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605-655 Level|   Word Problems|                                 
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imhimanshu
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Each year for 4 years, a farmer increased the number of trees in a 08 Jul 2012, 05:04
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75% (02:33) correct 25% (03:08) wrong based on 9516 sessions

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Each year for 4 years, a farmer increased the number of trees in a certain orchard by one fourth of the number of trees in the orchard the preceding year. If all 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


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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
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Bunuel
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Each year for 4 years, a farmer increased the number of trees in a 08 Jul 2012, 05:35
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imhimanshu
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
Show more

Say the number of trees at the beginning of the 4 year period was x, then:

At the end of the 1st year the number of trees would be \(x+\frac{1}{4}x=\frac{5}{4}*x\);
At the end of the 2nd year the number of trees would be \((\frac{5}{4})^2*x\);
At the end of the 3rd year the number of trees would be \((\frac{5}{4})^3*x\);
At the end of the 4th year the number of trees would be \((\frac{5}{4})^4*x\);
At the end of the \(n_{th}\) year the number of trees would be \((\frac{5}{4})^n*x\).

So, we have that \((\frac{5}{4})^4*x=6,250\) --> \(\frac{5^4}{4^4}*x=5^4*10\) --> \(x=4^4*10=2,560\).

Answer: D.

If the question were "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 we would have that: \((\frac{5}{4})^{15}*x=6,250\) --> \(x\neq{integer}\), so it would be a flawed question.

Hope it's clear.
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Each year for 4 years, a farmer increased the number of trees in a 07 Feb 2014, 18:28
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it can easily solved by compound interest formula
R = 25%
P = original number of trees
N = no of times of compounding within a year = 1
T = total number of years = 4

P+I at the end of compounding years = 6250
hence,

6250 = (1 + 0.25/1)^(4*1) * P
5^4*10 = 1.25^4 * P
(5/1/25)^4 *10 = P
4^4 * 10 = P
2560 = P


Another way to think is to pick ans choice which is exactly divisible by 4. A, B and C are not. Only D and E are.
Pick D and increase it by 25% for 4 times, you should get 6250.
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Each year for 4 years, a farmer increased the number of trees in a Updated on: 08 Oct 2022, 22:14
Originally posted by KarishmaB on 08 Jul 2012, 22:32.
Last edited by KarishmaB on 08 Oct 2022, 22:14, edited 1 time in total.
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imhimanshu
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
Show more

The number of trees increases by 1/4 i.e. 25% every year. It is just a matter of thinking in terms of successive percentage changes e.g. population increase. Here, we are talking about the increase of tree population.

If x increases by 25%, how we denote it? (5/4)*x
If next year, this new number increases by 25% again, how do we denote it? (5/4)*(5/4)*x
and so on...

So if we are taking into account 4 years, we simply get (5/4)^4 * x = 6250

As for your next question, the numbers given would be such that the calculation will not be tough.

Say, you have 8 years and 100% increase every year (population doubles every year). The final population will be divisible by 2^8 i.e. 256.
Something like 2^8 * x = 2560

and if you meant what you wrote (though I considered that the 4 was a typo because of the language of the question) "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", note that you still need to work with

(5/4)^4 * x = 6250
since you need the number of trees 4 yrs back only. The only thing is that the answer (2560) needs to be divisible by \(5^{11}\) which it isn't so there is a problem in this question. If it were an actual question, the answer would be divisible by \(5^{11}\) but you wouldn't really need to bother about it.
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Each year for 4 years, a farmer increased the number of trees in a 25 Oct 2013, 00:26
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Trees increase 1/4th every year, which means 100 trees become 125 after 1 yr & so on
So, 125/100 = 5/4 is the resultant (after adding 1/4 as the growth) for 1 year
So, for 4 yrs is 5^4/(4^4)
From the condition given, inital trees were = 6250 x 4^4 / 5^4 = 2560
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Each year for 4 years, a farmer increased the number of trees in a 27 Dec 2013, 09:31
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Bunuel
imhimanshu
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

Say the number of trees at the beginning of the 4 year period was x, then:
At the end of the 1st year the number of trees would be \(x+\frac{1}{4}x=\frac{5}{4}*x\);
At the end of the 2nd year the number of trees would be \((\frac{5}{4})^2*x\);
At the end of the 3rd year the number of trees would be \((\frac{5}{4})^3*x\);
At the end of the 4th year the number of trees would be \((\frac{5}{4})^4*x\);
At the end of the \(n_{th}\) year the number of trees would be \((\frac{5}{4})^n*x\);

So, we have that \((\frac{5}{4})^4*x=6,250\) --> \(\frac{5^4}{4^4}*x=5^4*10\) --> \(x=4^4*10=2,560\).

Answer: D.

If the question were "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 we would have that: \((\frac{5}{4})^{15}*x=6,250\) --> \(x\neq{integer}\), so it would be a flawed question.

Hope it's clear.
Show more


Isn't the question quite ambiguous, though? I mean the first scentence could be interpreted as "for the first year we have (4/4)x and for the second year (5/4)x and for the third..." etc.. With that reasoning one would have (5/4)^3 * x + x and then your approach doesnt work.

Obviously, I understand that this was a flaw in my reasoning but I cannot understand how they - with that wording - will assume that we totally understand that at the end of year one he has (5/4)x..

Is there a straightforward "word translation" way in knowing how to interpret wordings like this?
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aeglorre
Bunuel
imhimanshu
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

Say the number of trees at the beginning of the 4 year period was x, then:
At the end of the 1st year the number of trees would be \(x+\frac{1}{4}x=\frac{5}{4}*x\);
At the end of the 2nd year the number of trees would be \((\frac{5}{4})^2*x\);
At the end of the 3rd year the number of trees would be \((\frac{5}{4})^3*x\);
At the end of the 4th year the number of trees would be \((\frac{5}{4})^4*x\);
At the end of the \(n_{th}\) year the number of trees would be \((\frac{5}{4})^n*x\);

So, we have that \((\frac{5}{4})^4*x=6,250\) --> \(\frac{5^4}{4^4}*x=5^4*10\) --> \(x=4^4*10=2,560\).

Answer: D.

If the question were "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 we would have that: \((\frac{5}{4})^{15}*x=6,250\) --> \(x\neq{integer}\), so it would be a flawed question.

Hope it's clear.


Isn't the question quite ambiguous, though? I mean the first scentence could be interpreted as "for the first year we have (4/4)x and for the second year (5/4)x and for the third..." etc.. With that reasoning one would have (5/4)^3 * x + x and then your approach doesnt work.

Obviously, I understand that this was a flaw in my reasoning but I cannot understand how they - with that wording - will assume that we totally understand that at the end of year one he has (5/4)x..

Is there a straightforward "word translation" way in knowing how to interpret wordings like this?
Show more

Actually, it is not ambiguous. Read the statement:
Each year a farmer increased the number of trees by 1/4. He did this for 4 years. (In GMAT Verbal and Quant are integrated. You need Verbal skills (slash and burn) in Quant and Quant skills (Data Interpretation) in Verbal.

So in the first year, he increased it by 1/4
The next year, he again increased it by 1/4 (of preceding year)
Next year, again the same.
Next year, again the same.
So he did it for a total of 4 years.

So if initially the number of trees was x, in the first year he made them (5/4)x
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Each year for 4 years, a farmer increased the number of trees in a 27 Jul 2014, 10:08
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Hi Karishma,

I am not able to solve it by below method, where am I wrong:

6250*(3/4)*(3/4)*(3/4)

Regards,
Ravi
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email2vm
Hi Karishma,

I am not able to solve it by below method, where am I wrong:

6250*(3/4)*(3/4)*(3/4)

Regards,
Ravi
Show more

If I increase A by 25% and get B, I will not get A if I reduce B by 25%. The bases are different in the two cases.

e.g. A = 80
25% of A is 20 so I increase A by 25% to get B = 100

Now if I decrease B by 25%, I will decrease B by 25 (25% of 100). This will give me 75 which is not the same as A (which is 80).

In the first step, I found 25% of 80 and added that. In the second step, I found 25% of 100 and subtracted that. These two numbers are different.

You will get the correct answer if you do 6250*(4/5)*(4/5)*(4/5)*(4/5)
4/5 is the inverse of 5/4 (which is 25% increase in x).
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You can use the formula A=p(1 + r/100)^n with r = 25%

You will get P as 2560.
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Trees increase by 1/4 the number of trees in preceding year. Hence, correct answer must be divisible by 4. Based on divisibility rules, if last 2 digits are divisible by 4 then the number is divisible by 4. Thus, we can eliminate A, B, C. The answer has to be D or E.

Again, trees increase by 1/4 the number of trees in preceding year. Hence, the number of trees increase by 5/4 times the number of trees the preceding year.

If x = initial number of trees = Answer D or E = 2560 or 2752
Year 1 = 5/4x
Year 2 = (5/4)(5/4)x
Year 3 = (5/4)(5/4)(5/4)x
Year 4 = (5/4)(5/4)(5/4)(5/4)x

Only for Answer D: (5/4)(5/4)(5/4)(5/4) 2560 = 6250

Hence, correct answer = D
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@VeritasPrepKarishma
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imhimanshu
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

The number of trees increases by 1/4 i.e. 25% every year. It is just a matter of thinking in terms of successive percentage changes e.g. population increase. Here, we are talking about the increase of tree population.

If x increases by 25%, how we denote it? (5/4)*x
If next year, this new number increases by 25% again, how do we denote it? (5/4)*(5/4)*x
and so on...

For more on this, check: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/02 ... e-changes/

So if we are taking into account 4 years, we simply get (5/4)^4 * x = 6250

As for your next question, the numbers given would be such that the calculation will not be tough.

Say, you have 8 years and 100% increase every year (population doubles every year). The final population will be divisible by 2^8 i.e. 256.
Something like 2^8 * x = 2560

and if you meant what you wrote (though I considered that the 4 was a typo because of the language of the question) "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", note that you still need to work with

(5/4)^4 * x = 6250
since you need the number of trees 4 yrs back only. The only thing is that the answer (2560) needs to be divisible by \(5^{11}\) which it isn't so there is a problem in this question. If it were an actual question, the answer would be divisible by \(5^{11}\) but you wouldn't really need to bother about it.
Show more

Karisma,

Your explication is a spot-on!

However, it still isn't clear to me why we are keeping the same base for the increase during the 4-year period time. To me, this seems to be an example of successive % in which the base is shifting every year. Indeed every year we have an increase of 25% more on the increase of the previous year.

Why is it incorrect to compute the increase of the successive year on the increased base of the previous year?

Thank you a lot for your help!

Regards,
Eli.
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elisabettaportioli

However, it still isn't clear to me why we are keeping the same base for the increase during the 4-year period time. To me, this seems to be an example of successive % in which the base is shifting every year. Indeed every year we have an increase of 25% more on the increase of the previous year.

Why is it incorrect to compute the increase of the successive year on the increased base of the previous year?

Thank you a lot for your help!

Regards,
Eli.
Show more

The base for each year changes.

Say, you have 100 trees and you increase them by 20% each year.

At the end of the first year, you will have 100 + (20/100)*100 = 100*(1 + 20/100) = 100*120/100 = 100*(6/5) = 120 trees.

So if you have to increase a number by 20%, you just need to multiply it by 6/5 every time no matter what the number is.

By the same logic, next year, you will have 120 * (6/5) = 144 trees when you increase them by 20%.

You can write 120 as 100 * (6/5) and then multiply it by another 6/5 to increase 120 by 20%.

This is the same as 100 * (6/5) * (6/5) = 144 trees.

So every time you multiply it by 6/5, you increase the base. The next time you multiply it by 6/5, you are increasing the new base by 20%. This is the concept of successive percentage changes and I have discussed it in the link I mentioned in my post above.
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Each year for 4 years, a farmer increased the number of trees in a Updated on: 05 Jan 2021, 11:07
Originally posted by MHIKER on 13 Apr 2015, 08:38.
Last edited by MHIKER on 05 Jan 2021, 11:07, edited 1 time in total.
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\(\frac{1}{4} = 0.25 = 25\)%

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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Each year for 4 years, a farmer increased the number of trees in a 19 May 2015, 12:37
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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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Each year for 4 years, a farmer increased the number of trees in a 20 May 2015, 03:32
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BrainLab
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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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!

Japinder
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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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bimalr9
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
Show more

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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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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Each year for 4 years, a farmer increased the number of trees in a 29 Jan 2017, 23:57
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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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