The problems are on the mass and volume worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Mass is the amount of matter in an object, measured in kilograms or grams with a balance. These worked answers correct the 20 problems of the grade 6 science worksheet on mass and volume, from units and instruments to syrup bottle: each solution shows the formula, the substitution and the final answer with its unit.
Check each answer against your own working. Method marks matter: write the formula or unit conversion, substitute the numbers, then give the result with its unit. Re-read any problem where your answer differs.
Level: Grade 6 science (Year 7 · Class 6) · Curriculum: SEP measurement
Problem 1 Units and instruments
1. The base unit of mass is the kilogram (kg).2. Litres (L) and millilitres (mL) are common; cm³ is also correct.
3. Mass is measured with a balance.
Problem 2 Bag of rice
1. 2.5 kg × 1000 = 2500 g.2. 450 mg ÷ 1000 = 0.450 g.
Problem 3 Water bottle volumes
1. 1.5 × 1000 = 1500 mL.2. 250 ÷ 1000 = 0.25 L.
3. 1 cm³ = 1 mL, so the volume is 40 mL.
Problem 4 Pick the instrument
1. Balance.2. Graduated cylinder.
3. Balance.
4. Graduated cylinder. Mass uses a balance; volume of a liquid uses a cylinder.
Problem 5 Reading a small scale
1. Between 0 and 10 mL there are 5 spaces, so one division is 10 ÷ 5 = 2 mL.2. The surface is 2 divisions above 30 mL: 30 + 2 × 2 = 34 mL.
3. Eye level with the bottom of the meniscus.
Problem 6 Diego reads the top
1. The agreed reading point is the lowest point of the curve, not its top edge.2. The top edge lies higher on the scale, so his reading is too high.
Problem 7 Tare the beaker
1. 112.7 − 48.2 = 64.5 g.2. Tare removes the beaker’s mass from the display, so it shows 64.5 g.
Problem 8 Metal key
1. Start 40 mL, final 58 mL.2. 58 − 40 = 18 mL.
3. 1 mL = 1 cm³, so 18 cm³.
Problem 9 Rainwater tank
1. 1 m³ = 1000 L, so 0.35 × 1000 = 350 L.2. 350 × 1000 = 350 000 mL.
Problem 10 Baking day
1. Eggs: 3 × 55 = 165 g. Total = 450 + 120 + 250 + 165 = 985 g.2. 985 ÷ 1000 = 0.985 kg.
Problem 11 Pouring glasses
1. 2 × 1000 = 2000 mL.2. 2000 ÷ 250 = 8 glasses.
3. 5 × 250 = 1250 mL poured; 2000 − 1250 = 750 mL left.
Problem 12 Which cylinder?
1. The 10 mL cylinder.2. Its divisions are smaller, so the reading is more precise, and the narrow tube makes the level easier to judge.
Problem 13 Marbles in a cylinder
1. Each marble raises the level by 2 mL, so 2 cm³.2. 50 + 12 × 2 = 74 mL.
3. A floating marble would not sink, so it would displace less water and the pattern would break.
Problem 14 Coin on two scales
1. 6 − 5.87 = 0.13 g.2. The laboratory balance, because its smallest step is 100 times smaller than that of the kitchen scale.
Problem 15 Aquarium volume
1. 40 × 25 × 30 = 30 000 cm³.2. 30 000 cm³ = 30 000 mL = 30 L.
3. 0.80 × 30 = 24 L.
Problem 16 Cork and sinker
1. A: 30 mL; B: 38 mL; C: 50 mL.2. Sinker = 38 − 30 = 8 cm³.
3. Cork = 50 − 38 = 12 cm³.
4. The combined reading includes the sinker, so its volume must be known and subtracted.
Problem 17 Forgotten tare
1. 139.8 − 85.4 = 54.4 g.2. 139.8 − 54.4 = 85.4 g too high.
3. 85.4 ÷ 54.4 × 100 = 157 %.
Problem 18 Uncertainty in volume
1. 0.5 ÷ 25 × 100 = 2 %.2. 0.5 ÷ 5 × 100 = 10 %.
3. The same absolute uncertainty is a much larger fraction of a small reading, so use a smaller cylinder for small volumes.
Problem 19 Potato too big
1. 142 cm³, since 1 mL = 1 cm³.2. The potato pushes aside its own volume of water, and that water leaves through the spout.
3. Some water may stay in the spout or be left on the can, so less is collected.
Problem 20 Syrup bottle
1. 0.15 × 1000 = 150 mL.2. 3 × 5 = 15 mL per day.
3. 150 ÷ 15 = 10 days.
4. 150 − 7 × 15 = 150 − 105 = 45 mL.
