IBM
Company
Numerical Ability
Percentage
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)
-
- 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
- 8 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 - 5 years agoHelpfull: Yes(0) No(0)
IBM Other Question
In A,B,C are having some marbles with each of them. A has giben
B and C the same number of marbles they already have to each of them.
then, B gave C and A the same no. of marbles they have, then C gave
A and B the same no. of marbles they have. At the
end A,B,and C have equal no. of marbles.
(i) If x,y,z are the marbles initially with A,B,C respectively.
then the no of marbles B have at the end
(a) 2(x-y-z) (b) 4(x-y-z) etc.
(ii)If the total no. of marbles are 72, then the no. of marbles with A at the starting
a. 20 b. 30 c. 32
3.
A can run a kilometre in 4 min 54 sec and B in 5 min. How many metres start can A give B in a km race so that the race may end in a dead heat?
A. 25 m
B. 20 m
C. 15 m
D. 10 m