# Maharashtra Board Class 5 Math Chapter 16 Preparation for Algebra Solution

## Maharashtra Board Class 5 Math Solution Chapter 16 – Preparation for Algebra

Balbharati Maharashtra Board Class 5 Math Solution Chapter 16: Preparation for Algebra. Marathi or English Medium Students of Class 5 get here Preparation for Algebra full Exercise Solution.

 Std Maharashtra Class 5 Subject Math Solution Chapter Preparation for Algebra

### PROBLEM SET 54

1.) Using brackets, write three pairs of numbers whose sum is 13. Use them to write three equalities.

Three pairs of numbers whose sum is 13 are,

(7+6), (8+5), (11+2)

Equalities, (7+6) = (15 – 2 )

(8+5) = (26 ÷ 2)

(11+2) = (20 – 7)

2.) Find four pairs of numbers, one for each of addition, subtraction, multiplication and division that make the number 18. Write the equalities for each of them.

Subtraction, (24 – 6) =18

Multiplication, (9×2) = 18

Division, (36÷2) = 18

Equalities,

(10+8) = (3×6)

(24 – 6) = (9+9)

(9×2) = (54÷3)

(36÷2) = (19 – 1)

### PROBLEM SET 55

1. Say whether right or wrong.

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

It is right because (23+4) and (4+23) both results 27

(2) (9 + 4) > 12                                RIGHT

It is right because (9+4) equals to 13, which is bigger than 12.

(3) (9 + 4) < 12                              WRONG

it is wrong because (9+4) equals to 13, which is bigger than 12. But here it is showing 12 is bigger than 13, which is wrong

(4) 138 > 138                                  WRONG

It is wrong because both of the number is same and in the question it is showing one is bigger than the other.

(5) 138 < 138                                  WRONG

(6) 138 = 138                                   RIGHT

(7) (4 × 7) = 30 – 2                       RIGHT

It is right because

(4×7) = 28, (30 – 2) = 28

Both of them equals to 28, so the above statement is right.

(8) 25/5 > 5                                     WRONG

It is wrong because 25/5 results 5.

(9) (5 × 8) = (8 × 5)                    RIGHT

Right because (5×8) = 40 and (8×5) = 40, both of them equals to 40.

(10) (16 + 0) = 0                             WRONG

It is wrong because (16+0) = 16, not 0. So it isn’t equals to 0.

(11) (16 + 0) = 16                                RIGHT

(12) (9 + 4) = 12                            WRONG

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

(1) (45 ÷ 9) = (9 – 4)

(45 ÷ 9) = 5 and

(9 – 4) = 5, as they have equal results, the answer will be (45 ÷ 9) = (9 – 4)

(2) (6 + 1) > (3×2)

(6+1) = 7 and (3×2) = 6, as 7 is bigger than 6 so the answer will be (6+1) > (3×2).

(3) (12 × 2) < (25 + 10)

(12 × 2) = 24 and (25 + 10) = 35

As 35 is bigger than 24, so the answer will be (12 × 2) < (25 + 10)

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

(1) (1 × 7) = ( ? × 1)

In the right side (1×7) = 7, so we have to find a number that after multiplying with 1 equals to 7,

And that number is 7

Answer is (1 × 7) = ( 7 × 1)

(2) (5 × 4) > (7 × ? )

In the left side (5×4) = 20, as the ‘>’ suggests that 20 should be bigger than the number on the right side.

So, multiples of  7 that is smaller than 20 are (1×7) = 7, (2×7) = 14

So, we can write (5 × 4) > (7 ×1) or (5 × 4) > (7 × 2)

(3) (48 ÷ 3) < ( ? × 5)

In the left side (48÷3) = 16, as the ‘<’ suggests that 16 should be smaller than the number on the right side.

So, multiples of  5 that is bigger than 16 can be (5×4) = 20

So, the answer is (48 ÷ 3) < ( 4 × 5)

(4) (0 + 1) > (5 × ? )

In the left side (0+1) = 1, as the ‘>’ suggests that 1 should be bigger than the number on the right side.

If we multiply 5 with 0, (5×0) = 0, it is the only way to make a smaller number than 1

So, the answer is (0 + 1) > (5 × 0)

(5) (35 ÷ 7) = ( ?+? )

In the leftt side (35 ÷ 7) = 5, so we have to find two numbers that equals to 5, we can write (1+4), (2+3). Both of them equals to 5

So the answer is (35 ÷ 7) = ( 4+1 ), (35 ÷ 7) = ( 2+3 )

(6) (6 – ?) < (2 + 3)

In the right side (2+3) = 5, as the ‘<’ suggests that 5 should be bigger than the number on the left side.

5 would be only bigger if we have smaller numbers like 1,2,3,4

So we can use, (6 – 2) = 4 , (6 – 3) = 3, (6 – 4) = 2, (6 – 5) = 1

### PROBLEM SET 56

1.) Use a letter for ‘any number’ and write the following properties in short.

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

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

Take for example (5+3)

Take the value of a as 5 and b as 3.

Their sum will be a + b

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

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).’

We can prove it,

the value of a as 5 and b as 3.

Their sum will be a + b = (5+3) = 8 and for (b+a) = (3+5) = 8

So, we can say (a+b) = (b+a)

2.) Write the following properties in words :

(1) m – 0 = m

Subtraction of 0 from m gives us m.

The value of n can be any number except zero, take an example,

m = 4

So, if we apply value of m into the equation,

4 – 0 = 4

(2) n ÷ 1 = n

Division of n by 1 gives us n.

The value of n can be any number except zero, take an example,

n = 5

So, if we apply value of n into the equation,

5 ÷ 1 = 5

#### For More Solutions, Click Below:

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Part Two:

Updated: January 7, 2022 — 4:38 pm