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18 Jul 2017, 04:32
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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 \(4.7\).
(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 \(4.7\).
18 Jul 2017, 04:32
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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
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
General Discussion
03 Aug 2017, 19:07
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Hi Bunuel,
Please can you tell me if my below approach is correct.
15−9.x5 -->
15.00
- 9.x5
______
5.y5
where y is (9-x).
Given-When 15−9.x515−9.x5 is rounded to the nearest tenth, the result is 5.4-->5.x5=5.35
Thanks,
CP
Please can you tell me if my below approach is correct.
15−9.x5 -->
15.00
- 9.x5
______
5.y5
where y is (9-x).
Given-When 15−9.x515−9.x5 is rounded to the nearest tenth, the result is 5.4-->5.x5=5.35
Thanks,
CP
