Approximation and Estimation

Round the number 3728 to the nearest 10, 100, and 1000.

Solution:

When rounding to the nearest 10, we need to look at the digits from the 10s column onwards. In this case, we have 28. Now, we must ask ourselves, is 28 closer to 20 or 30? The answer is 30, since 28 is only 2 away from 30, whereas it is 8 away from 20. Thus, 28 rounds to 30 and so 3728 rounds up to 2730.

When rounding to the nearest 100, we look at the digits from the 100s column onwards. In this case, we have 728. Now, we must ask ourselves, is 728 closer to 700 or 800? Clearly, it is closer to 700 and so in this case, 3728 rounds down to 3700.

When rounding to the nearest 1000, we look at the digits from the 1000s column onwards. This is 3728. Now, we must ask ourselves, is 3728 closer to 3000 or 4000. In this case, it is closer to 4000 so we round up to 4000 when rounding to the nearest 1000.

Work out an estimate for $\frac{92.1\times 19.2}{98.1}$

Solution:

This calculation is quite difficult to work out without using a calculator. However, if we round each number to the nearest 10, we obtain$\frac{90\times 20}{100}=\frac{1800}{100}=18$. Thus, we can say that the estimation for the calculation is 18.

We could go one step further and work out the percentage error between the estimated value and the real value. Using a calculator, we can find that$\frac{92.1\times 19.2}{98.1}=18.0256881$. If we subtract the estimated value from the real value, we get what is called the absolute error. In this case, the absolute error is 0.0256881 (which is promising as it shows that our estimated value is close to the real value due to such a small absolute error).

Now, if we divide the absolute error by the actual value, and then multiply by 100, we get the percentage error. If we do this, we get the percentage error as$\frac{0.0256881}{18.0256881}\times 100\%=0.1425\%$. Since this number is small, we can see that we have a good estimation as we have such a small percentage error.

I buy 32 packets of crisps for a party. Each packet costs 21p. Estimate the total cost of the crisps.

Solution:

The total cost is 32 lots of 21p. Thus, we need to calculate$32\times 21$ to work out the cost (in pence). We can round both numbers to the nearest 10 to get$30\times 20$ which is far easier to work out. We get$20\times 30=600p=\pounds 6$. Thus, we can say that the total cost of the 32 packets of crisps is about £6. The true value of$32\times 21=672p=\pounds 6.72$and so we can see that our estimated value is close to the actual value.

If we were to go one step further and work out the percentage error, we would need to subtract the estimated value from the true value , divide by the actual value and then multiply by the true value as follows:

$\frac{6.72-6}{6.72}\times 100\%=10.7\%$. Thus, the percentage error of our estimate is 10.7%.

Estimate the cost of 123 paper plates that cost 11p each and 157 napkins that cost 9p each

Solution:

The total cost (in pence) of the paper plates is going to be$123\times 11$and the total cost of the napkins is$157\times 9$. Thus the total cost of both the paper plates and napkins is$123\times 11+157\times 9$. We can approximate the numbers in this calculation and instead work out $120\times 10+160\times 10=1200+1600=2800p=\pounds 28$. Thus, an estimate for the total cost is £28.

We could work out the percentage error by working out the true value of the cost. In this case, it is $123\times 11+157\times 9=2766p=\pounds 27.66$. Thus, the percentage error is$\frac{28-27.66}{27.66}\times 100\%=1.23\%$.

Estimate the value of $\frac{301\times 9.01}{0.499}$

Solution:

If we round 301 to the nearest 10, we get 300. If we round 9.01 to the nearest 10, we get 10. Now, 0.499 is approximately 0.5, so let's round it to that. Thus, we have $\frac{300\times 10}{0.5}=\frac{3000}{0.5}=6000$. Thus, 6000 is an estimate.

The true value of $\frac{301\times 9.01}{0.499}=5434.88978$ and so we can see that our estimated value is relatively close. The absolute error of our estimate is $6000-5434.88978=565.11022$ and our percentage error is $\frac{565.11022}{5434.88978}\times 100\%\approx 10.4\%$. Thus, we can say that our estimation is out by approximately 10.4%.

Estimate the value of $\frac{{(4.98)}^2}{0.482}$

Solution:

4.98 is quite close to 5, and it is easy to square 5, so let's approximate 4.98 as 5. 0.482 is close to 0.5, and it is fairly easy to divide by one half. Thus, we have the estimate $\frac{{(4.98)}^2}{0.482}\approx \frac{{\left(5\right)}^2}{0.5}=\frac{25}{0.5}=50$. Thus, the estimation for this calculation is 50.

We could work out the percentage error by working out the true value of$\frac{{(4.98)}^2}{0.482}=51.45$.

Thus, the percentage error is $\frac{51.45-50}{51.45}\times 100\%=2.82\%$.

Estimate the value of $\sqrt{\frac{51.3}{0.53}}$

Solution:

51.3 is approximately 50, and 0.53 is approximately 0.5. Thus, we have the estimation $\sqrt{\frac{51.3}{0.53}}\approx \sqrt{\frac{50}{0.5}}=\sqrt{100}=10$. Thus, an estimate for the value is 10.

We could work out the percentage error by working out the true value of $\sqrt{\frac{51.3}{0.53}}=9.84$.

Thus, the percentage error is $\frac{10-9.84}{9.84}\times 100\%=1.63\%$.