Class 6 Maths Chapter 1 Knowing Our Numbers
Directions: In questions 1 to 13, out of the four options, only one is correct. Write the correct answer.
1. The product of the place values of two 2’s in 428721 is
(1) 4
(2) 40000
(3) 400000
(4) 40000000
Solution : (3) Place values of 2’s in 428721 are 20000 and 20
The required product = 20000 x 20 = 400000
2. 3x 1 0000 + 7x 1 000 + 9×1 00 + 0x10 + 4 is the same as
(1) 3794
(2) 37940
(3) 37904
(4) 379409
Read and Learn More Class 6 Maths Exemplar Solutions
Solution (3) : 3 x 10000 + 7 x 1000 + 9 x 100 + 0 x 10 + 4
= 30000 + 7000 + 900 + 4 = 37904
3. If 1 is added to the greatest 7-digit number, it will be equal to
(1) 10 thousand
(2) 1 lakh
(3) 10 lakh
(4)1,00,00,000
Solution (4): The greatest 7-digit number = 99,99,999
Now, 99,99,999 +1 = 1,00,00,000
4. The expanded form of the number 9578 is
(1) 9×10000 + 5×1000 + 7×10 + 8×1
(2) 9x 1000 + 5x 100 + 7x 10 + 8×1
(3) 9x 1000 + 57x 10 + 8×1
(4) 9x 100 + 5x 100 + 7x 10 + 8×1
Solution (2) : Expanded form of 9578 = 9 x 1000 + 5 x 100 + 7 x 10 + 8×1
5. When rounded off to the nearest thousands, the number 85642 is
(1) 85600
(2) 85700
(3) 85000
(4) 86000
Solution (4): Round off 85642 to the nearest thousands = 86000
6. The largest 4-digit number, using any one digit twice, from digits 5, 9, 2 and 6 is
(1) 9652
(2) 9562
(3) 9659
(4) 9965
Solution: (4): The largest 4-digitnumber formed by 5, 9, 2 and 6 using digit ‘9’ twice = 9965
7. In Indian System of Numeration, the number 58695376 is written as
(1) 58,69,53,76
(2) 58,695,376
(3) 5,86,95,376
(4) 586,95,376
Solution: (3): In Indian System of Numeration, 58695376 can be written as 5,86,95,376
8. One million is equal to
(1) 1 lakh
(3) 1 crore
(2) 10 lakh
(4) 10 crore
Solution: (2): One million = 1,000,000
Also, 10,00,000 = 10 lakh
9. The greatest number which on rounding off to nearest thousands gives 5000, is
(1) 5001
(2) 5559
(3) 5999
(4) 5499
Solution: (4) : (1) Rounding off 5001 to nearest thousands = 5000
(2) Rounding off 5559 to nearest thousands = 6000
(3) Rounding off 5999 to nearest thousands = 6000
(4) Rounding off 5499 to nearest thousands = 5000
And 5499 > 5001
10. Keeping the place of 6 in the number 6350947 same, the smallest number obtained by rearranging other digits is
(1) 6975430
(2) 6043579
(3) 6034579
(4) 6034759
Solution (3): The new number formed = 6034579
11. Which of the following numbers in Roman numerals is incorrect?
(1) LXXX
(2) LXX
(3) LX
(4) LLX
Solution: (4) : LLX is incorrect because L can be used exactly once in Roman numerals.
12. The largest 5-digit number having three different digits is
(1) 98978
(2) 99897
(3) 99987
(4) 98799
Solution: (3) : The largest 5-digit number with three different digits is 99987
13. The smallest 4-digit number having three different digits is
(1) 1102
(2) 1012
(3) 1020
(4) 1002
Solution: (4) : The smallest 4-digit number with three different digits is 1002.
Directions: In questions 14 to 29 state whether the given statements are true (T) or false (F)
14. In Roman numeration, a symbol is not repeated more than three times.
Solution: True
15. In Roman numeration, if a symbol is repeated, its value is multiplied as many times as it occurs.
Solution: False
If a symbol is repeated, then its value is added as many times as it occurs in Roman numeration
16. 5555 = 5×1000 + 5×100 + 5×10 + 5×1
Solution: True
17. 39746 = 3 × 10000 + 9 × 1000 + 7 × 100 + 4×10 + 6
Solution: True
18. 82546 = 8 x 1000 + 2 × 1000 + 5 × 100 + 4 × 10 + 6
Solution: False
Since, 82546 = 8 × 10000 + 2× 1000 + 5×100 + 4×10 + 6
19. 532235 = 5 X 100000 + 3 x 10000 + 2 x 1000 + 2X 100 + 3X 10 + 5
Solution: True
20. XXIX = 31
Solution: False
Since, XXIX = 29
21. LXXIV = 74
Solution: True
22. The number LIV is greater than LVI.
Solution: False
Since, LIV = 54 and LVI = 56
LVI is greater than LTV
23. The numbers 4578, 4587, 5478, 5487 are in descending order.
Solution: False
Since, the numbers 4578, 4587, 5478, 5487 are in ascending order.
24. The number 85764 rounded off to nearest hundreds is written as 85700.
Solution: False
The number 85764 rounded off to nearest hundreds is 85800.
25. Estimated sum of 7826 and 1 2469 rounded off to hundreds is 20,000.
Solution: False
Round off 7826 to hundreds is 7800.
Round off 12469 to hundreds is 12500.
Required sum = 7800 + 12500 = 20300
26. The largest six digit telephone number that can be formed by using digits 5, 3, 4, 7, 0, 8 only once is 875403.
Solution: False
The largest six digit telephone number that can be formed by using digits 5, 3, 4, 7, 0, 8 only once is 875430.
27. The number 81652318 will be read as eighty one crore six lakh fifty two thousand three hundred eighteen.
Solution: False
The number 81652318 will be read as eight crore sixteen lakh fifty two thousand three hundred eighteen.
28. The largest 4-diglt number formed by the digits 6, 7, 0, 9 using each digit only once is 9760.
Solution: True
29. Among kilo, milli and centi, the smallest is centi.
Solution: False
Among kilo, milli and centi, the smallest is milli.
Directions: In questions 30 to 44, fill in the blanks to make the statements true
30. (1) 10 million =__________ crore
Solution:1
(2) 10 lakh =_______ million.
Solution: 1
31. (1) 1 metre = ____________ millimetres
Solution: 1000
(2) 1 centimetre =___________ millimetres
Solution: 10
(3) 1 kilometre =__________millimetres
Solution: 10,00,000
33. 100 thousands =________ lakh.
Solution: 1
34. Height of a person is 1 m 65 cm. His height in millimetres is_________.
Solution: 1650 : 1 m 65 cm = (1000 + 650) mm = 1650mm
35. Length of river ‘Narmada’ is about 1290 km. Its length in metres is___________.
Solution: 1290000 : 1290 km = (1290 x 1000) m = 1290000m
36. The distance between Srinagar and Leh is 422 km. The same distance in metres is___________
Solution: 422000: 422 km= (422 x 1000) m = 422000 m
37. Writing of numbers from the greatest to the smallest is called an arrangement in__________ order.
Solution: Descending
38. By reversing the order of digits of the greatest number made by five different non zero digits, the new number Is the________ number of five digits.
Solution: Smallest*
By reversing the order of digits of the greatest number made by five different non zero digits, the new number is the smallest number of these digits.
39. By adding 1 to greatest_______ digt number, we get ten lakh.
Solution: 6: Greatest 6-digit number = 999999 By adding1 to 999999, we get 1000000.
40. The number five crore twenty three lakh seventy eight thousand four hundred one can be written, using commas, in the Indian
System of Numeration as_______.
Solution: 5,23,78,401
41. In Roman Numeration, the symbol X can be subtracted from ________, M and C only.
Solution: L
42. The number 66 in Roman numerals is__________.
Solution: LXVI: 66 =LXVI
43. The population of Pune was 2,538,473 in 2001. Rounded off to nearest thousands, the population was____________.
Solution: 2,538,000
44. The smallest 6 digit natural number ending in 5 is__________
Solution: 100005
45. Arrange the following numbersin descending order: 8435, 4835, 13584, 5348, 25843
Solution: Descending order of given numbers is, 25843, 13584, 8435, 5348, 4835
46. Of the following numbers which is the greatest? Which is the smallest? 38051425, 30040700, 67205602
Solution: The greatest number is 67205602 and the smallest number is 30040700.
47. Write In expanded form :
(1) 74836
(2) 574021
(3) 8907010
Solution:(1) 74836- 7 x10000) x 4 x 1000 + 8 x 100 +3x 10+6×1
(2) 574021 =5x 1 00000 + 7x 1 0000 + 4x 1000 + 0x 100 + 2x 10+ 1 x1
(3) 8907010 = 8 x 1000000 + 9x 1 00000 + 0 x 10000 + 7 x 1000 + 0 x 100 + 1 x 10 + 0 x 1
48. As per the census of 2001, the population of four states are given below. Arrange the states in ascending and descending order of their population.
(1) Maharashtra 96878627
(2) Andhra Pradesh 76210007
(3) Bihar 82998509
(4) Uttar Pradesh 166197921
Solution: Ascending order -» (2), (3), (1), (4)
Descending order —» (4), (1), (3), (2)
49. The diameter of Jupiter is 142800000 metres. Insert commas suitably and write the diameter according to International System of Numeration.
Solution:
Given
The diameter of Jupiter is 142800000 metres.
The diameter of Jupiter is 142,800,000 metres.
50. India’s population has been steadily increasing from 439 millions in 1961 to 1028 millions in 2001. Find the total increase in population from 1961 to 2001. Write the increase in population in Indian System of Numeration, using commas suitably.
Solution:
Given
India’s population has been steadily increasing from 439 millions in 1961 to 1028 millions in 2001. Find the total increase in population from 1961 to 2001.
Total increaseinpopulation from 1961 to 2001= 1028 millions- 439 millions = 589 millions
According toIndian System of Numeration, the increase in population = 58,90,00,000
51. Radius of the Earth is 6400 km and that of Mars is 4300000 m. Whose radius is bigger and by how much?
Solution:
Given
Radius of the Earth is 6400 km and that of Mars is 4300000 m.
Radius of the Earth = 6400 km = 6400 x 1000 m = 6400000 m
And radius of the Mars = 4300000 m
Radius of the Earth is greater than the radius of the Mars by (6400(X)0- 4300000) m = 2100000 m.
52. In 2001, the populations of Tripura and Meghalaya were 3,199,203 and 2,318,822, respectively. Write the populations of these two states in words.
Solution:
Given
In 2001, the populations of Tripura and Meghalaya were 3,199,203 and 2,318,822, respectively.
The population of Tripura = 3,199,203 i.e., Three million one hundred ninety-nine thousand two hundred three.
And the population of Meghalaya = 2,318,822, i.e., Two million three hundred eighteen thousand eight hundred twenty two.
53. In a city, polio drops were given to 2,12,583 children on Sunday in March 2008 and to 2,16,813 children in the next month. Find the difference of the number of children getting polio drops in the two months.
Solution:
Given
In a city, polio drops were given to 2,12,583 children on Sunday in March 2008 and to 2,16,813 children in the next month.
Number of children getting polio drops in March 2008 = 2,12,583
Number of children getting polio drops in April 2008 = 2,16,813
The required difference of the number of children getting polio drops in the two months = 2,16,813- 2,12,583 = 4,230
54. A person had ? 1000000 with him. He purchased a colour T.V. for ? 16580, a motor cycle for ? 45890 and a flat for ? 870000. How much money was left with him?
Solution:
Given :
A person had 1000000 with him. He purchased a colour T.V. for 16580, a motor cycle for 45890 and a flat for 870000.
Total amount a personhad = 1000000
The amounthe spent on a colour T.V. = 16580
The amount he spent on a motorcycle = 45890
The amount he spent on a flat =870000
Total amount he spent = (16580 + 45890 + 870000) = 932470
Thus, the amount left with him = 1000000- 932470 = 67530
55. Out of 1 80000 tablets of Vitamin A, 1 8734 are distributed among the students in a district. Find the number of the remaining vitamin tablets.
Solution:
Given:
Out of 1 80000 tablets of Vitamin A, 1 8734 are distributed among the students in a district.
Total tablets of Vitamin A= 180000
Number of tablets distributed among the students in a district = 18734
The number of remaining vitamin tablets -180000-18734-161266
56. Chinmay had ? 610000. He gave 87500 to Jyoti, 126380 to Javed and 350000 to John. How much money was left with him?
Solution:
Chinmay had total amount = 610000
The amount he gave to Jyoti = 87500
The amount he gave to Javed = 126380
The amount he gave to John = 350000
Total amount given by Chinmay = (87500 + 126380+ 350000) = 563880
Thus, the amount left with him = 610000- 563880 = 46120
57. Find the difference between the largest number of seven digits and the smallest number of eight digits.
Solution: The smallest number of eight digits = 10000000
The largest number of seven digits = 9999999
The required difference = 10000000-9999999 =1
58. A mobile number consists of ten digits. The first four digits of the number are 9, 9, 8 and 7. The last three digits are 3, 5 and 5. The remaining digits are distinct and make the mobile number, the greatest possible number. What are these digits?
Solution:
Given
A mobile number consists of ten digits. The first four digits of the number are 9, 9, 8 and 7. The last three digits are 3, 5 and 5. The remaining digits are distinct and make the mobile number, the greatest possible number.
A mobile number consists of 10 digits.
If the first four digits of the number are 9, 9, 8 and 7 and the last three digits of the number are 3, 5 and 5.
Thus, for the greatest possible number, the remaining distinct digits are 6, 4 and 2.
59. A mobile number consists of ten digits. First four digits are 9, 9, 7 and 9. Make the smallest mobile number by using only one digit twice from 8, 3, 5, 6, 0.
Solution:
Given
A mobile number consists of ten digits. First four digits are 9, 9, 7 and 9.
A mobile number consists of 10 digits.
If the first four digits are 9, 9, 7 and 9.
Thus, the smallest mobile number by using only one digit twice from 8, 3, 5, 6, 0 is 9979003568.
60. In a five digit number, digit at ten’s place is 4, digit at unit’s place is one fourth of ten’s place digit, digit at hundred’s place Is 0, digit at thousand’s place is 5 times of the digit at unit’s place and ten thousand’s place digit is double the digit at ten’s place. Write the number.
Solution:
Given
In a five digit number, digit at ten’s place is 4, digit at unit’s place is one fourth of ten’s place digit, digit at hundred’s place Is 0, digit at thousand’s place is 5 times of the digit at unit’s place and ten thousand’s place digit is double the digit at ten’s place.
A number consists of 5 digits.
Now, the digit at ten’s place = 4,
the digit at unit’s place = 1/4×4=1,
tine digit at hundred’s place = 0,
the digit at thousand’s place = 5 x1 = 5
the digit at ten thousand’s place = 2×4 =8
Therefore, the number is 85041.
61. Find the sum of the greatest and the least six digit numbers formed by the digits 2, 0, 4, 7,6, 5 using each digit only once.
Solution: By using the digits 2, 0, 4, 7, 6, 5
The greatest number formed = 765420,
and the least number formed = 204567
The required sum= 765420 + 204567= 969987
62. A factory has a container filled with 35874 litres of cold drink. In how many bottles of 200 ml capacity each can it be filled?
Solution:
Given:
A factory has a container filled with 35874 litres of cold drink.
Quantity of cold drink in a container = 35874 litres = 35874 x 1000 ml = 35874000 ml
The capacity of one bottle = 200 ml
The required number of bottles = 35874000 + 200 = 179370
Therefore, 179370 bottles can be filledby cold drink.
63. The population of a town is 450772. In a survey, it was reported that one out of every 14 persons is illiterate. In all how many illiterate persons are there in the town?
Solution:
Given:
The population of a town is 450772. In a survey, it was reported that one out of every 14 persons is illiterate.
Total population of a town = 450772
Since, one out of every 14 personsis illiterate.
The number of illiterate persons in the town = 450772 + 14 = 32198
Therefore, 32198 persons are illiterate in the town.
64. Make the greatest and the smallest 5-digit numbers using different digits in which 5 appears at ten’s place.
Solution: By using the digit 5 at ten’s place, the greatest 5-digit number is 98756, and the smallest 5-digitnumber is 10253
65. How many grams should be added to 2 kg 300 g to make it 5 kg 68 g?
Solution: 5 kg 68 g = (5 x 1000 + 68) g = 5068 g and 2 kg 300 g = (2 x 1000 + 300) g = 2300 g
The required number of grams should be added = 5068 g- 2300 g = 2768 g or 2kg 768 g
2768 grams should be added to 2 kg 300 g to make it 5 kg 68 g
66. A box contains 50 packets of biscuits each weighing 1 20 g. How many such boxes can be loaded in a van which cannot carry beyond 900 kg?
Solution:
Given
A box contains 50 packets of biscuits each weighing 1 20 g.
The total weight of a box containing 50 packets ofbiscuits each weighing 120 g = 50 x 120 g = 6000 g
The capacity of a van = 900 kg = 900 x 1000 g = 900000 g
The required number of boxes = 900000/6000 = 150
Therefore, 150 boxes can be loaded in the van.
67. How many lakhs make five billions?
Solution: 50000 lakhs make 5 billions.
68. How many millions make 3 crores?
Solution: 30 millions make 3 crores.
69.Estimate each of the following by rounding off each number to nearest hundreds:
(1) 874 + 478
(2) 793 + 397
(3) 11244 + 3507
(4) 17677 + 13589
Solution:(1) 874 rounded off to The nearest hundreds-900
478 rounded off to the nearest hundreds -500
Estimated sum = 900 + 500 – 1400
(2) 793 rounded off to the nearest hundreds = 800
397 rounded off to the nearest hundreds = 400
Estimated sum = 800 + 400 = 1200
(3) 11244 rounded off to the nearest hundreds= 11200
3507 rounded off to the nearest hundreds = 3500
Estimated sum = 11200 + 3500 = 14700
(4) 17677roundedofftothenearesthundreds = 17700
13589 rounded off to the nearest hundreds = 13600
Estimated sum = 17700 + 13600 = 31300
70. Estimate each of the following by rounding off each number to nearest tens:
(1) 11963-9369
(2) 76877-7783
(3) 10732-4354
(4) 78203-16407
Solution: (1) 11963 rounded off to the nearest tens- 11960
9369 rounded off to the nearest tens = 9370
Estimated difference=11960- 9370= 2590
(2) 76877 rounded off to the nearest tens = 76880
7783 rounded off to the nearest tens = 7780
Estimated difference= 76880- 7780= 69100
(3) 10732 rounded off to the nearest tens = 10730
4354 rounded off to the nearest tens = 4350
Estimated difference = 10730- 4350 = 6380
(4) 78203 rounded off to the nearest tens = 78200
16407 rounded off to the nearest tens = 16410
Estimated difference = 78200 – 16410 = 61790
71. Estimate each of the following products by rounding off each number to nearest tens:
(1) 87×32
(2) 311 × 113
(3) 3239 × 28
(4) 1385×789
Solution: (1) 87 rounded off to the nearest tens = 90
32 rounded off to the nearest tens = 30
Estimated product = 90 * 30 = 2700
(2) 311 rounded off to the nearest tens = 310
113 rounded off to the nearest tens = 110
Estimated product = 310 * 110 = 34100
(3) 3239 rounded off to the nearest tens= 3240
28 rounded off to the nearest tens = 30
Estimated product = 3240 * 30 = 97200
(4) 1385 rounded off to the nearest tens=1390
789 rounded off to the nearest tens = 790
Estimated product= 1390 * 790= 1098100
72. The population of a town was 78787 in the year 1991 and 95833 in the year 2001. Estimate the increase in population by
rounding off each population to nearest hundreds.
Solution:
Given
The population of a town was 78787 in the year 1991 and 95833 in the year 2001.
78787 rounded off to the nearest hundreds = 78800
95833 rounded off to the nearest hundreds = 95800
The estimated increase in population = 95800-78800 = 17000
73. Estimate the product 758 x 6784 using the general rule.
Solution: 758 can be rounded off to 800 and 6784 can be rounded off to 7000 Estimated product= 800 × 7000 = 5600000
The product 758 x 6784 using the general rule = 5600000
74. A garment factory produced 216315 shirts, 182736 trousers and 58704 jackets in a year. What is the total production of all the three items in that year?
Solution:
Given
A garment factory produced 216315 shirts, 182736 trousers and 58704 jackets in a year.
Number of shirts produced by the factory = 216315
Number of trousers produced by the factory = 182736
Number of jackets produced by the factory = 58704
Total production of the factory = 216315 + 182736 + 58704 = 457755
Total production of the factory = 457755
75. A vessel has 13 litres 200 mL of fruit juice. In how many glasses each of capacity 60 mL can it be filled?
Solution:
Given:
A vessel has 13 litres 200 mL of fruit juice.
Quantity of fruit juice in a vessel = 13 L 200 mL
= (13 x 1000 + 200) mL = 13200 mL
Capacity of one glass = 60 mL
The required number of glasses = 13200 ÷ 60 = 220
Therefore, 220 glasses can be filled by fruit juice.
76. A loading tempo can carry 482 boxes of biscuits weighing 15 kg each, whereas a van can carry 518 boxes each of the same weight. Find the total weight that can be carried by both the vehicles.
Solution:
Given
A loading tempo can carry 482 boxes of biscuits weighing 15 kg each, whereas a van can carry 518 boxes each of the same weight.
Total weight can be carried by a tempo= (482×15) kg = 7230kg
and the total weight can be carried by a van = (518×15) kg = 7770 kg
Thus, the total weight that can be carried by both the vehicles = (7230 + 7770) kg = 15000 kg
The total weight that can be carried by both the vehicles = 15000 kg
77. In the marriage of her daughter, Lecla spent 216766 on food and decoration,122322 on jewellery, 88234 on furniture and 26780 on kitchen items. Find the total amount spent by her on the above items.
Solution:
Given :
In the marriage of her daughter, Lecla spent 216766 on food and decoration,122322 on jewellery, 88234 on furniture and 26780 on kitchen items.
Amount spent by Leela on food and decoration = 216766
Amount spent by her on jewellery = 122322
Amount spent by her on furniture = 88234
Amount spent by her on kitchen items = 26780
Total amount spent by her = (216766 + 122322 + 88234 + 26780) = 454102
Total amount spent by her = 454102
78. A box contains 5 strips each having 12 capsules of 500 mg medicine in each capsule. Find the total weight in grams of medicine in 32 such boxes.
Solution:
Given :
A box contains 5 strips each having 12 capsules of 500 mg medicine in each capsule.
Quantity of medicine in one capsule = 500 mg
Quantity ofmedicinein 12 capsules or 1 strip = (500 x 12) mg = 6000 mg- 6 g
Quantity of medicine in 5 strips or 1 box = (6 x 5) g = 30 g
Quantity of medicinein 32 boxes = (30 x 32)g = 960g
The total weight in grams of medicine in 32 such boxes = 960g