Question

Mr. Pandya told students to choose numbers and form two groups X and Y. The following numbers were chosen by the students. GROUP X: 0,-3, 2

GROUP Y: -6, 4, √3

He instructed students to form rational numbers of form where 'a' is to be chosen from group Y and 'b' is to be chosen from box X.

1. Although there are 6 numbers but the students form only 3 rational numbers. Why?

2. List the rational numbers formed by students.

3. How many positive rational numbers are there?

(i)1 (ii) 2 (iii) 3

(iv) 4 4. How many negative rational numbers are there? 2 (iii) 3 (iv) 4

(i)1

(ii)

5. How many integers are there? (i)1 (ii) 2 (iii) 3 (iv) 4

6. If 'a' is chosen from group X and 'b' is chosen from group Y then find the number of rational numbers formed.

7. In question 1 and 5 the numbers were chosen from same groups but still number of rational numbers formed is different. Why?​

Answer 1

Answer:

Answer is as follows

Step-by-step explanation:

Given :

X = 0, -3, 2

Y = -6, 4, $\sqrt{3}$

Solution :

1. Although there are 6 numbers but the students form only 3 rational numbers. Why?

From the given numbers possible pairs are :

$\frac{-6}{0}, \frac{-6}{-3}, \frac{-6}{2} ,\frac{4}{0} ,\frac{4}{-3} ,\frac{\sqrt{3} }{2} ,\frac{\sqrt{3} }{-3} ,\frac{\sqrt{3} }{2} ,\frac{4}{2}$

Here,

$\frac{-6}{0} ,\frac{4}{0} ,\frac{\sqrt{3} }{0} ,\frac{-4}{3} ,\frac{\sqrt{3} }{-3} ,\frac{\sqrt{3} }{2}$  are irrational numbers, so only 3 forms are possible rational numbers.

2. List the rational numbers formed by students.

The rational numbers formed are :

$\frac{-6}{-3} ,\frac{-6}{2} ,\frac{4}{2}$

All these are rational numbers as the value is not recurring.

3. How many positive rational numbers are there?

From given set of numbers, only two positive rational numbers are possible

Those are :

$\frac{-6}{-3} ,\frac{4}{2}$

4. How many negative rational numbers are there?

From the given set of numbers, only one negative rational number is possible.

That is :

$\frac{-6}{2}$

5. How many integers are there?

From the given set of numbers, three integers are possible.

Those are :

$\frac{-6}{-3} ,\frac{-6}{2} ,\frac{4}{2}$

6. If 'a' is chosen from group X and 'b' is chosen from group Y then find the number of rational numbers formed.

If the above condition is applied, then five rational numbers are possible

Those are :

$\frac{0}{-6} ,\frac{0}{4}, \frac{0}{\sqrt{3} } ,\frac{-3}{-6} ,\frac{2}{4}$

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