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vatsal323
Does the question stem say that it was 1/5 times taller or was 1/5 feet taller. It's not clear!

In the absence of the word "feet" following the fraction, the meaning is the same as "20% taller", or "6/5 as tall".
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Solution:

Let the tree increase by x units every year

Growth in 6 years are (4+x),(4+2x),(4+3x),(4+4x),(4+5x) ,(4+6x)

Given 6th -4th year = 1/5 (4+4x)

=> 4+6x-4-4x = 4/5 + 4x/5

=> 2x = 4/5 +4x/5

=>6x/5= 4/5

=>x=4/6 = 2/3 (option d)

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Hi avigutman - Tried doing this with ratios

Tried 3 attempts with ratios as the pic shows.

In All 3 attempts - the ratio is the same.

I was surprised to see how in every attempt -- the scale factor was coming up different.

It seems my take-away should be to
- Attempt # 2 - reduce down as much as possible so every cell is an integer but its reduced down as much as possible
OR
- Attempt # 1 - increase up so every cell with a decimal/ fraction has an integer

It seems like this reduction / integer constraint is vital for ratios (Which really surprised me)

Why do we need to (reduce down as much as possible) or (increase up so no decimals exists) in the ratio system when finding the scale factor

Technically - it should not matter if my starting ratio is 3x or 6x or 30x (as 3 ratio's are the same)
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jabhatta2
Hi avigutman - Tried doing this with ratios

Tried 3 attempts with ratios as the pic shows.

In All 3 attempts - the ratio is the same.

I was surprised to see how in every attempt -- the scale factor was coming up different.

It seems my take-away should be to
- Attempt # 2 - reduce down as much as possible so every cell is an integer but its reduced down as much as possible
OR
- Attempt # 1 - increase up so every cell with a decimal/ fraction has an integer

It seems like this reduction / integer constraint is vital for ratios (Which really surprised me)

Why do we need to (reduce down as much as possible) or (increase up so no decimals exists) in the ratio system when finding the scale factor

Technically - it should not matter if my starting ratio is 3x or 6x or 30x (as 3 ratio's are the same)

Note the exact wording of the question, jabhatta2:
Quote:
By how many feet did the height of the tree increase each year?
Your solutions all show that you're solving for x, but is the question asking for x? That depends on your definition of x. In attempt #1 you should have been solving for 0.5x. In attempt #2 you should have been solving for 5x.
I also noticed that you have inches in your solution, rather than feet. Your takeaway ought to be that you need to pay more attention to what is being asked, and be precise about it. In general, though, yes, best practice is to use the smallest integers possible when constructing a ratio, and avoiding fractions.
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anilnandyala
When a certain tree was first planted, it was 4 feet tall and the height of the tree increased by a constant amount each year for the next 6 years. At the end of the 6th year, the tree was 1/5 taller than it was at the end of the 4th year. By how many feet did the height of the tree increase each year?

A. 3/10
B. 2/5
C. 1/2
D. 2/3
E. 6/5
­Hi Can someone explain me why my approach is wrong..
 I used AP as it the height of the tree increased by a constant amount
4+5d=6/5(4+3k)
20+25k=24+18k
7k=4
k=4/7
 
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anushridi
anilnandyala
When a certain tree was first planted, it was 4 feet tall and the height of the tree increased by a constant amount each year for the next 6 years. At the end of the 6th year, the tree was 1/5 taller than it was at the end of the 4th year. By how many feet did the height of the tree increase each year?

A. 3/10
B. 2/5
C. 1/2
D. 2/3
E. 6/5
­Hi Can someone explain me why my approach is wrong..
 I used AP as it the height of the tree increased by a constant amount
4+5d=6/5(4+3k)
20+25k=24+18k
7k=4
k=4/7

 
­Should be 4 + 6x = 6/5(4 + 4x), which leads to x = 2/3.
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Bunuel
anushridi
anilnandyala
When a certain tree was first planted, it was 4 feet tall and the height of the tree increased by a constant amount each year for the next 6 years. At the end of the 6th year, the tree was 1/5 taller than it was at the end of the 4th year. By how many feet did the height of the tree increase each year?

A. 3/10
B. 2/5
C. 1/2
D. 2/3
E. 6/5
­Hi Can someone explain me why my approach is wrong..
 I used AP as it the height of the tree increased by a constant amount
4+5d=6/5(4+3k)
20+25k=24+18k
7k=4
k=4/7




 
­Should be 4 + 6x = 6/5(4 + 4x), which leads to x = 2/3.




 
­Ohkkk so year 0 4 ft
                  year 1 4+k ft

Got it! Thankss Bunnel!! :)
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Hi Bunuel,
Shouldn't we assume that since the tree was first planted, its height was 4 feet which increased by a constant amount in the consequent years? Therefore height at the 6 th year should be 4+5y?

Please help

Bunuel
When a certain tree was first planted, it was 4 feet tall and the height of the tree increased by a constant amount each year for the next 6 years. At the end of the 6th year, the tree was 1/5 taller than it was at the end of the 4th year. By how many feet did the height of the tree increase each year?

A. 3/10
B. 2/5
C. 1/2
D. 2/3
E. 6/5

Let the rate of increase be \(x\) feet per year.

At the end of the 4th year, the height was \(4+4x\).
At the end of the 6th year, the height was \(4+6x\), which was "1/5 taller than at the end of the 4th year."

Therefore, we have \(4+4x+\frac{1}{5}(4+4x)=4+6x\). Simplifying, we get \(\frac{1}{5}(4+4x)=2x\), leading to \(x=\frac{2}{3}\).

Answer: D.
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shvm_sin7
Hi Bunuel,
Shouldn't we assume that since the tree was first planted, its height was 4 feet which increased by a constant amount in the consequent years? Therefore height at the 6 th year should be 4+5y?

Please help

Bunuel
When a certain tree was first planted, it was 4 feet tall and the height of the tree increased by a constant amount each year for the next 6 years. At the end of the 6th year, the tree was 1/5 taller than it was at the end of the 4th year. By how many feet did the height of the tree increase each year?

A. 3/10
B. 2/5
C. 1/2
D. 2/3
E. 6/5

Let the rate of increase be \(x\) feet per year.

At the end of the 4th year, the height was \(4+4x\).
At the end of the 6th year, the height was \(4+6x\), which was "1/5 taller than at the end of the 4th year."

Therefore, we have \(4+4x+\frac{1}{5}(4+4x)=4+6x\). Simplifying, we get \(\frac{1}{5}(4+4x)=2x\), leading to \(x=\frac{2}{3}\).

Answer: D.

Starting height = 4 feet

Height after one years = 4 + x
Height after two years = 4 + 2x
Height after three years = 4 + 3x
Height after four years = 4 + 4x
Height after five years = 4 + 5x
height after six years = 4 + 6x
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