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One can think of the problem as a person would about a problem involving a car trying to catch up to a slower one, driving in the same direction.

The key trap is the question stem: how many years after VEERU’s investment do they catch up.


Veeru invests 10,000 at 5% interest per year.

He gets a 2 year head start.

So at end of year 2/beginning of year 3 he has 10,000 + 500 + 500 = 11,000.

He gains 500 a year

Joy starts at beginning of year 3 with 8,000 at 10% interest. She gains 800 a year.



The “gap” that Joy must catch up to in order to make the investments equals is: 11,000 - 8,000 = 3,000


Joy will gain 800 a year.

Veeru will gain 500 a year.

The “relative gain” (akin to relative speed) is +300 a year that Joy will catch up.

3,000 / 300 = 10 years for Joy to catch up.

But this is 10 years from when JOY started her investment.

Veeru started 2 years earlier.

Total time from the start of VEERU’s investment = 2 + 10 =

12 (C)

sarthak268
Veeru invested Rs 10000 at 5% simple annual interest, and exactly after two years, Joy invested Rs 8000 at 10% simple annual interest. How many years after Veeru’s investment, will their balances, i.e., principal plus accumulated interest, be equal?

A 10
B 11
C 12
D 13
E 14

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10^4 + 5/100*10^4 (x) = 8*10^3+8*10^3*10/100(x-2)

10 + x/2 = 8 + 4/5(x-2)

x = 12
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The catching up calculations using the difference in Principal and difference in interest is interesting.

But since this is just simple interest problem, I saw this problem as an Arithmetic Progression problem and used Tn = a + (n-1) d

Let n be the no of years of veeru's investment, so number years of Joy's investment is (n-2)

10000 + (n-1) 500 = 8000 + (n-3) 800

Solving for n, I get 13.

Bunuel @chetan4u - Does this mean that at the beginning of 13th year of the Veeru's investment, the amounts become equal and therefore it is 12 years after veeru's investment ?
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Vetrick
The catching up calculations using the difference in Principal and difference in interest is interesting.

But since this is just simple interest problem, I saw this problem as an Arithmetic Progression problem and used Tn = a + (n-1) d

Let n be the no of years of veeru's investment, so number years of Joy's investment is (n-2)

10000 + (n-1) 500 = 8000 + (n-3) 800

Solving for n, I get 13.

Bunuel @chetan4u - Does this mean that at the beginning of 13th year of the Veeru's investment, the amounts become equal and therefore it is 12 years after veeru's investment ?
Vetrick: Yes, your method and logic work — the beginning of the 13th year means that it is 12 years after Veeru's investment.

I find it simpler to just solve directly for what it's asking for — have "n" or "T" represent the years AFTER Veeru's investment.

Many people use this Simple Interest formula: I = PRT (where I = interest, P = Principal, R = Rate, T = Time in years)
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Veeru invested Rs 10000 at 5% simple annual interest, and exactly after two years, Joy invested Rs 8000 at 10% simple annual interest. How many years after Veeru’s investment, will their balances, i.e., principal plus accumulated interest, be equal?

The Simple interest Veeru gets every year is 5 % of 10,000 i.e 500

Joy gets 10 % of 8000 i.e Rs 800 every year as simple interest.

Assume that 'n' years after Veeru's investment, their balances (P + S.I) are equal. It's also given that Joy invested 2 years after Veeru.

The interest, Veeru got after 'n' years is 500*n, and Joy got 500(n-2) as interest after n-2 years.

Equating their total balance after 'n' years

10,000 + 500 n = 8000 + 800(n-2)

2000 = 800n - 1600 -500n
3600 = 300n
n = 12 years.

Option C is the answer.

Thanks,
Clifin J Francis,
GMAT SME
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GMATCoachBen
Vetrick
The catching up calculations using the difference in Principal and difference in interest is interesting.

But since this is just simple interest problem, I saw this problem as an Arithmetic Progression problem and used Tn = a + (n-1) d

Let n be the no of years of veeru's investment, so number years of Joy's investment is (n-2)

10000 + (n-1) 500 = 8000 + (n-3) 800

Solving for n, I get 13.

Bunuel @chetan4u - Does this mean that at the beginning of 13th year of the Veeru's investment, the amounts become equal and therefore it is 12 years after veeru's investment ?
Vetrick: Yes, your method and logic work — the beginning of the 13th year means that it is 12 years after Veeru's investment.

I find it simpler to just solve directly for what it's asking for — have "n" or "T" represent the years AFTER Veeru's investment.

Many people use this Simple Interest formula: I = PRT (where I = interest, P = Principal, R = Rate, T = Time in years)

Got it. Thank you GMATCoachBen.

And CrackVerbal 's reply above is also helpful. Instead of using n-1 & n-3, I could use n & n-2. Thank you.
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