\(\displaystyle 0.5=\frac{1}{2}\), so \(0.5\times18\) is half of \(18\).

\(18\div2=9\).

Answer

\(9\)

Use \(\displaystyle 0.04=\frac{4}{100}\), giving \(\displaystyle \frac{4}{100}\times700\).

Dividing \(700\) by \(100\) first leaves \(4\times7=28\).

Answer

\(0.04\times700=4\times7=28\).

Write \(9=10-1\). Thus \(6.3\times9=6.3\times10-6.3\times1\).

This gives \(63-6.3=56.7\). The amount removed must be one full copy of \(6.3\), not just one.

Answer

Option A: \(6.3\times10-6.3\).

Correct option: A

The denominator \(60\) was divided by \(10\), so divide the numerator by \(10\) too: \(23.76\div10=2.376\).

Then \(2.376\div6=0.396\). Check: \(0.396\times60=23.76\).

Answer

\(\square=2.376\); the quotient is \(0.396\).

Double \(2.50\) to get \(5\), and halve \(18\) to get \(9\). The new multiplication is \(5\times9=45\).

Doubling one factor would double the product, while halving the other halves it again. The two changes cancel, so the total remains \(45\) pounds.

Answer

\(45\) pounds.

The learner divided only the numerator by \(10\), making the quotient ten times smaller instead of keeping it equal.

Divide both numerator and denominator by \(10\): \(\displaystyle \frac{48.6}{30}=\frac{4.86}{3}\).

Now \(4.86\div3=1.62\). Check: \(1.62\times30=48.6\).

Answer

\(\displaystyle \frac{48.6}{30}=\frac{4.86}{3}=1.62\).

The first method uses \(19=20-1\): \(7.6\times20-7.6=152-7.6=144.4\).

The second uses \(7.6=7+0.6\): \(7\times19+0.6\times19=133+11.4=144.4\).

Both distribute multiplication across a sum or difference, but the first uses the especially easy multiplier \(20\).

Answer

Both methods give \(144.4\).

Multiplying the equation by \(100\) gives \(8M=640\), so \(M=80\).

The multiplier \(1.6\) is \(20\) times \(0.08\), so its product with the same \(M\) is \(20\times6.4=128\).

This uses the known product and its scale, rather than recalculating from the beginning.

Answer

\(M=80\) and \(1.6M=128\).

In the product, multiplying the first factor by \(10\) and dividing the second by \(10\) leaves the total factor unchanged.

In the division claim, the numerator becomes ten times larger while the divisor becomes ten times smaller. Each change makes the quotient ten times larger, so together they make it \(100\) times larger.

To keep the quotient unchanged, multiply both numerator and denominator by \(10\): \(\displaystyle \frac{0.6}{350}=\frac{6}{3500}\).

Answer

The correct division is \(0.6\div350=6\div3500\). The claimed quotient \(6\div35\) is \(100\) times the original.

At \(2.50\) pounds each, \(100\) items would cost \(250\) pounds. The cheaper items reduce this by \(0.42\), and the dearer items increase it by \(0.58\).

The actual total is \(250-0.42+0.58=250.16\) pounds.

Each replacement lowers the cost by \(2.51-2.49=0.02\) pounds. Removing the \(0.16\)-pound excess therefore takes \(8\) replacements.

This leaves \(50\) items at each price, whose equal one-penny adjustments cancel. Fewer replacements would leave part of the excess.

Answer

The total is \(250.16\) pounds. Replace \(8\) dearer items with cheaper ones.