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The mean of three numbers is 40. All the three numbers are different natural numbers. If lowest is 19, what could be highest possible number of remaining two numbers? - Mathematics

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प्रश्न

The mean of three numbers is 40. All the three numbers are different natural numbers. If lowest is 19, what could be highest possible number of remaining two numbers?

पर्याय

  • 81

  • 40

  • 100

  • 71

MCQ
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उत्तर

81

Explanation:

Mean of three numbers = 40 and lowest number = 19  ...[Given]

Let the three observations be 19, x and y, respectively.

∴ Mean = `"Sum of all observations"/"Total number of observations"`

⇒ `40 = (19 + x + y)/3`  ...[∵ Mean = 40, given]

⇒ 3 × 40 = 19 + x + y

⇒ 120 = 19 + x + y

⇒ x + y = 120 – 19

⇒ x + y = 101  ...(i)

Since, 19 is the lowest observation.

Hence, for the highest possible value of the remaining two numbers, one must be 20.

Let x = 20

From equation (i), we get

20 + y = 101

⇒ y = 101 – 20

⇒ y = 81

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पाठ 3: Data Handling - Exercise [पृष्ठ ७३]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 7
पाठ 3 Data Handling
Exercise | Q 13. | पृष्ठ ७३

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