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Application of basic arithmetic operations is fundamental for Science Laboratory Technology professionals in Kenya, where precise measurements and calculations are essential for reliable experimental results. Mastery of addition and subtraction across different number systems ensures accuracy in data recording, preparation of solutions, and interpretation of laboratory findings. This chapter focuses on the application of addition and subtraction with natural numbers, integers, fractions, and decimals, providing worked examples relevant to laboratory contexts.
Natural numbers are the positive whole numbers used for counting and ordering, essential in laboratory inventory management and sample counting. Operations on natural numbers form the basis for more complex calculations in data analysis and preparation of reagents.
The addition of two natural numbers \(a\) and \(b\) is given by:
$$S = a + b$$
The subtraction of two natural numbers \(a\) and \(b\), where \(a \geq b\), is given by:
$$D = a, b$$
Example 1: A laboratory technician counts 125 test tubes in one cabinet and 78 test tubes in another. Find the total number of test tubes.
Given: \(a = 125\), \(b = 78\)
$$S = a + b$$
$$S = 125 + 78$$
$$S = 203$$
Answer: 203 \text{ test tubes}
Example 2: A chemical supplier delivers 350 bottles of reagent. After distribution to various labs, 127 bottles remain. How many bottles were distributed?
Given: \(a = 350\), \(b = 127\)
$$D = a, b$$
$$D = 350 - 127$$
$$D = 223$$
Answer: 223 \text{ bottles distributed}
Example 3: A lab receives 240 samples on Monday and 185 samples on Tuesday. If 410 samples have been processed by the end of Tuesday, how many samples remain unprocessed?
Given: \(a = 240 + 185 = 425\), \(b = 410\)
$$D = a, b$$
$$D = 425 - 410$$
$$D = 15$$
Answer: 15 \text{ samples unprocessed}
Example 4: A laboratory has 560 ml of a solution. If 175 ml is used in an experiment, and then 230 ml is added later, what is the final volume?
Given: Initial volume \(a = 560\) ml, used volume \(b = 175\) ml, added volume \(c = 230\) ml
Step 1: Subtract used volume
$$V_1 = a, b$$
$$V_1 = 560 - 175$$
$$V_1 = 385 \text{ ml}$$
Step 2: Add added volume
$$V_f = V_1 + c$$
$$V_f = 385 + 230$$
$$V_f = 615 \text{ ml}$$
Answer: 615 \text{ ml final volume}
Example 5: A laboratory orders 1,200 units of consumables. The first shipment contains 850 units, and the second shipment contains 400 units. How many units are still outstanding to complete the order?
Given: Total order \(a = 1200\), shipment 1 \(b = 850\), shipment 2 \(c = 400\)
Step 1: Calculate total received
$$R = b + c$$
$$R = 850 + 400$$
$$R = 1250$$
Step 2: Calculate outstanding units
$$O = a - R$$
$$O = 1200 - 1250$$
Since \(O < 0\), this means 50 extra units were received.
Answer: 50 \text{ extra units received}
Integers extend natural numbers to include negative values, which are important in laboratory data analysis, such as temperature changes, pH variations, or weight differences. Managing both positive and negative values accurately is crucial in interpreting experimental results.
The addition of two integers \(a\) and \(b\) is:
$$S = a + b$$
The subtraction of two integers \(a\) and \(b\) is:
$$D = a, b$$
Example 1: The temperature in a lab was \(+25^\circ C\) at noon and dropped by \(15^\circ C\) by midnight. What was the temperature at midnight?
Given: Initial temperature \(a = +25^\circ C\), change \(b = -15^\circ C\)
$$S = a + b$$
$$S = 25 + (-15)$$
$$S = 10^\circ C$$
Answer: 10^\circ C
Example 2: A pH meter reading changed from -3 to +5 during calibration. What is the difference in readings?
Given: \(a = -3\), \(b = 5\)
$$D = b, a$$
$$D = 5 - (-3)$$
$$D = 5 + 3$$
$$D = 8$$
Answer: 8 \text{ units change}
Example 3: A lab balance shows a reading of -0.4 g (tare weight) and then a reading of 2.7 g. What is the net weight of the sample?
Given: \(a = 2.7\), \(b = -0.4\)
$$D = a, b$$
$$D = 2.7 - (-0.4)$$
$$D = 2.7 + 0.4$$
$$D = 3.1 \text{ g}$$
Answer: 3.1 g net weight
Example 4: The concentration of a solution decreased from +12 mol/L to -8 mol/L after dilution. Calculate the change in concentration.
Given: Initial \(a = 12\), final \(b = -8\)
$$D = b, a$$
$$D = -8 - 12$$
$$D = -20 \text{ mol/L}$$
Answer: -20 \text{ mol/L change}
Example 5: A temperature sensor records the following sequence: -5, +3, -7, +10 \(^{\circ}C\). Calculate the sum of these readings.
Given: \(a = -5\), \(b = 3\), \(c = -7\), \(d = 10\)
$$S = a + b + c + d$$
$$S = -5 + 3 + (-7) + 10$$
$$S = (-5 + 3) + (-7 + 10)$$
$$S = (-2) + 3$$
$$S = 1^\circ C$$
Answer: 1^\circ C
Fractions are common in laboratory measurements, such as volumes, concentrations, and dilutions. Being able to add and subtract fractions with unlike denominators is essential for accurate solution preparation and data analysis.
Addition of two fractions \(\frac{a}{b}\) and \(\frac{c}{d}\) requires a common denominator:
$$S = \frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd}$$
Subtraction follows similarly:
$$D = \frac{a}{b} - \frac{c}{d} = \frac{ad, bc}{bd}$$
Example 1: A solution volume of \(\frac{3}{4}\) litre is mixed with \(\frac{1}{4}\) litre. Find the total volume.
Given: \(a = \frac{3}{4}\), \(b = \frac{1}{4}\)
$$S = \frac{3}{4} + \frac{1}{4} = \frac{3+1}{4} = \frac{4}{4}$$
$$S = 1 \text{ litre}$$
Answer: 1 litre
Example 2: A lab technician adds \(\frac{2}{3}\) litre of reagent A and \(\frac{1}{6}\) litre of reagent B. Calculate the total volume.
Given: \(a = \frac{2}{3}\), \(b = \frac{1}{6}\)
$$S = \frac{2}{3} + \frac{1}{6} = \frac{(2 \times 2) + (1 \times 1)}{6} = \frac{4 + 1}{6} = \frac{5}{6}$$
Answer: \frac{5}{6} \text{ litre}
Example 3: A sample weighs \(\frac{7}{8}\) kg. After removing \(\frac{3}{8}\) kg, what is the remaining weight?
Given: \(a = \frac{7}{8}\), \(b = \frac{3}{8}\)
$$D = \frac{7}{8} - \frac{3}{8} = \frac{7 - 3}{8} = \frac{4}{8} = \frac{1}{2} \text{ kg}$$
Answer: \frac{1}{2} \text{ kg}
Example 4: A solution contains \(\frac{5}{6}\) litre. If \(\frac{1}{4}\) litre is removed, find the remaining volume.
Given: \(a = \frac{5}{6}\), \(b = \frac{1}{4}\)
$$D = \frac{5}{6} - \frac{1}{4} = \frac{(5 \times 4) - (1 \times 6)}{24} = \frac{20 - 6}{24} = \frac{14}{24} = \frac{7}{12}$$
Answer: \frac{7}{12} \text{ litre}
Example 5: Two solutions with volumes \(\frac{3}{5}\) litre and \(\frac{2}{7}\) litre are combined. Find the total volume.
Given: \(a = \frac{3}{5}\), \(b = \frac{2}{7}\)
$$S = \frac{3}{5} + \frac{2}{7} = \frac{(3 \times 7) + (2 \times 5)}{35} = \frac{21 + 10}{35} = \frac{31}{35}$$
Answer: \frac{31}{35} \text{ litre}
Decimal numbers are widely used in laboratory measurements for precision in volume, mass, concentration, and time. Accurate addition and subtraction of decimals ensure correct quantification and reporting of experimental data.
Addition of decimals \(a\) and \(b\) is:
$$S = a + b$$
Subtraction of decimals \(a\) and \(b\) is:
$$D = a, b$$
Example 1: A sample weighs 23.45 g and another weighs 12.78 g. Find the total weight.
Given: \(a = 23.45\), \(b = 12.78\)
$$S = 23.45 + 12.78$$
$$S = 36.23 \text{ g}$$
Answer: 36.23 g
Example 2: A solution volume of 15.75 ml is used from a 50.00 ml bottle. Find the remaining volume.
Given: \(a = 50.00\), \(b = 15.75\)
$$D = 50.00 - 15.75$$
$$D = 34.25 \text{ ml}$$
Answer: 34.25 ml
Example 3: A lab technician records temperatures of 23.6°C and 18.9°C. What is their sum?
Given: \(a = 23.6\), \(b = 18.9\)
$$S = 23.6 + 18.9$$
$$S = 42.5^\circ C$$
Answer: 42.5^\circ C
Example 4: The pH of a solution changes from 7.25 to 5.40. Calculate the change.
Given: Initial \(a = 7.25\), final \(b = 5.40\)
$$D = 7.25 - 5.40$$
$$D = 1.85$$
Answer: 1.85 \text{ pH units}
Example 5: A volume of 120.50 ml is increased by 35.75 ml, then decreased by 28.30 ml. Find the final volume.
Given: Initial \(a = 120.50\), increase \(b = 35.75\), decrease \(c = 28.30\)
Step 1: Add increase
$$V_1 = 120.50 + 35.75$$
$$V_1 = 156.25 \text{ ml}$$
Step 2: Subtract decrease
$$V_f = 156.25 - 28.30$$
$$V_f = 127.95 \text{ ml}$$
Answer: 127.95 ml
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Create a free accountThis chapter covered the fundamental arithmetic operations of addition and subtraction applied to natural numbers, integers, fractions, and decimals, emphasizing accurate calculation techniques across different number types. It then explored multiplication and division within the same categories, highlighting the importance of understanding sign rules and the handling of fractional and decimal values. The concepts of ratios, proportions, and percentages were introduced, distinguishing between direct and inverse proportion and their practical applications in problem solving. The chapter also addressed indices by defining bases, presenting the laws of indices for simplifying expressions, and solving indicial equations. Through these topics, students developed essential mathematical skills necessary for precise computation and analysis in scientific contexts. The progression from basic operations to more complex relationships and exponentiation ensures a comprehensive grasp of arithmetic principles. Mastery of these areas forms the foundation for advanced mathematical applications in various technical fields.
A laboratory technician has 125 test tubes and receives 78 more. How many test tubes are there in total? (2 marks)
A chemical sample weighs 150 g. If 75 g is removed, what is the remaining weight? (2 marks)
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