A certain insect population grows at a constant rate, increasing by

Question

A certain insect population grows at a constant rate, increasing by the same number of insects per hour. At the end of one hour (time t = 1), the population consists of p insects; at the end of five hours (t = 5), the number of insects has grown to 13p.

In the first column, indicate the number of insects by which the population grows in 1 hour. In the second column, indicate the projected population at time t = 8 if the growth rate remains constant.

Sajjad1994 · Accepted Answer

Official Explanation

The problem states that the population increases by the same number of insects each hour, and the first column asks for that value. Start there.

Assume that the population grows by x insects per hour:

t = 1 ➜ p
t = 2 ➜ p + x
t = 3 ➜ p + x + x = p + 2x
t = 4 ➜ p + x + x + x = p + 3x
t = 5 ➜ p + x + x + x + x = p + 4x

Since the population at t = 5 is given as 13p, set this equal to the expression shown above for t = 5 and solve for x.

13p = p + 4x
12p = 4x
x = 3p

Next, use the hourly growth to predict the population at t = 8. From t = 5 to t = 8 is a total of 3 hours, for a total of 3(3p) = 9p additional insects. The new population level is therefore 13p + 9p = 22p.

Column 1: The correct answer is (A).
Column 2: The correct answer is (F).

Sumi1010 · Answer

Let the initial population of the insect is  X 
 and, constant rate at which the insects grow (insects per hour) =  r

At the end of one hour, the number of insects
 p = X + r --- (1)
At the end of 5 hours, the number of insects
13p = X + 5r ---(2)

From (1) and (2),
p - r = 13p - 5r
4r= 12p
 r = 3p

At the end of 8 hours, the projected number of insects would be 
X + 8r = (X + r) + 7r
Since (X+r) = p and r = 3p
the projected number of insects would be 
p + 21p = 22p

Thus, 
 Growth per hour = 3p
Population at the end of 8 hours = 22p

kntombat · Answer

The growth per hour is 3p and the population when t is 8 is 22p.

Deepakjhamb · Answer

Growth per hour Population at t = 8

3p
4p
7p
12p
19p
22p

solution :

population grows at constant rate = k /hour

then pop at 1 hour = p and pop at 5 hours =13p

so in 4 hours population grew = 13p-1p =12p

so in hour population grew = 12p/4 = 3p

now at t=5 population = 13 p

so 3 hours later at t=8 , poulation will be = 13p + 3p *3 = 13p + 9p =22p

so answers are Growth per hour = 3p and Population at t = 8 is equal to 22p

rocky620 · Answer

Since, the population on insects is growing with constant number, this will be in Arithmetic Progression.
Let the common difference be x, so population at the end of hour,
1=p (given)
2=p+x
3=p+2x
4=p+3x
5=p+4x = 13p (given)...... .Eq1 
6=p+5x
7=p+6x
8=p+7x...... Eq2

From Eq1, p+4x=13,  x=3p 
From Eq2, Population at the end t8
 p+7*3p = 22p

In the first column, indicate the number of insects by which the population grows in 1 hour. In the second column, indicate the projected population at time t = 8 if the growth rate remains constant.

3p = Growth per hour  
4p
7p
12p
19p
  22p = Population at t = 8

Sajjad1994 · Answer

This is a medium level question, every one got correct, OA has been added and OE has also been posted.

adroitpensador · Answer

Posting for people ' who're wondering why didn't we use - p(t=0)*a^kt?' : Had the question been ' increasing by the same  rate'   instead of ' increasing by the same number' , we could've used the exponential formula.

Increasing by the same number = Arithmetic Progression

A certain insect population grows at a constant rate, increasing by

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A certain insect population grows at a constant rate, increasing by

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