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When the positive number a is rounded to the nearest tenth

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When the positive number a is rounded to the nearest tenth  [#permalink]

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New post 19 Sep 2012, 06:11
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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.

Any suggestions?

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Re: When the positive number a is rounded to the nearest tenth  [#permalink]

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New post 19 Sep 2012, 10:18
5
4
frankiegar wrote:
doraemonbeo wrote:
frankiegar wrote:

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. :lol: 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 :shock:


Rounding rules

Rounding is simplifying a number to a certain place value. To round the decimal drop the extra decimal places, and if the first dropped digit is 5 or greater, ROUND UP the last digit that you keep. If the first dropped digit is 4 or smaller, ROUND DOWN (keep the same) the last digit that you keep.

Example:
5.3485 rounded to the nearest tenth = 5.3, since the dropped 4 is less than 5.
5.3485 rounded to the nearest hundredth = 5.35, since the dropped 8 is greater than 5.
5.3485 rounded to the nearest thousandth = 5.349, since the dropped 5 is equal to 5.

For more check Number Theory chapter of Math Book: math-number-theory-88376.html

Hope it helps.
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Re: When the positive number a is rounded to the nearest tenth  [#permalink]

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New post 19 Sep 2012, 07:25
5
Assume a = x.yz, b = x.t
(1) When a is rounded to the nearest integer, the result is less than a => y is equal to or less than 4. INSUFFICIENT.
(2) When b is rounded to the nearest integer, the result is greater than b. => t must be at least 5. We cannot find y.

(1) & (2): y must be 4 so that t will be at least 5 when a is rounded to the nearest tenth.
=> The answer is C.
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Re: When the positive number a is rounded to the nearest tenth  [#permalink]

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New post 19 Sep 2012, 07:39
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doraemonbeo wrote:
Assume a = x.yz, b = x.t
(1) When a is rounded to the nearest integer, the result is less than a => y is equal to or less than 4. INSUFFICIENT.
(2) When b is rounded to the nearest integer, the result is greater than b. => t must be at least 5. We cannot find y.

(1) & (2): y must be 4 so that t will be at least 5 when a is rounded to the nearest tenth.
=> The answer is C.


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?
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Re: When the positive number a is rounded to the nearest tenth  [#permalink]

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New post 19 Sep 2012, 09:02
1
frankiegar wrote:
doraemonbeo wrote:
Assume a = x.yz, b = x.t
(1) When a is rounded to the nearest integer, the result is less than a => y is equal to or less than 4. INSUFFICIENT.
(2) When b is rounded to the nearest integer, the result is greater than b. => t must be at least 5. We cannot find y.

(1) & (2): y must be 4 so that t will be at least 5 when a is rounded to the nearest tenth.
=> The answer is C.


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.
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Re: When the positive number a is rounded to the nearest tenth  [#permalink]

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New post 19 Sep 2012, 10:11
doraemonbeo wrote:
frankiegar wrote:
doraemonbeo wrote:
Assume a = x.yz, b = x.t
(1) When a is rounded to the nearest integer, the result is less than a => y is equal to or less than 4. INSUFFICIENT.
(2) When b is rounded to the nearest integer, the result is greater than b. => t must be at least 5. We cannot find y.

(1) & (2): y must be 4 so that t will be at least 5 when a is rounded to the nearest tenth.
=> The answer is C.


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. :lol: 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 :shock:
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Re: When the positive number a is rounded to the nearest tenth  [#permalink]

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New post 19 Sep 2012, 10:22
Thanks bunuel.

Best,

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Re: When the positive number a is rounded to the nearest tenth  [#permalink]

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New post 19 Sep 2012, 22:32
frankiegar wrote:
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.

Any suggestions?

Best,


Let a = n.xy
Let b = n.r (r could be =x or =x+1)

What is x?

(1) a rounded becomes n. This means x is less than 1,2,3,4.
(2) b rounded becomes n+1. this means r is equal or greater than 5. Still haven't resolved whether r is equal to x or an increment of x.

Together,
(1) x is 1,2,3, or 4
(2) r is equal or greater than 5. But r is either an increment or equal x. It's obvious that r is not equal to its largest possible value 4. Then, r is an increment from 4. So x is 4.
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When the positive number a is rounded to the nearest tenth  [#permalink]

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New post 17 Oct 2012, 11:59
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
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New post 17 Oct 2012, 14:59
Jp27 wrote:
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.
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Re: When the positive number a is rounded to the nearest tenth  [#permalink]

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New post 18 Oct 2012, 04:19
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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.
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Re: When the positive number a is rounded to the nearest tenth  [#permalink]

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New post 18 Oct 2012, 10:37
Sorry, not getting this. if A=X.YZ, 'Y' is the unit place and 'Z' is the tenth place right?
if so, why not 'A' be say 5.61, then 'B' will be 5.6 and obviously 5.6<5.61
then 'C' will be 6 and obviously 6>5.6. So it is possible for the tenth place 'Z' from above to be anything from 0 through 4. Right? So the answer is 'E' right?
By looking at many of you confirming the answer to be 'C' I'm probably wrong. Please explain.
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New post 18 Oct 2012, 10:48
1022lapog wrote:
Sorry, not getting this. if A=X.YZ, 'Y' is the unit place and 'Z' is the tenth place right?
if so, why not 'A' be say 5.61, then 'B' will be 5.6 and obviously 5.6<5.61
then 'C' will be 6 and obviously 6>5.6. So it is possible for the tenth place 'Z' from above to be anything from 0 through 4. Right? So the answer is 'E' right?
By looking at many of you confirming the answer to be 'C' I'm probably wrong. Please explain.



Sorry, not getting this. if A=X.YZ, 'Y' is the unit place and 'Z' is the tenth place right? - NO

X - units
Y - tenths
Z - hundreths
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New post 25 Feb 2013, 04:19
i have understand explanation of statement 1 and statement 2 but not understanding how c is answer.can u explain me in details plz
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New post 25 Feb 2013, 13:28
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When the positive number a is rounded to the nearest tenth, the result is the number b. What is the tenths digit of a?

Let a=x.yz
b=x.n

Now, when a is rounded to the nearest tenth, the result is b. So, be must have only the tenth place in the decimal form and no hundreds place.

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

Now, if rounding a to the nearest integer should give a value less than a, then y<=4. Consider a=5.41. Rounding a to the nearest tenth would give, a=5. However this is INSUFFICIENT to find the value of the tenth digit of a.

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

Now, when b is rounded to the nearest integer, the result >b. This is possible only if n>=5. Try a similar example as stated above

Together,

We need to find a with y<=4 and b with n>=5 so that when we round a to the nearest tenth, the answer is b.

The tenth value of b >=5. Now what possible value could a's tenth place have so that when it is rounded off, the value must be >=5. We know that the maximum value y could have is 4 and if z>5, then y could be rounded off to 5 right?

Hence, c is the solution.

Hope this helps! Let me know if I could help any further.

mun23 wrote:
i have understand explanation of statement 1 and statement 2 but not understanding how c is answer.can u explain me in details plz
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New post 26 May 2014, 12:33
MacFauz wrote:
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.


Hi MacFauz,

In statement 2, how do you get a to be 4,5,6,7,8,9 from knowing B? Wouldn't the tenths digit of a just be 4,5,6,7, and 8. It can't have a 9 in there or am I missing something?
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New post 28 Aug 2016, 18:44
2
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
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When the positive number a is rounded to the nearest tenth  [#permalink]

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New post 10 Sep 2018, 16:08
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frankiegar wrote:
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.


? : 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).

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When the positive number a is rounded to the nearest tenth   [#permalink] 10 Sep 2018, 16:08
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