7 does not occur in 1000. So we have to count the number of times it appears between 1 and 999. Any number between 1 and 999 can be expressed in the form of xyz where 0 < x, y, z < 9.
1. The numbers in which 7 occurs only once. e.g 7, 17, 78, 217, 743 etc
This means that 7 is one of the digits and the remaining two digits will be any of the other 9 digits (i.e 0 to 9 with the exception of 7)
You have 1*9*9 = 81 such numbers. However, 7 could appear as the first or the second or the third digit. Therefore, there will be 3*81 = 243 numbers (1-digit, 2-digits and 3- digits) in which 7 will appear only once.
In each of these numbers, 7 is written once. Therefore, 243 times.
2. The numbers in which 7 will appear twice. e.g 772 or 377 or 747 or 77
In these numbers, one of the digits is not 7 and it can be any of the 9 digits ( 0 to 9 with the exception of 7).
There will be 9 such numbers. However, this digit which is not 7 can appear in the first or second or the third place. So there are 3 * 9 = 27 such numbers.
In each of these 27 numbers, the digit 7 is written twice. Therefore, 7 is written 54 times.
3. The number in which 7 appears thrice - 777 - 1 number. 7 is written thrice in it.
Therefore, the total number of times the digit 7 is written between 1 and 999 is
243 + 54 + 3 = 300
U ? A = {a, b, c, d, e, f} ? {a, b, c} = {a, b, c, d, e, f} = U
(U ? A)? = ?.
Sunil's income is less than Mukul's = [(10 / (100 + 10)] × 100 %
= ( 1000 / 110 )%
= 9.09 %
So, 39 is wrong.
So, 33 is wrong. It must be 35
∴ x = 7.
So, 40 is wrong.
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