s=36kmph
in meter per second is=36*5/18=>10
Three families
72 kmph
625
Remote Access Service
How will you know the odd is in lighter one or heavier one from only one weighing. It will require 2 weighing to find the odd set and one weighing for odd coin in that set i.e total 3 weighings.
Calculation:
⇒ If 1000 divided by 112, the remainder is 104. ⇒ 112 – 104 = 8 ⇒ If 8 is added to 1000 it will become the smallest four-digit number and a multiple of 112. ⇒ 1000 + 8 = 1008 ∴ The required result will be 1008.
Each PAIR of stations means a PAIR of tickets (A to B and B to A)
2(old stations)(new stations) + 2(new stations)(new stations – 1) = 46
(N * X) + (X * (X – 1)) = 23
factoring ___ X (N + X – 1) = 23
23 is a prime number with only two factors, 1 and 23
so N = 23 and X = 1
let the 4 digit number be ABCD.
First Digit is A :
Therefore; according to the question
A=B/3
B=3A
C=A+B=A+3A=4A
D=3B=3(3A)=9A
Since the last Digit is D and it can neither be double-digit nor 0 ;
Therefore ;
A=1;
B=3A=3(1)=3
C=A+B=1+3=4
D=9A=9(1)=9
Therefore, the 4 digit number is 1349.
264
GIVEN: 2A(B+C)+AC-2C(A-B)
THEREFORE 2AB+2AC+AC-2AC+2BC
2AB+AC+2BC
2(AB+BC)+AC
LET b1=AB b2=BC b3=AC
STEP1: b1 = b1+b2
so b1 = AB+BC
THEREFORE
NOW: b1 = AB+BC b2 = BC b3 = AC
STEP 2: b3 = b1+b3
so b3 = AB+BC+AC
THEREFORE
NOW: b1 = AB+BC b2 = BC b3 = AB+BC+AC
STEP3:
NOW: b1 = b1+b3
so b1 = AB+BC+AB+BC+AC
=2(AB+BC)+ AC
AB BC AC
STEP1 AB+BC BC AC
STEP2 AC BC AB+BC+AC
STEP3 AB+BC+AB+BC+AC BC AB+BC+AC
i.e 2(AB+BC)+AC BC AB+BC+AC
4.5
56.25%
Length of train = 125Mtrs
Speed of man = 5Km/Hr
=(5*1000)/(1*3600)
=1.4M/s
Time taken for train to pass = 10s
Spd = D/T
Total Distance (Man distance+ Train Length)
#Distance covered by Man in 10s
D=Spd*T
= (1.4M/s*10s)
= 14Mtrs
#Train length = 125Mtrs
Total D = 14Mtrs + 125Mtrs
= 139Mtrs
Time taken = 10s
Sp = D/t
=139M/10s
=13.9M/s
7 : 3
Total marbles are 15
Blue and yellow are 5
Probability is 5/15
20 %
speed = distance / time.
240/16=15m/sec=54km/hr