public class Tester { public static void main(String[] args) { byte x = 7, y = 6; // Line 1 System.out.println(z); } } Which expression when added at Line 1 will produce the output of 1.17?

Question

Options

• Afloat z = (float)(Math.round((int)y*100)/100);
• Bfloat z = Math.round((int)x/y)/2;
• Cfloat z = Math.round((float)x/y);
• Dfloat z = (float)Math.round((float)x/y*100)/100;

• A
  5% (2)
• B
  11% (4)
• C
  3% (1)
• D
  82% (31)

Explanation

Option D correctly produces 1.17 by following the right order of operations: (float)x/y yields 1.1666..., multiplying by 100 gives 116.666..., Math.round() produces 117 (a long), and finally casting to float before dividing by 100 gives 1.17. The key is that the cast to float happens before the final division, preventing integer division from truncating the result.

Why the distractors fail:

• A never divides x by y - it only uses y*100, so the result is 6.0, completely unrelated to the x/y ratio.
• B uses (int)x/y, which performs integer division (7/6 = 1), then 1/2 is also integer division yielding 0, so z = 0.0.
• C rounds 1.1666... to the nearest integer (1) before any scaling, so it loses all decimal precision and gives 1.0.

Memory tip: Think "Scale → Round → Unscale." To round to 2 decimal places, multiply by 100 first, round to a whole number, then cast to float and divide by 100. If you divide first without the float cast, Java's integer division silently discards everything after the decimal point.

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