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x=12;y=7
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
30,40
x+y+z
—– = 6800
3
x>=6400
6400 + y + z = 20400
y + z = 14000
to get the greatest of y and z, lets assume y = 6400
so, z = 7600
so ANS is 7600
A can copy 50 papers in 10 hrs
that means 5 papers in 1 hour
A & B can copy 70 papers in 10 hrs
that means both of them copy 7 papers in 1hour
we know A can copy 5 papers
there fore B can copy 2papers in 1 hour
so B can copu 26 papers in x hours
2*x=26
x=26/2=13 hours
D
#1: N = 1, f(N) = 1
#2: N = 199981, f(N) = 199981
#3: N = 199982, f(N) = 199982
#4: N = 199983, f(N) = 199983
#5: N = 199984, f(N) = 199984
#6: N = 199985, f(N) = 199985
#7: N = 199986, f(N) = 199986
#8: N = 199987, f(N) = 199987
#9: N = 199988, f(N) = 199988
#10: N = 199989, f(N) = 199989
#11: N = 199990, f(N) = 199990
#12: N = 200000, f(N) = 200000
#13: N = 200001, f(N) = 200001
#14: N = 1599981, f(N) = 1599981
#15: N = 1599982, f(N) = 1599982
1 ball – 4 runs (1st batter – 94+4=98)
2 ball – 3 runs and one run is short run during running between the wicket (1st batter- 98+3-1 =100)and strike changes for 2nd batter
3 ball – 6 runs ( 2nd batter 94+6=100)
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1>=y>x => y belongs to (-infinity,1]
x belongs to (-infinity,y)
if both x and y are negitive z can be greater than zero and
can be greater than y(1 and 4 are true)
if y equals 1 then x can be equal to z(2 is true)
therefore y=z is not true for any value of x and y
2nd