@pavimai said:a and b are natural numbers such that a>b>1 . If 8! is divisible by a^2 * b^2 then how many such sets (a,b) are possible ????
8! = 1*2*3*2^2*5*(2*3)*7*2^3 = 2^7*3^2*5*7
a^2*b^2 = (2,3); (4,3); (8,3) and (4,2)
so 8 such sets...
@ani6 said:plz helptuliram runs in a triathlon consisting of 3 phases in the following manner.running 12 km,cycling 24 kkm and swimming 5 km.his speeds in the 3 phases are in the ratio 2:6:1.he completes the race in n minutes.later he changes his stratgey so that the distances he covers in each phase are constant but his speeds are now in the ratio 3:8:1.the end result is that he completes the race taking 20 mins more than the earlier speed.it is also known that he has not changed his running speed when he changes his strategy.1)what is his intial speed? ans:0.5km/min
0.52 km/min
@tshaik said:how many 6-digit numbers contain exactly 4 different digits?(a)4536(b)294840(c)191520(d)None of theseGuys please post the solution,answer given is (b)
guys post the solution
@Logrhythm said:8! = 1*2*3*2^2*5*(2*3)*7*2^3 = 2^7*3^2*5*7a^2*b^2 = (2,3); (4,3); (8,3) and (4,2) so 8 such sets...
@Subhashdec2 said:(3,2)4,34,28,3this will come as 8 man
3,2
4,2
4,3
6,2
8,3
6,4
12,2
7 sets
Let the four different digits be a, b, c, d, then the number can be of two forms i.e. aabbcd or aaabcd.
So to calculate the total numbers; first we need to find the number of ways to select 4 digits (which is C(10, 4)); then number of ways to select the digits to be repeated in each case (which is C(4,2) and C(4,1) respectively); and then the number of ways of arrangement in both the cases (which is 6!/(2!)2 and 6!/3! respectively).
Hence the number of ways to form such 6-digit numbers become = C(10, 4)[C(4, 2)6!/(2!)2 + C(4, 1)6!/3!] = 10*9*8*7*5*13 = 327600 .
See in the initial selection for the digits, we have used all 10 digits (including 0) for selecting the four distinct digits. That means in the above arrangements there will be some numbers which'd be starting with zero. Now thinking alternately, first place of the number can be filled with any 10 digits equally. So removing the cases starting with zero, we are left with (9/10) of previously calculated result.
Thus the answer will be 9*9*8*7*5*13 = 294840
So to calculate the total numbers; first we need to find the number of ways to select 4 digits (which is C(10, 4)); then number of ways to select the digits to be repeated in each case (which is C(4,2) and C(4,1) respectively); and then the number of ways of arrangement in both the cases (which is 6!/(2!)2 and 6!/3! respectively).
Hence the number of ways to form such 6-digit numbers become = C(10, 4)[C(4, 2)6!/(2!)2 + C(4, 1)6!/3!] = 10*9*8*7*5*13 = 327600 .
See in the initial selection for the digits, we have used all 10 digits (including 0) for selecting the four distinct digits. That means in the above arrangements there will be some numbers which'd be starting with zero. Now thinking alternately, first place of the number can be filled with any 10 digits equally. So removing the cases starting with zero, we are left with (9/10) of previously calculated result.
Thus the answer will be 9*9*8*7*5*13 = 294840
@ishu1991 said:Let the four different digits be a, b, c, d, then the number can be of two forms i.e. aabbcd or aaabcd.So to calculate the total numbers; first we need to find the number of ways to select 4 digits (which is C(10, 4)); then number of ways to select the digits to be repeated in each case (which is C(4,2) and C(4,1) respectively); and then the number of ways of arrangement in both the cases (which is 6!/(2!)2 and 6!/3! respectively).Hence the number of ways to form such 6-digit numbers become = C(10, 4)[C(4, 2)6!/(2!)2 + C(4, 1)6!/3!] = 10*9*8*7*5*13 = 327600 .See in the initial selection for the digits, we have used all 10 digits (including 0) for selecting the four distinct digits. That means in the above arrangements there will be some numbers which'd be starting with zero. Now thinking alternately, first place of the number can be filled with any 10 digits equally. So removing the cases starting with zero, we are left with (9/10) of previously calculated result.Thus the answer will be 9*9*8*7*5*13 = 294840
awesome solution dude..:)
@ishu1991 said:A task is assigned to a group of 11 men, not all of whom have the same capacity to work. Every day exactly 2 men out of the group work on the task, with no pair of men working together twice. Even after all the possible pairs have worked once, all the men together had to work for exactly one day more to finish the task. What is the number of days that will be required for all the men working together to finish the job?(a) 11 (b) 21 (c) 33 (d) none of the foregoing
All possible pairs have worked for one day each. So each of the 11 people has worked once with each of the other 10 i.e on ten days. So before the last day, all the people have done 10 days worth of work each.
Since all together then worked 1 more day, effectively each did 11 days of work to finish the job so if they all work together it should get done in 11 days.
regards
scrabbler
Since all together then worked 1 more day, effectively each did 11 days of work to finish the job so if they all work together it should get done in 11 days.
regards
scrabbler
@saurav205 said:at the request of@DeAdLy :A positive integer is written on each faces of a cube. At each corner (vertex) of the cube, the product of the numbers on the faces that meet at the corner is written. The sum of the numbers written at all the corners is 2431. Find the sum of the numbers written on all the faces of the cube?OA by midnight courtesy Test Funda sorry if this question has already been posted by someone else...
OA : 41
probability of passing in physics maths and chem is p,m,c respectively
student has 75 percent chance in passing atleast one ,50 percent chance in passing atleast two and 40 percent chance in passing exactly 2 ...thenn which relation is true
plz put explanation too
p+m+c =19/20
p+m+c =27/20
pmc=1/10
pmc=1/4
student has 75 percent chance in passing atleast one ,50 percent chance in passing atleast two and 40 percent chance in passing exactly 2 ...thenn which relation is true
plz put explanation too
p+m+c =19/20
p+m+c =27/20
pmc=1/10
pmc=1/4
@ishu1991 said:probability of passing in physics maths and chem is p,m,c respectivelystudent has 75 percent chance in passing atleast one ,50 percent chance in passing atleast two and 40 percent chance in passing exactly 2 ...thenn which relation is trueplz put explanation toop+m+c =19/20p+m+c =27/20pmc=1/10pmc=1/4
I will go with 2nd option...
p + m + c = (passing in 1 alone) + 2(passing in 2 alone) +3(passing in all 3)
[I drew a rough Venn diagram to work this out, not included here :)]
Now passing in exactly 1 = 75 - 50 = 25%. Passing in exactly 2 is 40%. Passing in all 3 is 50 - 40 = 10%.
Hence p + m + c = 25 + 2 x 40 + 3 x 10 = 135% = 27/20.
3rd option also looks tempting, but I am not convinced that the three are independent so not going for that.
regards
scrabbler
p + m + c = (passing in 1 alone) + 2(passing in 2 alone) +3(passing in all 3)
[I drew a rough Venn diagram to work this out, not included here :)]
Now passing in exactly 1 = 75 - 50 = 25%. Passing in exactly 2 is 40%. Passing in all 3 is 50 - 40 = 10%.
Hence p + m + c = 25 + 2 x 40 + 3 x 10 = 135% = 27/20.
3rd option also looks tempting, but I am not convinced that the three are independent so not going for that.
regards
scrabbler
@ishu1991 said:probability of passing in physics maths and chem is p,m,c respectivelystudent has 75 percent chance in passing atleast one ,50 percent chance in passing atleast two and 40 percent chance in passing exactly 2 ...thenn which relation is trueplz put explanation toop+m+c =19/20p+m+c =27/20pmc=1/10pmc=1/4
Let
P=Event that he passed in Physics
M=Event that he passed in Maths
C=Event that he passed in Chemistry
Now,
P(P U M U C)=(p+m+c)-(P(PM)+P(PC)+P(CM))+P(P and C and M)
0.75=(p+m+c)-0.7+0.1
p+m+c=1.35=27/20
There are two stoves A and B both burn using sticks .Mr x want to cook the food fast on stove B so he blows air into it with his mouth using a pipe.He observes that the stove B burns 3 times faster than stove A due to the air blown by the person. The stove A cooks the food in 30 mins and the stove B cooks the same in 28 mins .if he takes 4 secs for a blow then How many blows will it take to cook the food in stove b ??
@veertamizhan said:
angle AED = angle ADE = 75 (as AE = AD)
so angle EDC = 15
similerly angle ECD = 15
so in triangle EDC
15+15+ DEC = 180
DEC = 150. ?
@veertamizhan said:
angle EAB = EBA = AEB = 60
angle EAD = 30
AE = AD
So angle ADE = angle AED = 75
Similarly angle BEC = 75
Angle DEC = 360 - 60 - 150 = 150
Among four people A,B,C,D, A takes thrice as much time as B to complete a piece of work. B takes thrice as much time as C. C takes thrice as much time as D.. One group of three of the four men can complete the work in 13 days while another group of three can do it in 31 days. Which group takes 13 days?
A,B,D
B,C,D
A,C,D
A,B,C
@nramachandran said:Among four people A,B,C,D, A takes thrice as much time as B to complete a piece of work. B takes thrice as much time as C. C takes thrice as much time as D.. One group of three of the four men can complete the work in 13 days while another group of three can do it in 31 days. Which group takes 13 days?A,B,DB,C,DA,C,DA,B,C
ABD??