Let us draw a figure below as per given question.
Let AB = CD = h meter be the heights of the towers. E is a point such that DE = 100 meter;
?CED = 60° and ?AEB = 30°
Now, BE = x meter (say)
From right triangle CDE.
h = 100 tan 60°
? h = 100?3 meter
From right triangle ABE,
x = h cot 30° put the value of h, we will get
x = 100?3 X ?3
x = 100 X 3 = 300 meters
Distance between the tower = DE + EB = 100 + 300 = 400 meters
Height of the tower = h = 100?3 meter
∴42437 is not the square of a natural number.
Let the number be N.
Then, N - N/3 = 48
2N/3 = 48.
C.P. of 1 orange = Rs. | ❨ | 350 | ❩ | = Rs. 3.50 |
100 |
S.P. of 1 orange = Rs. | ❨ | 48 | ❩ | = Rs. 4 |
12 |
∴ Gain% = | ❨ | 0.50 | x 100 | ❩% | = | 100 | % = 14 | 2 | % |
3.50 | 7 | 7 |
When E is fixed in the middle, then there are four places left to be filled by four remaining letters O, M, G and A and this can be done in 4! ways.
? Total number of ways = 4! = 24
Speed = | ❨ | 240 | ❩m/sec = 10 m/sec. |
24 |
∴ Required time = | ❨ | 240 + 650 | ❩sec = 89 sec. |
10 |
X's 1 day's work = 1/12
(X + Y)' s 1 day's work = 3/20
? Y's 1 day's work = (3/20) - (1/12) = 4/60 = 1/15
? Number of day's taken by Y to complete the work = 15 days
Then, | ❨ | 1200 x R x R | ❩ | = 432 |
100 |
⟹ 12R2 = 432
⟹ R2 = 36
⟹ R = 6.
Relative speed = (40 - 20) km/hr = | ❨ | 20 x | 5 | ❩ | m/sec = | ❨ | 50 | ❩ | m/sec. |
18 | 9 |
∴ Length of faster train = | ❨ | 50 | x 5 | ❩ | m = | 250 | m = 27 | 7 | m. |
9 | 9 | 9 |
Number of arrangements of beads = (12 - 1) !, but it is not mentioned that either it is clockwise or anti - clockwise .
So, required number of arrangements = [(12 - 1)! ] / 2 = 11 !/2
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