NCERT Solutions for Class 8 Science Chapter 16 Light

November 26, 202039 Insightful Publications

**NCERT Solutions For Class 8 Maths Chapter 8:** NCERT books are structured in a way that each chapter is followed by an exercise. Students need to go through all the topics in the chapter to be able to answer the questions in the exercise appropriately. The solution set for class 8 maths chapter 8 is prepared by a team of experts who understand the exam pattern and the marking scheme followed by NCERT. Students can follow these answers to understand the correct approach that needs to be followed to answer the questions appropriately.

CBSE Class 8 Maths Chapter 8 talks about Ratios, Converting Percentage into Fraction, Increase or Decrease in Percentage, Discounts, Tax Calculation for Purchase Goods and many more. Embibe offers 470+ practice questions for all the sub-topics of chapter 8. Students must practice all the in-text questions offered by Embibe to score good marks in the exams. Keep reading the article to know NCERT solutions for Class 8 Maths chapter 8.

Comparing Quantities may seem a bit difficult for the students to understand at first. But, practising the problems on a regular basis helps students to easily grasp the concepts and find the solutions to the problems. Students must solve as many practice questions as possible to develop proficiency in problem-solving.

Embibe offers solutions to each and every question mentioned in Class 8 Maths chapter 8 textbook considering the latest syllabus. Moreover, students can solve the problems for free at Embibe to score good marks. Embibe also offers NCERT 3D Videos, NCERT Exemplars, Embibe Explainers, etc., for free.

Below-mentioned is the list of topics present in Class 8 Maths chapter 8.

Some of the important points to remember for Class 8 Maths Chapter 8 have been mentioned below.

**Percent:**

(i) Percent means per hundred or for every hundred.

(ii) A fraction with its denominator as 100100 is called a percent and is equal to that percent as is the numerator.

(iii) A ratio with its second term 100100 is also called a percent.**Converting to Percent:**

(i) To convert a fraction into a percent, we multiply the fraction by 100100.

(ii) To convert a ratio into a percent, we write it as a fraction and multiply it by 100100.

(iii) To convert a decimal into a percent, we shift the decimal point two places to the right.**Converting from Percent**:

(i) To convert a percent into a fraction, we drop percent sign (%) and divide the remainder by 100100

(ii) To convert a percent into a ratio, we drop percent sign (%) and form a ratio with the remaining number as the first term and 100100 as the second term.

(iii) To convert a percent into a decimal, we drop percent sign (%) and shift the decimal point two places to the left.- If the borrower and the lender agree to fix up a certain interval of time (say, a year or a half-year or a quarter of a year etc.) so that the amount at the end of an interval becomes the principal for the next interval, then the total interest over all the intervals calculated in this way is called the compound interest and is denoted by C.I.

Also, C.I.=Amount−Principal

- 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 9: Algebraic Expressions and Identities
- Chapter 10: Visualising Solid Shapes
- Chapter 11: Mensuration
- Chapter 12: Exponents and Powers
- Chapter 13: Direct and Inverse Proportions
- Chapter 14: Factorisation
- Chapter 15: Introduction to Graphs
- Chapter 16: Playing with Numbers

**Ans:** Let the money with Chameli be ₹100

Money spent by her = 75% of 100

= 75/100 × 100 = ₹75

The money left with her = ₹100 – ₹75 = ₹25

₹25 are left with her out of ₹100

₹1 is left with her out of ₹100/25

₹600 will be left out of ₹100/25 × 600 = ₹2400

Hence, she had ₹2400 in beginning.

(i) Speed of a cycle 15 km per hour to the speed of a scooter 30 km per hour

(ii) 5 m to 10 km

(iii) 50 paise to ₹5

**Ans:** i) Ratio of the speed of cycle to the scooter = 15/30 = 1:2

ii) Since 1 km = 1000m, so the required ratio is:

5m/10000m = 1:2000

iii) Since ₹1 = 100 paisa

₹5 = 500 paisa

So, the required ratio is 50 paise/500 paise = 1/10 = 1:10

(i) 3:4

(ii) 2:3

**Ans:** i) 3:4 = 3/4 = 3/4 x 100/100 = 3/4 x 100% = 75%

ii) 2:3 = 2/3 = 2/3 x 100/100 = 2/3 x 100% = 66.6%

**Ans:** It is given that 72% of 25 students are good in mathematics.

Therefore,

Percentage of students who are not good in mathematics = (100 − 72)%

= 28%

∴Number of students who are not good in mathematics = 28/7 x 100

= 7

Thus, 7 students are not good in mathematics.

**Ans:** Percentage of people who like other games = (100 − 60 − 30)%

= (100 − 90)% = 10 %

Total number of people = 50 lakh

Therefore, the number of people who like cricket = (60/100 x 50) = 30 lakh

Number of people who like football = (30/100 x 50) = 15 lakh

Number of people who like other games = (10/100 x 50) = 5 lakh