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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
B
Let the total number of matches to played in the tournament be ‘x’.
Given that A county cricket team has won 10 matches and lost 4.
That means total number of matches played = 14.
So,
= > 70% of x = 14
= > (70/100) * x = 14
= > 70x = 1400
= > x = 20.
So, the number of matches to be played = 20.
They have a record of exactly 75% wins = 20 * 75/100
= > 15.
Therefore, the number of matches should the team win = 15 – 10 = 5.
Hope this helps!
If division of 3 is allowed then I might do 5 divisions in one go & do in less iterations. I would assume max 2 division as you have to hands, and compare the weight.
Statements :
All mangoes are golden in colour. No golden coloured things are cheap.
Conclusions :
I. All mangoes are cheap.
II. Golden coloured mangoes are not cheap.
A takes 6 hours to fill
Let ‘x’ be the number of perons in the group and let ‘y’ be
amount per head which they have to pay.
Then xy = 2400.
or y = 2400/x
Since two friends have forget the purse, x-2 persons should
share the total amount(2400).
If they share, they have to make an extra contribution of
Rs 100 to pay up the bill
That is x-2 persons should pay y+100 each
or (x-2)(y+100) = 2400
or y+100 = 2400/(x-2)
or y = 2400/(x-2) – 100
Therefore we have got two equations namely,
y= 2400/x and y = 2400/(x-2) – 100
Comparing these two, we get
2400/x = 2400/(x-2) – 100
Solving this we get
x^2 – 2x -48 = 0
or (x-8) (x+6) = 0
or x = 8 or x = -6
Since ‘x’ denote the number of persons, it should be
positive.
So, x = 8.
190 sec
24*4= x(x+4)
solve this
answer is x=8
X+4= 12
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