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Exercise 1.3 · Q1

Q.Make correct statements by filling in the symbols ⊂ or ⊄ in the blank spaces :

(i) { 2, 3, 4 } . . . { 1, 2, 3, 4,5 }
(ii) { a, b, c } . . . { b, c, d }
(iii) {x : x is a student of Class XI of your school}. . .{x : x student of your school}
(iv) {x : x is a circle in the plane} . . .{x : x is a circle in the same plane with radius 1 unit}
(v) {x : x is a triangle in a plane} . . . {x : x is a rectangle in the plane}
(vi) {x : x is an equilateral triangle in a plane} . . . {x : x is a triangle in the same plane}
(vii) {x : x is an even natural number} . . . { x : x is an integer}
Bihar BsebTextbookSubjective· 2mImportance★★★★★est
14% · 18/132 Questions
✓ Free question

The subset symbol ⊂\subset means "is a subset of" — every element of the first set must also belong to the second. We check each pair and fill ⊂\subset or ⊄\not\subset accordingly.

Concept first: Set membership and subsets

A set AA is a subset of BB (written A⊂BA \subset B) if every element of AA is also an element of BB. If even one element of AA is missing from BB, then A⊄BA \not\subset B. This is a pure "all-or-nothing" check — no partial credit in set theory.

The trick is to look at the elements, not the labels. For sets described by conditions, translate the condition into actual elements before comparing.

Let's go through each part.


  1. {2,3,4}…{1,2,3,4,5}\{2, 3, 4\} \dots \{1, 2, 3, 4, 5\}

    Every element of the first set — 2, 3, and 4 — appears in the second set. So the first is a subset of the second.

    ⊂\boxed{\subset}

  2. {a,b,c}…{b,c,d}\{a, b, c\} \dots \{b, c, d\}

    The first set contains aa, but aa is not in {b,c,d}\{b, c, d\}. Since one element fails, it is not a subset.

    ⊄\boxed{\not\subset}

  3. {x:x is a student of Class XI of your school}…{x:x is a student of your school}\{x : x \text{ is a student of Class XI of your school}\} \dots \{x : x \text{ is a student of your school}\}

    Every Class XI student is, by definition, a student of the school. So the first set is entirely contained in the second.

    ⊂\boxed{\subset}

  4. {x:x is a circle in the plane}…{x:x is a circle in the same plane with radius 1 unit}\{x : x \text{ is a circle in the plane}\} \dots \{x : x \text{ is a circle in the same plane with radius 1 unit}\}

    The first set includes circles of any radius — radius 2, radius 5, etc. The second set only contains circles of radius exactly 1. So a circle of radius 2 is in the first set but not in the second.

    ⊄\boxed{\not\subset}

    Watch out

    A common mistake is to reverse the direction: the set of radius‑1 circles is a subset of all circles, not the other way around. Here we are checking if all circles are radius‑1 circles — which is false.

  5. {x:x is a triangle in a plane}…{x:x is a rectangle in the plane}\{x : x \text{ is a triangle in a plane}\} \dots \{x : x \text{ is a rectangle in the plane}\}

    A triangle is never a rectangle, and a rectangle is never a triangle. The two sets have no overlap at all. So the first set is certainly not a subset of the second.

    ⊄\boxed{\not\subset}

  6. {x:x is an equilateral triangle in a plane}…{x:x is a triangle in the same plane}\{x : x \text{ is an equilateral triangle in a plane}\} \dots \{x : x \text{ is a triangle in the same plane}\}

    Every equilateral triangle is a triangle. So the first set is fully contained in the second.

    ⊂\boxed{\subset}

  7. {x:x is an even natural number}…{x:x is an integer}\{x : x \text{ is an even natural number}\} \dots \{x : x \text{ is an integer}\}

    Even natural numbers (2, 4, 6, …) are all integers. So the first set is a subset of the second.

    ⊂\boxed{\subset}


✓Final answer

The correct symbols are: (i) ⊂\subset,

(ii) ⊄\not\subset,

(iii) ⊂\subset,

(iv) ⊄\not\subset,

(v) ⊄\not\subset,

(vi) ⊂\subset,

(vii) ⊂\subset.

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