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ICSE Class 12 Computer Science Question 1 of 3

Solved 2017 Practical Paper ISC Computer Science — Question 1

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Question

Question 1

A company manufactures packing cartons in four sizes, i.e. cartons to accommodate 6 boxes, 12 boxes, 24 boxes and 48 boxes. Design a program to accept the number of boxes to be packed (N) by the user (maximum up to 1000 boxes) and display the break-up of the cartons used in descending order of capacity (i.e. preference should be given to the highest capacity available, and if boxes left are less than 6, an extra carton of capacity 6 should be used.)

Test your program with the following data and some random data:

Example 1

INPUT:
N = 726

OUTPUT:
48 * 15 = 720
6 * 1 = 6
Remaining boxes = 0
Total number of boxes = 726
Total number of cartons = 16

Example 2

INPUT:
N = 140

OUTPUT:
48 * 2 = 96
24 * 1 = 24
12 * 1 = 12
6 * 1 = 6
Remaining boxes = 2 * 1 = 2
Total number of boxes = 140
Total number of cartons = 6

Example 3

INPUT:
N = 4296

OUTPUT:
INVALID INPUT

Solution
import java.util.Scanner;

public class CartonBoxes
{
    public static void main(String args[]) {
        Scanner in = new Scanner(System.in);

        System.out.print("Enter number of boxes (N): ");
        int n = in.nextInt();
        
        if (n < 1 || n > 1000) {
            System.out.println("INVALID INPUT");
            return;
        }

        int cartonSizes[] = {48, 24, 12, 6};

        int total = 0;
        int t = n;
        for (int i = 0; i < cartonSizes.length; i++) {
            int cartonCount = t / cartonSizes[i];
            t = t % cartonSizes[i];
            total += cartonCount;
            if (cartonCount != 0) {
                System.out.println(cartonSizes[i] + " * " + cartonCount + 
                    " = " + (cartonSizes[i] * cartonCount));
            }
        }

        /*
         * This if check is for the case when
         * boxes left are less than 6. We need
         * one more carton of capacity 6 in this
         * case so total is incremented by 1.
         */
        if (t != 0) {
            System.out.println("Remaining boxes = " + t 
                + " * 1 = " + t);
            total++;
        }
        else {
            System.out.println("Remaining boxes = 0");
        }

        System.out.println("Total number of boxes = " + n);
        System.out.println("Total number of cartons = " + total);
    }
}
Output
BlueJ output of CartonBoxes.java
BlueJ output of CartonBoxes.java
BlueJ output of CartonBoxes.java
Answer

import
java.util.Scanner
;
public
class
CartonBoxes
{
public
static
void
main
(
String
args
[]) {
Scanner
in
=
new
Scanner
(
System
.
in);
System
.
out
.
print(
"
Enter number of boxes (N):
"
);
int
n
=
in
.
nextInt();
if
(n
<
1
||
n
>
1000
) {
System
.
out
.
println(
"
INVALID INPUT
"
);
return
;
}
int
cartonSizes[]
=
{
48
,
24
,
12
,
6
};
int
total
=
0
;
int
t
=
n;
for
(
int
i
=
0
; i
<
cartonSizes
.
length; i
++
) {
int
cartonCount
=
t
/
cartonSizes[i];
t
=
t
%
cartonSizes[i];
total
+=
cartonCount;
if
(cartonCount
!=
0
) {
System
.
out
.
println(cartonSizes[i]
+
"
*
"
+
cartonCount
+
"
=
"
+
(cartonSizes[i]
*
cartonCount));
}
}
/*
* This if check is for the case when
* boxes left are less than 6. We need
* one more carton of capacity 6 in this
* case so total is incremented by 1.
*/
if
(t
!=
0
) {
System
.
out
.
println(
"
Remaining boxes =
"
+
t
+
"
* 1 =
"
+
t);
total
++
;
}
else
{
System
.
out
.
println(
"
Remaining boxes = 0
"
);
}
System
.
out
.
println(
"
Total number of boxes =
"
+
n);
System
.
out
.
println(
"
Total number of cartons =
"
+
total);
}
}
Output