A pupil formed three-digit numbers using the digits 2, 1 and 6 without repeating a digit in each number formed. Write down the; (a) three-digit odd numbers formed. (b) three-digit numbers divisible by 4.

For a three-digit number to be odd, its last digit must be odd. Among the digits 2, 1, and 6, only 1 is an odd digit.

Place 1 in the units position. The remaining digits (2 and 6) can be arranged in the hundreds and tens positions.
The possible arrangements are 261 and 621.

The three-digit odd numbers formed are $\boxed{261, 621}$.

b) Write down the three-digit numbers divisible by 4.

The possible numbers are:
126, 162
216, 261
612, 621

A number is divisible by 4 if the number formed by its last two digits is divisible by 4.

• For 126, the last two digits form 26. $26 \div 4 = 6$ remainder 2. Not divisible by 4.
• For 162, the last two digits form 62. $62 \div 4 = 15$ remainder 2. Not divisible by 4.
• For 216, the last two digits form 16. $16 \div 4 = 4$. Divisible by 4.
• For 261, the last two digits form 61. $61 \div 4 = 15$ remainder 1. Not divisible by 4.
• For 612, the last two digits form 12. $12 \div 4 = 3$. Divisible by 4.
• For 621, the last two digits form 21. $21 \div 4 = 5$ remainder 1. Not divisible by 4.

The three-digit numbers divisible by 4 are $\boxed{216, 612}$.

22.

a) Round off 52.479 to the nearest tenth.

In 52.479, the tenths digit is 4.

The digit to the right of 4 is 7.

Since 7 is 5 or greater, we round up the tenths digit (4 becomes 5). All digits after the tenths place are dropped.

52.479 rounded to the nearest tenth is $\boxed{52.5}$.

b) Work out: $\frac{0.69 \times 1.2}{0.43 - 0.2}$

To simplify, multiply the numerator and denominator by 100 to remove decimals from the denominator:

Now, divide:

The value is $\boxed{3.6}$.