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NCERT Exemplar · Q30

Q.Let the relation RR be defined in N\mathbb{N} by aRbaRb if 2a+3b=302a + 3b = 30. Then R=R = ______.

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The relation RR consists of all ordered pairs (a,b)(a,b) of natural numbers satisfying 2a+3b=302a + 3b = 30. Solving for aa in terms of bb and restricting to N\mathbb{N} gives R={(3,8),(6,6),(9,4),(12,2)}R = \{(3,8), (6,6), (9,4), (12,2)\}.

We are working with the set of natural numbers N\mathbb{N}. Usually in Indian exams, N={1,2,3,… }\mathbb{N} = \{1, 2, 3, \dots\} (positive integers). The relation RR is defined by the condition 2a+3b=302a + 3b = 30, where aa and bb both belong to N\mathbb{N}.

The key idea: we need to find all pairs (a,b)(a,b) of natural numbers that satisfy this linear Diophantine equation. Since the equation is linear and the coefficients are small, we can solve by expressing one variable in terms of the other and then checking which values keep both variables natural.


  1. Express aa in terms of bb From 2a+3b=302a + 3b = 30, we get

2a=30−3b⇒a=30−3b2.2a = 30 - 3b \quad\Rightarrow\quad a = \frac{30 - 3b}{2}.

For aa to be a natural number, 30−3b30 - 3b must be positive and even, and aa must be at least 11.

  1. Determine the range of bb Since a≥1a \geq 1, we have

30−3b2≥1⇒30−3b≥2⇒3b≤28⇒b≤283≈9.33.\frac{30 - 3b}{2} \geq 1 \quad\Rightarrow\quad 30 - 3b \geq 2 \quad\Rightarrow\quad 3b \leq 28 \quad\Rightarrow\quad b \leq \frac{28}{3} \approx 9.33.

Also b≥1b \geq 1 (natural number). So bb can be 1,2,3,…,91, 2, 3, \dots, 9.

  1. Check which bb give integer aa

    aa is integer exactly when 30−3b30 - 3b is even. Since 3030 is even, 3b3b must be even. 3b3b is even only when bb is even (because 33 is odd). So bb must be even.

    From 11 to 99, the even numbers are 2,4,6,82, 4, 6, 8.

  2. Compute corresponding aa for each even bb

    • If b=2b = 2: a=30−62=242=12a = \frac{30 - 6}{2} = \frac{24}{2} = 12 → (12,2)(12,2)
    • If b=4b = 4: a=30−122=182=9a = \frac{30 - 12}{2} = \frac{18}{2} = 9 → (9,4)(9,4)
    • If b=6b = 6: a=30−182=122=6a = \frac{30 - 18}{2} = \frac{12}{2} = 6 → (6,6)(6,6) …

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