Back solving is a good technique but I like to keep it as the last option. Reasons for that:
1. It involves a lot of calculations so chances of error are relatively higher.
2. Sometimes, it is not possible to take an educated guess even after trying a couple of values so it could certainly be time consuming if the first numbers you try do not fall in place.

So, as an alternative approach, let me give you an equation that you can use.

Let the number be N.

"A number when divided by 5"
That is N/5

" gives a number which is 8 more than"
N/5 = 8 +

"the remainder obtained on dividing the same number by 34."
N = 34Q + R
So remainder obtained when N is divided by 34 is R which is (N - 34Q)
N/5 = 8 + (N - 34Q)
Isolate N from this equation to get:
N = 85Q/2 - 10

N will take the smallest value when Q = 2 (so that we get an integer value)
N = 85 - 10 = 75
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75/34 = rem 7 and 75/5 = 15, so Answer is B.
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I have one approach that is back solving.
see you have find out a number which divided by 5 but not by 34. and x/5=n will be 8 more than the remainder left after dividing an option by 34.
so 74 and is out because is not divided by 5.
Let 75. 75/5=15, and 75/34=2+remainder 7, which is 8 less than 15.
so ans is B.
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Excellent karishma
loved your approach its always simple and self explanatory.........
kudos to you.........
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do not hesitate me giving kudos if you like my post.

i followed exactly the same procedure. but initially lil bit nervous and confused in the wording of this question.

Nice one... Here we are told that a number divided by 5 gives a quotient with is 8 more than the remainder when number is divided by 34
hence let quotient be p
hence the remainder will be p+8
checking by plugging in the values we get that B is sufficient.
hence B
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let q=quotient for n/34
n/5-8=n-34q→
34q=4n/5+8
test values of 0 and 1 for q
lowest possible value for q=2
lowest possible value for n=75
B

let r=n/34 remainder
q=n/5
q=r+8
n/5=r+8
n=5r+40
n must be a multiple of 5
looking at 75, the least multiple of 5,
r=7; q=7+8=15
15*5=75=n
B