# Please show all the set up and work for these numerical problems. Please make sure the final answer shows the correct number of significant figures.

These are numerical problems testing your knowledge of unit conversions.

**1.** We can do it by using the fact that 1 acre contains 43560 ft^2.

Hence, area = `240 xx 10^6 acres xx 43560 ft^2/(acre) = 1.045 xx10^13 ft^2`

The same can also be solved by using the given information, 1 acre = 4046.86 m^2 and 1 m^2 = 10.76 ft^2

Since 240 million acres has only 2 significant figures, our answer will be rounded off to **1.0 x 10^13 ft^2**

**2. **1 lbs = 0.4536 kg

so the mass of the baby is 14 lbs = 14 x 0.4536 kg = 6.35 kg

Using the recommended dosage of 15 mg/kg, the baby will need 15 x 6.35 mg of drug = 95.25 mg

Using the given information that 80 mg drug is present in 0.8 ml

Using unitary method, 95.25 mg will be contained in 0.8 x 95.25/80 ml = **0.95 ml**. (two significant figures)

**3**. Here the numerator and denominator are solved separately and then divided. Remember, addition and subtraction does not change the units, while for multiplication, the units are also multiplied.

Hence, numerator : 2.54 g + 0.0028 g = 2.5428 g (we are not rounding off for significant figures, otherwise, this would be written as 2.54 gm, to account for least number of decimal places)

and denominator: 0.0105 cm^2 x 0.06 cm = 0.00063 cm^3 ( will be rounded off to 0.0006 to account for least number of significant figures in the LHS).

And the answer is: 2.54 gm/0.0006 cm^3 = 4233.33 gm/cm^3

and if we correct for significant figures, the answer would be **4000 gm/cm^3** (only 1 significant figure).

**4**. `(7.87 ml) / (16.1 ml-8.44 ml) = 7.87/7.66 = 1.03`

(two significant figures only since that is the minimum number of significant figures in numerator and denominator)

**5**. 100 m in 9.58 sec

Speed = distance/time = 100 /9.58 m/sec = 10.44 m/sec (if we use the rule of significant figures, the answer will be 10 m/sec, since answer will have only 1 significant figure)

1 mile contains 1609.344 m and 1 hour contains 3600 sec

`10.44 m/sec = 10.44 m/sec xx [(1 mi)/(1609.344 m)]/[(1 hr)/(3600 sec)] = 23.35 (mi)/(hr)`

(if we correct it for significant figures and round it off to 1 significant figure, the answer will be **20 mi/hr**).

Hope this helps.