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
TEACHER is written as VGCEJGT, how would DULLARD ?
T value is 19
V value is 21 ,
here we get to know that there is difference of 1 letter between them.
Answer: FWNNCTF.
Statements :
Some books are pens. No pen is pencil.
Conclusions :
I. Some books are pencils.
II. No book is pencil.
E
Let 10’splace digit is x and unit’s place digit y
First milestone : 10x+y
Second milestone : 10y+x
Third milestone: 100x+y
Since the speed is uniform so
Distance covered in first Hr = Distance covered in Second Hr
(10y+x)-(10x+y) = (100x+y)-(10y+x)
After solving, we get —-> y=6x but since x and y are digits so only possible combination is x=1 and y=6,
So average speed = 45 KM/HR
21
2.01
3, 7, 15, 39, 63, 127, 255, 511
C. 39
3*2+1 = 7
15*2+1 = *31*
31*2+1 = 63
63*2+1 = 127
127*2+1 = 255
255*2+1 = 511
D. Vice President
let the total no of breads be x.
1st man 2nd man
x- (x/2) – 1/2 – 1/2((x-1)/2) – 1/2 ….. so on.
ans is 31.
first ate : 15.5 + .5 = 16 remaining 15
second ate : 7.5+0.5 = 8 remaining 7.
third ate : 3.5 +0.5 = 4 remaining 3.
fourth ate : 1.5 + 0.5 = 2 remaining 1
fifth ate : 0.5 + 0.5 = 1 remaining 0
CASE 1: First we should take six balls divided equally and
then it is placed on the two pans.three on one and three on
other..
if the two pans are balanced then the defective ball is not
in the six..then we should the two and keep them one ball
on each.
CASE2: Again We should take any of the six balls and
divided equally and then it is placed on the two pans.. if
any of the pan weighs less than the other.. We should take
the three balls seperately..Now from that three we should
take any two and placed one on each.. fi both the pan
balances the ball which is left over is the defective.. if
one ball weighes less than the other,while keeping one on
each,then it is the defective one….
6/11 or 54.54%
40