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Each year for 4 years, a farmer increased the number of tree

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Re: Each year for 4 years, a farmer increased the number of tree [#permalink]

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New post 13 Apr 2015, 08:38
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1/4 = 0.25 = 25%
using the formula of compound interest calculation
p(1+r)^n= A

p = principal amount (to be identified), r = rate (0.25), n = numbers years (4), A= amount at the end of term(6250 given)

p(1+0.25)^4=6250
p(1.25)^4= 25*25*10 = 5^2*5^2*10=5^4*10
p(125/100)^4 = 5^4 x 10
p(5/4)^4 = 5^4 x 10
p(5^4)/4^4 = 5^4 x 10
p= (5^4 x 10 x 4^4)/5^4
p= 10 x 4^4
p= 2560
Ans D
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Re: Each year for 4 years, a farmer increased the number of tree [#permalink]

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New post 19 May 2015, 12:37
we have here 4 (4 years) successive increases of 25% or *1,25 --> 1,25^4 * X = 6250, X = 2560

See MGMAT (Percents) for detailed explanation of such question types.....
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Re: Each year for 4 years, a farmer increased the number of tree [#permalink]

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New post 20 May 2015, 03:32
BrainLab wrote:
we have here 4 (4 years) successive increases of 25% or *1,25 --> 1,25^4 * X = 6250, X = 2560

See MGMAT (Percents) for detailed explanation of such question types.....


Dear BrainLab

Perfect logic but for easier calculation, you may want to work with ratio here (1/4 increase per annum) instead of percentages (25% increase per annum). Both convey the same thing but the equation

\((\frac{5}{4})^4*X = 6250\)

will take lesser time to solve (especially if you know that \(5^4 = 625\)) than \((1.25)^4*X = 6250\)

Hope this was useful!

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Each year for 4 years, a farmer increased the number [#permalink]

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New post 18 Jan 2016, 23:28
Each year for 4 years, a farmer increased the number
of trees in a certain orchard by 1/4 of the number of
trees in the orchard the preceding year. If all of the
trees thrived and there were 6,250 trees in the
orchard at the end of the 4-year period, how many
trees were in the orchard at the beginning of the
4-year period?
(A) 1,250
(B) 1,563
(C) 2,250
(D) 2,560
(E) 2,752

I am going wrong somewhere.
Could someone help me, is something wrong with my method to solve this question.
Not getting the correct answer.
Please help.

My Approach:
Let number of trees at the beginning of the year be x.
2nd year no. of trees = x+x/4
3rd year no. of trees = (x+x/4) +1/4(x+x/4) 1.e. x + x/4 + x/4 + x/16 = x +x/2 + x/16
4th year no of trees = x + x/2 + x/16 + 1/4(x + x/2 + x/16)
6250 = x + x/2 + x/16 + x/4 + x/8 + x/64
6250 = (64x + 32x + 4x + 16x + 8x + x)/64
6250 = 125x/64
x = 3200

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Re: Each year for 4 years, a farmer increased the number of tree [#permalink]

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New post 18 Jan 2016, 23:36
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Sunil01 wrote:
Each year for 4 years, a farmer increased the number
of trees in a certain orchard by 1/4 of the number of
trees in the orchard the preceding year. If all of the
trees thrived and there were 6,250 trees in the
orchard at the end of the 4-year period, how many
trees were in the orchard at the beginning of the
4-year period?
(A) 1,250
(B) 1,563
(C) 2,250
(D) 2,560
(E) 2,752

I am going wrong somewhere.
Could someone help me, is something wrong with my method to solve this question.
Not getting the correct answer.
Please help.

My Approach:
Let number of trees at the beginning of the year be x.
2nd year no. of trees = x+x/4
3rd year no. of trees = (x+x/4) +1/4(x+x/4) 1.e. x + x/4 + x/4 + x/16 = x +x/2 + x/16
4th year no of trees = x + x/2 + x/16 + 1/4(x + x/2 + x/16)
6250 = x + x/2 + x/16 + x/4 + x/8 + x/64
6250 = (64x + 32x + 4x + 16x + 8x + x)/64
6250 = 125x/64
x = 3200


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Re: Each year for 4 years, a farmer increased the number of tree [#permalink]

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New post 19 Jan 2016, 01:25
Sunil01 wrote:
Each year for 4 years, a farmer increased the number
of trees in a certain orchard by 1/4 of the number of
trees in the orchard the preceding year. If all of the
trees thrived and there were 6,250 trees in the
orchard at the end of the 4-year period, how many
trees were in the orchard at the beginning of the
4-year period?
(A) 1,250
(B) 1,563
(C) 2,250
(D) 2,560
(E) 2,752

I am going wrong somewhere.
Could someone help me, is something wrong with my method to solve this question.
Not getting the correct answer.
Please help.

My Approach:
Let number of trees at the beginning of the year be x.
2nd year no. of trees = x+x/4
3rd year no. of trees = (x+x/4) +1/4(x+x/4) 1.e. x + x/4 + x/4 + x/16 = x +x/2 + x/16
4th year no of trees = x + x/2 + x/16 + 1/4(x + x/2 + x/16)
6250 = x + x/2 + x/16 + x/4 + x/8 + x/64
6250 = (64x + 32x + 4x + 16x + 8x + x)/64
6250 = 125x/64
x = 3200
...


Hi,
you have gone wrong in that you have calculated only for three years..
each year it increases by 1/4, so calculate for the end of the year,...
in this case the first year end would be x+x/4, which you have shown for second year..
By your approach you should check for % year begining to get the answer..

the answer you have got is at teh end of one year..
if it was y in begining, it will become5y/4..
so here 5y/4=3200...
or y=3200*4/5=4*640=2560..

Hope it helps you..
you have missed out on one year..
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Re: Each year for 4 years, a farmer increased the number of tree [#permalink]

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New post 16 Mar 2016, 03:23
i thought it as compound interest problem rate =1/4 =25 %
time =4 years amount =6250
therefore amount = p(1+r/100)^n= p(1+25/100)^4= 6250
solve for p
p= 2560 ...................(D)

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New post 16 Mar 2016, 04:56
imhimanshu wrote:
Each year for 4 years, a farmer increased the number of trees in a certain orchard by 1/4 of the number of trees in the orchard of the preceding year. If all of the trees thrived and there were 6250 trees in the orchard at the end of 4 year period, how many trees were in the orchard at the beginning of the 4 year period.

A. 1250
B. 1563
C. 2250
D. 2560
E. 2752

Can someone walk me through the logic behind this question. I am able to solve this by using options as well as by assuming the number of trees = x. However, had the question been, "If all of the trees thrived and there were 6250 trees in the orchard at the end of 15 year period, how many trees were in the orchard at the beginning of the 4 year period". then it would have been difficult to solve.

Thanks

Supposing that there were x number of trees in the beginning, at the end of 4 years the number of trees is (5/4)^4*x=6250
Solving we get x = 2560

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Re: Each year for 4 years, a farmer increased the number of tree [#permalink]

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New post 18 Apr 2016, 12:30
go on dividing the total number of trees i.e. 6250 by the 1.25^4 or (5/4)^4 to get the correct answer
1.25 is used because each year the number of trees increased by 1/4 or (25%)
so correct answer option D

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Re: Each year for 4 years, a farmer increased the number of tree [#permalink]

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New post 18 Apr 2016, 14:53
let t=# of original trees
(t)(5/4)^4=6250
t=6250/(5/4)^4
t=2560 trees

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Re: Each year for 4 years, a farmer increased the number of tree [#permalink]

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New post 22 Apr 2016, 14:55
This is actually a percent and interest question if you look closely ..
here the interest is compounded annually at 25% (1/4 )
so let trees at beginning be x => x(1+1/4)^4 = 6250 => x=2560
Smash that D
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Re: Each year for 4 years, a farmer increased the number of tree [#permalink]

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New post 15 May 2016, 15:15
If you can't (like myself) detect the pattern here off the bat, you're screwed. Pattern is that [(5/4)^4]*x = 5^4*10

5^4 cancel on both sides, and you are left with 4^4*10 = x , 16*16 = 256*10 = 2560

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Each year for 4 years, a farmer increased the number of tree [#permalink]

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New post 29 Jun 2016, 02:59
isn't this a GP and the 6250 is sum of the GP?

Can't we use S=\(\frac{a(1-r^n)}{1-r}\)

where r=1/4

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Re: Each year for 4 years, a farmer increased the number of tree [#permalink]

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New post 29 Jun 2016, 21:49
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bimalr9 wrote:
isn't this a GP and the 6250 is sum of the GP?

Can't we use S=\(\frac{a(1-r^n)}{1-r}\)

where r=1/4


Yes, there is a GP but r = 5/4. Also, 6250 is the last term, not sum of the GP.

a*(5/4), a*(5/4)*(5/4), ...

Every previous term is multiplied by 5/4 to get the next term. The 4th term is 6250.

\(6250 = a * (5/4)^4\)

\(a = 10 * 4^4\)

\(a = 2560\)

Answer (D)



a ( r^n - 1)/(r - 1) = a ((5/4)^4 - 1)/(5/4 - 1)

6250 = a * (5^4 - 4^4)/4^3

2 * 5^4 * 4^3 = a * 369
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Re: Each year for 4 years, a farmer increased the number of tree [#permalink]

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New post 05 Jul 2016, 12:05
converting 6250 into prime factorization form (625*10 = 5*5*5*5*5*2) can save some time in solving this question, as many 5's will cancel out nicely
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New post 29 Jan 2017, 23:57
What's wrong if we add up 1/4 for each year. Please help me understand this in detail.

First year begin: x------------- ending 5x/4
2nd year begin: 5x/4----------ending 3x/2(5x/4+x/4)
3year begin: 3x/2--------------ending 7x/4
4 year begin : 7x/4-----------ending 2x
I dont get the answer and iam not understanding what is wrong with this approch ... please help

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Re: Each year for 4 years, a farmer increased the number of tree [#permalink]

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imhimanshu wrote:
Each year for 4 years, a farmer increased the number of trees in a certain orchard by 1/4 of the number of trees in the orchard of the preceding year. If all of the trees thrived and there were 6250 trees in the orchard at the end of 4 year period, how many trees were in the orchard at the beginning of the 4 year period.

A. 1250
B. 1563
C. 2250
D. 2560
E. 2752


This problem is testing us on exponential growth. We are given that the number of trees increased by ¼ each year. To determine the number of trees in a certain year, we multiply the number of trees from the previous year by 1.25, or 5/4. Let’s let x equal the number of trees at the beginning (of the first year) of the 4-year period.

Start of year 1 = x

End of year 1 = x(5/4)

End of year 2 = x(5/4)(5/4)

End of year 3 = x(5/4)(5/4)(5/4)

End of year 4 = x(5/4)(5/4)(5/4)(5/4) = (625/256)x

We are given that there were 6,250 trees at the end of year 4, so we can set up the following equation:

(625/256)x = 6,250

625x = 6,250(256)

x = 10(256) = 2,560

Thus, there were 2,560 trees at the beginning of the 4-year period.

Answer: D
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Re: Each year for 4 years, a farmer increased the number of tree [#permalink]

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muthappashivani wrote:
What's wrong if we add up 1/4 for each year. Please help me understand this in detail.

First year begin: x------------- ending 5x/4
2nd year begin: 5x/4----------ending 3x/2(5x/4+x/4)
3year begin: 3x/2--------------ending 7x/4
4 year begin : 7x/4-----------ending 2x
I dont get the answer and iam not understanding what is wrong with this approch ... please help


This is what the question says:

"Each year for 4 years, a farmer increased the number of trees in a certain orchard by 1/4 of the number of trees in the orchard of the preceding year"

He increases the trees by 1/4 of the number in the preceding year. So in 3rd year, he increases the trees by a fourth of the number of trees in the 2nd year, not one fourth of those in the 1st year.
Hence, 2nd year end/3rd year beg, the number of trees will be 5x/4 + (1/4)*(5x/4)
and so on...
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Re: Each year for 4 years, a farmer increased the number of tree [#permalink]

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New post 25 Mar 2017, 06:24
let's make up an equation:
n*(5/4)^4=5250
n*625/256=6250
n=10*256
2560
Answer is D

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Each year for 4 years, a farmer increased the number of tree [#permalink]

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New post 21 Apr 2017, 05:18
All the above responses have been around trying to solve this algebraically. But we can do plug and play with the answer choices. We know what the end result trees are and we know what the annual increase. Given that the annual increase is 1/4 --> we know that the answer choices must be a factor of 4. So (B) is eliminated automatically since its an odd number.

The rest of the answers choices with the exception of D are not factors of 4. so we can eliminate that way.

Now if we had two answers that were divisible by 4, then we can easily do another step and recalculate.

Personally I find this much easier then having to the algebra formula but both can be done under 2 min easily.

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