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by aarzoo » Wed Sep 21, 2016 7:15 am
Rs.6,100/- was partly invested in scheme A at 10% per annum compound interest(compounded annually) for 2 years and partly in scheme B at 10% per annum simple interest for 4 years. Both the scheme pay equal interests. How much was invested in scheme A?
Options are:-
1.3,750
2.4,500
3.4,000
4.3,200
5.5,000
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Source: — Problem Solving |

by GMATGuruNY » Wed Sep 21, 2016 8:08 am
aarzoo wrote:Rs.6,100/- was partly invested in scheme A at 10% per annum compound interest(compounded annually) for 2 years and partly in scheme B at 10% per annum simple interest for 4 years. Both the scheme pay equal interests. How much was invested in scheme A?
Options are:-
1.3,750
2.4,500
3.4,000
4.3,200
5.5,000
We can PLUG IN THE ANSWERS, which represent the amount in invested in A.
Correct answers to compound interest problems are typically VERY ROUND NUMBERS.
Here, the correct answer is likely to be either C or E.

Answer choice C: 4000 in A, implying 2100 in B
Scheme A:
First-year interest = 10% of 4000 = 400.
Second-year interest = (10% of original investment) + (10% of first-year interest) = (10% of 4000) + (10% of 400) = 400+40 = 440.
Total interest = 400+440 = 840.

Scheme B:
Since 10% interest is earned each year, the total interest over 4 years = (4)(10% of 2100) = (4)(210) = 840.

Success!
The two schemes earn the same interest.

The correct answer is C.
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by [email protected] » Wed Sep 21, 2016 9:48 am
Hi aarzoo,

TESTing THE ANSWERS (the approach that Mitch used) is perfect for this type of question. There's an additional point to consider: since the two interest rates are the same, but the time frame for Scheme B is TWICE the time frame of Scheme A, it's likely that the total invested in Scheme A is pretty CLOSE to TWICE the amount invested in Scheme B - this is further evidence that the answer is probably 4,000 (or really close to it).

You can also solve this problem algebraically. We're told that the two sets of interest are EQUAL, so we can set up an equation that compares those two values.

X = total invested in Scheme A
(6100 - X) = total invested in Scheme B

Interest for Scheme A = (X)(1.1)^2 - X
Interest for Scheme B = (6100 - X)(.4)

.21X = 2440 - .4X
.61X = 2440
X = 4000

Final Answer: C

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by Arunkumar S » Sun Dec 18, 2016 8:14 am
*Both the scheme pay equal interests*(important data)
So S.I = C.I
Assume that Amount invested by "B" is "x"
So
A = 6100-x
B = X
S.I = PNR/100 = X(4)(10)/100 THIS IS FOR (B)
C.I = P[(1-R/100)^n-1] = (6100-X)[(1-1/100)^2 = 21(6100-X)/100
X(4)(10)/100 = 21(6100-X)/100
40X = 128100-21X = 61X =128100 ==> X = 2100
SO A= 6100-2100 = 4000
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