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													Ans. (d) double –5.0
Explanation:
The execution goes on like this:
– 4 + 1/2 + 2*-3 + 5.0;
– 4 + 0 + -6 + 5.0; // integer division: 1/2 truncates .5
– 10 + 5.0; // higher type is double 5.0, so -10 is casted to double
– 5.0; // finally, double -5.0.
Ans. (c) 1
Explanation:
It is printed one time only because break; will terminate the current loop.
Explanation:
z = ++x * (y– –)– y
 = 21 * (10) – 9
  = 210 – 9 = 201
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Download Now| Chapter No. | Chapter Name | 
|---|---|
| Chapter 1 | Revision of Class IX Syllabus | 
| Chapter 2 | Class as a Basis of all Computation | 
| Chapter 3 | User - defined Methods | 
| Chapter 4 | Constructors | 
| Chapter 5 | Library classes | 
| Chapter 6 | Encapsulation | 
| Chapter 7 | Arrays | 
| Chapter 8 | String handling | 
| Chapter Wise Important Questions for ICSE Board Class 10 Computer Applications | 
|---|
| Revision of Class IX Syllabus | 
| Class as a Basis of all Computation | 
| User - defined Methods | 
| Constructors | 
| Library classes | 
| Encapsulation | 
| Arrays | 
| String handling | 
CBSE Important Questions Class 10
ICSE Important Questions Class 10
CBSE Important Questions Class 10
ICSE Important Questions Class 10