3)
P= ? 12000, R-6% , T=2years then I=_____( )
A) Rs 1320 B) Rs 3120
C) Rs 1,220 D)Rs 1,440
Answers
Answer:
option D 1440
hope it's helpful
Answer:
Here, Principal (P) = Rs.12,000, Time (T) = 2 years, Rate of interest (R) = 6% p.a.
Simple Interest = \frac{P\times R\times T}{100}=\ \frac{12000\times6\times2}{100}=Rs.\ 1440
100
P×R×T
=
100
12000×6×2
=Rs. 1440
Had he borrowed this sum of 6% p.a., then
Compound interest = P\left(1+\frac{R}{100}\right)^n-PP(1+
100
R
)
n
−P
= 12000\left(1+\frac{6}{100}\right)^2-1200012000(1+
100
6
)
2
−12000
= 12000\left(1+\frac{3}{50}\right)^2-1200012000(1+
50
3
)
2
−12000
= 12000\left(\frac{53}{50}\right)-1200012000(
50
53
)−12000
= 12000\times\frac{53}{50}\times\frac{53}{50}-1200012000×
50
53
×
50
53
−12000
= Rs. 13,483.20 – Rs. 12,000
= Rs. 1,483.20
Difference in both interests
= Rs. 1,483.20 – Rs. 1,440.00 = Rs. 43.20