When the positive number a is rounded to the nearest tenth

Responding to a pm:

Here is a post on the rounding rules:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2014/0 ... or-a-five/

Say a = n.xyz
Then b = n.x or n.(x+1) depending on what y is.
(Say 1.342 will round to 1.3 (x stays the same) but 1.382 will round to 1.4 (x becomes x+1))

We need to know the value of x.

(1) When a is rounded to the nearest integer, the result is less than a.

When rounded, n.xyz gives n.
Since it is rounded down, it means x is one of 0/1/2/3/4.

(2) When b is rounded to the nearest integer, the result is greater than b.

b = n.x or n.(x+1)
For b to be rounded up, the tenths digit (x or x+1) is one of 5/6/7/8/9.

Using both statements, there is only one value of x that works.
x must be 4 which means x+1 must be 5 and that is how the tenths digit of b will lead to rounding up.

Hence x = 4

Answer (C)

Check this post for some more such questions: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2014/0 ... questions/

Thanks.

But when t is 5, if we round it to the nearest integer result will be higher than a.

Think about x.45

Round it to nearest tenth===> x.5
Round it to the nearest integer===>x+1

in that case result is higher than a. Am I missing some point?

t is the tenth digit of b, not a Moreover, you mustn't round it twice.
When x.45 is rounded to the nearest tenth, the result is x.5, but when it is rounded to the nearest integer, it is x.

Wow thanks my friend. Is it a rule in GMAT, the one about rounding twice?

Bc in statistics I always regard 0.45 as 1 if I round it

When the positive number a is rounded to the nearest tenth, the result is the number b. What is the tenths digit of a?

(1) When a is rounded to the nearest integer, the result is less than a.

(2) When b is rounded to the nearest integer, the result is greater than b.


OA

Let a =x.yz what is y?

1/ When a is rounded to the nearest integer, the result is less than a. ->

so a can be x.01 to x.49 so tenths digit of a (i.e y) can be anywer between 0 to 4 NS

2/ When b is rounded to the nearest integer, the result is greater than b.

b=x.y so y can be anywer between 5 to 9.

What's wrong with my interpretation of statement 2?
Stat1 gives 0<= y < 4 and stat 2 gives 5 <= y < 9

so a can be x.01 to x.49 - 0<= y <= 4 - y can also be 4

(2) b=x.y - b was obtained by rounding a to the nearest tenth. So, the tenths digit of b isn't necessarily the same as the tenths digit of a.
We can just say that the tenths digit of b is at least 5.

1) the tenths digit of a could be 0,1,2,3,4. Insufficient
2) the tenths digit of b could be 5,6,7,8,9 and hence the tenths digit of a could be 4,5,6,7,8,9. Insufficient

1 & 2 together. Only one number overlaps both sets. So answer is 4 and hence C.

Very intersting Question
Let a be ABC.PQR
we need the tens digit of b which is itself A rounded to tenths
Statement 1 => it basically is mentioning that P must be 0,1,2,4,5
But we dont know what is Q
e.g => a=123.299
b=123.300
and if a=123.488
b=123.500=> insuff
Statement 2 =>
the tenths digit of B must be 5,6,7,8
but we dont know the exact value
hence insuff


Combining them it becomes very very interesting
a must have a tenths digit of 0,1,2,3,4
and b must have a tenths digit of 5,6,7,8,9
but b is nothing but a rounded to tenths
so if a has 3 as its tenths then b would have 4=> not suff
and if a has 2,1,0 => not suff
but if a has 4 as its unit digit then => b would have 5 as its unit digit
hence the only case is => a has 4 as its unit digit and b has 5
suff

SMASH that C

? : tenth´s digit of A

Convention: we assume that the number 0.5 must be rounded to 1, when we are rounding numbers to the nearest integer.
(It´s really just a matter of convention. We explicit our convention so that we can deal with this problem without ambiguity.)

(1) (A rounded to nearest int) < A , i.e. , the tenth´s digit of A (our FOCUS) is less than 5 ...

:: Take A = 0.49 (rounded to the nearest integer is 0, that is less than A) , to answer 4
:: Take A = 0.39 (rounded to the nearest integer is 0, that is less than A), to answer 3

(2) (B rounded to nearest int) > B i.e., the tenth´s digit of B is no less than 5 ...

:: Take A = 0.56 (then B = 0.6 and B rounded to nearest integer is 1, that is greater than B) , to answer 5
:: Take A = 0.66 (then B = 0.7 and B rounded to nearest integer is 1, that is greater than B) , to answer 6

(1+2) We know that the tenth´s digit of A (our FOCUS) is less than 5 AND
when we round A to the nearest tenth´s digit (=B) , the tenth´s digit of this number (B) is no less than 5...

This is enough to guarantee that the tenth´s digit of A is 4 (our FOCUS)... SUFFICIENT!

Without loss of generality, we will still consider 0 < A < 1 only.
The general case (in which A>0 has any other integer part) is dealt in EXACTLY the same way.

In fact:
If 0.5 <= A < 1 , A does not satisfy statement (1) !!
If 0 < A < 0.4 , then B is 0.4 or less, therefore B does not satisfy statement (2) !!

The correct answer is therefore (C).

Regards,
fskilnik.

(1) INSUFFICIENT: According to this statement, a must be rounded down to the nearest integer. The tenths digit of a could thus be any of 0, 1, 2, 3, or 4. Since there are five possibilities for the tenths digit, this statement is insufficient.

Test numbers if you're not sure: if a is 14.0, it will round down to 14. If a is 14.4, it will also round down to 14. If a is 14.5, it will round up to 15, so 5 cannot be the tenths digit of a, nor can digits larger than 5.

(2) INSUFFICIENT: According to this statement, b, which is the result of rounding a to the nearest tenth, must be rounded up to the nearest integer. Therefore, the tenths digit of b could be any of 5, 6, 7, 8, or 9. Test numbers if you're not sure: if b is 11.9, it will round up to 12. If b is 11.5, it will round up to 12. If b is 11.4, however, it will round down to 11, so 4 cannot be the tenths digit of b, nor can anything smaller than 4.

Recall that the problem asks about a, not b; therefore, we need to figure out which possible values for a will then round to one of the tenths digits 5, 6, 7, 8, or 9. The lowest such value of a is xxx.45 (where xxx could be any value). This number will round to xxx.5, so a tenths digit of 4 for the number a can produce a tenths digit of 5 for the number b. At the other end, the highest such value is xxx.949999, which will round to xxx.9. Therefore, according to this statement, the tenths digit of a could be 4, 5, 6, 7, 8, or 9. There are six possibilities, so the statement is insufficient.

Test numbers if you're not sure or if the above method is too abstract. First, the question stem tells us that when a is rounded to the nearest tenth, the result is b. If a is 14.45, that will round to 14.5 (remember, a is allowed to round up here because we're ignoring statement 1). As a result, b = 14.5. If b is then rounded, it will round up, which is what statement 2 tells us, so the smallest possible value for the tenths digit of a is 4. At the opposite end, if a is 14.94, it will round to 14.9, which represents b. If b is rounded, it will also round up, so the largest possible value for the tenths digit of a is 9.

(1) AND (2) SUFFICIENT: According to statement (1) the tenths digit of a could be 0, 1, 2, 3, or 4. According to statement (2), the tenths digit of a could be 4, 5, 6, 7, 8, or 9. The only value common to the two statements is 4, so, taking both statements together guarantees that the tenths digit of a is 4.

The correct answer is C.

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