Sets, Relations and FunctionsMCQMTP Dec 22 - Series IQuestion 1926 of 217
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Two Finite sets have m\displaystyle m and n\displaystyle n elements. The total number of subsets of first set is 56 more than the total number of subsets of the second set. The value of m\displaystyle m and n\displaystyle n are

Options

A6, 3
B7, 6
C5, 1
D8, 7
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Correct Answer

Option a6, 3

All Options:

  • A6, 3
  • B7, 6
  • C5, 1
  • D8, 7

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Detailed Solution & Explanation

Let the first set have m\displaystyle m elements and the second set have n\displaystyle n elements.
The total number of subsets of a set with k\displaystyle k elements is given by 2k\displaystyle 2^k.
According to the given condition, the total number of subsets of the first set is 56 more than that of the second set:
2m2n=562^m - 2^n = 56
We can factor out 2n\displaystyle 2^n from the left side:
2n(2mn1)=562^n(2^{m-n} - 1) = 56
Expressing 56 as a product of a power of 2 and an odd number:
56=8×7=23×(231)56 = 8 \times 7 = 2^3 \times (2^3 - 1)
Comparing both sides, we get:
2n=23    n=32^n = 2^3 \implies n = 3
2mn1=231    mn=3    m3=3    m=62^{m-n} - 1 = 2^3 - 1 \implies m - n = 3 \implies m - 3 = 3 \implies m = 6
Checking the options:
- For m=6,n=3\displaystyle m = 6, n = 3, we have 2623=648=56\displaystyle 2^6 - 2^3 = 64 - 8 = 56, which is correct.
This corresponds to **Option A** (6,3\displaystyle 6, 3).
*Note: The textbook answer key incorrectly lists Option D (8,7\displaystyle 8, 7) as the correct answer, which yields 2827=256128=12856\displaystyle 2^8 - 2^7 = 256 - 128 = 128 \neq 56. Hence, Option A is the mathematically correct choice.*
Hence, **Option A** is the correct answer.

About This Chapter: Sets, Relations and Functions

Paper

Paper 3: Quantitative Aptitude

Weightage

3-5 Marks

Key Topics

Sets, Relations, Functions

This chapter covers Sets, Relations, Functions and is part of Paper 3: Quantitative Aptitude in the CA Foundation exam.

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This topic carries 3-5 Marks weightage. Focus on understanding core concepts rather than memorizing.

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