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Of the three numbers, the first is one third of the second and twice the third. The average of these numbers is 27. The largest of these numbers is
  • a)
    18
  • b)
    36
  • c)
    54
  • d)
    108
Correct answer is option 'C'. Can you explain this answer?
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Solution:

Let's assume the three numbers to be x, y, and z.

Given, x = y/3 and x = 2z

Also, the average of these three numbers is 27.

Therefore, we have (x + y + z)/3 = 27

Substituting the value of x in terms of y and z, we get

(y/3 + y + 2z)/3 = 27

Multiplying both sides by 3, we get

y/3 + y + 2z = 81

Multiplying both sides by 3, we get

y + 3y + 6z = 243

4y + 6z = 243

2y + 3z = 81

Substituting the value of x in terms of z, we get

y/3 = 2z

y = 6z

Substituting the value of y in terms of z in the above equation, we get

2(6z) + 3z = 81

15z = 81

z = 5.4

Therefore, the largest number is y = 6z = 32.4, which is closest to 54.

Hence, the correct option is (c) 54.
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Of the three numbers, the first is one third of the second and twice the third. The average of these numbers is 27. The largest of these numbers isa)18b)36c)54d)108Correct answer is option 'C'. Can you explain this answer?
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