Let the number be 'x'. Then, from given data

x/2 + x/3 + x/4 = x+22
13x/12 = x+22
x = 264

Let, Hema attempted 'k' sum correctly, then

k x 3 -2 x(35-k) = 60
5k = 130
k = 26

so 26 correct sums.

Given total rupees = 20 Rs
No. of one rupee stamps = 3
Now, remaining money = Rs. 17
With that he buys only 2 and 5 rupee stamps
Let number of Rs. 5 stamps = K
Let number of Rs. 2 stamps = L
5K + 2L = 17
K = 3, L = 1 (possible)
L = 6, K = 1 (possible)
=> But given that they are different in number so, K is not equal to 3
one rupee stamps = 3
2 two stamps = 6
5 rupee stamps = 1
Total number of stamps = 10.

Given numbers are 585 546 514 404 206 369
Then interchanging the 1st and 3rd digits of the given numbers, we get
585 645 415 404 602 963
Now, arranging them in ascending order
963 645 602 585 415 404
Then the last second number is 415.

Number = quotient x divisor + remainder;
so, here
number = 138 k + 26
=> (23 x 6k) + (23+3)
=> 23(6k+1)+3
so, remainder is 3.

Given series follows the rule of

1050/2.5 = 420
420/2.5 = 168
168/2.5 = 67.2
so 67.2/2.5 = 26.88

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Here, S = {1, 2, 3, 4, ...., 19, 20}=> n(s) = 20
Let E = event of getting a multiple of 4 or 15
=multiples od 4 are {4, 8, 12, 16, 20}
And multiples of 15 means multiples of 3 and 5
= {3, 6 , 9, 12, 15, 18, 5, 10, 15, 20}.
= the common multiple is only (15).
=> E = n(E)= 6
Required Probability = P(E) = n(E)/n(S) = 6/20 = 3/10.