Seven hundredths multiplied by \(6\) gives \(7\times6=42\) hundredths.

Forty-two hundredths is \(0.42\).

Answer

\(0.42\)

The factor \(4.7\) is one tenth of \(47\), so its product is one tenth of \(376\): \(37.6\).

The factor \(0.47\) is one hundredth of \(47\), so its product is one hundredth of \(376\): \(3.76\).

Answer

\(\square=37.6\) and \(\triangle=3.76\).

\(28\times4=112\). The original first factor is ten times smaller, so the product is \(112\div10=11.2\).

An estimate of \(3\times4=12\) also shows that \(11.2\) is the sensible scale.

Answer

Option B: \(11.2\).

Correct option: B

\(65\times18=65\times(10+8)=650+520=1170\).

The price was \(65\) hundredths of a pound, so the product is \(1170\) hundredths, or \(11.70\) pounds.

Answer

\(11.70\) pounds.

Type A contributes \(0.125\times24=3.000\,\mathrm{kg}\). For example, \(125\times24=3000\), then divide by \(1000\).

Type B contributes \(0.08\times15=1.20\,\mathrm{kg}\).

Together the packets have mass \(3.00+1.20=4.20\,\mathrm{kg}\).

Answer

\(4.2\,\mathrm{kg}\)

The estimate of scale already rejects \(90\): \(0.36\times25\lt 1\times25=25\).

\(36\times25=900\). Since \(0.36\) is \(36\) hundredths, the required product is \(900\div100=9\).

Answer

The correct product is \(9\). A product of \(90\) is impossible because \(0.36\lt 1\), so multiplying \(25\) by it must give less than \(25\).

\(2.4\) is half of \(4.8\), while \(30\) is twice \(15\). Halving one factor and doubling the other leaves the product unchanged.

\(4.8\times15=4.8\times10+4.8\times5=48+24=72\), so the second product is also \(72\).

Answer

The products are equal; both are \(72\).

Six jars and the box have mass \(6\times0.24+0.35=1.44+0.35=1.79\,\mathrm{kg}\), which is allowed.

Seven jars and the box have mass \(7\times0.24+0.35=1.68+0.35=2.03\,\mathrm{kg}\), above the limit.

Each additional jar increases the mass, so \(6\) is the greatest possible count.

Answer

\(6\) jars.

Each pouch has \(0.085-0.080=0.005\,\mathrm{kg}\) extra.

For twenty pouches, the extra amount is \(20\times0.005=0.100\,\mathrm{kg}\). The total difference is the sum of the twenty identical individual differences.

Answer

\(0.10\,\mathrm{kg}\) extra in total.

The product is the product of two two-digit whole numbers, divided by \(10\). For a fixed pairing of digits, placing the larger digit first in each factor maximises that product.

The three digit-pair partitions therefore have maximum whole-number products \(21\times63=1323\), \(31\times62=1922\), and \(61\times32=1952\).

The last is greatest, giving \(1952\div10=195.2\). Either whole-number factor can supply the decimal factor, giving the two stated arrangements. Reversing digits within a pair makes a factor smaller.

Answer

The greatest product is \(195.2\), achieved by \(6.1\times32\) and \(3.2\times61\).