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6/(2+3)-(2+3)/6
=6/5-5/6
=(36-25)/30
=11/30
ans is 11/30
9 sec=4 times
x sec=12 times
x=(9*12)/4=27 sec
14 km/h
Ans. let x is distance from A to B
and y is initial speed.
30/y+(x-30)5/4y -x/y = 3/4
=> 4x-12y=120 —-(1).
45/y + (x-45)5/4y -x/y=3/5
=> 5x-12y=225 ——-(2).
From equ (1) and equ (2) we will get.
x=25 and y=105
so initial speed is 25 km/hr
and Distance From A to B is 105 km
64
Question is not completed
The letters A, B, C, D, E, F and G, not necessarily in that order, stand for seven consecutive integers from 1 to 10
D is 3 less than A
B is the middle term
F is as much less than B as C is greater than D
G is greater than F
1. The fifth integer is
(a) A
(b) C
(c) D
(d) E
(e) F
ans:a
2.A is as much greater than F as which integer is
less than G
(a) A
(b) B
(c) C
(d) D
(e) E
ans:a
3. If A = 7, the sum of E and G is
(a) 8
(b) 10
(c) 12
(d) 14
(e) 16
4. An integer T is as much greater than C as C is
greater than E. T
can be written as A + E. What is D?
(a) 2
(b) 3
(c) 4
(d) 5
(e) Cannot be determined
ans:a
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
LCM of 4,6,8 and 10 are 120
So, the least number of four digits is 120*9=1080 ( <1000 )
we can draw 12 tangents, 4 for 2 circles and 12 for 3
circles