1. The ratio of 50cm to 2.5m is

(a) 10 : 1

(b) 5 : 1

(c) 1 : 5

(d) None of these

Ans: We know that, 1m = 100cm

${\text{2}}{\text{.5m = 2}}{\text{.5 }} \times {\text{ 100 = 250cm}}$

Ratio of 50cm to 2.5m $ = \,\dfrac{{50}}{{250}} = \,\dfrac{1}{5}\, = \,1:5$

2. There are 25 computers, 16 of them are out of order. Find the percentage of computers out of order?

Ans: Percentage = $\dfrac{{{\text{No}}{\text{. of computers out of order}}}}{{{\text{Total no}}{\text{. of computers}}}}\, \times \,100$

Percentage = $\dfrac{{16}}{{25}}\, \times \,100$ = 64%.

3. The number of unelectrified villages in India decreased from 18,000 to 12,000 in last 6 years. What is the percentage of decrease?

(a) 30%

(b) 50%

(c) $33\dfrac{1}{3}% $

(d) None of these.

Ans:

$ {\text{Percentage decrease = }}\dfrac{{{\text{Old value - New value}}}}{{{\text{Old value}}}}\, \times \,100 $

$ = \,\left( {\dfrac{{18,000\, - \,12,000}}{{18,000}}} \right)\, \times \,100 $

$ = \,\left( {\dfrac{{6,000}}{{18,000}}} \right)\, \times \,100\, = \,\dfrac{{100}}{3} = \,33\dfrac{1}{3}\% $

4. Nandini purchased a sweater and saved Rs. 20 when a discount voucher of 25% was provided. Find the price of sweater before discount?

Ans: Let the marked price of the sweater be ‘x’.

Then, 25% of x = 20

$\dfrac{{25}}{{100}}{\mkern 1mu} \times {\mkern 1mu} \,x{\mkern 1mu} {\text{ }} = {\text{ }}{\mkern 1mu} 20$

$ x\;{\text{ }} = \;{\text{ }}\dfrac{{20{\mkern 1mu} \times {\mkern 1mu} 100}}{{25}} $

$ x\;{\text{ }} = \;{\text{ }}20{\mkern 1mu} \times {\mkern 1mu} 4 $

$ x{\mkern 1mu} {\text{ }} = {\text{ }}{\mkern 1mu} 80 $

Therefore, the actual price of the sweater is Rs. 80.

5. Cost of an item is Rs. 50. It was sold with a profit of 12%. Find the selling price

(a) Rs.56

(b) Rs. 60

(c) Rs.70

(d) None of these.

Ans: We know that

Cost Price = Rs. 50

and, Profit % = 12

Therefore, Profit = $\dfrac{12}{100}\times 50$

⇒ Profit = 6

⇒ S.P. = C.P. + Profit

⇒ S.P. = 50 + 6

⇒ S.P. = Rs 56

6. The simple interest on Rs.6000 for 1 year at 4% per annum is

(a) Rs.126.50

(b) Rs.240

(c) Rs.43

(d) None of these

Ans: ${\text{S}}{\text{.I}}{\text{. = }}\dfrac{{{\text{PTR}}}}{{100}}\, = \,\dfrac{{6000\, \times \,1\, \times \,4}}{{100}} = \,240$

7. A school trip is being planned in a school for class VIII. Girls are 60% of the total strength and are 18 in number. Find the ratio of number of boys to number of girls.

Ans: Let ‘x’ be the total number of students.

Thus, number of girls = 60% of x = 18

$ \dfrac{{60}}{{100}}\, \times \,{\text{x = 18}} $

$ {\text{x = }}\dfrac{{1800}}{{60}} $

$ {\text{x = 30}} $

Number of boys = (Total number of students) - (Total number of girls)

= 30 – 18

= 12.

Hence ratio of number of boys to girls is

= 12 : 18

= 2 : 3.

8. In a constituency there are 120 voters 90 of them voted Yes. What percent voted Yes?

Ans: Given:

Number of voters = 120

Number of voters who voted Yes = 90

$ {\text{Voted percentage = }}\dfrac{{{\text{No}}{\text{. of voters voted Yes}}}}{{{\text{Total number of voters}}}}\, \times \,100 $

$ {\text{ = }}\dfrac{{90}}{{120}}\, \times \,100 $

$ = \,75\% $

9. If Rs. 250 is divided among Rakshith, Ravi and Raju. So that Rakshith gets 3 parts, Ravi gets 2 parts and Raju gets 5 parts. How much money will each get in percentages?

Ans: Given: total amount = 250

Total number of parts = 10

Names | No. of parts each get | Amount of money | Percentage |

Rakshith | 3 parts | $\dfrac{3}{{10}} \times 250\, = \,75$ | $\dfrac{{75}}{{250}} \times 100\, = \,30\% $ |

Ravi | 2 parts | $\dfrac{2}{{10}} \times 250\, = \,50$ | $\dfrac{{50}}{{250}} \times 100\, = \,20\% $ |

Raju | 5 parts | $\dfrac{5}{{10}} \times 250\, = \,125$ | $\dfrac{{125}}{{250}} \times 100\, = \,50\% $ |

10. My grandmother says in her childhood milk was at Rs.2 per litre. It was Rs.36 per litre today. By what percentage has the price gone up?

Ans: Given:

Old value = Rs. 2 per litre

New price = Rs. 36 per litre

${\text{Percentage increase = }}\dfrac{{{\text{New price - old price}}}}{{{\text{old price}}}} \times 100 = \dfrac{{36 - 2}}{{36}} \times 100 = \dfrac{{34}}{{36}} \times 100 = 94.44\% {\text{.}}$

11. The cost of a toy car is Rs. 140. If the shopkeeper sells it at a loss of 10%. Find the price at which it is sold.

Ans: Given:

C.P. of toy car = Rs. 140

Loss% = 10%

S.P. = ?

We know that,

${\text{Loss = }}\dfrac{{{\text{Lossn% }} \times {\text{ C}}{\text{.P}}{\text{.}}}}{{100}}\, = \,\dfrac{{10}}{{100}} \times 140 = {\text{Rs}}{\text{. 14}}$

Loss = C.P. – S.P.

S.P. = C.P. – Loss

S.P. = 140 – 14

S.P. = Rs.126

12. Rashida purchased an air-conditioner for Rs. 3400 including a tax of 10%. Find the actual price of the air conditioner before VAT was added.

Ans: Let ‘x’ be the cost before adding VAT.

VAT = 10% of x = 0.1x

Cost after adding VAT = x + 0.1x = 1.1x

Given: cost = Rs.3,400

$ {\text{1}}{\text{.1x = Rs}}{\text{. 3400}} $

$ {\text{x = }}\dfrac{{3400}}{{1.1}}\, = \,3090.9 $

Thus, the price of an air-conditioner = Rs. 3090.9.

13. Seema deals with second hand goods. She bought a second hand refrigerator for Rs. 5000. She spends RS.100 on transportation and Rs.600 on its repair. She sells the refrigerator for Rs. 7100. Find

(a) Total cost price

(b) Profit or loss percent.

Ans:

(a) From the given data,

Total cost price = Purchasing price + transportation charge+ repair charge

Total cost price = 5000 + 100 + 600

= Rs. 5700.

(b) Given: S.P. = Rs. 7100.

Since, S.P. > C.P., There is a profit.

Profit = S.P. – C.P.

= 7100 – 5700

= Rs. 1400

${\text{Profit % = }}\dfrac{{{\text{Profit }} \times {\text{ 100}}}}{{{\text{C}}{\text{.P}}{\text{.}}}}\, = \,\dfrac{{1400}}{{5700}} \times 100\, = \,24.5\% $

14. At what rate of simple interest will the sum double itself in 2 years?

Ans: We know that,

A = S.I. + P

Where, ${\text{S}}{\text{.I}}{\text{. = }}\dfrac{{{\text{PRT}}}}{{100}}$

Given: ${\text{A = 2 }} \times {\text{ principle = 2P}}$

Time = t = 2 years

R = ?

Formula becomes 2P = S.I. + P

${\text{2P = }}\dfrac{{{\text{PRT}}}}{{100}}\, + \,{\text{P}} $

$ {\text{2P - P = }}\dfrac{{{\text{PRT}}}}{{100}} $

$ {\text{P = }}\dfrac{{{\text{P}} \times {\text{2}} \times {\text{R}}}}{{100}}$

$ {\text{R = }}\dfrac{{100}}{2}\, = \,50\% $

$ {\text{R = 50% }} $

Therefore, at the rate of 50%, the sum will double.

15. In what time will Rs. 1600 amount to Rs. 1768 at 6% per annum simple interest?

Ans: Given:

Principle = Rs. 1600

Amount = Rs. 1768

Rate = 6% per year

Time = ?

A = S.I. + P

$ 1768\, = \,\dfrac{{{\text{PTR}}}}{{100}}\, + \,{\text{P}} $

$ {\text{1768}}\,{\text{ = }}\,\dfrac{{1600 \times {\text{T}} \times {\text{6}}}}{{100}}\, + \,1600$

$ 1768\, - \,1600\, = \,\dfrac{{1600 \times {\text{T}} \times {\text{6}}}}{{100}} $

$ 168\, \times \,100\, = \,1600 \times {\text{T}} \times {\text{6}}$

$ {\text{T = }}\dfrac{{168\, \times \,100}}{{1600 \times 6}} $

$ {\text{T = 1}}{\text{.75 years}} $

$ {\text{T = 1}}\dfrac{3}{4}{\text{years}} $

16. What amount Harish has to pay at the end of 3 years of Rs. 40,000 at an interest of 16% compounded annually?

Ans: We know that, formula for compound interest,

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

Where, P = principle

N = no. of years

P = Rs. 40,000, R = 16%, n = 2.

$ {\text{A = 40000}}{\left( {1\, + \,\dfrac{{16}}{{100}}} \right)^2} $

$ {\text{A = 40000}}{\left( {1\, + \,0.16} \right)^2} $

$ {\text{A = 40000(1}}{\text{.16}}{{\text{)}}^{\text{2}}} $

$ {\text{A = 40000 }} \times {\text{ 1}}{\text{.3456}} $

$ {\text{A = Rs}}{\text{. 53,824}} $

Amount paid by Harish at the end of 2 years is Rs. 53,824.

17. Mahesh sells two tables for Rs. 3000 each. He gains 20% on one table and on the other he loses 20%. Find his gain or loss percent on whole transaction.

Ans: For the first table: given:

S.P. = Rs. 3000

Gain% = 20% = $\dfrac{{20}}{{100}}$

Gain percent implies increased percent on cost price.

For Rs.100 cost price, the gain = Rs.20

S.P. = C.P. + gain

S.P. = 100 + 20 = Rs.120

Thus, S.P. is Rs. 120 when C.P. is Rs.100

Therefore, for S.P. of Rs. 3000, the cost price will be

$ = \dfrac{{3000 \times 100}}{{120}} = {\text{Rs}}{\text{.2500}}$

For second table,

S.P. = Rs.300

Loss percent = 20% = $\dfrac{{20}}{{100}}$

Loss percent decreases percent on cost price.

For Rs.100 of C.P., loss = Rs.20

S.P. = C.P. – loss = 100 – 20 = Rs.80.

Thus, S.P. is Rs.80 when C.P. is Rs.100

For S.P. of Rs.3000, the cost price is given by

$ = \dfrac{{3000 \times 100}}{{80}}\, = \,{\text{Rs}}{\text{.3750}}$

Total cost price = 2500 + 3750 = 6250

Total S.P. = 2500 + 3750 = 6000

Here, S.P. < C.P., Hence loss is Occured

Loss = C.P. – S.P. = 6250 – 6000 = 250

Loss percent = $\dfrac{{{\text{loss}}}}{{{\text{C}}{\text{.P}}{\text{.}}}} \times 100\, = \,\dfrac{{250}}{{6250}} \times 100\, = \,4\% $

Therefore, there is a loss of 4% on whole transaction.

18. Mary goes to a departmental store and buys the following goods.

Cosmetics worth of Rs. 345

Medicines worth of Rs. 228

Stationery worth of Rs. 170. If the sales tax is chargeable at the rate of 10% on cosmetics, 78% on medicines, 5% on stationary. Find the total amount to be paid by Mary.

Ans: Cost of cosmetics = Rs. 345

Sales tax on cosmetics = $

$ {\text{Rs}}\left[ {345 \times \dfrac{{10}}{{100}}} \right]$

$ = {\text{Rs}}{\text{. 34}}{\text{.50}} $

$Total cost of cosmetics = 345 + 34.5

=Rs. 379.50

Total cost of medicines = Rs. 228 + 7% of 228

$ = \,228\, + \,\dfrac{7}{{100}} \times 228 $

$ = \,228\, + \,15.96 $

$ = \,{\text{Rs}}{\text{. 243}}{\text{.96}} $

Total cost of stationary = Rs. 170 + 5% of 170

$ = 170\, + \,\dfrac{5}{{100}} \times 170 $

$ = \,170\, + \,8.50$

$ = \,{\text{Rs}}{\text{. 178}}{\text{.50}} $

Thus, total amount of money to be paid by Mary = 379.50 + 243.96 + 178.50

=Rs. 801.96

19. Prateeksha went to a shopping mall to purchase a saree. Marked price of the saree is Rs.2000. Shop owner gave a discount of 20% and then 5%.Find the single discount equivalent to these 2 successive discounts.

Ans: Marked price of the saree = Rs. 2000

First discount = 20% of 2000

$ = \dfrac{{20}}{{100}} \times 2000 $

$ = {\text{Rs}}.400 $

$= \,2000\, - \,400 $

After First Discount, Price = Rs.1600

Second Discount = 5% after first discount

$ = \dfrac{{5}}{{100}} \times 1600 $

$ = {\text{Rs}}.80 $

Price of Saree after second discount is Rs.1520

20. Rajanna purchased 25 dozen bananas for RS. 625. He spent Rs. 125 for transportation. He could not sell 5 dozen bananas as they were spoiled. He sold the remaining banana’s at Rs. 30 for each dozen. Find loss and profit percent.

Ans: Total cost price = Cost price of bananas + transportation charge

=${\text{Rs}}{\text{. 625 + Rs}}{\text{. 125 = Rs}}{\text{. 750}}$

Number of dozens of bananas sold = No. of purchased – No. of spoiled

= 25 – 5

= 20

Given: 1 dozen = Rs.30

Therefore, S.P. = $20 \times 30\, = \,{\text{Rs}}{\text{.600}}$

Since, S.P. < C.P., it is a loss

Loss = C.P. – S.P. = 750 – 600 = 150

${\text{Loss% = }}\dfrac{{{\text{loss}}}}{{{\text{C}}{\text{.P}}{\text{.}}}} \times 100\, = \,\dfrac{{150}}{{750}} \times 100\, = \,20\% $

21. A girl bought 16 dozen ball pens and sold them at a loss equal to S.P of 8 ball pens. Find her loss % and S.P of 1 dozen ball pens if she purchased these 16 dozen ball pen's for Rs 576.

Ans: Cost price of 16 dozen ball pens = Rs. 576

Cost price of one dozen = $\dfrac{{576}}{{16}}\, = \,36$

Cost price of 1 pen = $\dfrac{{36}}{{12}} = 3$

Let ‘x’ be the S.P. of each ball pen.

Total number of pens = $16\, \times \,12\, = \,192$

Thus, total S.P. of 192 pens = 192x.

Total S.P. = 576 – S.P. of 8 ball pens

$ {\text{192x}}\,{\text{ = }}\,576\, - \,{\text{8x}} $

$ {\text{192x + 8x = 576}} $

$ {\text{200x = 576}} $

$ {\text{x = }}\dfrac{{576}}{{200}} $

$ {\text{x = 2}}{\text{.88}} $

${\text{Loss = 8 }} \times {\text{ 2}}{\text{.88 = Rs}}{\text{. 23}}{\text{.04}} $

$ {\text{(a) Loss% = }}\dfrac{{23.04}}{{576}} \times 100 = 4\% $

$({\text{b) S}}{\text{.P}}{\text{. of 1 pen = 2}}{\text{.88}} $

${\text{S}}{\text{.P}}{\text{. of 1 dozen pen = 2}}{\text{.88 }} \times {\text{ 12 = Rs}}{\text{. 34}}{\text{.56}} $

22. In 1995, the price of 1litre of a certain kind of petrol was Rs. 54.9. By 1996, the price of 1 litre of the same kind of petrol has risen to Rs 56.3. The peroentage increase for each of the next four years is expected to be the same as in between 1995 to 1996. What is the price of 1 hr of petrol expected to be in the year 2000?

Ans:

Year: Price of 1 litre of petrol

1995: Rs. 54.9

1996: Rs. 56.3

$ {\text{Percentage increase = }}\dfrac{{{\text{Increase in value}}}}{{{\text{Old value}}}} \times 100 $

$ = \dfrac{{56.3\, - \,54.9}}{{54.9}} \times 100 $

$ = \dfrac{{1.4}}{{54.9}} \times 100 $

$ = 2.55\% $

${\text{Increase amount = }}\dfrac{{2.55}}{{100}} = {\text{Rs}}.0.0255$

Therefore, price in 2000 is

$ = 56.3\, \times \,{(1\, + \,0.0255)^4} $

$ = 56.3\,{(1.0255)^4} $

$ = 62.3\,{\text{per}}\,{\text{litre}} $

23. Simple interest on a sum of money for 3 years at 8% per annum is Rs.2400. What will be the compound interest on that sum at the same rate for the same period?

Ans:

Given: SI = 2400

T = 3 years

R = 8% per year

P = ?

CI = ?

$ {\text{SI}}{\text{ = }}\dfrac{{{\text{PTR}}}}{{100}} $

$ {\text{P = }}\dfrac{{2400 \times 100}}{{3 \times 8}} $

${\text{P = Rs}}{\text{. 10000}} $

${\text{A = P}}{\left( {1\, + \,\dfrac{{\text{R}}}{{100}}} \right)^3}\, = \,{\left( {1\, + \,\dfrac{8}{{100}}} \right)^3} $

$ {\text{A = 10000(1 + 0}}{\text{.08}}{{\text{)}}^{\text{3}}} $

$ {\text{A = 12,597}}{\text{.12}} $

We know that,

$ {\text{CI = A - P}} $

$ {\text{ = 12,597}}{\text{.12 - 10000}} $

$ {\text{ = 2597}}{\text{.12}} $

24. Find the compound interest on Rs. 320000 for one year at the rate of 20% per annum, if the interest is compounded quarterly.

Ans: We know that,

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

Since, compound interest should be computed quarterly, then n = 4n and r = r/4

Rewriting the formula, we get,

$ {\text{A = P}}{\left( {1\, + \,\dfrac{{({\text{R/4}})}}{{100}}} \right)^{{\text{4n}}}} $

$ {\text{A = 320000}}{\left( {1\, + \,\dfrac{{0.20}}{4}} \right)^4} $

$ {\text{A = 320000(1}}{\text{.05}}{{\text{)}}^{\text{4}}} $

$ {\text{A = 388962}} $

$ {\text{CI = A - P}} $

${\text{CI}}{\text{ = 388962 - 320000}} $

$ {\text{CI = 68962}} $

25. The simple interest on a certain amount of money for 3 years at 8% per annum is half the compound interest on Rs. 4,000 for 2 years at 10% per annum. What is the sum placed on simple interest?

Ans:

We know that, compound interest on Rs. 4,000 for 2 years at 10% = A – P

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

$ {\text{ = }}\,{\text{P}}\left[ {{{\left( {1\, + \,\dfrac{{\text{R}}}{{100}}} \right)}^{\text{n}}}\, - \,1} \right]$

$ = \,4000\left[ {{{\left( {1\, + \,\dfrac{{10}}{{100}}} \right)}^2}\, - \,1} \right] $

$ = \,4000\left[ {\dfrac{{121}}{{100}} - 1} \right] $

$ = \,4000\left[ {\dfrac{{21}}{{100}}} \right] $

$ = \,{\text{Rs}}{\text{. 840}} $

S.I. on unknown sum = $\dfrac{1}{2} \times 840\, = \,{\text{Rs}}.420$

Time = 3 years, Rate = 8% per annum

$ {\text{sum = }}\dfrac{{{\text{interest }} \times {\text{ 100}}}}{{{\text{rate }} \times {\text{ time}}}} $

$ {\text{sum = }}\dfrac{{420\, \times \,100}}{{8\, \times \,3}} $

$ {\text{sum = Rs}}{\text{. 1750}} $

## CBSE Important Questions for Class 8 Maths Chapter 8 - Free PDF Download

While practising mathematics questions, sometimes class 8 students get stuck and fail to solve their problems. In such a situation, it’s recommendable to consider the comparing quantities class 8 important questions. All the Important Questions are highlighted in an easy-to-understand plus straightforward way. The subject experts prepare the solutions to these questions by thorough research on the chapter. Hence, it offers genuine and appropriate information to class 8 students. They can secure a good percentage in their board exams by practising important questions of comparing quantities class 8.

### CBSE Important Questions Class 8 Maths Chapter 8

Practising important questions for class 8 maths comparing quantities is crucial for students to build their knowledge on the concept of tax, profit, loss, interest, and much more. Students will find questions on the discount formula, calculated by subtracting the market price’s sale price. The chapter introduces students to the concept of cost price and overhead expenses. By practising chapter 8 maths class 8 important questions, students will find how to solve the cost price when overhead expenses and buying price are given.

Students will become capable of solving problems involving applications on tax, percentages, and profit and loss. Solving important questions for class 8 maths chapter 8 make students able to find the difference between simple and compound interest, direct and indirect variation, time and work problems, and more. It enables students to get well-prepared throughout the academic year. Students will find important questions for comparing quantities class 8 from the below-mentioned exercises:

Exercise 8.1: Recalling Ratios and Percentages

Exercise 8.2: Finding the Increase or Decrease per cent

Exercise 8.3: Finding Discounts

Exercise 8.3.1: Estimation in Percentages

Exercise 8.4: Prices Related to Buying and Selling

Exercise 8.4.1: Finding Cost Price and Selling Price, Profit % and Loss %

Exercise 8.5: Sales Tax/ Value Added Tax

Exercise 8.6: Compound Interest

Exercise 8.7: Deducing a Formula for Compound Interest

Exercise 8.8: Rate Compounded Annually or Half Yearly

Exercise 8.9: Applications of Compound Interest Formula.

There are some situations where students can use the formula, which they learn in chapter 8 class 8. It’s the calculation of the compound interest amount. Exam preparation becomes much effortless by going through with class 8 maths ch 8 important questions.

### CBSE Class 8 Maths Chapter 8 - Weightage Marks

Out of 80, class 8 chapter 8 Maths hold a weightage of 16 marks in the final board exam.

Exercise 8.1 Include six questions (1 Long and 5 Short Answer Questions).

Exercise 8.2: Include ten questions (4 Long and 6 Short Answer Questions).

Exercise 8.3: Include 12 questions (6 Long and 6 Short Answer Questions).

### Benefits of CBSE Important Questions for Class 8 Maths Chapter 8

NCERT solutions serve as a great foundation for the CBSE board exams. That is why; it’s beneficial never to stop practising NCERT chapter 8 maths class 8 important questions from an exam point of view. Students who refer to the list of important questions get benefitted from the detailed methodology of the chapter. Apart from the vital questions, the step-by-step procedure of attempting the question encourages students to fetch good marks. Some of the benefits of referring to class 8 chapter 8 maths important questions provided by Vedantu include:

NCERT textbooks can be the best buddy a student can have to improve their knowledge base. Solving class 8 maths chapter 8 important questions with the solutions provided will help them brush up on their mathematical concepts.

Students can test their knowledge while preparing for exams and analyse their weak areas after solving these questions.

(Video) STD 8 MATHS CH 8 COMPARING QUANTITIES EXTRA QUESTIONS PART 1Comparing quantities class 8 important questions helps students do a quick revision one day before the exam.

Going through important questions of comparing quantities class 8 gives students a prior idea and reduces their fear of the exam.

## FAQs

### What I found challenging in chapter comparing quantities? ›

Answer: **Two ratios can be compared by converting them into like fractions**. If the two fractions are equal, we say that the two given ratios are equivalent. If two ratios are equivalent (or equal), then the involved four quantities are said to be in proportion.

### What is the formula of comparing quantities? ›

The important formulas covered under comparing quantities class 8 formulas are as below: **Discount = Marked Price – Sale Price**. **Discount percentage = ( Discount/ Marked Price ) x 100**. Overhead expenses are those additional expenses that are made after buying an article and are included in the cost price.

### How many exercises are there in comparing quantities? ›

Chapter 8 – Comparing Quantities contains **3 exercises**, and the NCERT Solutions for Class 7 Maths available on this page provide solutions to the questions present in the exercises. Following are the concepts covered in this chapter of Class 7 Maths NCERT Solutions.

### What are comparing quantities? ›

Comparing quantities is **the method of determining the amount of comparing units with respect to another standard or reference unit**. Two comparing units must have a common reference point; otherwise, they cannot be compared.

### What is P in comparing quantities? ›

• One of the ways of comparing quantities is percentage. **Per cent** is derived from Latin word 'per centum' meaning 'per hundred'. •

### What is r in comparing quantities? ›

Profit% = Profit/Cost Price X 100. Amount = P (1 + R/100)n where P is principal, R is the rate of interest, n is time period.

### What is the formula of CP? ›

CP = ( SP * 100 ) / ( 100 + percentage profit).

### How is percentage calculated? ›

If we have to calculate percent of a number, **divide the number by the whole and multiply by 100**. Hence, the percentage means, a part per hundred. The word per cent means per 100. It is represented by the symbol “%”.

### What is the ratio between 2 quantities? ›

The ratio of the two quantities of the same kind and in the same units is **a fraction that shows how many times one quantity is of the other**. Ratio is denoted by using the symbol ':'. The ratio of two quantities a and b (b ≠ 0) is a/b and is denoted by a : b.

### Which symbol is used for comparison of two quantities? ›

To represent two equal quantities, the symbol used is the equal to symbol **(=)** **Q**. is a number used in mathematics to describe no quantity or null quantity. Q.

### Are the ratio 1 ratio 2 and 2 ratio 3 equivalent? ›

Therefore, **The ratio 1:2 is not equivalent to the ratio 2:3**. Was this answer helpful?

### What is the another name for comparing quantities? ›

Ans: **A ratio** is a comparison of two or more quantities with the same units.

### Is the comparison between 2 or more quantities? ›

**A ratio** is a comparison of two quantities.

### How do you compare a percentage to a number? ›

Simply put, **the percentage value of X is X/100**. The percentage is denoted by the symbol %. For example, 2%, 7%, 15.5% and 30% etc. Therefore, 2% is nothing but 2 parts per 100 (2/100) or 30% is nothing but 30 out of 100 (30/100).

### What is a comparison of two different units? ›

**A rate**, like a ratio, is a comparison between two numbers or measurements, but the two numbers in a rate have different units. Eg: 2 miles per hour.

### How do you find the cost price? ›

**Cost price = Selling price − profit** ( when selling price and profit is given ) Cost price = Selling price + loss ( when selling price and loss is given )

### What is principal in math? ›

**The total amount of money borrowed (or invested), not including any interest or dividends**. Example: Alex borrows $1,000 from the bank. The Principal of the loan is $1,000. See: Interest.

### What is proportional quantity? ›

Jessica: When two quantities are proportional, it means that **as one quantity increases the other will also increase and the ratio of the quantities is the same for all values**. An example could be the circumference of a circle and its diameter, the ratio of the values would equal \pi.

### What is the formula for profit and loss? ›

**The profit or gain is equal to the selling price minus the cost price.** **Loss is equal to the cost price minus the selling price**.

### Is r the same as Q? ›

Generally speaking, **R refers to the real numbers, Q refers to the rationals**, and Z refers to the integers.

### What is principal in simple interest? ›

**The borrowed money which is given for a specific period** is called the principal. The extra amount which is paid back to the lender for using the money is called the interest. You calculate the simple interest by multiplying the principal amount by the number of periods and the interest rate.

### What does a CP of 1 mean? ›

A Cp of one indicates that **the width of the process and the width of the specification are the same**. A Cp of less than one indicates that the process spread is greater than the specification. This means that some of the data lies outside the specification.

### What is profit formula? ›

The formula to calculate profit is: **Total Revenue - Total Expenses = Profit**. Profit is determined by subtracting direct and indirect costs from all sales earned.

### What is a selling price? ›

noun. Britannica Dictionary definition of SELLING PRICE. [singular] : **the price for which something actually sells**. They asked $200,000 for the house, but the eventual selling price was $175,000.

### What number is 40% of 80? ›

Answer: 40% of 80 is **32**.

Let's find 40% of 80.

### How do you find 10 percent of a number? ›

Remember, to find 10% of a number means **dividing by 10** because 10 goes into 100 ten times. Therefore, to find 20% of a number, divide by 5 because 20 goes into 100 five times.

### What is a 2 to 4 ratio? ›

Note that the ratio 2 to 4 is said to be equivalent to the ratio **1 to 2**, that is 2:4 = 1:2.

### What is the 1 to 2 ratio? ›

They can also be written as "1 to 2" or as a fraction ½. The ratio **represents the number that needs to be multiplied by the denominator in order to yield the numerator**. In this case, ½. This is clearer if the first number is larger than the second, i.e. with the ratio 2:1, 2 can contain 1, 2 times.

### What is the ratio of 5 3? ›

The ratio 5 to 3 is the simplest form of the ratio **250 to 150**, and all three ratios are equivalent.

### What does ≥ mean? ›

The symbol ≥ means **greater than or equal to**.

### What does <= mean? ›

**Less than or equal to**.

### Is it greater or less than zero? ›

Numbers can be positive or negative. **Positive numbers are greater than 0**, and negative numbers are less than 0.

### What is the ratio of 750 m to 1 km? ›

The correct option is B **3 : 4**.

### What is a 1/3 ratio? ›

For example, if there is 1 boy and 3 girls you could write the ratio as: 1 : 3 (**for every one boy there are 3 girls)**

### What is a 2 3 ratio equal to? ›

Hence, **4:6** is the equivalent ratio of 2:3.

### What is the comparison of two numbers called? ›

**Ratios** are used to compare two numbers. The value of a ratio a:b is the quotient a ÷ b, or the result of dividing a by b.

### What is a comparison graph called? ›

A comparison diagram or can offer qualitative and/or quantitative information. This type of diagram can also be called **comparison chart or comparison chart**. The diagram itself is sometimes referred to as a cluster diagram.

### Which than is used for comparison? ›

**Than is used in comparisons as a conjunction (as in "she is younger than I am") and as a preposition ("he is taller than me")**. Then indicates time. It is used as an adverb ("I lived in Idaho then"), noun ("we'll have to wait until then"), and adjective ("the then-governor").

### Who much is 8% of 40 kg? ›

Thus, 8% of 40 kilograms weighs **3.2 kilograms**. Hence, 3.2 Kilograms is the correct answer.

### Which number comes first in ratio? ›

The order of the items in a ratio is very important, and must be respected; **whichever word came first in the ratio (when expressed in words), its number must come first in the ratio**.

### How do you solve 3 ratios? ›

To calculate a ratio of 3 numbers, we follow 3 steps: Step 1: Find the total number of parts in the ratio by adding the numbers in the ratio together. Step 2: Find the value of each part in the ratio by dividing the given amount by the total number of parts. Step 3: Multiply the original ratio by the value of each part.

### How do you calculate a 5% increase? ›

**Divide the number you wish to add 5% to by 100.** **Multiply this new number by 5.** **Add the product of the multiplication to your original number**.

### How do u calculate increase? ›

**To calculate the percentage increase:**

- First: work out the difference (increase) between the two numbers you are comparing.
- Increase = New Number - Original Number.
- Then: divide the increase by the original number and multiply the answer by 100.
- % increase = Increase ÷ Original Number × 100.

### How do I calculate decrease? ›

To calculate a percentage decrease, **first work out the difference (decrease) between the two numbers you are comparing.** **Next, divide the decrease by the original number and multiply the answer by 100**. The result expresses the change as a percentage—i.e., the percentage change.

### What are the 3 of comparison? ›

There are three degrees of comparison, **positive (or negative), comparative, and superlative**.

### What is the ratio of 2 and 5? ›

The ratio is 2 to 5 or **2:5 or 2/5**. All these describe the ratio in different forms of fractions. The ratio can consequently be expressed as fractions or as a decimal. 2:5 in decimals is 0.4.

### What is a unit rate? ›

A unit rate means **a rate for one of something**. We write this as a ratio with a denominator of one. For example, if you ran 70 yards in 10 seconds, you ran on average 7 yards in 1 second. Both of the ratios, 70 yards in 10 seconds and 7 yards in 1 second, are rates, but the 7 yards in 1 second is a unit rate.

### Why do we need to learn comparing quantities? ›

Why is Comparing Numbers Important? **Comparing numbers is an important part of building a student's number sense**. Number sense is the ability for a student to recognize a number, it's value and it's relationship with other numbers. It is this important last component that is built by comparing numbers.

### How we use comparing quantities in our daily life? ›

Comparing Quantities: In everyday life, we come across many situations, quantities, activities, etc. Many times, we need to compare and measure these quantities. The comparison is a daily activity among us. **Sometimes we compare our weight, height, marks, speed, sometimes distance, quantity, etc.**

### What are comparison problems? ›

COMPARISON problems are the type of problems looked at this week, which involve **figuring our similarities or differences between sets**. Difference Unknown: One type of compare problem involves finding out how many more are in one set than another.

### Is it greater or smaller than? ›

The symbol used to represent greater than is “>” and less than is “<”. **If one value is larger than the other value, we use greater than**. Similarly, if we want to represent one value that is less than the other value, we use less than. For example, 5 is greater than 5.

### What are the 2 methods for solving proportions? ›

Method I: **Draw a double-sided number line, label the parts, set up a proportion and solve.** **Method II: Using any method, calculate unit rate and then calculate how many pounds you can get for $30**.

### Why is greater than less than important? ›

Why do we use greater than and less than symbols? Greater than and less than symbols are used **to show the relationship between two numbers**. The wide open side of the sign always faces the number with the higher value. These symbols are especially useful in math problems where there is no clear answer.

### What is quantity used for? ›

Quantity, much like number, can be used for **singular or plural nouns that you can count or measure**. The main difference is that it's best to use quantity when you're talking about an inanimate object. However, there are times where you can use quantity and number interchangeably, specifically when the noun is plural.

### What is used for comparison? ›

**Adjectives and adverbs** can be used to make comparisons. The comparative form is used to compare two people, ideas, or things. The superlative form with the word "the" is used to compare three or more. Comparatives and superlatives are often used in writing to hedge or boost language.

### What are the three examples of quantities? ›

**Mass, time, distance, heat, and angle** are among the familiar examples of quantitative properties. Quantity is among the basic classes of things along with quality, substance, change, and relation.

### What are the 3 comparisons? ›

They are used to show what is different or similar about two or more things. There are three kinds of possible comparisons: **equal, comparative and superlative**.

### What number is 5 times as much as 6? ›

Any two factors and their product can be read as a comparison statement (5 × 6 = **30: 30** is 5 times as much as 6). In a multiplicative comparison problem, one quantity is always smaller or larger than the other quantity.

### What is a bar model? ›

In math a bar model is **a pictorial representation of a problem or concept where bars or boxes are used to represent the known and unknown quantities**. Bar models are most often used to solve number problems with the four operations – addition and subtraction, multiplication and division.

### What are the five points for comparison? ›

For the quantitative portion of the research, a five (5) point Likert scale was used as **Strongly Disagree, Disagree, Neutral, Agree, and Strongly Agree**, with numbers ranging from 1 to 5.

### What is method of substitution? ›

The substitution method is **the algebraic method to solve simultaneous linear equations**. As the word says, in this method, the value of one variable from one equation is substituted in the other equation.

### What are three types of faulty comparisons? ›

In this handout, we'll discuss several different kinds of comparison: **errors in degree of comparison, incomplete and ambiguous comparisons, and illogical comparisons**.