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In a hotel, 60% had vegetarian lunch while 30% had non-vegetarian lunch and 15% had both type of lunch. If 96 people were present, how many did not eat either type of lunch ?
27
26
25
24
Read Solution (Total 7)
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- According to question my ans is 53.1km/hr
9:36to 10:16 = 40min that is 2/3hr
Distance = 35.4/(2/3)= 53.1km/hr
- 9 years agoHelpfull: Yes(22) No(5)
- people ate either type of lunch = A union B = 60+30-15= 75%
people did not eat either type of lunch = 25%
so
96*25%=24
- 9 years agoHelpfull: Yes(17) No(0)
- time = 40/60= 2/3
dis= 35.4
same dis so, avg speed= 2*dis/t
= 2*35.4/(2/3)
=106.2 km/hr. - 9 years agoHelpfull: Yes(9) No(4)
- why it is twice of that..
2*(distance/time)
we know that avg speed = totALdistance/time - 9 years agoHelpfull: Yes(2) No(2)
- c.>
avg speed 2*51.2=106.2km/h - 9 years agoHelpfull: Yes(1) No(1)
- The avg speed= 2ab/a+b
so 53.1km/hr is the correct ans
- 9 years agoHelpfull: Yes(1) No(0)
- n(A)=(60/100∗96)=288/5
n(B)=(30/100∗96)=144/5
n(A∩B)=(15/100∗96)=72/5
People who have either or both lunch
n(A∪B)=288/5+144/5−72/5=360/5=72
So People who do no have either lunch were = 96 -72
= 24 - 6 years agoHelpfull: Yes(0) No(0)
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