Let the length of the 1st train = L mts
Speed of 1st train = 48 kmph
Now the length of the 2nd train = L/2 mts
Speed of 2nd train = 42 kmph
Let the length of the bridge = D mts
Distance = L + L/2 = 3L/2
Relative speed = 48 + 42 = 90 kmph = 90 x 5/18 = 25 m/s(opposite)
Time = 12 sec
=> 3L/2x25 = 12
=> L = 200 mts
Now it covers the bridge in 45 sec
=> distance = D + 200
Time = 45 sec
Speed = 48 x5/18 = 40/3 m/s
=> D + 200/(40/3) = 45
=> D = 600 - 200 = 400 mts
Hence, the length of the bridge = 400 mts.
According to question ,we can say that
From equation ?. x4 = 398 + 227 = 625
? x4 = 54
From equation ?. y2 = 346 - 321 = 25
? y2 = 52
But, Feb 2004 not included because we are calculating from March 2004 to March 2005. So it has 1 odd day only.
∴ The day on 6th March, 2005 will be 1 day beyond the day on 6th March, 2004.
Given that, 6th March, 2005 is Monday.
∴ 6th March, 2004 is Sunday (1 day before to 6th March, 2005).
Couldn't be determined, since the total amount of money is not given in either of the case
Given that a /b = b/c
? b2 = ac
? a4 : b4 = a4 : a2 c2 = a2 : c2
Given, A = 6 km/h, B = 3 km/ h
According to the formula, Average speed = 2AB/(A + B)
? Required average speed = 2 x 6 x 3/(6 + 3)
= 36/9 = 4 km/h
Given that, W + D = 6 ...(i)
[ w = Time taken while walking and
D = Time taken while driving ]
From Eq. (i)
5 +D = 6
? D = 1
2D = 2 x 1 = 2
? He will take 2 h to drive both ways.
175760 = 24 x 5 x 13 3
= 23 x 2 x 5 x 133
To make perfect cube, it must be divided by 2 x 5 = 10
On comparing it with ax2 + bx + c = 0 , we get a = 1, b = p, c = q
The roots of the equation x2 + px + q = 0 are equal if
b2 - 4ac = 0
? p2 - 4q = 0 ? p2 = 4q.
Thus , required answer is option B .
Let the volume be x3 and 27x3
? Their edges are x and 3x
Now Ratio of their surface area = 6x2 : 54x2 = 1 : 9
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