Mean and mode / linear system statements

CMAT 2021 · Quantitative Techniques & Data Interpretation · Medium · Statistics

Given below are two statements

Statement I : The set of numbers (5,6,7,p,6,7,8,q)(5, 6, 7, p, 6, 7, 8, q) has an arithmetic mean of 6 and mode (most frequently occurring number) of 7. Then p×q=16p \times q = 16. Statement II: Let pp and qq be two positive integers such that p+q+p×q=94p + q + p \times q = 94. Then p+q=20p + q = 20.

In light of the above statements, choose the correct answer from the options given below

  1. A.

    Both Statement I and Statement II are true

  2. B.

    Both Statement I and Statement II are false

  3. C.

    Statement I is true but Statement II is false

  4. D.

    Statement I is false but Statement II is true

Answer

B

Explanation

Statement I: The set has 8 elements. Mean = 6   ⟹  Sum=48\implies \text{Sum} = 48. 5+6+7+p+6+7+8+q=48  ⟹  p+q=95 + 6 + 7 + p + 6 + 7 + 8 + q = 48 \implies p + q = 9 Since 7 is the unique mode, both pp and qq (or at least one) must be 7. If p=7p=7, then q=2q=2, giving p×q=14≠16p \times q = 14 \neq 16. Thus Statement I is false.

Statement II: p+q+pq=94  ⟹  (p+1)(q+1)=95=5×19p + q + pq = 94 \implies (p+1)(q+1) = 95 = 5 \times 19. Since p,qp, q are positive integers, {p+1,q+1}={5,19}\{p+1, q+1\} = \{5, 19\}, so {p,q}={4,18}\{p, q\} = \{4, 18\}. Then p+q=4+18=22≠20p + q = 4 + 18 = 22 \neq 20. Thus Statement II is false.

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