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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GMATGuruNY
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We can PLUG IN THE ANSWERS, which represent the amount in invested in A.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
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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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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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
GMAT assassins aren't born, they're made,
Rich
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
GMAT assassins aren't born, they're made,
Rich
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Arunkumar S
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*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
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
