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Among three numbers, the first is twice the second and thrice the third. If the average of three numbers is 49.50, then what is the difference between the first and the third number?Options78655954

Question

Among three numbers, the first is twice the second and thrice the third. If the average of three numbers is 49.50, then what is the difference between the first and the third number?Options78655954

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Solution

Let's denote the three numbers as A, B, and C.

According to the problem, A is twice the second number (B) and thrice the third number (C). So, we can write this as:

A = 2B = 3C

The problem also states that the average of the three numbers is 49.50. The average is calculated by adding all the numbers together and dividing by the quantity of numbers. So, we can write this as:

(A + B + C) / 3 = 49.50

Substituting A = 2B = 3C into the equation, we get:

(2B + B + C) / 3 = 49.50 4B / 3 = 49.50 B = (49.50 * 3) / 4 B = 37.125

Substituting B = 37.125 into A = 2B, we get:

A = 2 * 37.125 A = 74.25

Substituting B = 37.125 into C = B / 2, we get:

C = 37.125 / 2 C = 18.5625

The difference between the first and the third number is:

A - C = 74.25 - 18.5625 = 55.6875

So, the difference between the first and the third number is approximately 56 when rounded to the nearest whole number. Therefore, the answer is 56.

This problem has been solved

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