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Assertion (A) If the median and mode of a frequency distribution are 150 and154 respectively, then its mean is 148. Reason (R) Mean, median and mode of a frequency distribution are related as:

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Question

Assertion (A) Reason (R)
If the median and mode of a
frequency distribution are 150 and
154 respectively, then its mean is 148.
Mean, median and mode of a
frequency distribution are related as:
mode = 3 median – 2 mean.

Options

  • Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).

  • Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).

  • Assertion (A) is true and Reason (R) is false.

  • Assertion (A) is false and Reason (R) is true.

MCQ
Assertion and Reasoning
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Solution

Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).

Explanation:

Reason (R) is clearly true.

Using the relation given in Reason (R), we have

2 mean = (3 median) – (mode)

= (3 × 150) – (154) 

= 450 – 154

= 296

∴ Mean = 148, which is true.

Thus, Assertion (A) and Reason (R) are both true and Reason (R) is the correct explanation of Assertion (A).

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Chapter 18: Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive - MULTIPLE-CHOICE QUESTIONS (MCQ) [Page 904]

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R.S. Aggarwal Mathematics [English] Class 10
Chapter 18 Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive
MULTIPLE-CHOICE QUESTIONS (MCQ) | Q 30. | Page 904
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