10. Find the sum, invested at 10% compounded
annually, on which the interest for the third
year exceeds the interest of the first year
by 252 (without using formula)
Answers
Answer:
Question 5:
Find the sum, invested at 10% compounded annually, on which the interest for the third year exceeds the interest of the first year by Rs.252.
solution:
Let the sum of money be Rs 100
Rate of interest= 10%p.a.
Interest at the end of 1st year= 10% of Rs100= Rs10
Amount at the end of 1st year= Rs100 + Rs10= Rs110
Interest at the end of 2nd year=10% of Rs110 = Rs11
Amount at the end of 2nd year= Rs110 + Rs11= Rs121
Interest at the end of 3rd year=10% of Rs121= Rs12.10
Difference between interest of 3rd year and 1st year
=Rs12.10- Rs10=Rs2.10
When difference is Rs2.10, principal is Rs100
When difference is Rs252, principal =
100×252/2.10 = 12000
Answer:
=12000
Answer:
let the sum=Rs.P
rate=10%
C.I for 3 year=Amount at the end of 3rd year-Amount at the end of 2nd year
=P(1+
100
10
)
3
−P(1+
100
10
)
2
=P[1+(
10
11
)
3
−(
10
11
)
2
]
=P(
10
11
)
2
−[
10
11
−1]
=P×
100
121
×
10
1
=
1000
121
P
C.P for 1 st year=
100
P×10×1
=
10
P
Then according to the question
⇒
1000
121
−
10
P
=252
⇒
1000
121P−100P
=252
⇒
1000
21P
=252
⇒P=
21
252×1000
=Rs.12000