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CBSE Questions for Class 11 Commerce Applied Mathematics Basics Of Financial Mathematics Quiz 7 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Basics Of Financial Mathematics
Quiz 7
Naman purchased an old bike for Rs.
20000
. If the cost of his bike is depreciated at a rate of
5
%
per annum, then find the cost of the bike after
2
years?
Report Question
0%
Rs.
18050
0%
Rs.
15000
0%
Rs.
1900
0%
Rs.
18000
Explanation
Initial price of bike is Rs.
20000
, time
=
2
years, rate od depreciation
=
5
%
Cost of bike after
2
years
=
initial price of bike
×
(
1
−
rate of depreciation
100
)
time
Therefore, cost of bike after
2
years
=
20000
(
1
−
5
100
)
2
=
20000
(
19
20
)
2
=
18050
Thus, cost of bike after
2
years is Rs.
18050
.
Identify the two statements by which the principal sum can be calculated?
P : The sum amounts to Rs.
690
in
3
yrs at Simple Interest.
Q : The sum amounts to Rs.
750
in
5
yrs at Simple interest.
R : The rate of interest is
5
%
per annum.
Report Question
0%
P & Q will give the answer
0%
R & Q will give the answer
0%
P & R will give the answer
0%
Any two will give the answer
Explanation
Any two of three will give the answer because if amount, rate and number of years are given, we can easily find out the principal sum.
Priya invested a certain amount of money and got back an amount of Rs.
15000
. If the bank paid an interest of Rs.
5000
, what is the amount Priya invested?
Report Question
0%
5000
0%
8000
0%
15000
0%
10000
Explanation
Amount
=
Rs.
15000
Simple interest
=
Rs.
5000
Principal
=
Amount
−
simple interest
=
15000
−
5000
Principal
=
10000
Therefore, priya invested Rs.
10000
.
Depreciation increase the value of asset.
Report Question
0%
True
0%
False
Explanation
Depreciation is the process of reduction in the value of asset.
Depreciation is the permanent and continuous decrease in the book value of a depreciable fixed asset due to use, effluxion of time, obsolescence, expiration of legal rights or any other cause.
The present population of a city is
18522000
. If it has been increased at the rate of
5
%
per annum, find the population(approx) after
3
years.
Report Question
0%
2522200
0%
16000000
0%
21441530
0%
252200
Explanation
Given, Principal amount
18522000
, rate of interest
=
5
%
and period
=
3
years
Total Population
=
18522000
[
1
+
5
100
]
3
=
18522000
[
1
+
1
20
]
3
=
21441530
(approx)
Oscar earned Rs.
2400
as simple interest on Rs. `
4500
for
3
months. What is the annual rate of interest?
Report Question
0%
16.66
%
0%
17.77
%
0%
18.88
%
0%
19.99
%
Explanation
Given, principal
=
4500
, Interest
=
Rs.
2400
, period
=
3
months
We know the formula,
Rate of interest,
R
=
I
×
100
P
T
⇒
R
=
2400
×
100
4500
×
3
⇒
R
=
17.77
%
Therefore, the annual rate of interest is
17.77
%
.
To start a grocery shop, a woman borrowed Rs.
1
,
500
. If the loan was for four years and the amount of interest was Rs.
150
, what simple interest rate was she charged?
Report Question
0%
1.5
%
0%
2.5
%
0%
3.5
%
0%
4.5
%
Explanation
Given, principal
=
Rs.
1500
, Interest
=
Rs.
150
, period
=
4
years
We know the formula,
Rate of interest,
R
=
I
×
100
P
T
⇒
R
=
150
×
100
1500
×
4
⇒
R
=
2.5
%
Therefore, the simple interest is
2.5
%
.
You borrow Rs.
4
,
000
from a loan shark. If you owe Rs.
7
,
200
in
4
years, what would be the simple interest rate?
Report Question
0%
10
%
0%
20
%
0%
30
%
0%
40
%
Explanation
Given:
Principal
=
Rs.
4
,
000
Interest
=
7200
−
4000
=
3200
We know the formula,
Rate of interest,
R
=
I
×
100
P
T
⇒
R
=
3200
×
100
4000
×
4
⇒
R
=
20
%
Thus, the simple interest rate is
20
%
.
Sharmila got a Rs.
1300
loan for
5
years. She paid Rs.
100
in interest. What was the interest rate?
Report Question
0%
1.1
%
0%
1.3
%
0%
1.5
%
0%
1.7
%
Explanation
Given, principal
=
Rs.
1300
, Interest
=
Rs.
100
, period
=
5
years
We know the formula,
Rate of interest,
R
=
I
×
100
P
T
⇒
R
=
100
×
100
1300
×
5
⇒
R
=
1.5
%
Therefore, the interest rate is
1.5
%
.
The present population of a town is
14000
. If it increases at the rate of
10
%
per annum, what will be its population after
4
years?
Report Question
0%
10497
0%
20497
0%
30497
0%
40497
Explanation
Population after
4
years
=
Present Population
(
1
+
r
100
)
n
Population after
4
years
=
14000
×
(
1
+
10
100
)
4
=
14000
×
1.4641
=
20497
Population after
4
years
=
20497
The present population of a city is
15000
. If it increases at the rate of
4
%
per annum. Find its population after
2
years.
Report Question
0%
16224
0%
26224
0%
36224
0%
46224
Explanation
Given,
P
=
15000
,
r
=
4
%
,
n
=
2
years
Population after
2
years
=
P
[
(
1
+
R
100
)
2
]
=
15000
[
(
1
+
4
100
)
2
]
=
15000
×
1.0816
=
16224
Population after
2
years is
16224
.
The cost of a machine is Rs.
24500
. After two year the value of that machine was depreciated by
6
%
. Find the value of machine after two year.
Report Question
0%
11560
0%
21560
0%
31560
0%
41560
Explanation
Given: Principal
=
24500
,
r
=
6
%
,
n
=
2
Reduction
=
6
%
of Rs.
24500
per year
=
24500
×
6
×
2
100
=
2940
Value at the end of
2
years
=
24500
−
2940
=
Rs.
21560
Jenn borrowed Rs.
5
,
000
for
5
years and had to pay Rs.
1
,
500
simple interest at the end of that time. What rate of interest did she pay?
Report Question
0%
4
%
0%
5
%
0%
6
%
0%
7
%
Explanation
Given,
I
=
1500
,
P
=
5000
,
T
=
5
years
We need to find rate of interest
R
.
We know the formula,
Rate of interest,
R
=
I
×
100
P
T
⇒
R
=
1500
×
100
5000
×
5
⇒
R
=
6
%
A machine was purchased
3
years ago. Its value decreases by
5
%
every year. Its present value is
R
s
.23000
. For how much money was the machine purchased?
Report Question
0%
R
s
.16828
.45
0%
R
s
.26826
.07
0%
R
s
.36828
.56
0%
R
s
.46828
.80
Explanation
Given,
Depreciated value
A
=
R
s
.
23
,
000
Rate of depreciation
R
=
5
%
Time
T
=
3
years
So,
n
=
3
let the machine was purchased in
R
s
.
P
Depreciated value,
A
=
P
(
1
−
R
100
)
n
⇒
23000
=
P
×
(
1
−
5
100
)
3
⇒
23000
=
P
×
(
1
−
1
20
)
3
⇒
23000
=
P
×
(
19
20
)
3
⇒
23000
=
P
×
19
20
×
19
20
×
19
20
⇒
P
=
23
,
000
×
20
19
×
20
19
×
20
19
⇒
P
=
26
,
826.07
Therefore, the machine was purchased for
R
s
.26
,
826.07
.
Calculate the compound interest on a sum of Rs.
20000
at the end of
3
years at the rate of
10
%
p.a. compounded annually.
Report Question
0%
6620
0%
5620
0%
3410
0%
2386
Explanation
Given,
P
=
20000
,
r
=
10
%
,
n
=
3
years
A
=
P
[
(
1
+
r
100
)
n
]
A
=
20000
[
(
1
+
10
100
)
3
]
A
=
20000
×
11
10
×
11
10
×
11
10
A
=
Rs.
26620
Compound interest
=
Amount
−
Principal
C
.
I
.
=
26620
−
20000
C
.
I
.
=
Rs.
6620
A machine depreciates at the rate of
10
%
of its value at the beginning of a year. If the present value of a machine is
Rs.
4000
, find its value after
3
years.
Report Question
0%
Rs.
1916
0%
Rs.
2916
0%
Rs.
3916
0%
Rs.
4916
Explanation
Given:-
P
=
Rs.
4000
r
=
10
%
n
=
3
years
Now, as the value of the machine depriciates every year by
10
%
,
A
=
P
(
1
−
r
100
)
n
=
4000
(
1
−
10
100
)
3
=
4000
(
0.9
)
3
=
Rs.
2916
Hence, the value of the machine after
3
years is
Rs.
2916.
A company bought a car that had a value of Rs.
10000
. each year the value of the car depreciates by
15
%
. What is the value of the car at the end of
2
years.
Report Question
0%
7225
0%
6225
0%
5225
0%
4225
Explanation
Given,
P
=
10000
,
r
=
15
%
depreciation,
n
=
2
years
We know
A
=
P
[
(
1
−
r
100
)
n
]
⇒
A
=
10000
[
(
1
−
15
100
)
2
]
⇒
A
=
Rs.
7225
Therefore, the value of car after
2
years is Rs.
7225
.
The annual growth of capital of a company is
12
%
. At present, the capital is Rs.
35000
. What will be the capital after
4
years?
Report Question
0%
55500
0%
55073
0%
45500
0%
56430
Explanation
Given:
P
=
35000
,
r
=
12
%
,
n
=
4
years
We know
A
=
P
[
(
1
+
r
100
)
n
]
Putting all the given values, we get
A
=
35000
[
(
1
+
12
100
)
4
]
A
=
Rs.
55073
The price of a new car is Rs.
2500
. The price depreciates by
12
%
each year (p.a). Find its value at the end of
3
years.
Report Question
0%
1703.68
0%
2703.68
0%
3703.68
0%
4703.68
Explanation
Given,
P
=
2500
,
r
=
12
%
,
n
=
3
years
We know,
A
=
P
[
(
1
−
r
100
)
n
]
⇒
A
=
2500
[
(
1
−
12
100
)
3
]
⇒
A
=
Rs.
1703.68
Thus, the value of a car at the end of
3
years is Rs.
1703.68
.
The population of a town increases every year at rate of
10
%
. Find the population of the town three years hence if it is
13000
now.
Report Question
0%
17303
0%
27303
0%
37303
0%
47303
Explanation
Given,
P
=
13000
,
r
=
10
%
,
n
=
3
years
We know,
A
=
P
[
(
1
+
r
100
)
n
]
⇒
A
=
13000
[
(
1
+
10
100
)
3
]
⇒
A
=
17303
Hence, the population of the town
3
years is
17303
.
Grace bought a new van for Rs.
15000
. Each year, the value of her car depreciated by
10
%
. Calculate the value of car before
2
years.
Report Question
0%
12150
0%
22150
0%
32150
0%
42150
Explanation
Given,
P
=
15000
,
r
=
10
%
depreciated,
n
=
2
years
We know,
A
=
P
[
(
1
−
r
100
)
n
]
⇒
A
=
15000
[
(
1
−
10
100
)
2
]
⇒
A
=
Rs.
12150
The value of car before
2
years is Rs.
12150
.
The population of an invasive species of moth doubles every
5
years. If the initial population is
300
, what will be the population after
15
years?
Report Question
0%
900
0%
1200
0%
2000
0%
2400
Explanation
Population of an invasive species of moth doubles every
5
years.
Let the initial population be
x
So after
5
years,
population
will be
2
×
x
=
2
x
After
10
years,
population
will be
2
×
2
x
=
4
x
After
15
years,
population
will be
2
×
4
x
=
8
x
.
Here, initial population is
x
=
300
.
After
15
years,
population
will be
8
x
=
8
×
300
=
2400
Hence, the population after
15
years is
2400
If a
12
,
000
c
a
r
l
o
s
e
s
10
3
$ years?
Report Question
0%
8748
0%
8750
0%
8752
0%
8744
Explanation
After
1
s
t
year the value of car is
12000
(
1
−
10
100
)
=
10800
The value of car after
2
n
d
year is
10800
(
1
−
10
100
)
=
9720
The value of car after
3
r
d
year is
9720
(
1
−
10
100
)
=
8748
Which of the following is true about annuity?
Report Question
0%
It is sequence of equal instalments.
0%
It is sequence of unequal instalments.
0%
It is paid at unequal interval of time.
0%
None of these
Explanation
⇒
The true statement about annuity is
I
t
i
s
s
e
q
u
e
n
c
e
o
f
e
q
u
a
l
i
n
s
t
a
l
m
e
n
t
s
.
⇒
Series of payments at fixed intervals, guaranteed for a fixed number of years or the lifetime of one or more individuals.
⇒
Annuities are insurance products that provide long-term income through a stream of future payments.
⇒
While investment annuities save money for retirement and beneficiaries, structured settlement annuities stem from personal-injury legal cases, wrongful-death claims or lottery payouts. When unexpected circumstances arise and require immediate funds, you can sell these payments for a lump sum of cash.
The compound interest on a certain sum of money for 2 years at 10
%
per annum is RsFind the simple interest on the same sum at the same rate and for the same time.
Report Question
0%
R
s
.
350
0%
R
s
.
375
0%
R
s
.
380
0%
R
s
.
400
Explanation
Given that, the compound interest on certain sum of money for
2
years at
10
%
per annum is
R
s
.
420
To find out: The simple interest on the same amount at the same rate and for the same time.
Let the principal amount be
R
s
.
x
We know that,
C
I
=
P
[
(
1
+
R
100
)
T
−
1
]
Here,
C
I
=
R
s
.
420
,
R
=
10
%
a
n
d
T
=
2
∴
420
=
x
[
(
1
+
10
100
)
2
−
1
]
⇒
420
=
x
[
(
11
10
)
2
−
1
]
⇒
420
=
x
[
(
121
100
)
−
1
]
⇒
420
=
x
(
121
−
100
100
)
⇒
x
=
420
×
100
21
∴
x
=
R
s
.
2000
Now, we also know that,
S
I
=
P
×
R
×
T
100
Here,
P
=
R
s
.
2000
,
R
=
10
%
a
n
d
T
=
2
∴
S
I
=
2000
×
10
×
2
100
⇒
S
I
=
40000
100
∴
S
I
=
R
s
.
400
Hence, the simple interest on the same amount, at the same rate of interest and for the same time is
R
s
.
400
.
Annuity where payments are made at the end of each payment period, i.e. 1st
payment is made at the end of the 1
st
payment interval, and so on, is known as
Report Question
0%
Perpetual annuity
0%
Contingent annuity
0%
Ordinary annuity
0%
Immediate annuity
Explanation
Answer is Ordinary or Immediate Annuity.
An ordinary annuity or immediate annuity is where payments are made at the end of each payment period, i.e. 1
st
payment is made at the end of the 1
st
payment interval, and so on. Examples are repayment of car loans, house mortgage etc.
A contingent annuity is one where the term depends upon some event whose occurrence is not fixed. An example is periodic payments of life insurance premiums which stop when the person dies.
A perpetual annuity is an annuity whose term does not end, i.e. it extends till infinity. Thus there is no last payment; they go on forever. An example is freehold property, where you can earn rent in perpetuity.
The C.I. on a certain sum for
2
years is Rs
410
and S.I. is Rs
400.
What is the rate of interest per annum?
Report Question
0%
10
%
0%
8
%
0%
5
%
0%
4
%
Explanation
⇒
It is given that simple interest in 2 years = Rs.400
⇒
Simple interest for 1 year = Rs.200
⇒
Hence, Compound interest for 1 year = Rs.200
⇒
Given compound interest for 2 years = Rs. 410
∴
Compound interest for 2 years = Compound interest 1st year + Compound interest 2nd year
⇒
Compound interest for 2nd year = Rs.410 - Rs.200 = Rs.210
⇒
That is interest on Rs.200 for 1 year = Rs.210-Rs.200=Rs.10
⇒
We know
I
=
P
×
T
×
R
100
⇒
10
=
200
×
1
×
R
100
⇒
2
R
=
10
∴
R
=
5
%
The price of commodity
X
increases by
40
paise every year, while the price of commodity
Y
increases by
15
paise every year. If in
2001
, the price of commodity
X
was Rs.
4.20
and that of
Y
was Rs.
6.30
, in which year commodity
X
will cost
40
paise more than the commodity
Y
?
Report Question
0%
2010
0%
2011
0%
2012
0%
2013
Explanation
Suppose commodity
X
will cost
40
paise more than
Y
after
z
years.
Then,
(
4.20
+
0.40
z
)
−
(
6.30
+
0.15
z
)
=
0.40
⇒
0.25
z
=
0.40
+
2.10
⇒
z
=
2.50
0.25
=
250
25
=
10
.
∴
X
will cost
40
paise more than
Y
10
years after
2001
which is
2011
.
Three types of annuities are
Report Question
0%
Annuity certain
0%
Annuity contingent
0%
Annuity perpetual
0%
All of the above
Explanation
⇒
Three types of annuities are :
(
1
)
A
n
n
u
i
t
y
c
e
r
t
a
i
n
- Annuity that, as a minimum, guarantees a fixed number of payments. It continues over the life of the annuitant, even if he or she lives beyond the number of payments specified in the annuity contract. In case the annuitant dies before exhausting the payments, a named beneficiary continues to receive the remaining number. Also called life annuity certain or life annuity certain and continuous.
(
2
)
A
n
n
u
i
t
y
c
o
n
t
i
n
g
e
n
t
- An annuity arrangement in which the beneficiary does not begin receiving payments until a specified event occurs. A contingent annuity may be set up to begin sending payments to a beneficiary upon the death of another individual who wishes to ensure financial stability for the beneficiary, or upon retirement or disablement of the beneficiary.
(
3
)
A
n
n
u
i
t
y
p
e
r
p
e
t
u
a
l
- Annuity derived from an asset (such as an income generating security) where the life span of the annuitant (security holder or his or her beneficiary) is of no consequence.
The difference between simple interest and compound on
R
s
.
1200
for one year at
10
% per annum reckoned half-yearly is:
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0%
R
s
.
2.50
0%
R
s
.
3
0%
R
s
.
3.75
0%
R
s
.
4
0%
None of these
Explanation
S
.
I
.
=
R
s
.
(
1200
×
10
×
1
100
)
=
R
s
.
120
.
C
.
I
.
=
R
s
.
[
1200
×
(
1
+
5
100
)
2
−
1200
]
=
R
s
.
123
.
∴
D
i
f
f
e
r
e
n
c
e
=
R
s
.
(
123
−
120
)
=
R
s
.
3
.
0:0:2
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Answered
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Incorrect : 0
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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers
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