At the beginning of 2010, 60% of the population of Town X

AAPL wrote:Manhattan Prep

At the beginning of 2010, 60% of the population of Town X lived in the south and the rest lived in the north. During 2010, the population of Town X grew by 5.5%. If the population in the south grew by 4.5%, by how much did the population in the north grow?

A. 1%
B. 3.5%
C. 6.5%
D. 7%
E. 13.75%

OA D

We can let the total people in the town at the beginning of 2010 = n.

Thus, 0.6n = south and 0.4n = north.

We can determine the growth in the north by creating the following equation where x is the percent increase in the north:

1.045(0.6n) + (1 + x/100)(0.4n) = 1.055n

Dividing both sides by n, we have:

0.627 + (1 + x/100)(0.4) = 1.055

(1 + x/100)(0.4) = 0.428

1+ x/100 = 1.07

x/100 = 0.07

x = 7

Alternate Solution:

Let’s assume the beginning population was 1000, so we know that 600 people lived in the south and 400 people lived in the north.

During 2010, the total population increased by 5.5%, so we have 1000 x 1.055 = 1055 citizens in the town at the end of 2010.

We know that the population in the south grew by 4.5%, so the population of the south was 600 x 1.045 = 627 at the end of 2010.

Thus, the population of the north at the end of 2010 had to be 1055 – 627 = 428. We use the formula for percent change: (New – Old)/Old x 100% , obtaining (428 – 400)/400 x 100% = 28/400 x 100% = 7%.

Answer: D