Ncert solutions

Ncert solutions

Grade 8

Comparing Quantities | Exercise 8.3

Question 1

Q1) Calculate the amount and compound interest on:

(a) Rs.10,800 for 3 years at 12\ \frac{1}{2}\% per annum compounded annually.

(b) Rs.18,000 for 2\ \frac{1}{2} years at 10% per annum compounded annually.

(c) Rs.62,500 for 1\ \frac{1}{2} years at 8% per annum compounded annually.

(d) Rs.8,000 for 1 years at 9% per annum compounded half yearly. (You could the year by year calculation using S.I. formula to verify).

(e) Rs.10,000 for 1 years at 8% per annum compounded half yearly.

Solution:

(a) Here, Principal (P) = Rs. 10800, Time (n) = 3 years Rate of interest (R) = 12\ \frac{1}{2}\%=\frac{25}{2\%}

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

= 10800\left(1+\frac{1}{8}\right)^3=10800\left(\frac{9}{8}\right)^3

= 10800\times\frac{9}{8}\times\frac{9}{8}\times\frac{9}{8}

= Rs. 15,377.34

Compound Interest (C.I.) = A – P

= Rs. 10800 – Rs. 15377.34 = Rs. 4,577.34

(b) Here, Principal (P) = Rs. 18,000, Time (n) = 2\ \frac{1}{2} years, Rate of interest (R) = 10% p.a.

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

= 18000\left(1+\frac{10}{100}\right)^2=18000\left(1+\frac{1}{10}\right)^2

= 18000\left(\frac{11}{10}\right)^2=18000\times\frac{11}{10}\times\frac{11}{10}

= Rs. 21,780

Interest for \frac{1}{2} years on Rs. 21,780 at rate of 10% = \frac{21780\times10\times1}{100}= Rs. 1089

Total amount for 2\ \frac{1}{2} years.

= Rs. 21,780 + Rs. 1089 = Rs. 22,869

Compound Interest (C.I.) = A – P

= Rs. 22869 – Rs. 18000 = Rs. 4,869

(c) Here, Principal (P) = Rs. 62500, Time (n) = 1\ \frac{1}{2}=\frac{3}{2} years = 3 years

Rate of interest (R) = 8% = 4% (compounded half yearly)

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

= 62500\left(1+\frac{4}{100}\right)^2

= 62500\left(1+\frac{1}{25}\right)^3

= 62500\left(\frac{26}{25}\right)^3

= 62500 \times\frac{26}{25}\times\frac{26}{25}\times\frac{26}{25}

= Rs. 70,304

Compound Interest (C.I.) = A – P

= Rs. 70304 – Rs. 62500 = Rs. 7,804

(d) Here, Principal (P) = Rs. 8000, Time (n) = 1 years = 2 years (compounded half yearly)

Rate of interest (R) = 9% = \frac{9}{2}\% (compounded half yearly)

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

= 8000\left(1+\frac{9}{2\times100}\right)^2

= 8000\left(1+\frac{9}{200}\right)^2

= 800\left(\frac{209}{200}\right)^2

= 8000\times\frac{209}{200}\times\frac{209}{200}

= Rs. 8,736.20

Compound Interest (C.I.) = A – P

= Rs. 8736.20 – Rs. 8000

= Rs. 736.20

(e) Here, Principal (P) = Rs. 10,000, Time (n) = 1 years = 2 years (compounded half yearly)

Rate of interest (R) = 8% = 4% (compounded half yearly)

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

= 10000\left(1+\frac{4}{100}\right)^2

= 10000\left(1+\frac{1}{25}\right)^2

= 10000\left(\frac{26}{25}\right)^2

= 10000\times\frac{26}{25}\times\frac{26}{25}

= Rs. 10,816

Compound Interest (C.I.) = A – P

= Rs. 10,816 – Rs. 10,000 = Rs. 816

Still have questions? Our expert teachers can help you out

Book a free class now

Want to top your mathematics exam ?

Learn from an expert tutor.

Book a free class now
subject-cta

Question 1

Q1) Calculate the amount and compound interest on:

(a) Rs.10,800 for 3 years at 12\ \frac{1}{2}\% per annum compounded annually.

(b) Rs.18,000 for 2\ \frac{1}{2} years at 10% per annum compounded annually.

(c) Rs.62,500 for 1\ \frac{1}{2} years at 8% per annum compounded annually.

(d) Rs.8,000 for 1 years at 9% per annum compounded half yearly. (You could the year by year calculation using S.I. formula to verify).

(e) Rs.10,000 for 1 years at 8% per annum compounded half yearly.

Solution:

(a) Here, Principal (P) = Rs. 10800, Time (n) = 3 years Rate of interest (R) = 12\ \frac{1}{2}\%=\frac{25}{2\%}

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

= 10800\left(1+\frac{1}{8}\right)^3=10800\left(\frac{9}{8}\right)^3

= 10800\times\frac{9}{8}\times\frac{9}{8}\times\frac{9}{8}

= Rs. 15,377.34

Compound Interest (C.I.) = A – P

= Rs. 10800 – Rs. 15377.34 = Rs. 4,577.34

(b) Here, Principal (P) = Rs. 18,000, Time (n) = 2\ \frac{1}{2} years, Rate of interest (R) = 10% p.a.

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

= 18000\left(1+\frac{10}{100}\right)^2=18000\left(1+\frac{1}{10}\right)^2

= 18000\left(\frac{11}{10}\right)^2=18000\times\frac{11}{10}\times\frac{11}{10}

= Rs. 21,780

Interest for \frac{1}{2} years on Rs. 21,780 at rate of 10% = \frac{21780\times10\times1}{100}= Rs. 1089

Total amount for 2\ \frac{1}{2} years.

= Rs. 21,780 + Rs. 1089 = Rs. 22,869

Compound Interest (C.I.) = A – P

= Rs. 22869 – Rs. 18000 = Rs. 4,869

(c) Here, Principal (P) = Rs. 62500, Time (n) = 1\ \frac{1}{2}=\frac{3}{2} years = 3 years

Rate of interest (R) = 8% = 4% (compounded half yearly)

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

= 62500\left(1+\frac{4}{100}\right)^2

= 62500\left(1+\frac{1}{25}\right)^3

= 62500\left(\frac{26}{25}\right)^3

= 62500 \times\frac{26}{25}\times\frac{26}{25}\times\frac{26}{25}

= Rs. 70,304

Compound Interest (C.I.) = A – P

= Rs. 70304 – Rs. 62500 = Rs. 7,804

(d) Here, Principal (P) = Rs. 8000, Time (n) = 1 years = 2 years (compounded half yearly)

Rate of interest (R) = 9% = \frac{9}{2}\% (compounded half yearly)

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

= 8000\left(1+\frac{9}{2\times100}\right)^2

= 8000\left(1+\frac{9}{200}\right)^2

= 800\left(\frac{209}{200}\right)^2

= 8000\times\frac{209}{200}\times\frac{209}{200}

= Rs. 8,736.20

Compound Interest (C.I.) = A – P

= Rs. 8736.20 – Rs. 8000

= Rs. 736.20

(e) Here, Principal (P) = Rs. 10,000, Time (n) = 1 years = 2 years (compounded half yearly)

Rate of interest (R) = 8% = 4% (compounded half yearly)

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

= 10000\left(1+\frac{4}{100}\right)^2

= 10000\left(1+\frac{1}{25}\right)^2

= 10000\left(\frac{26}{25}\right)^2

= 10000\times\frac{26}{25}\times\frac{26}{25}

= Rs. 10,816

Compound Interest (C.I.) = A – P

= Rs. 10,816 – Rs. 10,000 = Rs. 816

Still have questions? Our expert teachers can help you out

Book a free class now

Want to top your mathematics exam ?

Learn from an expert tutor.

Book a free class now
subject-cta
linkedin instgram facebook youtube
2020 © Quality Tutorials Pvt Ltd All rights reserved
linkedin instgram facebook youtube