The whole parts give \(2+3=5\), and the decimal parts give \(0.6+0.7=1.3\).

Combine them: \(5+1.3=6.3\).

Answer

\(6.3\)

The missing number is \(4.20-1.85\). Subtracting \(1.80\) leaves \(2.40\), then subtracting \(0.05\) leaves \(2.35\).

Check: \(2.35+1.85=4.20\).

Answer

\(2.35\)

\(10.00-0.48=9.52\). One way is to subtract \(0.50\) and then add back \(0.02\).

The inverse check \(9.52+0.48=10.00\) confirms it.

Answer

Option B: \(9.52\).

Correct option: B

The total cost is \(3.75+2.60=6.35\) pounds.

The change is \(10.00-6.35=3.65\) pounds. Check: \(6.35+3.65=10.00\).

Answer

\(3.65\) pounds.

\(3.450+2.800=6.250\), then \(6.250+4.075=10.325\,\mathrm{m}\).

Subtract the used length: \(12.000-10.325=1.675\,\mathrm{m}\).

Answer

Used: \(10.325\,\mathrm{m}\). Remaining: \(1.675\,\mathrm{m}\).

The number \(6.4\) is \(640\) hundredths, not \(64\) hundredths. Write it as \(6.40\) before adding.

\(6.40+0.85=7.25\). Also, adding a positive number to \(6.4\) cannot produce the smaller result \(1.49\).

Answer

The correct answer is \(7.25\).

Subtract the extra \(1.4\) from the sum: \(8.6-1.4=7.2\). This is twice the smaller number.

Half of \(7.2\) is \(3.6\), and adding \(1.4\) gives the larger number \(5.0\).

Check: \(3.6+5.0=8.6\) and \(5.0-3.6=1.4\).

Answer

\(3.6\) and \(5.0\).

The first addend increases by \(0.10\), while the second decreases by \(0.10\). These changes cancel, so the sum is unchanged.

Calculating one sum gives \(7.38+2.67=10.05\), which is therefore the value of both.

Answer

The sums are equal, and both equal \(10.05\).

Each measurement can range from \(1.25\) to \(1.34\), inclusive. Hence the extreme sums are \(1.25+1.25+1.25=3.75\) and \(1.34+1.34+1.34=4.02\).

The example \(1.33+1.33+1.34=4.00\) uses only allowed values, so it proves the requested total is possible.

Answer

Smallest total: \(3.75\). Largest total: \(4.02\). A total of \(4.00\) is possible, for example \(1.33+1.33+1.34\).

The hundredths digits add to \(A+C\) and must end in zero. Since both are non-zero digits, their sum must be \(10\), with a carry of \(1\).

The tenths column is then \(2B+1\), and its final digit must be \(1\). This allows \(B=0\) or \(B=5\).

If \(B=5\), an extra carry makes the whole-number total \(A+C+1=11\), which is wrong. Thus \(B=0\).

The distinct non-zero digit pairs summing to \(10\) are \(1,9\), \(2,8\), \(3,7\), and \(4,6\). The pair \(5,5\) is excluded, proving completeness.

Answer

\(1.09+9.01\), \(2.08+8.02\), \(3.07+7.03\), and \(4.06+6.04\), each equal to \(10.10\).