When I was 2 my sister was twice my age Now I’m 40

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  1. When I was 2 my sister was twice my age Now I’m 40, how old is my sister…
  2. Check all equations that are equivalent. A=(1)/(2)h(b^(1)+b^(2)) 2A=hb^(1)+b^(2) b^(1)=(2A)/(h)-b^(2) b^(1)=(2(A-(1)/(2)hb^(2)))/(h) RETRYV

The Correct Answer and Explanation is :

1. How old is your sister when you are 40?

When you were 2, your sister was twice your age, which means she was 4 years old at that time (since 2 * 2 = 4). The age difference between you and your sister is therefore 4 – 2 = 2 years.

Now that you are 40, your sister will be 40 + 2 = 42 years old.

2. Checking the Equivalence of Equations

We need to check if the given equations are equivalent to the original equation:

Original equation:

A=12h(b(1)+b(2))A = \frac{1}{2} h \left( b^{(1)} + b^{(2)} \right)

Equation 1: 2A=hb(1)+b(2)2A = h b^{(1)} + b^{(2)}

  • Multiply both sides of the original equation by 2: 2A=2⋅12h(b(1)+b(2))=h(b(1)+b(2))2A = 2 \cdot \frac{1}{2} h \left( b^{(1)} + b^{(2)} \right) = h \left( b^{(1)} + b^{(2)} \right)
  • This is not exactly the same as the given equation. In fact, it’s not possible to simplify to this form unless we disregard the grouping of terms (b^(1) + b^(2)) completely. So this equation is not equivalent.

Equation 2: b(1)=2Ah−b(2)b^{(1)} = \frac{2A}{h} – b^{(2)}

  • Solving the original equation for b(1)b^{(1)}: A=12h(b(1)+b(2))A = \frac{1}{2} h \left( b^{(1)} + b^{(2)} \right) Multiply both sides by 2: 2A=h(b(1)+b(2))2A = h \left( b^{(1)} + b^{(2)} \right) Subtract b(2)b^{(2)} from both sides: b(1)=2Ah−b(2)b^{(1)} = \frac{2A}{h} – b^{(2)}
  • This matches the second equation perfectly. Therefore, this equation is equivalent.

Equation 3: b(1)=2(A−12hb(2))hb^{(1)} = \frac{2 \left( A – \frac{1}{2} h b^{(2)} \right)}{h}

  • Start with the original equation: A=12h(b(1)+b(2))A = \frac{1}{2} h \left( b^{(1)} + b^{(2)} \right) Solve for b(1)b^{(1)}: 2A=h(b(1)+b(2))2A = h \left( b^{(1)} + b^{(2)} \right) Subtract b(2)b^{(2)} from both sides: b(1)=2Ah−b(2)b^{(1)} = \frac{2A}{h} – b^{(2)}
  • Now substitute: b(1)=2(A−12hb(2))hb^{(1)} = \frac{2 \left( A – \frac{1}{2} h b^{(2)} \right)}{h}
  • Expanding this: b(1)=2Ah−b(2)b^{(1)} = \frac{2A}{h} – b^{(2)}
  • This matches the second equation exactly, so this equation is equivalent as well.

Summary:

  • The second and third equations are equivalent to the original one.
  • The first equation is not equivalent.

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