W=(10!*11*12)-(10!-11)-10!
TAKING 10! COMMON
W=10!(11*12-11-1)
W=10!(132-12)
w=10!(120)
WE KNOW THAT ODD*ODD=ODD
EVEN*EVEN=EVEN
ODD*EVEN= EVEN
SO TO FIND IS Z(Y+1)IS ODD WE NEE TO KNOW IF ONE OF THE ie IF Z IAS ODD OR EVEN OR if y is even then y+1 is odd and if Y is odd then Y+1 is even
Statement 1: 10^ (any number) is always odd as it end with a zero and if we add 2 to it it will always be an even number.so wiyh statment 1 we can tell that Z(Y+1) is always even
statement 2: 2^X is the highest factor of W so we need to find the highest power of 2 in 10!(120)
tHE HIGHEST POWER OF 2 IN 10! IS 2^8 AND THE HIGEST POWER OF 2 IN 120 IS 2^3.TOGETHER IT ADDS UP TO 2^11
SO y IS 11 THEN y+1 IS ALWAYS EVEN SO Z(Y+1) IS ALWAYS EVEN
THEREFORE THE ANS IS D imo
This is clearly a cause and effect.When two events occur simultaneously.THe reasoning is drawn in a wrong way by assuming that one evnt actually caused the other.It could be that both the evnts might have been caused by an independent evnt or it might be that what is assumed as the cause is the effect and vice versa.Option C it is for me.OA please
here use the concept of manhours
man hours=number of men involved *number of hours each man works
so the manhours for the job=6*10=60 manhours
from 12:00 pm to 6:00 pm 6men work for 6 hours
so manhours spent is 6*6=36
the defeciet is 60-36=24
from 6:00 to 7:00 7 men work for 1 hour =7*1=7 manhours
from 7:00 to 8:00 8 men work for 1 hour =8*1=8 manhours
from 8:00 to 9:00 9 men work for 1 hour =9*1=9 manhours
total=24 manhours
the work is done by 9:00pm
oops thats should have been beaverhours i guess
the set {1,2,3,4,5,6} has 6 numbers so the event has 6P2 combinations which is 30 events
the favourable events (1,3)(3,1)(5,4)(4,5)(6,3)(3,9) which is a total of 6 events
probability 6/30=1/5
IMO 1/5
shingivyas
5" is not possible as the face ur placing the cylinder has length as 10" but breadth as 8" so a cylinder which has diameter as 10" will sit along the length but not the breadth as it is only 8".try drawing the figure it will be much clear
Volume of cylinder Pi * (d^2/4) * h
the cylinder can be placed on three faces of the box the details are as follows
The diameter of the cylinder that can be placed in the box on the 6" x 8" face is 6" and ht is 10" - Volume is 90Pi (always take the lowest of the 2 dimensions)
The diameter of the cylinder that can be placed in the box on the 6" x 10" face is 6" and ht is 8" - Volume is 72Pi
The diameter of the cylinder that can be placed in the box on the 8" x 10" face is 8" and ht is 6" - Volume is 96Pi
so IMO the ans is B