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13 May 2016, 05:07
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
59% (02:50) correct
41%
(02:59)
wrong
based on 252
sessions
History
Date
Time
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Not Attempted Yet
John invests $x at the semi-annual constant compounded rate of 2 percent and also does $10,000 at the quarterly constant compounded rate of 4 percent. If the interests are the same after 1 year, what is the value of x?
A. $20,001
B. $20,101
C. $20,201
D. $20,301
E. $20,401
A. $20,001
B. $20,101
C. $20,201
D. $20,301
E. $20,401
15 May 2016, 21:13
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From x(1+2%/2)^2-x=10000(1+4%/4)^4-10000, we get x(1.1^2-1)=10000(1.1^4-1)=10000(1.1^2-1)(1.1^2+1). (1.1^2-1) can be canceled out in either sides. So we get x=10000(1.1^2+1)=20,201. Hence, the correct answer is C.
15 May 2016, 23:31
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A = P(1+r/n)^nt
A= total amount accrued
P = principal deposited
r = rate of interest in decimal form
n = number of times per year, interest compounded
t = time in number of years.
.
x(1+0.02/2)^2 - x = 10,000(1+0.04/4)^4 - 10,000 [ when the principal is subtracted from the total amount accrued , the resulting difference is the interest portion and question states interests are equal)
=> x[(1.01)^2 - 1] = 10,000[(1.01)^4 - 1]
=> x[(1.01)^2 - 1] = 10,000[(1.01)^2+1][(1.01)^2-1] --> Using a^2-b^2 = a+b X a-b formula and cancel common expression on both sides
=> x = 10,000(1.0201+1) = 20,201.
.
Hence answer is C.
A= total amount accrued
P = principal deposited
r = rate of interest in decimal form
n = number of times per year, interest compounded
t = time in number of years.
.
x(1+0.02/2)^2 - x = 10,000(1+0.04/4)^4 - 10,000 [ when the principal is subtracted from the total amount accrued , the resulting difference is the interest portion and question states interests are equal)
=> x[(1.01)^2 - 1] = 10,000[(1.01)^4 - 1]
=> x[(1.01)^2 - 1] = 10,000[(1.01)^2+1][(1.01)^2-1] --> Using a^2-b^2 = a+b X a-b formula and cancel common expression on both sides
=> x = 10,000(1.0201+1) = 20,201.
.
Hence answer is C.
