Here is the solution to the given version of the puzzle (9 balls, one is heavier, need to identify oddball), where we label the balls A, B, …, I:
1. Weigh ABC versus DEF.
Scenario a: If these (1) balance, then we know the oddball is one of G, H, I.
2. Weigh G versus H.
Scenario a.i: If these (2) balance, the oddball is I.
Scenario a.ii: If these (2) do not balance, the heavier one is the oddball.
Scenario b: If these (1) do not balance, then the oddball is on the heavier side. For simplicity, assume the ABC side is heavier, so the oddball is one of A, B, C.
2. Weigh A versus B.
Scenario b.i: If these (2) balance, the oddball is C.
Scenario b.ii: If these (2) do not balance, the heavier one is the oddball.
answer is maximum of 2.
perimeter of rectangle = 2(l+b)
given l abd b as 18 cm and 26 cm
therefore perimeter of rectangle = 2(44)
perimeter of circle = 2*pi*r
given 2*pi*r=2(l+B)
i.e, 2*(22/7)*r= 2(44)
(22/7)*r=44
r=14 cms
area of the circle = pi*r*r
=(22/7)*14*14
=616 sqcm is the answer
2, 6, 12, 72, 824
2
6*2=12
12*6=72
72*12=864
1/4, 1/8
4*8=32
32-4=28
28-8=20
28*20/(4^2)=35.00
1min = 60 seconds
6 min=6*60=360seconds
4 sec=10
360sec=360*10 /4=900
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
Actually
carrots=> $12
celery => $3
X filling rate is 1/18 or 4/72 per hour
Y filling rate is 1/24 or 3/72 per hour
Combined as a pair it’s 7/72 in a complete 2 hour alternating sequence
72+ 7/72 =10 sequences = 70/72 on 20 hours, The outstanding balance is just 2/72
So at 20.5 hours with X turn filling is 100% complete with 72/72
Answer: It will take 20.5 hours to completely fill this tank.
Analysis of measured filling process, from X = 42/72 share & from Y = 30/72 share
Slice the cake
350m
Its very simple..
consider the fraction of s in the mixture = 1/3
So if we add one more R the the fraction wil be = 1/4
Automaticaly S becomes 25% of the mixture
a) who started with small amount of money?
b) who started with greatest amount of money?
c) what amount did B have?
status after 3 games
A-40
B-40
c-80
d-16
status after 2 games
A-20
B-20
C-40
D-96
status after 1 game
A-10
B-10
C-108
D-48
status after 0 games
A-5
B-93
C-54
D-24
so answers are
a)A
b)B
c)93
( a ) 28
0