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Preparation for Algebra Class 5 Problem Set 55 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 16 Preparation for Algebra Problem Set 55 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 16 Preparation for Algebra

Question 1.
Say whether right or wrong.

(1) (23 + 4) = (4 + 23)
Answer:
27 = 27 is right

(2) (9 + 4) > 12
Answer:
13 > 12 is right

(3) (9 + 4) < 12
Answer:
13 < 12 is wrong

(4) 138 > 138
Answer:
Wrong

(5) 138 < 138
Answer:
Wrong

(6) 138 = 138
Answer:
right

(7) (4 × 7) = 30 – 2
Answer:
28 = 28 is right

(8) \(\frac{25}{5}\) > 5
Answer:
5 > 5 is wrong.

(9) (5 × 8) = (8 × 5)
Answer:
40 = 40 is right

(10) (16 + 0) = 0
Answer:
16 + 0
= 16
16 = 0 is wrong

(11) (16 + 0) = 16
Answer:
16 = 16 is right.

(12) (9 + 4) = 12
Answer:
13 = 12 is wrong.

Question 2.
Fill in the blanks with the right symbol from <, > or =.

(1) (45 ÷ 9) [ ] (9 – 4)
Answer:
45 ÷ 9 = 5,
9 – 4 = 5 5
= 5
so, (45 + 9) = (9 – 4)

(2) (6 + 1) [ ] (3 × 2)
Answer:
6 + 1 = 7,
3 x 2 = 6
7 > 6
so, (6 + 1) > (3 x 2)

(3) (12 × 2) [ ] (25 + 10)
Answer:
12 x 2 = 24,
25 + 10 = 35
24 < 35
so, (12 x 2) < (25 + 10)

Question 3.
Fill in the blanks in the expressions with the proper numbers.

(1) (1 × 7) = ( [ ] × 1)
Answer:
1 x 7 = 7,
7 x 1 = 7
so, (1 x 7) = ( 7 x 1)

(2) (5 × 4) > (7 × [ ] )
Answer:
5 x 4 = 20, 7 x ………… must be less than 20.
7 x 2 = 14
so, (5 x 4) > ( 7 x 2)

(3) (48 ÷ 3) < ( [ ] × 5)
Answer:
48 – 3 = 16,
5 x 4 = 20
5 x 3 = 15
16 > 15 and 16 < 20 so, (48 + 3) <(4 x 5)

(4) (0 + 1) > (5 × [ ] )
Answer:
0 + 1 = 1,
5 x 1 = 5
5 x 0 = 0
1 < 5 and 1 > 0 so, (0 + 1) > (5 x Q)

(5) (35 ÷ 7) = ( [ ] + [ ] )
Answer:
35 ÷ 7 = 5,
3 + 2 = 5 so, (35 + 7) = (3 + 2)

(6) (6 – [ ] ) < (2 + 3)
Answer:
6 – < 2 + 3 = 5
5 > 6 – 2
so, (6 – 2) < (2 + 3)

Using letters
Symbols are frequently used in mathematical writing. The use of symbols makes the writing very short. For example, using symbols, ‘Division of 75 by 15 gives us 5’ can be written in short as ‘75 ÷ 15 = 5’. It is also easier to grasp.

Letters can be used like symbols to make our writing short and simple.

While adding, subtracting or carrying out other operations on numbers, you must have discovered many properties of the operations.

For example, what properties do you see in sums like (9 + 4), (4 + 9)?

The sum of any two numbers and the sum obtained by reversing the order of the two numbers is the same.

Now see how much easier and faster it is to write this property using letters.

• Let us use a and b to represent any two numbers. Their sum will be ‘a + b’.

Changing the order of those numbers will make the addition ‘b + a’. Therefore, the rule will be : ‘For all values of a and b, (a + b) = (b + a).’

Let us see two more examples.

• Multiplying any number by 1 gives the number itself. In short, a × 1 = a.
• Given two unequal numbers, the division of the first by the second is not the same as the division of the second by the first.

In short, if a and b are two different numbers, then (a ÷b) ≠ (b ÷a).

Take the value of a as 8 and b as 4 and verify the property yourself.

Preparation for Algebra Problem Set 54 Additional Important Questions and Answers

Say whether right or wrong.

(1) (15 ÷ 3) =5
Answer:
5 = 5 is right.

(2) (2 x 1) = 1
Answer:
2 = 1 is wrong.

(3) (16 ÷ 8) = (2 x 2)
Answer:
2 = 4 is wrong.

(4) (13 – 7) = 6
Answer:
6 = 6 is right.

(5) (1 x 0) = 1
Answer:
1 = 1 is wrong.

(6) (1 + 0) = 1
Answer:
1 = 1 is right.

Fill in the blanks with the right symbol from <, >, or =.

(1) (12 + 6) (10 X 2)
Answer:
12 + 6 = 18,
10 x 2 = 20
18 < 20
so, (12 + 6) < (10 x 2)

(2) (4 X 5) (10 X 2)
Answer:
4 x 5 = 20,
10 x 2 = 20
20 = 20
so, (4 x 5) = (10 x 2)

(3) (7 + 3) ………….. (3 X 3)
Answer:
7 + 3 = 10,
3 x 3 = 9
10 > 9
so, (7+ 3) > (3 x 3)

Fill in the blanks in the expressions with the proper numbers.

(1) (8 + ………….. ) = (8 x 1)
Answer:
8 + ……………. = 8,
8 x 1 = 8
8 + 0 = 8
so, (8 + 0) = (8 x 1)

(2) (5 x 6) > (14 x ……….. )
Answer:
5 x 6 = 30,
14 x 1 = 14
14 x 2 = 28
14 x 3 = 42
30 >28
so, (5 x 6) >(14 x 2)

(3) (6 X 7) < ( x 5)
Ans.
6 x 7 = 42,
9 x 5 = 45
42 < 45, 50, 55
so, (6 x 7) < (9 x 5)

Class 5 Maths Solution Maharashtra Board

• Three Dimensional Objects and Nets Problem Set 51 Class 5 maths Solutions
• Pictographs Problem Set 52 Class 5 maths Solutions
• Patterns Problem Set 53 Class 5 maths Solutions
• Preparation for Algebra Problem Set 54 Class 5 maths Solutions
• Preparation for Algebra Problem Set 55 Class 5 maths Solutions
• Preparation for Algebra Problem Set 56 Class 5 maths Solutions

Addition and Subtraction Class 5 Problem Set 9 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 9 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 3 Addition and Subtraction

Solve the following problems.

Question 1.
In a certain election, 13,47,048 women and 14,29,638 men cast their votes. How many votes were polled altogether?
Solution:
1 3 4 7 0 4 8 Women votes
+
1 4 2 9 6 3 8 Men votes
2 7 7 6 6 8 6 Total votes

Answer:
Altogether 27,76,686 votes were polled.

Question 2.
What will be the sum of the smallest and the largest six-digit numbers?
Solution:
1 0 0 0 0 0 Smallest six-digit No.
+
9 9 9 9 9 9 Largest six-digit No.
1 0 9 9 9 9 9 Total of six-digit No.

Answer:
Altogether 10,99,999 six-digit numbers.

Question 3.
If Surekhatai bought a tractor for ₹ 8,07,957 and a thresher for ₹ 32,609, how much money did she spend altogether?
Solution:
8 0 7 9 5 7 Tractor
+
3 2 6 0 9 Thresher
8 4 0 5 6 6 Total

Answer:
Surekhatai spend ₹ 8,40,566 altogether.

Question 4.
A textile mill produced 17,24,938 metres of cloth last year and 23,47,056 metres this year. What was the total production for the two years?
Solution:
1724938 m. prod, last year
+
2347056 m. prod, this year
4071994 m. prod, in 2 years

Answer:
40,71,994 metres was the total production for the two years.

Question 5.
If the Government gave ₹ 34,62,950 worth of computers and ₹ 3,26,578 worth of TV sets to the schools, what is the total amount it spent on this equipment?
Solution:
3 4 6 2 9 5 0 ₹ Computers
3 2 6 5 7 8 ₹ TV sets
3 7 8 9 5 2 8 Total ₹

Answer:
Total amount spent on equipments is ₹ 37,89,528

Subtraction : Revision

Study the following example.

Last year, 38,796 students took a certain exam. This year the number was 47,528. How many more students took the exam this year?

8,732 more students took the exam this year.

Addition and Subtraction Problem Set 9 Additional Important Questions and Answers

Question 1.
Solve the following problems.

(1) Goods of ₹ 14,08,345 was sold on last month and goods of ₹ 15,16,178 sold this month. What is the total sale of goods in these two months?
Solution:
1 4 0 8 3 4 5 Sale of last month
+
1 5 1 6 1 7 8 Sale in this month
2 9 2 4 5 2 3 Total sale of goods

Answer:
Total sale of good is ₹ 29,24,523

(2) Sundar purchased the flat of ₹ 28,15,000 and a Car of ₹ 12,05,500. What is the total amount spent by him?
Solution:
2 8 1 5 0 0 0 Cost of flat
1 2 0 5 5 0 0 Cost of Car
4 0 2 0 5 0 0 Total cost

Answer:
Total cost of ₹ 40,20,500

Class 5 Maths Solution Maharashtra Board

• Addition and Subtraction Problem Set 7 Class 5 Maths Solutions
• Addition and Subtraction Problem Set 8 Class 5 Maths Solutions
• Addition and Subtraction Problem Set 9 Class 5 Maths Solutions
• Addition and Subtraction Problem Set 10 Class 5 Maths Solutions
• Addition and Subtraction Problem Set 11 Class 5 Maths Solutions
• Addition and Subtraction Problem Set 12 Class 5 Maths Solutions
• Addition and Subtraction Problem Set 13 Class 5 Maths Solutions

Number Work Class 5 Problem Set 3 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 2 Number Work Problem Set 3 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 2 Number Work

Question 1.
Read the numbers and write them in words.
(1) 7,65,234
(2) 4,73,225
(3) 3,27,001
(4) 8,75,375
(5) 1,50,437
(6) 2,03,174
(7) 6,47,851
(8) 9,00,999
(9) 5,75,010
(10) 4,03,005
Answer:
(1) Seven lakh, sisxty-five thousand, two hundred and thirty-four.
(2) Four lakh, seventy-three thousand, two hundred and twenty-five.
(3) Three lakh, twenty-seven thousand and one
(4) Eight lakh seventy-five thousand three hundred and seventy-five
(5) One lakh fifty thousand four hundred and thirty seven
(6) Two lakh three thousand one hundred and seventy-four
(7) Six lakh forty seven thousand eight hundred and fifty-one
(8) Nine lakh nine hundred and ninety-nine
(9) Five lakh seventy-five thousand and ten
(10) Four lakh three thousand and five.

Question 2.
Read the numbers and write them in figures.
(1) One lakh thirty-five thousand eight hundred and fifty-five
(2) Seven lakh twenty-seven thousand
(3) Four lakh twenty-five thousand three hundred
(4) Nine lakh nine thousand ninety-nine
(5) Seven lakh forty-nine thousand three hundred and sixty-two
(6) Eight lakh one sixty tow
Answer:
(1) 1,35,008
(2) 7,27,1 55
(3) 4,25,003
(4) 9,09,099
(5) 7,49,003
(6) 8,00,162

Question 3.
Make five six-digit numbers, each time using any of the digits 0 to 9 only once.
Answer:

• 4,09,138
• 3,17,045
• 1,20,645
• 9,72,860
• 6,54,302

Introducing seven-digit numbers
Teacher : Now we shall learn about seven-digit numbers. Suppose 10 farmers borrow ₹ 1,00,000 each from a Co-operative Bank. Then, how much is the total loan given by the bank to them?

Ajit : We must find out what is ten times 1,00,000. That is, we multiply 1,00,000 by 10. That means we write one zero after the number to be multiplied.

Ajay : 1,00,000 × 10 = 10,00,000

Teacher : This becomes a seven-digit number. We read it as ‘Ten lakh’. We must make one more place for the 10 lakhs to the left of the lakhs place. In western countries, the term million is used. One million is equal to ten lakhs.

Thus, ten lakh = 10,00,000.

Just as we read ten thousands and thousands together, we read ten lakhs and lakhs together. So, we read 18,35,614 as ‘eighteen lakh, thirty-five thousand, six hundred and fourteen.

Study the seven-digit numbers given below in figures and in words.

• 31,25,745 : thirty-one lakh, twenty-five thousand, seven hundred and forty-five
• 91,00,006 : ninety-one lakh and six
• 63,00,988 : sixty-three lakh, nine hundred and eighty-eight
• 88,00,400 : eighty-eight lakh, four hundred
• seventy-two lakh and ninety-five : 72,00,095
• seventy lakh, two thousand, three hundred : 70,02,300

Roman Numerals Problem Set 2 Additional Important Questions and Answers

Question 1.
Read the numbers and write them in words:

(1) 4,00,527
Answer:
Four lakh five hundred and twenty-seven.

(2) 7,34,016
Answer:
Seven lakh thirty-four thousand and sixteen.

Question 2.
Read the numbers and write them in figures.
(1) Nine lakh three thousand and twenty-three.
(2) One lakh one thousand one hundred and one.
Answer:
(1) 9,03,023
(2) 1,01,101

Class 5 Maths Solution Maharashtra Board

• Roman Numerals Problem Set 1 Class 5 Maths Solutions
• Number Work Problem Set 2 Class 5 Maths Solutions
• Number Work Problem Set 3 Class 5 Maths Solutions
• Number Work Problem Set 4 Class 5 Maths Solutions
• Number Work Problem Set 5 Class 5 Maths Solutions 
• Number Work Problem Set 6 Class 5 Maths Solutions

Decimal Fractions Class 5 Problem Set 40 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 9 Decimal Fractions Problem Set 40 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 9 Decimal Fractions

Write the following fractions as decimal fractions.

(1) Two and a half
Answer:
2 \(\frac{1}{2}\) = 2.5 = 2.50

(2) Two and a quarter
Answer:
2 \(\frac{1}{4}\) = 2.25

(3) Two and three quarters
Answer:
2 \(\frac{3}{4}\) = 2.75

(4) Ten and a half
Answer:
10 \(\frac{1}{2}\) = 10.5 = 10.50

(5) Fourteen and three quarters
Answer:
14 \(\frac{3}{4}\) = 14.75

(6) Sixteen and a quarter
Answer:
16 \(\frac{1}{4}\) = 16.25

(7) Twenty-eight and a half
Answer:
28 \(\frac{1}{2}\) = 28.50 = 28.5

Adding decimal fractions

Sir : If the cost of one pencil is two and a half rupees and the cost of a pen is four and half rupees, what is the total cost?

Sumit : Two and a half rupees means two rupees and one half rupee. Similarly, four and a half rupees means four rupees and one-half rupee. 4 rupees and 2 rupees make 6 rupees and two half rupees make one rupee, so both objects together cost 6 + 1 = 7 rupees.

Sir : Correct ! Now, see how this is done using decimals.
The sum of the 0’s in the hundredths place is 0.
0.5 + 0.5 is the same as
\(\frac{5}{10}+\frac{5}{10}=\frac{5+5}{10}=\frac{10}{10}=\frac{1}{1}=1\)

This 1 is carried over to the units place. There is nothing in the tenths place, so we put a zero there. In the units place, 2 + 4 = 6 plus the carried over 1 makes 7.

So 2.50 rupees and 4.50 rupees add up to 7 rupees.

We use the decimal system to write whole numbers. We extend the same method to write fractions; therefore, we can add in the same way as we add whole numbers.

I will now show some more additions. Watch carefully.

Sumit : There is no carried over number in the first sum, but there are carried over numbers in the second and third sums.

Rekha : While adding whole numbers, we add units first. Similarly, here, tenths are added first. In the second example, the sum of the tenths place is 13. 13 tenths are 10 tenths + 3 tenths = 1 unit + 3 tenths.

Sumit : That is why, in the sum, 3 stayed in the tenths place and 1 was carried over to the units place. 6 + 5 plus 1 carried over makes 12.

Sir : Your observations are absolutely correct. We write digits one below the other according to their place values while adding whole numbers. We do the same thing here. Remember that while writing down an addition problem and the total, the decimal points should always be written one below the other.

Study the following additions. (Note that: 10 tenths = 1 unit. 10 hundredths = 1 tenth)

Example (1) Add : 7.09 + 54.93
First, add the digits in the 100ths place. 9 + 3 = 12.

The 1 from the sum 12 in the hundredths place is carried over to the tenths place and 2 is written in the hundredths place. Adding 1 + 9 gives 10 tenths or 1 unit. This 1 is carried over to the units place. 0 is left in the tenths place. Then, the addition is completed in the usual way.

Example (2) Add : 45.83 + 167.4
4 5 . 8 3 We arrange the numbers so that the places and
+
1 6 7. 4 decimal points come one below the other.

\(\frac{4}{10}=\frac{4 \times 10}{10 \times 10}=\frac{40}{100}\) Therefore, to make the denominators of the fractions equal, 167.4 is written as 167.40 and then the fractions are added.

As usual, the digits with the smallest place values are added first and then those with bigger place values are added serially.

Example (3) 10.46 Rupees

Example (4) 48.80 m

Example (5) 7.5 cm

Decimal Fractions Problem Set 40 Additional Important Questions and Answers

(1) Thirty and a quarter
Answer:
30 \(\frac{1}{4}\) = 30.25

(2) Thirty and a half
Answer:
30 \(\frac{1}{2}\) = 30.50 = 30.5

(3) Thirty and three quarters
Answer:
30 \(\frac{3}{4}\) = 30.75

Class 5 Maths Solution Maharashtra Board

• Decimal Fractions Problem Set 36 Class 5 Maths Solutions
• Decimal Fractions Problem Set 37 Class 5 Maths Solutions
• Decimal Fractions Problem Set 38 Class 5 Maths Solutions
• Decimal Fractions Problem Set 39 Class 5 Maths Solutions
• Decimal Fractions Problem Set 40 Class 5 Maths Solutions
• Decimal Fractions Problem Set 41 Class 5 Maths Solutions
• Decimal Fractions Problem Set 42 Class 5 Maths Solutions

Addition and Subtraction Class 5 Problem Set 13 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 13 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 3 Addition and Subtraction

Solve the following mixed word problems:

Question 1.
The Forest Department planted 23,078 trees of khair, 19,476 of behada besides trees of several other kinds. If the Department planted 50,000 trees altogether, how many trees were neither of khair nor of behada?
Solution:
2 3 0 7 8 Trees of khair
+
1 9 4 7 6 Trees of behada
4 2 5 5 4 Trees of khair and behada

5 0 0 0 0 Total trees planted
–
4 2 5 5 4 Khair and behada trees planted
7 4 4 6 Other kind of trees planted

Answer:
7,446 trees planted other than khair and behada trees.

Question 2.
A city has a population of 37,04,926. If this includes 11,24,069 men and 10,96,478 women, what is the number of children in the city?
Solution:
1 1 2 4 0 6 9 Men
+
1 0 9 6 4 7 8 Women
2 2 2 0 5 4 7 Total of men and women

3 7 0 4 9 2 6 Total population
–
2 2 2 0 5 4 7 Men and women
1 4 8 4 3 7 9 No. of children

Answer:
Number of children in the city is 14,84,379.

Question 3.
The management of a certain factory had 25,40,600 rupees in the labour welfare fund. From this fund, 12,37,865 rupees were used for medical expenses, 8,42,317 rupees were spent on the education of the workers’ children and the remaining was put aside for a canteen. How much money was put aside for the canteen?
Solution:
₹ 1 2 3 7 8 6 5 Medical expenses
₹ 8 4 2 3 1 7 Education for workers children
₹ 2080182 Spent for medical and education.
₹ 2 5 4 0 6 0 0 Labour welfare fund
₹ 2 0 8 0 1 8 2 Medical & education
₹ 4 6 0 4 1 8 Kept a side for canteen


Answer:
₹ 4,60,418 put aside for the canteen,

Question 4.
For a three-day cricket match, 13,608 tickets were sold on the first day and 8,955 on the second day. If, altogether, 36,563 tickets were sold in three days, how many were sold on the third day?
Solution:
1 3 6 0 8 Tickets sold on 1st day
+
8 9 5 5 Ticket sold on 2nd day
2 2 5 6 3 Tickets sold on 1st and 2nd day
3 6 5 6 3 Tickets sold in 3 days
–
2 2 5 6 3 Tickets sold in 2 days
1 4 0 0 0 Tickets sold on 3rd day


Answer:
₹ 14,000 tickets sold on the third day.

Addition and Subtraction Problem Set 13 Additional Important Questions and Answers

Solve the following mixed word problems:

Question 1.
A man had ₹ 1,65,346 in the bank. He deposited ₹ 2,47,190 in the bank, then he gave a cheque of ₹ 3,18,649 to Ashutosh. How much’is the balance in the bank now?
Solution:
₹ 1 6 5 3 4 6 Had in the bank
+
₹ 2 4 7 1 9 0 Deposited in the bank
₹ 4 1 2 5 3 6 Total balance
₹ 4 1 2 5 3 6 Total
–
₹ 3 1 8 6 4 9 Gave to Ashutosh
₹ 9 3 8 8 7 Balance in the bank

Answer:
Balance in the bank ₹ 93,887.

Question 2.
Vighanesh had ₹ 36,28,500 from this amount he gave ₹ 15,04,930 to his wife and ₹ 10,13,825to his son. How much amount left with him?
Solution:
₹ 3 6 2 8 5 0 0 Vighanesh had
–
₹ 1 5 0 4 9 3 0 given to wife
₹ 2 1 2 3 5 7 0 Total

₹ 2 1 2 3 5 7 0 Balance
–
₹ 1 0 1 3 8 2 5 Gave to son
₹ 1 1 0 9 7 4 5 Left with him

Answer:
₹ 11,09,745 left with Vighanesh.

Question 2.
Add the following:

(1) 3 0 5 8 3
+
1 2 3 2 9
_____________
_____________
Answer:
42912

(2) 4 5 3 7 8
+
4 4 6 2 2
_____________
_____________
Answer:
90000

(3) 7 5 0 3 8
+
1 7 4 1 8
_____________
_____________
Answer:
92456

(4) 2 2 1 0 5
+
3 9 6 5 1
_____________
_____________
Answer:
61756

Question 3.
Add the following:
(1) 63,348 + 74,35,631
(2) 9,65,247 + 3,28,925
(3) 7,61,856 + 1,45,437
(4) 33,23,057 + 35,28,436
(5) 3,451 + 62,507 + 3,40,678
(6) 48 + 38,41,705 + 98,314
(7) 25,38,781 + 328 + 16,508
(8) 29,145 + 40,37,615 + 8,70,469
Answer:
(1) 74,98,979
(2) 12,94,172
(3) 9,07,293
(4) 68,51,493
(5) 4,06,636
(6) 39,40,065
(7) 25,55,617
(8) 49,37,229

Question 4.
Match the equal numbers in three columns

Column (A)	Column (B)	Column (C)
(1) Thirteen thousand plus two hundred	(a) 304 + 500	(i) 80,704
(2) Eight thousand plus seventy	(b) 13,000 + 200	(ii) 804
(3) Three hundred and four plus five hundred	(c) 80,000 + 704	(iii) 8070
(4) Eighty thousand plus seven hundred and four	(d) 8,000 + 70	(iv) 13,200

Answer:
(1) b – iv
(2) d – iii
(3) a – ii
(4) c – i

Question 5.
Subtract the following:

(1) 7 6 3 8 5
–
5 7 6 3 7
____________
____________
Answer:
18,748

(2) 5 6 0 4 7
–
3 2 3 7 8
____________
____________
Answer:
23,669

(3) 8 2 3 5 6
–
4 1 5 6 3 9
____________
____________
Answer:
36,927

(4) 4 5 4 2 9
–
3 5 9 6 8
____________
____________
Answer:
04,788

(5) 7 4 3 5 0 8
–
4 1 5 6 3 9
____________
____________
Answer:
3,27,869

(6) 2 4 8 1 3 6 7
–
1 7 8 4 2 7 8
____________
____________
Answer:
6,97,089

(7) 5 9 3 1 6 6 5
–
4 3 6 5 7 4 9
____________
____________
Answer:
19,79,109

(8) 8 0 5 1 4 3 6
–
4 3 6 5 7 4 9
____________
____________
Answer:
36,85,687

Question 6.
Solve the following examples :
(1) (a) 64,83,217 – 23,94,128 + 16,84,579
(b) 36,94,523 + 28,17,689 – 50,49,876
(c) 83,47,215 – 38,58,386 – 25,74,978
(d) 3,72,190 + 2,18,310 – 1,56,900
(e) 36,00,800 – 27,91,978 – 3,01,005
(f) 51,51,515 – 5,55,555 + 6,66,006
Answer:
(a) 57,73,668
(b) 14,62,336
(c) 19,13,851
(d) 4,33,600
(e) 5,07,817
(f) 52,61,966

Class 5 Maths Solution Maharashtra Board

• Addition and Subtraction Problem Set 7 Class 5 Maths Solutions
• Addition and Subtraction Problem Set 8 Class 5 Maths Solutions
• Addition and Subtraction Problem Set 9 Class 5 Maths Solutions
• Addition and Subtraction Problem Set 10 Class 5 Maths Solutions
• Addition and Subtraction Problem Set 11 Class 5 Maths Solutions
• Addition and Subtraction Problem Set 12 Class 5 Maths Solutions
• Addition and Subtraction Problem Set 13 Class 5 Maths Solutions

Circles Class 5 Problem Set 30 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 7 Circles Problem Set 30 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 7 Circles

Question 1.
In the table below, write the names of the points in the interior and exterior of the circle and those on the circle.

Answer:

The circumference of a circle

Take a bowl with a circular edge.

Wind a string once around the bowl and make a full circle around it.
Unwind this circle and straighten it out as shown.
Measure the straightened part with a ruler. The length of that part is the circumference of the circle or of the bowl.

An arc of a circle
Shown alongside is a plastic bangle. If the bangle breaks at points A and B, it will split into two parts as shown in the picture.

Each of these parts is an arc of a circle.

On the given circle, there are two points P and Q. These two points have divided the circle into two parts. Each of these parts is an arc of the circle.

This means that P and Q have created two arcs. P and Q are the end points of both arcs.

From the name ‘arc PQ’, we cannot say which of the two arcs we are speaking of. So, an additional point is taken on each arc. This point is used to give each arc a three-letter name. In the figure, there are two arcs, arc PSQ and arc PRQ.

Circles Problem Set 30 Additional Important Questions and Answers

Question 1.
In the table below, write the names of the points in the interior and exterior of the circle and those on the circle.
Answer:

Question 2.
Draw a circle and take points A, B, C on the circle. L, M, N in the interior of the circle, P, Q, R in the exterior of the circle.
Answer:

Class 5 Maths Solution Maharashtra Board

• Circles Problem Set 28 Class 5 Maths Solutions
• Circles Problem Set 29 Class 5 Maths Solutions
• Circles Problem Set 30 Class 5 Maths Solutions
• Circles Problem Set 31 Class 5 Maths Solutions

Perimeter and Area Class 5 Problem Set 49 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 12 Perimeter and Area Problem Set 49 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 12 Perimeter and Area

Question 1.
How much wire will be needed to make a rectangle 7 cm long and 4 cm wide?
Solution:
Perimeter of a rectangle
= 2 x length + 2 x breadth
= 2 x 7 + 2 x 4
= 14 + 8
= 22 cm

∴ 22 cm wire will be needed to make a rectangle

Question 2.
If the length of a rectangle is 20 m and its width is 12m, what is its perimeter?
Solution:
Perimeter of a rectangle
= 2 x length + 2 x breadth
= 2×20 + 2×12
= 40 + 24
= 64 m
∴ Perimeter is 64 m

Question 3.
Each side of a square is 9 m long. Find its perimeter.
Solution:
Perimeter of a square
= 4 x length of one side
= 4 x 9
= 36 m
∴ Perimeter is 36 m

Question 4.
If we take 4 rounds around a field that is 160 m long and 90 m wide, what is the distance we walk in kilometres?
Solution:
Perimeter of a rectangular field
= 2 x length + 2 x breadth
= 2 x 160 + 2 x 90
= 320 + 180
= 500 m

In one round distance walked is 500 m, hence, distance walked in 4 rounds
= 500 x 4
= 2000 m
= 2km
∴ The distance walked in 4 rounds is 2 km

Question 5.
Sanju completes 12 rounds around a square park every day. If one side of the park is 120 m, find out in kilometres and metres the distance that Sanju covers daily.
Solution:
Perimeter of a square
= 4 x length of one side
= 4 x 120
= 480 m

So, in one round the distance can be covered is 480 m, hence in 12 rounds the distance can be covered is
= 480 x 12
= 5760 m
= 5000 m + 760 m

∴ Sanju covers 5 km 760 m daily

Question 6.
The length of a rectangular plot of land is 50 m and its width is 30 m. A triple fence has to be put along its edges. If the wire costs 60 rupees permetre, what will be the total cost of the wire needed for the fence?
Solution:
Perimeter of a rectangular plot
= 2 x length + 2 x breadth
= 2 x 50 + 2 x 30
= 100 + 60 – 160 m
For a triple fence, wire needed
= 3 x 160 = 480 m

Cost of the wire needed
= wire needed x rate
= 480 x 60
= 28800 rupees
∴ The total cost of the wire needed for the fence is ₹ 28,800

Question 7.
A game requires its players to run around a square playground. Each side of the playground is 20 m long. One player took 5 rounds around the playground. How many metres did he run altogether?
Solution:
Perimeter of a square
= 4 x length of one side
= 4 x 20
= 80 m

In one round 80 m.
So in 5 round
= 80 x 5
= 400
= 400 m

∴ He runs altogether = 400 m

Question 8.
Four rounds of wire fence have to be put around a field. If the field is 60 m long and 40 m wide, how much wire will be needed?
Solution:
Perimeter of rectangular field
= 2 x length + 2 x breadth
= 2 x 60 + 2 x 40
= 120 + 80
= 200 m
Hence, wire required for 4 rounds
= 200 x 4
= 800 m

∴ Wire required for 4 rounds
= 800 m

Question 9.
The sides of a triangle are 24.7cm, 20.4 cm and 10.5 cm respectively. What is the perimeter of the triangle?
Solution:
Perimeter of triangle
= 24.7 + 20.4 + 10.5
= 55.6

∴ The perimeter of a triangle
= 55.6 cm

Question 10.
Look at the figures on the sheet of graph paper. Measure their sides with the help of the lines on the graph paper. Write the perimeter of each in the right box.

(1) Perimeter of
rectangle ABCD
= [ ] cm
(2) Perimeter of
rectangle EFGH
= [ ] cm
(3) Perimeter of
square PQRS
= [ ] cm
(4) Perimeter of
rectangle STUV
= [ ] cm
Solution:

(1) Perimeter of a rectangle ABCD
= 2 x length + 2 x breadth
= 2 x 3.5 + 2 x 2.5
= 7 + 5
= 12 cm

∴ 12 cm

(2) Perimeter of a rectangle EFGH
= 2 x length + 2 x breadth
= 2 x 3.8 + 2 x 1.3
= 7.6 + 2.6
= 10.2 cm

∴ 10.2 cm

(3) Perimeter of a rectangle PQRS
= 2 x length + 2 x breadth
= 2 x 2.4 + 2 x 2.4
= 4.8+ 4.8
= 9.6 cm

∴ 9.6 cm

(4) Perimeter of a rectangle STUV
= 2 x length + 2 x breadth
= 2 x 3 + 2 x 2
= 6 + 4
= 10 cm

∴ 10 cm

(5) Perimeter of a triangle LMN
= 1.5 + 2.5 + 2
= 6 cm

∴ 6 cm

Area : Revision

Of the figures given above, figure ABCD has six squares of 1 cm each inside it. It means that its area is 6 sq cm.

In the same way, count the squares in each figure and write its area.
(1) Area of MNRS = [ ] sq cm
(2) Area of EFGH = [ ] sq cm
(3) Area of PQRS = [ ] sq cm
(4) Area of IJKL = [ ] sq cm

Atul : Sir, why is the unit for area written as sq cm? We measure the sides in centimetres.

Teacher : Centimetre is a standard unit of length. In order to measure area, we need a standard unit of area. For this, a square with a side 1 cm is taken as the standard unit. The area of this square is 1 square centimetre. That is why this unit is written as sq cm, in short.

To measure large areas like fields, parks and playgrounds, a square with side 1 m, that is, an area of 1 sq m, is taken as the standard unit.

To measure the areas oftalukas or districts, a square with side 1km, or 1sq km is the standard unit used.

Formula for the area of a rectangle

(1) In the rectangle ABCD given alongside, 1 cm divisions were marked off on each side. The points on opposite sides were joined as shown in the figure. The length of the sides of each square thus created is 1cm. Therefore, the area of each square is 1 sq cm.

In ABCD, 3 rows with 5 squares each have been created.
The number of squares in rectangle ABCD is 3 × 5 = 15.
Therefore, the area of rectangle ABCD is 15 sq cm.
Here, the length of the figure is 5 cm and its breadth is 3 cm.
Note that the product of 3 and 5 is 15.

(2) In the rectangle with sides 4 cm and 2 cm, make squares of 1 sq cm each as shown above. Count the number of squares.

Note that here too, the number of squares formed are the same as the product of the length and width of the rectangle.

Therefore, The area of a rectangle = length × breadth

Formula for the area of a square

(1) Look at the square given alongside. The side of the square is 3 cm long. 9 squares of 1 cm each are formed within this square.

Therefore, the area of this square is 9 sq cm.

Here, there are 3 rows with 3 squares each, i.e., there are 3 × 3 = 9 squares.
The length of each side of the square is 3 cm.
The product of two sides of the square is 3 × 3 = 9.

(2) Measure the area of a square with side 5 cm, in the same way.
The answer will be 25 sq cm.
Note that 5 × 5 = 25

Therefore, The area of a square = length of a side × length of a side

It is not necessary to divide a square or rectangle into small squares every time you calculate their area. The advantage of a formula is that you can calculate the area simply by substituting the appropriate values.

Word problems
Example (1) What is the area of a rectangle of length 20 cm and width 15 cm?
Area of a rectangle = length × breadth
= 20 × 15 = 300.
Therefore, the area of the rectangle is 300 sq cm.

Example (2) A wall that is 4 m long and 3 m wide has to be painted. If the labour charges are ₹ 25 per sq m, what is the cost of labour for painting this wall?

First let us calculate the area of the wall to be painted.
Area of the wall = length of the wall × breadth of the wall = 4 × 3 = 12
Thus, the area of the wall is 12 sq m.
Labour cost of 1 sq m is 25 rupees.
So the labour cost for 12 sq m will be = 12 × 25 = 300
The cost of labour for painting the wall will be 300 rupees.

Example (3) What will be the area of a square with sides 15 cm?
Area of a square = length of side × length of side
= 15 × 15 = 225
The area of the square is 225 sq cm.

Example (4) One side of a square room is 4 m. If the cost of labour for laying 1 sq m of the floor is 35 rupees, what will be the total cost of labour?
First we must find the area of the square room.
Area of the square room = length of side × length of side = 4 × 4 = 16
Therefore, the area of the square room is 16 sq m.
The labour cost of laying 1 sq m of flooring is 35 rupees.
Therefore, the cost of laying 16 sq m of flooring is 16 × 35 = 560 rupees.

Perimeter and Area Problem Set 49 Additional Important Questions and Answers

Question 1.
Devendra walks five rounds of a square garden everyday. If the side of the garden is 150 m, how many kilometres does Devendra walk every morning?
Solution:
Perimeter of a square garden
= 4 x one side of the garden
= 4 x 150
= 600 m

In 5 rounds walking
= 5 x 600
= 3000 m
= 3 km
3 km

Question 2.
The length of a rectangular play ground is 75 m and its breadth is 50 m. Rupali walks four rounds. How many kilometres did she walk?
Solution:
Perimeter of rectangle
= 2 x length + 2 x breadth
= 2 x 75 + 2 x 50
= 150 + 100
= 250 m

In 4 rounds walking
= 4 x 250
= 1000 m
= 1 km

∴ 1 km

Question 3.
Length of the rectangle is 10 cm and its breadth is 8 cm and one square is side 9 cm. Whose perimetre is more? By how much?
Solution:
Perimeter of a rectangle
= 2 x length + 2 x breadth
= 2 x 10 + 2 x 8
= 20 + 16
= 36 cm ……………….. (i)

Perimeter of a square
= 4 x length of side
= 4 x 9
36 cm ……………….. (ii)
From (i) and (ii) perimeter of both is equal.

∴ perimeter of both is equal

Class 5 Maths Solution Maharashtra Board

• Problems on Measurement Problem Set 46 Class 5 Maths Solutions
• Problems on Measurement Problem Set 47 Class 5 Maths Solutions
• Perimeter and Area Problem Set 48 Class 5 Maths Solutions
• Perimeter and Area Problem Set 49 Class 5 Maths Solutions
• Perimeter and Area Problem Set 50 Class 5 Maths Solutions

Decimal Fractions Class 5 Problem Set 39 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 9 Decimal Fractions Problem Set 39 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 9 Decimal Fractions

Question 1.
Write how many rupees and how many paise.

(1) ₹ 58.43
Answer:
58 rupees 43 paise.

(2) ₹ 9.30
Answer:
9 rupees 30 paise.

(3) ₹ 2.30
Answer:
2 rupees 30 paise.

(4) ₹ 2.3
Answer:
2 rupees 30 paise.

Question 2.
Write how many rupees in decimal form.

(1) 6 rupees 25 paise
Answer:
₹ 6.25

(2) 15 rupees 70 paise
Answer:
₹ 15.70

(3) 8 rupees 5 paise
Answer:
₹ 8:05

(4) 22 rupees 4 paise
Answer:
₹ 22.04

(5) 720 paise
Answer:
₹ 7.20

Question 3.
Write how many metres and how many centimetres.

(1) 58.75 m
Answer:
58 m 75 cm

(2) 9.30 m
Answer:
9 m 30 cm

(3) 0.30 m
Answer:
30 cm

(4) 0.3 m
Answer:
30 cm

(5) 1.62 m
Answer:
1 m 62 cm

(6) 91.4 m
Answer:
91 cm 40 cm

(7) 7.02 m
Answer:
7 m 2 cm

(8) 0.09 m
Answer:
9 cm

Question 4.
Write how many metres in decimal form.

(1) 1 m 50 cm
Answer:
1.5 m

(2) 50 m 40 cm
Answer:
50.40 m

(3) 50 m 4 cm
Answer:
50.04 m

(4) 734 cm
Answer:
7.34 m

(5) 10 cm
Answer:
0.1 m

(6) 2 cm
Answer:
0.02 m

Question 5.
Write how many centimetres and how many millimetres.

(1) 6.9 cm
Answer:
6 cm 9 mm

(2) 20.4 cm
Answer:
20 cm 4 mm

(3) 0.8 cm
Answer:
8 mm

(4) 0.5 cm
Answer:
5 mm

Question 6.
Write how many centimetres in decimal form.
(1) 7 cm 1 mm
Answer:
7.1 cm

(2) 16 mm
Answer:
1.6 cm

(3) 144 mm
Answer:
14.4 cm

(4) 8 mm
Answer:
0.8 cm

Writing half, quarter, three-quarters and one and a quarter in decimal form

‘Half’ is usually written as \(\frac{1}{2}\). To convert this fraction into decimal form, the denominator of \(\frac{1}{2}\) must be converted into an equivalent fraction with denominator 10.

\(\frac{1}{2}=\frac{1 \times 5}{2 \times 5}=\frac{5}{10}\) so the decimal form of \(\frac{1}{2}\) will be \(\frac{5}{10}\) or 0.5 Just as \(\frac{1}{2}=\frac{1 \times 5}{2 \times 5}=\frac{5}{10}\) = 0.5, note that \(\frac{1}{2}=\frac{1 \times 50}{2 \times 50}=\frac{50}{100}\) = 0.50

Therefore, ‘half’ is written as ‘0.5’ or 0.50’. ‘Quarter’ and ‘three quarters’ are written in fractions as \(\frac{1}{4}\) and \(\frac{3}{4}\) respectively. Let us convert them into decimal fractions. 10 is not divisible by 4. Therefore, the denominators of \(\frac{1}{4}\) and \(\frac{3}{4}\) cannot be made into fractions with multiples of 10. However, 4 × 25 = 100, so the denominator can be 100.

A quarter \(=\frac{1}{4}=\frac{1 \times 25}{4 \times 25}=\frac{25}{100}=0.25\)
and Three quarters \(=\frac{3}{4}=\frac{3 \times 25}{4 \times 25}=\frac{75}{100}=0.75\)
One and a quarter = 1 \(\frac{1}{4}\) = 1.25
One and a half = 1 \(\frac{1}{2}\) = 1.50 = 1.5
One and three quarters = 1 \(\frac{3}{4}\) = 1.75
Seventeen and a half = 17 \(\frac{1}{2}\) = 17.50 = 17.5

Decimal Fractions Problem Set 37 Additional Important Questions and Answers

Question 1.
Write how many rupees and how many paise.

(1) ₹ 147.5
Answer:
1 hundred and 47 rupees 50 paise.

(2) ₹ 40.4
Answer:
40 rupees and 40 paise.

Question 2.
Write how many rupees in decimal form.

(1) 105 paise
Answer:
₹ 1.05

(2) 6 rupees 6 paise
Answer:
₹ 6.06

(3) 20 rupees 20 paise
Answer:
₹ 20.2

Question 3.
Write how many metres and how many centimetres.

(1) 1.1m =
Answer:
1 m 10 cm

(2) 120 cm =
Answer:
1 m 20 cm

(3) 24.8 m =
Answer:
24 m 80 cm

(4) 0.5 m =
Answer:
50 cm

Question 4.
Write how many metres in decimal form.

(1) 110 cm =
Answer:
1.1m

(2) 60 cm =
Answer:
0.6 m

Question 5.
Write how many centimetres and how many millimetres.

(1) 0.1 cm =
Answer:
1 mm

(2) 10.5 cm =
Answer:
10 cm 5 mm

Question 6.
Write how form. many centimetres in decimal
(1) 1 mm =
Answer:
0.1 cm

(2) 100 mm =
Answer:
10.0 cm

Class 5 Maths Solution Maharashtra Board

• Decimal Fractions Problem Set 36 Class 5 Maths Solutions
• Decimal Fractions Problem Set 37 Class 5 Maths Solutions
• Decimal Fractions Problem Set 38 Class 5 Maths Solutions
• Decimal Fractions Problem Set 39 Class 5 Maths Solutions
• Decimal Fractions Problem Set 40 Class 5 Maths Solutions
• Decimal Fractions Problem Set 41 Class 5 Maths Solutions
• Decimal Fractions Problem Set 42 Class 5 Maths Solutions

Number Work Class 5 Problem Set 2 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 2 Number Work Problem Set 2 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 2 Number Work

Question 1.
Using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 write ten each of two-, three-, four- and five-digit numbers. Read the numbers.
Answer:

Two-digit numbers	Reading a number
37	Thirty-seven
80	Eighty
49	Forty-nine
65	Sixty-five
28	Twenty-eight
54	Fifty-four
92	Ninety-two
71	Seventy-one
16	Sixteen
22	Twenty-two
Three-digit numbers	Reading a number
504	Five hundred and four
386	Three hundred eighty-six
430	Four hundred thirty
891	Eight hundred ninety-one
615	Six hundred fifteen
267	Two hundred sixty-seven
900	Nine hundred
173	One hundred seventy-three
766	Seven hundred sixty-six
258	Two hundred and fifty-eight
Four-digit numbers	Reading a number
3,817	Three thousand eight hundred and seventeen
4,059	Four thousand fifty-nine
9,611	Nine thousand six hundred and eleven
7,413	Seven thousand four hundred thirteen
5,608	Five thousand six hundred and eight
Four-digit numbers	Reading a number
2,009	Two thousand and nine
6,420	Six thousand four hundred and twenty
1,357	One thousand three hundred and fifty-seven
8,172	Eight thousand one hundred and seventy-two
6,156     –	Six thousand one hundred and fifty-six
Five-digit numbers	Reading a number
41,309	Forty-one thousand, three hundred and nine
68,527	Sixty-eight thousand five hundred and twenty seven
50,348	Fifty thousand three hundred and forty eight
76,052	Seventy-six thousand and fifty-two
21,546	Twenty-one thousand five hundred and forty-six
10,358	Ten thousand three hundred and fifty-eight
94,215	Ninety-four thousand two hundred and fifteen
36,104	Thirty-six thousand one hundred and four
89,157	Eighty-nine thousand one hundred and fifty-seven
72,560	Seventy-two thousand five hundred and sixty

Question 2.
Fill in the blanks in the table below.

Answer:

	Devnagari numerals	International numerals	Number written in words
(1)	२,३५९	2,359	Two thousand three hundred and fifty nine
(2)	३२,७५६	32,756	Thirty two thousand seven hundred and fifty Six
(3)	६७,८५९	67,859	Sixty seven thousand eight hundred and fifty Nine
(4)	१,०३४	1,034	One thousand and thirty four
(5)	२७,८९५	27,895	Twenty seven thousand eight hundred and ninety five

Question 3.
As a part of the ‘Avoid Plastic Project’, Zilla Parishad schools made and provided paper bags to provision stores and greengrocers. Read the talukawise numbers of the bags and write the numbers in words.

Answer:

Talukas	No. of Bags	Numbers in words
Kopargaon	12,740	Twelve thousand seven hundred and forty
Shevgaon	28,095	Twenty-eight thousand and ninety-five
Karjat	31,608	Thirty-one thousand six hundred and eight
Sangamner	10,792	Ten thousand seven hundred and ninety-two

Question 4.
How many rupees do they make?
(1) 20 notes of 1000 rupees, 5 notes of 100 rupees and 14 notes of 10 rupees.
(2) 15 notes of 1000 rupees, 12 notes of 100 rupees, 8 notes of 10 rupees and 5 coins of 1 rupee.
Answer:

Question 5.
Write the biggest and the smallest five-digit numbers that can be made using the digits 4, 5, 0, 3, 7 only once.
Answer:
Biggest five digit number is 75,430 Smallest five digit number is 30,457

Question 6.
The names of some places and their populations are given below. Use this information to answer the questions that follow.

Tala : 40,642
Gaganbawada : 35,777
Bodhwad : 91,256
Moregaon : 87,012
Bhamragad : 35,950
Velhe : 54,497
Ashti : 76,201
Washi : 92,173
Morwada : 85,890

(1) Which place has the greatest population? What is its population?
(2) Which place, Morwada or Moregaon, has the greater population?
(3) Which place has the smallest population? How much is it?
Answer:
(1) Washi has the greatest population. Population of Washi is 92,173
(2) Moregaon has the greater population.
(3) Gaganbawada has the smallest population. Its population is 35,777

Introducing six-digit numbers
Teacher : How much, do you think, is the price of a four-wheeler?
Ajay : Maybe about six or seven lakh rupees.
Teacher : Do you know exactly how much one lakh is?
Ajay : It’s a lot, isn’t it? More than even ten thousand, right?
Teacher : Yes, indeed ! Let’s find out just how much. What is 999 + 1?
Ajay : One thousand.
Teacher : You have learnt to write 99000, too. Now, if you add 1000 to that, you will get one hundred thousand. That’s what we call one lakh.
Vijay : 9999+1 is 10,000 (ten thousand). We had made the ten thousands place for it. Can we make a place for one lakh too in the same way?
Teacher : Yes, of course. Carry out the addition 99,999 + 1 and see what you get.

Here we keep carrying over till we have to make a place for the ‘lakh’ on the left of the ten thousands place. And we write the last carried over one in that place. The sum we get is read as ‘one lakh’.
Vijay : Kishakaka bought a second-hand car for two and a half lakh rupees.
Ajay : How much is two and a half lakh?
Teacher : One lakh is 100 thousand. So, half a lakh is 50 thousand. Because, half of 100 is 50.

Vijay : That means two and a half lakh is 2 lakh 50 thousand.
Teacher : Now write this number in figures.
Vijay : 2,50,000.
Teacher : We have seen that a hundred thousand is 1 lakh. If we have 1000 notes of 100 rupees, how many rupees would they make?
Vijay : 1000 notes of 100 rupees would make 1 lakh rupees.

Reading six-digit numbers
(1) 2,35,705 : two lakh thirty-five thousand seven hundred and five
(2) 8,00,363 : eight lakh three hundred and sixty-three
(3) 3,07,899 : three lakh seven thousand eight hundred and ninety-nine
(4) 9,00,049 : nine lakh forty-nine
(5) 5,30,735 : five lakh thirty thousand seven hundred and thirty-five

Writing six-digit numbers in figures
(1) Eight lakh, nine thousand and forty-three : There are 8 lakhs in this number. There are no ten thousands, so we write 0 in that place. As there are 9 thousands, we write 9 in the thousands place. We write 0 in the hundreds place as there are no hundreds. Forty-three is equal to 4 tens and 3 units, so in the tens and units places we write 4 and 3 respectively. In figures : 8,09,043.

When writing numbers in figures, write the digit in the highest place first and then, in each of the next smaller places, write the proper digit from 1 to 9. Write 0, if there is no digit in that place. For example, if the number eight lakh, nine thousand and forty-three is written as ‘89043’, it is wrong. It should be written as 8,09,043. Here, we have to write zero in the ten thousands place.

(2) Four lakh, twenty thousand, five hundred : In this figure, there aren’t any thousands in the thousands place, so we write 0 in it. Since there are five hundreds, we write 5 in the hundreds place. There are no tens and units, hence, we write 0 in those places. In figures : 4,20,500.

Roman Numerals Problem Set 2 Additional Important Questions and Answers

Question 1.
Fill in the blanks in the table below:
Answer:

	Devnagari numerals	International numerals	The number written in words
(1)	५,५१८	5,518	Five thousand five hundred and eighteen
(2)	४९,८०९	49,809	Forty-nine thousand eight hundred and nine
(3)	७,२५६	7,256	Seven thousand two hundred and fifty-six

Question 2.
Solve the following:

(1) In an election, the First candidate received 58,735 votes, the Second candidate received 65,500, the Third candidate received 85,450 and the Fourth candidate got 09,689 votes. Read the numbers of the votes and write the numbers in words.
Answer:
First candidate – 58,735 – Fifty-eight thousand seven hundred and thirty-five
Second candidate – 65,500 – Sixty-five thousand five hundred
Third candidate – 85,450 – Eighty-five thousand four hundred and fifty
Fourth candidate – 09,689 – Nine thousand six hundred and eighty-nine

Question 3.
How many rupees do they make?
*(1) 10 notes of 2,000 rupees, 5 notes of 100 rupees and 14 notes of 10 rupees.
Solution:
10 notes of 2,000 rupees = 10 x 2,000 ₹ 20,000
5 notes of 100 rupees = 5 x 100 = ₹ 500
14 notes of 10 rupees = 14 x 10 = ₹ 140
Total = ₹ 20,640

∴ They make, twenty thousand, six hundred and forty.

*(2) 7 notes of 2,000 rupees, 12 notes of loo rupees, 8 notes of 10 rupees and 5 coins of 1 rupee
Solution:
7 notes of 2,000 rupees = 7 x 2,000 = ₹ 14,000
12 notes of 100 rupees = 12 x 100 = ₹ 1,200
8 notes of 10 rupees = 8 x 10 = ₹ 80
5 coins of 1 rupee = 5 x 1 = ₹ 5
Total = ₹ 15,285

∴ They make, fifteen thousand, two hundred and eighty five.

(3) 4 notes of 2,000 rupees, 6 notes of 100 rupees and 12 notes of 10 rupees
Solution:
4 notes of 2,000 rupees = 4 x 2,000 = ₹ 8,000
6 notes of 100 rupees = 6 x 100 = ₹ 600
12 notes of 10 rupees = 12 x 10 = ₹ 120
Total = ₹ 8,720

∴ They make, eight thousand, seven hundred and twenty.

(4) 5 notes of 2,000 rupees, 9 notes of 500 rupees, 8 notes of 100 rupees, 7 notes of 50 rupees, 6 notes of 20 rupees and 5 note of 10 rupees
Solution:
5 notes of 2,000 rupees 5 x 2,000 = ₹ 10,000
9 notes of 500 rupees = 9 x 500 = ₹ 4,500
8 notes of 100 rupees = 8 x 100 = ₹ 800
7 notes of 50 rupees = 7 x 50 = ₹ 350
6 notes of 20 rupees = 6 x 20 = ₹ 120
5 notes of 10 rupees = 5 x 10 = ₹ 50
Total = ₹ 15,820

∴ They make, fifteen thousand, eight hundred and twenty.

*Question 4.
Write the biggest and the smallest numbers using all the given digits in every number. Use each digit only once.
(1) 4, 8, 0, 2, 6, 5;
(2) 2, 6, 7, 1, 4;
(3) 5, 9, 6, 1, 4, 3;
(4) 9, 4, 1, 3, 6;
(5) 5, 3, 0, 0, 2
Answer:
(1) Biggest six digit number is 8,65,420 Smallest six digit number is 2,04,568
(2) Biggest five digit number is 76,421 Smallest five digit number is 12,467
(3) Biggest six digit number is 9,65,431 Smallest six digit number is 1,34,569
(4) Biggest five digit number is 96,431 Smallest five digit number is 13,469
(5) Biggest five digit number is 53,200 Smallest five digit number is 20,035

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Circles Class 5 Problem Set 31 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 7 Circles Problem Set 31 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 7 Circles

Question 1.
In the figure given alongside, points S, L, M, and N are on the circle. Answer the questions with the help of the diagram.

(1) Write the names of the arcs with end-points S and M.
Answer:
Arcs with the end-points S and M are, arc SLM and arc SNM.

(2) Write the names of the arcs with the end-points L and N.
Answer:
Arcs with the end-points L and N are, arc LMN and arc LSN.

Question 2.
Write the names of arcs that points A, B, C, and D in the given circle give rise to.
Answer:
Arcs with end-points A and C are, arc ABC and arc ADC.
Arcs with end-points B and D are, arc BAD and arc BCD.

Question 3.
Give the names of the arcs that are made by points P, Q, R, S, and T in the figure.
Answer:
Taking end-points : P and R, arc PQR, arc PTR.
Taking end-points: Q and S, arc QRS, arc QTS
Taking end-points : R and T, arc RST, arc RPT
Taking end-points : S and P, arc STP, arc SRP
Taking end-points: Q and T, arc QPT, arc QST

Question 4.
Measure and note down the circumference of different circular objects. (It is convenient to use a measuring tape for this purpose.)

Chapter 7 Circles Problem Set 31 Additional Important Questions and Answers

Question 1.
Draw circles with the radii given below:
(1) 1.2 cm
Answer:

(2) 2.5 cm
Answer:

(3) 3.3 cm
Answer:

Question 2.
Write true or false for the following statements:
(1) Longest chord is a diameter.
(2) Centre is not lying on the diameter.
(3) All chords are of equal length.
(4) All chords passes through the centre.
Answer:
(1) True
(2) False
(3) False
(4) False

Question 3.
Match the cplumns (A) and (B):

Answer:
(1) ↔ (c),
(2) ↔ (a),
(3) ↔ (d),
(4) ↔ (b)

Question 4.
Complete the following table by filling in the blanks:

Answer:
(1) 6 cm
(2) 10 cm
(3) 34 cm
(4) 9 cm

Question 5.
From the following figure, fill in the blanks:

(1) If OP = 4cm then AB = _________ cm, OA = _________ cm, OB = _________ cm.
(2) If AB = 10 cm then OA = _________ cm, OB = _________ cm, OP = _________ cm.
Answer:
(1) AB = 8 cm, OA = 4 cm, OB = 4 cm
(2) OA = 5 cm, OB = 5 cm, OP = 5 cm

Question 6.
In the table below, write the names of the points in the interior and exterior of the circle and those on the circle.

Answer:
Points in the exterior of the circle are A, F and G.
Points in the interior of the circle are O, E and B, and Points on the circle are C, D and H.

Question 7.
Radius of a circle with centre P is 4 cm. Fill in the blanks.
(1) The, point A is at a distance of 5 cm from the centre P. Flence the point A lies in the ________ of the circle.
(2) Point B is at a distance of 4 cm. from the centre P. Hence the point B lies ________ circle.
(3) Point C lies at a distance 3 cm from the centre P. Hence it lies in the ________ of the circle.
Answer:
(1) Exterior
(2) on the circle
(3) interior

Question 8.
Solve the following:
(1) What is the length of the diameter of a circle of radius 6 cm?
(2) What is the length of the radius of a circle of diameter 14 cm?
(3) Give the names of the arcs that are made by points X, Y, Z and W in this picture.

(4) Give the names of the arc that are made by points E, F, G and H, taking end points E and G.

(5) If the diameter of a circle is 7 cm. what is the length of the circumference? (Use measure tap)
Answer:
(1) 12 cm
(2) 7 cm
(3) Having end-points X and Z, arc XYZ and arc XWZ
Having end-points Y and W, arc YZW and arc YXW
(4) By end-points E and G, arc EFG and arc EHG
(5) 22 cm

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