WBBSE Solutions For Class 6 Maths Chapter 2 Concept Of Seven And Eight Digit Numbers

Class 6 Math WBBSE Solutions Chapter 2 Concept Of Seven And Eight Digit Numbers

Example: Population of Howrah District 48,41,638 Population of Kolkata District 44, 86, 679 Population of Howrah District
Solution:

48 lac 41 thousand 6 hundred thirty-eight

Population of Kolkata District

Forty-four lac eighty-six thousand six hundred seventy-nine

Population of Purulia District 29,27, 975

Population of Jalpaiguri District 38,69,675

Read and Learn More WBBSE Solutions For Class 6 Maths

Population of Purulia District

29 lac 27 thousand 9 hundred seventy-five

Population of Jalpaiguri District,

38 lac 69 thousand 9 hundred seventy-five

Let’s write a seven-digit number and expand it.

Using each digit only once, let us write four numbers of seven digits and then write the numbers in words.

Question 1. With the help of sticks and colored balls, a number of children who took the Polio Vaccine in four districts are shown below.
Solution:

28,22,000 (approx) children in North 24-Parganas

26,10,500 (approx.) children in South 24 – Parganas.

24,12,300 (approx.) children in Kolkata

Class 6 Math WBBSE Solutions

22,34,200 (approx.) children in Kolkata

Total number of children from Kolkata and Howrah who were given the Polio Vaccine.

Total number of children from 24 – Parganas, North, South, Kolkata, and Howrah, who all got Polio Vaccine.

Class 6 Math WBBSE Solutions

In Ten Lac stick, more than 10 balls can not be placed. Hence another strict or ‘place’ is needed. The name of this stick or ‘Place’ is “Crore”(c). Instead of 10 orange balls, I take 1□ ball.

Let us put balls in respective sticks and write the number:

1. One crore eleven lac six hundred five
2. Five crore two lac eigth thousand sixty-nine

Question 4. In the morning, people of our locality are very busy. They have queued up before our school to cast a vote. I could not do so, because my age is less than 18 years. I later came to know the results of the election.
Solution:

Party A got 46, 87, 905 votes

Party B got 44, 50, 896

Party C got 42, 95, 739 votes

The total number of votes parties have got is 13, 4, 34, 540 votes

Total number of voters are 135, 07, 375 in this locality

Total number of voters 135,07, 375

Votes that are cast 134,34, 540

Votes that are not cast or canceled = 72,835

Let us find which party won the election and by how many votes.

Class 6 Math WBBSE Solutions

Out of 46, 87, 905; 44, 50, 896, and 42, 95, 739, 46, 87, 905 is the greatest number.

Hence, Party A won the elections.

Let us find how many votes Party A got more than Party B.

Party A got 46, 87, 905 Party B got 44, 50, 896 Party A won by 2, 37, 009

WBBSE Class 6 Maths Solutions Chapter 2 Concept Of Seven And Eight Digit Numbers Exercise 2

Question 1. 30439872 ÷ 516
Solution:

⇒ \(\text { 516) } 30439872(58992 \\
2580 \\
4639 \\
\underline{4128} \\
5118 \\
\frac{4644}{4747} \\
\frac{4644}{1032} \\
\frac{1032}{\mathrm{x}}\)

Answer: 58992

Question 2. ▭ x 32=75285600
Solution: Required number =75285600 ÷ 32

⇒ \(\text { 32) } 75285600 \text { ( } 2352675 \\
\frac{64}{112} \\
\frac{96}{168} \\
\frac{160}{85} \\
\frac{64}{216} \\
\frac{192}{240} \\
\frac{224}{160} \\
\frac{160}{x}\)

Answer: 2352675

Question 3. 9128088 ÷ 388=
Solution:

⇒ \(\begin{aligned}
& 388) 9128088(23526 \\
& \frac{776}{1368} \\
& \frac{1164}{2040} \\
& \frac{1940}{1008} \\
& \frac{776}{2328} \\
& \frac{2328}{x}
\end{aligned}\)

Answer: 23526

Question 4. ▭ ÷ 297=65833
Solution: Required number = 65833 x 297

⇒ \(\begin{array}{r}
& 65833 \\
& 297 \\
& 460831 \\
& 592497 x \\
& 131666 x x \\
& \hline 19552401
\end{array}\)

WBBSE Class 6 Maths Solutions

Question 5. 30,08,00000 ÷ ▭=2,00000
Solution:

Required number =300800000 ÷ 200000 =1504

⇒ \(\begin{aligned}
& 200000) 300800000(1504 \\
& \frac{200000}{1008000} \\
& \frac{1000000}{800000} \\
& \frac{800000}{x}
\end{aligned}\)

Question 6. 752 x ▭ =3556208
Solution:

Required number =3556208 ÷ 752 =4729

⇒ \(\begin{aligned}
& \text { 752) } 3556208(4729 \\
& \frac{3008}{5482} \\
& \frac{5264}{2180} \\
& \frac{1504}{6768} \\
& \frac{6768}{x}
\end{aligned}\)

Class 6 Maths Solutions WBBSE Chapter 2 Concept Of Seven And Eight Digit Numbers Exercise 2.1

Question 1. Let us write in words

1. 782005
Solution: Seven lacs eighty-two thousand five

2. 4207029
Solution: Forty-two lacs seven thousand twenty-nine.

3. 30030030
Solution: Three crores thirty thousand thirty

4. 50505005
Solution: Five crores five lacs five thousand five

WBBSE Class 6 Maths Solutions

5. 42034047
Solution: Four crores twenty lacs thirty-four thousand forty-seven

Question 2. Let’s write these numbers using numerals

1. Seventy-eight lac eight hundred eight
Solution: 7800808

2. Ninety-three lac forty-four thousand six hundred five
Solution: 9344605

3. Three crore three lac three thousand three hundred three
Solution: 30303303

4. Thirty-three crore thirty-three lac thirty-three thousand thirty-three.
Solution: 333333033

5. Seventy-seven crore seven thousand seven
Solution: 770007007

Question 3. Let’s match the two sides

Solution: 1. D, 2. C, 3. A, 4. E, 5. B

Question 4. Let us write by choosing the correct answer

1. Twenty lac ten thousand eight

1. 2001008
2. 2010008
3. 2100008

Solution: 2. 2010008

WBBSE Class 6 Maths Solutions

2. One crore eleven lac eight thousand forty-one 

1. 11018041
2. 11010841
3. 11108041

Solution: 3. 11108041

3. Two crore three lac sixty thousand five hundred twenty-six

1. 20360526
2. 20365026
3. 20360562

Solution: 1. 20360526

Question 5. Let us write the expanded from of the following numbers

1. 4627593
Solution: 4000000 + 600000 + 20000 + 7000 + 500 + 90 + 3

2. 2213101
Solution: 2000000 + 200000 + 10000 + 3000 + 100 + 1

3. 9999999
Solution: 9000000 + 900000 + 90000 + 9000 + 900 + 90+9

4. 7007007
Solution: 7000000 + 7000 + 7

5. 2406739
Solution: 2000000 + 400000 + 6000 + 700 + 30 + 9.

Question 6. In the number 37452129, let us find the difference between the place values of the digit ‘2’.
Solution:

In the number 37452129,

Place values of 2 are 2000 and 20.

∴ The difference between 2000 and 20 is (2000-20) = 1980.

Question 7. In the number 27946138, let us find the difference between the place value and the actual value of ‘9′
Solution:

In the number 27946138, the place value of 9 is 900000.

∴ The difference between 900000 and 9 is (900000 – 9) = 899991.

Question 8. Let’s form greatest and least numbers of 8 digits with the digits given below:

1. 3, 5, 7, 9, 2, 6, 5, 6
Solution:

Greatest number = 9 7 6 6 5 5 3 2

Least number = 2 3 5 5 6 6 7 9

2. 6, 4, 8, 5,1,2, 0, 3
Solution:

Greatest number — 8 6 5 4 3 2 1 0

Least number- 1 0 2 3 4 5 6 8

3. 7, 3, 2,1, 9, 5, 6, 0
Solution:

Greatest number – 9 7 6 5 3 2 1 0

Least number – 1 0 2 3 5 6 7 9

4. 8, 9, 2, 4, 7, 3, 2, 1
Solution:

Greatest number – 9 8 7 4 3 2 2 1

Least number – 1 2 2 3 4 7 8 9

Class 6 Math Solutions WBBSE

Question 9. Let us write the following numbers in ascending (increasing) order:

1. 7525762,7525662,7526762,7525652
Solution: 7525652 < 7525662 < 7525762 < 7526762

2. 8705321,8702358, 8707341,8703741
Solution: 8705321 <8703741 <8705321 <8707341

3. 518896,872300, 27562, 300252
Solution: 27562 < 300252 < 518896 < 872300.

Question 10. Let us write the following numbers in descending (decreasing) order:

1. 4503210,4503201,4503120, 4502210
Solution: 4503210 > 4503201 > 4503120 > 4502210

2. 301516,8640051,302560, 6352289
Solution: 8640051 > 6352289 > 302560 >301516

3. 5102080, 5108200, 5100280, 5182000
Solution: 5182000 > 5108200 > 5102080 > 5100280

Question 11. The sum of two numbers is 82945195. If one number is 69100278, let us find the other number.
Solution: Other number = 82945195 – 69100278 = 13844917

Question 12. The difference between two numbers is 28351036. If one number is 30529179, let us find the other number.
Solution: Other number = 30529179 + 28351036 = 58880215

Question 13. From a factory in Bakultala, last year there was an income of Rs. 7521200. This year there is an additional earning of Rs. 3250325. Let us find the total earning in two years.
Solution:

The total earning in two years

= Rs. 7521200 + Rs. 3250325

= Rs. 10771525

Class 6 Math Solutions WBBSE

Question 14. The product of the numbers is 15050490. If one number is 5, let us find the other number.
Solution: Other number = 15050490 ÷ 5 = 3010098.

Question 15. By selling certain property Samirbabu got Rs. 35629850. He gave Rs. 1006200 to his wife and divided Rs. 13050000 among his three children. Rent of the money he donated for constructing a village school. Let us find
Solution:

1. Each child got = 13050000 ÷ 3 = 4350000

2. He donated for the school

= Rs. 35629850 – (1006200 + 13050000)

= Rs. (35629850-23112000)

= Rs. 12517850

Question 16. The population of a city is two crore ninety-eight lac seventy-two thousand six hundred. Of them 12500500 are men and 8872435 are women. Let us calculate the number of children.
Solution:

Total population = 29872600

⇒ \(\begin{array}{ll}
\text { Number of men } & =12500500 \\
\text { Number of women } & \equiv 8872435 \\
\hline \text { Total number of men and women } & \equiv 21372935
\end{array}\)

∴ Number of children = 29872600 – 21372935 = 8499665

Question 17. Let us find what least number must be added to 234567 so that it is divisible by 835.
Solution:

⇒ \(\begin{aligned}
& 835) 234567(280 \\
& \frac{1670}{6756} \\
& \frac{6680}{767}
\end{aligned}\)

∴ Required number = 835 – 767 = 68

Question 18. An organization bought two paintings of an artist for Rs. 900000 and Rs. 2000000 and two more paintings of another artist for Rs. 30,21636 and Rs. 1761084. Let us find how much has been spent on paintings by the organization.
Solution:

Total price of 4 paintings

= Rs. 900000 + Rs. 2000000 + Rs. 3021636 + Rs. 1761084

= Rs. 7682720

∴ Organisation spent on paintings = Rs. 7682720

Class 6 Math Solutions WBBSE

Question 19. The area of a country is 3287263 sq. km. It has 754740 sq. km of forest area and 2503000 sq. km area is river basin. Let us find the area of the remaining land leaving forest and river basin areas.
Solution:

Total area of the country = 3287263 sq. km.

Area of forest = 754740 sq. km.

and area of river basin = 2503000 sq. km.

Total area of forest = 3257740 sq. km.

and river basin

Remaining Land = (3287263 – 3257740) sq. km = 29523 Sqkm.