Question

An Amazing car's price $450,500.Price of the car increase at 3% of annual inflation rate .hence y wants to invest his money at a bank that offers 5.5% per year . how much annual amount y need to deposit at the end of each year to buy amazing car at five years.​

Answer 1

Answer:

sorry i am in sixth so I can't tell

Answer 2

Given:

Cost of the car =$450,500

inflation in the car price = 3%

bank rate = 5.5%

To find:

The amount of money you needs to deposit at the end of each year to buy the car at five years.

Solution:

First let's find the price of car in five years at 3% inflation.

1st year (start year)

Price = $450500

2nd year

Price = 3% of $450500 + $450500

$= \frac{3}{100} \times 450500 + 450500$

= 13515 + 450,500

= $464015

3rd year

Price = 3% of $464015 +464015

$= \frac{3}{100} \times 464015 + 464015$

= 13,920.45 + 464015

= $477,935.45

4th year = 3% of 477,935.45 + 477,935.45

$= \frac{3}{100} \times 477935.45 + 477935.45$

= 14,338.0635 + 477,935.45

= $492,273.5135

5th year = 3% of 492,273.5135 + 492,273.5135

$= \frac{3}{100} \times 492,273.5135 + 492273.5135$

= 14,768.205405 + 492,273.5135

= $507041.8

Therefore the cost of car in 5 years is $507041.8

Now the total amount needed by y in 5 years is known.

interst given by bank = 5.5%

now,

$inrterest = p \times \frac{r}{100} \times t$

let interest be 'x'

$x =(507041.8 - x) \times \frac{5.5}{100} \times 5$

$x = (507041.8 - x) \times \frac{55}{1000} \times 5$

$x = (507041.8 - x) \times \frac{11}{40}$

$\frac{40x}{11} = 507041.8 - x$

$3.64x = 507041.8 - x$

$3.64x + x = 507041.8$

$4.64x = 507041.8$

$x = 109,276.25$

Interest = $109276.25.

NOTE: Answers might slightly differ due to decimals.

Principle amount = 507041.8 - 109276.25

= $397,765.55

Payment needed to male at the end of each year

$= \frac{397,765.55}{5}$

= $79,553.11

The amount of money you needs to deposit at the end of each year to buy the car at five years

= $79,553.11