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Sets and Relations question

2021 · 16 Mar · Shift 2 · Q33
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  5. /2021 · 16 Mar · Shift 2 · Q33

Sets and Relations question

2021 · 16 Mar · Shift 2 · Q33

JEE MainMathematicsSets and RelationsMCQ+4 / −1
Let A = {2, 3, 4, 5, ....., 30} and '≃\simeq≃' be an equivalence relation on A ×\times× A, defined by (a, b) ≃\simeq≃ (c, d), if and only if ad = bc. Then the number of ordered pairs which satisfy this equivalence relation with ordered pair (4, 3) is equal to :
  1. A
    5
  2. B
    6
  3. C
    8
  4. D
    7
View written solutionFree

Correct answer: D

  1. Understand the equivalence relation

We are given: A={2,3,4,…,30}A = \{2,3,4,\dots,30\}A={2,3,4,…,30} and on A×AA \times AA×A the relation (a,b)≃(c,d)  ⟺  ad=bc.(a,b) \simeq (c,d) \iff ad = bc.(a,b)≃(c,d)⟺ad=bc.

We need the number of ordered pairs (a,b)∈A×A(a,b) \in A \times A(a,b)∈A×A such that (a,b)≃(4,3).(a,b) \simeq (4,3).(a,b)≃(4,3).

So put (c,d)=(4,3)(c,d)=(4,3)(c,d)=(4,3). Then (a,b)≃(4,3)  ⟺  a⋅3=b⋅4,(a,b) \simeq (4,3) \iff a\cdot 3 = b\cdot 4,(a,b)≃(4,3)⟺a⋅3=b⋅4, that is, 3a=4b.3a = 4b.3a=4b.


  1. Solve the Diophantine equation

We need integer solutions (a,b)(a,b)(a,b) with a,b∈{2,3,…,30}a,b \in \{2,3,\dots,30\}a,b∈{2,3,…,30} satisfying 3a=4b.3a=4b.3a=4b.

Since gcd⁡(3,4)=1\gcd(3,4)=1gcd(3,4)=1:

  • 4∣a4 \mid a4∣a, so let a=4ka=4ka=4k
  • 3∣b3 \mid b3∣b, so let b=3kb=3kb=3k

Thus all solutions are of the form (a,b)=(4k,3k)(a,b)=(4k,3k)(a,b)=(4k,3k) for some positive integer kkk.


  1. Apply the range restriction

Because a,b∈Aa,b \in Aa,b∈A, 2≤4k≤30,2≤3k≤30.2 \le 4k \le 30, \qquad 2 \le 3k \le 30.2≤4k≤30,2≤3k≤30.

From these:

  • 4k≤30⇒k≤7.54k \le 30 \Rightarrow k \le 7.54k≤30⇒k≤7.5
  • 3k≤30⇒k≤103k \le 30 \Rightarrow k \le 103k≤30⇒k≤10

So the effective restriction is k≤7.k \le 7.k≤7. Also k≥1k\ge 1k≥1 works, because then (4,3)∈A×A(4,3) \in A\times A(4,3)∈A×A.

Hence possible values are k=1,2,3,4,5,6,7.k=1,2,3,4,5,6,7.k=1,2,3,4,5,6,7.


  1. List the ordered pairs

These are:

(4,3),(8,6),(12,9),(16,12),(20,15),(24,18),(28,21).(4,3), (8,6), (12,9), (16,12), (20,15), (24,18), (28,21).(4,3),(8,6),(12,9),(16,12),(20,15),(24,18),(28,21).

Total number of ordered pairs =7=7=7.


  1. Check options

The correct option is: 7\boxed{7}7​ which is Option D.


  1. Compare with stored answer

Stored correct answer: D

Our derived answer: D

So they agree.

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