NCERT Solutions For Class 6 Maths Ex 1.3

NCERT Solutions For Class 6 Maths Ex 1.3 PDF Free Download

Estimation is very important to know the approximate values in various calculations. To make students familiar with these concepts, NCERT solutions are designed by faculty at BYJU’S with respect to the current NCERT syllabus. These solutions contain step by step explanations to help students understand the concepts that are covered under each exercise. To obtain a brief idea about the methods followed in solving problems, students can access NCERT Solutions Class 6 Maths Chapter 1 Knowing Our Numbers Exercise 1.3 PDF, which are available here.

NCERT Solutions for Class 6 Chapter 1: Knowing Our Numbers Exercise 1.3 Download PDF

 

ncert solutions class 6 maths chapter 1 ex 1.3
ncert solutions class 6 maths chapter 1 ex 1.3
ncert solutions class 6 maths chapter 1 ex 1.3
ncert solutions class 6 maths chapter 1 ex 1.3

 

Access NCERT Solutions for Class 6 Chapter 1: Knowing Our Numbers Exercise 1.3

1. Estimate each of the following using general rule:

(a) 730 + 998 (b) 796 – 314 (c) 12904 + 2888 (d) 28292 – 21496

Make ten more such examples of addition, subtraction and estimation of their outcome.

Solutions:

(a) 730 + 998

Round off to hundreds

730 is rounds off to 700

998 is rounds off to 1000

Hence, 730 + 998 = 700 + 1000 = 1700

(b) 796 – 314

Round off to hundreds

796 is rounds off to 800

314 is rounds off to 300

Hence, 796 – 314 = 800 – 300 = 500

(c) 12904 + 2888

Round off to thousands

12904 is rounds off to 13000

2888 is rounds off to 3000

Hence, 12904 + 2888 = 13000 + 3000 = 16000

(d) 28292 – 21496

Round off to thousands

28292 is round off to 28000

21496 is round off to 21000

Hence, 28292 – 21496 = 28000 – 21000 = 7000

Ten more such examples are

(i) 330 + 280 = 300 + 300 = 600

(ii) 3937 + 5990 = 4000 + 6000 = 10000

(iii) 6392 – 3772 = 6000 – 4000 = 2000

(iv) 5440 – 2972 = 5000 – 3000 = 2000

(v) 2175 + 1206 = 2000 + 1000 = 3000

(vi) 1110 – 1292 = 1000 – 1000 = 0

(vii) 910 + 575 = 900 + 600 = 1500

(viii) 6400 – 4900 = 6000 – 5000 = 1000

(ix) 3731 + 1300 = 4000 + 1000 = 5000

(x) 6485 – 4319 = 6000 – 4000 = 2000

2. Give a rough estimate (by rounding off to nearest hundreds) and also a closer estimate (by rounding off to nearest tens):

(a) 439 + 334 + 4317 (b) 108734 – 47599 (c) 8325 – 491 (d) 489348 – 48365

Make four more such examples.

Solutions:

(a) 439 + 334 + 4317

Rounding off to nearest hundreds

439 + 334 + 4317 = 400 + 300 + 4300

= 5000

Rounding off to nearest tens

439 + 334 + 4317 = 440 + 330 + 4320

= 5090

(b) 108734 – 47599

Rounding off to nearest hundreds

108734 – 47599 = 108700 – 47600

= 61100

Rounding off to nearest tens

108734 – 47599 = 108730 – 47600

= 61130

(c) 8325 – 491

Rounding off to nearest hundreds

8325 – 491 = 8300 – 500

= 7800

Rounding off to nearest tens

8325 – 491 = 8330 – 490

= 7840

(d) 489348 – 48365

Rounding off to nearest hundreds

489348 – 48365 = 489300 – 48400

= 440900

Rounding off to nearest tens

489348 – 48365 = 489350 – 48370

= 440980

Four more examples are as follows

(i) 4853 + 662

Rounding off to nearest hundreds

4853 + 662 = 4800 + 700

= 5500

Rounding off to nearest tens

4853 + 662 = 4850 + 660

= 5510

(ii) 775 – 390

Rounding off to nearest hundreds

775 – 390 = 800 – 400

= 400

Rounding off to nearest tens

775 – 390 = 780 – 400

380

(iii) 6375 – 2875

Rounding off to nearest hundreds

6375 – 2875 = 6400 – 2900

= 3500

Rounding off to nearest tens

6375 – 2875 = 6380 – 2880

3500

(iv) 8246 – 6312

Rounding off to nearest hundreds

8246 – 6312 = 8200 – 6300

1900

Rounding off to nearest tens

8246 – 6312 = 8240 – 6310

= 1930

3. Estimate the following products using general rule:

(a) 578 × 161

(b) 5281 × 3491

(c) 1291 × 592

(d) 9250 × 29

Make four more such examples.

Solutions:

(a) 578 × 161

Rounding off by general rule

598 and 161 rounded off to 600 and 200 respectively

600

× 200

____________

120000

_____________

(b) 5281 × 3491

Rounding off by general rule

5281 and 3491 rounded off to 5000 and 3500 respectively

5000

× 3500

_________

17500000

_________

(c) 1291 × 592

Rounding off by general rule

1291 and 592 rounded off to 1300 and 600 respectively

1300

× 600

_____________

780000

______________

(d) 9250 × 29

Rounding off by general rule

9250 and 29 rounded off to 9000 and 30 respectively

9000

× 30

_____________

270000

______________


Access other exercise solutions of Class 6 Maths Chapter 1: Knowing Our Numbers

Exercise 1.1 Solutions

Exercise 1.2 Solutions

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