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2, 4, 12, 48, 240, (…..)
C)1440
2*2=4
4*3=12
12*4=48
48*5=240
240*6=1440
copy cat
(3)^ 7.5 ÷ (27)^1.5 x (9)^2 = 3?
⇒(3)^7.5 ÷ {(3)^(3 x 1.5)} x {(3)^(2 x 2)} = 3 ?
⇒ 3^(7.5 – 4.5 + 4) = 3?
⇒ 3^7 = 3?
⇒ ? = 7
Ans = 8
Use simple box method
[1][2][2][2]= 1x2x2x2 =8
Logic >
once place have 2 number (2,6)
Tens place have also 2 number (7,3)
So that number is divisible by 4
Now, 2 numbers fixed at once and tens place left 2 numbers which is choice at hundredth place
Now 3 number fix left 1 number which is placed at thousands place
Total number form is 4×4 = 16
But divisible by 4 is 1x2x2x2 =8
30%
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Ok
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
167
17.9
4 years
68 is the least number when divided by 8, 12, 18 and 24 leaves the remainders 4, 8, 14 and 20
2hr 30min
– While the train is moving, the jogger will also be running in the same direction.
– for the head(engine) of the train to get to the current position of the jogger 240m away, it will take:
45km/hr => 12.5m/s => 240/12.5 = 19.2 seconds.
– But in the same period of time, the jogger will still be running and will have moved to a new location by: 9km/hr => 2.5m/s => 2.5 * 19.2 = 48m
To get to the new location at the speed of 12.5m/s will take the train:
48/12.5 = 3.84sec
In this additional time, the jogger will move forward by:
3.84 * 2.5 = 9.6m
at a speed of 12.5m/s, it will take the train less than a second to cover the additional 9.6m
If we add the distance the jogger will cover in 1 second to 9.6, it is still less than what the train can cover per second. let us see (9.6 + 2.5 = 12.1)
Therefore, the head of the train will pass the runner at approximately: 19.2 + 3.84 + 1 => 24.04 seconds.
For the train to completely pass the runner, it will need its whole length of 120m to be in front of the runner.
This will take an additional (9.6 + 2) seconds.
Therefore for the length of the train to be ahead of the runner it will take approx. 35.65 (24.04 + 9.6 + 2) seconds
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