{"id":76323,"date":"2021-10-20T10:30:41","date_gmt":"2021-10-20T05:00:41","guid":{"rendered":"https:\/\/www.cbselabs.com\/?p=76323"},"modified":"2021-10-20T11:46:10","modified_gmt":"2021-10-20T06:16:10","slug":"knowing-our-numbers-class-6-extra-questions","status":"publish","type":"post","link":"https:\/\/www.cbselabs.com\/knowing-our-numbers-class-6-extra-questions\/","title":{"rendered":"Knowing Our Numbers Class 6 Extra Questions Maths Chapter 1"},"content":{"rendered":"

Knowing Our Numbers Class 6 Extra Questions Maths Chapter 1<\/h2>\n

Extra Questions for Class 6 Maths Chapter 1 Knowing Our Numbers<\/strong><\/p>\n

Knowing Our Numbers Class 6 Extra Questions Very Short Answer Type<\/h3>\n

Class 6 Maths Chapter 1 Extra Questions Question 1.<\/strong>
\nWrite the smallest three digit number whose value does not change on reversing its digits.
\nSolution:
\nThe required number is 101.<\/p>\n

Knowing Our Numbers Class 6 Extra Questions Question 2.<\/strong>
\nWrite the greatest three digit number which does not change on reversing its digits.
\nSolution:
\nThe required number is 999.<\/p>\n

Knowing Our Numbers Class 6 Extra Questions Pdf Question 3.<\/strong>
\nWhat must be added to 203 to get a number whose digits are reversed of the given number?
\nSolution:
\nThe number obtained by reversing the digits of 203 = 302.
\n\u2234 Difference = 302 – 203 = 99
\nHence, the required number is 99.<\/p>\n

Knowing Our Numbers Class 6 Questions And Answers Question 4.<\/strong>
\nWrite the following in Roman numerals:
\n(a) 72
\n(b) 38
\nSolution:
\n(a) 72 = LXXII
\n(b) 38 – XXXVIII<\/p>\n

Extra Questions For Class 6 Maths Chapter 1 Question 5.<\/strong>
\nWrite 438 in its expanded form.
\nSolution:
\n438 = 4 x 100 + 3 x 10 + 8.<\/p>\n

Knowing Our Numbers Class 6 Test Paper Question 6.<\/strong>
\nWrite the greatest five-digit number using the digits 4, 2 and 0.
\nSolution:
\nThe greatest five-digit number using the digits 4, 2 and 0 is 44420.<\/p>\n

Knowing Our Numbers Class 6 Questions Question 7.<\/strong>
\nThe capacity of a water tank is 300 litres. Express its capacity in millilitres.
\nSolution:
\nWe know that
\n1 litre = 1000 mL
\n\u2234 300 litres = 300 x 1000 mL = 3,00,000 mL
\nHence, the capacity of water tank = 3 lakh millilitres.<\/p>\n

Class 6 Maths Ch 1 Extra Questions Question 8.<\/strong>
\nWhat is the successor of greatest 6-digit number?
\nSolution:
\nGreatest 6-digit number = 999999
\nSuccessor of it = 999999 + 1 = 1000000
\ni. e., smallest 7-digit number.
\nHence, the required successor = 10,00,000.<\/p>\n

Knowing Our Numbers Class 6 Worksheet Question 9.<\/strong>
\nWhat is the place value of 7 in 1743?
\nSolution:
\nLet us write 1743 in its expanded form
\n1743 = 1000 + 700 + 40 +3
\nPlace value of 7 = 700
\nHence, the place value of 7 = 700.<\/p>\n

Knowing Our Numbers Class 6 Extra Questions Short Answer Type<\/h3>\n

Chapter 1 Maths Class 6 Extra Questions Question 10.<\/strong>
\nOf 7,12,540 and 71,25,400 which number is greater and by how much?
\nSolution:
\nSince 71,25,400 is a seven-digit number and 7,12,540 is a six-digit number.
\nSo 71,25,400 is greater than 7,12,540.
\n\"Class
\nHence 71,25,400 is greater than 7,12,540 by 64,12,860.<\/p>\n

Class 6 Knowing Our Numbers Extra Questions Question 11.<\/strong>
\nWrite the smallest and the greatest 5-digit numbers using the digits 0,2,4,6,8 (Repetition of digits is not allowed).
\nSolution:
\nGiven digits are 0, 2, 4, 6, 8
\n5 – digit greatest number = 86420;
\n5 – digit smallest number = 20468.<\/p>\n

Knowing Our Numbers Class 6 Extra Questions And Answers Question 12.<\/strong>
\nWrite the following numbers in ascending order. How many of them are even numbers?
\n63,854, 63,584, 65,348, 68,543, 64,835
\nSolution:
\nThe given numbers are 63,854, 63,584, 65,348, 68,543 and 64,835.
\nAscending order is 63,584 ; 63,854 ; 64,835 ; 65,348 ; 68,543
\nEven numbers are 63,584, 63,854 and 65,348.<\/p>\n

Class 6 Chapter 1 Extra Questions Maths Question 13.<\/strong>
\nRound the given numbers to the nearest tens.
\n(a) 48
\n(b) 59
\n(c) 64
\n(d) 215
\nSolution:
\nGiven number Rounded off to tens
\n(a) 48 \u2192 50
\n(6) 59 \u2192 60
\n(c) 64 \u2192 60
\n(d) 215 \u2192 220<\/p>\n

Knowing Our Numbers Class 6 Practice Questions Question 14.<\/strong>
\nEstimate the following products:
\n(\u0430) 86 x 316
\n(b) 898 x 786
\nSolution:
\n(a) 86 x 316
\n\u2235 86 \u2192 90 [Rounding off to tens] and 316 \u2192 320 [Rounding off to tens]
\nSo, the estimated product is 90 x 320 = 28800<\/p>\n

(b) 898 x 786
\n\u2235 898 \u2192 900 [Rounding off to hundreds] and 786 \u2192 800 [Rounding off to hundreds]
\nSo, the estimated product is 900 x 800 = 720000.<\/p>\n

Ch 1 Maths Class 6 Extra Questions Question 15.<\/strong>
\nDivide 2,63,175 by 275.
\nSolution:
\nWe have
\n\"Knowing
\nHence, quotient = 957 and remainder = 0.<\/p>\n

Knowing Our Numbers Extra Questions Question 16.<\/strong>
\nA student multiplied 3759 by 231 instead of multiplying by 213. How much was his product greater than the correct product?
\nSolution:
\nFirst Method:
\n(3759 x 231)-(3759 x 213) = 868329 – 800667 = 67662
\nSecond Method: 3759 x (231 – 213) = 3759 x 18 = 67662
\nHence, the product difference is 67662.<\/p>\n

Question 17.
\nEstimate: 25,148 + 7394 + 9343 + 752
\nSolution:
\nEstimated values are
\n25,148\u00a0 \u00a0 \u00a0\u2192\u00a0 \u00a0 25100
\n7394\u00a0 \u00a0 \u00a0 \u00a0 \u2192\u00a0 \u00a0 \u00a07400
\n9343\u00a0 \u00a0 \u00a0 \u00a0\u2192\u00a0 \u00a0 \u00a0 9300
\n752\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u2192\u00a0 \u00a0 \u00a0 800
\nSo, the estimated sum is 25100 + 7400 + 9300 + 800 = 42600
\nHence, the estimated sum is 42600.<\/p>\n

Question 18.
\nWrite all the even numbers between 90 and 100 in Roman Numerals.
\nSolution:
\nEven numbers between 90 and 100, we have 92, 94, 96, 98.
\n\u2234 92 = XCII,
\n94 = XCIV,
\n96 = XCVI,
\n98 = XCVIII<\/p>\n

Knowing Our Numbers Class 6 Extra Questions Long Answer Type<\/h3>\n

Question 19.
\nWrite the missing digits in the following sums:
\n\"Knowing
\nSolution:
\n\"Knowing<\/p>\n

Question 20.
\nWrite Hindu-Arabic numerals for:
\n(a) LXXXVI
\n(b) LXXV
\n(c) XCIX
\n(d) XCI
\nSolution:
\n(a) LXXXVI = 50 + 30 + 6 = 86
\n(b) LXXV = 50 + 20 + 5 = 75
\n(c) XCIX = (100 – 10) + 9 = 99
\n(d) XCI = (100 – 10) + 1 = 91<\/p>\n

Question 21.
\nThe distance between the school and Reena\u2019s house is 1 km 480 m. Everyday she walks both ways. What distance does she cover in 6 days of a week?
\nSolution:
\nDistance covered when she walks one way = 1 km 480 m = 1480 m
\nTherefore, the distance covered when she walk both ways in a day = 1480 x 2 m = 2960 m
\nTotal distance covered by Reena in 6 days = 2960 x 6 m = 17760 m or 17 km 760 m.<\/p>\n

Question 22.
\nSimplify: 36 \u00f7 [5 + {4 x 5 \u00f7 2}]
\nSolution:
\nGiven:
\n36 \u00f7 [5 + {4 x 5 \u00f7 2}]
\nUsing B, O, D, M, A, S
\n\"Extra<\/p>\n

Question 23.
\nTo stitch a pant 1 m 15 cm cloth is needed. Out of 36 m cloth, how many pants can be stitched and how much cloth will remain?
\nSolution:
\n\"Knowing
\nCloth required to stitch 1 pant = 1 m 15 cm
\n= 100 cm + 15 cm [\u2235 1 m = 100 cm]
\n= 115 cm
\nTotal cloth = 36 m = 36 x 100 cm = 3600 cm
\nTherefore number of pants stitched = \\(\\frac { 3600 }{ 115 }\\)
\nHence, 31 pants can be stitched and cloth left over is 35 cm.<\/p>\n

Question 24.
\nWrite each of the following numbers in figures:
\n(a) Eighty-one million four hundred twelve thousand six hundred fifty.
\n(b) Twenty million three hundred eighty thousand one hundred.
\n(c) Ninety million nine.
\n(d) Forty-nine million seven hundred eighty two thousand fifty eight.
\n(e) Six millions three hundred fifty-two thousand nine hundred forty-six.
\n(f) Seven crore twenty-three lakh eighty-six thousand, five hundred ninety-four.
\n(g) Fifty crore forty lakh sixty thousand nine.
\n(h) Nineteen crore, ninety lakh, fourteen thousand, six hundred eighty.
\nSolution:
\nIn words\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0In figure
\n(\u0430) Eighty-one millions four hundred twelve thousand, six hundred fifty.\u00a0 \u00a0 \u00a081,412,650
\n(b) Twenty million three hundred eighty thousand one hundred\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a020,380,100
\n(c) Ninety million nine\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a090,000,009
\n(d) Forty-nine million seven hundred eighty-two thousand fifty-eight\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a049,782,058
\n(e) Six-millions three hundred fifty-two thousand nine hundred forty-six\u00a0 \u00a0 6,352,946
\n(f) Seven crore, twenty-three lakh eighty-six thousand five hundred ninety-four 7,23,86,594
\n(g) Fifty crore forty lakh sixty thousand nine\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 50,40,60,009
\n(h) Nineteen crore ninety lakh fourteen thousand six hundred eighty.\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a019,90,14,680<\/p>\n

Question 25.
\nWrite True\/False for the following statements:
\n(a) Roman symbol X cannot be repeated more than three times
\n(b) VXXX = 25 …….. .
\n(c) Estimate value of274 rounding off to nearest hundreds = 200 …….
\n(d) I and X can repeat at the most three times …….
\n(e) V, L and D are neither, repeated nor written to the left of greater value symbol ……..
\n(f) There are six basic symbols in Roman Numeration system ……..
\nSolution:
\n(a) True
\n(b) False
\n(c) False
\n(d) True
\n(e) True
\n(f) False.<\/p>\n

Knowing Our Numbers Class 6 Extra Questions Higher Order Thinking Skills (HOTS)<\/h3>\n

Question 26.
\nThere are two factories located at place P and the other at place Q. From these factories, a certain commodity is to be delivered to each of the depots situated at A, B and C. Weekly production of commodity by P and Q are 120 kg and 150 kg respectively. Weekly requirement of commodity by A, B and C are 80 kg, 90 kg and 100 kg respectively. P delivers 60 kg to A, 40 kg to B and 20 kg to C. How much amount of the commodity should Q deliver to A, B and C to meet their requirement? If the rate of the commodity is ? 20 per kg, find the total amount to be paid to P and Q.
\nSolution:
\n\"Knowing
\nAmount of commodity delivered by P to A = 60 kg
\nAmount of commodity delivered by Q to A = 80 – 60 = 20 kg
\nAmount of commodity delivered by P to B = 40 kg
\nAmount of commodity delivered by Q to B – 90 – 40 = 50 kg
\nAmount of commodity delivered by P to C = 20
\nAmount of commodity delivered by Q to C = 100 – 20 = 80 kg.
\nNow Amount of money to be paid to P by A, B and C = \u20b9( 60 x 20 + 40 x 20 + 20 x 20)
\n= \u20b9 (1200 + 800 + 400)
\n= \u20b9 2400
\nand amount of money to be paid to Q by A, B and C
\n= \u20b9 (20 x 20 + 50 x 20 + 80 x 20)
\n= \u20b9 (400 + 1000 + 1600) = \u20b9 3000
\nHence, the total amount
\n= \u20b9 2400 + \u20b9 3000 = \u20b9 5400.<\/p>\n

Extra Questions for Class 6 Maths<\/a><\/h4>\n

NCERT Solutions for Class 6 Maths<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"

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