Question

Coal from source a produces sulfur at a rate of 15g per 100kg of coal. Coal from source b produces sulfur at a rate of 50g per 400kg of coal. Source a sells for $0. 50 per kilogram and source b costs $1. 25 per kilogram.

a) How many grams of sulfur would be released from the combustion of $100 of each source of coal?

b) We use about 7 billion metric tons of coal worldwide per year for electricity generation. Using the least polluting coal source, how many kilograms of sulfur would be released per year?

Answer 1

The grams of sulfur released from Source A is 20,000 grams and 320,000 grams from source B.

The amount of kilograms of sulfur that will be released per year by using the least polluting coal source is 9.333 × 10¹⁵ kg

What is a word problem?

Word problem in mathematics involves crucial understanding and knowing the right procedure and mathematical concepts such as (fractions, ratio, algebraic expressions) to use with their arithmetic operations.

From the given information:

• Coal from source A = 15 g per 100kg of coal
• Coal from source B = 50g per 400 kg of coal

• Source A sells = $0.50 per kilogram
• Source B sells = $1.25 per kilogram

Let's first find the percentage of each coal from each source.

• Source A = (0.015 kg /100 kg) × 100% = 0.015 %
• Source B = (0.05 kg /400 kg) × 100% = 0.0125 %

Source A sells for $0.50 per kg, thus 0.015 kg will be:

$\mathbf{= \dfrac{\ 0.50 }{1 \ kg} \times 0.015 kg }$

= $0.075 in 0.015 kg

Source B sells for $1.25 per kg, thus 0.05 kg will be:

$\mathbf{= \dfrac{\ 1.25 }{1 \ kg} \times 0.0125 kg }$

=$0.015625 in 0.05 kg

In the combustion of $100 of each coal, we have the following amount in grams:

For Source A, we have:

$\mathbf{= \dfrac{\ 100 \times 0.015 \ kg }{\ 0.075}}$

= 20 kg of sulfur

= 20,000 grams of sulfur

For source B, we have:

$\mathbf{= \dfrac{\ 100 \times 0.05 \ kg }{\ 0.015625}}$

=320 kg of sulfur

= 320,000 grams of sulfur.

The grams of sulfur released from Source A is 20,000 grams and 320,000 grams from source B.

Using the least polluting coal source, i.e. source A, the number of kilograms of sulfur released per year will be:

• 1 metric ton = 1000 kg

Thus,

• 7 billion metric ton = 7000000000000 kg

$\mathbf{= \dfrac{\ 100 \times 7000000000000\ kg }{\ 0.075}}$

= 9.333 × 10¹⁵ kg

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