NCERT Solutions Class 8 Maths Chapter 7 Comparing Quantities Exercise 7.3

NCERT Solutions Class 8 Maths Chapter 7 Comparing Quantities Exercise 7.3 helps students understand the concept of compound interest. Unlike simple interest, compound interest is calculated on both the principal and the interest added over time. This exercise teaches students how money grows over time when interest is added regularly, which is a very useful skill for real-life money matters.

The solutions use easy methods and examples so that students can follow along without confusion. By practicing NCERT Solutions from Exercise 7.3, students will gain a strong understanding of compound interest and become more confident in solving financial maths problems.

1.0Download NCERT Solutions Class 8 Maths Chapter 7 Comparing Quantities Exercise 7.3: Free PDF

Download the free NCERT Solutions Class 8 Maths Chapter 7 Comparing Quantities Exercise 7.3 PDF with step-by-step answers to help you practice and understand concepts easily.

2.0Key Concepts in Exercise 7.3 of Class 8 Maths Chapter 7

Exercise 7.3 focuses on understanding ratios related to cost price, selling price, profit, and loss, making it essential for solving real-life business and shopping-related problems. The main concepts covered in this exercise include:

• Cost Price (CP) and Selling Price (SP): Students learn how to differentiate between the original price (cost price) and the price at which the item is sold (selling price).
• Profit and Loss:
• • Profit occurs when SP > CP
  • Loss occurs when CP > SP
  • These are calculated using the formulas:
  • • $$\begin{gathered} Profit=SP–CP \\ Loss=CP–SP \end{gathered}$$
• Percentage Profit and Loss: Students are taught to calculate profit or loss as a percentage using:
• $$\begin{gathered} \text{Profit}\%=\frac{\text{Profit}}{\text{CP}}\times 100 \\ \text{Loss}\%=\frac{\text{Loss}}{\text{CP}}\times 100 \end{gathered}$$
• Application Through Word Problems: The exercise includes real-life situations like buying and selling articles, helping students apply the concepts practically.
• Analytical Thinking: It strengthens the logical and step-by-step calculation skills needed for commercial mathematics.

Practical Use: These concepts are helpful in daily life activities like shopping, billing, and budgeting.

3.0NCERT Class 8 Maths Chapter 7: Other Exercises

4.0NCERT Class 8 Maths Chapter 7 Exercise 7.3: Detailed Solutions

1. The population of a place increased to 54000 in 2003 at a rate of $5\%$ per annum (i) Find the population in 2001. (ii) What would be its population in 2005? Sol. (i) Let $P$ be the population in 2001 , i.e., 2 years ago Then, Present population $=\mathrm{P}\times {\left(1+\frac{5}{100}\right)}^2$ $\Rightarrow 54000=\mathrm{P}\times \frac{105}{100}\times \frac{105}{100}$ $\Rightarrow \mathrm{P}=\frac{54000\times 100\times 100}{105\times 105}=48979.59$ Hence, the population in the year 2001 is 48980 (approx). (ii) Let $\mathrm{P}=$ Initial population $=54000$, i.e., in the year 2003. $\therefore$ Population after 2 years, i.e., in 2005 $=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{n}}=54000\times {\left(1+\frac{5}{100}\right)}^2$ $=54000\times \frac{105}{100}\times \frac{105}{100}=59535$
2. 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. Sol. We have, $\mathrm{P}=$ Original count of bacteria $=$ 506000, Rate of increase $=\mathrm{R}=2.5\%$ per hour, Time $=2$ hours. $\therefore$ Bacteria count after 2 hours $=506000\times {\left(1+\frac{2.5}{100}\right)}^2$ $=506000\times \frac{102.5}{100}\times \frac{102.5}{100}$ = 531616.25 = 531616 (approx.)
3. A scooter was bought at $₹42000$. Its value depreciated at the rate of $8\%$ per annum. find its value after one year. Sol. We have, ${\mathrm{V}}_0=$ Initial value $=₹42000$ $R=$ Rate of depreciation $=8\%$ p.a. $\therefore$ Value after 1 year $={\mathrm{V}}_0\left(1-\frac{\mathrm{R}}{100}\right)$ $=₹\left[42000\times \left(1-\frac{8}{100}\right)\right]$ $=₹\left(42000\times \frac{92}{100}\right)=₹38640$
4. Calculate the amount and compound interest on (i) ₹ 10,800 for 3 years at $12\frac{1}{2}\%$ per annum compounded annually. (ii) ₹ 18,000 for $2\frac{1}{2}$ years at $10\%$ per annum compounded annually. (iii) ₹ 62,500 for $1\frac{1}{2}$ years at $8\%$ per annum compounded half yearly. (iv) ₹ 8,000 for 1 year at $9\%$ per annum compounded half yearly. (You could use the year by year calculation using S.I. formula to verify). (v) ₹ 10,000 for 1 year at $8\%$ per annum compounded half yearly. Sol. (i) We have, $\mathrm{P}=₹10,800$ $\mathrm{R}=12\frac{1}{2}\%$ p.a. $=\left(\frac{25}{2}\right)\%$ per annum and $\mathrm{n}=3$ years. $\therefore$ Amount after 3 years $=₹\left[10800\times {\left(1+\frac{25}{2\times 100}\right)}^3\right]$ $=₹\left(10800\times \frac{9}{8}\times \frac{9}{8}\times \frac{9}{8}\right)$ = ₹ 15377.34 $\therefore$ C.I. $=₹(15377.34-10800)$ $\text{ = ₹ }4577.34$ (ii) We have, $\mathrm{P}=₹18,000$ $\mathrm{R}=10\%$ per annum $\mathrm{n}=2.5$ years or $2\frac{1}{2}$ years $\therefore$ Amount $\mathrm{A}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{n}}$ At the end of 2 years, $A=18000{\left(1+\frac{10}{100}\right)}^2=₹21780$ $\therefore$ Amount after 2.5 years $=21780\left(1+\frac{10\times \frac{1}{2}}{100}\right)=₹22869$ $\therefore$ Compound interest $=\mathrm{A}-\mathrm{P}$ $=22869-18000=₹4869$ (iii) Here, Principal = ₹ 62,500 Rate $=8\%$ per annum $=4\%$ per half year, Time $=1\frac{1}{2}$ years $=3$ half years $\therefore$ Amount $=₹\left[62500\times {\left(1+\frac{4}{100}\right)}^3\right]$ $=\left(62500\times \frac{104}{100}\times \frac{104}{100}\times \frac{104}{100}\right)$ = ₹ 70304 $\therefore$ C. $I=₹(70304-62500)$ = ₹ 7804 (iv) We have, $\mathrm{P}=₹8000$ $\mathrm{R}=9$ % per annum $=4.5$ % per half year $\mathrm{n}=1$ year = 2 half years So, amount $A=P{\left(1+\frac{R}{100}\right)}^n$ $=8000{\left(1+\frac{4.5}{100}\right)}^2=₹8736.20$ Hence, C.I. $=\mathrm{A}-\mathrm{P}=₹736.20$ (v) We have, $\mathrm{P}=₹10000$ $\mathrm{R}=8\%$ per annum $=4\%$ per half year $\mathrm{n}=1$ year $=2$ half years . Thus, amount $A=P{\left(1+\frac{R}{100}\right)}^n$ $=10000{\left(1+\frac{4}{100}\right)}^2=₹10816$ $\therefore$ Compound interest $=\mathrm{A}-\mathrm{P}=₹816$
5. Kamla borrowed ₹ 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? Sol. We have, $\mathrm{P}=₹26400$ R $=15\%$ per annum $\mathrm{n}=2$ years 4 months At the end of 2 years, $A=P{\left(1+\frac{R}{100}\right)}^n$ $A=26400{\left(1+\frac{15}{100}\right)}^2=₹34914$ Now, P = ₹ 34914 $\mathrm{R}=15\%$ per annum $\mathrm{n}=4$ months $=\frac{1}{3}$ year At the end of 2 years, 4 months $A=\left(1+\frac{15\times 1}{100\times 3}\right)34914=₹36659.70$
6. Fabina borrows ₹ 12500 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? Sol. In case of Fabina: $\mathrm{P}=$ Rs $12500,\mathrm{R}=12\%$ per annum and $\mathrm{T}=$ 3 years. Then, S.I. $=\frac{\mathrm{P}\times \mathrm{R}\times \mathrm{T}}{100}=₹\left(\frac{12500\times 12\times 3}{100}\right)$ = ₹ 4500 In case of Radha : Amount $=₹\left[12500\times {\left(1+\frac{10}{100}\right)}^3\right]$ $=₹\left(12500\times \frac{110}{100}\times \frac{110}{100}\times \frac{110}{100}\right)$ = ₹ 16637.50 $\therefore$ Compound interest $=₹$ (16637.50 12500) $=₹4137.50$ $₹(4500-4137.50)=₹362.50$ Hence, Fabina pays 362.50 more as interest as compared to Radha.
7. I borrowed ₹ 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? Sol. We have, $\mathrm{P}=₹12000$ $\mathrm{R}=6\%$ per annum $\mathrm{n}=2$ years For simple interest, $I_1=\frac{P\times R\times T}{100}$ $=\frac{12000\times 6\times 2}{100}=₹1440$ For compound interest, I2 $=\mathrm{P}\left[{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{n}}-1\right]$ $=12000\left[{\left(1+\frac{6}{100}\right)}^2-1\right]=₹1483.20$ So, he would have to pay $1483.20-1440$ = ₹ 43.20
8. Vasudevan invested $₹60,000$ at an interest rate of $12\%$ per annum compounded half-yearly. What amount would he get (i) After 6 months (ii) After 1 year. Sol. Here, Principal $=₹60000$, Rate $=12\%$ per annum $=6\%$ per half year. (i) Time $=6$ months $=1$ half-year $\therefore$ Amount after 6 months = Amount after 1 half-year $$\begin{gathered} =₹\left[60000\times {\left(1+\frac{6}{100}\right)}^1\right] \\ =₹\left(60000\times \frac{106}{100}\right)=₹63600 \end{gathered}$$ (ii) Time $=1$ year $=2$ half-years $\therefore$ Amount after 1 year $$\begin{gathered} =₹\left[60000\times {\left(1+\frac{6}{100}\right)}^2\right] \\ =₹\left(60000\times \frac{106}{100}\times \frac{106}{100}\right)=₹67416 \end{gathered}$$
9. Arif took a loan of ₹ 80,000 from a bank. If the rate of interest is $10\%$ per annum, find the difference in amounts he would be paying after $1\frac{1}{2}$ years if the interest is - (i) Compounded annually (ii) Compounded half yearly Sol. We have, $\mathrm{P}=₹80000$ $\mathrm{R}=10\%$ per annum $=5\%$ per half year (i) If interest is compounded annually $\mathrm{n}=1\frac{1}{2}$ years $=1$ years $+\frac{1}{2}$ year Amount after 1 year $A=80000\left(1+\frac{10}{100}\right)=₹88000$ Amount after $1\frac{1}{2}$ years $A=88000\left(1+\frac{10\times \frac{1}{2}}{100}\right)=₹92400$ (ii) If interest is compounded half-yearly $\mathrm{n}=1\frac{1}{2}$ years $=3$ half years $\therefore$ Amount $\mathrm{A}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{n}}$ $=80000{\left(1+\frac{5}{100}\right)}^3=₹92610$ $\therefore$ Difference in amounts $=92610-92400=₹210$
10. Maria invested ₹ 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. (ii) the interest for the 3rd year. Sol. (i) Here, $P=₹8000,R=5\%$ per annum and $\mathrm{n}=2$ years. $\therefore$ Amount after 2 years $=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{n}}$ $$\begin{gathered} =₹\left[8000\times {\left(1+\frac{5}{100}\right)}^2\right] \\ =₹\left(8000\times \frac{105}{100}\times \frac{105}{100}\right) \\ =₹8820 \end{gathered}$$ (ii) Principal for the 3rd year = ₹ 8820 Interest for the 3rd year $$\begin{gathered} =\left(\frac{8820\times 5\times 1}{100}\right) \\ =₹441 \end{gathered}$$
11. Find the amount and compound interest on ₹ 10000 for $1\frac{1}{2}$ years at $10\%$ per annum, compounded half yearly. Would this interest be more than the interest he would get if it was compounded annually? Sol. We have, $\mathrm{P}=₹10000$ $\mathrm{R}=10\%$ per annum $=5\%$ per half year $\mathrm{n}=1\frac{1}{2}$ year $=3$ half years If interest is compounded half yearly $A=10000{\left(1+\frac{5}{100}\right)}^3=₹11576.25$ Interest $=11576.25-10000$ = ₹ 1576.25 If interest is compounded annually. Amount after 1 year, $A=10000\left(1+\frac{10}{100}\right)=₹11000$ Amount after $1\frac{1}{2}$ year $A=11000\left(1+\frac{10\times 1}{100\times 2}\right)=₹11550$ Interest $=11550-10000=₹1550$ Thus, more interest would be generated if interest is calculated half yearly.
12. Find the amount which Ram will get on ₹ 4096, if he gave it for 18 months at $12\frac{1}{2}\%$ per annum, interest being compounded half yearly. Sol. Here, Principal = ₹ 4096, Time $=18$ months $=3$ half years Rate $=12\frac{1}{2}\%$ per annum $=\left(\frac{25}{4}\right)\%$ per half year. $\therefore$ Amount $=\left[4096\times {\left(1+\frac{25}{4\times 100}\right)}^3\right]$ $=\left(4096\times \frac{17}{16}\times \frac{17}{16}\times \frac{17}{16}\right)$ = ₹ 4913

5.0Key Features and Benefits of Class 8 Maths Chapter 7 Exercise 7.3

• Real-World Word Problems: The questions are based on situations students may encounter while buying or selling items which makes the learning more authentic.
• Easy to Follow Steps: The solutions consist of easy-to-follow steps that reduce the difficulty and time taken to complete the word problems for students.
• Helps Improve Logical Thinking: Working on these kinds of problems acts to reinforce decision making and analytical thought.
• As per NCERT Syllabus: The exercise is directly from the official NCERT book so it is really suitable for classwork, homework and exam preparation.