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Bunuel
Dan and Alex measure themselves against a tree in their backyard. The tree is 80% taller than Dan and 20% taller than Alex. Which of the following statements is true?

A. Dan is 60% taller than Alex.
B. Dan is 40% taller than Alex.
C. Alex is 33.3% taller than Dan.
D. Alex is 40% taller than Dan.
E. Alex is 50% taller than Dan.

We can let t = the tree’s height, d = Dan’s height, and a = Alex’s height.

Since the tree is 80% taller than Dan:

t = 1.8d

t/1.8 = d

10t/18 = d

5t/9 = d

Since the tree is 20% taller than Alex:

t = 1.2a

t/1.2 = a

10t/12 = a

5t/6 = a

Since the tree is 80% taller than Dan and only 20% taller than Alex, we see that Alex is taller than Dan. Let’s determine by how much, using the percent difference formula:

[(Alex’s height - Dan’s height)/Dan’s height] x 100

[(5t/6 - 5t/9)/(5t/9)] x 100

[(15t/18 - 10t/18)/(10t/18)] x 100

(5t/18)/(10t/18) x 100

1/2 x 100 = 50%

Answer: E
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Why do we look at the percent change in this case?
Why is the answer not simply 33% taller?

d/a = 12/18 = 2/3
d = 66.6%
a = 99.9%

99.9%-66.6% = 33.3%

Edit: On further reflection, it becomes clear that 99% is 50% MORE than 66% i.e 66+ 0.5(66) = 99
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Bunuel
Dan and Alex measure themselves against a tree in their backyard. The tree is 80% taller than Dan and 20% taller than Alex. Which of the following statements is true?

A. Dan is 60% taller than Alex.
B. Dan is 40% taller than Alex.
C. Alex is 33.3% taller than Dan.
D. Alex is 40% taller than Dan.
E. Alex is 50% taller than Dan.

Let Dan's height be 100,

Tree height=180

Alex height=A(6/5)=180=150

Clearly, Alex is 50% taller than Dan.

Answer D

OR

Let Dan's height be D and Alex's height be A

Tree= (18/10)D= (12/10)A

3D=2A
A=1.5D

Answer D
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The tree is 80% taller than Dan and 20% taller than Alex

Let Dan's height be d, Alex's height be a, and Tree's height be H

According to the given problem,

=>H=1.8d=1.2a

=> 1.8d=1.2a
=>a/d=1.5

hence Alex is 50% more taller than Dan

Hence E
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