Arithmetic expansion $(( )) in Bash Scripting - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to calculate the sum of 5 and 3 using arithmetic expansion.

Bash Scripting

result=$((5 [1] 3))
echo $result

Drag options to blanks, or click blank then click option'

A*

B-

C+

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using '-' will subtract instead of add.

Using '*' or '/' will multiply or divide, not add.

The plus sign (+) inside $(( )) adds the two numbers.

2fill in blank

medium

Complete the code to multiply 7 by 4 using arithmetic expansion.

Bash Scripting

result=$((7 [1] 4))
echo $result

Drag options to blanks, or click blank then click option'

A*

B-

C+

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using '+' or '-' will add or subtract instead of multiply.

Using '/' will divide instead of multiply.

The asterisk (*) inside $(( )) multiplies the two numbers.

3fill in blank

hard

Fix the error in the code to correctly subtract 2 from 10 using arithmetic expansion.

Bash Scripting

result=$((10 [1] 2))
echo $result

Drag options to blanks, or click blank then click option'

A/

B+

C*

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using '+' will add instead of subtract.

Using '*' or '/' will multiply or divide instead of subtract.

The minus sign (-) inside $(( )) subtracts the second number from the first.

4fill in blank

hard

Fill both blanks to calculate (17 % 5) - 2 using arithmetic expansion.

Bash Scripting

result=$((17 [1] 5 [2] 2))
echo $result

Drag options to blanks, or click blank then click option'

A%

B+

C-

D*

Attempts:

3 left

💡 Hint

Common Mistakes

Using '+' instead of '-' changes the result.

Using '*' instead of '%' does not give remainder.

The modulus operator (%) gives the remainder. The minus sign (-) is used to subtract 2 from the remainder.

5fill in blank

hard

Fill all three blanks to create a dictionary-like output with keys as numbers and values as their squares, but only for numbers greater than 3.

Bash Scripting

declare -A squares
for num in {1..5}; do
  if (( num [1] 3 )); then
    squares[[2]]=$((num [3] num))
  fi
done
for key in "${!squares[@]}"; do
  echo "$key: ${squares[$key]}"
done

Drag options to blanks, or click blank then click option'

A>

Bnum

C*

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' instead of '>' changes the filter condition.

Using wrong variable names causes errors.

Using '+' instead of '*' does not square the number.

The condition 'num > 3' filters numbers greater than 3. The key is 'num' and the value is 'num * num' (square).