Solutions For All Chapters Maths Class 6
NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Exercise 1.3
Ex 1.3 Class 6 Maths Question 1.
Estimate each of the following using general rule:
(a) 730 + 998
(b) 796 – 314
(c) 12,904 + 2,888
(d) 28,292 – 21,496
Make ten more such examples of addition, subtraction and estimation of their outcome.
Solution:
(a) 730 + 998
Rounding off 730 nearest to hundreds = 700
Rounding off 998 nearest to hundreds = 1,000
∴ 730 + 998 = 700 + 1000 = 1700
(b) 796 – 314
Rounding off 796 nearest to hundreds = 800
Rounding off 314 nearest to hundreds = 300
∴ 796 – 314 = 800 – 300 = 500
(c) 12,904 + 2,888
Rounding off 12,904 nearest to thousands = 13000
Rounding off 2888 nearest to thousands = 3000
∴ 12,904 + 2,888 = 13000 + 3000 = 16000
(d) 28,292 – 21,496
Rounding off 28,292 nearest to thousands = 28,000
Rounding off 21,496 nearest to thousands = 21,000
∴ 28,292 – 21,496 = 28,000 – 21,000 = 7,000
Ten more examples:
Example 1: 1210 + 2365 = 1200 + 2400 = 3600
Example 2: 3853 + 6524 = 4000 + 7000 = 11,000
Example 3: 8752 – 3654 = 9,000 – 4,000 = 5,000
Example 4: 4538 – 2965 = 5,000 – 3,000 = 2,000
Example 5: 1927 + 3185 = 2000 + 3,000 = 5,000
Example 6: 3258 – 1698 = 3000 – 2000 = 1,000
Example 7: 8735 + 6232 = 9000 + 6000 = 15,000
Example 8: 1038 – 1028 = 1000 – 1000 = 0
Example 9: 6352 + 5830 = 6,000 + 6,000 = 12,000
Example 10: 9854 – 6385 = 10,000 – 6000 = 4,000
Ex 1.3 Class 6 Maths Question 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 + 4,317
(b) 1,08,734-47,599
(c) 8,325-491
(d) 4,89,348-48,365
Make four such examples:
Solution:
(a)439 + 334 + 4,317
(i) Rough estimate (Rounding off to nearest hundreds)
439 + 334 + 4,317 = 400 + 300 + 4300 = 5,000
(ii) Closer estimate (Rounding off to nearest tens)
439 + 334 + 4317 = 440 + 330 + 4320 = 5090.
(b) 1,08,734 – 47,599
(i) Rough estimate (Rounding off to nearest hundreds)
1,08,734 – 47,599 = 1,08,700 – 47,600 = 61,100
(ii) Closer estimate (Rounding off to nearest tens)
1,08,734 – 47,599 = 1,08,730 – 47,600 = 61,130.
(c) 8325 – 491
(i) Rough estimate (Rounding off to nearest hundreds)
8325 – 491 = 8300 – 500 = 7800
(ii) Closer estimate (Rounding off to nearest tens)
8325 – 491 = 8330 – 490 = 7840.
(d) 4,89,348 – 48,365
(i) Rough estimate (Rounding off to nearest hundreds)
4,89,348 – 48,365 = 4,89,300 – 48,400 = 4,40,900
(ii) Closer estimate (Rounding off to nearest tens)
4,89,348 – 48,365 = 4,89,350 – 48,370 = 4,40,980
Four more examples:
Example 1:
384 + 562
Solution:
(i) Rough estimate (Rounding off to nearest hundreds)
384 + 562 = 400 + 600
= 1,000
(ii) Closer estimate (Rounding off to nearest tens)
384 + 562 = 380 + 560
= 940
Example 2:
8765 – 3820
Solution:
(i) Rough estimate (Rounding off to nearest hundreds)
8765 – 3820 = 8800 – 3900
= 4900
(ii) Closer estimate (Rounding off to nearest tens)
8765 – 3820 = 8770 – 3820
= 4950
Example 3:
6653 – 8265
Solution:
(i) Rough estimate (Rounding off to nearest hundreds)
6653 + 8265 = 6700 + 8300
= 15,000
(ii) Closer estimate (Rounding off to nearest tens)
6653 + 8265 = 6650 + 8270
= 14920
Example 4:
3826 – 1262
Solution:
(i) Rough estimate (Rounding off to nearest hundreds)
3826 – 1262 = 3800 – 1300
= 2500
(ii) Closer estimate (Rounding off to nearest tens)
3826 – 1262 = 3830 – 1260
= 2570
Ex 1.3 Class 6 Maths Question 3.
Estimate the following products using general rule:
(a) 578 x 161
(b)5281 x 3491
(c) 1291 x 592
(d) 9250 x 29
Make four more such examples.
Solution:
(a) 578 x 161 = 600 x 200 = 1,20,000
(b) 5281 x 3491 = 5000 x 3000 = 1,50,00,000
(c) 1291 x 592 = 1300 x 600 = 7,80,000
(d) 9250 x 29 = 9000 x 30 = 2,70,000
Four more examples:
Example 1.
382 x 1062
Solution:
382 x 1062 = 400 x 1000 = 4,00,000
Example 2.
6821 x 1291
Solution:
6821 x 1291 = 7000 x 1000 = 70,00,000
Example 3.
3858 x 9350
Solution:
3858 x 9350 = 4000 x 9000 = 3,60,00,000
Example 4.
3405 x 7502
Solution:
3405 x 7502 = 3000 x 8000 = 2,40,00,000