ML Aggarwal Solution Class 9 Chapter 2 Compound Interest Exercise 2.2

 Exercise 2.2

Question 1

Sol :

using the formula , we get

A=5000(1+6100)2  A=P(1+r100)n

=5000×(106100)2

=5000×53×5350×50

= 2×53×53

= 106×53

= 5618

C.I=5618-5000

=618


Question 2

Sol :

Using formula , we get

A=8000(1+10100)4

=8000(1+110)4

=8000(10+110)4

=8000(1110)4

=8000(11×11×11×1110×10×10×)

=585645

=11712.8

C.I=11712.8-8000

=3712.80


Question 3

Sol :

Since, rate of interest per annum is 5%, therefore,
rate of interest per conversion (half yearly)=2.5%
As the money compounded half yearly.
Number of conversion periods=2

A=P(1+r100)n
A=7400(1+2.5100)2
A=7400(1+140)2
A=7400(40+140)2
=7400×41×4140×40
=7774.625
∴Amount =7774.625

Question 4

Sol :

Since , rate of interest per annum 10% , therefore rate of interest per conversion=5%
As the money compounded semi annually,
n(number of conversion periods)=3

A=P(1+r100)n
=5000(1+5100)3
=5000(1+120)3
=5000(1+2120)3
=5000×21×21×2120×20×20
=2.5×21×21×214

Compound Interest=5788.125-5000
=788.125

Question 5

Sol :

Since, the rate of interest per 12 months (annum) therefore rate of interest per conversion =14×4%
=1 %

As the money compounded quarterly
n(number of conversions)=93
=3

A=P(1+r100)n
=10000(1+1100)3
=10000(100+1100)3
=100000×101×101×101100×100×100
=103030.10

Compound Interest=103030.10-100000
=3030.10

Question 6

Sol :

Principal=4800, n=2 and R=5%
Simple Interest=P×n×R100
=4800×2×5100
=480

A=(1+r100)nP
=4800(1+5100)2
=4800(1+120)2
=4800(20+120)2
=4800×21×2120×20
=12×21×21
=5292

C.I=5292-4800=492

∴Difference=492-480
=12(C.I-S.I)

Question 7

Sol :

Principal=2500, n=2 and R=4% per annum
Simple Interest=P×n×R100
=2500×2×4100
=200

Given that compound Interest revoked semi annually, therefore n=4 and R=42=2%

A=2500(1+2100)4 A=P(1+r100)n
=2500(1+2100)4
=2500(1+150)4
=2500(5150)4
=2500×51×51×51×5150×50×50×50
=67652012500
=2706.80

Compound Interest=2706.80-2500
=2206.80

∴Difference between simple interest and compound interest is
C.I-S.I=206.080-200
=6.08

Question 8

Sol :

Given that

Principal for the first year=2000

Interest for the first year=2000×1×4100

=80

Principal for the second year=2080

Interest for the second year=2080×1×3100

=62.40

Compound Interest=80+62.40

=142.40


Question 9

Sol :

A=P(1+r1100)(1+r2100)(1+r3100)

=3125(1+4100)(1+5100)(1+6100)

=3125(1+125)(1+120)(1+350)

=3125×26×21×5325×20×50

=144694

=3617.25

∴Compound Interest=3617.25-3125

=492.25


Question 12

Sol :

Given that A=132651, n=3, r=2

A=P(1+r100)n

132651=P(1+r100)3

P(1+150)3=132651

P×51×51×5150×50×50

=132651

P=132651×50×50×5051×51×51

P=132651×125000132651

P=125000


Question 13

Sol :

A=P(1+r100)n

=P(1+4100)2

=P(1+125)2

=P×26×2625×25

C.I.=26×2625×25(P)P

=676P625P625

=51P625

5712=51P625

5712×62551=P

P=70000


Question 14

Sol :

A=1275, P=1200, n=1 and R=?

1275=1200(1+r100)n

r=127512100

r=7512

r=6.25%

C.I for second year=1275×1×6.25100

=7968.75100

=79.68

∴Compound Interest for second year

=80


Question 15

Sol :

Given that A=2500, P=2304, n=2 and R=?
2500=2304(1+r100)2
25002304=(1+r100)2
100+r100=25002304
100+r100=(5048)2
100+r=100×5048
r=500048100
r=500048004800
r=2004800
r=124%
r=4.16%
r=416%

Question 16

Sol :

Given that
A=2516P , n=2 and r=?
A=P(1+r100)n
P=(1+r100)2=2516P
100+r100=2516
100+r100=54
r=125-100
r=25%

Question 17

Sol :

A=P(1+r100)n
2315.25=2000(1+r100)3
2315.252000=(100+r100)3
32.315252.000=100+r100
2315.25×100×5=(100+r)3
32315.25×500=100+r
31157625=100+r
3(105)3=100+r
r=105-100=5%


Question 18

Sol :

A=48620.25, P=40000, r=R2 , n=2×2=4
A=P(1+r100)n
48620.25=40000(1+R200)4
[(200+R)2]2=48620.25×200×200×200×20040000
(200+R)2=2(220.5)2(200)2
200+R=2(220.5)2(200)2
200+R=2200×220.5
R=210-200
R=10%

Question 19

Sol :

G/T A=216125P ,n=3
A=P(1+r100)n
216125P=P(1+r100)3
100+r100=3216125
100+r=65×100
r=120-100=20%
∴Rate of interest=20×2%
=40%


Question 20

Sol :

Given that A=88200, P=80000, n=2
A=P(1+r100)2
8820080000=(100+r100)2
221×21×220×20×2=100+r100
21×10020=100+r
r=105-100
r=5%

Principal= 88200

C.I=88200×1×5100
=882×5
=4410

∴Amount to be paid after 3 years=P+C.I
=88200+4410
=92610

∴A=92610

Question 21

Sol :

Let the rate of compound interest be r% per annum
The amount afters will be the principal for the third year

5556.50=5292(1+r100)

1+r100=5556.605292

r100=5556.6052925292

r=264605292

r=5%

∴The rate of compound Interest=5% per annum 
Let the original sum be P

5292=P(1+5100)2

P×21×2120×20=5292

P=5292×400441

P=12×400

P=4800

∴Original Sum is 4800

Question 22

Sol :

Let the rate of compound Interest be r% per annum 
The amount after 3 years will be the principal for the 4th year

=878.46

=798.60(1+r100)

878.46798.60=100+r100

r100=879.46798.60798.60

r=7986798.60

r=10%

∴The rate of compound interest = 10% per annum 
Let the original sum be P

798.6=P=(1+10100)3

798.6=P×1110×1110×1110

798.6×1000121×11=P

P=79860012×11

P=600

∴Original amount =600

Question 23

Sol :

A=P(1+r100)n

=17576=15625(1+4100)n

1757615625=(2625)n

(2625)3=(2625)n

Bases are equal then power should be equal n=3

Question 24

Sol :

P=12500, C.I.=3246.40 
Rate = 8%

A=12500+3246.40
=15746.40

Using 15746.40=12500(1+8100)n

15746.4012500=(2725)n

=(2725)3=(2725)n

∴Bases are equal then powers should be equal

∴Number of years=3

n=3

Question 25

Sol :

Here P=16000  , A=18522

As the interest is compounded semi annually
r crate of interest per conversion period 12 of 10%
=5%

Let the number of conversion periods be n, then

18522 =16000(1+5100)n

1852216000=(2120)n

(2120)4=(2120)n

∴Bases are equal then powers should be equal
n=4


Question 26

Sol :

Given that,
A=278250, 
R1=5 , r2=6

2782.50=P(1+5100)(1+6100)

2782.50=P(2120)(5350)

P=2782.50×20×5021×53

P=10000

∴Sum=10000



Question 27

Sol :

CI1=P1×1×r100

P1r=225×100

P1r=22500  P1=22500r

CI2=P2×1×r100

240=22500r×1×r100+r   [P2=22500r+2]

r=240225

Question 27

Sol :

G/I
Compound Interest for the first year CI1=225
Compound Interest for the second year CI2=240
CI1=P1×N×R100

225=P1×1×R100

P1R=22500

∴Principal  for the second year 
=22500R+CI1
=22500R+225

CI2=P2×N×R100
24×1000=P2×1×R
24000=22500R×R+225(R)
225R=24000-22500

R=1500225

(i) Rate pf interest =623 %

(ii) Original Sum
CI1=P1×N×R100

P1=225×100×320
=225×15
∴Original Sum=3375

(iii) Compound Interest for the third year CI3
Principal=3375+225+240
=3840

CI3=3840×20×13×100

Compound Interest for the third year=256

Question 28

Sol :

Simple Interest =P×N×R100

=P×2×5100=P10

Compound Interest =P[(1+5100)21]
=P[(2120)21]
=P[441400400]
=41P400

Given that
41P400P10=25
41P40P400=25
1P=2500×4
P=2500×4
Principal=10000

Question 29

Sol :

Simple Interest =P×1×10100=P10

Compound Interest =P[(1+5100)21]

=P[(1+120)21]

=P[(2120)21]

=P[441400400]=41P400


41P400P10=15

41P-40P=400×15

P=6000

Question 30

Sol :

Given that

Compound Interest for the first year =P[(1+r100)11]

1250=P[100+r-100]

1250=Pr

P=1250r

Compound Interest for the second year
=P[(1+r100)21]
1375=P[(1+1250P×100)21]
1375=P[(P+12.50)2P21]
1375=P[P2+25P+156.25P2P2]

Question 30

Sol :

Compound Interest of 2nd conversion period-Compound Interest of 1st conversion=S.I on the Interest of 1st conversion period

∴1375-1250=S.I on Interval on the Interest of 1st Conversion Period

125=1250×1×R100

∴Rate of Interest=10%


Question 31

Sol :

Simple Interest for three years=225

x×r×3100=225

x×r=225003 [∵x=Principal and r=rate of Interest]


Compound Interest for two years=153

P[(1+r100)21]=153

x[(100+r)2(100)2(100)2]=153

x[1002+r2+200r-1002]=153×10000

(xr)r+200xr=1530000

225003[r+200]=1530000

r=68×3-200
r=204-200=4

∴Rate of Interest=4%

∴Simple Interest on three years=225

P×3×4100=225

P=25×75

P=1875

Question 32

Sol :

Compounded annually

First, we found the amount after one year
A=P(1+r100)n
=P(1+10100)n
=8000(110100)1
=8000×1.1=8800

Principal for the second half year=8800

Interest for the next half year=8800×10×12100
=440

Compounded annually Interest=9240

As the money Invested for 112 years , therefore

n=number of conversion periods=3

A=P(1+r100)n
=8000(1+5100)3
=8000(2120)3
=8000(1.05)3
=9261

Interest for 112 years compounded semi annually=9261

difference between compounded semi annually and annually
=9261-9240
=21

Question 33

Sol :

Compounded Annually

Compound Interest for two years
=P[(1+r100)n1]
=P[(1+20100)n1]
=P[(1+15)21]
=P[(65)21]
=P[362525]
=11P25
=0.44P

Compounded half yearly,

Compound Interest for two years =P[(1+r100)n1]
=P[(1+10100)41]
=P[(11100)41]
=P[(1.1)41]
=P[1.4641-1]
=0.464P

∴Given that 0.464P-0.44P=482

0.024P=482

P=4820.0241

P=20000

∴Sum of money lent out=20000

Question 34

Sol :

Amount after one year=P(1+r100)n
13230=P(1+r/2100)2


Amount after 112 year =P(1+r100)n
13891.50=P(1+r/2100)3
13891.50=13230(1+r/2100)2(1+r/2100)2

[P=13230(1+r/2100)2]
[P=13230(1+5100)2]
=13230×20×2021×21
=132301.1025
∴Sum=12000


13891.5013230=1+r200

13891.5013230=200+r200

10.5×20=200+r
210=200+r
r=10
∴Rate of Interest=10%

Sum=1323011.10251
=12000

Comments

Popular posts from this blog

ML Aggarwal Solution Class 10 Chapter 15 Circles Exercise 15.1

ML Aggarwal Solution Class 9 Chapter 20 Statistics Exercise 20.2

ML Aggarwal Solution Class 9 Chapter 3 Expansions Exercise 3.2