ML Aggarwal Solution Class 9 Chapter 2 Compound Interest Exercise 2.1

 Exercise 2.1

Question 1

Sol :

Principal for the first year=8000

Interest for the first year $\frac{₹ 8,000 \times 5 \times 1}{100}$

=400

Amount after one year=8000+400=8400

Interest for the second year $\frac{8400 \times 5 \times 1}{100}$

=420

Amount after second year=8400+420

=8820

Compound interest after two years=final amount-principal

=8820-8000

=820


Question 2

Sol :

Principal for the first year =₹ 46,875

Rate of interest=4%

(i) Interest for the first year$\frac{46,875 \times 4 \times 1}{100}$

=1875

Amount after one year=46875+1875

=48750


(ii) Interest for the second year $\frac{48,750 \times 4 \times 1}{100}$

=1950

The amount standing to his credit at the end of

Second year=48750+1950

=50700


(iii) Interest for the third year$\frac{50,700 \times 4 \times 1}{100}$

=5007×4

=2028


Question 3

Sol :

Principal for the first year =28,000

Rate of Interest =10 %

Interest for the first year $\frac{8,000 \times 10 \times 1}{100}$

=800

Amount after one year=8000+800

=8800

Interest for the second year$\frac{8800 \times 10 \times 1}{100}$

=880

Amount after second year=8800+880

=9680

Compound interest for the second year=800+880

=1680

Interest for the third year$\frac{9,680 \times 10 \times 1}{100}$

=968

Sum due at the end of the third year

=9680+968

=10648


Question 4

Sol :

Principal for the first year =₹ 12,800

Rate of interest =10 %

Interest for the first year

$\frac{12,800 \times 10 \times 1}{100}$

=₹ 1,280

(i) Amount at the end of first year=12800+1280

=14080

(ii) Interest for the second year$\frac{14,080 \times 10 \times 1}{100}$

=1408

Amount at the end of second year=14080+1408

=15488

Interest for the third year$\frac{15,488 \times 10 \times 1}{100}$

=1548.8

(iii) The total amount due to him at the end of three years

=15488+1548.8

=17036.5


Question 5

Sol :

Interest for 2 years =1380

$\frac{x \times 2 \times 12}{100}=1380$

24x=13800

Sum of money$\frac{1,38,000}{24}$

=5,750

Principal for the first half year=5750

Interest for the first half year$\frac{5,750 \times 6 \times 1}{100}$

=57.5×6

=345

Amount after the first half year=5750+345

=6095

Interest for the second half year$\frac{6095 \times 6 \times 1}{100}$

=60.95×6

=365.7

∴Compound Interest on this Sum for one year payable half yearly at the same rate

=345+365.7

=710.70


Question 6

Sol :

Principal for the first year =10,000

Amount after one year=11200

Interest for the first year=11200-1000

=10200

Rate of Interest for the first year=?

Interest for the first year$=\frac{10,000 \times x \times 1}{100}$

$\frac{1200\times 100}{10000}=x$

x=12%

Interest for the second year$=\frac{11200\times 12\times 1}{100}$

=112×12

=1344

Amount at the end of second year=11200+1344

=12544


Question 7

Sol :

Principal for the first year=5000

Amount after one year=5600

Interest for first year=5600-5000

$\frac{5000\times x\times 1}{100}=600$

x=12

(i) Rate of Interest=12%

(ii) Interest acured in the second year$\frac{5,600 \times 12 \times 1}{100}$

=672

Amount after second year=5600+672

=6272

(iii) Interest for the first third year$=\frac{6272 \times 12 \times 1}{100}$

=62.72×12

=752.64

Amount after third year=62.72+752.64

=7024.64


Question 8

Sol :

Principal for the first year =₹ 2,000

Interest for the first year$\frac{2,000 \times 10}{100}$
$=\frac{2,000 \times 2}{20}$
=200

Amount after first year=2000+200
=2200

Interest for the second year$=\frac{2200 \times 10\times 1}{100}$

=220

Amount after second year=2200+220

=2420

Interest for the second half $2\frac{1}{2}$ year

$=\frac{2420\times 5\times 1}{100}$

=121

∴Compound Interest after $2\frac{1}{2}$ year=200+220+121

=541

Amount to be paid after $2\frac{1}{2}$ year=2000+541

=2541


Question 9

Sol :

Principal for the first half year =50,000

Interest for the first half year $\frac{50,000 \times 4 \times 1}{100}$

=2000

Amount after the first half year

=50000+2000

=52000

Principal after the second half year=52000

Interest for the second half year$\frac{52000 \times 4 \times 1}{100}$

=520×4

=2080

Amount after the second half year=52000+2080

=54080

Principal for third year=54080


Question 10

Sol :

Given that A=9261, n=3 and R=5

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

$9261=P\left(1+\frac{5}{100}\right)^{3}$

$9261=P\left(1+\frac{1}{20}\right)^{3}$

$9261=P\left(\frac{21}{20}\right)^{3}$

$\frac{9261 \times 20 \times 20 \times 20}{21 \times 21 \times 21}=P$

$\therefore P=\frac{9261 \times 8000}{9261}$

∴Principal=8000


Question 11

Sol :

Given that A=7803, n=2, r=2

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

$7803=P\left(1+\frac{2}{100}\right)^{2}$

$7803=\frac{P \times 51 \times 51}{50 \times 50}$

$P=\frac{19507,500}{2500}$

∴P=7500

Interest for the third half year$\frac{54080 \times 4 \times 1}{100}$

=540.8×4

=2163.2

Amount after third half year=54080+2163.2

=56.243.20

∴Compound Interest for $1 \frac{1}{2}$ years 

=56,243-50,000

=6243


Question 10

Sol :

Principal amount for the first year=5000

Interest for the first year$\frac{5,000 \times 6 \times 1}{100}$

=50×6

=300

Amount after first year=5000+300

=5300

Interest for the second year$\frac{5,300 \times 8 \times 1}{100}$

=53×8

=424

∴Amount after second year=5300+424

=5724

Compound interest for 2 years=5724-5000

=724


Question 11

Sol :

Principal for the first half year=17000

Interest for the first year at 10%$=\frac{17,000 \times 10 \times 1}{100}$

=1700

Amount after first year=17000+1700

=18700

Interest for the second year at 10% $\frac{18,700 \times 10 \times 1}{100}$

=1870

Amount after second year

=18700+1870

=20570

Interest for the third year at 14%

$=\frac{20,570 \times 14 \times 1}{100}$

=2879.8

Amount after third year=20570+2879.8

=23449.8

Compound Interest after three year=6449.80


Question 12

Sol :

Principal for the first year=9600

Interest for the first year$\frac{9,600 \times 10 \times 1}{100}$

=960

(i) Sum due at the end of first years=9600+960

=10560

Interest for the second year$\frac{10,560 \times 10 \times 1}{100}$

=1056

(ii) Sum due at the end of second year=10560+1056

=11626

(iii) Compound Interest Earned n 2 years

=11626-9600

=2026

(iv) Difference of (ii) and (i)=11626-10560

=1056

Interest for difference of amount for one year

$=\frac{1,056 \times 10 \times 1}{100}$

=105.60

compound Interest for third year=1056+105.6

=1161.60


Question 13

Sol :

Simple interest on certain money for 2 years at 10% is 1600

$\frac{x \times 10 \times 2}{100}=1,600$

x=8000

Principal=8000

Interest for the first year$\frac{8,000 \times 10 \times 1}{100}$

=800

Amount to be paid after one year=8000+800

=8800

Interest for the second year$\frac{8800 \times 10 \times 1}{100}$

$= \frac{8800}{10}$

=9680

Interest for the third year$\frac{9,680 \times 10 \times 1}{100}$

=968

Amount after three year =9680+968

=10648

Compound Interest for three year=800+880+968

=2648


Question 14

Sol :

Principal for the first year =6000

Interest for the first year$\frac{6,000 \times 5 \times 1}{100}$

=300

Amount after first year=6300

Money repaid=1200

Principal for the second year=6300-1200

=5100

Interest for the Second year$\frac{5,100 \times 5 \times 1}{100}$

=255

Principal amount after second year=5100+255

=5355

Amount outstanding at the beginning of third year

=5355-1200

=4155


Question 15

Sol :

Principal for the first year=100000

Interest for the first year$=\frac{1000000\times 11\times 1}{100}$

=11000

Amount after one year=100000+11000

=111000

Money repaid after one year=41000

Principal for second year=111000-41000

=70000

Interest for the second year$\frac{70,000 \times 11 \times 1}{100}$

=7700

Amount after second year=70000+7700

=77700

Principal for third year=77700-47700

=30000

Amount=30000


Question 16

Sol :

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

$=\frac{20000 \times 2.5 \times 10}{100}$

=200×2.5

=5000

Interest for the first year$=\frac{20000 \times 1 \times 10}{100}$

=2000

Amount after one year=20000+2000

=22000

Interest after the second year$\frac{22,000 \times 10 \times 1}{100}$

=2200

Amount after second year=22000+2200

=24200

Interest for second year 6 months$=\frac{24200\times 5}{100}$

=1210

∴Total amount=24200+1210

=25410

∴Gain=25410-25000

=410

Comments

Post a Comment

Popular posts from this blog

ML Aggarwal Solution Class 10 Chapter 15 Circles Exercise 15.1

Image
 Exercise 15.1 Question 1 Using the given information, find the value of x in each of the following figures : Sol : (i) ∠ADB and ∠ACB are in the same segment. ∠ADB = ∠ACB = 50° Now in ∆ADB, ∠DAB + X + ∠ADB = 180° (Angles of a triangles) $\Rightarrow 42^{\circ}+x+50^{\circ}=180^{\circ}$ $\Rightarrow x+92^{\circ}=180^{\circ}$ $\Rightarrow x=180^{\circ}-92^{\circ}=88^{\circ}$ (ii) ∠ABD=∠ACD (Angles in the same segment) But ∠ACD=32° ∴∠ABD=32° Now in ΔABD ∠ABD+∠ADB+∠DAB=180° ⇒32°+45°+x=180° ⇒77°+x=180° ⇒x=180°-77°=103° (iii) ∠BAD=∠BCD (Angles in the same segement) But ∠BAD=20° $\therefore \angle B C D=20^{\circ}$ $\because \angle \mathrm{CEA}=90^{\circ}$ $\therefore \angle \mathrm{CED}=90^{\circ}$ Now in $\Delta$ CED, $\angle C E D+\angle B C D+\angle C D E=180^{\circ}$ $\Rightarrow 90^{\circ}+20^{\circ}+x=180^{\circ}$ $\Rightarrow 110^{\circ}+x=180^{\circ}$ $\Rightarrow x=180^{\circ}-110^{\circ}=70^{\circ}$ (iv) In ΔABC ∠ABC+∠ACB+∠BAC=180° (∵Sum of angles of a triangle) $\Rightarrow 69...
Read more

ML Aggarwal Solution Class 9 Chapter 20 Statistics Exercise 20.2

  Exercise 20.2 Question 1 Sol : (i) discrete  (ii) continuous  (iii) discrete  (iv) continuous  (v) continuous  Question 2 Sol : Given data  13, 6, 10 ,5 ,11 ,14 ,2 ,8 ,15 ,16 ,9 ,13 ,17 ,11, 19 ,5 ,7 ,12 ,20 ,21 ,18 ,1 ,8 ,12 ,18 Classes 0-4 5-9 10-14 15-9 20-24 Frequency 2 7 8 6 2 Question 3 Sol : (i)  Classes 1-10 11-20 21-30 31-40 Frequency 8 7 6 6 (ii) Range = Maximum value - Minimum value  =40-2=38 (iii) Third class of frequency table $=\frac{21+30}{2}=\frac{51}{2}$ =25.5 Question 4 Sol : Variate : A particular value of a variable is called variate Class size :  The difference be...
Read more

ML Aggarwal Solution Class 9 Chapter 3 Expansions Exercise 3.2

Exercise 3.2 Question 1 Sol : Given : x-y=8..(i) xy=5...(ii) Squaring both sides in equation (i) ∴(x-y) 2 =8 2 ⇒x 2 -2.xy+y 2 =64 ⇒x 2 -2(5)+y 2 =64 ⇒x 2 +y 2 =64+10 ⇒x 2 +y 2 =74 Question 2 Sol : ⇒Given : x+y=10...(i) ⇒xy=21....(ii) Squaring both sides in equation (i) ⇒(x+y) 2 =10 2 ⇒x 2 +2xy+y 2 =100 ⇒x 2 +2(21)+y 2 =100 ⇒x 2 +4 2 +y 2 =100 ⇒x 2 +y 2 =100-42 ⇒x 2 +y 2 =58 2(x 2 +y 2 )=2×58=116 Question 3 Sol : ⇒Given : 2a+3b=7..(i) ⇒ab=2....(ii) Squaring both sides in equal (i) ⇒(2a+3b) 2 =7 2 ⇒(2a) 2 +2.2a+3b+(3b) 2 =49 ⇒4a 2 +12ab+9b 2 =49 ⇒4a 2 +12(2)+9b 2 =49 ⇒4a 2 +24+9b 2 =49 ⇒4a 2 +9b 2 =49-24 ⇒4a 2 +9b 2 =25 Question 4 Sol : ⇒Given : 3x-4y=16...(i) ⇒xy=4..(ii) Squaring on both sides in equation (i) ⇒(3x-4y) 2 =16 2 ⇒(3x) 2 -2.3x.4y+(4y) 2 =256 ⇒9x 2 -24xy+16y 2 =256 ⇒9x 2 -24(4)+16y 2 =256 ⇒9x 2 -96+16y 2 =256 ⇒9x 2 +16y 2 =256+96 ⇒9x 2 +16y 2 =352 Question 5 Sol : ⇒Given : x+y=8...(i) ⇒x-y=2...(ii) Since we know that ⇒2(x) 2 +2(y) 2 =(x+y) 2 +(x-y) 2 From (i) ⇒S...
Read more