A & B one day work= 1/18 + 1/30= 8/90
As we know that number of work is 1
1÷8/90= 1*90/8= 90/8.
Since they say “ twice the amount of work
Then ,2*90/8= 22.5 days
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1600 years contain 0 odd day.
300 years contain 1 odd day.
94 years = (23 leap years + 71 ordinary years)
= (46 + 71) odd days
= 117 odd days, i.e., 5 odd days
Days from 1st January 1995 to 28th February 1995
= (31 + 28) days = 59 days
= (8 weeks + 3 days) = 3 odd days
∴ Total number of odd days
= (0 + 1 + 5 + 3) = 9 odd days i.e., 2 odd days.
So, the required day is Tuesday.
151
132
upstream = 25km/hr
downstream=35km/hr
so…if the speed of current would be 5km/hr and boat speed = 30km/hr
so that it will be change towards upstream and downstram…
d=5
g=1
a=4, b=6,c=2,d=5,e=8,f=3,g=1,h=9,i=7
1/24S
Had is correct ! Anil went in right direction but ended up
with wrong answers
Soln: B lied ( see anil’s comment ) which implies B has not
stolen mule. So B could steal camel or sheep.
1) Lets assume B stole camel.
In that case, A lies ( refer – A says “B had stolen sheep”
), which implies A had stolen sheep ( becoz a lier cant
steal mule ). So A->sheep B->camel C->mule. But this cant
happen because C cant lie ( refer – C says ” B had stolen
mule” ).
2) Lets assume B stole sheep.
In that case, A is true ( refer – A says “B had stolen
sheep” ), which implies A had stolen mule. So A->Mule
B->Sheep C->camel. Here C lies ( refer – C says ” B had
stolen mule” )
SO ANS: answer A- mule, B-sheep, C camel
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
Correct Deven,
( C ) 6
c=a/b
c=a-1
=>b=a/c
=>b=a/a-1
C
1 – (-1) = 1 +1 = 2
c