ML Aggarwal Solution Class 9 Chapter 2 Compound Interest Exercise 2.2

 Exercise 2.2

Question 1

Sol :

using the formula , we get

$A=5000\left(1+\frac{6}{100}\right)^2$  $A=P\left(1+\frac{r}{100}\right)^n$

$=5000\times \left(\frac{106}{100}\right)^2$

$=5000\times \frac{53\times 53}{50\times 50}$

= 2×53×53

= 106×53

= 5618

C.I=5618-5000

=618


Question 2

Sol :

Using formula , we get

$A=8000\left(1+\frac{10}{100}\right)^4$

$=8000\left(1+\frac{1}{10}\right)^4$

$=8000\left(\frac{10+1}{10}\right)^4$

$=8000\left(\frac{11}{10}\right)^4$

$=8000\left(\frac{11\times 11\times 11\times 11}{10\times 10\times 10\times }\right)$

$=\frac{58564}{5}$

=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

$A=P\left(1+\frac{r}{100}\right)^n$
$A=7400\left(1+\frac{2.5}{100}\right)^2$
$A=7400\left(1+\frac{1}{40}\right)^2$
$A=7400\left(\frac{40+1}{40}\right)^2$
$=7400\times \frac{41\times 41}{40\times 40}$
=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

$A=P\left(1+\frac{r}{100}\right)^n$
$=5000\left(1+\frac{5}{100}\right)^3$
$=5000\left(1+\frac{1}{20}\right)^3$
$=5000\left(1+\frac{21}{20}\right)^3$
$=\frac{5000 \times 21\times 21 \times 21}{20\times 20\times 20}$
$=\frac{2.5\times 21\times 21\times 21}{4}$

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 $=\frac{1}{4}\times 4 \%$
=1 %

As the money compounded quarterly
n(number of conversions)$=\frac{9}{3}$
=3

$A=P\left(1+\frac{r}{100}\right)^n$
$=10000\left(1+\frac{1}{100}\right)^3$
$=10000\left(\frac{100+1}{100}\right)^3$
$=100000\times \frac{101\times 101\times 101}{100\times 100\times 100}$
=103030.10

Compound Interest=103030.10-100000
=3030.10

Question 6

Sol :

Principal=4800, n=2 and R=5%
Simple Interest$=\frac{P\times n \times R}{100}$
$=\frac{4800\times 2 \times 5}{100}$
=480

$A=\left(1+\frac{r}{100}\right)^n P$
$=4800\left(1+\frac{5}{100}\right)^2$
$=4800\left(1+\frac{1}{20}\right)^2$
$=4800\left(\frac{20+1}{20}\right)^2$
$=\frac{4800 \times 21 \times 21}{20\times 20}$
=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$=\frac{P\times n\times R}{100}$
$=\frac{2500\times 2\times 4}{100}$
=200

Given that compound Interest revoked semi annually, therefore n=4 and $R=\frac{4}{2}$=2%

$A=2500\left(1+\frac{2}{100}\right)^4$ $A=P\left(1+\frac{r}{100}\right)^n$
$=2500\left(1+\frac{2}{100}\right)^4$
$=2500\left(1+\frac{1}{50}\right)^4$
$=2500\left(\frac{51}{50}\right)^4$
$=\frac{2500 \times 51\times 51\times 51 \times 51}{50 \times 50 \times 50 \times 50}$
$=\frac{6765201}{2500}$
=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$=\frac{2000 \times 1\times 4}{100}$

=80

Principal for the second year=2080

Interest for the second year$=\frac{2080\times 1\times 3}{100}$

=62.40

Compound Interest=80+62.40

=142.40


Question 9

Sol :

$A=P\left(1+\frac{r_1}{100}\right)\left(1+\frac{r_2}{100}\right)\left(1+\frac{r_3}{100}\right)$

$=3125\left(1+\frac{4}{100}\right)\left(1+\frac{5}{100}\right)\left(1+\frac{6}{100}\right)$

$=3125\left(1+\frac{1}{25}\right)\left(1+\frac{1}{20}\right)\left(1+\frac{3}{50}\right)$

$=\frac{3125\times 26\times 21 \times 53}{25\times 20 \times 50}$

$=\frac{14469}{4}$

=3617.25

∴Compound Interest=3617.25-3125

=492.25


Question 12

Sol :

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

$A=P\left(1+\frac{r}{100}\right)^n$

$132651=P\left(1+\frac{r}{100}\right)^3$

$P\left(1+\frac{1}{50}\right)^3=132651$

$\frac{P\times 51\times 51 \times 51}{50\times 50\times 50}$

=132651

$P=\frac{132651\times 50 \times 50\times 50}{51\times 51\times 51}$

$P=\frac{132651\times 125000}{132651}$

P=125000


Question 13

Sol :

$A=P\left(1+\frac{r}{100}\right)^n$

$=P\left(1+\frac{4}{100}\right)^2$

$=P\left(1+\frac{1}{25}\right)^2$

$=\frac{P\times 26\times 26}{25\times 25}$

C.I.$=\frac{26\times 26}{25\times 25}(P)-P$

$=\frac{676P-625P}{625}$

$=\frac{51P}{625}$

$5712=\frac{51P}{625}$

$\frac{5712\times 625}{51}=P$

P=70000


Question 14

Sol :

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

$1275=1200\left(1+\frac{r}{100}\right)^n$

$r=\frac{1275}{12}-100$

$r=\frac{75}{12}$

r=6.25%

C.I for second year$=\frac{1275 \times 1\times 6.25}{100}$

$=\frac{7968.75}{100}$

=79.68

∴Compound Interest for second year

=80


Question 15

Sol :

Given that A=2500, P=2304, n=2 and R=?
$2500=2304\left(1+\frac{r}{100}\right)^2$
$\frac{2500}{2304}=\left(1+\frac{r}{100}\right)^2$
$\frac{100+r}{100}=\sqrt{\frac{2500}{2304}}$
$\frac{100+r}{100}=\sqrt{\left(\frac{50}{48}\right)^2}$
$100+r=\frac{100\times 50}{48}$
$r=\frac{5000}{48}-100$
$r=\frac{5000-4800}{4800}$
$r=\frac{200}{4800}$
$r=\frac{1}{24}$%
r=4.16%
$r=4\frac{1}{6}$%

Question 16

Sol :

Given that
$A=\frac{25}{16}P$ , n=2 and r=?
$A=P \left(1+\frac{r}{100}\right)^n$
$P=\left(1+\frac{r}{100}\right)^2=\frac{25}{16}P$
$\frac{100+r}{100}=\sqrt{\frac{25}{16}}$
$\frac{100+r}{100}=\frac{5}{4}$
r=125-100
r=25%

Question 17

Sol :

$A=P\left(1+\frac{r}{100}\right)^n$
$2315.25=2000\left(1+\frac{r}{100}\right)^3$
$\frac{2315.25}{2000}=\left(\frac{100+r}{100}\right)^3$
$\sqrt[3]{\frac{2.31525}{2.000}}=\frac{100+r}{100}$
2315.25×100×5=(100+r)3
$\sqrt[3]{2315.25\times 500}=100+r$
$\sqrt[3]{1157625}=100+r$
$\sqrt[3]{(105)^3}=100+r$
r=105-100=5%


Question 18

Sol :

A=48620.25, P=40000, $r=\frac{R}{2}$ , n=2×2=4
$A=P\left(1+\frac{r}{100}\right)^n$
$48620.25=40000\left(1+\frac{R}{200}\right)^4$
$[(200+R)^2]^2=\frac{48620.25\times 200 \times 200 \times 200 \times 200}{40000}$
$(200+R)^2=\sqrt[2]{(220.5)^2(200)^2}$
$200+R=\sqrt[2]{(220.5)^2}(200)^2$
$200+R=\sqrt[2]{200\times 220.5}$
R=210-200
R=10%

Question 19

Sol :

G/T $A=\frac{216}{125}P$ ,n=3
$A=P\left(1+\frac{r}{100}\right)^n$
$\frac{216}{125}P=P\left(1+\frac{r}{100}\right)^3$
$\frac{100+r}{100}=\sqrt[3]{\frac{216}{125}}$
$100+r=\frac{6}{5}\times 100$
r=120-100=20%
∴Rate of interest=20×2%
=40%


Question 20

Sol :

Given that A=88200, P=80000, n=2
$A=P\left(1+\frac{r}{100}\right)^2$
$\frac{88200}{80000}=\left(\frac{100+r}{100}\right)^2$
$\sqrt[2]{\frac{21\times 21 \times  2}{20\times 20 \times 2}}=\frac{100+r}{100}$
$\frac{21 \times 100}{20}=100+r$
r=105-100
r=5%

Principal= 88200

$C.I=\frac{88200 \times 1\times 5}{100}$
=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$=5292\left(1+\frac{r}{100}\right)$

$1+\frac{r}{100}=\frac{5556.60}{5292}$

$\frac{r}{100}=\frac{5556.60-5292}{5292}$

$r=\frac{26460}{5292}$

r=5%

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

$5292=P\left(1+\frac{5}{100}\right)^2$

$P\times \frac{21\times 21}{20\times 20}=5292$

$P=\frac{5292\times 400}{441}$

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

$=798.60\left(1+\frac{r}{100}\right)$

$\frac{878.46}{798.60}=\frac{100+r}{100}$

$\frac{r}{100}=\frac{879.46-798.60}{798.60}$

$r=\frac{7986}{798.60}$

r=10%

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

$798.6=P=\left(1+\frac{10}{100}\right)^3$

$798.6=P\times \frac{11}{10} \times  \frac{11}{10} \times  \frac{11}{10}$

$\frac{798.6\times 1000}{121\times 11}=P$

$P=\frac{798600}{12\times 11}$

P=600

∴Original amount =600

Question 23

Sol :

$A=P\left(1+\frac{r}{100}\right)^n$

=17576$=15625\left(1+\frac{4}{100}\right)^n$

$\frac{17576}{15625}=\left(\frac{26}{25}\right)^n$

$\left(\frac{26}{25}\right)^3=\left(\frac{26}{25}\right)^n$

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$=12500\left(1+\frac{8}{100}\right)^n$

$\frac{15746.40}{12500}=\left(\frac{27}{25}\right)^n$

$=\left(\frac{27}{25}\right)^3=\left(\frac{27}{25}\right)^n$

∴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 $\frac{1}{2}$ of 10%
=5%

Let the number of conversion periods be n, then

18522 $=16000\left(1+\frac{5}{100}\right)^n$

$\frac{18522}{16000}=\left(\frac{21}{20}\right)^n$

$\left(\frac{21}{20}\right)^4=\left(\frac{21}{20}\right)^n$

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


Question 26

Sol :

Given that,
A=278250, 
R1=5 , r2=6

$2782.50=P\left(1+\frac{5}{100}\right)\left(1+\frac{6}{100}\right)$

$2782.50=P\left(\frac{21}{20}\right)\left(\frac{53}{50}\right)$

$P=\frac{2782.50\times 20\times 50}{21\times 53}$

P=10000

∴Sum=10000



Question 27

Sol :

$CI_1=\frac{P_1 \times 1\times r}{100}$

$P_1 r=225\times 100$

$P_1 r=22500$  $P_1=\frac{22500}{r}$

${C_I}_2=\frac{P_2\times 1\times r}{100}$

$240=\dfrac{\frac{22500}{r}\times 1\times r}{100}+r$   $\left[\therefore P_2=\frac{22500}{r}+2\right]$

$r=\frac{240}{225}$

Question 27

Sol :

G/I
Compound Interest for the first year CI1=225
Compound Interest for the second year CI2=240
$CI_1=\frac{P_1\times N\times R}{100}$

$225=\frac{P_1\times 1\times R}{100}$

$P_1R=22500$

∴Principal  for the second year 
$=\frac{22500}{R}+CI_1$
$=\frac{22500}{R}+225$

$CI_2=\frac{P_2\times N\times R}{100}$
24×1000=P2×1×R
$24000=\frac{22500}{R}\times R +225 (R)$
225R=24000-22500

$R=\frac{1500}{225}$

(i) Rate pf interest $=6\frac{2}{3}$ %

(ii) Original Sum
$CI_1=\frac{P_1\times N \times R}{100}$

$P_1=\frac{225\times 100 \times 3}{20}$
=225×15
∴Original Sum=3375

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

$CI_3=\frac{3840\times 20\times 1}{3\times 100}$

Compound Interest for the third year=256

Question 28

Sol :

Simple Interest $=\frac{P\times N\times R}{100}$

$=\frac{P\times 2\times 5}{100}=\frac{P}{10}$

Compound Interest $=P\left[\left(1+\frac{5}{100}\right)^2-1\right]$
$=P\left[\left(\frac{21}{20}\right)^2-1\right]$
$=P\left[\frac{441-400}{400}\right]$
$=\frac{41P}{400}$

Given that
$\frac{41P}{400}-\frac{P}{10}=25$
$\frac{41P-40P}{400}=25$
1P=2500×4
P=2500×4
Principal=10000

Question 29

Sol :

Simple Interest $=\frac{P\times 1\times 10}{100}=\frac{P}{10}$

Compound Interest $=P\left[\left(1+\frac{5}{100}\right)^2-1\right]$

$=P\left[\left(1+\frac{1}{20}\right)^2-1\right]$

$=P\left[\left(\frac{21}{20}\right)^2-1\right]$

$=P\left[\frac{441-400}{400}\right]=\frac{41P}{400}$


$\frac{41P}{400}-\frac{P}{10}=15$

41P-40P=400×15

P=6000

Question 30

Sol :

Given that

Compound Interest for the first year $=P\left[\left(1+\frac{r}{100}\right)^1-1\right]$

1250=P[100+r-100]

1250=Pr

$P=\frac{1250}{r}$

Compound Interest for the second year
$=P\left[\left(1+\frac{r}{100}\right)^2-1\right]$
$1375=P\left[\left(1+\frac{1250}{P\times 100}\right)^2-1\right]$
$1375=P\left[\frac{(P+12.50)^2}{P^2}-1\right]$
$1375=P\left[\frac{P^2+25P+156.25}{P^2}-P^2\right]$

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

$125=\frac{1250\times 1\times R}{100}$

∴Rate of Interest=10%


Question 31

Sol :

Simple Interest for three years=225

$\frac{x\times r\times 3}{100}=225$

$x\times r=\frac{22500}{3}$ [∵x=Principal and r=rate of Interest]


Compound Interest for two years=153

$P\left[\left(1+\frac{r}{100}\right)^2-1\right]=153$

$x\left[\frac{(100+r)^2-(100)^2}{(100)^2}\right]=153$

x[1002+r2+200r-1002]=153×10000

(xr)r+200xr=1530000

$\frac{22500}{3}[r+200]=1530000$

r=68×3-200
r=204-200=4

∴Rate of Interest=4%

∴Simple Interest on three years=225

$\frac{P\times 3\times 4}{100}=225$

P=25×75

P=1875

Question 32

Sol :

Compounded annually

First, we found the amount after one year
$A=P\left(1+\frac{r}{100}\right)^n$
$=P\left(1+\frac{10}{100}\right)^n$
$=8000\left(\frac{110}{100}\right)^1$
=8000×1.1=8800

Principal for the second half year=8800

Interest for the next half year$=\frac{8800\times 10\times \frac{1}{2}}{100}$
=440

Compounded annually Interest=9240

As the money Invested for $1\frac{1}{2}$ years , therefore

n=number of conversion periods=3

$A=P\left(1+\frac{r}{100}\right)^n$
$=8000\left(1+\frac{5}{100}\right)^3$
$=8000\left(\frac{21}{20}\right)^3$
$=8000\left(1.05\right)^3$
=9261

Interest for $1\frac{1}{2}$ 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
$=P\left[\left(1+\frac{r}{100}\right)^n-1\right]$
$=P\left[\left(1+\frac{20}{100}\right)^n-1\right]$
$=P\left[\left(1+\frac{1}{5}\right)^2-1\right]$
$=P\left[\left(\frac{6}{5}\right)^2-1\right]$
$=P\left[\frac{36-25}{25}\right]$
$=\frac{11P}{25}$
=0.44P

Compounded half yearly,

Compound Interest for two years $=P\left[\left(1+\frac{r}{100}\right)^n-1\right]$
$=P\left[\left(1+\frac{10}{100}\right)^4-1\right]$
$=P\left[\left(\frac{11}{100}\right)^4-1\right]$
$=P[(1.1)^4-1]$
=P[1.4641-1]
=0.464P

∴Given that 0.464P-0.44P=482

0.024P=482

$P=\frac{482}{0.0241}$

P=20000

∴Sum of money lent out=20000

Question 34

Sol :

Amount after one year$=P\left(1+\frac{r}{100}\right)^n$
$13230=P\left(1+\dfrac{r/2}{100}\right)^2$


Amount after $1\frac{1}{2}$ year $=P\left(1+\frac{r}{100}\right)^n$
$13891.50=P\left(1+\frac{r/2}{100}\right)^3$
$13891.50=\frac{13230}{\left(1+\frac{r/2}{100}\right)^2} \left(1+\frac{r/2}{100}\right)^2$

$\left[\therefore P=\frac{13230}{\left(1+\frac{r/2}{100}\right)^2}\right]$
$\left[P=\frac{13230}{\left(1+\frac{5}{100}\right)^2}\right]$
$=\frac{13230\times 20\times 20}{21\times 21}$
$=\frac{13230}{1.1025}$
∴Sum=12000


$\frac{13891.50}{13230}=1+\frac{r}{200}$

$\frac{13891.50}{13230}=\frac{200+r}{200}$

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

Sum$=\frac{13230}{11.10251}$
=12000

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