Let us find the day on 1st July, 2004.
2000 years have 0 odd day. 3 ordinary years have 3 odd days.
Jan. Feb. March April May June July
31 + 29 + 31 + 30 + 31 + 30 + 1
= 183 days = (26 weeks + 1 day) = 1 t .
Total number of odd days = (0 + 3 + 1) odd days = 4 odd days. '
:. 1st July 2004 was 'Thursday',-,-
Thus, 1st Monday in July 2004 _as on 5th July.
Hence, during July 2004, Monday fell on 5th, 12th, 19th and 26th. .
Time taken by A to cover 200 m = 200/2 = 100 s
? B covers (200 - 10) = 190 m in (100 + 5) = 105 s
? B' s speed = 190/105 = 38/21 = 117/21 m/s
Required area = Length x Breadth
= 100 x 40 = 4000 sq m
= 4 x 103 sq m
(3 x 6) men = (5 x 18) women
18 men = 90 women
? 1 man = 5 women
? 4 men + 10 women
= 4 x 5 + 10 = 30 women
Given, M1 = 5, M2 = 20, D1 = 18 ,
W1 = W2 = 1 and D2 = ?
According to the formula, M1D1W2 = M2D2W1
? 5 x 18 x 1 = 30 x D2 x 1
? D2 = 5 x 18/30 = 3 days
Given that, area = 10 sq cm,
perpendicular = 20 cm and base = ?
area = (base x Perpendicular) / 2
? 10 = (base x 20) / 2
? base = 1 cm
Distance covered by B in 27 seconds = 100 x 27 / 30 = 90 m
Hence, A will beat B by (100 ? 90) = 10 m.
Clearly, X beats Y by 8 s.
Distance covered by Y in 600 s = 1000 m
(? 10 min = 600 s)
Distance covered by Y 80 s. = (1000/600) x 80 = 400/3 m = 1331/3 m
Let breadth = y meters,
Then, length = 2y meters
? Diagonal = ?y2 + (2y)2
= ?5y2 meters
So, ?5y2 = 9 ?5
? y= 9
Thus, breadth = 9 m and length = 18 m
? Perimeter = 2 (18 + 9) m = 54m.
1 |
2 |
2 |
5 |
8 |
15 |
9 |
20 |
9 |
20 |
Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.
∴ P(E) = | n(E) | = | 9 | . |
n(S) | 20 |
1 |
13 |
3 |
13 |
1 |
4 |
9 |
52 |
3 |
13 |
∴ P (getting a face card) = | 12 | = | 3 | . |
52 | 13 |
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