answer can be 1,2,3,4 can’t be determined exact number
just explaining a case:
there are 10 people in the party.
name of people no. of people with they made handshake list of those people(this can vary but showing the possibility)
1 1 9
2 2 8, 9
3 3 7, 8, 9
4 4 6, 7, 8, 9
5 5 6, 7, 8, 9. 10
6 6 4, 5, 7, 8, 9, 10
7 7 3, 4, 5, 6, 8, 9, 10
8 8 2, 3, 4, 5, 6, 7, 9, 10
9 9 1, 2, 3, 4, 5, 6, 7, 8, 10
10 4 5, 6, 7, 8, 9
jack got 9 different answer so jack can be either 4th number or 10th number and jack’s wife know jack very well so she can’t have handshake with jack so if 4th is jack then she can’t be handshake with 6,7,8,9, in this case she can be 1,2,3, 5, 10 and now depending upon which no is jack’s wife she can have hand shake with- 1- 4 people, and if jack is number 10 then she can’t be 5,6,7,8,9 so again depending upon her number she can handshake with people in range of 1-4
16, 33, 65, 131, 261, (…..)
523
let t = total no of students.. then
students who passed one or both subjects,
n(e U h) = n(e) + n(h) – n(e intersection h)
=> t = 0.8t + 0.7t – 144
=> t = 1.5t – 144
students who failed both subjects is 10% i.e. 0.1t
=>t-n(e U h) = 0.1t,
=>t -(1.5t – 144) = 0.1t
=>t- 1.5t- 0.1t = -144
=> -0.6t = -144
=>t = 240
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
d
c
2:1
suppose
pipe:
A -30 hours A’s effeciency (60/30) =2
60( lcm of 30 and 20)
B- 20 hours B’s effeciency (60/20)=3
time taken by both to fill = 60/5 =12 as given in question (effeciencies of both a+b =2+3=5)
time taken by faster pipe i.e b = 60/3 =20
3, 4, 9, 22.5, 67.5, 202.5, 810
A. 4
let the third no be x ,
then the first no is 3x ,
second no is 2*3x ie 6x
average is (3x+6x+x )/3 = 20
10x/3 = 20
x= 60/10 = 6
third no is 6 , second is 6x = 36, first no is 3x = 18
largest no is 36 i.e second no
b