TCS
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Numerical Ability
Probability
) In the class of 40 students, 30 speak Hindi and 20 speak English. What is the lowest possible number of students who speak both the languages?
(a) 5
(b) 20
(c) 15
(d) 10
(e) 30
Read Solution (Total 11)
-
- H | E=40
H=30
E=20
NOW-> H | E = H + E - H & E
SO--> H & E = 30+20-40
=50-40
=10(ANSWER)
- 13 years agoHelpfull: Yes(9) No(2)
- A℧B=A+B-AΩB
30+20-40=10 - 13 years agoHelpfull: Yes(7) No(1)
- 10...aUb=n(a)+n(b)-a(intesection)b
- 13 years agoHelpfull: Yes(4) No(1)
- ans is 10.becau the logic says .but y should it be the least.it should be unique
- 12 years agoHelpfull: Yes(4) No(2)
- AUB = A + B - A"intersection"B
40 = 30 + 20 - A"inters."B
A"inter"B =50-40 = 10 - 13 years agoHelpfull: Yes(3) No(3)
- 10 is maximum possible number of students who can speak both hindi and english according to options given the answere will be 5 .
- 10 years agoHelpfull: Yes(2) No(2)
- 40-30-20=-10 so who speak both the lang will b 10
- 8 years agoHelpfull: Yes(1) No(0)
- if out of 40 , only 30 speak hindi and 20 speak english
then both= ???????
Easy method is
30/40= 3/4
then
20/40=1/2
then
multiply 2*3+4=10 - 6 years agoHelpfull: Yes(1) No(1)
- 10
if first 30 speak hindi and the last 20 speak english then only 10 students remain in common.therefore the least number of students is 10 - 9 years agoHelpfull: Yes(0) No(1)
- (30+20)-40=10
- 9 years agoHelpfull: Yes(0) No(0)
- n=n(H)+n(E)-n(H^E)
40=30+20-n(H^E)
n(H^E)=10 - 3 years agoHelpfull: Yes(0) No(0)
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