First, we find out the day on 1.03.2005. Period upto 1.3.2005.
= (2004 yr + Period from 1.1.2005 to 1.3.2005)
? Odd day in 1600 yr = 0
Odd day in 400 yr = 0
4 yr = (1 leap year + 3 ordinary years)
= (1 x 2 + 3 x 1) = 5 odd days
Number of days between 1.1.2005 to 1.3.2005,
= January + February + March
= 31 + 28 + 1
= 60 days = (8 weeks + 4 days)
= 4 odd days
Total number of odd days
= (0 + 0 + 5 + 4) = 9
= 1 weak + 2 odd days
? 1.3. 2005 was Tuesday.
So, Saturday will fall on 5.3.2005.
Hence, Saturday lies on 5th, 12th, 19th, and 26th of March 2005.
Clearly, if A runs 600 meters, B runs = 540 m.
? If A runs 400 m, B runs = 540 x 400 / 600 = 540 x 4 / 6 = 90 x 4 = 360 m
Again, when B runs 500 m, C runs = 450 m
? When B runs 360 m, C runs = 450 x 360 / 500 = 45 x 36 / 5 = 9 x 36 = 324 m.
? A beats C by 400 - 324 = 76 m.
36.98276421 x 21.00002
Using nearest value in whole number for 36.9827421 is 37 and 21.0002 is 21
? 37 x 21 = 777 ? 775
= Unit digit in [(42)896 x 4]
= Unit digit in (6 x 4) = 4
Unit digit in (625)317 = Unit digit in (5)317 = 5
Unit digit in (341)491 = Unit digit in (1)491 = 1
Required digit = Unit digit in (4 x 5 x 1) = 0.
Speed of the train relative to man | = (63 - 3) km/hr | |||||||
= 60 km/hr | ||||||||
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∴ Time taken to pass the man |
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= 30 sec. |
(A's speed) : (B's speed) = √b : √a = √16 : √9 = 4 : 3.
Total distance covered |
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∴ Time taken |
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= 3 min. |
Unit digit must be 0 or 5 and sum of digits must be divisible by 9.
Among given numbers, such number is 202860.
Then, x = 7589 - 3434 = 4155
At 11 : 50 A.M. from the formula here P = 11 and Q = 50 so required angle is 11 x 50/2 - 30 x 11 = 275 - 330 = -55 or ignoring the negative sign the required angle is 55°
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