Three numbers can be selected and arranged out of 10 numbers in 10P3 ways 10!/7! = 10 x 9 x 8
Now, this arrangement is restricted to a given condition that first number is always less than the second number and second number is always than the third number. Thus, three numbers can be arranged among themselves in 3! ways.
Hence, required number of arrangement = (10 x 9 x 8)/(3 x 2)
= 120 ways
Here sum is put on compound interest,
? P.W. = A / (1 + r / 100)n = 2420 / (1 + 10 / 100)2 = 2420 x 100 / 121 = Rs. 2000
? T.D. = P.W. - P
? True discount = 2420 - 2000 = Rs. 420
The Sum of the digits in each number, Except 324 is 10.
The given number series follows the pattern that,
24×0 + 4 = 4
4×1 + 9 = 13
13×2 + 16 = 42
42×3 + 25 = 151
151×4 + 36 = 640
Therefore, the odd number in the given series is 41
From the beginning, the next term comes by adding prime numbers in a sequence of 2, 3, 5, 7, 9, 11, 13... to its previous term. But 165 will not be in the series as it must be replaced by 166 since 153+13 = 166.
The given number series follows a pattern that
196, 169, 144, 121, 100, 81, ?
-27 -25 -23 -21 -19 -17
=> 81 - 17 = 64
Therefore, the series is 196, 169, 144, 121, 100, 81, 64.
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