Question

A gardener is planting two types of trees: Type A is 7 feet tall and grows at a rate of 10 inches per year. Type B is 4 feet tall and grows at a rate of 14 inches per year. Algebraically determine exactly how many years it will take for these trees to be the same height.

Answer 1

Answer:

9 years

Step-by-step explanation:

Let the the number of years in which height of both the type of tree is same be n years

Initial length of type A tree = 7 feet

As rate of growth is given in inches to maintain uniformity of measuring unit lets convert feet to inches

1 feet = 12 inches

thus 7 feet = 12*7 inches = 84 inches.

length of type A tree in inches = 84 inches

Rate of growth of type A tree = 10 inches per year

Thus, actual growth of type a tree in "n" years = 10*n = 10n

Total height of type A tree in n years = initial length + growth in n years

= 84 inches+ 10n inches (1)

____________________________________________________

For type B

Initial length of type B tree = 4 feet

As rate of growth is given in inches to maintain uniformity of measuring unit lets convert feet to inches

1 feet = 12 inches

thus 4 feet = 12*4 inches = 48 inches.

length of type B tree in inches = 48 inches

Rate of growth of type A tree = 14 inches per year

Thus, actual growth of type a tree in "n" years = 14*n = 14n

Total height of type b tree in n years = initial length + growth in n years

= 48 inches+ 14n inches  (2)

_________________________________

Given condition that after n years height of both the type of tree is same

equation 1 should be equal to equation 2

84 inches+ 10n inches = 48 inches+ 14n inches

=>84 inches - 48 inches  = 14n inches - 10n inches

=> 36 inches = 4n inches

=> 36 = 4n

=> n = 36/4 = 9

Thus, after 9 years both of their height will be same which be equal to

84+10*9 = 84 + 90 = 174 inches.