(A) In this case first integer is 7 and last integer is 994 and as per Arithmetic Progression we can write 7+(n-1)*7=994 which gives n=142 which is the answer.
(B) Ingers which are divisible by 7 i.e these 142 nos inger are to be sorted out which are not divisible by 11. we can do that by taking all multiples of LCM (7,11) i.e 77.
77 *1=77,77*2=154,77*3=231, 77*4=308, 77*5=385, 77*6=462,77*7=539,77*8=616,77*9=693,77*10=710,77*11=847,77*12=924
these 12 integers will excluded from 142 nos integers i.e 143-12=131 nos integers.
(C) integers divisible both by 7 and 11 are the 12 nos as in (B) which are the multiples of LCM (7,11)
(D)Integers divisible by either 7 or 11.
Already we have got ingers divisible by 7 are 142 nos.
similarly we have to find out nos of integers divisible by 11 in which case first integer is 11 and last integer is 990. By A.P we can say
11+(n-1)*11=990 which gives n=90
so nos of integers divisible either by 7 or 11 is 142 +90=232
(E) Divisible exactly one of 7 and 11..
we know nos of ingers divisible by 7 is 142 nos
integers divisible by 11 is 90 nos
integers divisible both by 7 ans 11 is 12 nos,. hence nos of integers divisible one of 7 and 11 will 142+90-12=232-12=220 nos
(f) divisible neither by 7 nor 11
total nos of integers is 999
nos of integers divisible by 7 or 11 is 232.
so nos of integers divisible neither by 7 nor 11 is 999-232=767
(g) for distinct digit for two digit integers, total nos will be 10*9 =90 and for three digit integers it will be 10*9*8=720 and the total nos of integers will be =90+720=810 nos
(h) for distinct digits and even nos...in this case it will be 10*9