Which two statements at Line n1 independently enable you to print 1250?

Question

Options

• AInteger res = 250 + integer.parseInt(s)
• BInteger res = 250 + s;
• CInteger res = 250 + integer (s2);
• DInteger res = 250 + s2;
• EInteger res = 250 + integer .valueOf(s);
• FRes = s2;

• A
  72% (26)
• B
  3% (1)
• C
  3% (1)
• D
  17% (6)
• F
  6% (2)

Explanation

Options A and E both correctly convert the String variable s (which holds "1000") into a numeric type before performing addition. Integer.parseInt(s) returns an int primitive, and Integer.valueOf(s) returns an Integer object that is auto-unboxed - both yield 250 + 1000 = 1250.

Why the distractors fail:

• B (250 + s) - you cannot add an int directly to a String; Java would either concatenate ("2501000") or throw a compile error depending on context.
• C (integer(s2)) - lowercase integer is not a valid Java type or method; Java is case-sensitive and the correct wrapper class is Integer with a capital I.
• D (250 + s2) - s2 is a String, so direct arithmetic addition fails for the same reason as B.
• F (Res = s2) - Res with a capital R is an undeclared variable (Java is case-sensitive), and s2 is a String not assignable to Integer.

Memory tip: When you see a String needing arithmetic in Java, remember the two Integer doorways: parseInt → gives you a primitive int, valueOf → gives you a boxed Integer. Both sit on the Integer class with a capital I - spot any lowercase integer and you've found your trap answer.

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