Question

A store had 150 laptops in the month of January. Every month, 20% of the laptops were sold and 10 new laptops were stocked in the store. Which recursive function best represents the number of laptops in the store f(n) after n months? f(n) = 150 − 0.8 x f(n − 1) + 10, f(0) = 150, n > 0 f(n) = 0.2 x f(n − 1) + 10, f(0) = 150, n > 0 f(n) = 0.8 x f(n − 1) + 10, f(0) = 150, n > 0 f(n) = 150 + 0.8 x f(n − 1) + 10, f(0) = 150, n > 0

Answer 1

Answer:

f(n) =  0.8 × f(n -1) + 10, n > 0

Step-by-step explanation:

Given,

Original number of laptops, f(0) = 150, ( where, f(n) represents the number of laptops after n months )

∵ Every month, 20% of the laptops were sold and 10 new laptops were stocked in the store.

After 1 month,

The number of laptops, f(1) = original laptops - 20% of the original laptops + 10

= 150 - 20% of 150 + 10

= (100 - 20)% of 150 + 10

= 80% of 150 + 10

= 0.8 × f(0) + 10

Similarly, After 2 months,

The number of laptops, f(2) = 0.8 × f(1) + 10,

After 3 months, number of laptops, f(3) = 0.8 × f(2) + 10,

........ so, on

Thus, by following the pattern,

The number of laptops after n months,

f(n) =  0.8 × f(n -1) + 10, n > 0

Answer 2

Answer-

$\boxed{\boxed{f(n)=0.8(f(n-1))+10,\ f(0)=150,\ \ n>0}}$

Solution-

Here, n represents the number of months and f(n) represents the number of laptops in the store after n months.

As the store had 150 laptops in the month of January or at the beginning.

So $f(0)=150$

Every month, 20% of the laptops were sold and 10 new laptops were stocked in the store.

As 20% of laptops were sold, so 80% were in the store.

So, after one month total number of laptops in the store,

$\Rightarrow f(1)=0.8(150)+10$

$\Rightarrow f(1)=0.8(f(0))+10$     $(\because f(0)=150)$

Again after one month total number of laptops in the store,

$\Rightarrow f(2)=0.8\times (0.8(150)+10)+10$

$\Rightarrow f(2)=0.8(f(1))+10$     $(\because f(1)=0.8(150)+10)$

Analyzing the pattern, the recursive function f(n) will be,

$f(n)=0.8(f(n-1))+10,\ f(0)=150,\ \ n>0$