= (22 x 12) + (22 x 22) + (22 x 32) + ... + (22 x 102)
= 22 x [12 + 22 + 32 + ... + 102]
[ | Ref: (12 + 22 + 32 + ... + n2) = | 1 | n(n + 1)(2n + 1) | ] | |
6 |
= | ❨ | 4 x | 1 | x 10 x 11 x 21 | ❩ |
6 |
= (4 x 5 x 77)
= 1540.
∴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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