Let the universal set be \mathcal{U} = \{x \in \mathbb{Z}^+ : 1 \le x \le 200\}. Subsets A… — JAMB UTME

Let the universal set be $\mathcal{U} = \{x \in \mathbb{Z}^+ : 1 \le x \le 200\}$ . Subsets $A$ , $B$ , and $C$ of $\mathcal{U}$ are defined as follows:
- $A = \{x \in \mathcal{U} : x \text{ is a multiple of } 5\}$
- $B = \{x \in \mathcal{U} : x \text{ is a multiple of } 7\}$
- $C = \{x \in \mathcal{U} : x \text{ is a multiple of } 10\}$

Find the number of elements in the set $(A \cap B') \setminus C$ .

Cevap: 17 / seventeen

Cevap

The number of elements in the set

(A \cap B') \setminus C

is 17.

The set

(A \cap B') \setminus C

consists of elements in

A

that are neither in

B

nor in

C

. Taking

A \setminus C

isolates the 20 odd multiples of 5 between 1 and 200. From these 20 numbers, removing those divisible by 7 leaves only the odd multiples of 35 (35, 105, and 175) to be subtracted, giving

20 - 3 = 17

Adım Adım Çözüm

Simplify the set expression

(A \cap B') \setminus C

(A \cap B') \setminus C = (A \setminus C) \setminus B

By set algebra, taking elements in

A

that are not in

B

and then excluding elements in

C

is equivalent to first removing elements of

C

from

A

, and then removing any remaining elements that belong to

B

Determine the cardinality of

A \setminus C

n(A \setminus C) = 20

Set

A

contains all multiples of 5 up to 200, of which there are

\lfloor 200/5 \rfloor = 40

. Set

C

contains all multiples of 10 up to 200, which are the even multiples of 5, amounting to

\lfloor 200/10 \rfloor = 20

. Therefore,

A \setminus C

consists of the odd multiples of 5 up to 200, which gives

40 - 20 = 20

elements.

Find which elements of

A \setminus C

are also in set

B

The common elements are 35, 105, and 175 (3 elements).

Elements in both

A

and

B

are multiples of

\text{lcm}(5, 7) = 35

. The multiples of 35 up to 200 are 35, 70, 105, 140, and 175. However, 70 and 140 are multiples of 10 (belonging to set

C

) and have already been removed. Thus, only the odd multiples of 35 (35, 105, and 175) remain in

A \setminus C

and also belong to

B

Subtract these common elements to calculate

n((A \setminus C) \setminus B)

20 - 3 = 17

Removing the 3 common elements from the 20 elements of

A \setminus C

leaves 17 elements.

Anahtar Kavram

Set difference, intersection with complement, and evaluation of cardinalities using divisibility properties.

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