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NCERT Solutions for Class 8 Maths Chapter 8: Comparing Quantities - Exercise 8.3

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NCERT Solutions for Class 8 Maths Chapter 8 (EX 8.3)

Free PDF download of NCERT Solutions for Class 8 Maths Chapter 8 Exercise 8.3 and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 8 Maths Chapter 8 Comparing Quantities Exercise 8.3 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails.Vedantu is a platform that provides free NCERT Solution and other study materials for students.


Class:

NCERT Solutions for Class 8

Subject:

Class 8 Maths

Chapter Name:

Chapter 8 - Comparing Quantities

Exercise:

Exercise - 8.3

Content-Type:

Text, Videos, Images and PDF Format

Academic Year:

2023-24

Medium:

English and Hindi

Available Materials:

  • Chapter Wise

  • Exercise Wise

Other Materials

  • Important Questions

  • Revision Notes


 

Science Students who are looking for NCERT Solutions for Class 8 Science will also find the Solutions curated by our Master Teachers really Helpful.

Access NCERT Solution for Class 8 Maths Chapter 8 – Comparing Quantities

Exercise 8.3

Refer to page 10- 21 for exercise 8.3 in the PDF.

 

1. Calculate the amount and compound interest on

(a) Rs. 10800 for 3 years at $12\dfrac{1}{2}\% $ per annum compounded annually.

Ans: The amount invested is called Principal$\left( P \right)$$ = $ Rs. 10,800.

Rate at which money is invested is given by, $\left( R \right) = 12\dfrac{1}{2}\% $

$\left( R \right) = \dfrac{{25}}{2}\% $

The number of years money is invested is, $n = 3$.

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

Substitute the values and solve further,

$A = {\text{Rs}}{\text{. }}\left[ {10800{{\left( {1 + \dfrac{{25}}{{200}}} \right)}^3}} \right]$

$A = {\text{Rs}}{\text{. }}\left[ {10800{{\left( {\dfrac{{225}}{{200}}} \right)}^3}} \right]$

$A = {\text{Rs}}{\text{. }}\left( {10800 \times \dfrac{{225}}{{200}} \times \dfrac{{225}}{{200}} \times \dfrac{{225}}{{200}}} \right)$

$A = {\text{Rs}}{\text{. }}15377.34375$

$A = {\text{Rs}}{\text{. 15377}}{\text{.34 }}\left( {{\text{approximately}}} \right)$

The compound interest is given by, principle subtracted from the amount.

${\text{C}}{\text{.I}}{\text{.}} = {\text{ Rs}}{\text{. }}\left( {15377.34 - 10800} \right)$

${\text{C}}{\text{.I}}{\text{.}} = {\text{ Rs}}{\text{. }}4557.34$

 

(b) Rs 18000 for $2\dfrac{1}{2}$ years at $10\% $ per annum compounded annually.

Ans: The amount invested is called Principal$\left( P \right)$$ = $ Rs. 18,000.

Rate at which money is invested is given by, $R = 10\% $

The number of years money is invested is, $n = 2\dfrac{1}{2} = 2.5$ years.

Calculate the amount for 2 years and 6 months by calculating the amount for 2 years using the compound interest formula. After that, calculate the amount of simple interest for 6 months on the amount obtained at the end of 2 years.

For that, calculate the amount for 2 years.

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

Substitute the values and solve further,

$A = {\text{Rs}}{\text{. }}\left[ {18000{{\left( {1 + \dfrac{1}{{10}}} \right)}^2}} \right]$

$A = {\text{Rs}}{\text{. }}\left[ {18000{{\left( {\dfrac{{11}}{{10}}} \right)}^3}} \right]$

$A = {\text{Rs}}{\text{. }}\left( {18000 \times \dfrac{{11}}{{10}} \times \dfrac{{11}}{{10}}} \right)$

$A = {\text{Rs}}{\text{. 21780}}$

Consider Rs. 21780 as principal and calculate the S.I. for the next $\dfrac{1}{2}$ year.

${\text{S}}{\text{.I}}{\text{. =  Rs}}{\text{.}}\left( {\dfrac{{21780 \times \dfrac{1}{2} \times 10}}{{100}}} \right)$

${\text{S}}{\text{.I}} = {\text{Rs}}{\text{. }}1089$

The amount of interest earned for the first 2 years\[ = Rs.\left( {21780{\text{ }} - {\text{ }}18000} \right) = {\text{ }}Rs.3780\].

And interest for the next $\dfrac{1}{2}$ year\[ = Rs.1089\] .

The total compound interest is given by,

${\text{C}}{\text{.I}}{\text{.}} = {\text{ Rs}}{\text{. }}\left( {3780 - 1089} \right)$

${\text{C}}{\text{.I}}{\text{.}} = {\text{ Rs}}{\text{. }}4869$

Total amount is,

\[{\text{Amount  =  P + C}}{\text{.I}}{\text{.}}\]

$A = {\text{Rs}}{\text{. }}18000 + 4869$

$A = {\text{Rs}}{\text{. }}22869$

 

(c) Rs 62500 for $1\dfrac{1}{2}$ years at $8\% $ per annum compounded half yearly.

Ans: The amount invested is called Principal, $P = {\text{Rs}}{\text{. }}62,500$.

Rate at which money is invested is given by, $\left( R \right) = 8\% $ per annum or $4\% $ per half year.

The number of years money is invested is, $n = 1\dfrac{1}{2}$ years. There will be 3 half years in $1\dfrac{1}{2}$ years.

Calculate the amount for 3 half years, taking the rate as $4\% $ per half year.

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

Substitute the values and solve further,

$A = {\text{Rs}}{\text{. }}\left[ {62500{{\left( {1 + \dfrac{4}{{100}}} \right)}^3}} \right]$

$A = {\text{Rs}}{\text{. }}\left[ {62500{{\left( {\dfrac{{26}}{{25}}} \right)}^3}} \right]$

$A = {\text{Rs}}{\text{. }}\left( {62500 \times \dfrac{{26}}{{25}} \times \dfrac{{26}}{{25}} \times \dfrac{{26}}{{25}}} \right)$

$A = {\text{Rs}}{\text{. 70304}}$

The compound interest is given by, principle subtracted from the amount.

${\text{C}}{\text{.I}}{\text{.}} = {\text{ Rs}}{\text{. }}\left( {70304 - 62500} \right)$

${\text{C}}{\text{.I}}{\text{.}} = {\text{ Rs}}{\text{. 7804}}$

 

(d) Rs 8000 for 1 year at $9\% $ per annum compound half yearly.

Ans: The amount invested is called Principal$\left( P \right) = {\text{Rs}}{\text{. }}8000$.

Rate at which money is invested is given by, $\left( R \right) = 9\% $ per annum or $\dfrac{9}{2}\% $ per half year.

The number of years money is invested is, $n = 1$ years. There will be 2 half years in $1$ years.

Calculate the amount for 2 half years, taking the rate as $\dfrac{9}{2}\% $ per half year.

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

Substitute the values and solve further,

$A = {\text{Rs}}{\text{. }}\left[ {8000{{\left( {1 + \dfrac{9}{{200}}} \right)}^2}} \right]$

$A = {\text{Rs}}{\text{. }}\left[ {8000{{\left( {\dfrac{{209}}{{200}}} \right)}^2}} \right]$

$A = {\text{Rs}}{\text{. }}\left( {8000 \times \dfrac{{209}}{{200}} \times \dfrac{{209}}{{200}}} \right)$

$A = {\text{Rs}}{\text{. 8736}}{\text{.20}}$

The compound interest is given by, principle subtracted from the amount.

${\text{C}}{\text{.I}}{\text{.}} = {\text{ Rs}}{\text{. }}\left( {8736.20 - 8000} \right)$

${\text{C}}{\text{.I}}{\text{.}} = {\text{ Rs}}{\text{. 736}}{\text{.20}}$

 

(e) Rs 10000 for 1 year at $8\% $ per annum compounded half yearly. 

Ans: The amount invested is called Principal $\left( P \right) = {\text{Rs}}{\text{. }}10,000$.

Rate at which money is invested is given by, $R = 8\% $per annum or $4\% $ per half year.

The number of years money is invested is, $n = 1$ years. There will be 2 half years in $1$ years.

Calculate the amount for 2 half years, taking the rate as $4\% $ per half year.

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

Substitute the values and solve further,

$A = {\text{Rs}}{\text{. }}\left[ {10000{{\left( {1 + \dfrac{4}{{100}}} \right)}^2}} \right]$

$A = {\text{Rs}}{\text{. }}\left[ {10000{{\left( {\dfrac{{26}}{{25}}} \right)}^2}} \right]$

$A = {\text{Rs}}{\text{. }}\left( {10000 \times \dfrac{{26}}{{25}} \times \dfrac{{26}}{{25}}} \right)$

$A = {\text{Rs}}{\text{. 10816}}$

The compound interest is given by, principle subtracted from the amount.

${\text{C}}{\text{.I}}{\text{.}} = {\text{ Rs}}{\text{. }}\left( {10816 - 10000} \right)$

${\text{C}}{\text{.I}}{\text{.}} = {\text{ Rs}}{\text{. 816}}$

 

2. Kamala borrowed Rs. 26400 from a Bank to buy a scooter at a rate of $15\% $ p.a. compounded yearly. What amount will she pay at the end of 2 years and 4 months to clear the loan? (Hint: Find $A$ for 2 years with interest compounded yearly and then find SI on the ${2^{nd}}$ year amount for $\dfrac{4}{{12}}$ years.)

Ans: The amount invested is called Principal, $\left( P \right) = {\text{Rs}}{\text{. 26,400}}$

Rate at which money is invested is given by, $\left( R \right) = 15\% $.

The number of years money is invested is, $n = 2\dfrac{4}{{12}}$ years.

Calculate the amount for 2 years and 4 months by calculating the amount for 2 years using the compound interest formula. After that, calculate the amount of simple interest for 4 months on the amount obtained at the end of 2 years.

For this, calculate the amount for 2 years.

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

Substitute the values and solve further,

$A = {\text{Rs}}{\text{. }}\left[ {26400{{\left( {1 + \dfrac{{15}}{{100}}} \right)}^2}} \right]$

$A = {\text{Rs}}{\text{. }}\left[ {26400{{\left( {1 + \dfrac{3}{{20}}} \right)}^2}} \right]$

$A = {\text{Rs}}{\text{. }}\left( {26400 \times \dfrac{{23}}{{20}} \times \dfrac{{23}}{{20}}} \right)$

$A = {\text{Rs}}{\text{. 34914}}$

By taking Rs. 34914 as principal, the S.I. for the next $\dfrac{4}{{12}} = \dfrac{1}{3}$ year will be calculated.

${\text{S}}{\text{.I}}{\text{. =  Rs}}{\text{.}}\left( {\dfrac{{34914 \times \dfrac{1}{3} \times 15}}{{100}}} \right)$

${\text{S}}{\text{.I}} = {\text{Rs}}{\text{. 1745}}{\text{.70}}$

\[{\text{Interest for the first 2 years }} = {\text{ Rs}}{\text{. }}\left( {34914{\text{ }} - {\text{ }}26400} \right){\text{ }} = {\text{ Rs}}{\text{. }}8514\]

And interest for the next $\dfrac{4}{{12}}$ year = Rs. $1745.70$

The total compound interest is given by,

${\text{C}}{\text{.I}}{\text{.}} = {\text{ Rs}}{\text{. }}\left( {8415 + 1745.70} \right)$

${\text{C}}{\text{.I}}{\text{.}} = {\text{ Rs}}{\text{. 10259}}{\text{.70}}$

Total amount is,

${\text{Amount  =  Principal + C}}{\text{.I}}{\text{.}}$

$A = {\text{Rs}}{\text{. }}26400 + 10259.70$

$A = {\text{Rs}}{\text{. }}36659.70$

Therefore, the amount she will pay at the end of 2 years and 4 months to clear the loan is Rs. $36659.70$.

 

3. Fabina borrows Rs. 12,500 at $12\% $ per annum for 3 years at simple interest and Radha borrows the same amount for the same time period at $10\% $ per annum, compounded annually. Who pays more interest and by how much?

Ans: The amount invested is called Principal $(P) = {\text{Rs}}.12500$.

Rate at which money is invested is given by, $\left( R \right) = 12\% $

The number of years money is invested is, $n = 3$ years.

The interest paid by Fabina is, $ = \dfrac{{P \times R \times T}}{{100}}$.

$\dfrac{{12500 \times 12 \times 3}}{{100}} = {\text{Rs}}{\text{. }}4500$

For Radha, rate at which money is invested compounded annually is given by,$\left( R \right) = 10\% $

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

Substitute the values and solve further,

$A = {\text{Rs}}{\text{. }}\left[ {12500{{\left( {1 + \dfrac{{10}}{{100}}} \right)}^3}} \right]$

$A = {\text{Rs}}{\text{. }}\left( {12500 \times \dfrac{{110}}{{100}} \times \dfrac{{110}}{{100}} \times \dfrac{{110}}{{100}}} \right)$

$A = {\text{Rs}}{\text{. 16637}}{\text{.50}}$

The compound interest is given by, principle subtracted from the amount.

${\text{C}}{\text{.I}}{\text{.}} = {\text{ Rs}}{\text{. }}\left( {16637.50 - 12500} \right)$

${\text{C}}{\text{.I}}{\text{.}} = {\text{ Rs}}{\text{. 4137}}{\text{.50}}$

The interest paid by Fabina is Rs. 4,500 and by Radha is Rs. $4137.50$. 

Thus, Fabina pays more interest.

\[{\text{Rs }}4500{\text{ }} - {\text{ Rs }}4137.50{\text{ }} = {\text{ Rs }}362.50\]

Hence, Fabina will have to pay \[{\text{Rs }}362.50\] more.

 

4. I borrowed Rs. 12000 from Jamshed at $6\% $ per annum simple interest for 2 years. Had I borrowed this sum at $6\% $ per annum compound interest, what extra amount would I have to pay?

Ans: The amount invested is called Principal$\left( P \right) = 12,000$.

Rate at which money is invested is given by, $\left( R \right) = 6\% $

The number of years money is invested is, $n = 3$ years.

The simple interest on the given principle is, $ = \dfrac{{P \times R \times T}}{{100}}$.

$\dfrac{{12000 \times 6 \times 2}}{{100}} = {\text{Rs}}{\text{. 1}}440$

The amount if the sum is compounded annually,

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

Substitute the values and solve further,

$A = {\text{Rs}}{\text{. }}\left[ {12000{{\left( {1 + \dfrac{6}{{100}}} \right)}^2}} \right]$

$A = {\text{Rs}}{\text{. }}\left[ {12000{{\left( {1 + \dfrac{3}{{50}}} \right)}^2}} \right]$

$A = {\text{Rs}}{\text{. }}\left( {12000 \times \dfrac{{53}}{{50}} \times \dfrac{{53}}{{50}}} \right)$

$A = {\text{Rs}}{\text{. 13483}}{\text{.20}}$

The compound interest is given by, principle subtracted from the amount.

${\text{C}}{\text{.I}}{\text{.}} = {\text{ Rs}}{\text{. }}\left( {13483.20 - 12000} \right)$

${\text{C}}{\text{.I}}{\text{.}} = {\text{ Rs}}{\text{. 1483}}{\text{.20}}$

The simple interest is, Rs. 1,400 and compound interest is, Rs. $1483.20$. 

Thus, extra amount to be paid is,

\[{\text{Rs 1}}483.20{\text{ }} - {\text{ Rs 1440 }} = {\text{ Rs 4}}3.20\]

Hence, the extra amount paid is \[{\text{Rs 4}}3.20\].

 

5. Vasudevan invested Rs 60000 at an interest rate of $12\% $ per annum compounded half yearly. What amount would he get,

(i) after 6 months?

Ans: The amount invested is called Principal $\left( P \right) = {\text{Rs}}{\text{. }}60000$.

Rate at which money is invested is given by, $\left( R \right) = 12\%  = 6\% $.

The number of years money is invested is, $n = 6$ months, that is 1 half-year.

The amount if the sum is compounded annually,

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

Substitute the values and solve further,

$A = {\text{Rs}}{\text{. }}\left[ {60000{{\left( {1 + \dfrac{6}{{100}}} \right)}^1}} \right]$

$A = {\text{Rs}}{\text{. }}\left[ {60000\left( {\dfrac{{106}}{{100}}} \right)} \right]$

$A = {\text{Rs}}{\text{. 63600}}$

The amount he will get after 6 months is Rs. 63600.

 

(ii) after 1 year?

Ans: The amount invested is called Principal, $\left( P \right) = {\text{Rs}}{\text{. }}60,000$.

Rate at which money is invested is given by, $\left( R \right) = 12\%  = 6\% $  per half year.

The number of years money is invested is, $n = 1$ year, that is 2 half-year.

The amount if the sum is compounded annually,

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

Substitute the values and solve further,

$A = {\text{Rs}}{\text{. }}\left[ {60000{{\left( {1 + \dfrac{6}{{100}}} \right)}^2}} \right]$

$A = {\text{Rs}}{\text{. }}\left[ {60000 \times \dfrac{{106}}{{100}} \times \dfrac{{106}}{{100}}} \right]$

$A = {\text{Rs}}{\text{. 67416}}$

Therefore, the amount he will get after one year is Rs. 67416.

 

6. Arif took a loan of Rs 80,000 from a bank. If the rate of interest is $10\% $per annum, find the difference in amounts he would be paying after $1\dfrac{1}{2}$ years if the interest is,

(i) Compounded Annually

Ans: The amount invested is called Principal, $\left( P \right) = {\text{Rs}}{\text{. }}80,000$.

Rate at which money is invested is given by, $\left( R \right) = 10\% $  per year.

The number of years money is invested is, $n = 1\dfrac{1}{2}$ year.

Calculate the amount for 1 year and 6 months by calculating the amount for 1 year using the compound interest formula. After that, calculate the amount of simple interest for 6 months on the amount obtained at the end of 1year.

For this, calculate the amount for 1 year.

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

Substitute the values and solve further,

$A = {\text{Rs}}{\text{. }}\left[ {80000{{\left( {1 + \dfrac{{10}}{{100}}} \right)}^1}} \right]$

$A = {\text{Rs}}{\text{. }}\left[ {80000 \times \dfrac{{11}}{{10}}} \right]$

$A = {\text{Rs}}{\text{. 88000}}$

Consider Rs. 88,000 as principal and calculate the S.I. for the next $\dfrac{1}{2}$ year.

${\text{S}}{\text{.I}}{\text{. =  Rs}}{\text{.}}\left( {\dfrac{{88000 \times \dfrac{1}{2} \times 10}}{{100}}} \right)$

${\text{S}}{\text{.I}} = {\text{Rs}}{\text{. 4400}}$

\[{\text{Interest for the first year }} = {\text{ Rs}}{\text{. }}\left( {88000{\text{ }} - {\text{ }}80000} \right){\text{ }} = {\text{ }}Rs.{\text{ }}8,000\].

And interest for the next $\dfrac{1}{2}$ year = Rs. $4,400$.

The total compound interest is given by,

${\text{C}}{\text{.I}}{\text{.}} = {\text{ Rs}}{\text{. }}\left( {8000 + 4400} \right)$

${\text{C}}{\text{.I}}{\text{.}} = {\text{ Rs}}{\text{. 12400}}$

Total amount is,

${\text{Amount  =  Principal + C}}{\text{.I}}{\text{.}}$

$A = {\text{Rs}}{\text{. 80000 + 1}}2400$

$A = {\text{Rs}}{\text{. 92400}}$

Therefore, the difference in amounts he would be paying after $1\dfrac{1}{2}$ years if the interest is 92400 when compounded annually.

 

(ii) Compounded Half Yearly

Ans: The amount invested is called Principal, $\left( P \right) = {\text{Rs}}{\text{. }}80,000$

Rate at which money is invested is given by, $\left( R \right) = 10\% $ per year  which is $5\% $ per half year.

The number of years money is invested is, $n = 1\dfrac{1}{2}$ year, i.e. three half years.

For this, calculate the amount for 1 year.

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

Substitute the values and solve further,

$A = {\text{Rs}}{\text{. }}\left[ {80000{{\left( {1 + \dfrac{5}{{100}}} \right)}^3}} \right]$

$A = {\text{Rs}}{\text{. }}\left[ {80000 \times {{\left( {1 + \dfrac{1}{{20}}} \right)}^3}} \right]$

$A = {\text{Rs}}{\text{. }}\left[ {80000 \times \dfrac{{21}}{{20}} \times \dfrac{{21}}{{20}} \times \dfrac{{21}}{{20}}} \right]$

$A = {\text{Rs}}{\text{. 92610}}$

The total compound interest is given by,

${\text{C}}{\text{.I}}{\text{.}} = {\text{ Rs}}{\text{. }}\left( {92610 - 92400} \right)$

${\text{C}}{\text{.I}}{\text{.}} = {\text{ Rs}}{\text{. 210}}$

 

7. Maria invested Rs 8,000 in a business. She would be paid interest at 5% per annum compounded annually. Find.

(i) The amount credited against her name at the end of the second year.

Ans: The amount invested is called Principal$\left( P \right) = {\text{ Rs}}{\text{. }}8000$.

Rate at which money is invested is given by, $\left( R \right) = 5\% $per year.

The number of years money is invested is, $n = 2$ year.

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

Substitute the values and solve further,

$A = {\text{Rs}}{\text{. }}\left[ {8000{{\left( {1 + \dfrac{5}{{100}}} \right)}^3}} \right]$

$A = {\text{Rs}}{\text{. }}\left[ {8000 \times {{\left( {1 + \dfrac{1}{{20}}} \right)}^3}} \right]$

$A = {\text{Rs}}{\text{. }}\left[ {8000 \times \dfrac{{21}}{{20}} \times \dfrac{{21}}{{20}} \times \dfrac{{21}}{{20}}} \right]$

$A = {\text{Rs}}{\text{. 8820}}$

The required amount is Rs. 8820.

 

(ii) The interest for the 3rd year. 

Ans: The interest for the next one year which means for the third year needs to be calculated. Consider Rs. 8,820 as principal and calculate the S.I. for the next year.

Simple Interest, $ = \dfrac{{P \times R \times t}}{{100}}$

Substitute the values and solve further,

$ = {\text{Rs}}{\text{. }}\left[ {\dfrac{{8820 \times 5 \times 1}}{{100}}} \right]$

$ = {\text{Rs}}{\text{. 441}}$

The interest after $3$ years is $441Rs.$.

 

8. Find the amount and the compound interest on Rs 10,000 for years at $10\% $ per annum, compounded half yearly. Would this interest be more than the interest he would get if it was compounded annually?

Ans: The amount invested is called Principal$\left( P \right) = {\text{Rs}}{\text{. }}80,000$

Rate at which money is invested is given by, $\left( R \right)$= $10\% $ per year is $5\% $ per half year.

The number of years money is invested is, $n = 1\dfrac{1}{2}$ year, i.e. three half years.

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

Substitute the values and solve further,

$A = {\text{Rs}}{\text{. }}\left[ {10000{{\left( {1 + \dfrac{5}{{100}}} \right)}^3}} \right]$

$A = {\text{Rs}}{\text{. }}\left[ {10000 \times {{\left( {1 + \dfrac{1}{{20}}} \right)}^3}} \right]$

$A = {\text{Rs}}{\text{. }}\left[ {10000 \times \dfrac{{21}}{{20}} \times \dfrac{{21}}{{20}} \times \dfrac{{21}}{{20}}} \right]$

$A = {\text{Rs}}{\text{. 11576}}{\text{.25}}$

The total compound interest is given by,

${\text{C}}{\text{.I}}{\text{.}} = {\text{ Rs}}{\text{. }}\left( {11576.25 - 10000} \right)$

${\text{C}}{\text{.I}}{\text{.}} = {\text{ Rs}}{\text{. 1576}}{\text{.25}}$

Calculate the amount for 1 year and 6 months by calculating the amount for 1 year using the compound interest formula. After that, calculate the amount of simple interest for 6 months on the amount obtained at the end of 1year.

The amount for the first year has to be calculated first.

$A = {\text{Rs}}{\text{. }}\left[ {10000{{\left( {1 + \dfrac{{10}}{{100}}} \right)}^1}} \right]$

$A = {\text{Rs}}{\text{. }}\left[ {10000 \times \left( {\dfrac{{11}}{{10}}} \right)} \right]$

$A = {\text{Rs}}{\text{. 11000}}$

By taking Rs. 11,000 as principal, the S.I. for the next $\dfrac{1}{2}$ year will be calculated.

${\text{S}}{\text{.I}}{\text{. =  Rs}}{\text{.}}\left( {\dfrac{{11000 \times \dfrac{1}{2} \times 10}}{{100}}} \right)$

${\text{S}}{\text{.I}} = {\text{Rs}}{\text{. 550}}$

Interest for the first year = Rs. (11000 − 10000) = Rs. 1,000.

The total compound interest is given by,

${\text{C}}{\text{.I}}{\text{.}} = {\text{ Rs}}{\text{. }}\left( {1000 + 550} \right)$

${\text{C}}{\text{.I}}{\text{.}} = {\text{ Rs}}{\text{. 1550}}$

Therefore, the interest would be more when compounded half yearly than the interest when compounded annually.

 

9. Find the amount which Ram will get on Rs. 4,096, he gave it for 18 months at $12\dfrac{1}{2}\% $ per annum, interest being compounded half yearly.

Ans: The amount invested is called Principal$\left( P \right) = {\text{Rs}}{\text{. }}4,096$.

Rate at which money is invested is given by, $\left( R \right) = 12\dfrac{1}{2}\% $ per annum or $\dfrac{{25}}{4}\% $ per half year.

The number of years money is invested is, $n = 18$ months. There will be 3 half years in $18$ months.

Calculate the amount for 3 half years, taking the rate as $\dfrac{{25}}{4}\% $ per half year.

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

Substitute the values and solve further,

$A = {\text{Rs}}{\text{. }}\left[ {4096{{\left( {1 + \dfrac{{25}}{{400}}} \right)}^3}} \right]$

$A = {\text{Rs}}{\text{. }}\left[ {4096{{\left( {1 + \dfrac{1}{{16}}} \right)}^3}} \right]$

$A = {\text{Rs}}{\text{. }}\left( {4096 \times \dfrac{{17}}{{16}} \times \dfrac{{17}}{{16}} \times \dfrac{{17}}{{16}}} \right)$

$A = {\text{Rs}}{\text{. 4913}}$

Therefore, the required amount is, ${\text{Rs}}{\text{. }}4913$.

 

10. The population of a place increased to 54000 in 2003 at a rate of $5\% $ per annum

(i) Find the Population in 2001.

Ans: It is given that, population in the year 2003 is  54,000.

Rate of increase of population is $5\% $ per year.

Let $x$ be the population in 2001. Use the compound interest formula for finding the value of $x$.

$54000 = x{\left( {1 + \dfrac{5}{{100}}} \right)^2}$

Simplify further to solve for $x$.

$x = 54000 \times \dfrac{{20}}{{21}} \times \dfrac{{20}}{{21}}$

$x = 48979.59$

Thus, the population in the year 2001 was approximately 48,980.

 

(ii) What would be its population in 2005? 

Ans: It is given that, population in the year 2003  is  54,000.

Rate of increase of population is $5\% $ per year.

Let $x$ be the population in 2005. Use the compound interest formula for finding the value of $x$.

$x = 54000{\left( {1 + \dfrac{5}{{100}}} \right)^2}$

Simplify further to solve for $x$.

$x = 54000 \times \dfrac{{21}}{{20}} \times \dfrac{{21}}{{20}}$

$x = 59535$

Thus, the population in the year 2005 would be approximately 59,535.

 

11. In a laboratory, the count of bacteria in a certain experiment was increasing at the rate of $2.5\% $ per hour. Find the bacteria at the end of 2 hours if the count was initially 5,06,000.

Ans: The initial count of bacteria is given as 5,06,000.

Number of bacteria is increasing at the rate of $2.5\% $ per hour.

Bacteria at the end of 2 hours =$506000{\left( {1 + \dfrac{{2.5}}{{100}}} \right)^2}$.

Simplify further,

$506000{\left( {1 + \dfrac{{2.5}}{{100}}} \right)^2} = 506000 \times \dfrac{{41}}{{40}} \times \dfrac{{41}}{{40}} = 531616.25$

Thus, the count of bacteria at the end of 2 hours will be 5, 31,616 (approximately).

 

12. A scooter was bought at Rs 42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.

Ans: The principal is the cost price of the scooter, that is, ${\text{Rs}}{\text{. }}42000$.

The rate of depreciation of a scooter is $8\% $ per year, that means each year the price decreases at the rate of $8\% $.

In one year the price decreases, 

${\text{Rs}}{\text{. }}\dfrac{{42000 \times 8 \times 1}}{{100}} = {\text{Rs}}{\text{. }}3360$

The value of the scooter after 1 year is, ${\text{Rs}}{\text{. 42000 }} - {\text{ Rs}}{\text{. 3360 }} = {\text{ Rs}}{\text{. 38,460}}$.

 

NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities (Ex 8.3) Exercise 8.3

Opting for the NCERT solutions for Ex 8.3 Class 8 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 8.3 Class 8 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

 

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 8 students who are thorough with all the concepts from the Subject Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 8 Maths Chapter 8 Exercise 8.3 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

 

Besides these NCERT solutions for Class 8 Maths Chapter 8 Exercise 8.3, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

 

Do not delay any more. Download the NCERT solutions for Class 8 Maths Chapter 8 Exercise 8.3 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.


Exercise 8.3 of Comparing Quantities is based on following topics:

1. Compound Interest

Compound interest is a type of interest calculated on principal value along with interest, compounded at regular intervals. At regular intervals, the interest so far obtained will be added with the existing principal amount and then the interest is calculated for the new principal value. The new principal is the amount of the initial principal plus the interest earned thus far.


Formula of Compound Interest (C.I)= Interest on Principal (P) + Compounded Interest at Regular Intervals


2. Formula for Compound Interest

C.I = P(1 + r/n)nt - P

Here, 

  • P represents the Principal amount or initial amount

  • r is the interest rate per annum

  • n represents the compounding frequency or the number of times interest is compounded in a  year

  • t represents the number of years compounded


3. Rate Compounded Annually or Half Yearly

In the case of compound interest, the interest rate changes according to the computation time. If the interest is calculated every six months that is half-yearly, then the amount is compounded twice a year. Similarly, if the interest is calculated every 12 months (yearly), the amount is compounded once a year.


4. Applications of Compound Interest Formula

The compound interest formula is widely applied to various fields which measure increase and decrease, growth and decay, appreciation and depreciation and so on.


We Cover Other Exercises given in the Chapter 8 Comparing Quantities of Class 8:-

Chapter 8 - Comparing Quantities Exercises in PDF Format

Exercise 8.1

6 Questions & Solutions (1 Long Answers, 5 Short Answers)

Exercise 8.2

10 Questions & Solutions (4 Long Answers, 6 Short Answers)

FAQs on NCERT Solutions for Class 8 Maths Chapter 8: Comparing Quantities - Exercise 8.3

1. What are the advantages of NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Exercise 8.3?

NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities (Ex 8.3) are really advantageous in exam preparation. These solutions are available on e-learning platforms like Vedantu in a free to download PDF format. At Vedantu, these solutions are provided by subject matter experts. Students must download the PDF of Class 8 Maths Chapter 8 Comparing Quantities for Exercise 8.3 to find the solutions to the exercise questions. These solutions can help to clear any doubt regarding the exercise. Students can also get tips and tricks to solve the exercise in a faster manner with the help of these solutions. One can score well in the exam using exercise-wise NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities.

2. Where can I find NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Exercise 8.3?

Vedantu, a premier online learning platform, provides well-designed NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Exercise 8.3 and other exercises. Students can download this study material from Vedantu as these are 1005 authentic and accurate solutions. These solutions are prepared by subject experts who have years of teaching experience. Vedantu provides the free PDF NCERT Solutions for Class 8 Maths Chapter 8 Exercise 8.3 which include answers to the exercise questions. These solutions are helpful in exam preparation and provide effective revision during exams.

3. Why must students refer to Vedantu’s NCERT Solutions for Class 8 Maths Chapter 8 Exercise 8.3 to score well in exams?

Referring to NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.3 will provide complete coverage of the exercise questions. This is really helpful during exams as it allows students to have a thorough practice. Solutions are provided by experts who give detailed explanations to the questions. The free PDF of NCERT Solutions for CBSE Maths of Class 8 Chapter 8 can be used at the time of revision to learn the important concepts related to the chapter. These are really helpful in scoring well in the paper and must be downloaded by every CBSE Class 8 student.

4. What are the subtopics of Class 8 Maths Chapter 8 Comparing Quantities?

Following are the subtopics of Class 8 Maths Chapter 8 Comparing Quantities as per the NCERT textbook:

  1. Recalling Ratios and Percentages

  2. Finding the Increase or Decrease Per Cent

  3. Finding Discounts

  4. Prices Related to Buying and Selling (Profit and Loss)

  5. Sales Tax/Value Added Tax/Goods and Services Tax

  6. Compound Interest

  7. Deducing a Formula for Compound Interest

  8. Rate Compounded Annually or Half Yearly (Semi Annually)

  9. Applications of Compound Interest Formula

5. How many questions are there in Exercise 8.3 of Chapter 8 from Class 8 Maths and it deals with which concepts?

Exercise 8.3 of Chapter 8 from Class 8 Maths consists of six Long Answer Questions and six Short Answer Questions. Exercise 8.3 deals with the concepts based on 

  • Compound Interest

  • Deducing a Formula for Compound Interest

  • Rate Compounded Annually or Half Yearly 

  • Applications of Compound Interest Formula

Students can easily find the questions and solutions based on these concepts on the official site of Vedantu.

6. Where can I download the NCERT Solutions for Exercise 8.3 of Chapter 8 from Class 8 Maths PDF online?

Students can download the NCERT Solutions for Class 8 Maths Chapter 8 Exercise 8.3 for free on the Vedantu study portal. All the solutions from the exercise are curated by expert teachers having vast conceptual knowledge and experience. NCERT solutions are one of the best-rated solutions available online. All the questions are solved in a stepwise manner based on the CBSE latest syllabus and the exam pattern designed by the CBSE board.

7. Are NCERT Solutions for Exercise 8.3 of Chapter 8 from Class 8 Maths PDF enough to score well in the exams?

Following NCERT Solutions is absolutely enough for the students to score the highest marks in the exam. Students can view the solutions online or download them for free from the Vedantu website. They can access it anywhere and anytime without any time constraints. Practising regularly and revising thoroughly will help the students to achieve the marks they want. It will also help them to improve their time management and problem-solving abilities. Exercise-wise solutions are provided to help the students gain knowledge of the concepts.

8. Do NCERT Solutions for Exercise 8.3 of Chapter 8 from Class 8 Maths have answers for all the textbook questions?

Vedantu provides NCERT Solutions for all the questions present in Exercise 8.3 of Chapter 8 from Class 8 Maths. All the solutions are well explained in a detailed stepwise manner by the expert teachers of Vedantu. Students can clear all their doubts instantly by using these solutions. We provide accurate solutions making it easier for the students to save their time during homework as well as with their exam preparation. The solutions are free of cost and available on Vedantu Mobile app

9. Do I need to make notes while referring to NCERT Solutions for Exercise 8.3 of Chapter 8 of Class 8 Maths?

Students are highly recommended to make notes in their own words while studying from the NCERT Solutions. Making notes is a best practice as it will help the students to quickly recall the concepts they have studied earlier. For some problems, writing down the methods in an easy way will help the students to remember for a long time. Therefore, this will help the students to complete the revision at twice the pace and memorise twice the amount of information compared to studying without notes.