Obsrving the given arrangement, we find the following pattern:
If we arrange the words in the given input in the alphabetical order and then label them as 1,2,3,4,5,6, then the last step contains the words in the order 6,1,5,2,4,3. However, the position of only one word is alteredat each step.
Input:car on star quick demand fat
Step1: star car on quick demand fat
Step2: star car quick on demand fat
Step3: star car quick demand on fat
Hence, the answer is C
∴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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