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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