Exercise 3B
Solutions to Exercise 3B
Approximation and Estimation
Basic Level
1. State the number of significant figures in each of the following:
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(a) $39\,018$
All non-zero digits are significant, and the zero between them is also significant. Therefore, the number of significant figures is 5.
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(b) $0.028\,030$
The leading zeros are not significant. The zero between non-zero digits is significant, and the trailing zero after the decimal point is also significant. Therefore, the number of significant figures is 5.
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(c) $2900$ (to the nearest 10)
The instruction "to the nearest 10" implies that the last zero is a significant digit. The first two digits are non-zero and are significant. Therefore, the number of significant figures is 3.
2. Round off each of the following to the number of significant figures as given in brackets:
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(a) $728$ (2 s.f.)
The second significant figure is 2. The next digit is 8, which is $\geq 5$, so we round up. The number becomes 730.
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(b) $503.88$ (4 s.f.)
The fourth significant figure is 8. The next digit is 8, which is $\geq 5$, so we round up. The number becomes 503.9.
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(c) $0.003\,018\,5$ (4 s.f.)
The first two zeros are not significant. The fourth significant figure is 8. The next digit is 5, which is $\geq 5$, so we round up. The number becomes 0.003\,019.
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(d) $6396$ (2 s.f. and 3 s.f.)
To 2 s.f.: The second significant figure is 3. The next digit is 9, which is $\geq 5$, so we round up. The number becomes 6400.
To 3 s.f.: The third significant figure is 9. The next digit is 6, which is $\geq 5$, so we round up. The number becomes 6400.
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(e) $9.9999$ (3 s.f.)
The third significant figure is 9. The next digit is 9, which is $\geq 5$, so we round up. This causes a cascade of rounding. The number becomes 10.0.
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(f) $8.076$ (3 s.f.)
The third significant figure is 7. The next digit is 6, which is $\geq 5$, so we round up. The number becomes 8.08.
Intermediate Level
3. The number $143\,200$ is correct to $x$ significant figures. Write down the possible values of $x$.
The number $143\,200$ has 4 non-zero digits. The two trailing zeros may or may not be significant. The possible values of $x$ are the number of significant figures.
- 4 s.f. (1, 4, 3, 2). The zeros are placeholders.
- 5 s.f. (1, 4, 3, 2, 0). The first zero is significant.
- 6 s.f. (1, 4, 3, 2, 0, 0). Both zeros are significant.
Possible values of $x$ are 4, 5, and 6.
4. Evaluate each of the following and correct your answers to the number of significant figures as given in brackets:
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(a) $\frac{1}{99}$ (4 s.f.)
$1 \div 99 = 0.01010101...$
The first significant figure is the first digit that is not zero, which is the 1 in the hundredths place. Rounding to 4 s.f., we get 0.01010.
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(b) $871 \times 234$ (2 s.f.)
$871 \times 234 = 203\,754$
Rounding to 2 s.f., we get 200\,000.
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(c) $\frac{21^2}{0.219}$ (5 s.f.)
$21^2 = 441$.
$\frac{441}{0.219} \approx 2013.6986...$
Rounding to 5 s.f., we get 2013.7.
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(d) $3.91^2 - \frac{2.1}{6.41}$ (2 s.f.)
$3.91^2 = 15.2881$
$\frac{2.1}{6.41} \approx 0.3276...$
$15.2881 - 0.3276... \approx 14.9605...$
Rounding to 2 s.f., we get 15.
Advanced Level
5. What is the greatest number of sweets that can be bought with $2 if each sweet costs 30 cents?
First, convert the cost to a single unit. $2 is equal to 200 cents.
Number of sweets = $\frac{200 \text{ cents}}{30 \text{ cents/sweet}} = \frac{20}{3} = 6$ with a remainder of 2.
The greatest number of sweets is 6.
6. The area of a square is $264 \text{ cm}^2$. Find (i) the length, (ii) the perimeter of the square.
(i) Length of the square, $s = \sqrt{\text{Area}} = \sqrt{264} \approx 16.248 \text{ cm}$.
(ii) Perimeter of the square, $P = 4 \times s = 4 \times 16.248 \approx 64.992 \text{ cm}$.
7. The circumference of a circle is $136 \text{ m}$. Find (i) the radius, (ii) the area of the circle.
(i) Circumference, $C = 2 \pi r$. So, $136 = 2 \pi r$.
$r = \frac{136}{2\pi} = \frac{68}{\pi} \approx 21.64 \text{ m}$.
(ii) Area, $A = \pi r^2 = \pi \left(\frac{68}{\pi}\right)^2 = \frac{68^2}{\pi} = \frac{4624}{\pi} \approx 1471.9 \text{ m}^2$.
8. The number $21X09$ is equal to $22\,000$, correct to 2 significant figures. Find the value of X if $21X09$ is a perfect square.
$22\,000$ correct to 2 s.f. means the number is in the range $[21\,500, 22\,500)$.
We need to find a perfect square of the form $21X09$ in this range. A perfect square ending in 9 must have a square root ending in 3 or 7. Let's test numbers whose squares are in this range and end in 3 or 7.
$\sqrt{21500} \approx 146.6$. So we can check $147^2$.
$147^2 = 21609$. This number has the form $21X09$, with $X=6$. It is also within the required range.
Let's check the next possible square root ending in 3 or 7: $153^2 = 23409$, which is outside the range.
Therefore, the value of X is 6.
9. The number of people at a concert is stated as $21\,200$, correct to 3 significant figures. What is the largest and the smallest possible number of people at the concert?
The number $21\,200$ is rounded to the nearest hundred (3 s.f. is the digit 2). The rounding was applied to the hundreds place.
To find the smallest possible number, we look at the midpoint below $21\,200$, which is $21\,150$. Any number from $21\,150$ up to but not including $21\,250$ would round to $21\,200$.
The smallest possible number is 21,150.
The largest possible number is the one just before it would round up to $21\,300$. This would be $21\,249$.
The largest possible number is 21,249.
10. (i) Without using a calculator, evaluate $987\,654\,321 + 0.000\,007 - 987\,654\,321$.
(ii) Use a calculator to evaluate $987\,654\,321 + 0.000\,007 - 987\,654\,321$.
(iii) Do you get the same answer for (i) and (ii)? Explain your answer.
(i) Without a calculator, the two large numbers cancel each other out, as they are a positive and a negative version of the same number.
$(987\,654\,321 - 987\,654\,321) + 0.000\,007 = 0 + 0.000\,007 = 0.000\,007$
The answer is 0.000\,007.
(ii) Using a standard calculator, the result is likely to be **0**.
(iii) You will likely get a different answer. This is because calculators and computers use floating-point arithmetic with limited precision. When a very large number ($987\,654\,321$) is added to a very small number ($0.000\,007$), the smaller number's significant digits can be lost during the calculation due to rounding or truncation, resulting in the sum being treated as equal to the large number itself. The subsequent subtraction then yields zero. The manual calculation avoids this loss of precision.