Interest Problem

Question

An amount of $12820 due 3 yrs hence, is fully repaid in three annual installments starting after 1 year. The 1st instalment is 1/2 the 2nd instalment and 2nd instalment is 2/3rd of the 3rd. If the rate of interest is 10% per annum, find the 1st installment.

1.$2400
2.$1800
3.$2000
4.$2200
5.$1950

gurpreetsingh · Answer

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.

shrivastavarohit · Answer

Wouldnt this problem be much simpler if we choose 3rd installment as x:

3rd I = x
2nd I = 2/3x
1st I = 1/2 (2/3x) = 1/3x

1st + 2nd + 3rd = total amount

solve for x and then Find the value of :

1st installment + 10% to it.

Get a rough estimate rather than doing so much calculation.

Answer shall still remain the same.

If this works, kudos shall follow.

Posted from my mobile device

fruit · Answer

shrivastavarohit, as I understand, the interest rate charge after each installment, isn it?

3rd installment is x, so the equation will be:

1.1*x + 2/3*x*1.1 + 1.1*1/2*2/3*x=12820

after long and boring calculations I get the answer close to 1950

Hussain15 · Answer

Kudos for the input!!

OA is "C".

suramya26 · Answer

Can you give a solution to this problem???

I have tried the following approach
Please let me know where I am wrong
I have assumed the 3rd installment to be x
Therefore,
2nd inst.=2x 
1st inst.=x
since the repaying started after the 1st year, hence interest will be added from 2nd year onwards
>> x*(1.1)^2+2x*(1.1)+3x=12820
>> 6.41x=12820

>> x=2000

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

Interest Problem

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards