a man deposited Rs.10,000 on saving account at 5%p.a interest compounded yearly and another sum on fixed deposit account at 8%p.a interest compounded half early. After one year the interest on fixed deposit account was Rs.152.80 more than the interest on the saving account, find the total amount of money in his two accounts at the end of the year.
Answers
Answer:
According to the given question
Intersect per year =10,000×5 %
=10,000×
100
5
=500
Now,
Amount in I
st
years =10,000
Amount in 2
nd
year =10,000+intersect
=10,000+500=10,500
Amount in 3
rd
year =10,500+Intersect
=10,500+500=11,000
Then our series becomes
10,000,10,500,11000.
This is an AP
where a=10,000
d=10,500−10,000=500
time (n)=15 year
Then,
Amount in 15
th
year =a+(n−1)d
=10,000+(15−1)×500
=10,000+14×500
10,000+7000
17,000
Thus amount in 15
th
year in Rs.17,000
Amount of per 20 year =a
n
=a+(n−1)d
10,000+(21−1)×500
10,000+10,000
20,000
Thus, the amount in 20
th
years is
Rs.⇒20,000/−
Hence, this is the answer.
Answer:
the total amount of money in his two accounts at the end of the year.
C.A1 + C. A2 = 10500 + 8652.80
=Rs.19152.80
Step-by-step explanation:
given
deposited
saving account
p = Rs 10000
R = 5%
c.i = ?
According to the question.
c.i = p [ ( 1+ R\100) ^T-1]
= 10000 [ ( 1+ 5\100) ^1-1]
= 10000 [ 1.05-1]
= 10000 X 0.05
=500
;. C. I =Rs. 500
Again,
According to the question.
let p= Rs. x
R = 8%
semi - c. I =p[ ( 1+ R\200) ^2T-1]
= x[1( 1+8\200) ^2×1 -1]
= x[(1.04) ^2-1]
= x[1.0816 - 1]
= 0.0816x
Now,
semi - c.i. - yearly c. I =Rs.152.80
0.0816x - 500 = 152.80
0.0816x = 652.80
x = 652.80\0.0816
X = Rs. 8,000
so the principal pries (p) =Rs 8000
Agai
amount saving account (C.A1)
= 10000+500
= Rs. 10500
and
fixed amount saving account (c. A2)
= 8000 +0.0816 ×8000
= 8652.80
the total amount of money in his two accounts at the end of the year.
C.A1 + C. A2 = 10500 + 8652.80
=Rs.19152.80