“x” amounts to “y” when invested for “n” years is compound interest compounded annually. “y” amounts to “z” when invested for “2n” years in compound interest compounded annually at the same interest as before. Which of the following is true?
1. “xz” amount to “y^2” when invested for “n” years in compound interest compounded annually at the same interest rate.
2. “x^2” amount to “yz” when invested for “n” years in compound interest compounded annually at the same interest rate.
3. “y^2” amount to “xz” when invested for “n” years in compound interest compounded annually at the same interest rate.
4. “y^2” amount to “xz” when invested for “2n” years in compound interest compounded annually at the same interest rate
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Last update on October 12, 2:30 am by Rahil Choudhary.
Given
Y=X * p^N P =(1+r%)
Z=Y*p^2N
Let us analyze each option
Option 1) x*z*p^N = Y. Z and not Y^2
Option 2) x^2p^N =y^2/p^2N *p^N not equal to YZ
Option 3) y^2*P^n =y*y*p^N =x*y*p^2N =x*z (matches)
Hence option 3
Answer is 3)
For the first case,
P=x
A=y
T=n
=> y=x(1+r/100)^n
Similarly,
x=y(1+r/100)^2n
From both the equations,
z=x(1+r/100)^3n
In option 3,
P=y^2
T=n
=>A=y^2(1+r/100)^n
Put the value of y^2
A=x^2(1+r/100)^3n
Also, xz=x^2(1+r/100)^3n
And in option 3 A is xz
Therefore, right answer is 3
Option 3
rate of interest be a% pa
duration be b years.
From first statement, y=x * [1+(a/100)]^b
From second statement, z = y * [1+(a/100)]^2b
Dividing 2 by the 1, we get (z/y) = (y/x) * [1+(a/100)]^b
Hence, xz = y^2 * [1+(a/100)]^b
Answer is 3
duration be d years.
rate of interest be i% pa
From 1, y=x * [1+(i/100)]^d
From second statement, z = y * [1+(i/100)]^2d
Dividing 2 by the 1, we get (z/y) = (y/x) * [1+(i/100)]^d
Hence, xz = y^2 * [1+(i/100)]^d
Statement 1, y=x * [1+(r/100)]^n
Statement 2, z = y * [1+(r/100)]^2n
2/1,
We get (z/y) = (y/x) * [1+(r/100)]^n
Multiply by y both side,
Hence, 3
Answer is 3