Class 5

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 21 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 21

Question 1.
Subtract the following.

\(\text { (1) } \frac{5}{7}-\frac{1}{7}\)
Answer:

\(\text { (2) } \frac{5}{8}-\frac{3}{8}\)
Answer:

\(\text { (3) } \frac{7}{9}-\frac{2}{9}\)
Answer:

\(\text { (4) } \frac{8}{11}-\frac{5}{11}\)
Answer:

\(\text { (5) } \frac{9}{13}-\frac{4}{13}\)
Answer:

\(\text { (6) } \frac{7}{10}-\frac{3}{10}\)
Answer:

\(\text { (7) } \frac{9}{12}-\frac{2}{12}\)
Answer:

\(\text { (8) } \frac{10}{15}-\frac{3}{15}\)
Answer:

Question 2.
\(\frac{7}{10}\) of a wall is to be painted. Ramu has painted 410 of it. How much more needs to be painted?
Solution:
To be painted – painted

Answer:
\(\frac{3}{10}\) more needs to be painted.

Addition and subtraction of unlike fractions

Example (1) Add : \(\frac{2}{3}+\frac{1}{6}\)

First let us show the fraction \(\frac{2}{3}\) by coloring two of the three equal parts on a strip.

You have learnt to add and to subtract fractions with common denominators. Here, we have to add the fraction \(\frac{1}{6}\) to the fraction \(\frac{2}{3}\).

So let us divide each part on this strip into two equal parts. \(\frac{4}{6}\) is a fraction equivalent to \(\frac{2}{3}\). Now, as 16 is to be added to \(\frac{2}{3}\) i.e. to \(\frac{4}{6}\), we shall colour one more of the six parts on the strip. Now, the total coloured part is \(\frac{5}{6}\).

Therefore,
That is,

Example (2) Add : \(\frac{1}{2}+\frac{2}{5}\)
Here, the smallest common multiple of the two denominators is 10. So, we shall change the denominator of both fractions to 10.

Example (3) Add : \(\frac{3}{8}+\frac{1}{16}\)
Here, 16 is twice 8. So, we shall change the denominator of both fractions to 16.

Example (4) Subtract : \(\frac{3}{4}-\frac{5}{8}\)
Let us make 8 the common denominator of the given fractions.

Example (5) Subtract : \(\frac{4}{5}-\frac{2}{3}\)
The smallest common multiple of the denominators is 15. So, we shall change the denominator of both fractions to 15.

Addition and Subtraction Problem Set 13 Additional Important Questions and Answers

\(\text { (1) } \frac{9}{14}-\frac{3}{14}\)
Answer:

\(\text { (2) } \frac{5}{6}-\frac{3}{6}\)
Answer:

\(\text { (3) } \frac{9}{16}-\frac{5}{16}\)
Answer:

\(\text { (4) } \frac{7}{8}-\frac{3}{8}-\frac{1}{8}\)
Answer:

(5) \(\frac{7}{9}\) part of the work done by Neha and Supriya together. \(\frac{5}{9}\) part of this work was done by Neha. How much work done by Supriya?
Solution:
Total work done – work done by Neha = work done by Supriya

Answer:
\(\frac{2}{9}\) work done by Supriya

(6) Mr. Sharma is \(\frac{14}{9}\) m tall. Mrs. Sharma is \(\frac{4}{9}\) shorter than him. What is Mrs. Sharma’s height?
Solution:
Mrs. Sharma’s height

Answer:
Mrs. Sharma’s height = \(\frac{10}{9}\)

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 15 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 15

Question 1.
Solve the following and write the quotient and remainder.
(1) 1284 ÷ 32
Solution :

Quotient = 40
Remainder = 4

(2) 5586 ÷ 87
Solution :

Quotient = 64
Remainder =18

(3) 1207 ÷ 27
Solution:

Quotient = 44
Remainder =19

(4) 8543 ÷ 41
Solution :

Quotient = 208
Remainder =15

(5) 2304 ÷ 43
Solution:

Quotient = 53
Remainder = 25

(6) 56,741 ÷ 26
Solution:

Quotient =2182
Remainder = 9

Question 2.
How many hours will it take to travel 336 km at a speed of 48 km per hour?
Solution:
Time = Distance ÷ Speed

Answer:
It will take 7 hours.

Question 3.
Girija needed 35 cartons to pack 1400 books. There are an equal number of books in every carton. How many books did she pack into each carton?
Solution:
No. of cartons x No. of books in each carton = Total no. of books 35 x No. of books in each carton = 1400 No. of books in each carton = 1400 35

Answer:
She packs 40 books in each carton.

Question 4.
The contribution for a picnic was 65 rupees each. Altogether, 2925 rupees were collected. How many had paid for the picnic?
Solution:

Answer:
45 persons paid for the picnic.

Question 5.
Which number, on being multiplied by 56, gives a product of 9688?
Solution:

Answer:
173

Question 6.
If 48 sheets are required for making one notebook, how many notebooks at the most will 5880 sheets make and how many sheets will be left over?
Solution:

Answer:
122 notebooks can be made and 24 sheets left over.

Question 7.
What will the quotient be when the smallest five-digit number is divided by the smallest four-digit number?
Solution:
Smallest five-digit number is 10,000 and smallest four-digit number is 1,000.
So, 10000 ÷ 1000 = 10

Answer:
Quotient = 10

Mixed examples

A farmer brought 140 trays of chilli seedlings. Each tray had 24 seedlings. He planted all the seedlings in his field, putting 32 in a row. How many rows of chillies did he plant?

Let us find out the total number of seedlings when there were 24 seedlings in each of the 140 trays. We shall multiply 140 and 24.

Total number of seedlings 3,360.
To find out how many rows were planted with 32 seedlings in each row, we shall divide 3,360 by 32.
The quotient is 105.
Therefore, the number of rows is 105.
Carry out the multiplication of 105 × 32 and verify your answer.

Multiplication and Division Problem Set 15 Additional Important Questions and Answers

Solve the following and write the quotient and remainder.

(1) 9148 ÷ 37
Solution

Quotient = 247
Remainder = 9

(2) 1175 ÷ 15
Solution :

Quotient =78
Remainder = 5

Solve the following word problems:

(1) If 45 kg of sugar cost 1305 rupees, what is the rate of sugar per kg?
Solution:

Answer:
The rate per kg of sugar is 29 rupees.

(2) 17 people spent ₹ 83,475. How much did each person spend and what is the amount left?
Solution:

Answer:
Each person spent ₹ 4,910 and the amount left is ₹ 5

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 14 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 14

Question 1.
Multiply the following:

(1) 327 × 92
Solution:
3 2 7
x
9 2
_____
6 5 4
+
2 9 4 3 0
Answer:
3 0 0 8 4

(2) 807 × 126
Solution:
8 0 7
x
1 2 6
______
4 8 4 2
+
1 6 1 4 0
+
8 0 7 0 0
Answer:
1 0 1 6 8 2

(3) 567 × 890
Solution:
5 6 7
x
8 9 0
______
0 0 0
+
5 1 0 3 0
+
4 5 3 6 0 0
Answer:
5 0 4 6 3 0

(4) 4317 × 824
Solution:
4 3 1 7
8 2 4
1 7 2 6 8
+ 8 6 3 4 0
3 4 5 3 6 0 0
Answer:
3 5 5 7 2 0 8

(5) 6092 × 203
Solution:
6 0 9 2
x
2 0 3
______
1 8 2 7 6
+
0 0 0 0 0
+
1 2 1 8 4 0 0
Answer:
1 2 3 6 6 7 6

(6) 1177 × 99
Solution:
1 1 7 7
x
9 9
1 0 5 9 3
+
1 0 5 9 3 0
Answer:
1 1 6 5 2 3

(7) 456 × 187
Solution:
4 5 6
x
1 8 7
3 1 9 2
+
3 6 4 8 0
+
4 5 6 0 0
Answer:
8 5 2 7 2

(8) 6543 × 79
Solution:
6 5 4 3
x
7 9
5 8 8 8 7
+
4 5 8 0 1 0
Answer:
5 1 6 8 9 7

(9) 2306 × 832
Solution:
2 3 0 6
x
8 3 2
______
4 6 1 2
+
6 9 1 8 0
+
1 8 4 4 8 0 0
______
Answer:
1 9 1 8 5 9 2

(10) 6429 × 509
Solution:
6 4 2 9
x
5 0 9
______
5 7 8 6 1
+
0 0 0 0 0
+
3 2 1 4 5 0 0
Answer:
3 2 7 2 3 6 1

(11) 4,321 × 678
Solution:
4 3 2 1
x
6 7 8
_____
3 4 5 6 8
3 0 2 4 7 0
2 5 9 2 6 0 0
_______
Answer:
2 9 2 9 6 3 8

(12) 20,304 × 87
Solution:
2 0 3 0 4
x
8 7
1 4 2 1 2 8
+
1 6 2 4 3 2 0
_________
Answer:
1 7 6 6 4 4 8

Question 2.
As part of the ‘Avoid Plastic’ campaign, each of 745 students made 25 paper bags. What was the total number of paper bags made ?
Solution:
7 4 5 Number of students
x
2 5 bags made by each
___________
3 7 2 5
+
1 4 9 0 0
__________
1 8 6 2 5
__________

Answer:
18,625 bags made

Question 3.
In a plantation, saplings of 215 medicinal trees have been planted in each of the 132 rows of trees. How many saplings are there in the plantation altogether ?
Solution:
2 1 5 Saplings in each now
x
1 3 2 Number of rows
__________
4 3 0
+
6 4 5 0
+
2 1 5 0 0
__________
2 8 3 8 0
__________

Answer:
Altogether there is 28,380 saplings.

Question 4.
One computer costs 27,540 rupees. How much will 18 such computers cost ?

Solution:
2 7 5 4 0 Cost of 1 computer
x
1 8 No. of computers
__________
2 2 0 3 2 0
+
2 7 5 4 0 0
___________
4 9 5 7 2 0
__________

Answer:
4,95,720 rupees cost of 18 computers.

Question 5.
Under the ‘Inspire Awards’ scheme, 5000 rupees per student were granted for the purchase of science project materials. If 154 students in a certain taluka were covered under the scheme, find the total amount granted to that taluka.

Solution:
₹ 5 0 0 0 Granted per student
x 1 5 4 Number of students
__________
2 0 0 0 0
+
2 5 0 0 0 0
+
5 0 0 0 0 0
______________
₹ 7 7 0 0 0 0
______________

Answer:
7,70,0000 granted totally

Question 6.
If a certain two-wheeler costs 53,670 rupees, how much will 35 such two-wheelers cost ?
Solution:
5 3 7 6 0 Cost of 1 two-wheeler
x 3 5 No. of two-wheelers
___________
2 6 8 8 0 0
+
1 6 1 2 8 0 0
_____________
1 8 8 1 6 0 0
______________

Answer:
18,81,600 is the total cost of 35- two-wheelers

Question 7.
One hour has 3,600 seconds. How many seconds do 365 hours have ?
Solution:
3 6 0 0 Seconds of 1 hour
x
3 6 5 No. of hours
_________
1 8 0 0 0
+
2 1 6 0 0 0
+
1 0 8 0 0 0 0
______________
1 3 1 4 0 0 0
______________

Answer:
13,14,000 seconds for 365 hours.

Question 8.
Frame a multiplication word problem with the numbers 5473 and 627 and solve it.

Solution:
Cost of one mobile is 5,473. What is the cost of such 627 mobiles?
5 4 7 3 Cost of 1 mobile
x 6 2 7 Number of mobiles
__________
3 8 3 1 1
+
1 0 9 4 6 0
+
3 2 8 3 8 0 0
_____________
3 4 3 1 5 7 1
_____________

Answer:
34,31,571 cost for 627 mobiles.

Question 9.
Find the product of the biggest three-digit number and the biggest four-digit number.

Solution:
9 9 9 9 Biggest four digit no.
x 9 9 9 Biggest three-digit no.
_________
8 9 9 9 1
+
8 9 9 9 1 0
+
8 9 9 9 1 0 0
_____________
9 9 8 9 0 0 1
_____________

Answer:
99,89,001

Question 10.
One traveller incurs a cost of 7,650 rupees for a certain journey. What will be the cost for 26 such travellers?
Solution:
7 6 5 0 Cost of one traveller
x 2 6 No. of travellers
______
4 5 9 0 0
+
1 5 3 0 0 0
_____________
1 9 8 9 0 0
_____________

Answer:
1,98,900 cost of 26 travellers.

Pairing off objects from two groups in different ways

(1) Ajay wants to travel light. So he took with him three shirts – one red, one green and one blue and two pairs of trousers – one white and one black. How many different ways does he have of pairing off a shirt with trousers?

Writing ‘S’ for shirt and ‘T’ for trousers, the possible different pairs are :

(2) Suresh has three balls of different colours marked A, B and C and three bats marked P, Q and R. He wishes to take only one bat and one ball to the playground. In how many ways can he pair off a ball and a bat to take with him?

How many different pairs have been shown here?

(3) The three friends, Sanju, John and Ali went to the fair. A shop there, had five different types of hats. Each of the boys had photos taken of himself, wearing every type of hat, in turn. Find how many photographs were taken in all.

How many different pairs were formed ? That is, how many photos were taken ?

Take two collections, each containing the given number of objects. Make as many different pairs as possible, taking one object from each collection every time. Thus, complete the table below.

What does this table tell us ?
The number of different pairs formed by pairing off objects from two groups is equal to the product of the number of objects in the two groups.

Division
Teacher : You have learnt some things about division. For example, we know that division means making equal parts of a given number, or, subtracting a number repeatedly from a given number. What else do you know ?
Shubha : We know that we get two divisions from one multiplication. From 9 × 4 = 36, we get the divisions 36 ÷ 4 = 9 and 36 ÷ 9 = 4.
Teacher : Very good! Right now, there’s nothing new to learn about division. Only the number of digits in the dividend and the divisor will grow. Tell me what is 354 ÷ 6 ?
Sarang : 354 = 300 + 54. 300 divided by six is 50. And 54 ÷ 6 = 9. Hence the quotient is 50 + 9 = 59.
Teacher : Right! Now let’s learn, step by step, how to divide a four-digit number by a one-digit number. So now, divide 4925 by 7 and tell me the quotient and the remainder.
Shubha : We cannot divide 4 thousands by 7 into whole thousands. Now, 4 Th = 40 H. So let us instead take the 40 hundreds together with 9 hundreds and divide 49 hundreds. 49 ÷ 7 = 7. So, everyone gets 7 hundreds. Now, we cannot divide 2T equally among 7 people. So we must write 0 in the tens place in the quotient. Then on dividing 25 by seven, we get quotient 3 and the remainder is 4. Thus, the answer is quotient 703, remainder 4.

Teacher : Very good ! Now divide 7439 by 9.
Sarang : It’s difficult to do this mentally. I’ll write it down on paper. The quotient is 826 and the remainder, 5.
Teacher : We use the same method to divide a four-digit number by a two-digit number. If necessary, we can prepare the table of the divisor before we start.

Study the solved examples shown below.
Example (1)

Quotient 170, Remainder 4

Example (2)

Quotient 305, Remainder 23

Example (3) Divide. 9842 ÷ 45

We can prepare the 45 times table to do this division.
But when the divisor is a big number, we can solve the example by first guessing what the quotient will be. Let us see how to do that.
We have 0 in the thousands place in the quotient.
Now, to guess the quotient when dividing 98 by 45, look at the first digits in both – the dividend and the divisor. These are 9 and 4, respectively.
Dividing 9 by 4, we will get 2 in the quotient. Let us see if 2 times 45 can be subtracted from 98. 45 × 2 = 90. 90 < 98. So, we write 2 in the hundreds place in the quotient.
Next, dividing 84 by 45 we can easily see that as 90 > 84, we have to write 1 in the tens place in the quotient.
Now, we have to divide 392 by 45. As 3 < 4, let us look at 39, the number formed by the first 2 digits, to guess the next digit in the quotient.
4 × 9 = 36 and 36 < 39. Let us check if the next digit can be 9. 45 × 9 = 405 and 405 > 392. Therefore, 9 cannot be the next digit in the quotient.
Let us check for 8. 45 × 8 = 360. 360 < 392. So, we write 8 in the units place of the quotient.
We subtract 8 × 45 from 392 and complete the division.
The quotient is 218 and the remainder, 32.

Example (4)
If 35 kilograms of wheat cost 910 rupees, what is the rate of wheat per kg?

Weight of wheat in kg × rate of wheat per kg = cost of wheat Hence, 35 × rate per kg = 910
Therefore, when we divide 910 by 35, we will get the per kg rate of wheat.
The rate per kilogram of wheat is 26 rupees.

Multiplication and Division Problem Set 14 Additional Important Questions and Answers

Multiply the following:

(1) 2132 x 231
Solution:
2 1 3 2
x
2 3 1
2 1 3 2
+
6 3 9 6 0
+
4 2 6 4 0 0
____________
Answer:
4 9 2 4 9 2

(2) 1863 x 432
Solution:
1 8 6 3
x
4 3 2
3 7 2 6
+
5 5 8 9 0
+
7 4 5 2 0 0
___________
Answer:
8 0 4 8 1 6

Solve the following word problems:

(1) A factory manufactures 34,796 pairs of socks in one hour. How many pairs will the factory manufacture in one day?
Solution:
3 4 7 9 6 Pairs of socks
x 2 4 No. of hours
______
1 3 9 1 8 4
+
6 9 5 9. 2 0
_____________
8 3 5 1 0 4
_____________

Answer:
8,35,104 pairs of socks manufactured in one day

(2) There are 375 toffees in a box. How many toffees will be there in 632 such boxes?
Solution:
3 7 5 No. of toffees
x 6 3 2 No. of boxes
_______
7 5 0
+
1 1 2 5 0
+
2 2 5 0 0 0
___________
2 3 7 0 0 0
_____________

Answer:
There will be 2,37,000 toffees.

(3) There are 144 articles in a gross. How many articles are there in 2174 gross?
Solution:
2 1 7 4 No. of gross
x 1 4 4 Articles in I gross
______
8 6 9 6
+
8 6 9 6 0
+
2 1 7 4 0 0
_____________
3 1 3 0 5 6
_____________

Answer:
There are 3,13,056 articles in 2174 gross.

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