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The sum of three numbers is 132. If the first number be twice the second and third number is one third of the first, then the second number is

Question

The sum of three numbers is 132. If the first number be twice the second and third number is one third of the first, then the second number is

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Solution

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

  1. According to the problem, the sum of the three numbers is 132: A+B+C=132 A + B + C = 132

  2. The first number A A is twice the second number B B : A=2B A = 2B

  3. The third number C C is one third of the first number A A : C=A3 C = \frac{A}{3}

Now, we substitute the expressions for A A and C C into the sum equation:

  1. Substitute A=2B A = 2B and C=2B3 C = \frac{2B}{3} into A+B+C=132 A + B + C = 132 : 2B+B+2B3=132 2B + B + \frac{2B}{3} = 132

  2. Combine the terms on the left side: 2B+B+2B3=132 2B + B + \frac{2B}{3} = 132 (2+1+23)B=132 \left(2 + 1 + \frac{2}{3}\right)B = 132 (32+31+23)B=132 \left(\frac{3 \cdot 2 + 3 \cdot 1 + 2}{3}\right)B = 132 (6+3+23)B=132 \left(\frac{6 + 3 + 2}{3}\right)B = 132 (113)B=132 \left(\frac{11}{3}\right)B = 132

  3. Solve for B B : B=132×311 B = 132 \times \frac{3}{11} B=12×3 B = 12 \times 3 B=36 B = 36

Therefore, the second number is 36 36 .

This problem has been solved

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