Skip to main content

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 2 Estimation

Selina Concise Mathematics Class 6 ICSE Solutions – Estimation

Selina Concise Mathematics Class 6 ICSE Solutions – Estimation

EXERCISE 2(A)


Question 1.
Round off each of the following to the nearest ten :
(i) 62
(ii) 265
(iii) 543
(iv) 8261
(v) 6294
(vi) 3008
(vii) 72326
Solution:
If the digit at ones place is less than 5, replace ones digit by 0, and keep the other digit as they are :
And if the digit at ones place is 5 or more than 5, increase tens digit by 1, and replace ones digit by 0.
(i) 62 – 60
(ii) 265 – 270
(iii) 543 – 540
(iv) 8261 – 8260
(v) 6294 – 6290
(vi) 3008-3010

Question 2.
Round off each of the following to the nearest hundred :
(i) 748
(ii) 784
(iii) 2667
(iv) 5432
(v) 6388
(vi) 59237
Solution:
If the tens digit is less than 5, replace each one of tens and once digits by 0 and keep the other digits as they are :
(i) 748 – 700
(ii) 784 – 800
(iii) 2667 – 2700
(iv) 5432 – 5400
(v) 6388 – 6400
(vi) 59237 – 59200

Question 3.
Round off each of the following to the nearest thousand :
(i) 6475
(ii) 6732
(iii) 5352
(iv) 32568
(v) 9248
(vi) 83294
Solution:
If the tens digit is 5 or more than 5, increase the hundreds digit by 1 and replace each of tens digit and ones digit by 0.
(i) 6475 – 6000
(ii) 6732 – 7000
(iii) 25352 – 25000
(iv) 32568 – 33000
(v) 9248 – 9000
(vi) 83294 – 83000

Question 4.
Round off
(i) 578 to the nearest ten.
(ii) 578 to the nearest hundred.
(iii) 4327 to the nearest thousand.
(iv) 32974 to the nearest ten-thousand.
(v) 27487 to the nearest ten-thousand.
Solution:
(i) 578 – 580
(ii) 578 – 600
(iii) 4327 – 4000
(iv) 32974 – 30000
(v) 27487 – 30000

Question 5.
Round off each of the following to the nearest tens, hundreds and thousands.
(i) 864
(ii) 1249
(iii) 54, 547
(iv) 68, 076
(v) 56, 293
(vi) 7, 293
(vii) 89, 24, 379
Solution:
(i) 864 – 860
864 – 900
864 – 1000
(ii) 1249 – 1250
1249 – 1200
1249 – 1000
(iii) 54, 547 – 54550
54, 547 – 54500
54, 547 – 55000
(iv) 68, 076 – 68, 080
68, 076 – 68, 100
68, 076 – 68, 000
(v) 56, 293 – 56, 290
56, 293 – 56, 300
56, 293 – 56, 000
(vi) 7, 293 – 7290
7, 293 – 7, 300
7, 293 – 7000
(vii) 89, 24, 379 – 89, 24, 380
89, 24, 379 – 89, 24, 400
89, 24, 379 – 89, 24, 000

Question 6.
Round off the following to the nearest tens ;
(i) ₹ 562
(ii) 837 m
(iii) 545 cm
(iv) ₹ 27
Solution:
(i) ₹ 562
₹ 562 – ₹ 560
(ii) 837 m
837 – 840 m
(iii) 545 cm
545 – 550
(iv) ₹ 21
₹ 27 – ₹ 30

Question 7.
List all the numbers which can be round off to 30.
Solution:
The numbers that can be rounded off to 30 are:
26, 27, 28, 29, 31, 32, 33, 34

Question 8.
List all the numbers which can be rounded off to 50.
Solution:
The numbers that can be rounded off to 50 are:
46, 47, 48, 49, 51, 52, 53, 54

Question 9.
Write the smallest and the largest numbers which are rounded off to 80.
Solution:
The smallest number which is rounded off to 80 is 75
and the largest number which is rounded off to 80 is 84.

Question 10.
Write the smallest and the largest numbers which are rounded off to 130.
Solution:
The smallest number which is rounded off to 130 is 125
and The largest number which is rounded off to 130 is 134.

EXERCISE 2(B)

Question 1.
Estimate the sum of each pair of numbers to the nearest ten :
(i) 67 and 44
(ii) 34 and 87
(iii) 23 and 66
(iv) 78 and 18
(v) 96 and 55
(vi) 76 and 62
(vii) 457 and 175
(viii) 474 and 173
(ix) 527 and 267
Solution:
(i) 67 and 44
67 to the nearest ten 70 and, 44 to the nearest ten = 40
∴ Required sum (70 + 40) = 110
(ii) 34 and 87
34 to the nearest ten = 30 and, 87 to the nearest ten = 90
∴ Required sum = (30 + 90) = 120
(iii) 23 and 66
23 to the nearest ten = 20 and, 66 to the nearest ten = 70
∴ Required sum = (20 + 70) = 90
(iv) 78 and 18
78 to the nearest ten = 80 and, 18 to the nearest ten = 20
∴ Required sum = (80 + 20) = 100
(v) 96 and 55
96 to the nearest ten = 100 and, 55 to the nearest ten = 60
∴ Required sum = (100 + 60) = 160
(vi) 76 and 62
76 to the nearest ten = 80 and, 62 to the nearest ten = 60
∴ Required sum = (80 + 60) = 140
(vii) 457 and 175
457 to the nearest ten = 460 and, 175 to the nearest ten = 180
∴ Required sum = (460 + 180) = 640
(viii) 474 and 173
474 to the nearest ten = 470 and, 173 to the nearest ten = 170
∴ Required sum = (470 + 170) = 640
(ix) 527 and 267
527 to the nearest ten = 530 and, 267 to the nearest ten = 270
∴ Required sum = (530 + 270) = 800

Question 2.
Estimate the sum of each pair of numbers to the nearest hundred :
(i) 336 and 782
(ii) 546 and 342
(iii) 270 and 495
(iv) 4280 and 5295
(v) 4230 and 2410
(vi) 30047 and 39287
Solution:
(i) 336 and 782
336 to the nearest hundred = 300 and, 782 to the nearest hundred = 800
∴ Required sum = (300 + 800) = 1100
(ii) 546 and 342 and, 342 to the nearest hundred = 300
∴ Required sum = (500 + 300) = 800
(iii) 270 and 495
270 to the nearest hundred = 300 and, 495 to the nearest hundred = 500
∴ Required sum = (300 + 500) = 800
(iv) 4280 and 5295
4280 to the nearest hundred = 4300 and, 5295 to the nearest hundred = 5300
∴ Required sum = (4300 + 5300) = 9600
(v) 4230 and 2410
4230 to the nearest hundred = 4200 and, 2410 to the nearest hundred = 2400
∴ Required number = (4200 + 2400) = 6600
(vi) 30047 and 39287
30047 to the nearest hundred = 30000 and, 39287 to the nearest hundred = 39300
∴ Required sum = (30000 + 39300) = 69, 300

Question 3.
Estimate the sum of the following pair of numbers to the nearest thousand:
(i) 53826 and 36455
(ii) 56802 and 22475
Solution:
(i) 53826 and 36455
53826 to the nearest hundred = 54000 and, 36455 to the nearest hundred = 36000
∴ Required sum = (54000 + 36000) = 90000
(ii) 56802 and 22475
56802 to the nearest thousand = 57000 and, 22475 to the nearest thousand = 22000
∴ Required sum = (57000 + 22000) = 79000

Question 4.
Estimate the following differences correct to nearest ten :
(i) 82 – 27
(ii) 96 – 36
(iii) 508 – 248
Solution:
(i) 82 – 27
82 to the nearest ten = 80 and, 27 to the nearest ten = 30
∴ Required difference = (80 – 30) = 50
(ii) 96 – 36
96 to the nearest ten = 100 and, 36 to the nearest ten = 40
∴ Required difference = (100 – 40) = 60
(iii) 508 – 248
508 to the nearest ten = 510 and, 248 to the nearest ten = 250
∴ Required difference = (510 – 250) = 260

Question 5.
Estimate each difference to the nearest hundred :
(i) 769 – 314
(ii) 856 – 687
(iii) 6352 – 2086
Solution:
(i) 769 – 314
769 to the nearest hundred = 800 and, 314 to the nearest hundred = 300
∴ Required difference = (800 – 300) = 500
(ii) 856 – 687
856 to the nearest hundred = 900 and, 687 to the nearest hundred = 700
∴ Required difference = (900 – 700) = 200
(iii) 6352 – 2086
6352 to the nearest hundred = 6400 and, 2086 to the nearest hundred = 2100
∴ Required difference = (6400 – 2100) = 4300

Question 6.
Estimate each difference to the nearest thousand :
(i) 45974 – 38766
(ii) 76003 – 48399
Solution:
(i) 45974 – 38766
45974 to the nearest thousand = 46000 and, 38760 to the nearest thousand = 39000
∴ Required difference = (46000 – 39000) = 7000
(ii) 76003 – 48399
76003 to the nearest thousand = 76000 and, 48399 to the nearest thousand = 48000
∴ Required difference = (76000 – 48000) = 28000

Question 7.
Estimate each of the following products by rounding of each number to the nearest ten :
(i) 49 x 52
(ii) 63 x 38
(iii) 27 x 54
(iv) 53 x 85
(v) 74 x 67
(vi) 25 x 33
Solution:
(i) 49 x 52
49 to the nearest ten = 50 and, 52 to the nearest ten = 50
∴ Required product = (50 x 50) = 2500
(ii) 63 x 38
63 ta the nearest ten = 60 and, 38 to the nearest ten = 40
∴ Required product = (60 x 40) = 2400
(iii) 27 x 54
27 to the nearest ten = 30 and, 54 to the nearest ten = 50
∴ Required product = (30 x 50) = 1500
(iv) 53 x 85
53 to the nearest ten = 50 and, 85 to the nearest ten = 90
∴ Required product = (50 x 90) = 4500
(v) 74 x 67
74 to the nearest ten = 70 and, 67 to the nearest ten = 70
∴ Required product = (70 x 70) = 4900
(vi) 25 x 33
25 to the nearest ten = 30 and, 33 to the nearest ten = 30
∴ Required product = (30 x 30) = 900

Question 8.
Estimate each of the following products by rounding off each number to the nearest hundred :
(i) 477 x 213
(ii) 624 x 236
(iii) 333 x 247
(iv) 537 x 283
(v) 382 x 127
(vi) 427 x 328
Solution:
(i) 477 x 213
477 to the nearest hundred = 500 and, 213 to the nearest hundred = 200
∴ Required product = (500 x 200) = 100000
(ii) 624 x 236
624 to the nearest hundred = 600 and, 236 to the nearest hundred = 200
∴ Required product = (600 x 200) = 120000
(iii) 333 x 247
333 to the nearest hundred = 300 and, 247 to the nearest hundred = 200
∴ Required product = (300 x 200) = 60000
(iv) 537 x 283
537 to the nearest hundred = 500 and, 283 to the nearest hundred = 300
∴ Required product = (500 x 300) = 150000
(v) 382 x 127
382 to the nearest hundred = 400 and, 127 to the nearest hundred = 100
∴ Required product = (400 x 100) = 40000
(vi) 427 x 328
472 to the nearest hundred = 500 and, 328 to the nearest hundred = 300
∴ Required product = (500 x 300) = 150000

Question 9.
Estimate each of the following products by rounding off the first number correct to nearest ten and the other number correct to nearest hundred :
(i) 28 x 287
(ii) 432 x 128
(iii) 48 x 165
(iv) 72 x 258
(v) 83 x 664
(vi) 44 x 250
Solution:
(i) 28 x 287
28 to the nearest ten = 30 and, 287 to the nearest hundred = 300
∴ Required product = (30 x 300) = 9000
(ii) 432 x 128
432 to the nearest ten = 430 and, 128 to the nearest hundred = 100
∴ Required product = (430 x 100) = 43000
(iii) 48 x 165
48 to the nearest ten = 50 and, 165 to the nearest hundred = 200
∴ Required product = (50 x 200) = 10000
(iv) 72 x 258
72 to the nearest ten = 70 and, 258 to the nearest hundred = 300
∴ Required product = (70 x 300) = 21000
(v) 83 x 664
83 to the nearest ten = 80 and, 664 to the nearest hundred = 700
∴ Required product = (80 x 700) = 56000
(vi) 44 x 250
44 to the nearest ten = 40 and, 250 to the nearest hundred = 300
∴ Required product = (40 x 300) = 12000

Question 10.
Estimate each of the following quotients by converting each number to the nearest ten :
(i) 87 ÷ 28
(ii) 84 ÷ 23
(iii) 77 ÷ 22
(iv) 198 ÷ 24
(v) 355 ÷ 26
(vi) 444 ÷ 42
(vii) 843 ÷ 33
Solution:
(i) 87 ÷ 28
(87 ÷ 28) is (approximately to the nearest 10) equal to 90 ÷ 30 = 3
(ii) 84 ÷ 23
84 ÷ 23 is (approximately to the nearest 10) equal to 80 ÷ 20 = 4
(iii) 77 ÷ 22
77 ÷ 22 is (approximately to the nearest 10) equal to 80 ÷ 20 = 4
(iv) 198 ÷ 24
198 ÷ 24 is (approximately to the nearest 10) equal to 200 ÷ 20 = 10
(v) 355 ÷ 26
355 ÷ 26 is (approximately to the nearest 10) equal to 360 ÷ 30 = 12
(vi) 444 ÷ 42
444 ÷ 42 is (approximately to the nearest 10) equal to 440 ÷ 40 = 11
(vii) 843 ÷ 33
843 ÷ 33 is (approximately to the nearest 10) equal to 840 ÷ 30 = 28

Popular posts from this blog

ICSE Solutions for Class 9 History and Civics | Indian History World Developments and Civics ICSE Class IX Question Answers Total Solutions APC Avichal Publishing Company BB Tayal

📚  ICSE Solutions for Class 9 History and Civics ICSE Solutions for Class 9 History and Civics Indian History, World Developments and Civics for ICSE Class- IX by BB-Tayal of Avichal Publishing Company (APC) Buy ICSE Total History & Civics For Class 9 (Latest Syllabus 2022 ) Online Icse Total History & Civics For Class 9 (Latest Syllabus 2022) HISTORY The Harappan Civilization Early Vedic Civilization The Later Vedic Age India in the 6th Century BC: Rise of Jainism and Buddhism The Mauryan Empire The Sangam Age: Kingdoms and The Social and Economic Conditions The Age of the Guptas South India and the Cholas The Delhi Sultanate The Mughal Empire The Composite Culture: Bhakti Movement, Sufism and Influence of Christianity on Indian Society The Renaissance The Reformation Industrial Revolution and Capitalism and Socialism CIVICS Our Constitution and Its Preamble Fundamental Rights, Fundamental Duties and Dir

ICSE Solutions for Class 9 History and Civics - The Harappan Civilization

ICSE Solutions for Class 9 History and Civics – The Harappan Civilization ICSE Solutions for Class 9 History and Civics – The Harappan Civilization Exercises Question 1. Mention any two sources to reconstruct the Harappan Civilization. Answer: The remains of the two towns, Mohenjo-daro and Harappan reveal and remarkable sense of town planning—the drainage system, the Great Bath, the Assembly Hall and other public buildings. From Seals we come to know about the physical features, dress, ornaments and religious beliefs of the people. Question 2. Why did the Indus Valley Civilization come to be known as Harappan Civilization? Answer: Indus Valley Civilization came to be known as Harappan Civilization because this Civilization flourished in the pre-historic cities of Harappan in West Punjab and Mohenjo-daro in Sind. Question 3. Mention any two important centres of the Indus Valley Civilization. Answer: Northern and Western parts of India and the present day Pakistan.

ICSE Solutions for Class 8 History and Civics - A Period of Transition

ICSE Solutions for Class 8 History and Civics – A Period of Transition I. FILL IN THE BLANKS:   1. The Renaissance thinkers believed in life in this World.   2. The term Reformation refers to two major developments, the Protestant Reformation and the Catholic Reformation.   3. Vasco-da-Gama reached Calicut on the West Coast of India.   4. The Industrial Revolution began in England in about 1750 .   5. In 1793, Eli Whitney invented a Cotton gin . II. MATCH THE CONTENTS OF COLUMN A AND COLUMN B: Answer:   III. STATE WHETHER THE FOLLOWING STATEMENTS ARE TRUE OR FALSE: 1. The Renaissance and the Reformation along with new voyages ushered in the Modern Age. True. 2. The Industrial Revolution began in Germany.   False. 3. Me Adam devised railway tracks. False. 4. The Rise of capitalism and imperialism can be attributed to the industrial Revolution. True. 5. The East India Company gradually became rulers from being traders. True.