Class 5

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 19 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 19

Write the proper symbol from < , > , or = in the box.

Answer:
=

Answer:
>

Answer:
<

Answer:
=

Answer:
>

Answer:
>

Answer:
>

Answer:
>

Answer:
>

Answer:
>

Answer:
=

Answer:
=

Answer:
>

Answer:
>

Answer:
<

Answer:
>

Addition of like fractions
Example (1) 3/7 + 2/7 = ?
Let us divide a strip into 7 equal parts. We shall colour 3 parts with one colour and 2 parts with another.
The part with one colour is 3/7, and that with the other colour is 2/7.
The total coloured part is shown by the fraction 5/7.
It means that, \(\frac{3}{7}+\frac{2}{7}=\frac{3+2}{7}=\frac{5}{7}\)

Example (2) Add : \(\frac{3}{8}+\frac{2}{8}+\frac{1}{8}\)
The total coloured part is \(\frac{3}{8}+\frac{2}{8}+\frac{1}{8}=\frac{3+2+1}{8}=\frac{6}{8}\)

When adding like fractions, we add the numerators of the two fractions and write the denominator as it is.
Example (3) Add : 2/6 + 4/6 \(\frac{2}{6}+\frac{4}{6}=\frac{2+4}{6}=\frac{6}{6}\)
However, we know that 6/6 means that all 6 of the 6 equal parts are taken. That is, 1 whole figure is taken. Therefore, 6/6 = 1.

Note that:
If the numerator and denominator of a fraction are equal, the fraction is equal to one.
That is why, \(\frac{7}{7}=1 ; \frac{10}{10}=1 ; \frac{2}{5}+\frac{3}{5}=\frac{2+3}{5}=\frac{5}{5}=1\)
Remember that, if we do not divide a figure into parts, but keep it whole, it can also be written as 1.
This tells us that \(1=\frac{1}{1}=\frac{2}{2}=\frac{3}{3}\) and so on.
You also know that if the numerator and denominator of a fraction have a common divisor, then the fraction obtained by dividing them by that divisor is equivalent to the given fraction.
\(\frac{5}{5}=\frac{5 \div 5}{5 \div 5}=\frac{1}{1}=1\)

Fractions Problem Set 19 Additional Important Questions and Answers

Question 1.

Answer:
>

Question 2.

Answer:
=

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 16 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 16

Question 1.
From a total of 10,000 rupees, Anna donated 7,000 rupees to a school. The remaining amount was to be divided equally among six students as the ‘all-round student’ prize. What was the amount of each prize?
Solution:
1 0 0 0 0 Total rupees
–
7 0 0 0 rupees donated
_______
3 0 0 0 remained
_______
This amount was divided among 6 students 500

Answer:
Amount of the prize is ₹ 500.

Question 2.
An amount of 260 rupees each was collected from 50 students for a picnic. If 11,450 rupees were spent for the picnic, what is the amount left over?
Solution:
₹ 2 6 0 Collected from 1 student
x
5 0 No. of students
0 0 0
+
1 3 0 0 0
1 3 0 0 0 Rupees, collected amount
–
1 1 4 5 0 Rupees spent
1 5 5 0 Rupees left over

Answer:
1,550 Rupees leftover

Question 3.
A shopkeeper bought a sack of 50kg of sugar for 1750 rupees. As the price of sugar fell, he had to sell it at the rate of 32 rupees per kilo. How much less money did he get than he had spent?
Solution:
₹ 3 2 Sale price of 1 kg
x
5 0 kg sold
0 0
+
1 6 0 0
1 6 0 0 Amount received
₹ 1 7 5 0 Purchased price
–
₹ 1 6 0 0 Obtained price
₹ 1 5 0 Lesshegot

Answer:
₹ 150 less he got than he had spent

Question 4.
A shopkeeper bought 7 pressure cookers at the rate of 1870 rupees per cooker. He sold them all for a total of 14,230 rupees. Did he get less or more money than he had spent?
Solution:
₹ 1 8 7 0 Purchase price of 1 cooker
x
7 No. of cookers
₹ 1 3 0 9 0 Purchase price
₹ 1 4 2 3 0 Sell price
–
₹ 1 3 0 9 0 Purchase price
₹ 0 1 1 4 0 he got more

Answer:
₹ 1,140 he got more

Question 5.
Fourteen families in a Society together bought 8 sacks of wheat, each weighing 98 kilos. If they shared all the wheat equally, what was the share of each family?
Solution:
9 8 Kilo weight of 1 sack
x
8 No. of sacks
7 8 4 Kilo

Answer:
Share of each family = 56 kilo

Question 6.
The capacity of an overhead water tank is 3000 litres. There are 16 families living in this building. If each family uses 225 litres every day, will the tank filled to capacity be enough for all the families? If not, what will the daily shortfall be?
Solution:
₹ 2 2 5 Litres uses 1 family
x 1 6 No. of families
1 3 5 0
+
2 2 5 0
3 6 0 0 Litres required
–
3 0 0 0 Litres capacity
6 0 0 Litres daily shortfall

Answer:
600 litres is daily shortfall

Multiplication and Division Problem Set 16 Additional Important Questions and Answers

Solve the following word problems:

(1) A farmer brought 250 trays of tomatoes seedlings. Each tray had 48 seedlings. He planted all the seedlings in his field, putting 25 in a row. How many rows of tomatoes did he plant?
Solution:
₹ 2 5 0 Tray of tomatoes seedlings
x
4 8 Seedlings in 1 tray
2 0 0 0
+
1 0 0 0 0
1 2 0 0 0 Total no. of seedlings


Answer:
The number of rows is 480.

Q.l. Solve the following :
Multiply :
(1) 438 x 76
(2) 594 x 208
(3) 3467 x 926
(4) 3581 x 87
(5) 425 x 87
(6) 579 x 49
Answer:
(1) 33,288
(2) 1,23,552
(3) 32,10,442
(4) 3,11,547
(5) 36,975
(6) 28,371

Solve the following and write the quotient and remainder:

(1) 1345 ÷ 37
(2) 9682 ÷ 83
(3) 6371 ÷ 42
(4) 72534 ÷ 23
(5) 1284 ÷ 32
(6) 63240 ÷ 37
Answer:
(1) Quotient 36, Remainder 13
(2) Quotient 116, Remainder 54
(3) Quotient 151, Remainder 29
(4) Quotient 3153, Remainder 15
(5) Quotient 40, Remainder 4
(6) Quotient 1709, Remainder 7

Fill in the blanks :

(1) 88 x 17; 17 is called …………………………. and 88 is called
(2) Product of the greatest three-digit number and smaller two-digit number is ………………………… .
(3) Multiplicand and multiplier are interchanged the product remains the ………………………… .
(4) While multiplying by tens digit, we have to put in the units place ………………………… .
Answer:
(1) multiplier, multiplicand
(2) 99,900
(3) same
(4) zero

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 18 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 18

Convert the given fractions into like fractions.

Solution :
8 is the multiple of 4 So, make 8, the common denominator \(\frac{3}{4}=\frac{3 \times 2}{4 \times 2}=\frac{6}{8}\).Thus 6/8 and 5/8are the required like fractions.

Solution :
The number 35 is a multiple of both 5 and 7 So, making 35 as the common denominater \(\frac{3}{5}=\frac{3 \times 7}{5 \times 7}=\frac{21}{35}, \frac{3}{7}=\frac{3 \times 5}{7 \times 5}=\frac{15}{35}\) Therefore, 21/35 and 15/35 are required like fractions.

Solution :
Here 10 is the multiples of 5. So make 10 as the common denominator \(\frac{4}{5}=\frac{4 \times 2}{5 \times 2}=\frac{8}{10}\). Thus 8/10 and 3/10 are required like fractions.

Solution :
Least common multiple of 9 and 6 is 18. So, make, 18 as the common denominator. \(\frac{2}{9}=\frac{2 \times 2}{9 \times 2}=\frac{4}{18}, \frac{1}{6}=\frac{1 \times 3}{6 \times 3}=\frac{3}{18}\). Thus, 4/18 and 3/18 are the required like fractions.

Solution :
Least common multiple of 4 and 3 is 12 So, make 12 as common denominator \(\frac{1}{4}=\frac{1 \times 3}{4 \times 3}=\frac{3}{12}, \frac{2}{3}=\frac{2 \times 4}{3 \times 4}=\frac{8}{12}\). so, \(\frac{3}{12}, \frac{8}{12}\) are required like fractions.

Solution :
Least common multiple of 6 and 5 is 30 So, make 30 as common denominator \(\frac{5}{6}=\frac{5 \times 5}{6 \times 5}=\frac{25}{30}, \frac{4}{5}=\frac{4 \times 6}{5 \times 6}=\frac{24}{30}\) So, \(\frac{25}{30}, \frac{24}{30}\) are required like fractions.

Solution :
Least common multiple of 8 and 6 is 24 So, make 24 as common denominator \(\frac{3}{8}=\frac{3 \times 3}{8 \times 3}=\frac{9}{24}, \frac{1}{6}=\frac{1 \times 4}{6 \times 4}=\frac{4}{24}\) So, \(\frac{9}{24}, \frac{4}{24}\) are required like fractions.

Solution :
Least common multiple of 6 and 9 is 18 So, make 18 as common denominator \(\frac{1}{6}=\frac{1 \times 3}{6 \times 3}=\frac{3}{18}, \frac{4}{9}=\frac{4 \times 2}{9 \times 2}=\frac{8}{18}\) So, 3/18 and 8/18 are the required like fractions.

Comparing like fractions
Example (1) A strip was divided into 5 equal parts. It means that each part is 1/5 . The coloured part is \(\frac{3}{5}=\frac{1}{5}+\frac{1}{5}+\frac{1}{5}\).

The white part is \(\frac{2}{5}=\frac{1}{5}+\frac{1}{5}\). The coloured part is bigger than the white part. This tells us that 3/5 is greater than 2/5. This is written as 3/5 > 2/5.

Example (2) This strip is divided into 8 equal parts. 3 of the parts have one colour and 4 have another colour. Here, 3/8 < 8/4.

In like fractions, the fraction with the greater numerator is the greater fraction.

Comparing fractions with equal numerators
You have learnt that the value of fractions with numerator 1 decreases as the denominator increases.

Even if the numerator is not 1, the same rule applies so long as all the fractions have a common numerator. For example, look at the figures below. All the strips in the figure are alike.
2 of the 3 equal parts of the strip
2 of the 4 equal parts of the strip
2 of the 5 equal parts of the strip
The figure shows that 2/3 > 2/4 > 5/2.

Of two fractions with equal numerators, the fraction with the greater denominator is the smaller fraction.

Comparing unlike fractions
Teacher : Suppose we have to compare the unlike fractions 3/5 and 4/7. Let us take an example to see how this is done. These two boys are standing on two blocks. How do we decide who is taller?

Sonu : But the height of the blocks is not the same. If both blocks are of the same height, it is easy to tell who is taller.

Nandu : Now that they are on blocks of equal height, we see that the boy on the right is taller.

Teacher : The height of the boys can be compared when they stand at the same height. Similarly, if fractions have the same denominators, their numerators decide which fraction is bigger.

Nandu : Got it! Let’s obtain the same denominators for both fractions.

Sonu : 5 × 7 can be divided by both 5 and 7. So, 35 can be the common denominator.

To compare unlike fractions, we convert them into their equivalent fractions so that their denominators are the same.

Fractions Problem Set 18 Additional Important Questions and Answers

Question 1.
\(\frac{5}{9}, \frac{17}{36}\)
Solution :
36 is the multiple of 9 So, make 36 the common denominator \(\frac{5}{9}=\frac{5 \times 4}{9 \times 4}=\frac{20}{36}\), Thus 20/36 and 17/36 are the required like fractions.

Question 2.
\(\frac{5}{6}, \frac{7}{9}\)
Solution:
Least common multiple of 6 and 9 is 18 So, make 18 as the common denominator \(\frac{5}{6}=\frac{5 \times 3}{6 \times 3}=\frac{15}{18}, \quad \frac{7}{9}=\frac{7 \times 2}{9 \times 2}=\frac{14}{18}\) So, 15/18 and 14/18 are the required like fractions.

Question 3.
\(\frac{7}{11}, \frac{3}{5}\)
Solution:
Least common multiple of 11 and 5 is 55 So, make 55 as the common denominator. \(\frac{7}{11}=\frac{7 \times 5}{11 \times 5}=\frac{35}{55}, \frac{3}{5}=\frac{3 \times 11}{5 \times 11}=\frac{33}{55}\). Thus 35/55 and 33/55 are required like fractions.

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 17 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 17

Question 1.
Write the proper number in the box.

Answer:
Here 20 = 2 x 10

Answer:
Here 15 = 3 x 5

Answer:
Here 18 = 9 x 2
hehce,

Answer:
Here 40 + 5 = 8,
hence,

Answer:
Here 26 ÷ 2 = 13,
hence,

Answer:
Here 6 ÷ 2 = 3,
hence,

Answer:
Here 4 ÷ 4 = 1,
hence,

Answer:
Here 25 ÷ 5 = 5,
hence,

Question 2.
Find an equivalent fraction with denominator 18, for each of the following fractions.

Solution:

Question 3.
Find an equivalent fraction with denominator 5, for each of the following fractions.

Solution:

Question 4.
From the fractions given below, pair off the equivalent fractions.

Solution:

Question 5.
Find two equivalent fractions for each of the following fractions.

Solution:

Like fractions and unlike fractions

Fractions such as \(\frac{1}{7}, \frac{4}{7}, \frac{6}{7}\) whose denominators are equal, are called ‘like fractions’.
Fractions such as \(\frac{1}{3}, \frac{4}{8}, \frac{9}{11}\) which have different denominators are called unlike fractions’.

Converting unlike fractions into like fractions

Example (1) Convert 5/6 and 7/9 into like fractions.
Here, we must find a common multiple for the numbers 6 and 9.
Multiples of 6 : 6, 12, 18, 24, 30, 36, ……..
Multiples of 9 : 9, 18, 27, 36, 45 ……..
Here, the number 18 is a multiple of both 6 and 9. So, let us make 18 the denominator of both fractions.

Thus, 15/18 and 1418 are like fractions, respectively equivalent to 5/6 and 7/9.
Here, 18 is a multiple of both 6 and 9. We could also choose numbers like 36 and 54 as the common denominators.

Example (2) Convert 4/8 and 5/16 into like fractions.
As 16 is twice 8, it is easy to make 16 the common denominator.
\(\frac{4}{8}=\frac{4 \times 2}{8 \times 2}=\frac{8}{16}\) Thus, 8/16 and 5/16 are the required like fractions.

Example (3) Find a common denominator for 4/7 and 3/4.
The number 28 is a multiple of both 7 and 4. So, make 28 the common denominator. . Therefore, 16/28 and 21/28 are the required like fractions.

Fractions Problem Set 17 Additional Important Questions and Answers

Question 1.
Find two equivalent fractions for each of the following fraction:

Solution:

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 11 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 11

Question 1.
Subtract the following:

(1) 8,57,513 – 4,82,256
Solution:

Answer:
3,75,257

(2) 13,17,519 – 10,07,423
Solution:

Answer:
3,10,096

(3) 68,34,501 – 23,57,823
Solution:

Answer:
44,76,678

(4) 45,43,827 – 12,05,938
Solution:

Answer:
33,37,889

(5) 70,12,345 – 28,64,547
Solution:

Answer:
41,47,798

(6) 38,01,213 – 37,54,648
Solution:

Answer:
46,565

Study the following word problem.

In 2001, the population of a city was 21,43,567. In 2011, it was 28,09,878. By how much did the population grow?

The population grew by 6,66,311.

Addition and Subtraction Problem Set 11 Additional Important Questions and Answers

Subtract the following:

(1) 53,14,018 – 43,14,019
Solution:

Answer:
9,99,999

(2) 67,05,136 – 34,56,789
Solution:

Answer:
32,48,347

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

Maharashtra State Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 13

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

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 12 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 12

Question 1.
Solve the following word problems:

1. Prathamesh wants to buy a laptop worth 27,450 rupees. He has 22,975 rupees. What is the amount he still needs to be able to buy the laptop?
Solution:
₹ 2 7 4 5 0 Laptop worth
–
₹ 2 2 9 7 5 Prathmesh has
0 4 4 7 5 Require more
Answer:
₹ 04,475 amount require to buy the laptop,

2. A company produced 44,730 scooters in a certain year and 43,150 in the next. How many more scooters did they produce in the previous year?
Solution:
4 4 7 3 0 In previous year
–
4 3 1 5 0 In next year
0 1 5 8 0
Answer:
1,580 scooters produced more in the previous year.

3. In a certain city, the number of men is 16,37,856 and the number of women is 16,52,978. By how many does the number of women exceed the number of men?
Solution:
1 6 5 2 9 7 8 Women
–
1 6 3 7 8 5 6 Men
0 0 1 5 1 2 2
Answer:
15,122 women more than number of men.

4. An organization decided to collect 25,00,000 rupees for a certain project. They collected 26,57,340 through donations and other kinds of aid. By how much did they exceed their target?
Solution:
2 6 5 7 3 4 0 collected
–
2 5 0 0 0 0 0 decided to collect
0 1 5 7 3 4 0 collected more
Answer:
1,57,340

5. Use the numbers 23,849 and 27,056 to make a subtraction problem. Solve the problem.
Solution:
In a certain shop the price of a computer was ₹ 23,849 and that of TV set is ₹ 27,056. Price of TV set is how much more than that of a computer.
₹ 2 7 0 5 6 Price of T.V. sets
₹ 2 3 8 4 9 Price of computer
0 3 2 0 7 Price is more
Answer:
Price of TV set is more than computer by ₹ 3,207

Mixed examples
Study the following solved examples.

Example (1)
4,13,758 + 2,09,542 – 5,16,304
4,13,758 + 2,09,542 – 5,16,304 = 1,06,996

Example (2)
345678 – 162054 + 600127
345678 – 162054 + 600127 = 7,83,751

In these examples, both operations, addition and subtraction, have to be done. They are done in the order in which they are given. In actual cases, we need to consider the specific problem to decide which operation must be done first.

Example (3)
The total amount spent on building a certain house was ₹ 87,14,530. Of this amount, ₹ 24,72,615 were spent on buying the plot of land, ₹ 50,43,720 on the construction material and the rest on labour charges. What was the amount spent on labour?

Method : 1
8 7 1 4 5 3 0 → Total amount spent
–
2 4 7 2 6 1 5 → Cost of plot
6 2 4 1 9 1 5 → Cost of material and labour

6 2 4 1 9 1 5 → Cost of material and labour
–
5 0 4 3 7 2 0 → Cost of material
1 1 9 8 1 9 5 → Amount spent on labour

Method : 2
2 4 7 2 6 1 5 Cost of plot
+
5 0 4 3 7 2 0 Cost of material
7 5 1 6 3 3 5 Cost of plot and material

8 7 1 4 5 3 0 Total amount spent
–
7 5 1 6 3 3 5 Cost of plot and material
1 1 9 8 1 9 5 Amount spent on labour

Let us verify our answer.
2 4 7 2 6 1 5 Cost of plot
+
5 0 4 3 7 2 0 Cost of material
+
1 1 9 8 1 9 5 Amount spent on labour
8 7 1 4 5 3 0 Total cost

The sum total of all the amounts spent tallies with the given total cost. It means that our answer is correct.

Addition and Subtraction Problem Set 12 Additional Important Questions and Answers

Solve the following word problems:

Question 1.
Jethalal purchased goods for ₹ 53,25,675 sold it for ₹ 62,14,563. How much he obtained more in this transaction?
Solution:
6 2 1 4 5 6 3 Sold price
5 3 2 5 6 7 5 Purchased price
8 8 8 8 8 8 Obtained more
Answer:
₹ 8,88,888

Question 2.
In an election candidate A got 13,90,211 votes and candidate B got 8,57,143 votes. By how many votes the winner A defeated the looser B?
Solution:
1 3 9 0 2 1 1 Votes obtained by A
–
8 5 7 1 4 3 Votes obtained by B
0 5 3 3 0 6 8 Votes more obtained by A
Answer:
5,33,068

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

Maharashtra State Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 9

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

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 10 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 10

Question 1.
Subtract the following:

(1) 64293
–
28547
________
________
Solution:
6 4 2 9 3
–
2 8 5 4 7
Answer:
3 5 7 4 6

(2) 37058
–
23469
________
________
Solution:
3 7 0 5 8
–
2 3 4 6 9
Answer:
1 3 5 8 9

(3) 71540
–
58628
________
________
Solution:
7 1 5 4 0
–
5 8 6 2 8
Answer:
1 2 9 1 2

(4) 50432
–
48647
________
________
Solution:
5 0 4 3 2
–
4 8 6 4 7
Answer:
0 1 7 8 5

Subtraction of six and seven-digit numbers

You have learnt to carry out subtractions of five-digit numbers. Using the same method, we can do subtractions of numbers with more than five digits. Study the following examples.

Subtract :
Example (1) 65,07,843 – 9,25,586

Example (2) 34,61,058 – 27,04,579

As shown in the above example, learn to subtract by keeping the borrowed numbers in your mind without writing them down.

Subtraction by another method Before subtracting one number from another, if we add 10 or 100 to both of them, the difference remains the same. Let us use this fact.

Example : Subtract : 724 – 376
As we cannot subtract 6 from 4, we shall add a ten to both the numbers. For the upper number, we untie one tens. We add those ten units to 4 units to get 14 units.

We write the tens added to the lower number below it, in the tens place.

We subtract 6 units from 14.

Now, we cannot subtract ( 7 + 1) i.e. 8 tens from 2 tens. So, we add one hundred to both the numbers. For the upper number, we untie the hundred and add the ten tens to 2 tens. To add the hundred to the lower number, we write it below, in the hundreds place. 12 tens minus 8 tens is 4 tens. And 7 hundreds minus (3 + 1) i.e. 4 hundreds is 3 hundreds. Hence, the difference is 348.

Example (1)

Example (2)

Roman Numerals Problem Set 5 Additional Important Questions and Answers

Question 1.
Subtract the following:

(1) 81,345 – 35,667
Solution:
8 1 3 4 5
–
3 5 6 6 7
Answer:
4 5 6 7 8

(2) 64,723 – 52,378
Solution:
6 4 7 2 3
–
5 2 3 7 8
Answer:
1 2 3 4 5

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 2 Number Work Problem Set 6 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 2 Number Work Problem Set 6

Question 1.
Write the proper symbol, ‘<’ or ‘>’ in the box.
(1) 5,705 [ < ] 15,705
(2) 22,74,705 [  ] 12,74,705
(3) 35,33,302 [  ] 35,32,302
(4) 99,999 [  ] 9,99,999
(5) 4,80,009 [  ] 4,90,008
(6) 35,80,177 [  ] 35,88,172
Answer:
(1) <
(2) >
(3) >
(4) <
(5) <
(6) <

Question 2.
Solve the problems given below.

(1) The Swayamsiddha Savings Group made 3,45,000 papads while the Swabhimani Group made 2,95,000. Which group made more papads?
Answer:
Here, 3,45,000 > 2,95,000
Hence, the Swayamsiddha saving group made more papads.

(2) Children of the Primary School in Ahmadnagar District collected 2,00,000 seeds while those in Pune District collected 3,25,000. Which children collected more seeds?
Answer:
Here, 3,25,000 > 2,00,000
Hence, Pune District children collected more seeds.

(3) The number of people who took part in the Republic Day flag hoisting ceremony was 2,01,306 in Pandharpur taluka and 1,97,208 in Malshiras taluka. In which taluka did a larger number of people participate?
Answer:
Here, 2,01,306 > 1,97,208
Hence, people of Pandharpur taluka participated in larger number

(4) At an exhibition, the Annapoorna Savings Group sold goods worth 5,12,345. The Nirman Group sold goods worth 4,12,900. This figure was 4,33,000 for the Srujan Group and 5,11,937 for the Savitribai Phule group.

Which group had the largest sales?

Which group had the smallest?

Write the sales figures in ascending order.
Answer:
Among the numbers 5,12,345; 4,12,900; 4,33,000; 5,11,937

5,12,345 is largest and 4,12,900 is smallest. Hence, Annapoorna group had the largest sale and Nirman Group had the smallest sales.

Sales in ascending order

4,12,900 < 4,33,000 < 5,11,937 < 5,12,345

Introducing crores

99,99,999 is the biggest seven-digit number. On adding the number 1 to it, we get the smallest eight-digit number, 1,00,00,000. We read this number as ‘one crore’. The new place created to write this number is called the ‘crores’ place.

From the following examples, you can learn to read eight-digit numbers.

Number – Reading

8,45,12,706 – Eight crore forty-five lakh twelve thousand seven hundred and six
5,61,63,589 – Five crore sixty-one lakh sixty-three thousand five hundred and eighty-nine
6,09,04,034 – Six crore nine lakh four thousand and thirty-four

Something more

On the left of the crores place are the places for ten crores, abja, and ten abja in that order. The place value of each of these is ten times the value of the one on its right. According to the Census of the year 2011, the population of our country is 1,21,01,93,422. We read this as ‘one Abuja twenty-one crore one lakh ninety-three thousand four hundred and twenty-two.

Roman Numerals Problem Set 4 Additional Important Questions and Answers

Question 1.
Write the proper symbol, ‘<‘ or ‘>’ in the box.
(1) 68,34,170 [     ] 8,43,170
(2) 5,04,132 [     ] 5,04,123
(3) 1,01,001 [     ] 1,00,101
(4) 14,55,432 [     ] 4,54,532
Answer:
(1) >
(2) >
(3) >
(4) >

Question 2.
Write the numbers in words.

(1) 15,97,21,409
Answer:
Fifteen crore, ninety-seven lakh, twenty-one thousand, four hundred and nine

(2) 99,99,99,999
Answer:
Ninety-nine crore, ninety-nine lakh, ninety- nine thousand, nine hundred and ninety nine.

(3) 7,54,21,607
Answer:
Seven crore, fifty-four lakh, twenty-one thousand, six hundred and seven.

(4) 5,16,36,854
Answer:
Five crore, sixteen lakh, thirty-six thousand, eight hundred and fifty four.

Question 3.
Write in figures.

(1) One crore, fifteen lakh, fifty-nine thousand, seven hundred and four
Answer:
1,15,59,704

(2) Sixty-five crore, seventy lakh, fifty thousand and thirty nine
Answer:
65,70,50,039

(3) Four crore, fifty-nine lakh, fourty-three thousand, five hundred and thirty four
Answer:
4,59,43,534

(4) Eighteen crore, seventy-six lakh, fifty-four thousand and one
Answer:
18,76,54,001

Question 4.
Fill in the blanks in the table below:

Answer:

Question 5.
Write the following numbers in words.
(1) 17,301
(2) 45,019
(3) 40,018
(4) 28,740
Answer:
(1) Seventeen thousand, three hundred and one.
(2) Forty-five thousand and nineteen.
(3) Forty thousand and eighteen
(4) Twenty-eight thousand seven hundred and forty

Question 6.
How many rupees do they make?
(1) 8 notes of rupees 2,000, 3 notes of rupees 100,11 notes of rupees 10.
Answer:
16,410

(2) 9 notes of rupees 2,000, 18 notes of rupees 100,18 notes of rupees 50,18 notes of rupees 10.
Answer:
20,880

(3) Write the smallest and the biggest five-digit numbers that can be made using the digits only once.
(a) 6, 8, 0,1, 9
(b) 3, 5,1,2, 8
Answer:
Smallest number : (i) 10,689 (ii) 12358
Biggest number : (i) 98,610 (ii) 85321

(4) Write the smallest and the biggest number from the following numbers.
(a) 35,798
(b) 39,785
(c) 39,587
(d) 35,789
Answer:
Smallest number : 35,789
Biggest number : 39,785

(5) Write the number from the given number which is neither biggest nor smallest.
(a) 45, 798
(b) 45, 789
(c) 45, 897
Answer:
45,798.

(6) Write the biggest and the smallest three-digit numbers that can be made using the digits 0,1, 2, 3, 4, 5, 6, 7, 8, 9 only once.
Answer:
Biggest three-digit number : 987
Smallest three-digit number : 102

Question 7.
Read the numbers and write them in words.
(1) 2,65,048
(2) 1,80,794
(3) 1,06,709
(4) 8,80,006
Answer:
(1) Two lakh sixty-five thousand and forty- eight,
(2) One lakh eighty thousand seven hundred and ninety-four.
(3) One lakh six thousand seven hundred and nine.
(4) Eight lakh eighty thousand and six.

Question 8.
Read the numbers and write them in figures.
(1) Two lakh five thousand three hundred and six.
(2) Six lakh and six
(3) Nine lakh forty thousand and thirty seven.
(4) Five lakh ninety-nine thousand and fifteen.
Answer:
(1) 2,05,306
(2) 6,00,006
(3) 9,40,037
(4) 5,99,015

Question 9.
Write six, six-digit numbers using the digits 0,.3,5,7,9,1 only once with 9 lakh fifty-seven thousand in all numbers.
Answer:
(1) 9,57,301
(2) 9,57,310
(3) 9,57,103
(4) 9,57,130
(5) 9,57,013
(6) 9,57,031

Question 9.
(A) Match the columns:

(A)	(B)
(1) Nine lakh nine thousand nine	(a) 9,09,090
(2) Nine lakh june thousand nine hundred nine	(b) 9,90,090
(3) Nine lakh nine thousand ninety	(c) 9,09,009
(4) Nine lakh ninety thousand ninety	(d) 9,09,909

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

(B) Match the columns:

(A)	(B)
(1) Thirty-three lakh, three thousand and three	(a) 33,30,300
(2) Thirty-three lakh, thirty thousand, three hundred	(b) 33,03,003
(3) Thirty lakh, three thousand and thirty.	(c) 30,30,003
(4) Thirty lakh, thirty thousand and three	(d) 30,03,030

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

Question 10.
Read the numbers and write them in words.
(1) 34,87,569
(2) 70,85,039
(3) 48,07,102
(4) 67,40,960
(5) 88,00,080
(6) 40,40,004
Answer:
(1) Thirty-four lakh, eighty-seven thousand, five hundred and sixty-nine.
(2) Seventy lakh, eight-five thousand and thirty-nine.
(3) Forty-eight lakh, seven thousand, one hundred and two.
(4) Sixty-seven lakh, forty thousand, nine hundred and sixty.
(5) Eighty-eight lakh and eighty.
(6) Forty lakh, forty thousand and four.

Question 11.
Read the numbers and write them in figures.
(1) Fifty-nine lakh, seven thousand, seventeen.
(2) Twenty-two lakh, ten thousand, five hundred.
(3) Fifty-two lakh, twenty-five thousand, four hundred and fifteen.
(4) Thirty lakh, thirty thousand and thirty.
Answer:
(1) 59,07,017
(2) 22,10,500
(3) 52,25,415
(4) 30,30,030

Question 12.
Write the place value of the underlined digit.
(1) 68,03,512
(2) 3,42,157
(3) 84,52,170
(4) 79,345
(5) 38,14,093
(6) 8,10,618
(7) 35,10,387
Answer:
(1) 8,00,000
(2) 40,000
(3) 2,000
(4) 5
(5) 90
(6) 600
(7) 30,00,000

Question 13.
Write the numbers in their expanded form.
(1) 78,15,692
(2)50,95,182
(3)6,40,078
(4) 9,58,802
Answer:
(1) 70,00,000 + 8,00,000 + 10,000 + 5,000 + 600 + 90 + 2
(2) 50,00,000 + 90,000 + 5,000 + 100 + 80 + 2
(3) 6,00,000 + 40,000 + 70 + 8
(4) 9,00,000 + 50,000 + 8,000 + 800 + 2

Question 14.
Write the place name and place value of each digit in the following numbers.
(1) 27,306
(2) 1,70,425
(3) 75,68,041
(4) 55,555
Answer:
(1) 27,306
(2) 1,70,425
(3) 75,68,041
(4) 55,555

Question 15.
The expanded form of the number is given. Write the number.
(1) 70,000 + 6,000 + 500 + 40 + 8
(2) 8,00,000 +-30,000 + 5,000 + 400 + 3
(3) 60,00,000 + 2,00,000 + 70 + 4
(4) 20,00,000 + 5,00,000 + 900 + 5
Answer:
(1) 76,548
(2) 8,35,403
(3) 62,00,074
(4) 25,00,905

Question 16.
Considering the number 50,43,176.
Fill in the blanks.
(1) The digit in the ten thousand place is ……………………………………….. .
(2) Place value of 1 is ……………………………………….. .
(3) The digit in the lakhs place is ……………………………………….. .
(4) Place value of 5 is ……………………………………….. .
(5) The digit 7 is in ……………………………………….. place.
Answer:
(1) 4
(2) 100
(3) 0
(4) 50,00,000
(5) tens

Question 17.
Write the proper symbols ‘<‘ or ‘>’ in the box.
(1) 12,625 [     ] 21,526
(2) 23,564 [     ] 23,546
(3) 36,60,660 [     ] 36,60,606
(4) 89,14,507 [     ] 89,15,407
Answer:
(1) <
(2) >
(3) >
(d) <

Question 18.
Solve the problems given below.
(1) Population of city A is 8,57,238 and that of city B is 8,75,461. Population of which city is more?
Answer:
city B

(2) Yearly income of Rajnikant is? 3,48,600 and that of Shashikant is? 3,46,500. Whose income is less?
Answer:
Shashikant

Question 19.
Profit of the four companies A, B, C, D is as follows.

Now, answer the following questions.
(1) Which company made maximum profit?
(2) Which company made minimum profit?
(3) Write the profit of the companies in the descending order.
Answer:
(1) B
(2) C
(3) profit of company B > D > A > C

Question 20.
In a certain election, candidates : Tavade, Patel, Chauhan, and Shinde got the votes as follows.

Now, answer the following questions.
(1) Who got the highest number of votes?
(2) Who got the least number of votes?
(3) Write the number of votes obtained in the ascending order.
Answer:
(1) Patel
(2) Shinde
(3) 34,67,008 < 37,51,386 < 43,51,239 < 48,00,173

Question 21.
Compare the following using >, < or = signs.
(1) 3,97,48,632 [     ] 3,97,58,632
(2) 1,50,15,178 [     ] 1,50,15,780
(3) 3,74,98,561 [     ] 96,42,748
(4) 30,49,75,831 [     ] 30,49,00,831
Answer:
(1) <
(2) <
(3) >
(4) >

Question 22.
Circle the correct answer:
(1) Mark periods 617231801 according to the Indian Number system.
(a) 61,72,31,801
(b) 16,172,31
(c) 617,231,801
Answer:
(a) 61,72,31,801

(2) Mark periods 90289164 according to the international Number system.
(a) 9,0289,164
(b) 902891,64
(c) 90,289,164
Answer:
(c) 90,289,164

(3) 1,00,00,000 is read as ……………………………….. .
(a) ten crore
(b) one crore
(c) hundred thousand
Answer:
(b) one crore