Interest Problem

IMO C - 2000

Whats OA?

Solution :

let x , y, z be the instalments for 1st 2nd and 3rd year

given x=y/2 , y =2z/3 => y =2x and z =3y/2 = 3x

Let P be the present amount

after 1 year, amount = P(1+10/100) = 11p/10
after 1st instalment, amount = 11P/10 - x = P'

after 2nd year, amount = P'(1+10/100)
=(11P/10 - x) (1+10/100) = (11P/10 - x) *11/10
after 2nd instalment , amount = { (11P/10 - x) *11/10 } - y = { (11P/10 - x) *11/10 } - 2x
= 121P/100 - 1.1x - 2x = 121P/100 - 3.1x = P'''

After 3rd year, amount = P'''(1+10/100) = 11P'''/10 = 11/10 *( 121P/100 - 3.1x )
since it is fully paid, last instalment must be equal to 11P'''/10

Thus Z = 11P'''/10 => 3x = 11P'''/10 = 11/10 *( 121P/100 - 3.1x )

=> 30x = 121*11P/100 - 3.1x*11 = 2321P/100 - 34.1x
=> 64.1 x = 11^3*P/100 ---------------------------------------------------1
here P is the present amount and amount after 3 years is 12820
=> 12820 = P(11/10)^3 => p = 12820* (10/11)^3 -----------------------2

using 1 and 2 we get

\(64.1 x = \frac{11^3}{100}* 12820 * \frac{1000}{11^3} = 128200\\
x = \frac{128200}{64.1} = 2000\)

If you have to guess.... total instalment paid < 12820

=. x+y+z < 12820 => 6x < 12820 => x < 2133...... so 1950 and 2000 seems to be eligible as 1800 is quite low.
On Gmat i wouldnt have solved it, i would have guessed either 2000 or 1950.

Best was to take one complete equation as

\([ (1.1p - x) *1.1 - 2x ] * 1.1 = 3x\) => this will save a lot of time.