At the beginning of the year, a library had 800 history books. In the first half of the ye… — SAT

At the beginning of the year, a library had $800$ history books. In the first half of the year, the number of history books increased by $15\%$ . In the second half of the year, the library acquired more history books, representing a percent increase of $x\%$ over the number of history books at the midyear point. If the library had a total of $1{}012$ history books at the end of the year, what is the value of $x$ ?

Answer: 10

Answer

The value of

x

10

To find the second percent increase, we must first calculate the intermediate midyear value. A

15\%

increase on

800

800 \times 1.15 = 920

. The subsequent increase of

x\%

is based on this midyear value of

920

. Setting up the equation

920 \times (1 + \frac{x}{100}) = 1{}012

and solving for

x

yields

1 + \frac{x}{100} = 1.10

, which gives

x = 10

Step-by-Step Solution

Calculate the number of history books at the midyear point after the first increase of

15\%

920

books

An increase of

15\%

on the initial

800

books is calculated as

800 \times (1 + 0.15) = 920

Set up an equation for the second percent increase of

x\%

from the midyear value to the final value of

1{}012

920 \times (1 + \frac{x}{100}) = 1{}012

The second increase is

x\%

of the midyear value of

920

, resulting in the final value of

1{}012

Solve the equation for

x

x = 10

Divide both sides of the equation by

920

to get

1 + \frac{x}{100} = 1.10

, subtract

1

to get

\frac{x}{100} = 0.10

, and multiply by

100

to find

x = 10

Key Concept

Calculating successive percent increases by applying the percent change to the intermediate base value.

Alternative Method

Instead of calculating the intermediate number of books, we can express the final number of books as

800 \times 1.15 \times (1 + \frac{x}{100}) = 1{}012

. Simplifying

800 \times 1.15

gives

920

, so

920 \times (1 + \frac{x}{100}) = 1{}012

, leading to the same result of

x = 10

Estimated Time:1m 30s

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

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