NCERT Exercise 9.1 Try These:1-2
Rational Numbers 7th Math
NCERT Exercise 9.1 Try These:1-2
In the book of ncert chapter rational numbers before exercise 9.1 some questions have been given named try these. These questions are given after the solved examples. These questions are equally important. In the section questions given in the try these 1.1 and 1.2 have been solved.
Solution of Try These NCERT Exercise 9.1 questions 1 and 2
Try These 1.1 (1) Is the number 2/– 3 rational? Think about it.
Solution
Yes, 2/– 3 is a rational number.
Explanation A number in the form of p/q where p and q are integers and q ≠ 0, is called a rational number.
Here, since 2/– 3 is in the form of p/q and 2 and – 3 are integers and –3 ≠ 0, thus it is a rational number
Try These 1.1 (2) List ten rational numbers.
Solution
We know that a number in the form of p/q where p and q are integers and q ≠ 0, is called a rational number.
Thus, ten rational numbers are
1/2, – 1/2, 2/3, – 2/3, 4/5, 5/6, 6/7, 7/8, 8/9, 9/10
Try These 1.2 Fill in the boxes
Try These 1.2 (i) 5/4 = ☐/16 = 25/☐ = – 15/☐
Solution
Given,
5/4 = ☐/16 = 25/☐ = – 15/☐
(a) 5/4 = ☐/16
By multiplying the numerator and denominator of a rational number by the same non-zero integer, we obtain another rational number equivalent to the given rational number. Similarly, the division of the numerator and denominator by the same non-zero integers also gives equivalent rational numbers. This is exactly like obtaining equivalent fractions.
In this given set of two rational numbers, if we multiply the 4 by 4 we get 16, thus, to fill the box, we need to multiply the numerator 5 by 4 to get the number required in the blank box.
Thus, 5/4 = [5 ×4]/4 × 4
= 5/4 = [ 20 ]/16
Thus, number in the box = 20
Alternate method to find the missing number in the given set of two rational numbers
(a) 5/4 = ☐/16
Let the number in the box = a
∴ 5/4 = a/16
After cross multiplication, we get
4 a = 16 × 5
⇒ a = 16 × 5/4
⇒ a = 16 4 × 5/4
⇒ a = 4 × 5 = 20
Thus, missing number in the box = 20
Thus, 5/4 = [ 20 ]/16
(b) 5/4 = 25/☐
By multiplying the numerator and denominator of a rational number by the same non-zero integer, we obtain another rational number equivalent to the given rational number.
In this given set of two rational numbers, if we multiply the numerator 5 by 5 we get 25, thus, to fill the box, we need to multiply the denominator 4 by 5 to get the number required in the blank box.
Thus, 5/4 = [5 ×5]/[ 4 × 5 ]
= 5/4 = 25/[ 20 ]
Thus, number in the box = 20
Alternate method to find the missing number in the given set of two rational numbers
(b) 5/4 = 25/☐
Let the number in the box = a
∴ 5/4 = 25/[ a ]
After cross multiplication, we get
5 a = 4 × 25
⇒ a = 4 × 25/5
⇒ a = 4 × 25 5/5
⇒ a = 4 × 5 = 20
Thus, missing number in the box = 20
Thus, 5/4 = 25/[ 20 ]
(c) 5/4 = – 15/☐
By multiplying the numerator and denominator of a rational number by the same non-zero integer, we obtain another rational number equivalent to the given rational number.
In this given set of two rational numbers, if we multiply the numerator 5 by – 3 we get –15, thus, to fill the box, we need to multiply the denominator 4 by – 3 = – 12 to get the number required in the blank box.
= 5/4 = 5 × (– 3)/[ 4 × (– 3)]
= 5/4 = – 15/[ – 12]
Thus, missing number in the box = – 12
Alternate method to find the missing number in the given set of two rational numbers
(c) 5/4 = – 15/☐
Let the missing number in the box = a
∴ 5/4 = – 15/[ a ]
After cross multiplication, we get
5 × a = 4 × (– 15)
a = 4 × (– 15)/5
⇒ a = 4 × (– 15 3)/5
⇒ a = 4 × (– 3) = – 12
Thus, missing number in the box = – 12
Thus, 5/4 = – 15/[ – 12 ]
Therefore, 5/4 = [ 20 ]/16 = 25/[20] = – 15/[ – 12 ]
Try These 1.2 (ii) – 3/7 = ☐/14 = 9/☐ = – 6/☐
Solution
(a) Given, – 3/7 = ☐/14
By multiplying the numerator and denominator of a rational number by the same non-zero integer, we obtain another rational number equivalent to the given rational number.
If the denominator 14 of the second rational number by the denominator 7 of the first rational number we get 2, i.e. 14 ÷ 7 = 2
This, means if we multiply the 7 by 2, we will get 14, i.e. 7 × 2 = 2
Thus, we need to multiply the numerator – 3 of the first rational number by 2 to get the missing number in the box.
Therefore, – 3/7 = [ – 3 × 2 ]/7 × 2
= – 3/7 = [ – 6]/14
Thus, the missing number in the box = – 6
Alternate method to find the missing number in the given set of two rational numbers
(a) Given, – 3/7 = ☐/14
Let the missing number in the box = a
= – 3/7 = [a]/14
After cross multiplication, we get
7 a = – 3 × 14
After transposing 7 to the RHS
⇒ a = – 3 × 14 2/7
⇒ a = – 3 × 2 = – 6
Thus, missing number = – 6
Therefore, – 3/7 = [ – 6]/14
(b) – 3/7 = 9/☐
if 9 is divided by – 3, we get – 3
This means the numerator of the rational number situated in the RHS can be obtained by multiplying – 3 by – 3
Thus, the denominator situated in the RHS can also be obtained by multiplying the denominator of the rational number in LHS by – 3
= – 3/7 = – 3 × (– 3)/[ 7 × (– 3) ]
= – 3/7 = 9/[ – 21 ]
Thus, the missing number = – 21
Alternate method to find the missing number in the given set of two rational numbers
(b) – 3/7 = 9/☐
Let the missing number in the box = a
= – 3/7 = 9/[a]
After cross multiplication, we get
– 3 × a = 7 × 9
After transposing – 3 to RHS, we get
a = 7 × 9/– 3
⇒ a = 7 × (– 3)
⇒ a = – 21
Thus, the missing number = – 21
= – 3/7 = 9/[ – 21 ]
(c) – 3/7 = – 6/☐
if – 6 is divided by – 3, we get 2
This means the numerator of the rational number situated in the RHS can be obtained by multiplying – 3 and 2
Thus, the denominator situated in the RHS can also be obtained by multiplying the denominator of the rational number in LHS by 2
– 3/7 = – 3 × 2/[7 × 2]
– 3/7 = – 6/[ 14 ]
Thus, the missing number in the box = 14
Alternate method to find the missing number in the given set of two rational numbers
(c) – 3/7 = – 6/☐
Let the missing number in the box = a
= – 3/7 = – 6/[a]
After cross multiplication, we get
– 3 × a = – 6 × 7
By transposing – 3 to the RHS, we get
a = – 6 × 7– 3
⇒ a = 2 × 7 = 14
Thus, the missing number in the box = 14
– 3/7 = – 6/[ 14 ]
Thus, the missing numbers in the box are
– 3/7 = [ – 2 ]/14 = 9/[ – 21 ] = – 6/[ 14 ] Answer
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