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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
X-340 = X x 4 x 8/ 100
100X – 34000 = 32X
68X = 34000
X = 34000/68
X= 500.
b
5A=3B,
*3 SO,
15A=9B
The correct Answer is 69
a)0
b)206
c)0
d)250
e)39
f)92
1.5 hr
Matches played: 60.
Matches won: 30% of 60 => (60*(30/100)) = 18 matches.
Iterative approach:
On adding 1 to matches played and matches won, on every iteration until the win percentage gets to 50. So
19 / 61 = 0.3114754098360656
20 / 62 = 0.3225806451612903
21 / 63 = 0.3333333333333333
22 / 64 = 0.34375
…
…
…
…
Similarly,
41 / 83 = 0.4939759036144578
42 / 84 = 0.5
So, after 60th match 24 more matches has to be played and won to get 50% average winning rate.
C. 4
Is it not 24?
Fathers before age=3*before sons age—-(1)
After 2.5 years..
Fathers before age+2.5 years =2.5*sons age—-(2)
Substitute (1) in (2),
3*sons before age+2.5 years=2.5*sons age
0.5*sons age=2.5 years
Sons before age=5years
So current age after 2.5 years,
Sons before age+2.5 =5+2.5=7.5 years
It is assumed to be constant. Now, to cross past the pole, the train should cover a distance of x meters. Now, the time taken by the train to cross a platform of length 100 m is 25 seconds. Hence, the length of the train is 150 m.
3, 7, 15, 39, 63, 127, 255, 511
C. 39
3*2+1 = 7
15*2+1 = *31*
31*2+1 = 63
63*2+1 = 127
127*2+1 = 255
255*2+1 = 511
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