Let's simplify things a bit. What you want to know is why 450d / 420 and 450d / 7 / 60 (specifically) give different results.
Let's see how division is performed in IEE double-precision floating point format. Without going deep into implementation details, it's basically XOR-ing the sign bit, subtracting the exponent of the divisor from the exponent of the dividend, dividing the mantissas, and normalizing the result.
First, we should represent our numbers in the proper format for double:
450 is 0 10000000111 1100001000000000000000000000000000000000000000000000
420 is 0 10000000111 1010010000000000000000000000000000000000000000000000
7 is 0 10000000001 1100000000000000000000000000000000000000000000000000
60 is 0 10000000100 1110000000000000000000000000000000000000000000000000
Let's first divide 450 by 420
First comes the sign bit, it's 0 (0 xor 0 == 0).
Then comes the exponent. 10000000111b - 10000000111b + 1023 == 10000000111b - 10000000111b + 01111111111b == 01111111111b
Looking good, now the mantissa:
1.1100001000000000000000000000000000000000000000000000 / 1.1010010000000000000000000000000000000000000000000000 == 1.1100001 / 1.101001. There are a couple of different ways to do this, I'll talk a bit about them later. The result is 1.0(001) (you can verify it here).
Now we should normalize the result. Let's see the guard, round and sticky bit values:
0001001001001001001001001001001001001001001001001001 0 0 1
Guard bit's 0, we don't do any rounding. The result is, in binary:
0 01111111111 0001001001001001001001001001001001001001001001001001
Which gets represented as 1.0714285714285714 in decimal.
Now let's divide 450 by 7 by analogy.
Sign bit = 0
Exponent = 10000000111b - 10000000001b + 01111111111b == -01111111001b + 01111111111b + 01111111111b == 10000000101b
Mantissa = 1.1100001 / 1.11 == 1.00000(001)
Rounding:
0000000100100100100100100100100100100100100100100100 1 0 0
Guard bit is set, round and sticky bits are not. We are rounding to-nearest (default mode for IEEE), and we're stuck right between the two possible values which we could round to. As the lsb is 0, we add 1. This gives us the rounded mantissa:
0000000100100100100100100100100100100100100100100101
The result is
0 10000000101 0000000100100100100100100100100100100100100100100101
Which gets represented as 64.28571428571429 in decimal.
Now we will have to divide it by 60... But you already know that we have lost some precision. Dividing 450 by 420 didn't require rounding at all, but here, we already had to round the result at least once. But, for completeness's sake, let's finish the job:
Dividing 64.28571428571429 by 60
Sign bit = 0
Exponent = 10000000101b - 10000000100b + 01111111111b == 01111111110b
Mantissa = 1.0000000100100100100100100100100100100100100100100101 / 1.111 == 0.10001001001001001001001001001001001001001001001001001100110011
Round and shift:
0.1000100100100100100100100100100100100100100100100100 1 1 0 0
1.0001001001001001001001001001001001001001001001001001 1 0 0
Rounding just as in the previous case, we get the mantissa: 0001001001001001001001001001001001001001001001001010.
As we shifted by 1, we add that to the exponent, getting
Exponent = 01111111111b
So, the result is:
0 01111111111 0001001001001001001001001001001001001001001001001010
Which gets represented as 1.0714285714285716 in decimal.
Tl;dr:
The first division gave us:
0 01111111111 0001001001001001001001001001001001001001001001001001
And the last division gave us:
0 01111111111 0001001001001001001001001001001001001001001001001010
The difference is in the last 2 bits only, but we could have lost more - after all, to get the second result, we had to round two times instead of none!
Now, about mantissa division. Floating-point division is implemented in two major ways.
The way mandated by the IEEE long division (here are some good examples; it's basically the regular long division, but with binary instead of decimal), and it's pretty slow. That is what your computer did.
There is also a faster but less accrate option, multiplication by inverse. First, a reciprocal of the divisor is found, and then multiplication is performed.
