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How to solve this type of question??

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How to solve this type of question??

by alivapriyada » Mon Aug 09, 2010 11:54 am
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 increases each year?

(1) 3/10
(2) 2/5
(3) 1/2
(4) 2/3
(5) 6/5

OA is (4)
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by kvcpk » Mon Aug 09, 2010 12:10 pm
alivapriyada wrote: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 increases each year?

(1) 3/10
(2) 2/5
(3) 1/2
(4) 2/3
(5) 6/5

OA is (4)
Let the tree grow by k amount each year.
At the end of first year height = 4+k
At the end of second year -= 4+2k
third year = 4+3k
fourth = 4+4k
fifth = 4+5k
sixth = 4+6k

Now, 4+6k - (4+4k) = (4+4k)*1/5
10k = 4+4k
k = 4/6 = 2/3

hope this helps
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by indiantiger » Mon Aug 09, 2010 6:44 pm
time period--->1-------------2--------3---------4---------5---------6
initial height =4
height increase per year = x
height---------(4+x)-----(4+2x)---(4+3x)---(4+4x)--(4+5x)----(4+6x)

now given to us
=>(4+4x)+(4+4x)/5 = 4+6x
=>24x+24 = 20 + 30x
=>4 = 6x
=>x = 2/3 (answer)
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by aarati » Mon Aug 09, 2010 10:41 pm
kvcpk wrote:
alivapriyada wrote: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 increases each year?

(1) 3/10
(2) 2/5
(3) 1/2
(4) 2/3
(5) 6/5

OA is (4)
Let the tree grow by k amount each year.
At the end of first year height = 4+k
At the end of second year -= 4+2k
third year = 4+3k
fourth = 4+4k
fifth = 4+5k
sixth = 4+6k

Now, 4+6k - (4+4k) = (4+4k)*1/5
10k = 4+4k
k = 4/6 = 2/3

hope this helps
thank you very much...... is there any another way to solve.......
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by kvcpk » Mon Aug 09, 2010 11:06 pm
aarati wrote: thank you very much...... is there any another way to solve.......
I think this is the easy approach aarati.

Another approach would be:

Since tree grew by 1/5 than 4th year, tree height in sixth year would be (1+1/5) = 6/5 times the height at fourth year.

4+6H = 6/5(4+4H)
Hence H = 2/3

This turns out ot be paraphrased version of the above method.

So it doesnt matter. You can apply whichever is comfortable for you.
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by sdotcruz » Tue Aug 31, 2010 10:57 am
I was thrown off by this question. The reason i was that I thought the tree was growing exponentially. Did the words "Constant Amount" signal that it was increasing by the same amount each year? What words would signal that the object is growing exponentially?
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by Arcane66 » Tue Aug 31, 2010 12:47 pm
sdotcruz wrote:I was thrown off by this question. The reason i was that I thought the tree was growing exponentially. Did the words "Constant Amount" signal that it was increasing by the same amount each year? What words would signal that the object is growing exponentially?
I did this problem the same way kvcpk's second post illustrates. I think this reasoning is sound. A constant amount means it is growing by the same amount, in this case, for six straight years. Think of it like you're going to a bar and drinking five shots every night. This is a constant amount of alcohol every night for a period of time. The number or amount does not change. It would say it is growing exponentially. A good example of exponential growth is when you have a number that is being raised to a certain power each time it grows. This is not the case with this problem.
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by Stuart@KaplanGMAT » Tue Aug 31, 2010 2:46 pm
alivapriyada wrote: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 increases each year?

(1) 3/10
(2) 2/5
(3) 1/2
(4) 2/3
(5) 6/5
We can also solve this problem by backsolving, i.e. working with the choices.

Generally when we backsolve we start with either (2) or (4). Let's start with (4) here.

If the tree is growing by 2/3 of a foot per year, then its height is:

yr 0: 4 feet
yr 1: 4 2/3
yr 2: 5 1/3
yr 3: 6
yr 4: 6 2/3
yr 5: 7 1/3
yr 6: 8

Is 8 (year 6 height) 1/5th greater than 6 2/3 (hear 4 height)?

Converting to thirds:

Is 24/3 1/5th greater than 20/3?

(24/3)/(20/3) = 24/20 = 6/5 = 1 1/5

Yes! Therefore, (4) is the correct answer.

If (4) gave us a result less than 1 1/5, then we'd need a bigger number; if (4) gave us a result greater than 1 1/5, then we'd need a smaller number.
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by josemaus » Tue Aug 31, 2010 4:29 pm
When I first read this question, I interpreted the phrase "1/5 taller" to mean 6th year - 4th year = 1/5; it seems more appropriate if the phrasing were, "4th year was 1/5 as tall". Regardless of the phrasing, I tried to solve the question using a different method; however, was unable to get to the desired answer.

Since a 4 foot tree grows at a constant rate, the tree should grow as follows:
0: 4
1: 4f (f being the growth)
2: 4f^2
...
6: 4f^6

So solve you set up the equation 4f^6 = 4f^4*(6/5). I cannot get f to equal 2/3. What am I doing wrong???!!!!
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by josemaus » Tue Aug 31, 2010 4:42 pm
As a follow-up, this is where my logic is coming from:

OG 12 #81
Bacteria Table
Time Amount

1pm 10g
4pm x grams
7pm 14.4 g

Data for a certain biology experiment are given in the table above. If the amount of bacteria present increased by the same factor during each of the two 3-hour periods shown, how many grams of bacteria were present at 4pm?

a) 12.0
b) 12.1
c) 12.2
d) 12.3
e) 12.4

OA: A

The way to solve this is 1st 3-hour period 10*f, and 2nd 3-hour period 10*f^2, not by doing 10 + x and 10 + 2x.
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