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Official Solution:


If \(x\) is the tenths digit in the decimal \(9.x5\), what is the value of \(x\)?

(1) When \(15 - 9.x5\) is rounded to the nearest tenth, the result is \(5.4\).

This implies that \(15 - 9.x5\) must be between \(5.35\) (inclusive) and \(5.45\) (not inclusive). Any number from this range when rounded to the nearest tenth will be \(5.4\). So, we can write the following inequality:

\(5.35 \leq (15 - 9.x5) < 5.45\);

Subtract 15 from all parts: \(-9.65 \leq -9.x5 < -9.55\);

Multiply by -1 and flip the signs: \(9.65 \geq 9.x5 > 9.55\), which is the same as \(9.55 < 9.x5 \leq 9.65\). From this we can deduce that \(x\) can be only \(6\). Sufficient.

(2) When \(9.x5 – 5\) is rounded to the nearest tenth, the result is \(4.7\).

This implies that \(9.x5 – 5\) must be between \(4.65\) (inclusive) and \(4.75\) (not inclusive). Any number from this range when rounded to the nearest tenth will be \(4.7\). So, we can write the following inequality:

\(4.65 \leq (9.x5 – 5) < 4.75\);

Add 5 to all parts: \(9.65 \leq 9.x5 < 9.75\). From this we can deduce that \(x\) can be only \(6\). Sufficient.


Answer: D
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number = 9.x5

from [1]:

9.x5 => 15 - 5.35 => 9.65 => x = 6
9.x5 => 15 - 5.35 => 9.65 => x = 6
9.x5 => 15 - 5.36 => 9.64 => x = 6
9.x5 => 15 - 5.37 => 9.63 => x = 6
9.x5 => 15 - 5.38 => 9.62 => x = 6
9.x5 => 15 - 5.39 => 9.61 => x = 6
9.x5 => 15 - 5.40 => 9.60 => x = 6
9.x5 => 15 - 5.41 => 9.59 => x = 5
9.x5 => 15 - 5.42 => 9.58 => x = 5
9.x5 => 15 - 5.43 => 9.57 => x = 5
9.x5 => 15 - 5.44 => 9.56 => x = 5

so x can be 6 or 5 : not sufficient

from [2]

9.x5 => 5 + 4.65 => 9.65 => x = 6
9.x5 => 5 + 4.66 => 9.66 => x = 6
9.x5 => 5 + 4.67 => 9.67 => x = 6
9.x5 => 5 + 4.68 => 9.68 => x = 6
9.x5 => 5 + 4.69 => 9.69 => x = 6
9.x5 => 5 + 4.70 => 9.70 => x = 7
9.x5 => 5 + 4.71 => 9.71 => x = 7
9.x5 => 5 + 4.72 => 9.72 => x = 7
9.x5 => 5 + 4.73 => 9.73 => x = 7
9.x5 => 5 + 4.74 => 9.74 => x = 7

so x can be 6 or 7 : not sufficient

from [1] & [2] x = 6 ==> C
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Leo8
number = 9.x5

from [1]:

9.x5 => 15 - 5.35 => 9.65 => x = 6
9.x5 => 15 - 5.35 => 9.65 => x = 6
9.x5 => 15 - 5.36 => 9.64 => x = 6
9.x5 => 15 - 5.37 => 9.63 => x = 6
9.x5 => 15 - 5.38 => 9.62 => x = 6
9.x5 => 15 - 5.39 => 9.61 => x = 6
9.x5 => 15 - 5.40 => 9.60 => x = 6
9.x5 => 15 - 5.41 => 9.59 => x = 5
9.x5 => 15 - 5.42 => 9.58 => x = 5
9.x5 => 15 - 5.43 => 9.57 => x = 5
9.x5 => 15 - 5.44 => 9.56 => x = 5

so x can be 6 or 5 : not sufficient

from [2]

9.x5 => 5 + 4.65 => 9.65 => x = 6
9.x5 => 5 + 4.66 => 9.66 => x = 6
9.x5 => 5 + 4.67 => 9.67 => x = 6
9.x5 => 5 + 4.68 => 9.68 => x = 6
9.x5 => 5 + 4.69 => 9.69 => x = 6
9.x5 => 5 + 4.70 => 9.70 => x = 7
9.x5 => 5 + 4.71 => 9.71 => x = 7
9.x5 => 5 + 4.72 => 9.72 => x = 7
9.x5 => 5 + 4.73 => 9.73 => x = 7
9.x5 => 5 + 4.74 => 9.74 => x = 7

so x can be 6 or 7 : not sufficient

from [1] & [2] x = 6 ==> C

for statement 2, in your solution the format should be 9.x5 which only occurs in the first line where you are adding 4.65, so x can only be 6, the case is similar for statement 1 as well.
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Answer is D
here how i solved it from 1 we have When 15−9.x5 is rounded to the nearest tenth, the result is 5.4
15-5.4 will give us the number 9.6=9.x5 so we have x=1 hence sufficient .
From 2 When 9.x5–5 is rounded to the nearest tenth, the result is 4.7 we have 9.x5-5 will equal 4.x5 that is equal to 4.7 hence x=2
sufficient.
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DIII
Leo8
number = 9.x5

from [1]:

9.x5 => 15 - 5.35 => 9.65 => x = 6
9.x5 => 15 - 5.35 => 9.65 => x = 6
9.x5 => 15 - 5.36 => 9.64 => x = 6
9.x5 => 15 - 5.37 => 9.63 => x = 6
9.x5 => 15 - 5.38 => 9.62 => x = 6
9.x5 => 15 - 5.39 => 9.61 => x = 6
9.x5 => 15 - 5.40 => 9.60 => x = 6
9.x5 => 15 - 5.41 => 9.59 => x = 5
9.x5 => 15 - 5.42 => 9.58 => x = 5
9.x5 => 15 - 5.43 => 9.57 => x = 5
9.x5 => 15 - 5.44 => 9.56 => x = 5

so x can be 6 or 5 : not sufficient

from [2]

9.x5 => 5 + 4.65 => 9.65 => x = 6
9.x5 => 5 + 4.66 => 9.66 => x = 6
9.x5 => 5 + 4.67 => 9.67 => x = 6
9.x5 => 5 + 4.68 => 9.68 => x = 6
9.x5 => 5 + 4.69 => 9.69 => x = 6
9.x5 => 5 + 4.70 => 9.70 => x = 7
9.x5 => 5 + 4.71 => 9.71 => x = 7
9.x5 => 5 + 4.72 => 9.72 => x = 7
9.x5 => 5 + 4.73 => 9.73 => x = 7
9.x5 => 5 + 4.74 => 9.74 => x = 7

so x can be 6 or 7 : not sufficient

from [1] & [2] x = 6 ==> C

for statement 2, in your solution the format should be 9.x5 which only occurs in the first line where you are adding 4.65, so x can only be 6, the case is similar for statement 1 as well.


Yeah both statement 1 & 2 end up with the value of x to be 6.
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arvind910619
Answer is D
here how i solved it from 1 we have When 15−9.x5 is rounded to the nearest tenth, the result is 5.4
15-5.4 will give us the number 9.6=9.x5 so we have x=1 hence sufficient .
From 2 When 9.x5–5 is rounded to the nearest tenth, the result is 4.7 we have 9.x5-5 will equal 4.x5 that is equal to 4.7 hence x=2
sufficient.

hi arvind910619

In GMAT DS both statement result in same answer (if they are sufficient). so value of "x" has to be same.

for statement 1 you have directly subtracted 5.4 from 15 whereas 5.4 is the rounded figure. Original number could be 5.41, 5.42, 5.43, 5.44, 5.35, 5.36, 5.37....and so on
Hence for your approach you need to subtract exact value from 15.
Per your solution x=1 which means our original number is 9.15. But 15-9.15 = 5.85 and if you round this number to the nearest tenth digit it will be 5.9 and not 5.4
similarly for statement 2.
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DIII
Leo8
number = 9.x5

from [1]:

9.x5 => 15 - 5.35 => 9.65 => x = 6
9.x5 => 15 - 5.35 => 9.65 => x = 6
9.x5 => 15 - 5.36 => 9.64 => x = 6
9.x5 => 15 - 5.37 => 9.63 => x = 6
9.x5 => 15 - 5.38 => 9.62 => x = 6
9.x5 => 15 - 5.39 => 9.61 => x = 6
9.x5 => 15 - 5.40 => 9.60 => x = 6
9.x5 => 15 - 5.41 => 9.59 => x = 5
9.x5 => 15 - 5.42 => 9.58 => x = 5
9.x5 => 15 - 5.43 => 9.57 => x = 5
9.x5 => 15 - 5.44 => 9.56 => x = 5

so x can be 6 or 5 : not sufficient

from [2]

9.x5 => 5 + 4.65 => 9.65 => x = 6
9.x5 => 5 + 4.66 => 9.66 => x = 6
9.x5 => 5 + 4.67 => 9.67 => x = 6
9.x5 => 5 + 4.68 => 9.68 => x = 6
9.x5 => 5 + 4.69 => 9.69 => x = 6
9.x5 => 5 + 4.70 => 9.70 => x = 7
9.x5 => 5 + 4.71 => 9.71 => x = 7
9.x5 => 5 + 4.72 => 9.72 => x = 7
9.x5 => 5 + 4.73 => 9.73 => x = 7
9.x5 => 5 + 4.74 => 9.74 => x = 7

so x can be 6 or 7 : not sufficient

from [1] & [2] x = 6 ==> C

for statement 2, in your solution the format should be 9.x5 which only occurs in the first line where you are adding 4.65, so x can only be 6, the case is similar for statement 1 as well.

Thanks DIII.. I totally overlooked it
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[quote="Bunuel"]If x is the tenths digit in the decimal \(9.x5\), what is the value of x?

(1) When \(15 - 9.x5\) is rounded to the nearest tenth, the result is 5.4.
(2) When \(9.x5 – 5\) is rounded to the nearest tenth, the result is 4.7.


Both statements tell us about subtracting the given number from an integer and rounding it to the nearest 10th.
Now, subtracting 9.x5 from any integer will always result in the format of Y.Z5. If the Round off resulted in 5.4, most definitely the result of subtraction was 5.35 (bcoz 5.25 round off will be 5.3, 5.45 will become 5.5 and so on). Now that we know 5.35 was the result, we can calculate value of 'x' by simply subtracting 5.35 from 15 (Therefore SUFFICIENT)

On the same lines the result of statement 2 would have been 4.65. And value of 'x' can be determined by adding 5 to 4.65 (Therefore SUFFICIENT)

Ans: D
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