Let "a" be the constant increment in height of the tree, then height of the tree after six years is (4+6a) and height of the tree after four years is (4+4a).
Now we know that height of the tree at six years was 1/5 greater than height at 4 years. Thus we have the following equation
(4+6a) = (6/5) (4+4a) { 6/5 is 1/5 greater than one}
solving we get a = 2/3
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Stuart@KaplanGMAT
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We can solve this algebraically or by backsolving.
First, the algebraic solution:
Let's let the amount of growth per year be x.
So, at the end of 4 years, the tree will be 4 + 4x feet tall.
At the end of 6 years, the tree is is 4 + 6x feet tall, but since we know the relationship between yr 6 and yr 4, we also know that at the end of 6 years the tree will be 6/5 (4 + 4x) feet tall.
So, 4 + 6x = 6/5(4+4x)
4 + 6x = 24/5 + 24/5(x)
(let's multiply both sides by 5 to elimiate the fractions)
20 + 30x = 24 + 24x
30x - 24x = 24 - 20
6x = 4
x = 4/6 = 2/3
Backsolving involves working backward from the answer choices. We generally want to start with either (b) or (d) (so we don't have to check as many choices).
If we start with (d), we're assuming that the tree grows 2/3 of a foot per year.
So, after 4 years the tree would be 4 + 4(2/3) = 12/3. + 8/3 = 20/3 feet tall.
After 6 years, the tree would be 4 + 6(2/3) = 12/3 + 12/3 = 24/3 = 8 feet tall.
Now we ask, is 8 = (6/5)(20/3)?
6/5(20/3) = 120/15 = 8.
So, if we pick 2/3 of a foot per year, we get the correct ending to the story (i.e. 6/5(yr4) = yr6) - therefore answer (d) is correct.
First, the algebraic solution:
Let's let the amount of growth per year be x.
So, at the end of 4 years, the tree will be 4 + 4x feet tall.
At the end of 6 years, the tree is is 4 + 6x feet tall, but since we know the relationship between yr 6 and yr 4, we also know that at the end of 6 years the tree will be 6/5 (4 + 4x) feet tall.
So, 4 + 6x = 6/5(4+4x)
4 + 6x = 24/5 + 24/5(x)
(let's multiply both sides by 5 to elimiate the fractions)
20 + 30x = 24 + 24x
30x - 24x = 24 - 20
6x = 4
x = 4/6 = 2/3
Backsolving involves working backward from the answer choices. We generally want to start with either (b) or (d) (so we don't have to check as many choices).
If we start with (d), we're assuming that the tree grows 2/3 of a foot per year.
So, after 4 years the tree would be 4 + 4(2/3) = 12/3. + 8/3 = 20/3 feet tall.
After 6 years, the tree would be 4 + 6(2/3) = 12/3 + 12/3 = 24/3 = 8 feet tall.
Now we ask, is 8 = (6/5)(20/3)?
6/5(20/3) = 120/15 = 8.
So, if we pick 2/3 of a foot per year, we get the correct ending to the story (i.e. 6/5(yr4) = yr6) - therefore answer (d) is correct.
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desihokie
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[quote="simplyjat"]Let "a" be the constant increment in height of the tree, then height of the tree after six years is (4+6a) and height of the tree after four years is (4+4a).
Now we know that height of the tree at six years was 1/5 greater than height at 4 years. Thus we have the following equation
(4+6a) = (6/5) (4+4a) { 6/5 is 1/5 greater than one}
solving we get a = 2/3[/quote]
Thank you very much to both of you for your reply. I was doing a silly mistake. Instead of multiplying by 6/5 i was doing this:
(4+6a) = 4+4a + 1/5, which is obviously wrong.
also, could you please answer to another question i posted here (on geometry)....just a couple of posts below? TIA
Now we know that height of the tree at six years was 1/5 greater than height at 4 years. Thus we have the following equation
(4+6a) = (6/5) (4+4a) { 6/5 is 1/5 greater than one}
solving we get a = 2/3[/quote]
Thank you very much to both of you for your reply. I was doing a silly mistake. Instead of multiplying by 6/5 i was doing this:
(4+6a) = 4+4a + 1/5, which is obviously wrong.
also, could you please answer to another question i posted here (on geometry)....just a couple of posts below? TIA
