Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 16 Preparation for Algebra Problem Set 54 Textbook Exercise Important Questions and Answers.

## Maharashtra State Board Class 5 Maths Solutions Chapter 16 Preparation for Algebra Problem Set 54

Question 1.

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

Answer:

(7 + 6), (8 + 5), (9 + 4). since 7 + 6

= 13,8 + 5

= 13, 9 + 4

= 13.

(7 + 6)

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

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

= (9 + 4).

Question 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.

Answer:

(9 + 9), (20 – 2), (9 x 2), (36 ÷ 2).

since 9 + 9

= 18, 20 – 2

= 18, 9 x 2

= 18 and 36 + 2

= 18, so (9 + 9)

= (20 – 2)

= (9 x 2)

= (36 ÷ 2).

Inequality

The values of 7 + 5 and 7 × 5 are 12 and 35 respectively. It means that they are not equal. To represent ‘not equal’, the symbol ‘≠’ is used.

To show that (7 + 5) and (7 × 5) are not equal, we write (7 + 5) ≠ (7 × 5) in short.

This kind of representation is called an ‘inequality’.

(9 – 5) ≠ (15 ÷ 3) means that the expressions (9 – 5) and (15 ÷ 3) are not equal.

If two expressions are not equal, one of them is greater or smaller than the other.

To show greater or lesser values, we use the symbols ‘<’ and ‘>’. Therefore, these symbols can also be used to show inequalities.

The value of (9 – 5) is 4 and the value of (15 ÷ 3) is 5. 4 < 5, so the relation between (9 – 5) and (15 ÷ 3) can be shown as (9 – 5) < (15 ÷ 3) or (15 ÷ 3) > (9 – 5).

Fill in the boxes between the expressions with <, = or > as required.

(1) (9 + 8) [ ] (30 ÷ 2)

9 + 8 = 17,

30 ÷ 2 = 15

17 > 15

Therefore (9 + 8) [ > ] (30 ÷ 2)

(2) (16 × 3) (4 × 12)

16 × 3 = 48,

4 × 12 = 48,

48 = 48

Therefore (16 × 3) [ = ] (4 × 12)

(3) (16 – 5) [ ] (2 × 7)

16 – 5 = 11,

2 × 7 = 14,

11 < 14

Therefore (16 – 5) [ < ] (2 × 7)

Write a number in the box that will make this statement correct.

(1) (7 × 2) = ( [ ] – 6)

The value of the expression 7 × 2 is 14, so the number in the box has to be one that gives 14 when 6 is subtracted from it. Subtracting 6 from 20 gives us 14.

Therefore (7 × 2) = ( [ 20 ] – 6 )

(2) (24 ÷ 3) < (5 + [ ] )

The value of the expression 24 ÷ 3 is 8, so the number in the box has to be such that when it is added to 5, the sum is greater than 8.

Now, 5 + 1 = 6, 5 + 2 = 7, 5 + 3 = 8. So the number in the box has to be greater than 3.

Therefore, writing any number like 4, 5, 6 … onwards will do. It means that this problem has several answers. (24 ÷ 3) < (5 + [ 4 ] ) is one among many answers. Even if that is true, writing only one answer will be enough to complete this statement.

**Preparation for Algebra Problem Set 54 Additional Important Questions and Answers**

Question 1.

Fill in the blanks.

(1) 7 + 3 = …………….. – ……………..

(2) 7 + 3 = …………….. x ……………..

(3) 7 + 3 = …………….. + ……………..

Answer:

(1) 7 + 3 = 10 and 12 – 2 = 10 or 15 – 5 = 10

(2) 7 + 3 = 10 and 10 x 1 = 10 or 5 x 2 = 10

(3) 7 + 3 = 10 and 20 + 2 = 10 or 30 + 3 = 10

Question 2.

Write the proper number in the box.

(1) 7 + 8 = 10 + [ ]

(2) 7 + 8 = 20 – [ ]

(3) 7 + 8 = 30 + [ ]

(4) 7 + 8 = 5 x [ ]

Answer:

(1) 7 + 8 = 15 so, 10 + [ ] = 15.

∴ [ ] = 15 – 10 = 5

(2) 7 + 8 = 15 s0, 20 – [ ] = 15.

∴[ ] = 20 – 15 = 5

(3) 7 + 8 = 15 so, 30 + [ ] = 15.

∴ [ ] = 30 + 15 = 2

(4) 7 + 8 = 15 so, 5 x [ ] = 15.

∴[ ] = 15 + 5 = 3