Get the most accurate GSEB Solutions for Class 6 Mathematics Chapter 01 સંખ્યા પરિચય here. Updated for the 2026-27 academic session, these solutions are based on the latest GSEB textbooks for Class 6 Mathematics. Our expert-created answers for Class 6 Mathematics are available for free download in PDF format.
Detailed Chapter 01 સંખ્યા પરિચય GSEB Solutions for Class 6 Mathematics
For Class 6 students, solving GSEB textbook questions is the most effective way to build a strong conceptual foundation. Our Class 6 Mathematics solutions follow a detailed, step-by-step approach to ensure you understand the logic behind every answer. Practicing these Chapter 01 સંખ્યા પરિચય solutions will improve your exam performance.
Class 6 Mathematics Chapter 01 સંખ્યા પરિચય GSEB Solutions PDF
Gujarat Board Textbook Solutions Class 6 Maths Chapter 1 સંખ્યા પરિચય Ex 1.3
Question 1. Estimate the following sums and differences:
(a) 730 + 998
(b) 796-314
(c) 12,904 + 2888
(d) 28,292 – 21,496
Answer:
(a) 730 + 998
730 → 700 (Approximate value to the nearest hundred)
998 → 1000 (Approximate value to the nearest hundred)
\( \implies \) Approximate sum = 700 + 1000 = 1700
(b) 796-314
796 → 800 (Approximate value to the nearest hundred)
314 → 300 (Approximate value to the nearest hundred)
\( \implies \) Approximate difference = 800 – 300 = 500
(c) 12,904 + 2888
12,904 → 13,000 (Approximate value to the nearest thousand)
2888 → 3000 (Approximate value to the nearest thousand)
\( \implies \) Approximate sum = 13,000 + 3000 = 16,000
(d) 28,292 – 21,496
28,292 → 28,000 (Approximate value to the nearest thousand)
21,496 → 21,000 (Approximate value to the nearest thousand)
\( \implies \) Approximate difference = 28,000 – 21,000 = 7000
In simple words: We round each number to its closest hundred or thousand, then perform the addition or subtraction with these rounded numbers to find an approximate result.
Exam Tip: When estimating, always clearly state to which place value you are rounding each number (e.g., nearest hundred, nearest thousand).
Question 2. Give a rough estimate to the nearest hundred. Also, give a rough estimate to the nearest ten:
(a) 439 + 334 + 4317
(b) 1,08,734 – 47,599
(c) 8325-491
(d) 4,89,348 – 48,365
Answer:
(a) 439 + 334 + 4317
(i) Rough estimate to the nearest hundred:
439 → 400 (Approximate value to the nearest hundred)
334 → 300 (Approximate value to the nearest hundred)
4317 → 4300 (Approximate value to the nearest hundred)
\( \implies \) Approximate sum = 400 + 300 + 4300 = 5000
(ii) Rough estimate to the nearest ten:
439 → 440 (Approximate value to the nearest ten)
334 → 330 (Approximate value to the nearest ten)
4317 → 4320 (Approximate value to the nearest ten)
\( \implies \) Approximate sum = 440 + 330 + 4320 = 5090
(b) 1,08,734 – 47,599
(i) Rough estimate to the nearest hundred:
1,08,734 → 1,08,700 (Approximate value to the nearest hundred)
47,599 → 47,600 (Approximate value to the nearest hundred)
\( \implies \) Approximate difference = 1,08,700 – 47,600 = 61,100
(ii) Rough estimate to the nearest ten:
1,08,734 → 1,08,730 (Approximate value to the nearest ten)
47,599 → 47,600 (Approximate value to the nearest ten)
\( \implies \) Approximate difference = 1,08,730 – 47,600 = 61,130
(c) 8325-491
(i) Rough estimate to the nearest hundred:
8325 → 8300 (Approximate value to the nearest hundred)
491 → 500 (Approximate value to the nearest hundred)
\( \implies \) Approximate difference = 8300 – 500 = 7800
(ii) Rough estimate to the nearest ten:
8325 → 8330 (Approximate value to the nearest ten)
491 → 490 (Approximate value to the nearest ten)
\( \implies \) Approximate difference = 8330 – 490 = 7840
(d) 4,89,348 – 48,365
(i) Rough estimate to the nearest hundred:
4,89,348 → 4,89,300 (Approximate value to the nearest hundred)
48,365 → 48,400 (Approximate value to the nearest hundred)
\( \implies \) Approximate difference = 4,89,300 – 48,400 = 4,40,900
(ii) Rough estimate to the nearest ten:
4,89,348 → 4,89,350 (Approximate value to the nearest ten)
48,365 → 48,370 (Approximate value to the nearest ten)
\( \implies \) Approximate difference = 4,89,350 – 48,370 = 4,40,980
In simple words: To estimate, we first round each number to the nearest hundred, then to the nearest ten. After rounding, we perform the given operation (addition or subtraction) to get the estimated result for each specific place value.
Exam Tip: Remember that rounding to the nearest ten generally gives a more precise estimate than rounding to the nearest hundred.
Question 3. Using the general rule, estimate the product of the following:
(a) 578 × 161
(b) 5281 × 3491
(c) 1291 × 592
(d) 9250 × 29
Answer:
(a) 578 × 161
578 → 600 (Approximate value to the nearest hundred)
161 → 200 (Approximate value to the nearest hundred)
\( \implies \) Approximate product = 600 × 200 = 1,20,000
(b) 5281 × 3491
5281 → 5000 (Approximate value to the nearest thousand)
3491 → 3500 (Approximate value to the nearest hundred)
\( \implies \) Approximate product = 5000 × 3500 = 1,75,00,000
(c) 1291 × 592
1291 → 1300 (Approximate value to the nearest hundred)
592 → 600 (Approximate value to the nearest hundred)
\( \implies \) Approximate product = 1300 × 600 = 7,80,000
(d) 9250 × 29
9250 → 9000 (Approximate value to the nearest thousand)
29 → 30 (Approximate value to the nearest ten)
\( \implies \) Approximate product = 9000 × 30 = 2,70,000
In simple words: To estimate a product, we round each number to its greatest place value or a significant place value before multiplying them. This helps us get a quick, rough answer.
Exam Tip: For multiplication, it is usually best to round each number to its highest place value to get a reasonable estimate, unless specified otherwise.
Free study material for Mathematics
GSEB Solutions Class 6 Mathematics Chapter 01 સંખ્યા પરિચય
Students can now access the GSEB Solutions for Chapter 01 સંખ્યા પરિચય prepared by teachers on our website. These solutions cover all questions in exercise in your Class 6 Mathematics textbook. Each answer is updated based on the current academic session as per the latest GSEB syllabus.
Detailed Explanations for Chapter 01 સંખ્યા પરિચય
Our expert teachers have provided step-by-step explanations for all the difficult questions in the Class 6 Mathematics chapter. Along with the final answers, we have also explained the concept behind it to help you build stronger understanding of each topic. This will be really helpful for Class 6 students who want to understand both theoretical and practical questions. By studying these GSEB Questions and Answers your basic concepts will improve a lot.
Benefits of using Mathematics Class 6 Solved Papers
Using our Mathematics solutions regularly students will be able to improve their logical thinking and problem-solving speed. These Class 6 solutions are a guide for self-study and homework assistance. Along with the chapter-wise solutions, you should also refer to our Revision Notes and Sample Papers for Chapter 01 સંખ્યા પરિચય to get a complete preparation experience.
FAQs
The complete and updated GSEB Class 6 Maths Solutions Chapter 1 સંખ્યા પરિચય Exercise 1.3 is available for free on StudiesToday.com. These solutions for Class 6 Mathematics are as per latest GSEB curriculum.
Yes, our experts have revised the GSEB Class 6 Maths Solutions Chapter 1 સંખ્યા પરિચય Exercise 1.3 as per 2026 exam pattern. All textbook exercises have been solved and have added explanation about how the Mathematics concepts are applied in case-study and assertion-reasoning questions.
Toppers recommend using GSEB language because GSEB marking schemes are strictly based on textbook definitions. Our GSEB Class 6 Maths Solutions Chapter 1 સંખ્યા પરિચય Exercise 1.3 will help students to get full marks in the theory paper.
Yes, we provide bilingual support for Class 6 Mathematics. You can access GSEB Class 6 Maths Solutions Chapter 1 સંખ્યા પરિચય Exercise 1.3 in both English and Hindi medium.
Yes, you can download the entire GSEB Class 6 Maths Solutions Chapter 1 સંખ્યા પરિચય Exercise 1.3 in printable PDF format for offline study on any device.