Question

On a map, the distance between Milvale and Dracut is 114 inches. If the map uses a scale of 14 inch represents 1 mile, what is the actual distance in miles from Milvale to Dracut?

Answer 1

To solve the problem, we need to use the scale provided in the question:

14 inches represents 1 mile

We can set up a proportion:

14 inches / 1 mile = 114 inches / x miles

where x is the actual distance in miles from Milvale to Dracut.

To solve for x, we can cross-multiply:

14 inches * x miles = 1 mile * 114 inches

Simplifying the right side:

114 inches = 114/1 inches = 114 miles / 14

Multiplying both sides by 14:

14 inches * x miles = (114 miles / 14) * 14 inches

Simplifying the right side:

14 inches * x miles = 114 miles

Dividing both sides by 14 inches:

x miles = 114 miles / 14 inches

x miles = 8.14 miles

Therefore, the actual distance in miles from Milvale to Dracut is 8.14 miles.

Answer 2

A food company reduced the amount of salt in one of their food products from 700 milligrams to 630 milligrams. What is the percent decrease in the amount of salt in this food product?

Answer 3

To find the percent decrease, we need to calculate the difference between the original amount of salt and the new amount of salt, divide that by the original amount, and then multiply by 100 to express the result as a percentage.

The difference between the original amount and the new amount is:

700 milligrams - 630 milligrams = 70 milligrams

To find the percentage decrease, we divide this difference by the original amount:

70 milligrams / 700 milligrams = 0.1

Multiplying by 100, we get:

0.1 * 100 = 10%

Therefore, the percent decrease in the amount of salt in the food product is 10%.