Solve problems 1124, 1125, and 1126 regarding discounts, rectangle area, and alloy composition.

The original price of one toy car is 420 tg. The discount is 15%.

$$\text{Discount per car} = 0.15 \times 420 \text{ tg} = 63 \text{ tg}$$

The kindergarten buys 7 toy cars.

$$\text{Total savings} = 7 \times 63 \text{ tg} = 441 \text{ tg}$$

The kindergarten saves 441 tg.

$$\boxed{\text{441 tg}}$$

1125.

The width of the rectangle is 6 cm. The length is 75% longer than the width.

$$\text{Length} = 6 \text{ cm} + (0.75 \times 6 \text{ cm})$$ $$\text{Length} = 6 \text{ cm} + 4.5 \text{ cm} = 10.5 \text{ cm}$$

The area of a rectangle is given by the formula $\text{Area} = \text{Length} \times \text{Width}$.

$$\text{Area} = 10.5 \text{ cm} \times 6 \text{ cm} = 63 \text{ cm}^2$$

The area of the rectangle is 63 cm$^2$.

$$\boxed{\text{63 cm}^2}$$

1126.

Bronze is 85% copper, and the rest is tin.

$$\text{Percentage of tin} = 100\% - 85\% = 15\%$$

$$\text{Difference} = \text{Mass of copper} - \text{Mass of tin}$$ $$\text{Difference} = 4.25 \text{ kg} - 0.75 \text{ kg} = 3.5 \text{ kg}$$

There is 3.5 kg more copper than tin.

$$\boxed{\text{3.5 kg}}$$