Exercise 8E
Q1 | Ex-8E | Class 8 | S.Chand | New Learning Composite maths |Percentage and its application | myhelper
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
Q2 | Ex-8E | Class 8 | S.Chand | New Learning Composite maths |Percentage and its application | myhelper
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
Q3 | Ex-8E | Class 8 | S.Chand | New Learning Composite maths |Percentage and its application | myhelper
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
Q4 | Ex-8E | Class 8 | S.Chand | New Learning Composite maths |Percentage and its application | myhelper
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
Q5 | Ex-8E | Class 8 | S.Chand | New Learning Composite maths |Percentage and its application | myhelper
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
Q6 | Ex-8E | Class 8 | S.Chand | New Learning Composite maths |Percentage and its application | myhelper
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
=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
=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
Q7 | Ex-8E | Class 8 | S.Chand | New Learning Composite maths |Percentage and its application | myhelper
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
Q8 | Ex-8E | Class 8 | S.Chand | New Learning Composite maths |Percentage and its application | myhelper
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
Q9 | Ex-8E | Class 8 | S.Chand | New Learning Composite maths |Percentage and its application | myhelper
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
Q10 | Ex-8E | Class 8 | S.Chand | New Learning Composite maths |Percentage and its application | myhelper
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
