ML AGGARWAL CLASS 8 CHAPTER 8 Simple and Compound Interest Exercise 8.2

Exercise 8.2


Question 1

Sol :
Given 

principal p = 6000

Rate R = 10 % 

Time T = 2 Year 

Interest for first year = $\frac{P T R}{100}$

$\frac{6000 \times 4 \times 10}{100}$

$I_{1}$ = 6000

Amount at the end of the First year =6000+600

Principal for second year =266001

Interest for Second year $=\frac{P T R}{100}$

$=\frac{6600 \times 1 \times 10}{100}$

$I_{2}=660$

$\begin{aligned} \text { Amount at the end of second yean } &=6600+660 \\ &=7260 /-\end{aligned}$

∴ Compound interest for 2 year = final amount - (original) principal
= 7260 - 6000
= 1260

Question 2

Sol- 
$\begin{aligned} \text { Principal for first } \operatorname{year}\left(P_{1}\right)=& ₹ 1875 /-\\ \text { Rate }(R)=4 \% & \text { Time }(T)=2 \text { years. } \\ \text { Interest for first year }\left(I_{1}\right)=\frac{P_{1} \times T_{1} \times R}{100} \\=& \frac{1875 \times 1 \times 4}{100} \\ I_{1}=& ₹ 75\end{aligned}$

Amount at the end of first year =1875+75
=1950

Interest for Second your $I_{2}=\frac{1950 \times 1 \times 4}{100}$

$I_{2}=78 /-$

$\therefore$ Compound Interest after 2 year $=I_{1}+I_{2}$

$\begin{array}{l}=75+78 \\=153\end{array}$

 Hence, Salma has to pay 2153 as compound Interest to the Mahila samit

Question 3

Sol :
Principal for first year $\left(R_{1}\right)=12000$

$Rate=10 \%, \text { Time }=3 \text { years }$

$\begin{aligned} \text { interest for first year }\left(I_{1}\right) &=\frac{12000 \times 1 \times 10}{100} \\ I_{1} &=₹ 1200\end{aligned}$

Total amount at the end of first year 12000+1200
=13200

principal for second year = Rs 13200 

interest for second year $I_{2}=$ $\frac{13200 \times 1 \times 10}{100}$
=₹ 1320 

Total amount at the end of the second year is 13200 + 1320 = Rs 14520

Principal amount for third year = Rs 14520

Interest for third year $I_{3}=\frac{14520 \times 1 \times 10}{100}$

$I_{3}=₹ 1452$

$\begin{aligned} \text { Total amount at the end of third year } &=14520+1452 \\ &=₹ 15972 \end{aligned}$

$\begin{aligned} \text { Compound Interest for } 3 \text { years } &=\text { Final amount }-\text { Principal }\left(P_{1}\right) \\ &=15972-12000 \\ &=23972 \end{aligned}$

Hence , At the end of 3 year jacob will get total 

Amount as Rs 15972 and compound interest Rs 3972 

Question 4

Sol :

Given

Principal $\left(P_{1}\right)=46875$

Rate (R)=4 %, Time =3 years.

principal amount for first year $\left(p_{1}\right)$ = Rs 46875 

Rate (R) = 4 % time = 3 years


(i) Interest for first year $\left(I_{1}\right)$

$I_{1}=1875$

$\begin{aligned} \text { Total amount at the end of Girst year } &=46875+1875 \\ &=48750 \end{aligned}$

Principal amount for second $\operatorname{year}\left(R_{2}\right)=$ 2 48750

Interest for Second year $\left(I_{2}\right)=\frac{48750 \times 4 \times 1}{100}$

$I_{2}$= Rs 21950


(ii) Total amount at the end of second year
= 48750 + 1950
= Rs 50700

Principal for third year $\left(P_{3}\right)$=  Rs 50700


(iii) $\begin{aligned} \text { Interest for third year }\left(I_{3}\right) &=\frac{50700 \times 1 \times 4}{100} \\ &=2028 \end{aligned}$

Hence ,

(i) The interest for first year $=21875$

(ii) The Amount standing to his credit at the end of
second year is  Rs 50700

(iii) The interest for third year =Rs  2028

Question 5

Sol :
Given 

Principal =260001 , Time =3 years

Rate =10

Principal amount for first $\operatorname{year}\left(P_{1}\right)=26000$

Interest for first year $\left(I_{1}\right)=\frac{6000 \times 1 \times 10}{100}$

Total amount at the end of first year 6000+600
=6600

Principal for second year $\left(P_{2}\right)=6600$


(i)  Interest for second year $\left(I_{2}\right)=\frac{6600 \times 1 \times 10}{100}$

$I_{2}=₹ 660$

Total amount at the end of second year =6600+660
=7260

$\begin{array}{ll}\text { Principal for third year } & \left(P_{3}\right)=27260 \\ \text { Interest for Third year } & \left(-I_{3}\right)=\frac{7260 \times 1 \times 10}{100} \\ & I_{3}=726 /-\end{array}$


(ii) Total amount at the end of third year = 7260 + 726
= Rs 7986

 hence 
(i) Compound interest for second year = Rs 660

(ii) Total amount at the end of the third year = Rs 7986

Question 6

Sol :
principal amount for first year $\left(P_{1}\right)$=Rs 5000

Rate of interest for first year $\left(R_{1}\right)=6 \%$

$\begin{aligned} \text { Interest for first year } &=\frac{5000 \times 1 \times 6}{100} \\\left(I_{1}\right) &=300 \end{aligned}$

$\begin{aligned} \text { Total amout at the end of first year } &=5000+300 \\ &=5300\\ \text { Principal amourt for second year }\left(P_{2}\right)=& 5300 \\ \text { Rate of Interest for Second year} &=8 \% . \\ \text { Compomd Interest for Second year } &=\frac{5300 \times 1 \times 8}{100} \end{aligned}$

$\left(I_{2}\right)=424$

$\begin{aligned} \text { Total amount at the end of second year } &=5300+424 \\ &= 5724 \end{aligned}$

$\begin{aligned} \text { Compound interest for } 2 \text { years } &=\text { final amount }-\text { Principal }\left(P_{1}\right) \\ &=5724-5000 \\ &=₹ 724 \end{aligned}$

Hence total amount after 2 year = Rs 5724

Compound interest for 2 year = Rs 724


Question 7

Sol :

Given principal = Rs 20000

Time = 2 year 

Rate of interest = 8%

Simple Interest $(I)=\frac{P T R}{100}$

$=\frac{20000 \times 2 \times 8}{100}$

Simple Interest (I.)=33200

Let compound interest as 'C'

Interest for first year = $\frac{20000 \times 1 \times 8}{100}$

$I_{1}=1600$

Total amount at the end of first year = 20000+ 1600
= 21600

Compound interest for 2 year = $I_{1}+I_{2}$
=1600+1728
C=3328

∴ Difference between the compound interest and 

simple interest is (c - I ) =  3328 - 3200

 = Rs 128

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