Quantitative Aptitude > Interest
SIMPLE & COMPOUND INTEREST MCQs
Compound Interest, Simple Interest, Interest (combined)
Compound Interest (CI) is the interest earned on the principal amount and also on the accumulated interest of the previous periods.
Formula for Compound Interest (CI) = P (1 + r/n) ^ nt
Where, P = Principal Amount
r = Rate of Interest
n = Number of times the interest is compounded
t = Time in years
Given,
P = Rs 6950
r1 = 6% p.a. for the first two years
r2 = 9% p.a. for the third year
n = 2 (half-yearly)
t = 3 years
Calculation:
Compound Interest (CI) for first two years
= P (1 + r1/n) ^ nt
= Rs 6950 x (1+ 6/2) ^ (2 x 3)
= Rs 6950 x (1 + 3) ^ 6
= Rs 6950 x 729
= Rs 50,530
Compound Interest (CI) for third year
= P (1 + r2/n) ^ nt
= Rs 6950 x (1 + 9/2) ^ (2 x 1)
= Rs 6950 x (1 + 4.5) ^ 2
= Rs 6950 x 20.25
= Rs 140,612.50
Total Compound Interest (CI)
= CI for first two years + CI for third year
= Rs 50,530 + Rs 140,612.50
= Rs 191,142.50
Compound Interest (CI) for 3 years
= Total CI - Principal Amount
= Rs 191,142.50 - Rs 6950
= Rs 184,192.50
Compound Interest (CI) for 3 years
= CI for 3 years - CI for first two years
= Rs 184,192.50 - Rs 50,530
= Rs 133,662.50
Therefore, the compound interest on Rs 6950 for 3 years if interest is payable half-yearly, at the rate of 6% p.a. for the first two years and at the rate of 9% p.a. for the third year is Rs 1590.
If you think the solution is wrong then please provide your own solution below in the comments section .
Amount = Rs.\(\left[1600\times\left(1+\frac{5}{2\times100}\right)^{2}+1600\times\left(1+\frac{5}{2\times100}\right)\right]\)
= Rs. \(\left[1600\times\frac{41}{40}\times\frac{41}{40}+1600\times\frac{41}{40}\right]\)
= Rs. \(\left[1600\times\frac{41}{40}\left(\frac{41}{40}+1\right)\right]\)
= Rs. \(\left[\frac{1600\times41\times81}{40\times40}\right]\)
= Rs. 33.21
So, C.I. = Rs. (3321 - 3200) = Rs. 121
Let the sum be Rs. x. Then,
C.I. = \(\left[x\left(1\frac{4}{100}\right)^{2}-x\right] = \left(\frac{676}{925}x-x\right) = \frac{51}{625}x.\)
S.I. = \(\left(\frac{x\times4\times2}{100}\right)= \frac{2x}{25.}\)
So, \(\frac{51x}{625}-\frac{2x}{25}=1\)
x=625.
Let P = Rs. 100. Then, S.I. Rs. 60 and T = 6 years.
So, R = \(\left(\frac{100\times60}{100\times6}\right)\) =10% p.a.
Now, P = Rs. 12000. T = 3 years and R = 10% p.a.
So, C.I. = Rs. \(\left[12000\times\left\{\left(1+\frac{10}{100}\right)^{3}-1\right\}\right]\)
= Rs. \(\left(12000\times\frac{331}{1000}\right)\)
= 3972
C.I. when interest compounded yearly = Rs. \(\left[5000\times\left(1+\frac{4}{100}\right)\times\left(1+\frac{\frac{1}{2}\times4}{100}\right)\right]\)
= Rs. \(\left(5000\times\frac{26}{25}\times\frac{51}{50}\right)\)
= Rs 5304.
C.I. when interest compounded half- yearly = Rs. \(\left[5000\times\left(1+\frac{2}{100}\right)^{3}\right]\)
= Rs \(\left(5000\times\frac{51}{50}\times\frac{51}{50}\times\frac{51}{50}\right)\)
= Rs. 5306.4
So, Difference = Rs. (5306.04 - 5304) = Rs. 2.04
Amount = Rs. (30000 + 4347) = Rs. 34347.
Let the time be n years.
Then , \(30000\left(1+\frac{7}{100}\right)^{n}= 34347\)
\(\Rightarrow\left(\frac{107}{100}\right)^{n}=\frac{34347}{30000}=\frac{11449}{10000}=\left(\frac{107}{100}\right)^{2}\)
So, n = 2 years.