NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities

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# NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities

Edited By Ramraj Saini | Updated on Oct 09, 2023 04:55 PM IST

## NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities

Comparing Quantities Class 8 Questions And Answers provided here. These NCERT Solutions are created by expert team at craeers360 keeping the latest syllabus and pattern of CBSE 2023-23. This chapter includes several important topics, like ratios, percentages, discount, profit & loss, and simple and compound interest. The ratio of two quantities is known as comparing quantity. In this chapter, we are going to compare quantities like marks of two students, the height of two persons, a comparison of profit, sales value, etc.

In the NCERT solutions for Class 8 Maths chapter 8 Comparing Quantities, there are 3 exercises with 26 questions. The first exercise of CBSE Class 8 chapter Comparing Quantities is on percentage and ratio. The second exercise is on finding the increase or decrease percent and finding discounts. Through these NCERT solutions for Class 8 Maths chapter 8 Comparing Quantities, students can cover all the problems related to ratios, percentages, discount, profit-loss, and simple-compound interest. Here you will get the detailed NCERT Solutions for Class 8 Maths by clicking on the link.

## Comparing Quantities Class 8 Solutions - Important Formulae

• Profit: Profit = Selling price - Cost price

• Loss: Loss = Cost price - Selling price

• Profit or Loss Determination:

• If SP > CP, then it is a profit.

• If SP = CP, then it is neither profit nor loss.

• If CP > SP, then it is a loss.

• Discount: Discount = Marked Price - Sale Price

• Discount Percentage: Discount % = (Discount × 100) / Marked Price

• Profit Percentage: Profit Percentage = (Profit / Cost Price) × 100

• Loss Percentage: Loss Percentage = (Loss / Cost Price) × 100

• Percentage Increased: Percentage Increased = (Change in Value / Original Value) × 100

• Simple Interest: Simple Interest = (Principal × Rate × Time) / 100

• Compound Interest Formula: Compound Interest = Amount - Principal

• Sales Tax or VAT: Sales tax or VAT = (Cost Price × Rate of Sales Tax) / 100

• Billing Amount: Billing Amount = Selling price + VAT

Free download NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities for CBSE Exam.

## Comparing Quantities Class 8 NCERT Solutions (Intext Questions and Exercise)

Comparing quantities class 8 NCERT solutions - Topic 8.1

(i)

Using this, answer the following :

(i) How many parents were surveyed?

Let the number of parents surveyed be X.

30% of them helped for $\frac{1}{2}$ hour to $1\frac{1}{2}$ hours.

In total, there were 90 such parents.

$\therefore$ 30% of X = 90

$\Rightarrow \frac{30}{100}\times X = 90 \\ \Rightarrow X = 90\times100 /30 = 300$

Therefore, there were a total of 300 parents who were surveyed.

(ii)

Using this, answer the following :

(ii) How many said that they did not help?

from the previous question number of parents surveyed X = 300.

Given, 50% did not help.

Therefore, number of parents that did not help = 50% of 300 = 150.

(iii)

to $1\frac{1}{2}$ hours. The distribution of parents according to the time for which they said they helped is given in the adjoining figure; 20% helped for more than $1\frac{1}{2}$ hours; 30% helped for $\frac{1}{2}$ hour to $1\frac{1}{2}$ hours; 50 % did not help at all.

Using this, answer the following :

(iii) How many said that they helped for more than $1\frac{1}{2}$ hours?

from the first question number of parents surveyed X = 300

Given, 20% helped for more than $1\frac{1}{2}$ hours.

Therefore, number of such parents = 20% of 300 = 2 x 30 = 60.

Class 8 comparing quantities NCERT solutions - Topic 8.3

(a) A dress marked at Rs 120.

The marked price of dress = Rs120

Discount rate = 20%

$\\\\ \\Total\ discount =Marked\ Price\times \frac{Discount\ rate}{100}\\ \\Total\ discount=120\times \frac{20}{100}\\ \\Total\ discount=Rs.24\\$

Selling price (SP) = Marked price - Total discount = Rs (120- 24) = Rs 96.

(b) A pair of shoes marked at Rs 750

The marked price of shoes=Rs 750

Discount rate = 20%

$\\Total\ discount =Marked\ Price\times \frac{Discount\ rate}{100}\\ \\Total\ discount=750\times \frac{20}{100}\\ \\Total\ discount=Rs.150\\$

Selling price= Marked price- Total discount =Rs (750- 150) = Rs 600.

(c) A bag marked at Rs 250

The marked price of bag = Rs 250

Discount rate=20%

$\\Total\ discount =Marked\ Price\times \frac{Discount\ rate}{100}\\ \\Total\ discount=250\times \frac{20}{100}\\ \\Total\ discount=Rs.50\\$

Selling price = Marked price - Total discount = Rs (250- 50) = Rs 200.

Marked price = Rs 15000

Selling Price = Rs 14400

We know, Discount amount = Marked price - Selling price

Discount amount = Rs 15000-Rs 14400 = Rs 600

$\\Discount\ percent=\frac{Discount}{Marked\ Price}\times 100\\ \therefore Discount\ percent=\frac{600}{15000}\times 100=4\%\\$

Therefore, Discount percentage = 4%

Selling price = Rs 5225

Discount percent = 5%

$\\ Selling\: price = Marked \:price - Discount \:Amount\\ Selling\: price=Marked\ price-Marked\ price\times \frac{5}{100}\\ Selling\: price=Marked price\times \frac{95}{100}\\ \\Marked\ price= Selling\ price\times \frac{100}{95}\\ \\Marked\ price=\frac{100}{95} \times 5225\\$

Therefore, the marked price = Rs 5500

NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Topic 8.4.1

(a) A cycle of Rs 700 with Rs 50 as overhead charges.

$\therefore$ Cost price = Rs 700 + Rs 50 = Rs 750

Profit percent = 5%

We know,

$\dpi{100} \\Profit= Cost\ price\times \frac{Profit\ percent}{100} \\Profit=750\times \frac{5}{100}$

$\therefore$ Profit = Rs 37.5

Now, Selling price (SP) = Cost price + Profit =Rs 750 + Rs 37.5 = Rs 787.50

Therefore, the selling price (SP) is Rs 787.50

(b) a lawnmower bought at Rs 1150 with Rs 50 as transportation charges.

If a profit of 5% is made on a lawnmower bought at Rs 1150 with Rs 50 as transportation charges then selling price is:

$\\Cost\ price=Rs\ 1150+Rs\ 50\\ \\Cost\ price=Rs\ 1200\\ \\Profit\ percent=5 \\Profit= Cost\ price\times \frac{Profit\ percent}{100} \\Profit=1200\times \frac{5}{100} \\Profit=Rs\ 60 \\Selling\ price= Cost\ price+Profit \\Selling\ price=Rs\ 1200+Rs\ 60 \\Selling\ price=Rs\ 1260$

(c) a fan bought at Rs 560 and expenses of Rs 40 made on its repairs.

$\therefore$ Cost price = Rs 560 + Rs 40 = Rs 600

Profit percent = 5%

$\dpi{100} \\Profit= Cost\ price\times \frac{Profit\ percent}{100} \\Profit=600\times \frac{5}{100}$

$\therefore$ Profit = Rs 30

Now, Selling price (SP) = Cost price + Profit =Rs 600 + Rs 30 = Rs 630

Therefore, the selling price (SP) is Rs 630

$\\Profit\ on\ first\ TV=10000\times \frac{10}{100}=Rs\ 1000\\ \\Loss\ on\ second\ TV=10000\times \frac{10}{100}=Rs\ 1000\\$

Net profit = Rs 1000 - Rs 1000 = 0

He neither made an overall profit or loss since the profit made on the first tv equals the loss suffered on the second one.

Let the number be x.

Half the number = $\frac{x}{2}$

$\\Decrease\ in\ percent=\frac{Number-Half\ the\ number}{Number}\times 100\\ =\frac{x-\frac{x}{2}}{x}\times 100\\ =\frac{1}{2}\times 100\\$

= 50 %

Therefore, If we take half the number, the decrease in percent is 50%

Difference between Rs 2400 and Rs 2000 = Rs 2400- Rs 2000 = Rs 400

$\therefore$ The per cent by which Rs 2,000 is less than Rs 2,400 is

$\\ \frac{400}{2400}\times 100=16.66 \%\\$ ( Less with respect to 2400. Hence, 2400 will be in the denominator! )

Rs 2000 is less than Rs 2400 by 16.66%

Now,

$\therefore$ The per cent by which Rs 2,400 is more than Rs 2,000 is

$\frac{400}{2000}\times 100=20 \%$ ( More with respect to 2000. Hence, 2000 will be in the denominator! )

Rs 2400 is more than Rs 2000 by 20%

Therefore, they are not the same.

NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Topic 8.6

$\\Interest\ =Principal\ amount\times\frac{Interest\ rate}{100}\times time\\ \:\:=15000\times \frac{5}{100}\times 2\\$

=Rs1500

Amount to be paid = Interest + Principal Amount

=Rs 15000 + Rs 1500

=Rs 16500

Find the time period and rate for each.

1. A sum taken for $1\frac{1}{2}$ years at 8% per annum is compounded half yearly.

Since the sum taken is compounded half yearly:

$\\Since\ the\ sum\ taken\ is\ compounded\ half\ yearly:\\ Time\ period=2\times 1\frac{1}{2}\\ Time\ period=3\\ Rate=\frac{8}{2}=4\%$

Time period = 3 half years

Rate = 4% per half year

NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Topic 8.8

Find the time period and rate for each.

A sum taken for 2 years at 4% per annum compounded half yearly.

Since the sum taken is compounded half yearly:

$\\Since\ the\ sum\ taken\ is\ compounded\ half\ yearly:\\ Time\ period=2\times 2\\ Time\ period=4\\ Rate=\frac{4}{2}=2\%$

Time period = 4 half years.

Rate = 2% per half year

Find the amount to be paid

$\\A_{n}=P_{1}(1+\frac{R}{100})^{n}\\ P_{1}=Rs\ 2400\\ n=2\\ R=5\\ A_{2}=2400\times (1+\frac{5}{100})^2\\ A_{2}=Rs\ 2646\\$

The amount to be paid at the end of 2 years is Rs 2646

Find the amount to be paid

At the end of 1 year on Rs. 1800 at 8% per annum compounded quarterly.

$\\A_{n}=P_{1}(1+\frac{R}{100})^{n}\\ P_{1}=Rs\ 1800\\ n=4\\ R=\frac{8}{4}=2\ percent\ per\ quarter\ year\\ A_{4}=1800\times (1+\frac{2}{100})^4\\ A_{4}=Rs\ 1948.37\\$

The Amount to be paid at the end of 4 years is Rs 1948.37

NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Topic 8.9

Price of machinery = Rs. 10,500

$Depriciation=10500\times \frac{5}{100}=Rs.\ 525$

Value after one year = Rs 10,500- Rs 525 = Rs 9975

The current population of city

= 12 lakh

1200000

Population after two years =

$P\left ( 1+\frac{R}{100} \right )^{n}$

$= 1200000\left ( 1+\frac{4}{100} \right )^{2}$

$= 1200000\left ( 1+0.04 \right )^{2}$

$= 1200000\left ( 1.04 \right )^{2}$

$= 1200000\times 1.0816$

$= 1297920$

Thus, the population after two years is 1297920.

Class 8 maths chapter 8 question answer - Exercise 8.1

Speed of a cycle 15 km per hour to the speed of scooter 30 km per hour.

The ratio of the speed of cycle to the speed of scooter =

$\frac{15\ km/ hour}{30\ km/ hour} = \frac{15}{30} = \frac{1}{2}$

= 1:2

5m to 10km

To find the ratio, we need to make both quantities of the same unit.

We know, 1 km = 1000 m.

$\therefore$ 10 km = 10x1000 m = 10000 m

Therefore the required ratio =

$\frac{5m}{10\times1000m} = \frac{5}{10000} = \frac{1}{2000}$

= 1 : 2000

50 paise to Rs. 5

To find the ratio, we first make the quantities of the same unit.

We know, Rs. 1 = 100 paisa. $\therefore$ Rs. 5 = 5x100 = 500 paisa

Therefore, the required ratio =

$\frac{50 paisa}{Rs. 5}= \frac{50 paisa}{5\times100 paisa} = \frac{1}{10}$

= 1:10

(a) 3:4

To convert a ratio to a percentage, we multiply the ratio by 100%.

$\therefore$ The required percentage of the ratio 3:4 =

$\frac{3}{4}\times100\ \%= 75\%$

(b) $2:3$

To convert a ratio to a percentage, we multiply the ratio by 100%.

$\therefore$ The required percentage of ratio 2:3 =

$\frac{2}{3}\times100\ \%= 66.67 \%$

Given;

72% of 25 students are interested in mathematics

$\therefore$ Percentage of students not interested in mathematics = (100 - 72)% = 28 %

$\therefore$ The number of students not interested in mathematics = 28% of 25

$=\frac{28}{100}\times25 = \frac{28}{4}$

$= 7$

Therefore, 7 students (out of 25) are not interested in mathematics.

Given,

Win percentage of the team = 40%

This means that they won 40 matches out of 100 matches played.

$\therefore$ They won 10 matches out of $\frac{100}{40}\times10$ = 25 matches played.

Therefore, they played a total of 25 matches in all.

Let the amount of money Chameli had in the beginning = Rs. X

She spends 75% of the money.

$\therefore$ Percentage of money left = (100 - 75)% = 25%

Since she has Rs. 600 left.

$\therefore$ 25% of X = 600

$\Rightarrow \frac{25}{100}\times X = 600 \\$

$\Rightarrow \frac{X}{4} = 600$

$\\ \Rightarrow X = 600\times4$

X =2400

Therefore, Chameli had Rs. 2400 with her in the beginning.

Given,

Total number of people = 50 lakhs

Percentage of people who like cricket = 60%

Percentage of people who like football = 30%

Since remaining people like other games,

$\therefore$ Percentage of people who like other games = {100 - (60 + 30) } = (100 - 90) = 10%

10% of the people like other games.

Now,

$\therefore$ Number of people who like cricket = 60% of 50 lakhs

$=\frac{60}{100}\times50= 30 lakhs$

$\therefore$ Number of people who like football = 30% of 50 lakhs

$=\frac{30}{100}\times50= 15 lakhs$

$\therefore$ Number of people who like other games = 10% of 50 lakhs

$=\frac{10}{100}\times50 = 5 lakhs$

Class 8 maths chapter 8 NCERT solutions - Exercise 8.2

Given,

Percentage increase in the salary = 10%

Therefore, if the original salary was Rs. 100, the new salary is Rs. 110

If new salary is Rs 1,54,000, the original salary was

$=Rs. \frac{100}{110}\times154000= Rs. 140000$

Therefore, the original salary was Rs. 14,00,00.

Given,

Number of people who went to the zoo on Sunday = 845

Number of people who went to the zoo on Monday = 169

$\therefore$ The decrease in the number of people visiting the zoo = (845 - 169) = 676

$\therefore$ Percentage decrease

$=\frac{676}{845}\times100 \%$

$=80\%$ (Decrease from the original number)

The percent decrease in the people visiting the Zoo is 80%

Given,

Cost price (CP) of the 80 articles = Rs. 2,400

Profit percentage = 16%

$\therefore$ Profit amount on all 80 articles = 16% of 2400 = 16/100 x 2400 = Rs. 384

$\therefore$ The selling price of the 80 articles = Rs. (2400 + 384) = Rs. 2784

$\therefore$ Selling price (SP) of each item = Rs. 2784/80 = Rs. 34.8

Therefore, the selling price of one article is Rs.34.8

Given,

Cost of the article = Rs. 15500

Cost of repair = Rs. 450

$\therefore$ Cost price(CP) of the article = Rs. (15500 + 450) = Rs. 15950

Profit percentage = 15%

$\therefore$ Profit amount = 15% of Rs.15950 = Rs. 2392.50

$\therefore$ Selling Price = Rs.(15950 + 2392.50) = Rs. 18342.50

OR

$\therefore$ Selling price (SP) of the article = CP + Profit = CP + (15% of CP)

= 115% of CP

$=\frac{115}{100}\times15950= Rs. 18342.50$

Given,

The cost price of the VCR = Rs. 8000

The cost price of TV = Rs. 8000

Now, He made a loss of 4% on VCR.

$\therefore$ Selling price (SP) of the VCR = (100-4)% of CP = 96% x Rs. 8000 = Rs. 7680

Again, He made a profit of 8% on TV.

$\therefore$ Selling price (SP) of the TV = (100+8)% of CP = 108% x Rs. 8000 = Rs. 8640

Net Selling price = Rs. (7680 + 8640) = Rs. 16320

Net Cost price = Rs. (2x8000) = Rs. 16000

Since, SP > CP, he made a net profit (gain)

Now, Net Gain = CP - SP = Rs. (16320-16000) = Rs. 320

$\therefore$ Gain % = $\frac{Profit}{CP}\times100\% = \frac{320}{16000}\times100\%$ = 2%

$\therefore$ The shopkeeper made a gain of 2% on the whole transaction.

Since the 10% discount is on all the items, we can calculate the Selling price by totalling the Cost price of all item bought.

Now,

Total Cost price(CP) of the items he bought = CP of a pair of jeans + CP of two shirts = Rs.(1450 + 850 + 850) = Rs. 3150

$\therefore$ The selling price of these items = (100- 10)% of Rs. 3150 = 90% x Rs.3150 = Rs. 2835

$\therefore$ The customer has to pay Rs. 2835.

Given,

The milkman sold two of his buffaloes for Rs 20,000 each. This is the Selling price of the buffaloes.

Let CP of one of the buffalo be Rs. X and the other be Rs. Y

Since he made a profit of 5% on one of them.

$\therefore$ 105% of X = Rs. 20000

$\\ \implies \frac{105}{100}\times X = 20000 \\ \implies X = \frac{100}{105}\times20000$

$\therefore$ X = Rs. 19047.6

Similarly, since he made a loss of 10% on the other.

$\therefore$ 90% of Y = Rs. 20000

$\\ \implies \frac{90}{100}\times Y = 20000 \\ \implies Y = \frac{100}{90}\times20000$

$\therefore$ Y = Rs. 22222.2

$\therefore$ Net CP = Rs.(19047.6 + 22222.2) = Rs. 41269.8

And net SP = 2 x Rs.20000 = Rs. 40000

Since SP < CP, he made a net loss.

$\therefore$ His overall loss = Rs. (41269.8 - 40000) = Rs. 1269.8 = Rs. 1270 (approx)

Given,

Cost price of the TV = Rs. 13000

Sales tax at the rate of 12%

$\therefore$ Selling price = CP + Sales Tax = CP + 12% of CP

= 112% of CP =

$\frac{112}{100}\times 13000 = Rs. 14560$

$\therefore$ Vinod will have to pay Rs. 14,560.

The discount given was 20% which means if CP is Rs. 100 then the SP is Rs. 80

$\therefore$ If SP is Rs. 80 then CP is Rs. 100

For SP of Rs. 1600, CP

$=\frac{100}{80}\times1600 = Rs. 2000$

$\therefore$ The marked price is Rs. 2000.

Given,

VAT = 8%

Let the original price be Rs. 100

Original price + VAT = Rs. 100 + Rs. $\left ( 8\%\, of\, 100 \right )$

Original price + VAT = Rs. 100 + Rs. 8 = Rs. 108

$\therefore$ If the price after VAT is Rs.5400, then the price before VAT is $Rs. \frac{100}{108}\times5400 = Rs. 5000$

$\therefore$ The price before VAT was added is Rs. 5000.

Given,

GST = 18%

Cost with GST included = Rs. 1239

Cost without GST = x Rs.

$x + ( \frac{18}{100} \times x ) = 1239$

cost before GST+ GST = cost with GST

$x + ( \frac{9x}{50} ) = 1239$

x = 1050

Price before GST = 1050 rupees

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

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

Given,

Principal,P = Rs 10800

Compound Interese Rate,R = $12\frac{1}{2} \% = \frac{25}{2}\%$ p.a.

Time period,n = 3 years.

We know,

Amount when interest is compounded annually, A = $P(1+\frac{R}{100})^n$

Therefore, the required amount = $10800(1+\frac{25}{2\times100})^{3}$ = Rs. 15377.34

And Compound Interest, CI = Amount - Principal = Rs. (15377.34 - 10800) = Rs. 4577.34

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

Given,

Principal,P =Rs.18000, Rate,R = 10% and time period,n = 2.5 years.

We know, Amount when interest is compounded annually = $P(1+\frac{R}{100})^n$

Amount after 2 years at 10% , A = $18000(1+\frac{10}{100})^2$ = Rs.21780

This acts as the principal amount for the next half year.

SI on next 1/2 year at = $\frac{21780\times\frac{1}{2}\times10}{100}$ = Rs. 1089

Therefore, Total amount to be paid after 2.5 years = Rs. (21780+1089) = Rs.22869

Now, Compound Interest after 2 years = A - P = Rs.(21780-18000) = Rs. 3780

Therefore, Compound Interest after 2.5 years, CI = Rs. 3780 + SI = Rs.4869

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

Given,

Principal,P =Rs 62500,

Compound interest Rate,R = 8% compounded half yearly for 1.5 years.

Since it is compounded half yearly, R becomes half = 4%, and time period doubles, n = 3 years.

We know, Amount when interest is compounded annually, A = $P(1+\frac{R}{100})^n$

Therefore, the required amount = $62500(1+\frac{4}{100})^3$ = Rs.70304

And Compound Interest, CI = Amount - Principal = Rs. (70304 - 62500) = Rs. 7804

(d) Rs 8000 for 1 year at 9% per annum compounded half yearly. (You could use the year by year calculation using SI formula to verify)

Given,

Principal,P =Rs.8000, Rate, R = 9% per annum compounded half yearly for 1 year.

Now, there are two half years in a year. Therefore compounding has to be 2 times.

And rate = half of 9% = 4.5% half yearly.

Therefore, the required amount = $8000(1+\frac{9}{2\times100})^{2}$ = Rs. 8736.20

And Compound Interest, CI = Amount - Principal = Rs. (8736.20 - 8000) = Rs. 736.20

Rs 10,000 for 1 year at 8% per annum compounded half yearly.

Given,

Principal, P =Rs.10000, Rate, R = 8% per annum compounded half yearly for 1 year.

Now, there are two half years in a year. Therefore compounding has to be 2 times.

And rate = half of 10% = 5% half yearly.

Therefore, the required amount = $10000(1+\frac{4}{100})^{2}$ = Rs. 10816

And Compound Interest, C.I. = Amount - Principal = Rs. (10816 - 10000) = Rs. 816 .

The amount borrowed from the bank = Principal amount, P = Rs 26400

Compound interest rate, R = 15% p.a.

Time period = 2 years 4 months = $2\frac{4}{12} = 2\frac{1}{3} years$

We know, Amount when interest is compounded annually, A = $P(1+\frac{R}{100})^n$

Therefore, for the first 2 years, amount, A = $26400(1+\frac{15}{100})^2$ = Rs 34914

Now, this would act as principal for the next 1/3 year. We find the SI on Rs 34914 for 1/3 year.

SI = $\frac{34914\times\frac{1}{3}\times15}{100}$ = Rs 1745.70

Therefore, Required amount at the end of 2 years and 4 months = A + SI = Rs (34914 + 1745.70) = Rs 36659.70

For Fabina,

Principal,P =Rs 12500

Simple interest Rate,R = 12% p.a.

Time period,n = 3 years.

$\therefore$ Simple Interest, SI at 12% for 3 years = $3\times\frac{12500\times12}{100}$ = Rs 4500

Principal,P =Rs 12500

Compound interest Rate,R = 10% p.a.

Time period,n = 3 years.

We know, Amount when interest is compounded annually,

$A =P(1+\frac{R}{100})^n$

$A=12500(1+\frac{10}{100})^3 = Rs 16637.50$

$\therefore Compound \:\:Interest, CI = A - P = Rs\: (16637.50 - 12500) = Rs \:4137.50$

$\therefore$ Fabina pays more interest and Rs (4500 - 4137.50) = Rs 362.50 more.

Given,

Principal,P =Rs 12000

Simple interest Rate,R = 6% p.a.

Time period,n = 2 years.

$\therefore$ Simple Interest, SI at 6% for 2 years =

$=2\times\frac{12000\times6}{100} = Rs 1440$

If he would have borrowed it at a compound interest rate, R = 6% p.a.

We know, Amount when interest is compounded annually, A =

$A = P(1+\frac{R}{100})^n$

$\therefore A = 12000(1+\frac{6}{100})^2= Rs 13483.20$

$\therefore Compound\:\: Interest, CI = A - P = Rs \:(13483.20 - 12000) = Rs\: 1483.20$

$\therefore$ He would have to pay Rs (1483.20 - 1440) = Rs 43.20 extra.

(i) after 6 months?

Given,

Principal, P = Rs 60,000

Compound interest rate, R = 12% p.a

= 6 % half yearly

For a period of 6 months. $\therefore$ Time period, n = 1 half year (As there is 1 half year in 6 months.)

We know, Amount when interest is compounded annually, A =

$A = P(1+\frac{R}{100})^n$

$A =60000(1+\frac{6}{100})^1= Rs 63600$

After 6 months, Vasudevan would get an amount Rs. 63600.

(ii) after 1 year?

Given,

Principal, P = Rs 60,000

Compound interest rate, R = 12% p.a

= 6 % half-yearly

For a period of 1 year. $\therefore$ Time period, n = 2 half years (As there are 2 half years in a year.)

We know, Amount when interest is compounded annually, A =

$A = P(1+\frac{R}{100})^n$

$A = 60000(1+\frac{6}{100})^2= Rs 67416$

After 1 year, Vasudevan would get an amount Rs. 67416.

(i) compounded annually.

(ii) compounded half yearly.

(i) Given,

Principal amount, P = Rs 80000

Rate of interest, R = 10% p.a.

Time period = $1\frac{1}{2}$ years.

We know, Amount when interest is compounded annually, A =

$A =P(1+\frac{R}{100})^n$

Now, For the first year, A=

$A =80000(1+\frac{10}{100})^1= Rs. 88000$

For the next half-year, this will act as the principal amount.

$\therefore$ Interest for 1/2 year at 10% p.a =

$=\frac{88000\times\frac{1}{2}\times10}{100}= Rs 4400$

Required total amount = Rs (88000 + 4400) = Rs 92400

(ii) If it is compounded half-yearly, then there are 3 half years in $1\frac{1}{2}$ years.

$\therefore$ n = 3 half years.

And, Rate of interest = half of 10% p.a = 5% half yearly

$\therefore A =80000(1+\frac{5}{100})^3= Rs.\: 92610$

$\therefore$ The difference in the two amounts = Rs (92610 - 92400) = Rs 210

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

Given,

Principal amount, P = Rs 8,000

Compound rate of interest, R = 5% p.a.

Time period, n = 2 years

We know, Amount when interest is compounded annually,

$A =P(1+\frac{R}{100})^n$

$A =8000(1+\frac{5}{100})^2= Rs.\: 8,820$

Therefore, the amount credited against her name at the end of the second year is Rs 8,820

(ii) The interest for the 3rd year.

Now, the amount after 2nd year will become the principal amount for the 3rd year

$\therefore$ Principal amount, P = Rs 8,820

Compound rate of interest, R = 5% p.a.

Time period, n' = 1 year

$\therefore Interest \:for\: the \:3rd \:year =\frac{8820\times1\times5}{100} = Rs.\ 441$

Therefore, the interest for the 3rd year is Rs 441.

Principal = Rs.10,000

Time = $1\frac{1}{2}$ years

Rate = 10% per annum

CASE 1 Interest on compounded half yearly.

Rate = 10% per annum = 5 % per half yearly

$T= 1\frac{1}{2} years,n=3$

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

$A= 10000\left ( 1+\frac{5}{100} \right )^{3}$

$A= 10000\left ( 1+0.05 \right )^{3}$

$A= 10000\left ( 1.05 \right )^{3}$

$A= 11576.25$ = Amount

CI = Amount - principal

CI = $11576.25-10000$

CI = 1576.25

CASE 2 Interest on compounded anually

Rate = 10% per annum

$T= 1\frac{1}{2} years ,n=1$

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

$A= 10000\left ( 1+\frac{10}{100} \right )^{1}$

$A= 10000\left ( 1+0.1\right )^{1}$

$A= 10000\left ( 1.1\right )^{1}$

$A= 11000$ = Amount

CI = Amount - principal

CI = $11000-10000$

CI = 1000

Interest for half years on 11000 =

$\frac{P\times R\times T}{100}$

$=\frac{11000\times 10\times \frac{1}{2}}{100}$

$= 55\times 10$

= 550

Total interest = $1000+550$

= RS 1550

Since 1576.25 $>$ 1000

Thus, interest would be more in CASE 1 i.e. compounded half yearly

Given,

Principal amount, P = Rs 4,096

Rate of interest, R

$R =12\frac{1}{2} \% = \frac{25}{2} \% p.a. =\frac{25}{4} \% \ half\ yearly$

Time period, n = 18 months = (12 + 6) months = 1.5 years = 3 half years

(There are 3 half years in 1.5 years)

We know,

Amount when interest is compounded annually, (A)

$A =P(1+\frac{R}{100})^n$

Therefore, the required amount

$=4096(1+\frac{25}{4\times100})^{3} = 4913$

$\therefore$ Ram will get Rs 4,913 after 18 months.

(i) find the population in 2001.

Let the population in 2001 be P

Compound rate of increase = 5% p.a.

The population in 2003 will be more than in 2001

Time period, n = 2 years (2001 to 2003)

$\therefore$ $\\ 54000 = P(1 + \frac{5}{100})^2 \\ \implies P = \frac{54000\times100\times100}{105\times105} \approx 48980$

Therefore, the population in 2001 was 48980 (approx)

(ii) what would be its population in 2005?

Let the population in 2001 be P'

Compound rate of increase = 5% p.a.

The population in 2005 will be more than in 2003

Time period, n = 2 years (2003 to 2005)

$\therefore$ $\\ P' = 54000(1 + \frac{5}{100})^2 \\ \implies P' = \frac{54000\times105\times105}{100\times100} \approx 59535$

Therefore, the population in 2005 will be 59535 (approx)

Given,

Initial count of bacteria, P = 5, 06,000 (Principal Amount)

Rate of increase, R = 2.5% per hour.

Time period, n = 2 hours

(This question is done in a similar manner as compound interest)

Number of bacteria after 2 hours = $P(1+\frac{R}{100})^n$

$= 506000(1+\frac{2.5}{100})^2 \approx 531616$

Therefore, the number of bacteria at the end of 2 hours will be 531616 (approx)

Given,

Principal = Rs 42,000

Rate of depreciation = 8% p.a

$\therefore$ Reduction = 8% of Rs 42000 per year

$=\frac{42000\times 8\times 1}{100} = Rs\ 3360$

$\therefore$ Value at the end of 1 year = Rs (42000 – 3360) = Rs 38,640

## NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities - Topics

• Recalling Ratios and Percentages
• Finding the Increase or Decrease Percent
• Finding Discounts
• Prices Related to Buying and Selling (Profit and Loss)
• Sales Tax/Value Added Tax/Goods and Services Tax
• Compound Interest
• Deducing a Formula for Compound Interest
• Rate Compounded Annually or Half Yearly (Semi-Annually)
• Applications of Compound Interest Formula

## NCERT Solutions for Class 8 Maths - Chapter Wise

 Chapter -1 Rational Numbers Chapter -2 Linear Equations in One Variable Chapter-3 Understanding Quadrilaterals Chapter-4 Practical Geometry Chapter-5 Data Handling Chapter-6 Squares and Square Roots Chapter-7 Cubes and Cube Roots Chapter-8 Comparing Quantities Chapter-9 Algebraic Expressions and Identities Chapter-10 Visualizing Solid Shapes Chapter-11 Mensuration Chapter-12 Exponents and Powers Chapter-13 Direct and Inverse Proportions Chapter-14 Factorization Chapter-15 Introduction to Graphs Chapter-16 Playing with Numbers

## Key Features Of class 8 comparing quantities NCERT solutions

Complete Coverage: Solutions for maths chapter 8 class 8 cover all topics and concepts related to comparing quantities as per the Class 8 syllabus.

Step-by-Step Solutions: Detailed, step-by-step explanations for each problem, making it easy for students to understand and apply mathematical concepts related to ratios, percentages, and discounts.

Illustrations and Diagrams: The inclusion of diagrams, figures, and illustrations to visually explain concepts and methods for comparing quantities ch 8 maths class 8.

## NCERT Solutions for Class 8 - Subject Wise

Also Check NCERT Books and NCERT Syllabus here:

1. What are the important topics of NCERT book chapter Comparing Quantities ?

Increase or decrease percentage, discounts, profit and loss, simple and compound interest are the important topics of this chapter.

2. Does CBSE NCERT syllabus class 8 maths is tough ?

CBSE class 8 maths is damn simple and basic. Most of the chapters are related to the previous classes.

3. How does the NCERT solutions are helpful ?

NCERT solutions are helpful for the students if they are not able to NCERT problems on their own. These solutions are provided in a very detailed manner which will give them conceptual clarity.

4. How many chapters are there in the CBSE class 8 maths ?

There are 16 chapters starting from rational number to playing with numbers in the CBSE class 8 maths.

5. Where can I find the complete solutions of NCERT for class 8 ?

Here you will get the detailed NCERT solutions for class 8 by clicking on the link.

6. Where can I find the complete solutions of NCERT for class 8 maths ?

Here you will get the detailed NCERT solutions for class 8 maths by clicking on the link.

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3 Jobs Available
##### QA Manager

Quality Assurance Manager Job Description: A QA Manager is an administrative professional responsible for overseeing the activity of the QA department and staff. It involves developing, implementing and maintaining a system that is qualified and reliable for testing to meet specifications of products of organisations as well as development processes.

2 Jobs Available

A QA Lead is in charge of the QA Team. The role of QA Lead comes with the responsibility of assessing services and products in order to determine that he or she meets the quality standards. He or she develops, implements and manages test plans.

2 Jobs Available
##### Reliability Engineer

Are you searching for a Reliability Engineer job description? A Reliability Engineer is responsible for ensuring long lasting and high quality products. He or she ensures that materials, manufacturing equipment, components and processes are error free. A Reliability Engineer role comes with the responsibility of minimising risks and effectiveness of processes and equipment.

2 Jobs Available
##### Safety Manager

A Safety Manager is a professional responsible for employee’s safety at work. He or she plans, implements and oversees the company’s employee safety. A Safety Manager ensures compliance and adherence to Occupational Health and Safety (OHS) guidelines.

2 Jobs Available
##### Corporate Executive

Are you searching for a Corporate Executive job description? A Corporate Executive role comes with administrative duties. He or she provides support to the leadership of the organisation. A Corporate Executive fulfils the business purpose and ensures its financial stability. In this article, we are going to discuss how to become corporate executive.

2 Jobs Available
##### ITSM Manager

ITSM Manager is a professional responsible for heading the ITSM (Information Technology Service Management) or (Information Technology Infrastructure Library) processes. He or she ensures that operation management provides appropriate resource levels for problem resolutions. The ITSM Manager oversees the level of prioritisation for the problems, critical incidents, planned as well as proactive tasks.

3 Jobs Available
##### Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack

3 Jobs Available
##### Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available
##### Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

3 Jobs Available
##### IT Consultant

An IT Consultant is a professional who is also known as a technology consultant. He or she is required to provide consultation to industrial and commercial clients to resolve business and IT problems and acquire optimum growth. An IT consultant can find work by signing up with an IT consultancy firm, or he or she can work on their own as independent contractors and select the projects he or she wants to work on.

2 Jobs Available
##### Data Architect

A Data Architect role involves formulating the organisational data strategy. It involves data quality, flow of data within the organisation and security of data. The vision of Data Architect provides support to convert business requirements into technical requirements.

2 Jobs Available
##### Security Engineer

The Security Engineer is responsible for overseeing and controlling the various aspects of a company's computer security. Individuals in the security engineer jobs include developing and implementing policies and procedures to protect the data and information of the organisation. In this article, we will discuss the security engineer job description, security engineer skills, security engineer course, and security engineer career path.

2 Jobs Available
##### UX Architect

A UX Architect is someone who influences the design processes and its outcomes. He or she possesses a solid understanding of user research, information architecture, interaction design and content strategy.

2 Jobs Available