Let A, B and C be the three 6-faced dice.
Then, according to the question,
Since two dices has to be equal, that value can be any of the 6 faces, i.e., 6C1 cases.
Now for each case, 2 equal dices can be selected from 3 dices in 3C2 i.e., 3 ways.
And for each of the above, the third dice can have any of the 5 remaining faces
The possible outcomes are P(A)=61,P(B)=61,P(C)=65,P(A)=61,P(B)=65,P(C)=61 and P(A)=65,P(B)=61,P(C)=61
Hence the required probability = 61×61×65×6×3=21690=125
3606
area is given by Area of Rectangle = Length of Rectangle*(sqrt((Diagonal of Rectangle^2)-(Length of Rectangle^2)))
so substituting we get area=4*(sqrt((25-16)))
area=4*(sqrt(9))
since area cannot be negative hence we take sqrt of 9 as 3 (as sqrt of 9 is 3 and -3 as well)
area=4*3=12m^2
5 litre jug contains 4 litres of salt water solution
with 15% of salt, that means it has 4000*.15 = 600 ml salt
in it.
If 1.5 litres solution is spills out, remaining solution
is 2.5 litres, then the salf content is 2500*.15 = 375ml.
Now, the jug is filled to full capacity with water i.e.,
now the jug has 5 litre solution in it.
Now, the salt content is 375/5000 = 7.5%
3×8=24=2xt ===> t=12hours
180*2/3 = 60degree
Number of persons between Vijay and Jack = 48 – (14 + 17) = 17.
Now, Mary lies in middle of these 17 persons i.e. at the eight position.
So, number of persons between Viji and mary = 7.
In this question only one data is given ie., hole length - 6". In the question there is no mistake.
Find the answer.
volume of a sphere is 4/3 * pi * r^3
by drilling a hole 6 inches long the volume of the sphere is
not affected bcos the hole is a 1 dimension quantity..(say a
straight line) n so it does not have any value w.r.t volume…
hence the volume of the sphere does not change..
877.4
A & B one day work= 1/18 + 1/30= 8/90
As we know that number of work is 1
1÷8/90= 1*90/8= 90/8.
Since they say “ twice the amount of work
Then ,2*90/8= 22.5 days
( d ) 8404
X= pigeon
Y= hares
X+Y= 200
2x+4y=580
By solving y= 90
X=110
So Hares= 90
Let ‘N’ is the smallest number which divided by 13 and 16 leaves respective remainders of 2 and 5.
Required number = (LCM of 13 and 16) – (common difference of divisors and remainders)
= (208) – (11) = 197.