Mohammad Gauri
A perfect square is a square number of a digit. eg 64 is a perfect number, a square of 8
Now digits AB9 is a square number of a number.
AB9 Can also be written as A multiply by B multiply by 9
Get the Square root of AB9
Assumption, A=1, B=1
1*1=1=A, 1*1=1=B Therefore,
Square root of A = A, B = B and 9=3
Therefore
An odd number is a number indivisible by 2.
for example 1,3,5,7…….
Therefore Squares A*B*9= AB9
Where a=1, b=1, 3 as digits.
Conclusion
A=1 is an odd number
28th February,1956
C
North
540 meters or 0.54 km
total distance to travel the end of the train = 360 + 140 = 500 m
speed of train 45000m/60min = 750m/min or 750m/60sec
(500m*60sec)/750m=40sec the time will take to pass
4, 5, 15, 49, 201, 1011, 6073
Correct option is A)
2nd term =(1stterm×1+2)=(4×1+2)=6
3rd term =(2ndterm×2+3)=(6×2+3)=15
4th term =(3rdterm×3+4)=(15×3+4)=49
5th term =(4thterm×4+5)=(49×4+5)=201 and so on
∴5 is wrong.
32 Km/hr
5, 15, 30, 135, 405, 1215, 3645
5 is the answer….
Remaining all are the multiples of 3 expect 5
108
If Vijay gives ‘x’ marbles to Ajay then Vijay and Ajay would have V – x and A + x marbles.
V – x = A + x — (1)
If Ajay gives 2x marbles to Vijay then Ajay and Vijay would have A – 2x and V + 2x marbles.
V + 2x – (A – 2x) = 30 => V – A + 4x = 30 — (2)
From (1) we have V – A = 2x
Substituting V – A = 2x in (2)
6x = 30 => x = 5.
Let’s assume the length of each train is ‘L’ and the speeds of the two trains are ‘V₁’ and ‘V₂’ respectively.
When the trains are moving in the opposite direction, their relative speed is the sum of their individual speeds. The total distance they need to cover is the sum of their lengths. Since they cross each other completely in 5 seconds, we can set up the following equation:
(V₁ + V₂) × 5 = 2L
When the trains are moving in the same direction, their relative speed is the difference between their individual speeds. The total distance they need to cover is the difference between their lengths. Since they cross each other completely in 15 seconds, we can set up the following equation:
(V₁ – V₂) × 15 = 2L
Now, let’s solve these equations to find the ratio of their speeds.
From the first equation, we have:
(V₁ + V₂) × 5 = 2L
V₁ + V₂ = (2L) / 5
From the second equation, we have:
(V₁ – V₂) × 15 = 2L
V₁ – V₂ = (2L) / 15
Let’s add these two equations together:
V₁ + V₂ + V₁ – V₂ = (2L) / 5 + (2L) / 15
2V₁ = (6L + 2L) / 15
2V₁ = (8L) / 15
V₁ = (4L) / 15
So, the speed of the first train is (4L) / 15.
Now, let’s substitute this value back into the first equation to find V₂:
(4L) / 15 + V₂ = (2L) / 5
V₂ = (2L) / 5 – (4L) / 15
V₂ = (6L – 4L) / 15
V₂ = (2L) / 15
Therefore, the speed of the second train is (2L) / 15.
The ratio of their speeds is given by:
(V₁ / V₂) = ((4L) / 15) / ((2L) / 15)
(V₁ / V₂) = 4L / 2L
(V₁ / V₂) = 2
So, the ratio of their speeds is 2:1.
C. Rs. 6000