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