M32-15

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