If n is a positive integer and the tens digit of n+8 is 6, w

Could you state exactly which part was not clear?

Anyways, here is another approach : From F.S 2, we know that \(40\leq{n-8}\leq{49} \to 48\leq{n}\leq{57}\)

Also, the question stem states that : \(60\leq{n+8}\leq{69} \to 52\leq{n}\leq{61}\)

The common intersection of both the in-equalities is \(\to 52\leq{n}\leq{57}\). So , yes, "c" as in your example can range only from 2 to 7.

Hope this is clear.

Note: When I say \(52\leq{n}\leq{57}\), it doesn't mean that n is a 2 digit number. It is just a scalable inequality for its last 2 digits.

Similar questions to practice:
if-z-is-a-three-digit-positive-integer-what-is-the-value-of-69999.html
if-the-units-digit-of-the-three-digit-positive-integer-k-is-85443.html
both-a-b-and-c-are-3-digits-integers-where-a-b-c-is-the-126570.html
if-k-is-a-3-digit-positive-number-what-is-the-hundred-digit-104529.html
if-k-is-a-positive-integer-and-the-tens-digit-of-k-5-is-100041.html

Hope it helps.

Ques :- If n is a positive integer and the tens digit of n+8 is 6, what is the tens digit of n?

Tens digit of n+8 is 6, So lets assume that n+8 is 6b, where b is units digit.
We are asked the tens digit of n. that means ques is indirectly asking that if we subtract 8 from 6b (i.e. from n+8) would the tens digit reduce from 6 to 5?

Further examination would tell us that if b takes any value from 0 to 7 tens digit will become 5 (i.e. will change) and it b takes value of 8 or 9 tens digit will remain same because when smaller units digit subtracted from larger units digit, the subtraction would not affect the tens digit.

So the question is basically asking us What is the units digit of n?


(1) The units digit of n is a prime number :- Units digit is one from 1,2,5,7. All these values are below 8. So we can say tens digit of n is 5. Sufficient

(2) The tens digit of n-8 is 4 :- This is tricky. (n+8) and (n-8) have difference of 16. n+8 has tens digit as 6 and n-8 has tens digit as 4. So these numbers must be of the form 6b (59<6b<70) and 4a (39<4a<50)
The only pairs of the numbers that obeys above conditions and differed by 16 are as follows
44 and 60
45 and 61
46 and 62
47 and 63
48 and 64
49 and 65
In all the cases we can see the units digit of all 6b (i.e. of n+8) lies between 0 and 5. That means it is below 8. So tens digit of n is 5. Sufficient

Answer D

@Narenn shouldn't the tens digit be 5? The total sum is 6 ( when its carried over)

Yeah, You were correct. Ten's digit should be 5. There was a typing mistake, which I just corrected.

Thanks

There's no need for all this. From statement 2 we have that n - 8 tens digit 4. We had that n+8 tens digit 6. Therefore, only tens digit that is possible is 5, since both are 16 apart.

Hope this clarifies
Cheers
J

Guys I think the answer should be B. For A what about 28+8 or 38+8 we can never find out what the tens digit is.what do you say?

Let ....yx be he number n where y is ten's digit and x is unit'd digit.

Given information : ten's digit of (n+8) =6
It means two possibilites :-
<1> y=6 ,x=0,1
<2>y=5,x>=2
Now let's go to options:-

<A> Unit digit is a prime number
It means x=2 or 3 or 5 or 7
This imples y=5 Hence ,ten's digit of n can be uniquely determined as 5
<B> ten's digit of (n-8)=4
It means two possibilities :-
<1> y=4,x=8,9
<2> y=5,x=1,2,3,4,5,6,7
This means y=5 as from given information y=5 or 6.

Hence , option D ,Each statement is suffiecient alone

Notice that we are told that the tens digit of n+8 is 6. In your examples, the tens digit of 28+8=36 is 3, not 6 and the tens digit of 38+8=46 is 4, not 6. Also, the first statement says that the units digit of n is a prime number, and 8 (the units digit of 28 and 38), is NOT a prime.

Consider number 1234.567

1 - THOUSANDS
2 - HUNDREDS
3 - TENS
4 - UNITS
. - decimal point
5 - TENTHS
6 - HUNDREDTHS
7 - THOUSANDTHS

Hope this helps.

Just consider the tens and units digit of n let it be a,b

In ab+8 ,the tens digit of n+8 is 6 that means

then 52,53,54,55,56,57,58,59,60,61 can be considered as the values of n since n+8 10"s digit is 6 --->X

as per the option 1

unit digit is prime that rules out few options from statement x

answers left are 52,53,55,57 --(Y)

So 10's digit is common for all that is 5

as per the second option

10's digit of n-8 is four which removes few options from the statement x

n is 52 solves the criterion ---(Z)

Now in both the statements Y and Z tens digit remains same that is 5 hence either of the two is sufficient to answer the Q

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