# Balbharati Solutions Class 5 Mathematics Preparation for Algebra

## Balbharati Solutions Class 5 Mathematics Preparation for Algebra

Welcome to NCTB Solutions. Here with this post we are going to help 5th class students for the Solutions of Balbharati Class 5 Math Book, Problem Set 54, 55 and 56, Preparation for Algebra. Here students can easily find step by step solutions of all the problems for Preparation for Algebra. Also our Expert Mathematic Teacher’s solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get Preparation for Algebra solutions. Here all the solutions are based on Maharashtra State Board latest syllabus.

Preparation for Algebra all Question Solutions :

Problem Set 54 :

Question no – (1)

Solution :

We know,

= 7 + 6 = 13

= 8 + 5 = 13

= 9 + 4 = 13

Three pairs of numbers whose sum is 13 are,

= (7 + 6) , (8 + 5) , (9 + 4)

Using this three equalities are,

= (7 + 6) = (8 + 5)

= (7 + 6) = (9 + 4)

= (8 + 5) = (9 + 4)

Question no – (2)

Solution :

We know,

= 10 + 8 = 18

= 25 –7 = 18

= 9 × 2 = 18

= 36/2 = 18

Four pairs of numbers, one for each of addition, subtraction, multiplication and division that make the number 18 are,

= (10 + 8) , (25 – 7) , (9 × 2 ) , (36 / 2 )

Equalities for each of them,

= (10 + 8) = (25 – 7)

= (9 × 2 )

= (36/2)

Problem Set 55 :

Question no – (1)

Solution :

(1) (23 + 4) = (4 + 23)

We know,

23 + 4 = 27

And 4 + 23 = 27

So, they are equal

(23 + 4) = (4 + 23) It is right

(2) (9 + 4) > 12

We know,

9 + 4 = 13

13 > 12

(9 + 4) > 12 It is right

(3) (9 + 4) < 12

We know,

9 + 4 = 13

13<12

(9 + 4) 138

We know,

138 = 138

138 > 138 It is wrong

(5) 138 < 138

We know,

138 = 138

138 5

We know,

25/5 = 5

So, 25/5 > 5 It is wrong

(9) (5 × 8) = (8 × 5)

We know,

5×8 = 40

8×5 = 40

40 = 40

(5 × 8) = (8 × 5) It is right

(10) (16 + 0) = 0

We know,

16 + 0 = 16

So,

(16 + 0) = 0 It is wrong

(11) (16 + 0) = 16

We know,

16 + 0 = 16

16 = 16

Thus, (16 + 0) = 0 It is right

(12) (9 + 4) = 12

We know,

9 + 4 = 13

So, (9 + 4) = 12 It is wrong

Question no – (2)

Solution :

(1) (45 ÷ 9) ___ (9-4)

We know,

49 ÷ 9 = 5

9 – 4 = 5

5 = 5

Hence, (45 ÷ 9 = (9-4)

(2) (6 + 1) ___ (3 × 2)

We know,

6 + 1 = 7

3 × 2 = 6

7 is greater than 6

(6 + 1) > (3 × 2)

(3) (12 × 2) ____ (25 + 10)

We know,

12 × 2 = 24

25 + 10 = 35

24 is less than 35

(12 × 2) < (25 + 10)

Question no – (3)

Solution :

(1) (1 × 7) = ( ___× 1)

We know,

1 × 7 = 7

7 × 1 = 7

7 = 7

So, the answer is – (1×7) = (7×1)

(2) (5 × 4) > (7 ×___)

We know,

5 × 4 = 20

7 × 2 = 14

20 is greater than 14

Therefore, the answer is – (5 × 4) > (7 × 2)

(3) (48 ÷ 3) < ( ___× 5)

We know,

48 ÷ 3 = 16

5 × 4 = 20

16 is less than 20

So, the answer is – (48 ÷ 3) (5 ×___ )

(4) (0 + 1) > (5 × ___)

We know,

0 + 1 = 1

5 × 0 = 0

20 is greater than 14

So, the answer is – (0 + 1) > (5 × 0)

(5) (35 ÷ 7) = (___ +___ )

We know,

35 ÷ 7 = 5

3 + 2 = 5

5 = 5

So, The answer is (35 ÷ 7) = (3 + 2)

(6) (6 – ___) < (2 + 3)

We know,

6 – 3 = 3

2 + 3 = 5

3 is less than 5

So, the answer is – (6-3) < (2+3)

Problem Set 56 :

Question no – (1)

Solution :

(1) The sum of any number and zero is the number itself.

Let any number be A

A + 0 = A

(2) The product of any two numbers and the product obtained after changing the order of those numbers is the same.

Let one number be A and another number be B

A × B = AB

change the order

B × A = BA

AB = BA

A × B = B × A

(3) The product of any number and zero is zero.

Let any number be A

A × 0 = 0

Question no – (2)

Solution :

(1) m – 0 = m

Let m be any number

Subtracting zero from any number gives the number itself

(2) n ÷ 1 = n

Let n be any number

Dividing any number by one gives the number itself.

More Solutions :

👉 Three Dimensional Objects and Nets

👉 Patterns

Updated: July 1, 2023 — 7:52 am