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M07-21

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M07-21  [#permalink]

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New post 16 Sep 2014, 00:35
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  45% (medium)

Question Stats:

68% (01:16) correct 32% (01:17) wrong based on 59 sessions

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If a 5-year construction project is structured to double the investment every succeeding year, approximately what percentage of the total cost was invested in the first year?

A. 12.5%
B. 10%
C. 8.25%
D. 5%
E. 3%

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Re M07-21  [#permalink]

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New post 16 Sep 2014, 00:35
Official Solution:

If a 5-year construction project is structured to double the investment every succeeding year, approximately what percentage of the total cost was invested in the first year?

A. 12.5%
B. 10%
C. 8.25%
D. 5%
E. 3%

Let's say we invested $10 the first year. The second year it will be $20 more, third - $40 more, fourth - $80 more, and fifth - $160 more. Now, add them: 10+20+40+80+160 = 310. So, the original investment was \(\frac{10}{310}\) or \(\frac{1}{31}\) or about 3%.

Answer: E
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: M07-21  [#permalink]

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New post 28 Jul 2015, 09:36
Bunuel wrote:
Official Solution:

If a 5-year construction project is structured to double the investment every succeeding year, approximately what percentage of the total cost was invested in the first year?

A. 12.5%
B. 10%
C. 8.25%
D. 5%
E. 3%

Let's say we invested $10 the first year. The second year it will be $20 more, third - $40 more, fourth - $80 more, and fifth - $160 more. Now, add them: 10+20+40+80+160 = 310. So, the original investment was \(\frac{10}{310}\) or \(\frac{1}{31}\) or about 3%.

Answer: E


hi..!

this question could be interpreted as follows also..?

say first year we invested 10$
so 2nd year : 20 $

Now 3rd year we invest double the invested amount : so it could be 2*(10+20) = 60 $.

i assumed it this way, solved it and then saw that there were no matching choices. So had to do it again in the above mentioned manner..:-)
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Re: M07-21  [#permalink]

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New post 31 Oct 2015, 19:51
I interpreted it as follows: if A is initial investment, then after 5 years, this will compound to A*2^5. i.e. doubling every year for 5 years. Then we have A/A*2^5 = 1/32 = 3.125% ~ = 3%
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Re M07-21  [#permalink]

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New post 17 Dec 2015, 10:44
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I think this is a poor-quality question and I agree with explanation. the question is ambiguous and it is difficult to understand what is asked.
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Re: M07-21  [#permalink]

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New post 06 Aug 2016, 09:44
As the investment is doubled every year, if you want to get the initial amount you have to divide final by 2, and repeat 5 times

\(100 * 2^-5\)

\(\frac{100}{2^5}\)

\(\frac{100}{32}\)

more or less 3%
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Re: M07-21  [#permalink]

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New post 11 Aug 2016, 14:49
I think this is a poor quality question. The wording makes it sound as though every year it is providing a return of double the amount invested. Plus, "cost" and "investment" are used interchangeably. Good concept, bad wording.
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Re: M07-21  [#permalink]

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New post 07 Feb 2017, 21:16
riyazgilani wrote:
Bunuel wrote:
Official Solution:

If a 5-year construction project is structured to double the investment every succeeding year, approximately what percentage of the total cost was invested in the first year?

A. 12.5%
B. 10%
C. 8.25%
D. 5%
E. 3%

Let's say we invested $10 the first year. The second year it will be $20 more, third - $40 more, fourth - $80 more, and fifth - $160 more. Now, add them: 10+20+40+80+160 = 310. So, the original investment was \(\frac{10}{310}\) or \(\frac{1}{31}\) or about 3%.

Answer: E


hi..!

this question could be interpreted as follows also..?

say first year we invested 10$
so 2nd year : 20 $

Now 3rd year we invest double the invested amount : so it could be 2*(10+20) = 60 $.

i assumed it this way, solved it and then saw that there were no matching choices. So had to do it again in the above mentioned manner..:-)


Although I got this correct, I agree with this dude. It would be clearer if the question stated something along the lines of, "Over a 5 year period, if the amount invested in a construction project each year is double the previous year's amount, approximately what percentage of the total cost was invested in the first year?"
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Re M07-21  [#permalink]

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New post 16 Sep 2017, 06:23
I think this is a poor-quality question and the explanation isn't clear enough, please elaborate. I am not sure if the terms such as, 'cost' & 'investment' are interchangeable.
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Re M07-21  [#permalink]

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New post 08 Feb 2018, 21:59
I think this is a poor-quality question and I agree with explanation. The language used is difficult to understand.
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Re: M07-21  [#permalink]

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New post 11 May 2018, 07:02
I got it right but awkward question. Is every year a "succeeding year"? :/
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Re M07-21  [#permalink]

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New post 06 Jun 2018, 03:39
I think this is a poor-quality question and I agree with explanation. While the concept testing is nice, the question wording is very unclear.
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Re: M07-21  [#permalink]

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New post 25 Jun 2018, 09:35
This is how I solved the question algebraically.

Let the initial investment in year 1 be X. We are told that the investment doubles every succeeding year for 5 years. This means:

1st year = x
2nd year = 2*X = 2x
3rd year = 2* (2x) = 4x
4th year = 2*4x = 8x
5th year = 2*8x = 16x

Sum up all the investments from year 1 to year 5. x + 2x + 4x + 8x + 16x = 31x
The initial investment/ total investment = x/31x = 1/31 = ~3% Choice E
Re: M07-21 &nbs [#permalink] 25 Jun 2018, 09:35
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