JEE Main 2021 (March) Q1: 16 March (Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Sets and Relations: Let $A = {2, 3, 4, 5, ..., 30}$ and ' $\sim$ ' be an equivalence relation on $A \times A$ , defined by $(a, b) \sim (c, d)$ , if and only if $a d = b c$ . Then the number of ordered pairs which satisfy this equivalence relation with ordered pair $(4, 3)$ is equal to

Q: Wait, why do we use $c = 4k$ and $d = 3k$? Can't we just pick random numbers?

Great question! If $\frac{c}{d} = \frac{4}{3}$, then $c$ must be a multiple of 4 and $d$ must be a multiple of 3, provided the fraction is in its simplest form. By introducing the parameter $k$, we ensure that the ratio is maintained while allowing us to scale the pair up to find all possible candidates within the set $A$.

Q: Why is $k$ restricted to integers? Can't $k$ be a fraction like $1.5$?

Remember, the set $A$ contains only integers. If $k$ were $1.5$, then $c = 4 \times 1.5 = 6$ and $d = 3 \times 1.5 = 4.5$. Since $4.5$ is not in the set $A$, it cannot be part of an ordered pair $(c, d)$ where both elements belong to $A$. Thus, $k$ must be an integer to ensure $c$ and $d$ remain within the defined set.

Q: What if the ratio was something like $\frac{6}{9}$? Would the number of pairs change?

That is a brilliant observation! If the ratio were $\frac{6}{9}$, you would first simplify it to $\frac{2}{3}$. If you used $c=6k$ and $d=9k$, you would miss pairs like $(2, 3)$ or $(4, 6)$. Always simplify the ratio to its lowest terms before introducing the multiplier $k$ to ensure you capture every possible pair.

Q: Is there a faster way to find the number of pairs without writing out all the inequalities?

Once you have the simplified ratio $\frac{a}{b} = \frac{p}{q}$, the constraints are $p \cdot k \le 30$ and $q \cdot k \le 30$. You simply need to find the largest possible $k$ such that $\max(p, q) \cdot k \le 30$. In our case, $\max(4, 3) = 4$, so $4k \le 30$ gives $k \le 7.5$. Since $k \ge 1$, there are exactly 7 values.

Select Answer:

Visualized Solution

Understanding the Set $A$ and Relation $\sim$

Given set $A = {2, 3, 4, 5, \dots, 30}$
Equivalence relation $\sim$ on $A \times A$ is defined as:
$(a, b) \sim (c, d) ⟺ a d = b c$

Targeting the Pair $(4, 3)$

We need to find $(c, d)$ such that $(4, 3) \sim (c, d)$
Applying the definition: $4 \times d = 3 \times c$

Converting to a Ratio

Rearranging the equation: $\frac{c}{d} = \frac{4}{3}$
This represents a line passing through the origin.

Introducing Multiplier $k$

Let $c = 4 k$ and $d = 3 k$ for some integer $k$ .
Since $c, d \in A$ , they must be integers, so $k$ must be an integer.

Bounding $c$

Constraint for $c$ : $2 \leq 4 k \leq 30$
Dividing by 4: $\frac{2}{4} \leq k \leq \frac{30}{4}$
$0.5 \leq k \leq 7.5$

Bounding $d$

Constraint for $d$ : $2 \leq 3 k \leq 30$
Dividing by 3: $\frac{2}{3} \leq k \leq \frac{30}{3}$
$0.66 \leq k \leq 10$

Finding Valid $k$ Values

Intersection of bounds: $0.66 \leq k \leq 7.5$
Since $k$ is an integer: $k \in {1, 2, 3, 4, 5, 6, 7}$

Generating the Ordered Pairs

For $k = 1, 2, \dots, 7$ , we get 7 distinct pairs.
Example: $k = 2 ⟹ (8, 6)$
Total number of ordered pairs is 7.

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The Sigma Insight: Equivalence Relations

Solution Diagram

Analyzing the Setup

The problem defines a set

A = {2, 3, 4, \dots, 30}

and an equivalence relation on

A \times A

given by

(a, b) \sim (c, d) ⟺ a d = b c

We are tasked with finding the number of ordered pairs

(c, d)

A \times A

that are equivalent to the pair

(4, 3)

Decoding the Relation

The condition

a d = b c

can be rewritten as:

\frac{a}{b} = \frac{c}{d}

For the specific pair

(4, 3)

, the relation implies that any equivalent pair

(c, d)

must satisfy:

\frac{c}{d} = \frac{4}{3}

This indicates that all such pairs

(c, d)

must represent the same ratio, effectively lying on the same line passing through the origin in the Cartesian plane.

The Power of the Multiplier

To find all such pairs, we express

c

and

d

in terms of a common multiplier

k

. Since the ratio is

4 : 3

, we set:

c = 4 k, d = 3 k

Because

c

and

d

must be integers belonging to the set

A

k

must also be a positive integer. This substitution allows us to reduce the problem to finding the valid range of

k

Navigating the Constraints

The elements

c

and

d

are constrained by the set

A = {2, 3, \dots, 30}

. We apply these bounds to our expressions for

c

and

d

For

c

2 \leq 4 k \leq 30 ⟹ 0.5 \leq k \leq 7.5

For

d

2 \leq 3 k \leq 30 ⟹ 0.66 \leq k \leq 10

To satisfy both conditions simultaneously,

k

must lie in the intersection of these intervals:

0.66 \leq k \leq 7.5

The Final Count

Since

k

must be an integer, the possible values for

k

are

{1, 2, 3, 4, 5, 6, 7}

Each value of

k

corresponds to a unique ordered pair

(c, d)

that satisfies the equivalence relation. For example:

k = 1

(c, d) = (4, 3)

k = 7

(c, d) = (28, 21)

Since there are 7 valid values for

k

, there are exactly 7 such ordered pairs in the equivalence class of

(4, 3)

Common Traps & FAQs

Wait, why do we use

c = 4 k

and

d = 3 k

? Can't we just pick random numbers?

Great question! If

\frac{c}{d} = \frac{4}{3}

, then

c

must be a multiple of 4 and

d

must be a multiple of 3, provided the fraction is in its simplest form. By introducing the parameter

k

, we ensure that the ratio is maintained while allowing us to scale the pair up to find all possible candidates within the set

A

Why is

k

restricted to integers? Can't

k

be a fraction like

1.5

Remember, the set

A

contains only integers. If

k

were

1.5

, then

c = 4 \times 1.5 = 6

and

d = 3 \times 1.5 = 4.5

. Since

4.5

is not in the set

A

, it cannot be part of an ordered pair

(c, d)

where both elements belong to

A

. Thus,

k

must be an integer to ensure

c

and

d

remain within the defined set.

Do I need to check if the pair

(4, 3)

itself is included in the count?

Absolutely! The problem asks for the number of ordered pairs

(c, d)

that satisfy the relation with

(4, 3)

. Since

(4, 3)

satisfies the relation with itself (reflexive property), it corresponds to the case where

k = 1

. Our method naturally includes this, so you don't need to worry about missing it.

What if the ratio was something like

\frac{6}{9}

? Would the number of pairs change?

That is a brilliant observation! If the ratio were

\frac{6}{9}

, you would first simplify it to

\frac{2}{3}

. If you used

c = 6 k

and

d = 9 k

, you would miss pairs like

(2, 3)

(4, 6)

. Always simplify the ratio to its lowest terms before introducing the multiplier

k

to ensure you capture every possible pair.

Is there a faster way to find the number of pairs without writing out all the inequalities?

Once you have the simplified ratio

\frac{a}{b} = \frac{p}{q}

, the constraints are

p \cdot k \leq 30

and

q \cdot k \leq 30

. You simply need to find the largest possible

k

such that

max (p, q) \cdot k \leq 30

. In our case,

max (4, 3) = 4

, so

4 k \leq 30

gives

k \leq 7.5

. Since

k \geq 1

, there are exactly 7 values.

Select Answer:

Visualized Solution

Understanding the Set $A$ and Relation $\sim$

Targeting the Pair $(4, 3)$

Converting to a Ratio

Introducing Multiplier $k$

Bounding $c$

Bounding $d$

Finding Valid $k$ Values

Generating the Ordered Pairs

The Sigma Insight: Equivalence Relations

Analyzing the Setup

Decoding the Relation

The Power of the Multiplier

Navigating the Constraints

The Final Count

Similar Questions

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Select Answer:

Visualized Solution

Understanding the Set .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A and Relation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∼

Targeting the Pair .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (4,3)

Converting to a Ratio

Introducing Multiplier .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k

Bounding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c

Bounding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d

Finding Valid .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k Values

Generating the Ordered Pairs

The Sigma Insight: Equivalence Relations

Analyzing the Setup

Decoding the Relation

The Power of the Multiplier

Navigating the Constraints

The Final Count

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Targeting the Pair $(4, 3)$

Introducing Multiplier $k$

Bounding $c$

Bounding $d$

Finding Valid $k$ Values

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