# Basic Arithmetic : Percents and Decimals

## Example Questions

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### Example Question #1 : Finding Percent Change

In 2012, a university received 150,249 applicants. In 2013, the same university received 190,201 applicants. By what percent did the total number of applicants increase from 2012 to 2013?

Explanation:

To find the percent change, use the following formula:

Our new number of applicants is 190,201, while the original number of applicants is 150,249.

### Example Question #1 : Finding Percent Change

In January, Widget Company made , widgets. In February, the same company made , widgets. By what percentage did the amount of widgets made increase?

Explanation:

To find the percent increase, use the following formula:

In the case of Widget Company,

Therefore, the percent increase is

### Example Question #2 : Finding Percent Change

Derek ran his first mile in  minutes and his second mile in  minutes.

What is the percent increase between the time it took Derek to run the first and second mile?

Explanation:

The correct answer is a percent change (increase) of .

This can be calculated using the following equation:

Therefore, the answer is a percent increase of .

### Example Question #1 : Finding Percent Change

In May of the past year, Harry's company made  in profit, and in June of that same year, his company made  in profit. By what percent did his profit increase from May to June?

Explanation:

To find percent change, use the following formula:

Our new value is the profit in June, $210,219. Our old value is the profit in May,$189,281.

### Example Question #1 : Finding Percent Change

In March, an athlete weighed  pounds. In April, the same athlete weighed  pounds. What is the athlete's percentage of weight loss?

Explanation:

The formula to find percentage change is

.

So for our athlete, our new weight is 170 pounds, and the original weight is 195 pounds.

Because the question already tells us that we need to find the percent lost, or percent decrease, we can drop the negative sign in the answer.

### Example Question #1 : Finding Percent Change

Last month the shirt you wanted to buy cost  and this month it costs . What is the percent decrease?

Explanation:

1. Subtract the lower price from the higher price.

2. Divide the difference or amount of change by the initial price:

or 10% price decrease

### Example Question #3 : Finding Percent Change

is decreased by . What is the new number?

Explanation:

We first find 18% of 76:

Since 76 is being decreased by this amount, we subtract 13.68:

### Example Question #1 : Percents And Decimals

At age  you are  inches taller than you were  years ago, when you were  inches tall. What is the percent change in your height from  years ago to now? Round your answer to the nearest tenth.

Explanation:

The questions throws a lot of numbers at us, but the only ones we are concerned with are those having to do with height. We use the following formula:

We want to know the percent change. We know the amount change, 6 inches, and the original amount, 52 inches. Plugging these into the formula, we get:

Rounding this number to the nearest tenth gives us 11.5%.

### Example Question #4 : Finding Percent Change

is decreased to . What is the percent change?

Explanation:

Percent change is found by the following formula:

Amount change is the difference between the old amount and the new amount.

Here, the original amount is 30, and the new amount is 27. The difference between them is:

We plug in 3 for amount change and 30 for original amount and simplify:

### Example Question #2 : Percents And Decimals

What is the perimeter of a semicircle with an area of ?

Explanation:

This is a multi-step problem.  The perimeter of a semicircle is the sum of the circumference and diameter.  First, since we are given the area, we will need to find the radius.

For the area of a semicircle:

Since the area is , substitute this into A to find radius r.

Since the radius is 2, the diameter is 4.

The circumference for a semicircle is:

The perimeter is the sum of the circumference and diameter:

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