Given that 10^x + 14^x / 15^x + 21^x = 2 1/4, find the value of x.

The jet flies at an average speed of $850 \text{ km/h}$.
The distance covered is $3,400 \text{ km}$.

$$\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{3400 \text{ km}}{850 \text{ km/h}} = 4 \text{ hours}$$

The jet consumes $5,000 \text{ litres}$ of fuel per hour.

$$\text{Fuel consumed} = \text{Fuel consumption rate} \times \text{Time}$$ $$\text{Fuel consumed} = 5000 \text{ litres/h} \times 4 \text{ h} = 20000 \text{ litres}$$

The quantity of fuel consumed is $\boxed{\text{20000 litres}}$.

1. (b)

The original fuel consumption rate is $5,000 \text{ litres/hour}$.
To reduce fuel consumption by $10\%$:

$$\text{Reduction} = 10\% \text{ of } 5000 \text{ litres/hour} = 0.10 \times 5000 = 500 \text{ litres/hour}$$ $$\text{New fuel consumption rate} = 5000 - 500 = 4500 \text{ litres/hour}$$

Given that fuel consumption is proportional to speed, we can write $\frac{\text{Fuel consumption rate}}{\text{Speed}} = \text{constant}$.
Let $F_1$ and $S_1$ be the initial fuel consumption rate and speed, and $F_2$ and $S_2$ be the new values.

$$\frac{F_1}{S_1} = \frac{F_2}{S_2}$$ $$\frac{5000 \text{ litres/h}}{850 \text{ km/h}} = \frac{4500 \text{ litres/h}}{S_2}$$ $$S_2 = \frac{4500 \text{ litres/h} \times 850 \text{ km/h}}{5000 \text{ litres/h}}$$ $$S_2 = \frac{3825000}{5000} = 765 \text{ km/h}$$

The average speed required is $\boxed{\text{765 km/h}}$.