Cars for sale!

Let $x$ be the original price. The first discount is $x-0.3x=0.7x$ . The second discount is $0.7x-0.3(0.7x)=0.49x$ . The third discount is $0.49x-0.3(0.49x)=0.343x$ .

Then

$0.343x=24300$

$x=100000$

$total~amount=100000+34300=$ $\color{#D61F06}\boxed{134300}$

Let the actual cost of the car before discount be $x$ .

After the 1st discount the cost is $x$ $-$ $\frac{30}{100}x$ $=$ $\frac{70}{100}x$ .

Similarly after the 2nd discount the cost is $\frac{70}{100}x$ $-$ $\frac{30}{100}x \times \frac{70}{100}x$ $=$ $\frac{49}{100}x$ .

Same way is followed and after the 3rd discount the cost will be $\frac{343}{1000}x$ .( After this no further discount is implemented.)

Now, we get $\frac{343}{1000}x$ $=$ $34300$ .

Thus, the actual cost of the car was $\$100000$ .

Therefore, the total money spent on the cars is $34300 + 100000 = \$134300$ ( It is because the father bought the car without any discount.)

If a 30% discount has been applied, we can assume that our price is 0.7 of the original. therefore we can divide by 7 and multiply by ten, We do this three times for the three successive discounts of 30% which gives us a price of $100000. We add these together to get $134300