S.chand publication New Learning Composite mathematics solution of class 8 Chapter 8 Percentage and Its Application Exercise 8E

 Exercise 8E

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Q1 | Ex-8E | Class 8 | S.Chand | New Learning Composite maths |Percentage and its application | myhelper

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Without using C.I. formula, calculate compound interest on:

Question 1

Rs. 6000 for 2 years at 10% p.a.

Sol :

P=6000, n=2 , R=10%

⇒$A=P\left(1+\frac{R}{100}\right)^n$

$=6000\left(1+\frac{10}{100}\right)^2$

=7260

∴CI=A-P

=7260-6000

=1260

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Q2 | Ex-8E | Class 8 | S.Chand | New Learning Composite maths |Percentage and its application | myhelper

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Question 2

Rs.15000 for 2 years at 8% p.a.

Sol :

P=15000 , n=2 , R=8%

⇒$A=P\left(1+\frac{R}{100}\right)^n$

$=15000\left(1+\frac{8}{100}\right)^2$

=17496

∴CI=A-P

=17496-15000

=2496

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Q3 | Ex-8E | Class 8 | S.Chand | New Learning Composite maths |Percentage and its application | myhelper

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Question 3

Rs. 50,000 for $2\dfrac{1}{2}$ years at 10% p.a.

Sol :

P=50000, $n=2\frac{1}{2}=\frac{5}{2}$ years , R=10%

⇒$A=P\left(1+\frac{R}{100}\right)^n$

$=50000\left(1+\frac{10}{100}\right)^{\frac{5}{2}}$

=63440

∴CI=A-P

=63440-50000

=13440

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Q4 | Ex-8E | Class 8 | S.Chand | New Learning Composite maths |Percentage and its application | myhelper

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Question 4

Rs. 16,000 for $1\dfrac{1}{2}$ years at 10% p.a., the interest being compounded half-yearly.

Sol :

P=16000 , $n=1\frac{1}{2}=\frac{3}{2}$ years , R$=\frac{10}{2}=5$%

⇒$A=P\left[1+\frac{\frac{R}{2}}{100}\right]^{2n}$

$=16000\left(1+\frac{5}{100}\right)^{\frac{3}{2}\times 2}$

=18522

∴CI=P-A

=18522-16000

=2522

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Q5 | Ex-8E | Class 8 | S.Chand | New Learning Composite maths |Percentage and its application | myhelper

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Question 5

Rs.15000 for 2 years when the rate of interest for the consecutive years is 8% and 9% p.a. respectively.

Sol : 

P=15000 , n=2 , R1=8% , R2=9

⇒A$=P\left(1+\frac{R_1}{100}\right)\left(1+\frac{R_2}{100}\right)$

$=15000\left(1+\frac{8}{100}\right)\left(1+\frac{9}{100}\right)$

=17658

∴CI=A-P

=17658-15000

=2658

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Q6 | Ex-8E | Class 8 | S.Chand | New Learning Composite maths |Percentage and its application | myhelper

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Question 6

Using C.I. formula, calculate the amount and compound interest on

(a) Rs. 40,000 for 2 years at 9% p.a. interest being paid annually.

(b) Rs. 5000 for 2 years at 10% p.a. interest being paid annually.

(c) Rs. 10,752 for 3 years at 12 1 by 2% p.a., interest being payable yearly.

Sol :

(a) P=40000 , n=2, R=9% [paid annually]

⇒A$=P\left(1+\frac{R}{100}\right)^n$

⇒A$=40000\left(1+\frac{9}{100}\right)^2$

=47524

∴CI=A-P

=47524-40000

=7524

(b) P=5000 , n=2, R=10% [paid annually]

⇒A$=P\left(1+\frac{R}{100}\right)^n$

⇒A$=5000\left(1+\frac{10}{100}\right)^2$

=6050

∴CI=A-P

=6050-5000

=1050

(c) P=10752 , n=3 , $R=12\frac{1}{2}=\frac{25}{2}$%

⇒A$=P\left(1+\frac{R}{100}\right)^n$

$=107520\left(1+\frac{1}{8}\right)^3$

=15309

∴CI=A-P

=15309-10752

=4557

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Q7 | Ex-8E | Class 8 | S.Chand | New Learning Composite maths |Percentage and its application | myhelper

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Question 7

How much will Rs. 256 amount to in one year at 12 1 by 2% per annum when the interest is compounded half-yearly?

Sol :

P=256 , n=1 , 

$R=\frac{R}{2}=\dfrac{12\frac{1}{2}}{2}$ $=\frac{25}{2\times 2}=\frac{25}{4}$

∵Half yearly

⇒$A=P\left(\dfrac{1+\frac{R}{2}}{100}\right)^{2n}$

$=256\left(1+\dfrac{\frac{25}{4}}{100}\right)^{2\times 1}$

$=256\left(1+\frac{25}{4\times 100}\right)^2$

=289

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Q8 | Ex-8E | Class 8 | S.Chand | New Learning Composite maths |Percentage and its application | myhelper

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Question 8

Find the amount and the compound interest on Rs. 24,000 at 10% p.a. for 1 1 by 2 years, compound interest being reckoned half yearly.

Sol :

P=24000 , $n=2n=2\times \left(1\frac{1}{2}\right)=2\times \frac{3}{2}$=3 years , $R=\frac{10}{2}=5$%

⇒$A=P\left(1+\dfrac{\frac{R}{2}}{100}\right)^{2n}$

$=24000\left(1+\frac{5}{100}\right)^3$

=27783

∴CI=A-P

=27783-2400

=3783

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Q9 | Ex-8E | Class 8 | S.Chand | New Learning Composite maths |Percentage and its application | myhelper

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Question 9

Anirudh took a loan of Rs. 8192 from Praveen to buy a washing machine at 12.5% p.a. compunded half yearly. The loan was rapid after 1 1 by 2 years. Calculate the amount that Anirudh had to pay to clear the loan.

Sol :

P=8192 , $n=1\frac{1}{2}=\frac{3}{2}$ , $R=12.5=\frac{125}{2\times 10}=\frac{25}{4}$%

⇒$A=P\left(1+\dfrac{\frac{R}{2}}{100}\right)^{2n}$

$=8192\left(1+\dfrac{\frac{25}{4}}{100}\right)^{2\times \frac{3}{2}}$

$=8192\left(1+\frac{1}{16}\right)^3$

=9826

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Q10 | Ex-8E | Class 8 | S.Chand | New Learning Composite maths |Percentage and its application | myhelper

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Question 10

Calculate the compound interest on Rs. 80,000 for 3 years if the rate for the 3 years are 2%, 5% and 10% respectively.

Sol :

P=80000 , n=3 , R1=4% , R2=5% , R3=10%

⇒$A=P\left(1+\frac{R_1}{100}\right)\left(1+\frac{R_2}{100}\right)\left(1+\frac{R_3}{100}\right)$

$=80000\left(1+\frac{2}{100}\right)\left(1+\frac{5}{100}\right)\left(1+\frac{10}{100}\right)$

$=80000\times \frac{102}{100}\times \frac{105}{100}\times \frac{110}{100}$

=94248

∴I=A-P

=94248-80000

=14248