Class 6 Mathematics solutions for Chapter 1 Structure of Numbers (সংখ্যাৰ গঠন)
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Structure of Numbers (সংখ্যাৰ গঠন)
NCERT Solutions for Class 6 Chapter 1 Structure of Numbers
Page Number: 2-3
1. Write the following numbers in words. (তলৰ সংখ্যা কেইটা আখৰেৰে লিখা)
(a) 3102 .
(b) 4300 .
(c) 1266 .
(d) 29,008 .
(e) 15,927 .
Answer:
(a) Three thousand one hundred two
(b) Four thousand three hundred
(c) One thousand two hundred sixty-six
(d) Twenty-nine thousand eight
(e) Fifteen thousand nine hundred twenty-seven
2. Write in numbers (অংকৰে লিখা)
(a) Three thousand two hundred forty five
(b) Twelve thousand seven hundred eighty
(c) Twenty five thousand three hundred four
(d) One lakh thirty four thousand seven hundred four
Answer:
(a) 3245
(b) 12780
(c) 25304
(d) 134704
3. Write the largest (আটাইতকৈ ডাঙৰ) and the smallest (আটাইতকৈ সৰু number from the following.
(a) 7372, 4927, 317, 69875, 650,600
Answer: The largest number 69875 The smallest number 317
(b) 2853, 8691, 9999, 13001, 123,600
Answer: The largest number 13001 The smallest number 123
4. Write the largest and smallest number formed by the digits 7, 8, 3, 5
Answer: The largest number 8753 The smallest number 3578
5. Write the largest and the smallest four digit number from the digits 2, 8 and 5 using any of the digits more than once.
Answer: The largest number 8852 The smallest number 2258
6. Write the largest and the smallest five digits number from the digits 3,7,0 and 5 using any of the digit more than once.
Answer: The largest number 77530 The smallest number 30075
7. Write the following numbers in ascending order (ঊৰ্ধ্বক্ৰম).
96,259; 20,635; 96,025; 87,562; 70,025
Answer: 20635 70025 87562 96025 96259
8. Write the following numbers in descending order (অধঃক্ৰম)
50,000; 74,002; 78,162; 85,715; 60,035
Answer: 85715 78162 74002 60025 50000
9. Write the following numbers in expanded form (বিস্তৃত ৰূপ).
(a) 3520 (b) 2222 (c) 2098
(d) 98,810 (e) 65,006 (D) 99,999
Answer:
(a) 3520= 3×1000+5×100+2×10+0×1
(b) 2222= 2×1000+2×100+2×10+2×1
(c) 2098= 2×1000+0×100+9×10+8×1
(d) 98,810= 9×10000+8×1000+8×100+1×10+0×1
(e) 65,006= 6×10000+5×1000+0×100+0×10+6×1
(D) 99,999= 9×10000+9×1000+9×100+9×10+9×1
10. Write the place value (স্থানীয় মান) of the following under lined digits.
(a) 9146 (b) 4362 (c) 7405
(d) 5651 (e) 6065 (f) 7050
(g) 9209 (h) 24652 (i) 86,702
(j) 30,725 (k) 5 8982 (l) 79012
Answer:
(a) 9146 = 6×1= 6
(b) 4362 = 6×10= 60
(c) 7405 = 4×100= 400
(d) 5651 = 1×1= 1
(e) 6065= 6×1000=6000
(f) 7050= 0×100= 0
(g) 9209= 9×1000= 9000
(h) 24652= 4×1000= 4000
(i) 86,702= 2×1=2
(j) 30,725= 0×1000=0
(k) 5 8982= 5×10000=50000 AND 8×1000=8000
(l) 79012= 7×10000 = 70000 AND 0×100=0
Try yourself
1. Write the place value of 8 and 5 from the number 58,972.
Answer:
58,972 = 58×10000=58000
2. Write the number 99,999 in expanded form.
Answer:
99,999=9×10000+9×1000+9×100+9×10+9×1
3. Write the number 80,029 in words.
Answer: Eighty thousand twenty-nine.
NCERT Solutions for Class 6 Chapter 1 Structure of Numbers
Page Number: 4-5
Try yourself
1. Write the following numbers in expanded form
4,57,628 = .
5,48,393 = .
3,740,156 = .
Answer:
4,57,628 = 4×100000+5×10000+7×1000+6×100+2×10+8×1
5,48,393 = 5×100000+4×10000+8×1000+3×100+9×10+3×1
3,740,156 = 3×1000000+7×100000+4×10000+0×1000+1×100+5×10+6×1
2. Write the following numbers in Indian System of Numeration.
|
Number |
Ten Lakhs |
Lakhs |
Ten Thousand |
Thousand |
Hundreds |
Ten |
Ones |
|
72,34,560 |
7 |
2 |
3 |
4 |
5 |
6 |
0 |
|
62,31,315 |
|
|
|
|
|
|
|
|
86,45,234 |
|
|
|
|
|
|
|
Answer:
|
Number |
Ten Lakhs |
Lakhs |
Ten Thousand |
Thousand |
Hundreds |
Ten |
Ones |
|
72,34,560 |
7 |
2 |
3 |
4 |
5 |
6 |
0 |
|
62,31,315 |
6 |
2 |
3 |
1 |
3 |
1 |
5 |
|
86,45,234 |
8 |
6 |
4 |
5 |
2 |
3 |
4 |
3. Fill the entries in the blanks left
|
Number |
Ten Lakhs |
Lakhs |
Ten Th |
Th |
H |
T |
O |
Number Name |
Expansion |
|
8,25,431 |
- |
8 |
2 |
5 |
4 |
3 |
1 |
Eight Lakh twenty-five thousand four hundred thirty-one |
8×1,00,000+2×10,000+5×1,000+4×100+3×10+1×1 |
|
51,76,432 |
5 |
1 |
7 |
6 |
4 |
3 |
2 |
……………. |
|
Answer:
|
Number |
Ten Lakhs |
Lakhs |
Ten Th |
Th |
H |
T |
O |
Number Name |
Expansion |
|
8,25,431 |
- |
8 |
2 |
5 |
4 |
3 |
1 |
Eight Lakh twenty-five thousand four hundred thirty-one |
8×1,00,000+2×10,000+5×1,000+4×100+3×10+1×1 |
|
51,76,432 |
5 |
1 |
7 |
6 |
4 |
3 |
2 |
Fifty one lakh seventy-six thousand four hundred thirty-two |
5×1000000+1×100000+7×10000+6×1000+4×100+3×10+2×1 |
4. What are the place values of 7, 9 and 0 in the number 7, 91, 207.
Answer:
Place values of 7 = 7 lakhs
= 7× 100000
Place values of 9 = 9×10,000
= 90000
Place values of 0 = 0×10
= 00
5. Write the number name of 6, 42, 039 in words.
Answer: Six lakhs forty-two thousand thirty-nine
6. Write the expanded form of 39, 40, 444
Answer:
39, 40, 444 = 3×1000000+9×100000+4×10000+0×1000+4×100+4×10+4×1
7. Write the place value of 8, 6, 7 of the number 86, 47, 903 and write its number name.
Answer:
The place value of 8 = 8×1000000=80,00,000
The place value of 6 = 6×100000 = 6,00,000
The place value of 7 = 7×1000 = 7000
And,
86, 47, 903 = Eighty-six lakhs forty-seven thousand nine hundred three
8. Write 74, 09, 777 in expanded form.
Answer:
74, 09, 777 = 7×1000000+4×100000+0×10000+9×1000+7×100+7×10+7×1
(Let us include numbers up to crore as shown below)
|
Number |
Crores |
Ten Lakhs |
Lakhs |
Ten Th |
Th |
Hun |
Tens |
Ones |
Number Name |
|
8,90,20,120 |
8 |
9 |
0 |
2 |
0 |
1 |
2 |
0 |
Eight crore ninety lakh twenty thousand one hundred twenty |
|
7,53,07,608 |
|
|
|
|
|
|
|
|
|
Answer:
|
Number |
Crores |
Ten Lakhs |
Lakhs |
Ten Th |
Th |
Hun |
Tens |
Ones |
Number Name |
|
8,90,20,120 |
8 |
9 |
0 |
2 |
0 |
1 |
2 |
0 |
Eight crore ninety lakh twenty thousand one hundred twenty |
|
7,53,07,608 |
7 |
5 |
3 |
0 |
7 |
6 |
0 |
8 |
Seven crore fifty-three lakh seven thousand six hundred eight |
NCERT Solutions for Class 6 Chapter 1 Structure of Numbers
Page Number: 6
Let us do
1. Write the following numbers in the expanded form
(i) 8,90,20,120 = 8×10000000+9×1000000+0×100000+2×10000+ 0×1000 +1x100+2×10+0×1
(ii) 5,05,53,624=
(iii) 6,84,65,804=
(iv) 4, 30,26,532 =
Answer:
(i) 8,90,20,120 = 8×10000000+9×1000000+0×100000+2×10000+ 0×1000 +1x100+2×10+0×1
(ii) 5,05,53,624
= 5×10000000+0×1000000+5×100000+5×10000+3×1000+6×100+2×10+4×1
(iii) 6,84,65,804
= 6×10000000+8×1000000+4×00000+6×10000+5×1000+8×100+0×10+4×1
(iv) 4,30,26,532
= 4×10000000+3×1000000+0×100000+2×10000+6×1000+5×100+3×10+2×1
2. Write the number name of 8, 05, 64, 021.
Answer: Eight crore five lakh sixty-four thousand twenty-one
3. Write the place value of 5,9 and 2 for the number 5,93,20,067
Answer:
The place value of 5 = 5 Crores
= 5×10000000
= 500,00,000
The place value of 9 = 9 ten lakhs
= 9 × 10,00,000
= 90,00,000
The place value of 2 = 2 ten thousand
= 2 × 10,000
= 2 0,000
4. Complete the following
10 – 1 =…………………
100 – 1 =………..………....
1000 – 1 =………………….....
10000 – 1 =………………….......
100000 – 1 =………………………...
1000000 – 1 =…………………………...
10000000 – 1 =........................................
100000000 – 1 =..........................................
Answer:
10 – 1 = 9
100 – 1 = 99
1000 – 1 = 999
10000 – 1 = 9999
100000 – 1 = 99999
1000000 – 1 = 999999
10000000 – 1 = 9999999
100000000 – 1 = 99999999
NCERT Solutions for Class 6 Chapter 1 Structure of Numbers
Page Number: 7
Let us know
Do you know?
How many lakhs make one million?
Answer: Yes I know, one million is equal to 10 lakhs.
How many millions make one crore?
Answer: Yes I know, One crore is equal to 10 million.
NCERT Solutions for Class 6 Chapter 1 Structure of Numbers
Page Number: 9
Do it yourself
(a) Write in numbers by inserting commas (,) suitably.
(i) Seventy five lakh sixty nine thousand three hundred seven
Answer: 75,69,307
(ii) Nine crore eleven lakh twenty one thousand two hundred two.
Answer: 9,11,21,202
(iii) Fifty million eight hundred one thousand five hundred ninety two.
Answer: 50,801,592
(iv) Fifty eight million twenty six thousand two hundred two.
Answer: 58,026,202
(b) Insert commas suitably according to Indian System of Numeration-
(i) 97057201 (ii) 99990046 (iii) 98423107
Answer:
(i) 9,70,57,201 (ii) 9,99,90,046 (iii) 9,84,23,107
(c) Insert commas suitably according to International System of Numeration-
(i) 79821902 (ii) 99958102 (iii) 48094381
Answer:
(i) 79,821,902 (ii) 99,958,102 (iii) 48,094,381
(d) Write appropriate numbers in the blank space given below-
1,00,000= Ten Thousand = Hu= Tens
94,35,992= + + + + + +
Answer:
1,00,000= 10 Ten Thousand = 1000Hu= 10000 Tens
94,35,992= 90,00,000+4,00,000+ 30,000+5000+900+90+2
(e) Write the expanded forms-
(i) 2,76,36,708 (ii) 54,30,36,706
Answer:
(i) 2,76,36,708= 2×10000000+7×1000000+6×100000+3×10000+6×1000+7×100+0×10+8×1
(ii) 54,30,36,706= 5×100000000+4×10000000+3×1000000+0×100000+3×10000+6×1000+7×100+0×10+6×1
(f) Write the place value of the underlined digits given below in both Indian and International numeration system.
(i) 87456210 (ii) 789999 (iii) 60 200547
Answer:
(i) 87456210
Indian System
The place value of 8 = 8×10000000=8,00,00,000
The place value of 4 = 4×100000=4, 00,000
The place value of 2 = 2×100=200
International System
The place value of 8 = 8×10000000=80,000,000
The place value of 4 = 4×100000=400,000
The place value of 2 = 2×100=200
(ii) 789999
Indian System
The place value of 8 = 8×10000=80,000
The place value of 9 = 9×100=900
International System
The place value of 8 = 8×10000=80,000
The place value of 9 = 9×100=900
(iii) 60 200547
Indian System
The place value of 0 = 0×1000000=00,00,000
The place value of 2 = 2×100000=2,00,000
The place value of 5 = 5×100=500
International System
The place value of 0 = 0×1000000=0,000,000
The place value of 2 = 2×100000=200,000
The place value of 5 = 5×100=500
(g) Insert commas (,) suitably in the numbers (Indian and International system)
42613287 3000090 4654489 1145275
Answer:
(i) 42613287
Indian System = 4,26,13,287
International System = 42,613,287
(ii) 3000090
Indian System = 30,00,090
International System = 3,000,090
(iii) 4654489
Indian System = 46,54,489
International System = 4,654,489
(iv) 1145275
Indian System = 11,45,275
International System = 1,145,275
(h) How many digits are there in 10 crores?
Answer: There are 9 digits in 10 crore.
(i) What is the place value of 10 lakh in International System?
Answer: 10 lakhs = 1 million.
(j) Population of a city is 11528735. Out of them 1357632 are service holders and remaining are businessmen. How many people of the city are businessmen?
Solution:
Given,
Population of a city = 11528735
No of service holders = 1357632
No of businessmen = 11528735 – 1357632 = 10171103
∴ There are 10171103 businessmen in the city
(k) Cost of a bicycle is Rs. 5,300. What is the cost of 20 such bicycles?
Solution:
Given,
The cost of a bicycle is = 5,300.
∴ cost of 20 bicycles = 5300 × 20
= 106000
(l) Cost of one box of tea leaves is Rs. 5,000. What is the cost of 100 such boxes? Write the total cost in Indian System on Numeration.
Solution:
Given,
Cost of 1 box of tea leaves = 5000
∴ Cost of 100 box of tea leaves = 5000×100
= 500000
∴ The total cost in Indian System is Rs. 5, 00,000
(m) If the expenses of a person for a tour to Europe is Rs. 2,50,000 then what will be the total expenditure for a group of 5 persons. Write the total expenditure in International System of Numeration.
Solution:
Given,
The expenses of a person for a tour to Europe = Rs. 2,50,000
The total expenditure = 5 ×2,50,000
= 1250000
∴ The total expenditure in International System is Rs. 1,250,000
NCERT Solutions for Class 6 Chapter 1 Structure of Numbers
Page Number: 12-13
Let us estimate the sum (আনুমানিক যোগফল উলিয়াও আহা)
628+15820 (estimated to hundreds)
Firstly, in 628, 28→00 So, 628→ 600+00= 600
In 15820, 20→ 00 So, 15820→ 15800 + 00 = 15800
So, the estimated sum is 600+15800=16400
Actual sum is 628+15820=16448
Approximate value (আসন্নমান):
Example 1: To find the approximate value of 178 estimating to the nearest hundreds.
Solution:
178 = 100+78
50<78, So, 78→100
∴ Required approximate value of 178 = 100+ 100=200
Example 2: To find the approximate value of 223 estimating to the nearest hundreds.
Solution:
223 = 200+23
23<50, so, 23 →00
∴223 Required approximate value of 200+00=200
Example 3: To find the approximate value of 4973 estimating to the nearest hundreds.
Solution:
4973 = 4900+73
50<73, So, 73 →100
∴ Required approximate value of 4900+ 100=5000
Example 4: To find the approximate value of 3209 estimating to the nearest hundreds.
Solution:
3209 = 3200+9
9<50, So, 9→00
∴ Required approximate value of = 3200+00=3200
Example 5: To find the approximate value of 2560 estimating to the nearest thousands.
Solution:
2560 = 2000+ 560
500<560, So, 560→1000
∴ Required approximate value of 2000+ 1000=3000
Example 6: To find the approximate values of the following, estimating to the nearest thousands
15318, 14892, 16489
Solution:
15318 = 15000+318
318<500, So, 318 →000
∴ Required approximate value of 15000+ 000=15000
14892 = 14000+892
500<892, So, 892→1000
∴ Required approximate value of 14000+ 1000=15000
16489 = 16000+489
489<500, So, 489 →000
∴ Required approximate value of 16000+ 000=16000
Example 7: Find the approximate value of 3289 nearest to tens/hundred thousands
Solution:
For estimating to the nearest tens
3289 = 3280+9, 5<9, 9→10
So, the required approximate value nearest to the tens
3280+10=3290
For estimating to the nearest hundreds
3289=3200+89, 50<89, 89 →100
So, the required approximate value nearest to the hundreds
3200+100 = 3300
For estimating to the nearest thousands
3289 = 3000+289, 289<500, 289=000
So, the required approximate value of 3289 nearest to the thousands
3000+000 = 3000
NCERT Solutions for Class 6 Chapter 1 Structure of Numbers
Page Number: 14
Do it yourself
1. Match the following :
|
Number |
Nearest Approximate Value |
|
28 (nearest to tens) |
100 |
|
99 (nearest to tens) |
100 |
|
97 (nearest to tens) |
3200 |
|
243 (nearest to tens) |
370 |
|
351 (nearest to tens) |
30 |
|
367 (nearest to tens) |
4000 |
|
4050 (nearest to hundreds) |
240 |
|
3222 (nearest to hundreds) |
350 |
|
49,630 (nearest to thousands) |
900 |
|
889 (nearest to hundreds) |
50,000 |
|
2223 (nearest to thousands) |
2000 |
Solution:
|
Number |
Nearest Approximate Value |
|
28 (nearest to tens) |
30 |
|
99 (nearest to tens) |
100 |
|
97 (nearest to tens) |
100 |
|
243 (nearest to tens) |
240 |
|
351 (nearest to tens) |
350 |
|
367 (nearest to tens) |
370 |
|
4050 (nearest to hundreds) |
4000 |
|
3222 (nearest to hundreds) |
3200 |
|
49,630 (nearest to thousands) |
50,000 |
|
889 (nearest to hundreds) |
900 |
|
2223 (nearest to thousands) |
2000 |
2. Find the value of the following numbers nearest to their tens.
23, 7, 5, 4, 27, 39, 44, 273, 302, 5432, 1490, 1555
Solution:
|
Number |
Nearest Approximate Value of tens |
|
23 |
20 |
|
7 |
10 |
|
5 |
10 |
|
4 |
0 |
|
27 |
30 |
|
39 |
40 |
|
44 |
40 |
|
273 |
270 |
|
302 |
300 |
|
5432 |
5430 |
|
1490 |
1490 |
|
1555 |
1560 |
3. Find the value of the following numbers nearest to their hundreds.
410, 481, 9547, 5650, 4650, 49630, 69999, 799999
Solution:
|
Number |
Nearest Approximate Value of hundreds |
|
410 |
400+10 = 400 |
|
481 |
400+81 = 400+100 = 500 |
|
9547 |
9500+47 = 9500 |
|
5650 |
5600+50 = 5600+100 = 5700 |
|
4650 |
4600+50 = 4600+100 = 4700 |
|
49630 |
49600+30 = 49600 |
|
69999 |
69900+99 = 69900+100 = 70,000 |
|
799999 |
499900+99 = 799900+100 = 8,00,000 |
4. Find the value of the following numbers nearest to their thousands.
9562, 53554, 64347, 873412, 8694591
Solution:
|
Number |
Nearest Approximate Value of thousands |
|
9562 |
9000+562 = 9000+1000 = 10,000 |
|
53554 |
53000+554 = 53000+1000 = 54000 |
|
64347 |
64000+347 = 64000+000 = 64000 |
|
873412 |
873000+412 = 873000+000 = 873000 |
|
8694591 |
8694000+591 = 8694000+1000 = 8695000 |
NCERT Solutions for Class 6 Chapter 1 Structure of Numbers
Page Number: 15
Example 1: Compare the numbers 3872 and 2524
Solution:
Both the numbers have the same number of digits
The digit at the thousands place in 3872 is 3
The digit at the thousands place in 2572 is 2
As, 3>2
So, 3872 is greater than 2574
Example 2: Compare the numbers 5432 and 5368
Solution:
The digits at the thousands place are same in both is 5, 5
So, we move to the digits at the hundreds place
The digit at the hundreds place in 5432 is 4
The digit at the hundred place in 5368 is 3.
So, 4> 3
So, 5432 >5368 That is 5432 is greater than 5368
Example 3: Compare the numbers 3874 and 3872.
Solution:
Here, the digits at the thousands, hundreds and tens place are same in both.
So, we move to the digits at the ones place for both of them
As, 4>2
∴ 3874> 3872
That is, 3874 is greater
Do it yourself
1. Put ‘>' or ‘=’ or '<' sign in blank boxes.
(i) 38640 48694 (ii) 20643 22000
(iii) 100200 300100 (iv) 2000200 200002000
(v)27856145 27840561 (vi) 87456210 91024562
Solution:
(i) 38640 < 48694 (ii) 20643 < 22000
(iii) 100200 < 300100 (iv) 2000200 < 200002000
(v) 27856145 > 27840561 (vi) 87456210 < 91024562
2. Arrange the following in increasing order.
5245678, 4365710, 2378671, 4638690, 3676532, 4276549
Solution:
Here are the numbers arranged in increasing order:
2378671<3676532<4276549<4365710<4638690<5245678
3. Arrange the following in decreasing order.
6478546, 4756435, 2186458, 3653870, 7416543, 4739841
Solution:
Here are the numbers arranged in decreasing order:
7416543>6478546>4756435>4739841>3653870>2186458
NCERT Solutions for Class 6 Chapter 1 Structure of Numbers
Page Number: 16
Do it yourself:
1. A merchant earns Rs. 33,669 and Rs. 25,784 respectly from his two business in a year. Find his total estimate earning for the year.
33,669 rounds off to thousands = .
25,784 rounds off to thousands = .
__________________________________________________
∴ Total earning = .
Solution:
33,669 rounds off to thousands = 34000
25,784 rounds off to thousands = 26000
__________________________________________________
∴ Total earning = Rs. 60000
2. Let us find the estimated sum
(i) 7380 and 29,786 (ii) 525 and 995 (iii) 1274 and 1896
Solution:
(i) Given,
7380 and 29,786
7380 rounds off to thousands = 7000
29,786 rounds off to thousands = 30,000
Now, add the rounded numbers:
7,000 + 30,000 = 37,000
So, the estimated sum of 7380 and 29,786 is approximately 37,000.
(ii) Given,
525 and 995
525 rounds off to hundred = 500.
995 rounds off to hundred = 1000.
Now, add the rounded numbers:
500 + 1000 = 1500
So, the estimated sum of 525 and 995 is approximately 1500.
(iii) Given
1274 and 1896
1,274 rounds off to thousands = 1,000.
1,896 rounds off to thousands = 2,000.
Now, add the rounded numbers:
1,000 + 2,000 = 3,000
So, the estimated sum of 1,274 and 1,896 is approximately 3,000.
3. Let us find the estimated difference
(i) 38925 – 23473 (ii) 875 – 521 (iii) 697 – 521
Solution:
(i) 38925 – 23473
38925 rounds to off to thousands = 39000
23473 rounds to off to thousands = 23000.
Now, subtract the rounded numbers:
39000 - 23000 = 16000
So, the estimated difference is 16,000.
(ii) 875 – 521
875rounds to off to hundred = 900
521 rounds to off to hundred = 500.
Now, subtract the rounded numbers:
900 - 500 = 400
So, the estimated difference is 400.
(iii) 697 - 521
697 rounds to off to hundred = 700
521 rounds to off to hundred = 500.
Now, subtract the rounded numbers:
700 - 500 = 200
So, the estimated difference is 200.
NCERT Solutions for Class 6 Chapter 1 Structure of Numbers
Page Number: 17
To Estimate Products (পুৰণৰ মোটামুটি হিচাপ):
Example 1: Let us estimate for 218 × 68
Solution:
Approximate value of 218 nearest to hundreds = 200
Approximate value of 68 nearest to tens= 70
∴ estimated product = 200 × 70
= 14,000
Example 2: Estimate for 479 × 192
Solution:
Approximate value of 479 nearest to hundred = 500
Approximate value of 192 nearest to hundred = 200
∴ estimated product = 500×200
= 1, 00,000
Estimate the following
1. 5267+369 (Round off to tens/hundreds/thousands)
2. 30295+4320 (Round off to tens/hundreds/thousands)
3. 8999+7025 (Round off to tens/hundreds/thousands)
4. 72 × 13 (Round off to tens)
5. 60 × 21 (Round off to tens)
6. 111 × 11 (Round off to tens)
7. 119 × 29 (Round off to tens)
8. 349 × 59 (Round off to tens/hundreds)
Solution:
1. 5267+369 (Round off to tens/hundreds/thousands)
5267 rounds off to tens = 5270.
369 rounds off to tens = 370.
Now, add the rounded numbers:
5270 + 370 = 5640
∴ The estimated sum of 5267 and 369 is approximately 5640.
Again,
5267 rounds off to hundreds = 5300.
369 rounds off to hundreds = 400.
Now, add the rounded numbers:
5300 + 400 = 5700
∴ The estimated sum of 5267 and 369 is approximately 5700.
Again,
5267 rounds off to thousands = 5000.
369 rounds off to thousands = 000.
Now, add the rounded numbers:
5000 + 000 = 5000
∴ The estimated sum of 5267 and 369 is approximately 5000.
2. 30295+4320 (Round off to tens/hundreds/thousands)
30295 rounds off to tens = 30300.
4320 rounds off to tens = 4320.
Now, add the rounded numbers:
30300 + 4320 = 34620
∴ The estimated sum of 30295 and 4320 is approximately 34620.
Again,
30295 rounds off to hundreds = 30300.
4320 rounds off to hundreds = 4300.
Now, add the rounded numbers:
30300 + 4300 = 34600
∴ The estimated sum of 30295 and 4320 is approximately 34600.
Again,
30295 rounds off to thousands = 30000.
4320 rounds off to thousands = 4000.
Now, add the rounded numbers:
30000 + 4000 = 34000
∴ The estimated sum of 30295 and 4320 is approximately 34000.
3. 8999+7025 (Round off to tens/hundreds/thousands)
8999 rounds off to tens = 9000.
7025 rounds off to tens = 7030.
Now, add the rounded numbers:
9000 + 7030 = 16030
∴ The estimated sum of 8999 and 7025 is approximately 16030.
Again,
8999 rounds off to hundreds = 9000.
7025 rounds off to hundreds = 7000.
Now, add the rounded numbers:
9000 + 7000 = 16000
∴ The estimated sum of 8999 and 7025 is approximately 16000.
Again,
8999 rounds off to thousands = 9000.
7025 rounds off to thousands = 7000.
Now, add the rounded numbers:
9000 + 7000 = 16000
∴ The estimated sum of 8999 and 7025 is approximately 16000.
4. 72 × 13 (Round off to tens)
Approximate value of 72 nearest to tens = 70
Approximate value of 13 nearest to tens = 10
∴ estimated product = 70×10
= 700
5. 60 × 21 (Round off to tens)
60 rounds off to tens = 60.
21 rounds off to tens = 20.
∴ estimated product = 60×20
= 1200
6. 111 × 11 (Round off to tens)
111 rounds off to tens = 110.
11 rounds off to tens = 10.
∴ estimated product = 110×10
= 1100
7. 119 × 29 (Round off to tens)
119 rounds off to tens = 120.
29 rounds off to tens = 30.
∴ estimated product = 120×30
= 3600
8. 349 × 59 (Round off to tens/hundreds)
349 rounds off to tens = 350.
59 rounds off to tens = 60.
∴ estimated product = 350×60
= 2100
Again
349 rounds off to hundreds = 300.
59 rounds off to hundreds = 100.
∴ estimated product = 300×100
= 30000
Example 1:
Solution:
105 × 6 = (100+5) ×6 (It is called distribution of multiplication over addition)
= 100×6+5×6
= 600+30
= 630
26 × 103 = 26× (100+3)
= 26×100+26×3
= 2600+78
= 2678
NCERT Solutions for Class 6 Chapter 1 Structure of Numbers
Page Number: 18
Example 2: During summer vacation Chumu and Rumu works in their garden for 7 days, Chumu works 4 hours a day and Rumu 5 hours a day, How many hours do both of them work in 7 days?
Solution:
In one day Chumo works = 4 hours
In one day Rumu works = 5 hours
In one day both of them works = 4+5 = 9 hours
∴ They worked 7 day in all
= (4+5) ×7 hours
= 9 ×7 hours
= 63 hours
Alternatively,
Chumu works in one day = 4 hours
∴ Chumu works in 7 days = 4×7 hours
= 28 hours
Again Rumu works in one day = 5 hours
∴ Rumu works in 7 days = 5×7 hours
= 35 hours
∴ Total hours both of them worked
= (4×7+5×7) hours
= (28+35) hours
= 63 hours
NCERT Solutions for Class 6 Chapter 1 Structure of Numbers
Page Number: 19
Example 3: 305 × 105
Solution:
305 × 105 = (300+5) (100+5)
= 300 × (100+5) +5 (100+5)
= 300×100+ 300×5+5×100+ 5×5
= 30000 + 1500+ 500+25
= 32025
Solve using brackets (বন্ধনী ব্যৱহাৰ কৰি সমাধান কৰা)
(a) 5 × 19 (b) 7 × 108 (c) 16×102 (d) 102×103
Solution:
(a) 5 × 19 = 5×(10+9)
= 5×10+5×9
= 50+45
= 95
(b) 7×108 = 7×(100+8)
= 7×100+7×8
= 700+56
= 756
(c) 16×102 = 16×(100+2)
= 16×100+16×2
= 1600+32
= 1632
(d) 102×103 = (100+2)(100+3)
= 100× (100+3) +2× (100+3)
= 100×100+100×3+2×100+2×3
= 10000+300+200+6
= 10506
Roman Numerals (ৰোমান সংখ্যা)
|
Hindu-Arabic numerals |
Roman numerals |
|
1 |
I |
|
2 |
II |
|
3 |
III |
|
4 |
IV |
|
5 |
V |
|
6 |
VI |
|
7 |
VII |
|
8 |
VIII |
|
9 |
IX |
|
10 |
X |
|
50 |
L |
|
100 |
C |
|
500 |
D |
|
1000 |
M |
The rules for the Roman system are (ৰোমান প্ৰণালীত ব্যবহৃত নিয়মবোৰ):
1. A symbol is not repeated more than three times but the symbols V, L, D are never repeated.
2. If a symbol is repeated, its value is added as many times as it occurs.
For example:
III = 1+1+1= 3
XX = 10+10 = 20
3. If a symbol of smaller value is written to the right of a symbol of greeter value, its value gets added to the value of greater symbol.
For example:
VI = 5+1 = 6
XII = 10+1+1 = 12
4. If a symbol of smaller value is written to the left of a symbol of greater value, its value subtracted from the value of the greater symbol.
For example:
IX = 10 – 1 = 9
XC = 100 – 10 = 90
5. The symbol V, L, D is never written to the left of a symbol of greater value that is VX is not correct. The correct form is XV
6. The symbol I can be subtracted from V and X only. The symbol X can be subtracted from L, M and C only. The symbol C can be subtracted from D and M only.
7. If a symbol of smaller value is written in between the symbols of two greater values, its value is subtracted from the value of the greater symbol to the right
For example:
XIX = 10+ (10-1) =10+9 = 19
DIX = 500+ (10-1) = 500+9 = 509
8. If a horizontal line is drawn above the any of the symbol, the value of that symbol is increased by thousand times.
For example:
=
5×1000 = 5000
= 21×1,000 = 21000
NCERT Solutions for Class 6 Chapter 1 Structure of Numbers
Page Number: 20
Try These:
1. Write 49, 91, 206, 587, 1490, 1449, 2019 and 5000 in Roman numerals.
Solution:
49 = 40+ (10– 1)
= XLIX
91 = (100 – 10) +1
= XCI
206 = 100+100+5+1
= CCVI
587 = 500+50+10+10+10+5+1+1
= DLXXXVII
1490 = 1000+ (500-100) + (100-10)
= MCDXC
1449 = 1000+ (500-100) + (50-10) + (10-1)
= MCDXLIX
2019 = 1000+1000+10+(10-1)
= MMXIX
5000 = 5×1,000
=
Or
Here are the Roman numeral representations of the given numbers:
49 in Roman numerals is XLIX.
91 in Roman numerals is XCI.
206 in Roman numerals is CCVI.
587 in Roman numerals is DLXXXVII.
1490 in Roman numerals is MCDXC.
1449 in Roman numerals is MCDXLIX.
2019 in Roman numerals is MMXIX.
5000 in Roman numerals is V.
(Note that 5000 is represented by the letter "V" with a horizontal line or bar on top, which multiplies its value by 1,000.)
Please note that the traditional Roman numeral system doesn't have a way to represent the number zero, and it is primarily an additive system where numbers are formed by combining various symbols.
2. Write the following Roman numerals are Hindu-Arabic numerals.
XXX, XL, LX, LXX, XC, LXV, DCXXXV, MCMXII, MDCCCXLVI, LXIX, XCVIII
Solution:
XXX = 10+10+10 = 30
XL = 50 – 10 = 40
LX = 50+10 = 60
LXX = 50+10+10 = 70
XC = 100 – 10 = 90
LXV = 50 +10+5 = 65
DCXXXV = 500+100+10+10+10+5 = 635
MCMXII = 1000+ (1000-100) +10+1+1 =1912
MDCCCXLVI = 1000+500+100+100+100+(50-10)+5+1 = 1846
LXIX = 50+10+ (10-1) = 69
XCVIII = (100-10) +5+1+1+1 = 98
Or
Here are the given Roman numerals converted into Hindu-Arabic numerals:
XXX = 30
XL = 40
LX = 60
LXX = 70
XC = 90
LXV = 65
DCXXXV = 635
MCMXII = 1912
MDCCCXLVI = 1846
LXIX = 69
XCVIII = 98
NCERT Solutions for Class 6 Chapter 1 Structure of Numbers
Page Number: 21 – 22
Try These:
1. How many millimeters are there in 2 miters?
Solution:
We know that,
1 Meter = 1000 millimeter
∴ 2 meter = 2×1000 = 2000 millimeters
Or
There are 2,000 millimeters in 2 meters because 1 meter is equal to 1,000 millimeters.
2. Do 26 centimeters make 260 millimeters?
Solution:
We know that,
1 centimeters = 10 millimeters
∴ 26 centimeters = 26×10= 260 millimeters
Or
No, 26 centimeters do not make 260 millimeters. 26 centimeters are equal to 260 millimeters because there are 10 millimeters in 1 centimeter.
3. Express 13 kilograms in grams.
Solution:
We know that,
1 kilogram = 1000 grams
∴ 13 kilograms = 13×1000 grams = 13000 grams
Or
13 kilograms is equal to 13,000 grams because 1 kilogram is equal to 1,000 grams.
4. How many centimeters are there in 9 kilometers?
Solution:
We know that,
1 kilometer = 1000×100 = 100,000 centimeters
∴9 kilometer = 9×100000 = 900,000 centimeters
Or
There are 900,000 centimeters in 9 kilometers because 1 kilometer is equal to 100,000 centimeters.
5. How many centimeters are there in 5 millimeters?
Solution:
We know that,
10 millimeter = 1 centimetres
∴ 5 millimeter = 
× 5
= 0.5 centimetres
Or,
There are 500 centimeters in 5 millimeters because there are 10 millimeters in 1 centimeter, and 5 multiplied by 10 equals 50.
6.
Is 1 milligram =
centigram?
Answer: Yes.
7. How many meters are there in 2 meters 70 centimetres ?
Answer: 2.70 metres
8. 400 metres = kilometre
Answer:
400 metres = 0.4 kilometre
9. 25 litres = kilolitre
= millilitre
Answer:
25 litres = 0.025 kilolitre
= 25000 millilitre
**********************
NCERT Solutions for Class 6 Chapter 1 Structure of Numbers
Page Number: 22 – 25
Structure of Numbers (সংখ্যাৰ গঠন)
Exercise:
1. Write the number name of-
(a) 87595762 (b) 48049831
(Write in both Indian and International System of Numeration)
Solution:
(a) 87595762
Indian System of Numeration: Eight Crore Seventy-Five Lakh Fifty-Nine Thousand Seven Hundred Sixty-Two
International System of Numeration: Eighty-Seven Million Five Hundred Ninety-Five Thousand Seven Hundred Sixty-Two
(b) 48,049,831
Indian System of Numeration: Four Crore Eight Lakh Fourty-Nine Thousand Eight Hundred Thirty-One
International System of Numeration: Forty-Eight Million Forty-Nine Thousand Eight Hundred Thirty-One
2. Write the following numbers in both Indian and International System of Numeration by inserting commas suitably.
3018982, 82160000, 58042513, 34561897, 6015008
Solution:
Indian System of Numeration:
30,18,982; 8,21,60,000; 5,80,42,513; 3,45,61,897; 60,15,008
International System of Numeration:
3,018,982; 82,160,000; 58,042,513; 34,561,897; 6,015,008
3. According to 2011 census, population of Assam is 31168272. Using comma write this number in Indian System of Numeration appropriately write its number name. What will you write for this number according to International system?
Solution:
According to the 2011 census, the population of Assam is 31168272.
In the Indian System of Numeration, this number is written as follows:
Number: 31,168,272
Number Name: Three Crore Eleven Lakh Sixty-Eight Thousand Two Hundred Seventy-Two
In the International System of Numeration, this number remains the same:
Number: 31,168,272
4. Write in numerals-
(i) Seventy three lakh seventy five thousand three hundred seven.
(ii) Twenty three lakh thirty thousand ten
(iii) Three crore five lakh forty.
(iv) Seven crore forty seven lakh forty seven thousand four hundred forty seven.
(v) Five hundred sixty one thousand two hundred ninety.
(vi) Six million two hundred eighty eight thousand seven hundred ninety seven.
(vii) One million nine hundred fourteen thousand two hundred nine.
Solution:
(i) Seventy three lakh seventy five thousand three hundred seven.
Numerals: 73,75,307
(ii) Twenty three lakh thirty thousand ten.
Numerals: 23,30,010
(iii) Three crore five lakh forty.
Numerals: 3,05,00,040
(iv) Seven crore forty seven lakh forty seven thousand four hundred forty seven.
Numerals: 7,47,47,447
(v) Five hundred sixty one thousand two hundred ninety.
Numerals: 5,61,290
(vi) Six million two hundred eighty eight thousand seven hundred ninety seven.
Numerals: 6,288,797
(vii) One million nine hundred fourteen thousand two hundred nine.
Numerals: 19,14,209
5. In 1991, population of a city was 2, 35, 471 and in 2001, population was increased by 72, 950. What is the total population of that city in 2001?
Solution:
To find the total population of the city in 2001, you need to add the increase in population (72,950) to the population in 1991 (2,35,471).
Population in 2001 = Population in 1991 + Increase
Population in 2001 = 2,35,471 + 72,950
Population in 2001 = 3,08,421
So, the total population of the city in 2001 was 3,08,421.
6. Number of bicycles sold in a state in the year 2002-2003 was 7, 43,000 and in the year 2003-2004. 8,00,100. In which year more bicycles were sold and how many more bicycles were sold?
Solution:
Given,
Number of bicycles sold in 2002-2003= 7,43,000
Number of bicycles sold in 2003-2004= 8,00,100
Now, let's find the difference:
Difference = Number of bicycles sold in 2003-2004 - Number of bicycles sold in 2002-2003
Difference = 8,00,100 - 7,43,000
Difference = 57,100
So, in the year 2003-2004, more bicycles were sold, and the difference in the number of bicycles sold was 57,100.
7. A daily newspaper of a city consists of 12 pages. 11,980 copies of that news paper is printed every day. Find the total number of pages of the newspaper printed every day.
Solution:
Given,
Number of pages in each copy = 12 pages
Number of copies printed every day = 11,980 copies
Total number of pages printed every day = Number of pages in each copy × Number of copies printed
Total number of pages printed every day = 12 pages ×11,980 copies
Total number of pages printed every day = 143,760 pages
So, the total number of pages of the newspaper printed every day is 143,760 pages.
8. There are 75,000 sheets of paper for making note books. 4 pages of the note book can be made from a single piece of paper. Each notebook consists of 200 pages. How many note books can be made from 75,000 sheets of paper?
Solution:
The number of sheets of paper available for making notebooks = 75,000
Total number of pages available = 4 × 75,000
= 300000 pages
Given,
Each notebook contains 200 pages
Total no. of notebooks can be made = 300000/200
= 6000/2
= 1500 note books can be made.
9. Write the place value of the underlined digits of the following numbers.
(a) 45367 (b) 708453
(c) 27636708 (d) 614001
Solution:
(a) 45367
Place value of 5 is 5 thousand = 5×1000 = 5000
(b) 708453
Place value of 8 is 8 thousand = 8×1000 = 8000
(c) 27636708
Place value of 0 is 0 tens = 0×10 = 0
Place value of 7 is 7 ten lakhs = 7×1000000 = 7000000
(d) 614001
Place value of 1 is 1 ones = 1 and
Place value of 1 is 1 ten thousand = 1×10000 = 10000
10. Insert commas suitably according to the Indian System of Numeration and the fill the blank boxes using (>,=,<)
(a) 124054 2140589
(b) 24456122 34547892
(c) 90000001 9000001
(d) 1111111 1111111
Solution:
(a) 124054 < 2140589
(b) 24456122 < 34547892
(c) 90000001 > 9000001
(d) 1111111 = 1111111
11. Write the expanded form of the following-
(a) 792456 (b) 8000670 (c) 4279
(d) 5690 (e) 90007000
Solution:
(a) 792456 = 7×100000+9×10000+2×1000+4×100+5×10+6×1
(b) 8000670 = 8×1000000+0×100000+0×10000+0×1000+6×100+7×10+0×1
(c) 4279=4×1000+2×100+7×10+9×1
(d) 56090 =5×10000+6×1000+0×100+9×10+0×1
(e) 90007000 = 9×10000000+0×1000000+0×100000+0×10000+7×1000+0×100+0×10+0×1
12. Manisha purchased 12 kilogram 250 grams of sugar, 25 kilogram 750 grams of rice, 2 kilogram salt and 350 grams of cumin (Jeera) from a shop. What is the total weight of the items she purchesed?
Solution:
Given,
Manisha purchased Sugar = 12 kilograms 250 grams
Rice = 25 kilograms 750 grams
Salt = 2 kilograms
Cumin (Jeera) = 0 kilograms 350 grams
Total = 40 kilograms 350 grams
So, the total weight of the items Manisha purchased is 40.35 kilograms.
13. The capacity of a water tank is 200 kilolitres. On the first day 84 kilolitres 950 litres of water was drained out. How much water remained in the tank? (Express in litre)
Solution:
Given,
Total capacity of the tank = 200 kilolitres
Amount of water drained out on the first day = 84 kilolitres 950 litres
Remaining water = Total capacity - Amount drained out
Remaining water = 200,000 litres - (84,000 litres + 950 litres) (1 kilolitre = 1000 litres)
Remaining water = 200,000 litres - 84,950 litres
Remaining water = 115,050 litres
So, there are 115,050 litres of water remaining in the tank.
14. Find the approximate value of the following to the nearest places mentioned.
(i) 27, 41, 95, 38, 59, 72, 64, 135, 4931, 51630, 7325114 (nearest to tens)
Solution:
27 = 20+7
5<7, so, 7→ 10
∴ Required approximate value of 27 = 20+10 = 30
41 = 40+1
1<5, so, 1→ 0
∴ Required approximate value of 41 = 40+0 = 40
95 = 90+5
5 = 5, so, 5→10
∴ Required approximate value of 95 = 90+10 = 100
38 = 30+8
5<8, so, 8→ 10
∴ Required approximate value of 38 = 30+10 = 40
59 = 50+9
5<9, so, 9→ 10
∴ Required approximate value of 59 = 50+10 = 60
72 = 70+2
2<5, so, 2→ 0
∴ Required approximate value of 72 = 70+0 = 70
64 = 60+4
4<5, so, 4→ 0
∴ Required approximate value of 64 = 60+0 = 60
135 = 130+5
5 = 5, so, 5→10
∴ Required approximate value of 135 = 130+10 = 140
4931 = 1930+1
1<5, so, 1→ 0
∴ Required approximate value of 4931 = 4930+0 = 4930
51630 = 51630+0
0<5, so, 0→ 0
∴ Required approximate value of 51630 = 51630+0 = 51630
7325114 = 7325110+4
4<5, so, 4→ 0
∴ Required approximate value of 7325114 = 7325110+0 = 7325110
Or
Here are the values rounded to the nearest tens place:
27 is approximately 30.
41 is approximately 40.
95 is approximately 100.
38 is approximately 40.
59 is approximately 60.
72 is approximately 70.
64 is approximately 60.
135 is approximately 140.
4931 is approximately 4930.
51630 is approximately 51630 (already in tens).
7325114 is approximately 7325110.
So, the values rounded to the nearest tens place are as follows:
30, 40, 100, 40, 60, 70, 60, 140, 4930, 51630, 7325110.
(ii) 360, 720, 110, 690, 555,839, 281,2457,6666, 615320 (nearest to hundreds)
Solution:
360 = 300+60
50<60, so, 60→ 100
∴ Required approximate value of 360 = 300+100 = 400
720 = 700+20
20<50, so, 20→ 00
∴ Required approximate value of 720 = 700+00 = 700
110 = 100+10
10<50, so, 10→ 00
∴ Required approximate value of 110 = 100+00 = 100
690 = 600+90
50<90, so, 90→ 100
∴ Required approximate value of 690 = 600+100 = 700
555 = 500+55
50<55, so, 55→ 100
∴ Required approximate value of 555 = 500+100 = 600
839 = 800+39
39<50, so, 39→ 00
∴ Required approximate value of 839 = 800+00 = 800
281 = 200+81
50<81, so, 81→ 100
∴ Required approximate value of 281 = 200+100 = 300
2457 = 2400+57
50<57, so, 57→ 100
∴ Required approximate value of 2457 = 2400+100 = 2500
6666 = 6600+66
50<66, so, 66→ 100
∴ Required approximate value of 6666 = 6600+100 = 6700
615320 = 615300+20
20<50, so, 20→ 00
∴ Required approximate value of 615300 = 615300+00 = 615300
Or,
Here are the values rounded to the nearest hundreds place:
360 is approximately 400.
720 is approximately 700.
110 is approximately 100.
690 is approximately 700.
555 is approximately 600.
839 is approximately 800.
281 is approximately 300.
2457 is approximately 2500.
6666 is approximately 6700.
615320 is approximately 615300.
So, the values rounded to the nearest hundreds place are as follows:
400, 700, 100, 700, 600, 800, 300, 2500, 6700, 615300.
(iii) 8100, 1900, 4500, 9300, 6630, 5295, 4444, 61620 (nearest to thousands)
Solution:
8100 = 8000+100
100<500, so, 100→ 000
∴ Required approximate value of 8100 = 8000+000 = 8000
1900 = 1000+900
500<900, so, 900→ 1000
∴ Required approximate value of 1900 = 1000+1000 = 2000
4500 = 4000+500
500=500, so, 500→ 1000
∴ Required approximate value of 4500 = 4000+1000 = 5000
9300 = 9000+300
300<500, so, 300→ 000
∴ Required approximate value of 9300 = 9000+000 = 9000
6630 = 6000+630
500<630, so, 630→ 1000
∴ Required approximate value of 6630 = 6000+1000 = 7000
5295 = 5000+295
295<500, so, 295→ 000
∴ Required approximate value of 5295 = 5000+000 = 5000
4444 = 4000+444
444<500, so, 444→ 000
∴ Required approximate value of 4444 = 4000+000 = 4000
61620 = 61000+620
500<620, so, 620→ 1000
∴ Required approximate value of 61620 = 61000+1000 = 62000
Or
Here are the values rounded to the nearest thousands place:
8100 is approximately 8000.
1900 is approximately 2000.
4500 is approximately 5000.
9300 is approximately 9000.
6630 is approximately 7000.
5295 is approximately 5000.
4444 is approximately 4000.
61620 is approximately 62000.
So, the values rounded to the nearest thousands place are as follows:
8000, 2000, 5000, 9000, 7000, 5000, 4000, 62000.
(iv) 140000, 770000, 320000, 890000, 435000, 651230, 7325114, 64296045 (nearest to lakhs)
Solution:
140000 = 100000+40000
40000<50000, so, 40000→ 00000
∴ Required approximate value of 140000 = 100000+00000 = 100000
770000 = 700000+70000
50000<70000, so, 70000→ 100000
∴ Required approximate value of 770000 = 700000+100000 = 800000
320000 = 300000+20000
20000<50000, so, 20000→ 00000
∴ Required approximate value of 320000 = 300000+00000 = 300000
890000 = 800000+90000
50000<90000, so, 90000→ 100000
∴ Required approximate value of 890000 = 800000+100000 = 900000
435000 = 400000+35000
35000<50000, so, 35000→ 00000
∴ Required approximate value of 435000 = 400000+00000 = 400000
651230 = 600000+51230
50000<51230, so, 51230→ 100000
∴ Required approximate value of 651230 = 600000+100000 = 700000
7325114 = 7300000+25114
25114<50000, so, 25114→ 00000
∴ Required approximate value of 7325114 = 7300000+00000 = 7300000
64296045 = 64200000+96045
50000<96045, so, 96045→ 100000
∴ Required approximate value of 64296045 = 64200000+100000 = 64300000
(v) Approximate value of 7, 34, 21, 846 nearest to its crores.
Solution:
73421846 = 70000000+3421846
3421846<5000000, so, 3421846→ 0000000
∴ Required approximate value of 73421846 = 70000000+0000000 = 70000000
15.Write 5 numbers between 640 and 740 which can be approximated to 700 nearest to their hundreds.
Solution:
Certainly! Here are five numbers between 640 and 740 that can be approximated to the nearest hundred as 700
664, 678, 711, 725, 739
16. Find the approximate value of 157249 nearest to tens. Then find the approximate value of the new number nearest to thousands. Also find the approximate value of the new number found nearest to lakhs.
Solution:
The number 157249 rounded to the nearest tens is 157250.
∴ required value of the new number = 157250
The number 157250 rounded to the nearest thousands is 157000.
The number 157000 rounded to the nearest lakhs is 200000.
So, the approximate values are as follows:
Nearest tens = 157250
Nearest thousands = 157000
Nearest lakhs = 200000
17. Write the following numbers in Roman Numerals
(a) 225 (b) 600 (c) 906 (d) 598 (e) 56
(f) 555 (g) 53 (h) 69 (i) 999
Solution:
18. Write the following Roman Numerals in Hindu Arabic system.
(a) CMXCIV (b) CLXX (c) CDX (d) DCXLIX (e) CCCXXV
(f) DCCCLXXXVIII (g) XLV (h) XXIII (i) XCVIII (j) CCLVI
Solution:
19. Express the sum in Indian system of Numeration.
(a) (7296325+4173243)+45639
(b) (7281639+568548) + 4251408
(c) 4253798+61271037+3821570
Solution:
20. Express the product in Indian System of Numeration.
(i) 6250×625 (ii) 40003×203
(iii) 39625×75 (iv) 48062×243
Solution:
21. Estimate the sums.
(a) 6290+19986 (b) 108734+47599
(c) 5672 +536 (d) 439 +334+4317
(e) 7245506+1068297 (f) 6404759+5869422
Solution:
22. Estimate the difference
(a) 95 – 91
(b) 439 – 334
(c) 798 – 232
(d) 8325 – 48365
(e) 489348 – 48365
(f) 107634 – 67599
(g) 3638775 – 540060
Solution:
23. Estimate the products-
(a) 378×62
(b) 578×61
(c) 7250×29
(d) 24725×31
(e) 59986×245
(f) 360345×3456
Solution:
24. A man bought a piece of land for Rs.1263501 to build a house. He spent some amount of money to build the house. His total expenditure was Rs. 1860257. Find the amount of money he spent for building the house.
Solution:
25. In a book shop 32307 books were sold in first six months of the year 2011 and 22690 books in next six months. Find the total amount of books sold in 2011? Compare the total number of books with the number of books sold in each six months by approximating their values nearest to thousands.
Solution:
26. There are 226458 books in your school library. During the first week of the begining of the year 49957 books were borrowed by the students. How many books were left in the library? (Compare the number of books by finding their approximate values of nearest to thousands).
Solution:
27. A motor cycle dealer has 85 motor cycles in his shop. Cost of a motor cycle is Rs. 57690, find the total cost of 85 motor cycles.
Solution:
28. Cost of a bus ticket from Guwahati to Jorhat is Rs. 403. If the total cost of all the tickets against the seats (আসন) of the bus is Rs. 24180, find the total number of seats on the bus.
Solution:
29. A person travelled 13 kilometres by bicycle then 44 kilometres by bus and remaining 5 kilometres on foot to reach his destination. Find the total distance covered by him.
Solution:
30. A shopkeeper sells 25 kg of rice at the rate of Rs.35 per kg on the first day 40 kg of rice at the rate of Rs.30 per kg on the second day and 30 kg of rice at the rate of Rs. 25 per kg on the three days respectively. Find the total quantity of rice sold during three days along with the rate at which the rice was sold. Find the total quantity of rice sold by him in three days with what amount of money?
Solution:
31. There are 40 students in a school. Find how many litres of milk is required if 250 millilitres of milk is distributed to each of the students.
Solution:
32. A milkman (গুৱাল) distributes milk in hotels everyday. In first hotel he distributes 13 litres, in second he 15 litres 250 milliliter and in the third one 17 litres 750 millilitres. Find how many litres of milk the milkman distributes everyday.
Solution:
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