Q1.
When interest is compounded monthly, then rate of interest divided by _____________ .
Solution.
12
Solution
Q2.
Amount of 3rd year (in the case of compound interest) will be principal of 4th year . (True/False)
Solution
True
Solution
Q3.
Find the difference between SI and CI for two years on Rs.1500 at 15% p.a.
Solution.
Difference of SI & CI = P(
r/100
)
2
= 1500(
15/100
)
2
=
1500 x 225/100 x 100
=
3375/100
= 33.75
Solution
Q4.
For two consecutive year compound interest on a particular sum is Rs.60 and Rs.70 respectively , find the rate of C.I .
Solution.
C.I(rate) =
(c2 - c1)/
c1
x 100
(70 - 60)/60
x 100
=
1000/60
= 16.6%
Solution
Q5.
The SI and CI for two years on a certain sum is Rs.310 and Rs.340 respectively find the sum.
Solution.
2pr/100
=310 ⇒ pr = 15500
P(
r2/100
) =
Difference of SI & CI ⇒ P(
r/100
)
2
= 30
pr(r) = 3,00,000
15500(r) = 3,00,000 ⇒ r = 19.35%
Solution
Q6.
On a specific money the ratio of CI for 3 year and 2 year is 21:20 respectively, find the rate of interest .
Solution.
21 x = P(1 +
r/100
)
3
⇒ 20x = P(1 +
r/100
)
2
21/20
= 1 +
r/100
⇒
r/100
=
1/20
⇒ r = 5%
Solution
Q7.
A sum of money becomes Rs760 in 2 years and becomes Rs836 in 3 years , find the compound interest .
Solution.
836 = P(1 +
r/100
)
3 ⇒
760 = P(1 +
r/100
)
2
836/760
= 1 +
r/100
⇒
76/760
=
r/100
⇒
r = 10%
Short Cut :- rate of interest =
A2 - A1/A1
x 100
=
76/760
x 100 = 10%
Solution
Q8.
In CI a sum of money becomes double in 6 years , in how many years it will be 8 times .
Solution.
(n
1)
1/t1 = (n
2)
1/t2 ⇒ 2
1/5 = (8)
1/t2
2
1/5 = 2
3/t2 ⇒
1/5
=
3/t2
⇒ t
2
= 15
Solution
Q9.
A sum of money becomes
36/25
times in 2 years
find the rate of CI .
Solution.
A = P(1 +
r/100
)
n ⇒
36P/25
= P(1 +
r/100
)
2
6/5
= 1 +
r/100
⇒
1/5
=
r/100
⇒ r = 20%
Solution
Q10.
12% compounded halfyearly = _____________% compounded per annum .
Solution.
i
e = (1 +
1/m
)
m
- 1
i =
r/100
,
m = no. of compounding
i
e = (1 +
0.12/2
)
2 - 1 ⇒
i
e = (1.06)
2 - 1 = 0.1236
r = 12.36%
Solution
