A newborn baby weighs 2.7 kg. Find her weight in pounds. The weight of the baby is 2.7 kg = 2700 g. 453.6 g = 1 lb. 1 lb = (453.6 g * 2700) lb = 5.95 lb. Now, how was this solved?
The Correct Answer and Explanation is:
Correct Answer:
A newborn baby weighs 2.7 kg.
We are given:
- 1 kilogram (kg) = 1000 grams (g)
- 1 pound (lb) = 453.6 grams (g)
Step-by-step conversion:
- Convert kilograms to grams:
2.7 kg=2.7×1000=2700 g2.7 \text{ kg} = 2.7 \times 1000 = 2700 \text{ g}2.7 kg=2.7×1000=2700 g
- Convert grams to pounds:
2700 g÷453.6 g/lb≈5.95 lb2700 \text{ g} \div 453.6 \text{ g/lb} \approx 5.95 \text{ lb}2700 g÷453.6 g/lb≈5.95 lb
Final Answer:2.7 kg≈5.95 lb2.7 \text{ kg} \approx 5.95 \text{ lb}2.7 kg≈5.95 lb
Explanation
To convert a baby’s weight from kilograms to pounds, we need to understand the relationship between these units. A kilogram is a unit of mass in the metric system, and a pound is a unit of weight commonly used in the imperial system, especially in the United States. Since both units measure the same physical quantity (mass/weight), they can be converted using standard conversion factors.
The problem gives us two key facts:
- 1 kilogram = 1000 grams – this is a basic metric conversion.
- 1 pound = 453.6 grams – this is the bridge between the metric and imperial systems.
The baby’s weight is 2.7 kilograms. First, we convert this into grams because the conversion factor to pounds is given in grams. We multiply 2.7 by 1000, which gives us 2700 grams.
Now, to find how many pounds this is, we divide the number of grams by how many grams are in one pound:2700÷453.6≈5.952700 \div 453.6 \approx 5.952700÷453.6≈5.95
This means that 2.7 kilograms is approximately equal to 5.95 pounds.
Note that the incorrect step in the original explanation was multiplying instead of dividing by 453.6. The correct operation is division because we are converting from grams to a larger unit (pounds), and therefore we need fewer units — that is, fewer pounds than grams.
Understanding the logic behind conversion ensures accuracy and helps apply this method to other units of measure.
