmph. By what percent should he increase his speed on the return journey so that his average speed for the round trip is 48 kmph."
@veertamizhan said:"A person travels from P to Q at the speed of 40 kmph. By what percent should he increase his speed on the return journey so that his average speed for the round trip is 48 kmph."
20kmph increase?
@veertamizhan said:"A person travels from P to Q at the speed of 40 kmph. By what percent should he increase his speed on the return journey so that his average speed for the round trip is 48 kmph."
50%??
20 km/hr
@Enceladus said:when 4 dices are rolled. we have the following sums possible:4,5,6,7,8,9,10,11,12,13,[14],15,16,17,18,19,20,21,22,23,24. Now 14 is the mid point. This implies the probability of getting a sum on each side of the mid point is similar as these both are mirror images.Hence, ways(of getting 18) = ways(of getting 10) = C(9,6) = 84. but we need to remove those 4 cases jispe max sum aya hai. Hence, prob = 84-4/6^4.^^ ye acha shortcut hai for ppl who do not know binomial. (mere jaise).
could not understand this... which are the 4 cases
but we need to remove those 4 cases jispe max sum aya hai.
@veertamizhan said:"A person travels from P to Q at the speed of 40 kmph. By what percent should he increase his speed on the return journey so that his average speed for the round trip is 48 kmph."
48 = 2*40*b/(40+b)
3/5 = b/(40+b)
b = 24 + 0.6b
0.4b = 24
b = 60kmph
so he has to inc by 20kmph
"The average speed of a bus excluding all stoppage time is 54 kmph and including all stoppage time is 45 kmph. Find the duration that the bus stops every hour"
@veertamizhan said:"The average speed of a bus excluding all stoppage time is 54 kmph and including all stoppage time is 45 kmph. Find the duration that the bus stops every hour"
54 = 9*6
45 = 9*5
lcm = 9*6*5 = 270
so lets say 270 is the total distance.
so with 54kmph the train would take 5hrs
with 45kmph the train would take 6hrs
so 1/6*60 = 10 mins
@anytomdickandhary : bhai, you had earlier given a wonderful way of thinking for this problem. forgot it 😞 could you post it again
@veertamizhan said:"The average speed of a bus excluding all stoppage time is 54 kmph and including all stoppage time is 45 kmph. Find the duration that the bus stops every hour"
@chandrakant.k said:54 = 9*6 45 = 9*5lcm = 9*6*5 = 270so lets say 270 is the total distance.so with 54kmph the train would take 5hrswith 45kmph the train would take 6hrsso 1/6*60 = 10 mins @anytomdickandhary : bhai, you had earlier given a wonderful way of thinking for this problem. forgot it could you post it again
Difference in speeds created due to stoppages =9km/hr
Now, 9km=54km/hr * time of stoppages
So, time of stoppages=10mins. Is my logic right??
Now, 9km=54km/hr * time of stoppages
So, time of stoppages=10mins. Is my logic right??