Assuming Sripad has scored the least marks in subject other than science,
Then the marks he could secure in other two are 58 each.
Since the average mark of all the 3 subject is 65.
i.e (58+58+x)/3 = 65
116 + x = 195
x = 79 marks.
Therefore, the maximum marks he can score in maths is 79.
∴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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