Manhattan Prep
Two trees have a combined height of 60 feet, and the taller tree is \(x\) times the height of the shorter tree. How tall is the shorter tree, in term of \(x\)?
A. \(\frac{60}{1+x}\)
B. \(\frac{60}{x}\)
C. \(\frac{30}{x}\)
D. \(60-2x\)
E. \(30-5x\)
OA A
Two trees have a combined height of 60 feet, and the taller
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We can create the equation in which s = the height of the shorter tree, and xs = the height of the taller tree.AAPL wrote:Manhattan Prep
Two trees have a combined height of 60 feet, and the taller tree is \(x\) times the height of the shorter tree. How tall is the shorter tree, in term of \(x\)?
A. \(\frac{60}{1+x}\)
B. \(\frac{60}{x}\)
C. \(\frac{30}{x}\)
D. \(60-2x\)
E. \(30-5x\)
OA A
s + xs = 60
s(1 + x) = 60
s = 60/(1 + x)
Answer: A
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Let the taller tree = T
Let the shorter tree = S
$$T+S=60ft---eqni$$
$$T=x\cdot S---equ\ ii$$
$$substituting\ \left(X\cdot S\right)for\ T\ in\ equ\ i$$
$$\left(x\cdot S\right)+S=60$$
$$xS+S=60$$
$$S\left(x+1\right)=60$$
divide through by coefficient of S
$$\frac{S\left(x+1\right)}{x+1}=\frac{60}{x+1}$$
$$S=\frac{60}{x+1}$$
$$Answer\ is\ Option\ A$$
Let the shorter tree = S
$$T+S=60ft---eqni$$
$$T=x\cdot S---equ\ ii$$
$$substituting\ \left(X\cdot S\right)for\ T\ in\ equ\ i$$
$$\left(x\cdot S\right)+S=60$$
$$xS+S=60$$
$$S\left(x+1\right)=60$$
divide through by coefficient of S
$$\frac{S\left(x+1\right)}{x+1}=\frac{60}{x+1}$$
$$S=\frac{60}{x+1}$$
$$Answer\ is\ Option\ A$$