NCERT Solutions for Class 6 Mathematics: Chapter 01 સંખ્યા પરિચય
Access comprehensive textbook solutions for Chapter 01 સંખ્યા પરિચય using the official curriculum guides for Class 6 Mathematics. Designed to align with the 2026-27 GSEB standards, these detailed answers help students reinforce core academic concepts.
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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.
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Free GSEB Textbook Explanations: Class 6 Mathematics Chapter 01 સંખ્યા પરિચય
Chapter Exercise Answers for Class 6 Mathematics
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Detailed Answer Guides for Chapter 01 સંખ્યા પરિચય
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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.
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