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445, 221, 109, 46, 25, 11, 4
C.46
4+7
11+(7*2)
25+(7*2*2)
53+(7*2*2*2)
109+(7*2*2*2*2)…..
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
C
FYI
21
Answer is 400.(14^2=196, 16^2=256, 18^2=324, 20^2=400.)
8400
1 tank will fill in 10 minutes.
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1/x=1/5+1/10+1/30=lcm is 30.
->30/5=6
->30/10=3
->30/30=1
1/x=6+3+1=10
1/x=10/30=1/3=x=3
Ans is 3 hr.
The number is greater than the number obtained in reversing the digits and the ten’s digit is greater than the unit’s digit
Let ten’s and unit’s digit be 2x and x respectively
Then, (10×2x+x)−(10x+2x)=36
9x=36
x=4
Required difference=(2x+x)−(2x−x)=2x=8
Circular Track sis of 11 km
Speed of Mens are
4 , 5.5 , 8 km/hr
Time at Which they are at starting Points again
11/4 , 11/5 , 11/8
11/4 , 2.2 , 11/8
We need to find LCM of these
to find at what time they Meet again at starting point
LCM 11 * 2 = 22
After 22 Hrs they will meet at starting point