When a certain tree was first planted, it was 4 ft tall

[guest]

Can anyone help answer the following:

1). When a certain tree was first planted, it was 4 ft tall, and the height of the tree increased by a constant amt each yr for the next 6 yrs. At the end of the 6th year, the tree was 1/5 taller than it was at the end of the 4th year. By how many feet did the height of the tree increase each yr?

Answer: 2/3

2). For every positive even integer n, the function h(n) is defined to be the product of all even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100)+1, then p is
(a) between 2 and 10
(b) between 10 and 20
(c) between 20 and 30
(d) between 30 and 40
(e) greater than 40

Answer: E

3). A boat traveled upstream a distance of 90 miles at an average speed of (v-3) miles per hour and then traveled the same distance downstream at an avg speed of (v+3) miles per hour. If the upstream trip took half an hour longer than the trip downstream, how many hours did it take the boat to travel downstream?

Answer: 2.3

Thank you!

[Harish Dorai]

Problem #1:

Let us assume the height at the tree at the time planting as A which is equal to 4 Ft. Let us assume each year the tree grows by h feet.

So after

1st year height is A + h
2nd year height is A + 2h
.....
4th year A + 4h
6th year A + 6h

Given that after 6 years the height is 1/5 (which is 0.2) more than what it was in 4th year. So this can be converted into an equation as:

A + 6h = 1.2(A + 4h)

1.2h = 0.2A
h = A/6

= 4/6

= 2/3 feet.

Problem #2:

There is already a discussion in this forum that details the solution to the same problem. Please refer that.

Problem #3:

Let us assume for simplicity the speed (v-3) as X. So (v+3) = v-3 + 6 = X + 6

Time taken to travel upstream = 90/X
Time taken to travel downstream = 90/(X+6)

Given that upstream took 1/2 hour more.

So 90/X = 90/(X+6) + 0.5

If you simplify this, it becomes a Quadratic equation Square(X) + 30X - 1080 = 0

This is (X + 36) (X - 30) = 0. Which means X = -36 or X = 30
Since speed cannot be negative, the correct solution is X = 30

That is v-3 = 30
Which means v+3 = 36 mph.

Now time taken to travel downstream = 90/36 = 2 hours 18 minutes OR 2.3 hours.

Hope it helps.

[Guest]

Problem #1:

Let us assume the height at the tree at the time planting as A which is equal to 4 Ft. Let us assume each year the tree grows by h feet.

So after

1st year height is A + h
2nd year height is A + 2h
.....
4th year A + 4h
6th year A + 6h

Given that after 6 years the height is 1/5 (which is 0.2) more than what it was in 4th year. So this can be converted into an equation as:

A + 6h = 1.2(A + 4h)

1.2h = 0.2A
h = A/6

= 4/6

= 2/3 feet.

Problem #2:

There is already a discussion in this forum that details the solution to the same problem. Please refer that.

Problem #3:

Let us assume for simplicity the speed (v-3) as X. So (v+3) = v-3 + 6 = X + 6

Time taken to travel upstream = 90/X
Time taken to travel downstream = 90/(X+6)

Given that upstream took 1/2 hour more.

So 90/X = 90/(X+6) + 0.5

If you simplify this, it becomes a Quadratic equation Square(X) + 30X - 1080 = 0

This is (X + 36) (X - 30) = 0. Which means X = -36 or X = 30
Since speed cannot be negative, the correct solution is X = 30

That is v-3 = 30
Which means v+3 = 36 mph.

Now time taken to travel downstream = 90/36 = 2 hours 18 minutes OR 2.3 hours.

Hope it helps.

How is 90/36 not 2.5 hrs?

[RonPurewal]

please post these problems in separate threads, one problem per thread, per the forum rules (see rule #3).