class 6 math assignment 18 week

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Sixth Grade Math Worksheets

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4 operations.

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Class 6 Assignments Download Pdf

Please refer to Assignments for Class 6 for all subjects given below. We have given below free printable Assignments for Class 6 for easy and free download in PDF. The assignments have been developed based on the current year’s NCERT Books for Class 6. These Assignments for Grade 6 cover all critical topics which can come in your standard 6 class tests and exams. Free printable Assignments for CBSE Class 6 , school and class worksheets, and practice test papers have been developed by our experienced class 6 teachers. You can free download CBSE NCERT printable assignments for Class 6 with solutions and answers. All assignments and test sheets have been prepared by expert faculty as per the latest curriculum in Class 6. Students can simply click on the links below and download all Pdf assignments for class 6 for free. Latest Kendriya Vidyalaya Class 6 assignments and test papers are provided below.

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Class 6 Math Assignment 2022

Class 6 Math 1st, 7th, 13th, 18th week assignment answer (ষষ্ঠ শ্রেণীর অংক / গণিত এসাইনমেন্ট সমাধান, সকল প্রশ্নের উত্তর) have been published. For those who are thinking about mathematics, especially the students of class 6 and their parents, we have solved the whole question of this assignment. So 1st, 7th, 13th, 18th graders will be able to complete assignments by looking at our solutions. Read the entire post carefully to get the solution of the class 6 1st, 7th, 13th, 18th week math assignment. This question contains the answers to all the questions in mathematics.

Class 6 Math 1st Week Assignment 2022

Assignment : লটারির সাহায্যে স্বাভাবিক সংখ্যা বিষয়ক গাণিতিক সমস্যার সমাধান।

Class Six 1st Week Math Assignment Answer

Class 6 Math 18th Week Assignment 2021

Class 6 Math 18th Week Assignment 2021 Answer

Class 6 Math 13th Week Assignment 2021

Class 6 13th week math assignment has been published. This is the third determining task for Class 6 students in math. The third determining task has been selected from the second chapter. The title of the second chapter of the Class 6 students’ mathematics subject is Proportion and Percentage. This means that students in the Class 6 13th week will be able to provide mathematical solutions to proportions and percentages. Class 6 students are asked to solve three mathematical equations in terms of proportions and percentages as part of the 13th week math assignment.

Students are given a number of instructions to solve the Class 6 13th week math assignment. The Class 6 math assignment needs to be solved by following the instructions correctly. So all the students who are solving Class 6 13th week math assignments follow us. Below are some of the topics that must be mentioned in the answer sheet to solve the Maths Assignment for the 13th week of Class 6. Assignments need to be solved with a clear idea of ​​proportions and percentages. Percentages have to be converted to fractions and ratios. You also need to determine the percentage. Lastly, you need to determine the selling price by percentage.

Class 6 Math 13th Week Assignment Answer

7th Week Math Assignment of Class 6

We know that for class six students, assignments seem to be submitted in a total of six weeks. Of these, the 7th week means math in the last week. A total of three math assignments have to be submitted for 7th grade students. Two math assignments have already been completed and this is number three and the last assignment. So in order to get good marks, you have to look at the third assignment with equal importance and answer the assignment correctly.

The 7th grade math assignment for 7th grade was released today, Wednesday, December 2nd. Although the assignment work will start in two to three days. The reason for publishing the assignment a few days ago is so that the school teachers can assign the assignment to their students at the specified time.

7th Soptaher Gonit Assignment Topics

Since, in class six, the students will be passed in the next class only by submitting the assignment instead of the annual examination. So every week assignment questions are asked about certain topics or chapters. The third math assignment in the 7th grade means that all the chapters that were asked in the 7th week are: Students need to answer in the light of these chapters.

Class 6 Math 7th Week Assignment Question & Answer

Class 6 Math Assignment

The 7th grade math assignment question of the 7th grade has been published. According to the notice, the math questions of the 7th week are given below and the answers to all the questions are linked with the questions. Click on the question to get the answer.

চতুর্থ অধ্যায় :

২. বীজগণিতীয় রাশির সদৃশ ও বিসদৃশ পদ

৩. এক বা একাধিক পদ বিশিষ্ট বীজগণিতীয় রাশি বর্ণনা

৪. বীজগণিতীয় রাশির যোগ ও বিয়ােগ

সপ্তম অধ্যায় :

১. রেখাংশ পরিমাপ

২. রেখাংশ অঙ্কন

৩. কোণের চিত্র অঙ্কন

অষ্টম অধ্যায় :

১. তথ্য ও উপাত্ত

২. অবিন্যস্ত উপাত্তের গড়, মধ্যক ও প্রচুরক নির্ণয়

এ্যাসাইনমেন্ট / নির্ধারিত কাজ ২

প্রশ্ন : ০১

5x 2 -2xy+3y 2 , x 2 -3xy, -y 2 +5xy তিনটি বীজগণিতীয় রাশি হলে –

ক) প্রথম রাশিটির পদ সংখ্যা কয়টি এবং কী কী?

খ) রাশি তিনটির যােগফল নির্ণয় কর।

গ) x=3, y=2 হলে, ১ম রাশি থেকে ৩য় রাশির বিয়ােগফলের মান নির্ণয় কর।

প্রশ্ন : ০২

ক) ∠ABC কে চাঁদার সাহায্যে অংকন কর।

খ) ∠ABC কে সমদ্বিখন্ডিত কর (রুলার ও কম্পাসের সাহায্যে)।

গ) অংকনের চিহ্ন ও বিবরণ দাও।

প্রশ্ন : ০৩

একজন শিক্ষার্থী ৪০ থেকে ৬০ পর্যন্ত সংখ্যাগুলাের মধ্যে নিমের সংখ্যাগুলাে লিখল। ৫০, ৪৫, ৪৮, ৪৯. ৬০, ৫৮, ৫৭, ৪৫, ৪৭, ৪৫, ৪৩, ৪২, ৪৭.

ক) উপাত্তগুলােকে বিন্যস্ত করা।

খ) উপাত্তগুলাের গড় নির্ণয় কর।

গ) উপাত্তগুলাের মধ্যক ও প্রচুরক নির্ণয় কর।

মূল্যায়ন নির্দেশক

১. বীজগণিতীয় রাশির যােগ বিয়োগ করতে পারা।

২, রুলার ও কম্পাসের সাহায্যে কোণের সমদ্বিখন্ডক এঁকে বিবরণ দিতে পারা।

৩. অবিন্যস্ত উপাত্ত বিন্যস্ত করে গড়, মধ্যক ও প্রচুরক নির্ণয় করতে পারা।

Class 6 M ath Assignment Solution

We answered all the questions in the 7th grade math assignment of 7th grade. Here all the answers are given in the form of pictures. If you benefited from the math solution, you must share the post with your friends and name our website.

Instructions For Answering The 7th Week Math Assignment Of 7th Grade

Here are some guidelines for answering the 7th week math question. The question should be well read and understand how to answer any question. Below are the 7th week class six assignment guidelines:

1. To be able to subtract of an algebraic sum.

2, with the help of ruler and compass to be able to draw the isosceles of the angle to give details.

3. To be able to determine average, mean and abundance by arranging unstructured data.

ষষ্ঠ শ্রেণীর সপ্তম সপ্তাহের গণিত অ্যাসাইনমেন্ট প্রকাশিত হয়েছে। গণিত নিয়ে যাদের চিন্তা, বিশেষ করে ক্লাস ৬ এর স্টুডেন্ট ও তাদের অভিভাবক, তাদের জন্য আমরা এই এসাইনমেন্ট এর সম্পূর্ণ প্রশ্নের সমাধান করেছি। সুতরাং ষষ্ঠ শ্রেণীর শিক্ষার্থীরা আমাদের সমাধান দেখে এসাইনমেন্ট সম্পন্ন করতে পারবেন। ষষ্ঠ শ্রেণীর শেষ গণিত এসাইনমেন্টের সমাধান পেতে সম্পূর্ণ পোস্টটি ভালভাবে পড়ুন। এই প্রশ্নের মধ্যেই গণিতের সকল প্রশ্নের উত্তর রয়েছে।

ষষ্ঠ শ্রেণীর সপ্তম সপ্তাহের গণিত অ্যাসাইনমেন্ট

আমরা জানি ক্লাস সিক্সের ছাত্র ছাত্রীদের জন্য মোট ছয় সপ্তাহে এসাইনমেন্ট জমা দেওয়া লাগছে। এরমধ্যে সপ্তম সপ্তাহে মানে শেষ সপ্তাহে গণিত রয়েছে। ষষ্ঠ শ্রেণীর ছাত্রছাত্রীদের জন্য মোট তিনটি গণিত অ্যাসাইনমেন্ট জমা দিতে হবে। ইতিপূর্বেই দুইটি গণিত অ্যাসাইনমেন্ট শেষ হয়ে গেছে এবং এটি তিন নম্বর এবং সর্বশেষ অ্যাসাইনমেন্ট। সুতরাং ভালো মার্ক পেতে হলে তৃতীয় এসাইনমেন্টকেও সমান গুরুত্ব সহকারে দেখতে হবে এবং সঠিকভাবে এসাইনমেন্টের উত্তর সম্পন্ন করতে হবে।

ষষ্ঠ শ্রেণীর সপ্তম গণিত অ্যাসাইনমেন্টটি আজ বুধবার 2 ডিসেম্বর তারিখে প্রকাশ পেয়েছে। যদিও অ্যাসাইনমেন্ট এর কাজগুলো শুরু হবে দুই তিন দিন পরে। কিছুদিন পূর্বে এসাইনমেন্ট প্রকাশের কারণ হলো স্কুল শিক্ষকরা যেন তাদের ছাত্র-ছাত্রীদের নির্দিষ্ট সময়ে অ্যাসাইনমেন্ট অর্পণ করতে পারে।

ক্লাস সিক্সের সপ্তম সপ্তাহের গণিত অ্যাসাইনমেন্ট এর আলোচ্য বিষয়

যেহেতু, ক্লাস সিক্সে এবার বাষিক পরীক্ষা না হয়ে অ্যাসাইনমেন্ট জমা নেওয়ার মাধ্যমেই শিক্ষার্থীদের পরবর্তী শ্রেণীতে উত্তীর্ণ করা হবে। তাই প্রতি সপ্তাহেই কিছু নির্দিষ্ট বিষয় বা অধ্যায় নিয়ে এসাইনমেন্টের প্রশ্ন করা হয়। ষষ্ঠ শ্রেণীর তৃতীয় গণিত অ্যাসাইনমেন্টে মানে সপ্তম সপ্তাহে যে সকল অধ্যায় থেকে প্রশ্ন করা হয়েছে সেগুলো হলোঃ। শিক্ষার্থীদের এইসকল অধ্যায়ের আলোকে উত্তর করতে হবে।

ষষ্ঠ শ্রেণীর সপ্তম সপ্তাহের গণিত অ্যাসাইনমেন্টের প্রশ্ন সমূহ

ষষ্ঠ শ্রেণীর সপ্তম সপ্তাহের গণিত অ্যাসাইনমেন্ট এর প্রশ্ন প্রকাশিত হয়েছে। নোটিশ অনুযায়ী সপ্তম সপ্তাহের গণিত প্রশ্ন গুলো নিচে তুলে ধরা হলো এবং সকল প্রশ্নের উত্তর প্রশ্নের সাথে লিংক করে দেওয়া হল। প্রশ্নের উপর ক্লিক করলে উত্তর পেয়ে যাবেন।

সপ্তম সপ্তাহের ষষ্ঠ শ্রেণীর গণিত অ্যাসাইনমেন্ট সমাধান

আমরা ষষ্ঠ শ্রেণীর সপ্তম সপ্তাহের গণিত অ্যাসাইনমেন্ট এর সকল প্রশ্নের উত্তর করেছি। এখানে সকল উত্তর ছবি আকারে দেওয়া হল। গণিত সমাধান দ্বারা আপনি যদি উপকৃত হন তাহলে পোস্টটি অবশ্যই আপনাদের বন্ধুদের নিকট শেয়ার করবেন এবং আমাদের ওয়েবসাইটের নাম বলবেন।

ষষ্ঠ শ্রেণীর সপ্তম সপ্তাহের গণিত অ্যাসাইনমেন্ট এর উত্তর করার নির্দেশাবলী

সপ্তম সপ্তাহের অংক প্রশ্নের উত্তর করার জন্য কিছু নির্দেশনা দেওয়া রয়েছে। প্রশ্নটি ভালভাবে পড়তে হবে এবং বুঝতে হবে যে কোন প্রশ্নের উত্তর কিভাবে করতে হবে। নিচে সপ্তম সপ্তাহের ক্লাস সিক্সের এসাইনমেন্ট নির্দেশনাগুলো তুলে ধরা হলো:

4 thoughts on “Class 6 Math Assignment 2022”

Tnx bro… Keep going to teach or give us knowladge

Thanks a lot man. I like these answers.

7th week math assignment tar গ number vul ache. oita 11.22 hobena.

here is a mistake i notice , 18th week assignment’s math in খ answer is wrong the mistake is (20+x)=250/2 (20+x)=105 here is the mistake , the correct answer is 125 please fix up this problem

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Maths Assignment 18 Worksheet for Class 6 PDF with Answers

These Maths Assignment 18 Worksheet for Class 6 PDF can be helpful for both teachers and students. Teachers can track their student’s performance in the chapter Maths Assignment 18. Students can easily identify their strong points and weak points by solving questions from the worksheet. Accordingly, students can work on both weak points and strong points. 

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Maths Assignment 18 Worksheet for Class 6 with Solutions

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Maths Assignment 18 Worksheet PDF

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Regular solving questions from the #Maths Assignment 18 Worksheet for Class 6 Worksheet PDF can help students to build a strong foundation for the chapter Maths Assignment 18. Strong foundation of the chapter Maths Assignment 18 can help students to understand further chapters. 

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Class 6 Mathematics 18th Week Assignment 2021 Answer PDF Download

Class 6 Mathematics 18th Week Assignment 2021 Answer PDF Download Has been published today on My website. Although classes have started in the schools, assignment activities for students of class VI to IX are going on. The 17th week assignment of these students has been published by the Department of Secondary and Higher Education. The assignment was published on the department’s website on Tuesday (September 26).

The Department of Secondary and Higher Education has asked all schools to follow the hygienic rules and regulations in distributing and submitting assignments to all students.

Table of Contents

৬ষ্ট/ষষ্ঠ শ্রেণির ১৮তম সপ্তাহের এসাইনমেন্ট ২০২১ উত্তর/সমাধান গণিত (এসাইনমেন্ট-৪) | ২০২১ সালের ৬ষ্ট/ষষ্ঠ শ্রেণির (১৮তম) সপ্তাহের গণিত এসাইনমেন্ট সমাধান/উত্তর

All the primary, secondary and higher secondary level educational institutions of the country have been opened since last 12 September. In this situation, the assignment of the 18th week of these students was published.

For the readers of our website, the assignments of 6th to 9th grade students for the 16th week are highlighted.

Click to view assignment: Class 9,8,7,6 students 18th week Assignment Answer 2021 All Subjects

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• NCERT Solutions
• NCERT Class 6
• NCERT 6 Maths
• Chapter 12: Ratio Proportion

NCERT Solutions for Class 6 Maths Chapter 12: Ratio and Proportion

NCERT Solutions for Class 6 Maths Chapter 12 Ratio and Proportion comprises exercise-wise solved questions of Ratio and Proportion concept. These study materials, prepared by the subject-matter experts at BYJU’S are available in PDF for greater assistance. Students can easily download the PDF file and study offline as well. These NCERT Solutions for Class 6   Maths are prepared as per the CBSE Board syllabus.

• Chapter 1 Knowing Our Numbers
• Chapter 2 Whole Numbers
• Chapter 3 Playing with Numbers
• Chapter 4 Basic Geometrical Ideas
• Chapter 5 Understanding Elementary Shapes
• Chapter 6 Integers
• Chapter 7 Fractions
• Chapter 8 Decimals
• Chapter 9 Data Handling
• Chapter 10 Mensuration
• Chapter 11 Algebra
• Chapter 12 Ratio and Proportion
• Chapter 13 Introduction to Symmetry
• Chapter 14 Practical Geometry
• Exercise 12.1 Chapter 12 Ratio and Proportion
• Exercise 12.2 Chapter 12 Ratio and Proportion
• Exercise 12.3 Chapter 12 Ratio and Proportion

NCERT Solutions for Class 6 Maths Chapter 12 Ratio and Proportion

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Further, students can have a look at other NCERT Solutions , notes and preparation tips available on the website to prepare well for the exams. To get an idea of the types of questions asked from Ratio and Proportion, students are advised to solve sample papers and previous years’ question papers as well.

Exercises of NCERT Solutions Class 6 Maths Chapter 12 – Ratio and Proportion

Exercise 12.1 Solutions

Exercise 12.2 Solutions

Exercise 12.3 Solutions

Access NCERT Solutions for Class 6 Chapter 12: Ratio and Proportion

Exercise 12.1 Page No. 251

1. There are 20 girls and 15 boys in a class.

(a) What is the ratio of the number of girls to the number of boys?

(b) What is the ratio of the number of girls to the total number of students in the class?

Number of girls = 20 girls

Number of boys = 15 boys

Total number of students = 20 + 15

(a) Ratio of the number of girls to the number of boys = 20 / 15 = 4 / 3

(b) Ratio of the number of girls to the total number of students = 20 / 35 = 4 / 7

2. Out of 30 students in a class, 6 like football, 12 like cricket and the remaining like tennis. Find the ratio of

(a) The number of students liking football to the number of students liking tennis.

(b) The number of students liking cricket to the total number of students.

Number of students who like football = 6

Number of students who like cricket = 12

Number of students who like tennis = 30 – 6 – 12

(a) Ratio of the number of students liking football to the number of students liking tennis

= 6 / 12 = 1 / 2

(b) Ratio of the number of students liking cricket to the total number of students

3. See the figure and find the ratio of

(a) Number of triangles to the number of circles inside the rectangle.

(b) Number of squares to all the figures inside the rectangle.

(c) Number of circles to all the figures inside the rectangle.

Given in the figure

Number of triangles = 3

Number of circles = 2

Number of squares = 2

Total number of figures = 7

(a) Ratio of the number of triangles to the number of circles inside the rectangle

(b) Ratio of the number of squares to all the figures inside the rectangle

(c) Ratio of the number of circles to all the figures inside the rectangle

4. The distance travelled by Hamid and Akhtar in an hour is 9 km and 12 km, respectively. Find the ratio of the speed of Hamid to the speed of Akhtar.

We know that the speed of a certain object is the distance travelled by that object in an hour

Distance travelled by Hamid in one hour = 9 km

Distance travelled by Akhtar in one hour = 12 km

Speed of Hamid = 9 km/hr

Speed of Akhtar = 12 km/hr

The ratio of the speed of Hamid to the speed of Akhtar = 9 / 12 = 3 / 4

5. Fill in the blanks:

15 / 18 = ☐ / 6 = 10 / ☐ = ☐ / 30 [Are these equivalent ratios?]

15 / 18 = (5 × 3) / (6 × 3)

5 / 6 = (5 × 2) / (6 × 2)

5 / 6 = (5 × 5) / (6 × 5)

Hence, 5, 12 and 25 are the numbers which come in the blanks, respectively.

Yes, all are equivalent ratios.

6. Find the ratio of the following:

(a) 81 to 108

(b) 98 to 63

(c) 33 km to 121 km

(d) 30 minutes to 45 minutes

(a) 81 / 108 = (3 × 3 × 3 × 3) / (2 × 2 × 3 × 3 × 3 )

(b) 98 / 63 = (14 × 7) / (9 × 7)

(c) 33 / 121 = (3 × 11) / (11 × 11)

(d) 30 / 45 = (2 × 3 × 5) / (3 × 3 × 5)

7. Find the ratio of the following:

(a) 30 minutes to 1.5 hours

(b) 40 cm to 1.5 m

(c) 55 paise to ₹ 1

(d) 500 ml to 2 litres

30 min = 30 / 60

= 0.5 hours

Required ratio = (0.5 × 1 ) / (0.5 × 3)

1.5 m = 150 cm

Required ratio = 40 / 150

₹ 1 = 100 paise

Required ratio = 55 / 100 = (11 × 5) / (20 × 5)

1 litre = 1000 ml

2 litre = 2000 ml

Required ratio = 500 / 2000 = 5 / 20 = 5 / (5 × 4)

8. In a year, Seema earns ₹ 1,50,000 and saves ₹ 50,000. Find the ratio of

(a) Money that Seema earns to the money she saves

(b) Money that she saves to the money she spends.

Money earned by Seema = ₹ 150000

Money saved by her = ₹ 50000

Money spent by her = ₹ 150000 – ₹ 50000 = ₹ 100000

(a) Ratio of money earned to money saved = 150000 / 50000 = 15 / 5

(b) Ratio of money saved to money spent = 50000 / 100000 = 5 / 10

9. There are 102 teachers in a school of 3300 students. Find the ratio of the number of teachers to the number of students.

Number of teachers in the school = 102

Number of students in the school = 3300

The ratio of the number of teachers to the number of students = 102 / 3300

= (2 × 3 × 17) / (2 × 3 × 550)

10. In a college, out of 4320 students, 2300 are girls. Find the ratio of

(a) Number of girls to the total number of students.

(b) Number of boys to the number of girls.

(c) Number of boys to the total number of students.

Total number of students = 4320

Number of girls = 2300

Number of boys = 4320 – 2300

(a) Ratio of the number of girls to the total number of students = 2300 / 4320

= (2 × 2 × 5 × 115) / (2 × 2 × 5 × 216)

= 115 / 216

(b) Ratio of the number of boys to the number of girls = 2020 / 2300

= (2 × 2 × 5 × 101) / (2 × 2 × 5 × 115)

= 101 / 115

(c) Ratio of the number of boys to the total number of students = 2020 / 4320

= (2 × 2 × 5 × 101) / (2 × 2 × 5 × 216)

= 101 / 216

11. Out of 1800 students in a school, 750 opted for basketball, 800 opted for cricket, and the remaining opted for table tennis. If a student can opt for only one game, find the ratio of

(a) The number of students who opted for basketball to the number of students who opted for table tennis.

(b) The number of students who opted for cricket to the number of students opting for basketball.

(c) The number of students who opted for basketball to the total number of students.

Number of students in the school = 1800

The number of students who opted for basketball = 750

The number of students who opted for cricket = 800

The number of students who opted for table tennis = 1800 – (750 + 800) = 1800 – 1550 = 250

( a) Ratio of the number of students who opted for basketball to the number of students who opted for table tennis = 750 / 250 = 3 / 1

(b) Ratio of the number of students who opted for cricket to the number of students who opted for basketball

= 800 / 750 = 16 / 15

(c) Ratio of the number of students who opted for basketball to the total number of students

= 750 / 1800 = 25 / 60 = 5 / 12

12. Cost of a dozen pens is ₹ 180, and the cost of 8 ball pens is ₹ 56. Find the ratio of the cost of a pen to the cost of a ball pen.

Cost of a dozen pens = ₹ 180

Cost of 1 pen = 180 / 12

Cost of 8 ball pens = ₹ 56

Cost of 1 ball pen = 56 / 8

Hence, the required ratio is 15 / 7.

13. Consider the statement: The ratio of breadth and length of a hall is 2:5. Complete the following table that shows some possible breadths and lengths of the hall.

(i) Length = 50 m

Breadth / 50 = 2 / 5

By cross multiplication

5× breadth = 50 × 2

Breadth = (50 × 2) / 5

(ii) Breadth = 40 m

40 / Length = 2 / 5

2 × Length = 40 × 5

Length = (40 × 5) / 2

Length = 200 / 2

Length = 100 m

14. Divide 20 pens between Sheela and Sangeeta in a ratio of 3:2.

Terms of 3:2 = 3 and 2

The sum of these terms = 3 + 2

Now, Sheela will get 3 / 5 of the total pens, and Sangeeta will get 2 / 5 of the total pens

Number of pens with Sheela = 3 / 5 × 20

Number of pens with Sangeeta = 2 / 5 × 20

15. Mother wants to divide ₹ 36 between her daughters Shreya and Bhoomika in the ratio of their ages. If the age of Shreya is 15 years and the age of Bhoomika is 12 years, find how much Shreya and Bhoomika will get.

Ratio of ages = 15 / 12

Hence, the mother wants to divide ₹ 36 in the ratio of 5:4

Terms of 5:4 are 5 and 4

The sum of these terms = 5 + 4

Here, Shreya will get 5 / 9 of the total money, and Sangeeta will get 4 / 9 of the total money

The amount Shreya get = 5 / 9 × 36

The amount Sangeeta get = 4 / 9 × 36

Therefore, Shreya will get ₹ 20, and Sangeeta will get ₹ 16.

16. Present age of the father is 42 years, and that of his son is 14 years. Find the ratio of

(a) Present age of the father to the present age of the son

(b) Age of the father to the age of the son, when the son was 12 years old

(c) Age of father after 10 years to the age of son after 10 years

(d) Age of father to the age of son when father was 30 years old

(a) Present age of father = 42 years

Present age of son = 14 years

Required ratio 42 / 14

(b) 2 years ago, the son was 12 years old. So, 2 years ago, the age of the father was

= 42 – 2 = 40 years

Required ratio = 40 / 12 = (4 × 10) / (4 × 3) = 10 / 3

(c) After ten years, the age of the father = 42 + 10 = 52 years

After 10 years, the age of the son = 14 + 10 = 24 years

Required ratio = 52 / 24 = (4 × 13) / (4 × 6)

(d) 12 years ago, the age of the father was 30

At that time, the age of the son = 14 – 12

Required ratio = 30 / 2 = (2 × 15) / 2

Exercise 12.2 Page No. 255

1. Determine if the following are in proportion.

(a) 15, 45, 40, 120

(b) 33, 121, 9, 96

(c) 24, 28, 36, 48

(d) 32, 48, 70, 210

(e) 4, 6, 8, 12

(f) 33, 44, 75, 100

15 / 45 = 1 / 3

40 / 120 = 1 / 3

Hence, 15:45 = 40:120

∴ They are in proportion.

33 / 121 = 3 / 11

9 / 96 = 3 / 32

Hence 33:121 ≠ 9: 96

∴ They are not in proportion.

24 / 28 = 6 / 7

36 / 48 = 3 / 4

Hence, 24:28 ≠ 36:48

32 / 48 = 2 / 3

70 / 210 = 1 / 3

Hence, 32:48 ≠ 70:210

4 / 6 = 2 / 3

8 / 12 = 2 / 3

Hence 4: 6 = 8: 12

∴ These are in a proportion

33/ 44 = 3/ 4

75 / 100 = 3 / 4

Hence, 33:44 = 75:100

∴ These are in proportion.

2. Write True (T) or False ( F ) against each of the following statements :

(a) 16:24 :: 20:30

(b) 21:6 :: 35:10

(c) 12:18 :: 28:12

(d) 8:9 :: 24:27

(e) 5.2:3.9 :: 3:4

(f) 0.9:0.36 :: 10:4

16 / 24 = 2 / 3

20 / 30 = 2 / 3

Hence, 16:24 = 20:30

Therefore, True.

21 / 6 = 7 / 2

35 / 10 = 7 / 2

Hence, 21:6 = 35:10

12 / 18 = 2 / 3

28 / 12 = 7 / 3

Hence, 12:18 ≠ 28:12

Therefore, False,

(d) 8:9:: 24:27

We know that = 24 / 27 = (3 × 8) / (3 × 9)

Hence, 8:9 = 24:27

(e) 5.2:3.9 :: 3: 4

As 5.2 / 3.9 = 4 / 3

Hence, 5.2: 3.9 ≠ 3: 4

0.9 / 0.36 = 90 / 36

Hence, 0.9: 0.36 = 10: 4

Therefore, True,

3. Are the following statements true?

(a) 40 persons: 200 persons = ₹ 15 : ₹ 75

(b) 7.5 litres: 15 litres = 5 kg : 10 kg

(c) 99 kg: 45 kg = ₹ 44 : ₹ 20

(d) 32 m: 64 m = 6 sec : 12 sec

(e) 45 km : 60 km = 12 hours : 15 hours

(a) 40 persons : 200 persons = ₹ 15 : ₹ 75

40 / 200 = 1 / 5

15 / 75 = 1 / 5

Hence, True.

(b) 7.5 litres : 15 litres = 5 kg : 10 kg

7.5 / 15 = 1 / 2

5 / 10 = 1 / 2

(c) 99 kg : 45 kg = ₹ 44 : ₹ 20

99 / 45 = 11 / 5

44 / 20 = 11 / 5

(d) 32 m : 64 m = 6 sec : 12 sec

32 / 64 = 1 / 2

6 / 12 = 1 / 2

45 / 60 = 3 / 4

12 / 15 = 4 / 5

Hence, False.

4. Determine if the following ratios form a proportion. Also, write the middle terms and extreme terms where the ratios form a proportion.

(a) 25 cm : 1 m and ₹ 40 : ₹ 160

(b) 39 litres : 65 litres and 6 bottles : 10 bottles

(c) 2 kg : 80 kg and 25 g : 625 g

(d) 200 mL : 2.5 litre and ₹ 4 : ₹ 50

25 cm = 25 / 100 m

0.25 / 1 = 1 / 4

40 / 160 = 1 / 4

Yes, these are in proportion.

Middle terms are 1 m, ₹ 40, and extreme terms are 25 cm, ₹ 160.

39 / 65 = 3 /5

6 / 10 = 3 / 5

Middle terms are 65 litres, 6 bottles, and extreme terms are 39 litres, 10 bottles.

2 / 80 = 1 / 40

25 / 625 = 1 / 25

No, these are not in proportion.

2.5 litre = 2500 ml

200 / 2500 = 2 / 25

4 / 50 = 2 / 25

Middle terms are 2.5 litres, ₹ 4 and extreme terms are 200 ml, ₹ 50.

Exercise 12.3 Page No. 259

1. If the cost of 7 m of cloth is ₹ 1470, find the cost of 5 m of cloth.

Cost of 7 m cloth = ₹ 1470

Cost of 1 m cloth = 1470 / 7

So, cost of 5 cloth = 210 × 5 = 1050

∴ The cost of 5 m cloth is ₹ 1050.

2. Ekta earns ₹ 3000 in 10 days. How much will she earn in 30 days?

Money earned by Ekta in 10 days = ₹ 3000

Money earned by her in one day = 3000 / 10

So, money earned by her in 30 days = 300 × 30

3. If it has rained 276 mm in the last 3 days, how many cm of rain will fall in one full week (7 days)? Assume that the rain continues to fall at the same rate.

The measure of rain in 3 days = 276 mm

The measure of rain in one day = 276 / 3

So, the measure of rain in one week, i.e. 7 days = 92 × 7

4. Cost of 5 kg of wheat is ₹ 91.50.

(a) What will be the cost of 8 kg of wheat?

(b) What quantity of wheat can be purchased for ₹ 183?

(a) Cost of 5 kg wheat = ₹ 91.50.

Cost of 1 kg wheat = 91.50 / 5

So, the cost of 8 kg wheat = 18.3 × 8

(b) Wheat purchased for ₹ 91.50 = 5 kg

Wheat purchased for ₹ 1 = 5 / 91.50 kg

So, wheat purchased for ₹ 183 = (5 / 91.50) × 183

5. The temperature dropped 15 0 C in the last 30 days. If the rate of temperature drop remains the same, how many degrees will the temperature drop in the next ten days?

Temperature drop in 30 days = 15 0 C

Temperature drop in 1 day = 15 / 30

= (1 / 2) 0 C

So, the temperature drop in next 10 days = (1 / 2) × 10

∴ The temperature drop in the next 10 days will be 5 0 C

6. Shaina pays ₹ 15000 as rent for 3 months. How much does she have to pay for a whole year if the rent per month remains the same?

Rent paid by Shaina in 3 months = ₹ 15000

Rent for 1 month = 15000 / 3

So, rent for 12 months, i.e. 1 year = 5000 × 12

∴ Rent paid by Shaina in 1 year is ₹ 60,000

7. Cost of 4 dozen bananas is ₹ 180. How many bananas can be purchased for ₹ 90?

Number of bananas bought  for ₹ 180 = 4 dozens

= 48 bananas

Number of bananas bought for ₹ 1 = 48 / 180

So, number of bananas bought for ₹ 90 = (48 / 180) × 90

= 24 bananas

∴ 24 bananas can be purchased for ₹ 90

8. The weight of 72 books is 9 kg. What is the weight of 40 such books?

Weight of 72 books = 9 kg

Weight of 1 book = 9 / 72

So, weight of 40 books = (1 / 8) × 40

∴ The weight of 40 books is 5 kg.

9. A truck requires 108 litres of diesel to cover a distance of 594 km. How much diesel will be required by the truck to cover a distance of 1650 km?

Diesel required for 594 km = 108 litres

Diesel required for 1 km = 108 / 594

= 2 / 11 litre

So, diesel required for 1650 km = (2 / 11) × 1650

= 300 litres

∴ The diesel required by the truck to cover a distance of 1650 km is 300 litres.

10. Raju purchases 10 pens for ₹ 150, and Manish buys 7 pens for ₹ 84. Can you say who got the pens cheaper?

Pens purchased by Raju for ₹ 150 = 10 pens

Cost of 1 pen = 150 / 10

Pens purchased by Manish for ₹ 84 = 7 pens

Cost of 1 pen = 84 / 7

∴ Pens purchased by Manish are cheaper than those of Raju.

11. Anish made 42 runs in 6 overs, and Anup made 63 runs in 7 overs. Who made more runs per over?

Runs made by Anish in 6 overs = 42

Runs made by Anish in 1 over = 42 / 6

Runs made by Anup in 7 overs = 63

Runs made by Anup in 1 over = 63 / 7

∴ Anup scored more runs than Anish.

Frequently Asked Questions on NCERT Solutions for Class 6 Maths Chapter 12

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