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Case Study Questions for Class 7 Maths

Case Study Questions for Class 7 Maths

Table of Contents

Here in this article, we are providing case study questions for class 7 maths.

Maths Class 7 Chapter List

Latest chapter list (2023-24).

There is total 13 chapters.

Chapter 1 Integers Case Study Questions Chapter 2 Fractions and Decimals Case Study Questions Chapter 3 Data Handling Case Study Questions Chapter 4 Simple Equations Case Study Questions Chapter 5 Lines and Angles Case Study Questions Chapter 6 The Triangles and its Properties Case Study Questions Chapter 7 Comparing Quantities Case Study Questions Chapter 8 Rational Numbers Case Study Questions Chapter 9 Perimeter and Area Case Study Questions Chapter 10 Algebraic Expressions Case Study Questions Chapter 11 Exponents and Powers Case Study Questions Chapter 12 Symmetry Case Study Questions Chapter 13 Visualising Solid Shapes Case Study Questions

Old Chapter List

Chapter 1 Integers Chapter 2 Fractions and Decimals Chapter 3 Data Handling Chapter 4 Simple Equations Chapter 5 Lines and Angles Chapter 6 The Triangles and its Properties Chapter 7 Congruence of Triangles Chapter 8 Comparing Quantities Chapter 9 Rational Numbers Chapter 10 Practical Geometry Chapter 11 Perimeter and Area Chapter 12 Algebraic Expressions Chapter 13 Exponents and Powers Chapter 14 Symmetry Chapter 15 Visualising Solid Shapes

Deleted Chapter:

  • Chapter 7 Congruence of Triangles
  • Chapter 10 Practical Geometry

Tips for Answering Case Study Questions for Class 7 Maths in Exam

Tips for Answering Case Study Questions for Class 7 Maths in Exam

1. Comprehensive Reading for Context: Prioritize a thorough understanding of the provided case study. Absorb the contextual details and data meticulously to establish a strong foundation for your solution.

2. Relevance Identification: Pinpoint pertinent mathematical concepts applicable to the case study. By doing so, you can streamline your thinking process and apply appropriate methods with precision.

3. Deconstruction of the Problem: Break down the complex problem into manageable components or steps. This approach enhances clarity and facilitates organized problem-solving.

4. Highlighting Key Data: Emphasize critical information and data supplied within the case study. This practice aids quick referencing during the problem-solving process.

5. Application of Formulas: Leverage pertinent mathematical formulas, theorems, and principles to solve the case study. Accuracy in formula selection and unit usage is paramount.

6. Transparent Workflow Display: Document your solution with transparency, showcasing intermediate calculations and steps taken. This not only helps track progress but also offers insight into your analytical process.

7. Variable Labeling and Definition: For introduced variables or unknowns, offer clear labels and definitions. This eliminates ambiguity and reinforces a structured solution approach.

8. Step Explanation: Accompany each step with an explanatory note. This reinforces your grasp of concepts and demonstrates effective application.

9. Realistic Application: When the case study pertains to real-world scenarios, infuse practical reasoning and logic into your solution. This ensures alignment with real-life implications.

10. Thorough Answer Review: Post-solving, meticulously review your answer for accuracy and coherence. Assess its compatibility with the case study’s context.

11. Solution Recap: Before submission, revisit your solution to guarantee comprehensive coverage of the problem and a well-organized response.

12. Previous Case Study Practice: Boost your confidence by practicing with past case study questions from exams or textbooks. This familiarity enhances your readiness for the question format.

13. Efficient Time Management: Strategically allocate time for each case study question based on its complexity and the overall exam duration.

14. Maintain Composure and Confidence: Approach questions with poise and self-assurance. Your preparation equips you to conquer the challenges presented.

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case study problems for class 7

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  • Important Questions for CBSE Class 7 Maths Chapter 1 - Integers (Free PDF Download)

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Important Practice Problems for CBSE Class 7 Maths Chapter 1: Integers

Mathematics, an intriguing yet demanding subject, has been an integral part of every student's education since childhood. However, over time, the academic curriculum has evolved, incorporating higher standards and increased difficulty levels to encompass various branches of Mathematics, promoting holistic development among students. Despite being a captivating and logic-oriented subject, many students struggle to excel in Mathematics.

This struggle often instills fear and discouragement among students, especially those who perceive Mathematics as a formidable challenge. The fear of Mathematics gradually engulfs them, impacting their overall academic performance and impeding their educational growth. It is during Class 7 that students lay the foundation for modern Mathematics, making it a crucial phase in their academic journey.

To alleviate the complexities and challenges associated with Mathematics, Vedantu offers a comprehensive support system designed to accelerate students' academic success. Through this system, Vedantu aims to simplify the intricate concepts of Mathematics, providing efficient guidance and assistance to students, enabling them to overcome obstacles and achieve their full potential.

Vedantu is a platform that provides free CBSE Solutions (NCERT) and other study materials for students. Maths Students who are looking for better solutions can download Class 7 Maths NCERT Solutions to help you to revise the complete syllabus and score more marks in your examinations. Register Online for NCERT Solutions Class 7 Science tuition on Vedantu.com to score more marks in the CBSE board examination.

Study Important Questions for Class 7 Maths Chapter 1- Integers

A.Very short answer question – 1 marks

1. Define Integers.

Ans: The numbers range from negative infinity to positive infinity including zero. They are denoted by I i.e.  $ \text{I=}\left\{ \left. .....\text{-3,-2,-1,0,1,2,3}..... \right\} \right. $ .

2. We move to the left in the number line when we__________ or __________.

Ans: We move to the left in the number line when we add a negative integer or subtract a positive integer .

3. Additive inverse of  $ \text{-25} $  is ____.

Ans:  $ 25 $ .

4. Fill the blanks for  $ \text{-228+96+125} $  ___  $ \text{-451+197+76}\left( \text{use,,=} \right) $ 

Ans:  $  >  $ 

5. What would come in place of ? in  $ \text{-11+0=?} $ 

Ans:  $ \text{-11} $ 

6. Fill the blanks for  $ \text{-22 }\!\!\times\!\!\text{ -13 }\!\!\times\!\!\text{ 5=} $  _____

Ans:  $ \text{1430} $ 

7.Fill the blanks for  $ \text{-3 }\!\!\times\!\!\text{ 125=} $  ___

Ans: $ \text{-375} $ 

B. Short Answer Questions – 2 marks

8. Verify  $ \text{a-}\left( \text{-b} \right)\text{=a+b} $  for the following values of  $ \text{a} $  and  $ \text{b} $ 

a.  $ \text{a=25,b=12} $ 

Ans: Substituting value of  $ \text{a} $  and  $ \text{b} $  in given equation

$ \text{a-}\left( \text{-b} \right)\text{=a+b} $ 

 $ \text{25-}\left( \text{-12} \right)\text{=25+12} $ 

 $ \,\text{25+12=25+12} $ 

 $ \text{37=37} $ 

Hence, verified.

b. $ \text{a=113,b=16} $ 

 $ \text{a-}\left( \text{-b} \right)\text{=a+b} $ 

 $ \text{113-}\left( \text{-16} \right)\text{=113+16} $ 

 $ \text{113+16=113+16} $ 

 $ \text{129=129} $ 

9. Use  $ \text{,} $   or  $ \text{=} $  sign for the below statements to make it true 

a. $ \left( \text{-9} \right)\text{+}\left( \text{-28} \right) $ ____  $ \left( \text{-9} \right)\text{-}\left( \text{-28} \right) $ 

Ans: Solving both sides-

$ \left( \text{-9} \right)\text{+}\left( \text{-28} \right)=-37 $ 

$ \left( \text{-9} \right)\text{-}\left( \text{-28} \right)=19 $ 

Thus,  $ \left( \text{-9} \right)\text{+}\left( \text{-28} \right) < \left( \text{-9} \right)\text{-}\left( \text{-28} \right) $  

b. $ \text{25+}\left( \text{-14} \right)\text{-18} $  ____   $ \text{25+}\left( \text{-14} \right)\text{-}\left( \text{-18} \right) $ 

 $  \text{2}5+\left( -14 \right)-18=11-18  $ 

 $  =\text{-7} $ 

 $  \text{25+}\left( \text{-14} \right)\text{-}\left( \text{-18} \right)=11+18  $ 

 $  =29 $ 

Thus,  $ \text{2}5+\left( -14 \right)-18 < \text{25+}\left( \text{-14} \right)\text{-}\left( \text{-18} \right) $.

10. Write down a pair of integers for the following 

a. Sum gives   $ \text{-9} $ 

Ans:  A pair of integers that gives sum  $ \text{-9} $  is  $ \left( -6,-3 \right) $.

b. Difference gives  $ \text{-11} $ 

Ans:  A pair of integers that gives sum  $ \text{-11} $  is  $ \left( -14,3 \right) $.

11.  a. Write a positive and negative integer whose sum is  $ \text{-4} $ .

Ans: $ \left( 4,-8 \right) $  is a positive and negative integer whose sum is  $ \text{-4} $  .

b.Write a negative integer and a positive integer whose difference is  $ \text{-2} $ .

Ans:  $ \left( -1,1 \right) $  is a positive and negative integer whose sum is  $ \text{-2} $  .  

12. Fill in the blanks

a.  $ \left( \text{-4} \right)\text{+}\left( \text{-11} \right)\text{=}\left( \text{-11} \right)\text{+ }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{ } $ 

Ans: $ \left( \text{-4} \right)\text{+}\left( \text{-11} \right)\text{=}\left( \text{-11} \right)\text{+ }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{ } $ 

\[\Rightarrow \left( \text{-4} \right)\text{+}\left( \text{-11} \right)\text{+11}\]

\[\Rightarrow -4\]

Thus,  $ \left( \text{-4} \right)\text{+}\left( \text{-11} \right)\text{=}\left( \text{-11} \right)\text{+-4} $ 

b.  $ \left[ \text{22+}\left( \text{-9} \right) \right]\text{+}\left( \text{-2} \right)\text{=22+}\left[ \text{ }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{ +}\left( \text{-2} \right) \right] $ 

Ans: \[\left[ \text{22+}\left( \text{-9} \right) \right]\text{+}\left( \text{-2} \right)\text{=22+}\left[ \text{ }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{ +}\left( \text{-2} \right) \right]\]

\[\Rightarrow \text{13-2=22+}\left[ \text{ }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{ +}\left( \text{-2} \right) \right]\]

\[\Rightarrow \text{11-22=}\left[ \text{ }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{ +}\left( \text{-2} \right) \right]\]

\[\Rightarrow \text{-11= }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{ +}\left( \text{-2} \right)\]

\[\Rightarrow \text{-11+2}\]

\[\Rightarrow \text{-9}\]

Thus, \[\left[ \text{22+}\left( \text{-9} \right) \right]\text{+}\left( \text{-2} \right)\text{=22+}\left[ \text{-9+}\left( \text{-2} \right) \right]\]

13.  Verify  $ \text{7 }\!\!\times\!\!\text{ }\left[ \left( \text{22} \right)\text{+}\left( \text{-9} \right) \right]\text{=}\left[ \left( \text{7} \right)\text{ }\!\!\times\!\!\text{ 22} \right]\text{+}\left[ \text{7 }\!\!\times\!\!\text{ -9} \right] $ 

Ans:  On solving both sides

$ \text{7 }\!\!\times\!\!\text{ }\left[ \left( \text{22} \right)\text{+}\left( \text{-9} \right) \right]\text{=}\left[ \left( \text{7} \right)\text{ }\!\!\times\!\!\text{ 22} \right]\text{+}\left[ \text{7 }\!\!\times\!\!\text{ -9} \right] $ 

$ 7\times \left[ 13 \right]=154-63 $ 

$ 91=91 $ 

14.Find the product of

a.  $ \text{63 }\!\!\times\!\!\text{ 0 }\!\!\times\!\!\text{ -7} $ 

Ans: The product of  $ \text{63 }\!\!\times\!\!\text{ 0 }\!\!\times\!\!\text{ -7} $  is  $ 0 $  .

b. $ \text{5 }\!\!\times\!\!\text{ }\left( \text{-3} \right)\text{ }\!\!\times\!\!\text{ -2} $ 

Ans: So, $ 5\times \left( -3 \right)\times -2=5\times 6 $ 

$ =30 $ 

The product of   $ 5\times \left( -3 \right)\times -2 $  is  $ 30 $  .

a.  $ \text{-2 }\!\!\times\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{ =14} $ 

Ans: So,   $ \text{-2 }\!\!\times\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{ =14} $ 

$ \Rightarrow \dfrac{\text{14}}{\text{-2}} $ 

$ \Rightarrow \text{-7} $ 

Hence,  $ \text{-2 }\!\!\times\!\!\text{ 7 =14} $  .

b. \[\text{ }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\times\!\!\text{ -8=-32}\]

Ans:  So, \[\text{ }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{   }\!\!\times\!\!\text{ -8=-32}\]

\[\Rightarrow \dfrac{-32}{-8}\]

\[\Rightarrow 4\]

Hence, \[\text{4 }\!\!\times\!\!\text{ -8=-32}\]

16. Evaluate 

a. $ \text{-39 }\!\!\div\!\!\text{ 13} $ 

Ans:  $ -39\div 13 $ 

$ \Rightarrow \dfrac{-39}{13} $ 

$ \Rightarrow -3 $ 

Hence,  $ -39\div 13=-3 $ 

b. $ \text{-64 }\!\!\div\!\!\text{ }\left[ \text{-8 }\!\!\times\!\!\text{ -8} \right] $ 

Ans:   $ -64\div \left[ -8\times -8 \right] $ 

$ \Rightarrow \dfrac{-64}{\left[ -8\times -8 \right]} $ 

$ \Rightarrow \dfrac{-64}{64} $ 

$ \Rightarrow -1 $ 

Hence,  $ -64\div \left[ -8\times -8 \right]=-1 $

17. Write two pairs of integers such that  $ \text{a }\!\!\div\!\!\text{ b=-5} $ 

Ans: The two pairs of integers such that  $ \text{a }\!\!\div\!\!\text{ b=-5} $  are:

> $ \left( 10,-2 \right) $ 

> $ \left( -70,14 \right) $ 

C. Short answer questions – 3 marks

18.  Manvita deposits Rs.  $ \text{5000} $   in her bank account after two days. She withdraws Rs.  $ \text{3748} $   from it. If the amount deposited is a positive integer. How will you represent the amount withdrawn and also find the balance amount in the account?

Ans: The amount withdrawn should always be represented as a negative integer.

Thus, it would be  $ -3748 $.

Since, Total balance  $ = $   Amount deposited  $ - $  Amount withdrawn

Total balance  $ =5000-3748 $ 

 $ \text{=Rs}\text{. 1252} $ .

Hence, the amount withdrawn would be negative integer i.e.,  $ -3748 $  and the balance amount in the account is  $ \text{Rs}\text{. 1252} $ .

19. In a game Mishala scored  $ \text{20,}\,\text{-40,}\,\text{10} $  and Meera scored  $ \text{-40,10,}\,\text{20} $ . Who scored more and can we add scores (integers) in any order?

Ans: Since, Mishala scored  $ \text{20,}\,\text{-40,}\,\text{10} $ .

Therefore, total score of Mishala is 

 $ \text{=20-40+}\,\text{10} $ 

 $ \text{=-20+10} $ 

 $ \text{=-10} $ 

And since, Meera scored  $ \text{-40,}\,1\text{0,}\,2\text{0} $ .

Therefore, total score of Meera is 

 $ \text{=-40+}\,\text{10+20} $ 

Hence, both scored the same points in a game but in a different order.

Yes, we can add integers in any order.

20. Find the product with suitable properties for the following-

a. $ \text{16 }\!\!\times\!\!\text{ }\left( \text{-34} \right)\text{+}\left( \text{-34} \right)\text{ }\!\!\times\!\!\text{ }\left( \text{-18} \right) $ 

Ans: Given 

$ \text{16 }\!\!\times\!\!\text{ }\left( \text{-34} \right)\text{+}\left( \text{-34} \right)\text{ }\!\!\times\!\!\text{ }\left( \text{-18} \right) $ 

By distributive property-

 $ \text{a }\!\!\times\!\!\text{ b+a }\!\!\times\!\!\text{ c=a}\left[ \text{b+c} \right] $ 

Thus, 

 $ \text{=-34}\left[ \text{16-18} \right] $ 

 $ \text{=-34 }\!\!\times\!\!\text{ -2} $ 

 $ \text{=68} $ 

Hence,  $ \text{16 }\!\!\times\!\!\text{ }\left( \text{-34} \right)\text{+}\left( \text{-34} \right)\text{ }\!\!\times\!\!\text{ }\left( \text{-18} \right)=68 $ .

b. $ \text{23 }\!\!\times\!\!\text{ -36 }\!\!\times\!\!\text{ 10} $ 

$ 23\times -36\times 10 $ 

By commutative property-

 $ \left( \text{a }\!\!\times\!\!\text{ b} \right)\text{ }\!\!\times\!\!\text{ c=a }\!\!\times\!\!\text{ }\left( \text{b }\!\!\times\!\!\text{ c} \right) $ 

 $ =23\times \left[ -36\times 10 \right] $ 

 $ =23\times -360 $ 

 $ =-8280 $ 

21.  A fruit merchant earns a profit of Rs. $ \text{6} $  per bag of orange sold and a loss of Rs. $ \text{4} $  per bag of grapes sold.

a. Merchant sells  $ \text{1800} $  bags of orange and  $ \text{2500} $  bags of grapes. What is the profit or loss?

Ans: Since, profit is denoted by a positive integer and a loss is denoted by a positive integer.

Therefore, profit earned by selling  $ \text{1} $  bag of orange is Rs.  $ 6 $ 

Profit earned by selling  $ \text{1800} $  bags or orange is 

 $ \text{6 }\!\!\times\!\!\text{ 1800} $  

 $ \text{=Rs}\text{. 10,800} $  

Loss incurred by selling  $ 1 $  bag of grapes is Rs.  $ -4 $  

Loss incurred by selling  $ 2500 $  bags of grapes is

 $ =-4\times 2500 $ 

 $ =10,000 $  

Total profit or loss earned  $ = $  Profit  $ + $ Loss

 $  =10,800+10,000  $ 

 $  =800 $  

Hence, a profit of Rs. $ 800 $   will be earned by a merchant.

b. What is the number of bags of oranges to be sold to have neither profit nor loss if the number of grapes bags are sold is  $ \text{900} $  bags?

Ans: Since profit is denoted by a positive integer and a loss is denoted by a positive integer.

Therefore, Loss incurred while selling  $ 1 $   bag of grapes  $ \text{=-Rs}\text{.4} $  

Loss incurred while selling 900 bags of grapes be 

 $ =-4\times 900 $ 

 $ =-3600 $  

Let the number of bags of oranges to be sold  $ \text{=x} $  

Profit earned when  $ 1 $  bag of orange is sold  $ \text{=Rs}\text{.6} $  

Profit earned while selling x bags of orange  $ \text{=6x} $  

Condition for no profit, no loss

Profit earned  $ + $   Loss incurred  $ =0 $  

 $ \text{6x-3600=0} $ 

 $ \text{6x=3600} $ 

 $ \text{x=}\dfrac{\text{3600}}{\text{6}} $ 

\[\text{x=600}\]

Hence, to have neither profit nor loss \[\text{600}\] number of bags of oranges to be sold.

22.  Verify that \[\text{a }\!\!\div\!\!\text{ }\left( \text{b+c} \right)\ne \left( \text{a }\!\!\div\!\!\text{ b} \right)\text{+}\left( \text{a }\!\!\div\!\!\text{ c} \right)\] for each of the following values of  $ \text{a,b} $  and  $ \text{c} $ .

a. $ \text{a=8,}\,\text{b=4,}\,\text{c=2} $ 

Ans: For equation \[\text{a }\!\!\div\!\!\text{ }\left( \text{b+c} \right)\ne \left( \text{a }\!\!\div\!\!\text{ b} \right)\text{+}\left( \text{a }\!\!\div\!\!\text{ c} \right)\].

L.H.S  \[\text{=a }\!\!\div\!\!\text{ }\left( \text{b+c} \right)\]

\[\text{=8 }\!\!\div\!\!\text{ }\left( \text{-4+2} \right)\]

\[\text{=8 }\!\!\div\!\!\text{ }\left( \text{-2} \right)\]

\[\text{=-4}\]

R.H.S  \[=\left( \text{a }\!\!\div\!\!\text{ b} \right)\text{+}\left( \text{a }\!\!\div\!\!\text{ c} \right)\]

 $ \text{=}\left( \text{8 }\!\!\div\!\!\text{ -4} \right)\text{+}\left( \text{8 }\!\!\div\!\!\text{ 2} \right) $ 

 $ \text{=-2+4} $ 

 $ \text{=2} $ 

Hence,  $ \text{L}\text{.H}\text{.S}\ne \text{R}\text{.H}\text{.S} $  .

Thus, \[\text{a }\!\!\div\!\!\text{ }\left( \text{b+c} \right)\ne \left( \text{a }\!\!\div\!\!\text{ b} \right)\text{+}\left( \text{a }\!\!\div\!\!\text{ c} \right)\] for  $ \text{a=8,}\,\text{b=4,}\,\text{c=2} $ .

b. $ \text{a=-15,}\,\text{b=2,}\,\text{c=1} $ 

 $ \text{=-15 }\!\!\div\!\!\text{ }\left( \text{2+1} \right) $ 

 $ \text{=-15 }\!\!\div\!\!\text{ 3} $ 

 $ \text{=-5} $ 

 $ \text{=}\left( \text{-15 }\!\!\div\!\!\text{ 2} \right)\text{+}\left( \text{-15 }\!\!\div\!\!\text{ 1} \right) $ 

 $ \text{=-7}\text{.5+}\left( \text{-15} \right) $ 

 $ \text{=-22}\text{.5} $ 

Hence,  $ \text{L}\text{.H}\text{.S}\ne \text{R}\text{.H}\text{.S} $ 

Thus, \[\text{a }\!\!\div\!\!\text{ }\left( \text{b+c} \right)\ne \left( \text{a }\!\!\div\!\!\text{ b} \right)\text{+}\left( \text{a }\!\!\div\!\!\text{ c} \right)\] for  $ \text{a=-15,}\,\text{b=2,}\,\text{c=1} $  .

23. In a CET Examination  $ \left( \text{+2} \right) $  marks are given for every current answer and  $ \left( \text{-0}\text{.5} \right) $   marks are given for every wrong answer and  $ 0 $  for non-attempting any question.

a. Likitha scores  $ \text{30} $  marks. If she got  $ \text{20} $ correct answers, how many questions she has attempted incorrectly?

Ans: Marks obtained for  $ 1 $  correct answer  $ \text{=+2} $  

Marks obtained for  $ 1 $  wrong answer  $ \text{=-0}\text{.5} $  

So, Marks scored by Likitha =  $ 30 $  

Marks obtained by  $ 20 $  correct answers $ =20\times 2=40 $  

Marks obtained for incorrect answer  $ = $  Total score  $ - $  Marks obtained by  $ 20 $  correct answer

 $ =30-40 $  

 $ =-10 $ 

Marks obtained for  $ 1 $  wrong answer $ =-0.5 $  

 $ \therefore  $ The number of incorrect answers $ =\dfrac{-10}{-0.5} $  

 $ =20 $  

Hence, she attempted  $ 20 $  questions wrongly.

b. Saara scores  $ \text{-4} $  marks if she got  $ \text{3} $  correct answers. How many were incorrect?

So, Marks scored by Saara  $ =-4 $  

Marks obtained for 3 correct answers $ =3\times 2=6 $  

Marks obtained for incorrect answers  $ = $  Total score  $ - $  Marks obtained for  $ 3 $  correct answer

 $ =-4-6=-10 $  

Marks obtained for 1 wrong answer $ =-0.5 $  

 $ \therefore  $ The number of incorrect questions  $ =\dfrac{-10}{-0.5} $  

Hence,  $ 20 $  questions were incorrect. 

Important Questions for CBSE Class 7 Maths Chapter 1 - Integers

Students who have not developed a solid base feel that Maths is hard to excel in and may perceive the subject as a burden. As they have not worked on the subject or failed to grasp its theories and logic, they must start well in class 7 to lead a strong foundation in Mathematics, to conquer good marks in the examination. The magic wand of a good Mathematician is practice. Therefore, to secure a good score in the examination, the students must practice questions based on every theory and formula of Maths.

They may begin with class 7 Maths Chapter 1 important questions to start their practice from the first chapter. If the students find the topics challenging or cannot deal with complex issues, they may use efficient study material to dismantle diplomatic problems into manageable solutions. The critical questions for Class 7 Maths Chapter 1 are available in text and PDF formats online, which is accessible to them for free.

Details of Class 7 Maths Chapter 1 Integers

In the first chapter of class 7 Maths, the students get a sneak peek into the integer number system. The students must ensure a deep understanding of the topic to proceed with the upcoming chapters. To grasp the topic’s in-depth knowledge, they must practice the important questions for Class 7 Maths Chapter 1. Now, we will highlight some of the important concepts that are included in chapter 1:

An integer is one of the fundamental parts of Mathematics. It can be quoted as a number that can be depicted without any fractional components. For instance, 3, 61, 70, 5 all are integers, while 6.54, 5.89 are non-integer numbers.

We can easily blend out an integer from a series of counting numbers. Let's make it clear with an example, suppose if a counting number is subtracted from itself, the result will be zero. If a larger counting number is removed from a smaller whole number, the output becomes a negative integer. When we subtract the smaller number from the larger whole number, it results in a positive integer. Applying this methodology, we can derive many integers ranging from negative to positive. A set of integers are depicted by 'Z.'

Z = {….., -4,-3,-2,-1,0,1,2,3,4,…..n}.

Properties of Integers

Some of the properties of an integer are as follows:

Commutative Property

The commutative properties of an integer depict that if we perform any operation like multiplication or some numbers, the numbers’ position can be swapped without differing in the output.

Let's make the property clear with an example:

Suppose X and Y are two non-zero integers,

Therefore, the commutative property of addition is X + Y = Y + X.

And, the commutative property of multiplication is X x Y = Y x X.

Associative Property

The associative property of integers depicts that if we perform an addition or multiplication operation on any set of numbers, the result will be identical, irrespective of the grouping of the multiplicands or addends. Some of the traits of associative properties are mentioned below:

The associative property of integers involves a minimum of 3 numbers.

Generally, the integers are grouped using parenthesis.

The numbers defined within the parenthesis are depicted by one unit.

The associative property can only be implemented for addition and multiplication operations and not for division or subtraction.

Let's take an example to make the associative property clear. According to the property, 7 + (8 + 2) = 2 + (7 + 8).

Distributive Property

The distributive property depicts that if two or more numbers are added and multiplied with another number, it will be identical to the current output if each addend is individually multiplied and then added together.

Here's an example to clear the distributive property of integers:

(7 + 1 + 2) x 5

This equation can be simplified to 10 x 5 = 50,

While if we dismantle the equation as 7 x 5 + 1 x 5 + 2 x 5, the result will be equal, i.e., 50.

Arithmetical Operations Using Integers

Addition of integers.

There are a set of rules to add integer with same and different signs:

During the addition of two integer numbers with the same sign, the output generated also depicts the same sign. Example: 7 + 8 = 15.

For the addition of two integers, one with positive and one with negative signs, the result must retain the largest integer sign. The operation must be performed by subtracting the two integers. Example: 8 + (-14) = -6.

Subtraction of Integers

The rules to perform subtraction of integers are as follows:

If the subtraction is to be performed between two integers with different signs, i.e., one negative and one positive, the output will retain the largest integer's sign. Example: 9 - 5 = 4.

If the subtraction is performed between two negative integers, the result can be obtained by adding the same number with the opposite sign. Example: -7 - (-9) = -7 + 9 = 2.

Multiplication of Integers

Multiplication of a positive and a negative integer - The result of the multiplication of a positive and a negative integer can be generated simply by multiplying both the numbers and denoting the output with a minus(-) sign. For example: -7 x 6 = -42.

Multiplication of Two Negative Integers - The product of two negative integers is always a positive integer. For example: -2 x -7 = 14.

Multiplication of Three or More Negative Integers - If the total integers to be multiplied is even, then the output will carry plus (+) sign. The total number of integers to be multiplied is odd. The result will carry the minus(+) sign.

Division of Integers

Division of a Negative Integer by a Positive Integer - When a negative number is divided by a positive number, the quotient comes with a negative sign.

Division of a Negative Integer by Another Negative Integer - when a negative number is divided by another negative number, the quotient comes with a positive sign.

To get in-depth details about the topics, students can practice the important questions in class 7 Maths chapter 1.

Important Questions for Class 7 Maths Chapter 1 Integers

Thorough practicing of the important questions for Class 7 Maths integers may help the students grip the topics efficiently. Here we present you the vital questions on the chapter: 

What is an Integer? Explain with examples.

What are the dissimilarities between Integers and Real numbers?

What is the commutative property of integers? Explain with examples.

Define the additive identity of an integer.

What is the distributive property of an integer?

Explain the multiplication properties of integers.

What are the division properties of the integers?

A lift is moving towards a mine craft at a rate of 8m/min. How much time will it take to reach -300m if the descent began from 15m above the ground level?

Explain the associative property of integers with an example.

What are the five integer operations?

Benefits of Important Questions for Class 7 Maths Chapter 1

Students deserve the best study material to guide them on the topics. Therefore, they can unleash the actual benefits of the Class 7 Maths Chapter 1 important questions by thoroughly practicing to secure good marks in the examination. Here are some of the fruitfulness that the students can enjoy from the questions:

The questions are carefully designed under the strict guidelines of CBSE and are reviewed by top-notch professionals.

The students must practise these important questions to understand every concept and theory of the chapter.

In case the students find any difficulty to solve problems or understand any concept, they may refer to the in-detailed solutions. Vedantu, with its team of expert educators, has framed every important question of Class 7 Maths Integers after efficient analysis of the past year’s question papers.

Students who find it challenging to study Mathematics can build their logic to score well in the examination. To kick start the class 7 academic journey, the students must efficiently practice the Important Questions of Class 7 Maths Chapter 1 Integers, which they can access for free on Vedantu’s website. Students just have to download the PDF and then they can solve the extra questions from the comfort of their homes anytime.

For more study materials related to different chapters of Class 7 Mathematics, students can visit Vedantu’s website and explore the huge collection of free resources available with us.

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FAQs on Important Questions for CBSE Class 7 Maths Chapter 1 - Integers (Free PDF Download)

1. What are the various types of integers?

The various types of Integers are as follows:

Positive Numbers - are those numbers which have a plus sign (+). They are always found on the right side of zero on the number line. Most of the time positive numbers are represented simply as a whole number without the plus sign (+). Every positive number is greater than zero as well as negative numbers.

Example: 1,2, 234, 5667,99999999, etc.

Negative Numbers

Negative numbers are numbers symbolized with a minus sign (-). Negative numbers are represented to the left of zero on a number line.

Example: …., -99, -158, -110, -1.

Zero is neither a positive nor a negative integer. It is a neutral number i.e. zero has no sign (+ or -).

2: What are the operations that can be done on Integers? What are its main properties?

The various operations that are performed on Integers are as follows: 

Subtraction

Multiplication

The main properties of integers are as follows: 

Closure Property

Identity Property

3: What do you mean by commutative property?

The commutative property of addition and multiplication states that irrespective of the order of terms the result will not change. Swapping of terms will have no effect on the sum or product.

Ex:  x and y are any two integers, then

⇒ x + y = y + x

⇒ x × y = y × x

14 + (−6) = 8 = (−6) +1 4;

10 × (−3) = −30 = (−3) × 10

Note: Subtraction (x − y ≠ y − x) and division (x ÷ y ≠ y ÷ x) are not commutative for integers and whole numbers.

4: Fill in the blanks to make the following statements true:

(–5) + (– 8) = (– 8) + (…………)

A4: Let us assume the missing integer be x,

= (–5) + (– 8) = (– 8) + (x)

= – 5 – 8 = – 8 + x

= – 13 = – 8 + x

By sending – 8 from RHS to LHS it becomes 8,

= – 13 + 8 = x

Now substitute the x value in the blank place,

(–5) + (– 8) = (– 8) + (- 5) … (This equation is in the form of Commutative law of Addition)

5. What is Chapter 1 of Class 7 Maths about?

The first Chapter of Class 7 Maths begins with a review of the fundamentals of integers that were covered in the previous classes. These recollected ideas are used in the first exercise in the Class 7 Mathematics Chapter 1 answers. You will learn more about integers, their characteristics, and operations in Chapter 1 of Class 7 Mathematics. Expert teachers prepare the important questions to assist students. For a student, the Important Questions of Chapter 1 of Class 7 Mathematics provided by Vedantu are very useful.

6. How many questions are there in Chapter 1 -Integer of Class 7 Maths?

There are four exercises in Chapter 1 “Integers” of Class 7 Maths. These four exercises contain a total of 30 questions. For more practice of Chapter 1 of Class 7 Maths, students can refer to the Important Questions of Chapter 1 of Class 7 Maths, prepared by experts at Vedantu for the benefit of the students.

7. Do I need to practice all questions provided in Important Questions for Chapter 1 of Class 7 Maths?

Yes, you need to practise all the questions. All the questions in this chapter deal with a unique concept and in order to get clarity on all the concepts, you must practice every sum. A short review of integers and their representation on the number line, operations such as addition, subtraction, and others that may be done on them, additive inverse, and negative integers are just a few of the subjects covered at the beginning. For important questions of Chapter 1 of Class 7 Maths, visit Vedantu website or mobile app and download the PDF free of cost.

8. What is the associative property of integers?

The associative property of integers states that no matter how the multiplicands or addends are grouped, the result of an addition or multiplication operation on any set of numbers will be the same. The following are some of the characteristics of associative properties:

In general, parentheses are used to arrange the numbers.

One unit represents the numbers defined within the parentheses.

Integers have an associative characteristic that requires a minimum of three numbers.

Q9. What is Associative Property?

Ans: The associative property of integers states that no matter how the multiplicands or addends are grouped, the result of an addition or multiplication operation on any set of numbers will be the same. The following are some of the characteristics of associative properties:

Chapterwise Important Questions for CBSE Class 7 Maths

Ncert solutions for class 7 cbse.

Net Explanations

Class 7 Science Case Study Question

Case study question class 7 science (cbse / ncert board).

Class 7 Science Case Study Question and Answer: CBSE / NCERT Board Class 7 Science Case Study Question prepared by expert Science Teacher. Students can learn Case Based Question / Paragraph Type Question for NCERT Class 7 Science.

There are total 18 chapter Nutrition in Plants, Nutrition in Animals, Fibre to Fabric, Heat, Acids, Bases and Salts, Physical and Chemical Changes, Weather, Climate and Adaptations of Animals to Climate, Winds, Storms and Cyclones, Soil, Respiration in Organisms, Transportation in Animals and Plants, Reproduction in Plants, Motion and Time, Electric Current and Its Effects, Light, Water: A Precious Resource, Forests: Our Lifeline, Wastewater Story

For any problem during learning any Case or any doubts please comment us. We are always ready to help You.

CBSE Class 7 Science Case Study Question

  • Chapter 1 Nutrition in Plants Case Study Question
  • Chapter 2 Nutrition in Animals Case Study Question
  • Chapter 3 Fibre to Fabric Case Study Question
  • Chapter 4 Heat Case Study Question
  • Chapter 5 Acids, Bases and Salts Case Study Question
  • Chapter 6 Physical and Chemical Changes Case Study Question
  • Chapter 7 Weather, Climate and Adaptations of Animals to Climate Case Study Question
  • Chapter 8 Winds, Storms and Cyclones Case Study Question
  • Chapter 9 Soil Case Study Question
  • Chapter 10 Respiration in Organisms Case Study Question
  • Chapter 11 Transportation in Animals and Plants Case Study Question
  • Chapter 12 Reproduction in Plants Case Study Question
  • Chapter 13 Motion and Time Case Study Question
  • Chapter 14 Electric Current and Its Effects
  • Chapter 15 Light
  • Chapter 16 Water: A Precious Resource
  • Chapter 17 Forests: Our Lifeline
  • Chapter 18 Wastewater Story

What is Case Study Question?

Ans. At case Study there will one paragraph and on the basis of that concept some question will made. Students have to solve that question.

How many marks will have at case based question?

Most of time 5 questions will made from each case. There will 1 or 2 marks for each question.

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Simple Equations

Simple Equations

  • An equation which has only one variable is called Linear Equation

Maths class 8 Linear equations and variables

Solving equations which have linear equations on one side and numbers on the other side

EXAMPLE 1: Solve 3x + 15 = 1

  • We need to keep the variable on the left side and numerical terms on the right side. Because of this, we have to transport 15 to the right side by changing its sign i.e., -15. => 3x + 15 = 1 => 3x = 1-15

Maths class 8 Linear equations and variables

  • We need to keep the variable on the left side and numerical terms on the right side. Because of this, we have to transport -12 to the other side by changing its sign i.e., +12.

Maths class 8 Linear equations and variables

Solving equations having variables having the variable on both sides

EXAMPLE 1: Solve .

  • Simplify the given equation. => =>
  • Bring the variable terms on the left side of the equation and the other numerical terms on the right side of the equation. => =>
  • Now, to find the value of ‘x’ we need to divide both sides of the equation by 6 to maintain equality.

Maths class 8 Linear equations and variables

Application of Linear Equations

Linear equations are used to find the value of an unknown quantity. Have a look at the following examples:

EXAMPLE 1: The sum of the digits of a 2 digit number 13. The numbers obtained by interchanging the digits is 14 more than the given number. Find the number.

SOLUTION: Let the digit at units place be x and the number at tens place be y.

=> y + x = 13 [sum is 13 given] => y = 13 – x [Transposing x to the other side by changing its sign] Thus, the formed number is= [Since x is at ones place and y=13-x is at tens place] After interchanging the digits, the number is= [Now x is at tens place & y=13-x is at one's place] The interchanged number is greater than the original number by 14. [Given]

New number Old number Difference

  • Simplify => => => =>
  • Transpose the variable term ‘x’ on the left side of the equation and other numerical terms on the right side of the equation by changing their sign. => 18x – 117 = 14 => 18x = 131

Maths class 8 Linear equations and variables

EXAMPLE 2: The distance between town A and town B is 123 km. Two buses begin their journey from these towns and move directly toward each other. From town A, the bus is moving at a speed of 45 km per hour and from town B, the bus is moving at 67 km per hour. Assuming the buses start at the same time, find how far is their meeting point from town A. SOLUTION: Let the buses meet after t hours.

Maths class 8 Linear equations and variables

We know that distance= speed X time Distance covered by bus 1 = 45 X t Distance covered by bus 2 = 67 X t Therefore, 45t + 67t = 123

The distance travelled by bus 1 from city A to the meeting point= speed of bus 1 X time taken by it to reach the meeting point. =45 X 1.098= 49.41 km Thus, the distance of reaching point from town A is 49.41km. [ANS]

Reducing Equation to Simpler Form

Maths class 8 Linear equations and variables

Equations reducible to linear form

Maths class 8 Linear equations and variables

  • Now, simplify => 6x + 12 + 12x + 15 = 14x + 16 => 18x + 27 = 14x +16
  • Transpose the variable term ‘x’ to the left side and the numerical terms on the right side of the equation by changing their sign. => 18x – 14x = 16 – 27 => 4x = 11

Maths class 8 Linear equations and variables

Practice these questions

Maths class 8 Linear equations and variables

Q3) Three numbers are in the ratio 1:2:3. If the sum of the largest and the smallest equals the second and 45. Find the numbers.

Q4) Find the number whose 1/6 th part decreased 7 equals its 8/9 th part diminished by 1.

Q5) The difference between two numbers is 23. And the quotient obtained by dividing the larger number by the smaller one is 4. Find the numbers.

Q6) A man cycles to the office from his house at a speed of 5km per hour and reaches 6 minutes late. If he cycles at a speed of 7km/hr, he reaches 8 minutes early. What is the distance between the office and his house? Q7) Suraj is now half as old as his father. 20 years ago, Suraj’s father was six times Suraj’s age. What are their ages now? Q8) The perimeter of an isosceles triangle is 91cm. If the length of each equal side is 2cm more than the length of its base. Find the lengths of the sides of the triangle. Q9) The age of a boy in months is equal to the age of his grandfather in years. If the difference between their ages is 66 years, find their ages.

  • The basic principle used in solving any linear equation is that any operation performed on one side of the equation must also be performed on the other side of the equation.
  • Any term in an equation can be transposed from one side to other side by changing its sign.
  • In cross multiplication, we multiply the numerator of LHS by the denominator of RHS and the denominator of LHS by the numerator of RHS and the resultant expression are equal to each other.
  • Practical problems are based on the relations between some known and unknown quantities. We convert such problems into equations and then solve them.

Quiz for Simple Equations

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  • CBSE Notes For Class 7
  • CBSE Notes Class 7 Maths
  • Chapter 11: Perimeter And Area

Perimeter and Area Class 7 Notes: Chapter 11

According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 9.

Perimeter of a Quadrilateral    

For more information on Perimeter, watch the below video.

case study problems for class 7

To know more about Perimeter, visit here .

Areas of a closed figure

For more information on Visualising Area, watch the below video.

case study problems for class 7

The perimeter of Square and Rectangle

  • Perimeter of a square = a + a + a + a = 4a, where a is the length of each side.

The perimeter of Square and Rectangle

  • Perimeter of a rectangle = l + l + b + b = 2(l + b), where l and b are length and breadth, respectively.

Rectangle with length 'l' units and breadth 'b' units

To know more about Perimeter Formula’s of All geometrical Figures, visit here .

Area of Square & Rectangle

Area of square = 4 a 2

Here a is the length of each side

Square with the length of each side 'a' units

Area of rectangle = Length(l) × Breadth(b) = l × b

Area of rectangle = Length(l) × Breadth(b) = l×b

Area of a Parallelogram

Area of a Parallelogram

  • Area of parallelogram ABCD =  ( b a s e × h e i g h t )
Area of parallelogram ABCD  =  ( b × h )

Triangle as Part of Rectangle

  • The rectangle can be considered as a combination of two congruent triangles.

Triangle as Part of Rectangle

  • Area of each triangle = 1 2 (Area of the rectangle). =  1 2 ( l e n g t h × b r e a d t h ) =  1 2 ( 10 c m × 5 c m ) =  25 c m 2

Area of a Triangle

  • Consider a parallelogram ABCD.
  • Draw a diagonal BD to divide the parallelogram into two congruent triangles.

Area of a Triangle

  • Area of triangle ABD = 1/2 (Area of parallelogram ABCD)

Area of triangle ABD  = 1/2  ( b × h )

To know more about Area and Perimeter, visit here .

Conversion of Units

  • Kilometres, metres, centimetres, millimetres are units of length.
  • 10 millimetres = 1 centimetre
  • 100 centimetres = 1 metre
  • 1000 metres = 1 kilometre

Terms Related to Circle

  • A circle is a simple closed curve which is not a polygon.
  • A circle is  a collection of points which are equidistant from a fixed point .

Circle

  • The fixed point in the middle is called the  centre .
  • The fixed distance is known as  radius .
  • The perimeter of a circle is also called as the  circumference of the circle.

For more information on Terms Related to Circle, watch the below video.

case study problems for class 7

Circumference of a Circle

  • The circumference of a circle ( C )  is the total path or total distance covered by the circle. It is also called a perimeter of the circle.
Circumference of a circle = 2 × π × r ,

where r   is the radius of the circle.

Visualising Area of a Circle

Area of circle.

  • Area of a circle is the total region enclosed by the circle.
Area of a circle = π × r 2 , where r is the radius of the circle.

For more information on Area of Circle, watch the below video.

case study problems for class 7

To know more about Circles, visit here .

Introduction and Value of Pi

  • Pi ( π )    is the constant which is defined as the ratio of a circle’s circumference  ( 2 π r )   to its   diameter(2r) .
π = Circumference ( 2 π r )/Diameter (2r)
  • The value of pi is approximately equal to 3.14159 or 22/7.

For more information on The Value Of Pi, watch the below video.

case study problems for class 7

To know more about Value of Pi, visit here .

Problem Solving

Cost of framing, fencing.

  • Cost of framing or fencing a land is calculated by finding its perimeter.
  • Example: A square-shaped land has length of its side 10m. Perimeter of the land = 4 × 10 = 40m Cost of fencing 1m = Rs 10 Cost of fencing the land = 40 m × Rs 10 = Rs 400

Cost of Painting, Laminating

  • Cost of painting a surface depends on the area of the surface.
  • Example: A wall has dimensions 5 m × 4 m . Area of the wall = 5 m × 4 m = 20 m 2 Cost of painting 1 m 2 of area is R s   20 . Cost of painting the wall = 20 m 2 × R s   20 = R s   400

Area of Mixed Shapes

  • Find the area of  the shaded portion using the given information.

Area of Mixed Shapes

Solution: Diameter of the semicircle = 10cm Radius of semicircle = 5cm Area of the shaded portion = Area of rectangle ABCD – Area of semicircle Area of the shaded portion  = ( l × b)  − ( πr 2 /2) =  30×10  − ( π ×5 2 /2) = 300  −  ( π ×25/2) = (600 – 25 π) /2 = (600 – 78.5)/2 = 260.7   c m 2

Frequently Asked Questions on CBSE Class 7 Maths Notes Chapter 11 Perimeter and Area

What is value of pi.

The value of pi is 3.14159.

What is meant by area of a substance?

The amount of space taken up by any 2D shape is known to be the area covered by that substance.

What is the purpose of Unit conversion?

Unit conversion helps in the expression of the same property in different units of measurement. For example, time can be expressed in hours, minutes or seconds.

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Trump Fraud Trial Penalty Will Exceed $450 Million

The ruling in Donald J. Trump’s civil fraud case could cost him all his available cash. The judge said that the former president’s “complete lack of contrition” bordered on pathological.

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Donald Trump, wearing a blue suit and blue tie, sits at the defendant’s table in a courtroom.

By Jonah E. Bromwich and Ben Protess

A New York judge on Friday handed Donald J. Trump a crushing defeat in his civil fraud case , finding the former president liable for conspiring to manipulate his net worth and ordering him to pay a penalty of nearly $355 million plus interest that could wipe out his entire stockpile of cash .

The decision by Justice Arthur F. Engoron caps a chaotic, yearslong case in which New York’s attorney general put Mr. Trump’s fantastical claims of wealth on trial. With no jury, the power was in Justice Engoron’s hands alone, and he came down hard: The judge delivered a sweeping array of punishments that threatens the former president’s business empire as he simultaneously contends with four criminal prosecutions and seeks to regain the White House.

Justice Engoron barred Mr. Trump for three years from serving in top roles at any New York company, including portions of his own Trump Organization. He also imposed a two-year ban on the former president’s adult sons and ordered that they pay more than $4 million each. One of them, Eric Trump, is the company’s de facto chief executive, and the ruling throws into doubt whether any member of the family can run the business in the near term.

The judge also ordered that they pay substantial interest, pushing the penalty for the former president to $450 million, according to the attorney general, Letitia James.

In his unconventional style, Justice Engoron criticized Mr. Trump and the other defendants for refusing to admit wrongdoing for years. “Their complete lack of contrition and remorse borders on pathological,” he said.

He noted that Mr. Trump had not committed violent crimes and also conceded that “Donald Trump is not Bernard Madoff.” Still, he wrote, “defendants are incapable of admitting the error of their ways.”

case study problems for class 7

The Civil Fraud Ruling on Donald Trump, Annotated

Former President Donald J. Trump was penalized $355 million plus interest and banned for three years from serving in any top roles at a New York company, including his own, in a ruling on Friday by Justice Arthur F. Engoron.

Mr. Trump will appeal the financial penalty but will have to either come up with the money or secure a bond within 30 days. The ruling will not render him bankrupt, because most of his wealth is in real estate, which altogether is worth far more than the penalty.

Mr. Trump will also ask an appeals court to halt the restrictions on him and his sons from running the company while it considers the case. In a news conference from his Palm Beach, Fla., home, Mar-a-Lago, on Friday evening, he attacked Ms. James and Justice Engoron, calling them both “corrupt.”

Alina Habba, one of Mr. Trump’s lawyers, described the ruling in her own statement as “a manifest injustice — plain and simple.” She added that “given the grave stakes, we trust that the Appellate Division will overturn this egregious verdict.”

But there might be little Mr. Trump can do to thwart one of the judge’s most consequential punishments: extending for three years the appointment of an independent monitor who is the court’s eyes and ears at the Trump Organization. Justice Engoron also strengthened the monitor’s authority to watch for fraud and second-guess transactions that look suspicious.

Mr. Trump’s lawyers have railed against the monitor, Barbara Jones, saying that her work had already cost the business more than $2.5 million; the decision to extend her oversight of the privately held company could enrage the Trumps, who see her presence as an irritant and an insult.

Ms. James had sought an even harsher penalty, asking for Mr. Trump to be permanently barred from New York’s business world. In the 2022 lawsuit that precipitated the trial, she accused Mr. Trump of inflating his net worth to obtain favorable treatment from banks and other lenders, attacking the foundation of his public persona as a billionaire businessman.

Even though the lenders made money from Mr. Trump, they were the purported victims in the case, with Ms. James arguing that without his fraud, they could have made even more.

The financial penalty reflects those lost profits, with nearly half of the $355 million — $168 million — representing the interest that Mr. Trump saved, and the remaining sum representing his profit on the recent sale of two properties, money that the judge has now clawed back from Mr. Trump and corporate entities he owns.

Before the trial began, Justice Engoron ruled that the former president had used his annual financial statements to defraud the lenders, siding with the attorney general on her case’s central claim. The judge’s Friday ruling ratified almost all of the other accusations Ms. James had leveled against Mr. Trump, finding that the former president had conspired with his top executives to violate several state laws.

The judge’s decision for now grants Ms. James, a Democrat, a career-defining victory. She campaigned for office promising to bring Mr. Trump to justice, and sat calmly in the courtroom as the former president attacked her, calling her a corrupt politician motivated solely by self-interest.

“This long running fraud was intentional, egregious, illegal,” Ms. James said during a Friday evening news conference, adding that “there cannot be different rules for different people in this country, and former presidents are no exception.”

New York Attorney General Speaks on Trump Fraud Case Decision

“today we are holding donald trump accountable,” said the attorney general, letitia james, after a new york judge found donald j. trump’s claims of wealth fraudulent..

No matter how rich, powerful or politically connected you are, everyone must play by the same rules. We have a responsibility to protect the integrity of the marketplace. And for years, Donald Trump engaged in deceptive business practices and tremendous fraud. Donald Trump falsely, knowingly, inflated his net worth by billions of dollars to unjustly enrich himself, his family, and to cheat the system. After 11 weeks of trial, we showed the staggering extent of his fraud and exactly how Donald Trump and the other defendants deceived banks, insurance companies and other financial institutions for their own personal gain. We prove just how much Donald Trump, his family and his company unjustly benefited from his fraud. White-collar financial fraud is not a victimless crime. When the powerful break the law and take more than their fair share, there are fewer resources available for working people, small businesses and families. Today we are holding Donald Trump accountable.

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Her win is Mr. Trump’s second major courtroom loss in two months, following a January jury verdict in a defamation case brought by E. Jean Carroll, a writer whom he was found liable of sexually abusing. The jury penalized him $83.3 million.

Friday’s ruling comes as Manhattan prosecutors are set to try Mr. Trump on criminal charges late next month . He is also contending with 57 more felony counts across three other criminal cases.

But none of his legal troubles seem to have anguished Mr. Trump quite like the fraud case. During the trial, he protested its premise, pleading, “This has been a persecution of somebody that’s done a good job in New York.”

Mr. Trump’s lawyers argued that the fraud did not have a victim in the traditional sense , daring the attorney general to find someone who was harmed. And in a statement on Friday, a Trump Organization spokeswoman noted that the company had “never missed any loan payment or been in default on any loan” and that the lenders “performed extensive due diligence prior to entering into these transactions.”

At trial, Mr. Trump’s lawyers called as witnesses the president’s former bankers, who testified that they had been delighted to have Mr. Trump as a client.

Eric Trump and his brother Donald Trump Jr. also testified, but their efforts to distance themselves from their father’s financial statements fell flat with the judge. Justice Engoron’s decision to bar them from running any New York business for two years — and Mr. Trump for three — will likely strike a nerve with the Trump family.

Before the trial, the fallout from the case seemed to threaten the Trump Organization’s very existence. When Justice Engoron first ruled that Mr. Trump had committed fraud, he ordered the dissolution of much of the former president’s New York empire.

But legal experts had questioned the judge’s ability to do that , and in his ruling on Friday, Justice Engoron pulled back. Instead, the judge said any “restructuring and potential dissolution” would be up to Ms. Jones, the independent monitor.

The judge also granted Ms. Jones new authority as part of an “enhanced monitorship,” and asked her to recommend an independent compliance director who will oversee the company’s financial reporting from within its ranks.

The monitorship and other penalties, including a three-year ban on Mr. Trump and his company seeking loans from banks registered in New York, could hamstring the company as it seeks to compete in the state’s crowded real estate market.

However, nothing will hurt quite as much as the financial penalty. If upheld on appeal, it could erase the cushion of liquidity — cash, stocks and bonds — that Mr. Trump built in his post-presidential life.

Mr. Trump claimed under oath last year that he was sitting on more than $400 million in cash, but between Justice Engoron’s $355 million punishment, the interest Mr. Trump owes and the $83.3 million payout to Ms. Carroll, that might all be gone. If so, Mr. Trump might have to sell one of his properties or another asset to cover the payouts.

The symbolism of the punishments cannot be overlooked, either. Mr. Trump is synonymous with the company he ran for decades, and by severing him from its operations, the judge has written an embarrassing epilogue to the former president’s story of his career as a New York mogul.

For now, Mr. Trump has spun his legal misfortunes into what he sees as political gold. He has used the cases to falsely portray himself as a victim of a Democratic cabal led by President Biden, and he has campaigned at every courthouse he has visited.

In Justice Engoron’s courtroom, Mr. Trump delivered a rally-made rant from the witness stand, marking the climax of a monthslong proceeding that was alternately stultifying and scintillating. The former president attacked one of Ms. James’s lawyers, saying: “You and about every other Democrat, district attorney, A.G. and U.S. attorney were coming after me from 15 different sides. All Democrats, all Trump haters.”

He did not spare Ms. James herself, or the judge, calling the attorney general a “political hack” and Justice Engoron an “extremely hostile judge.”

Mr. Trump later delivered his own closing statement, calling Ms. James’s fraud accusation a “fraud on me” and saying that the attorney general was the one who “should pay me.”

He generated drama even when not in the spotlight, rolling his eyes at the defense table and muttering to his lawyers. He was particularly enraged by the testimony of his former fixer, Michael D. Cohen , who linked Mr. Trump directly to the fraud scheme.

Mr. Trump’s lawyers succeeded in rattling Mr. Cohen, and asked, based on apparent contradictions in his testimony, that Justice Engoron throw out the case. When the judge declined, Mr. Trump abruptly stood up and stormed out of the courtroom.

The judge largely tolerated Mr. Trump’s behavior, but early on, he barred the former president from attacking his staff members, most prominently his law clerk, who sat near the judge throughout the trial so they could confer. Mr. Trump twice violated that order, prompting $15,000 in fines from the judge.

Courtroom theatrics notwithstanding, the evidence presented was often tedious, consisting of years-old emails and spreadsheets. Through that documentary evidence, Ms. James’s lawyers showed that Mr. Trump’s company had ignored appraisals and manipulated numbers to inflate the value of properties such as golf clubs and office buildings, sometimes to absurd heights.

The most blatant exaggeration was the listed size of Mr. Trump’s triplex apartment in Trump Tower on Fifth Avenue. For years, the former president had valued it as if it were 30,000 square feet, when it was actually 10,996.

In his ruling, Justice Engoron blasted Mr. Trump and the other defendants, saying that misstating the apartment’s size was the only error to which they would admit.

Justice Engoron wrote that he was not looking to “judge morality” — only to find facts and apply the law.

“The court intends to protect the integrity of the financial marketplace and, thus, the public as a whole,” he wrote.

Justice Engoron added that Mr. Trump’s refusal to admit error left him with no choice but to conclude that the former president would continue to commit fraud unless he was stopped.

William K. Rashbaum , Claire Fahy and Maggie Haberman contributed reporting.

Jonah E. Bromwich covers criminal justice in New York, with a focus on the Manhattan district attorney's office, state criminal courts in Manhattan and New York City's jails. More about Jonah E. Bromwich

Ben Protess is an investigative reporter at The Times, writing about public corruption. He has been covering the various criminal investigations into former President Trump and his allies. More about Ben Protess

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    Her win is Mr. Trump's second major courtroom loss in two months, following a January jury verdict in a defamation case brought by E. Jean Carroll, a writer whom he was found liable of sexually ...