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# X pounds of birdseed are left out in a pile in the forest

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X pounds of birdseed are left out in a pile in the forest  [#permalink]

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07 Jul 2017, 22:05
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Question Stats:

85% (01:12) correct 15% (01:50) wrong based on 76 sessions

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X pounds of birdseed are left out in a pile in the forest. If the pile loses half of its total every day, then what fraction of the original pile will remain after 5 days?

A. 1/64
B. 1/32
C. 1/10
D. 5/32
E. 63/64

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X pounds of birdseed are left out in a pile in the forest  [#permalink]

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07 Jul 2017, 22:55
Lets presume the total to be 100 pounds of Birdseed in the initial period.

Since the pile loses $$\frac{1}{2}$$ of its total on a daily basis, we will be left with $$\frac{1}{2^5}$$ or $$\frac{1}{32}$$ of the total fraction.

So $$\frac{1}{32}$$th of $$100 = \frac{100}{32}$$ of the initial mixture will be left.

Since we have been asked to find what fraction of the original pile remains,
we will have $$\frac{100}{32}/\frac{1}{100}$$ = $$\frac{1}{32}$$(Option B)
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X pounds of birdseed are left out in a pile in the forest  [#permalink]

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07 Jul 2017, 22:58
1
Skywalker18 wrote:
X pounds of birdseed are left out in a pile in the forest. If the pile loses half of its total every day, then what fraction of the original pile will remain after 5 days?

A. 1/64
B. 1/32
C. 1/10
D. 5/32
E. 63/64

Let the Total pounds Pile of birdseed be $$= 100$$

Pile of birdseed loses half of its total every day.

Therefore, Pile of birdseed left is half of its total every day.

Total Pile of birdseed left after $$5$$ days $$= \frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*100 = \frac{25}{8}$$

Required Fraction = (Fraction of Pile of birdseed left after $$5$$ days) / (Total pile of birdseed)

Required Fraction $$= \frac{(25/8)}{100} = \frac{25}{8 * 100} = \frac{1}{32}$$

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Re: X pounds of birdseed are left out in a pile in the forest  [#permalink]

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07 Jul 2017, 23:38
Let x = 100.

Fraction lost every day for 5 days = $$(\frac{1}{2})^{5}$$
Amount lost = $$\frac{1}{32} * 100 = \frac{100}{32}$$

Fraction remaining = $$\frac{100}{32*100} = \frac{1}{32}$$. Ans - B.
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X pounds of birdseed are left out in a pile in the forest  [#permalink]

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08 Jul 2017, 17:50
Skywalker18 wrote:
X pounds of birdseed are left out in a pile in the forest. If the pile loses half of its total every day, then what fraction of the original pile will remain after 5 days?

A. 1/64
B. 1/32
C. 1/10
D. 5/32
E. 63/64

Looking at the denominators in the answer choices, let the number of pounds of bird seed in the pile be 64. Each day the pile loses half its total.

Day 1: 64 --> 32
Day 2: 32 --> 16
Day 3: 16 ---> 8
Day 4: 8 ----> 4
Day 5: 4 ----> 2

Fraction of original that remains: $$\frac{2}{64}$$ = $$\frac{1}{32}$$

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Re: X pounds of birdseed are left out in a pile in the forest  [#permalink]

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12 Jul 2017, 08:25
1
Isn't the given information representing a series of G.P of common ratio 1/2 ?
x/2 , x/4, x/8, x/16 & x/32 ?
So the the fifth day will have 1/32 fraction of the pile?
Ans : B
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Re: X pounds of birdseed are left out in a pile in the forest  [#permalink]

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12 Jul 2017, 09:32
Imp thing to note here is that a
day has 2 nos start of the day and end of the day. If the quantity is X at the start of the day then by end of the day it will become X/2. This will be the opening quantity for the next day.

It is easier to start from the 5th day to avoid Decimals or fraction when you half the quantity.

End Start
Day 5 1 2
Day 4 2 4
Day 3 4 8
Day 2 8 16
Day 1 16 32

fraction Day5/Day1=1/32

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Re: X pounds of birdseed are left out in a pile in the forest  [#permalink]

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10 Sep 2017, 20:04
Skywalker18 wrote:
X pounds of birdseed are left out in a pile in the forest. If the pile loses half of its total every day, then what fraction of the original pile will remain after 5 days?

A. 1/64
B. 1/32
C. 1/10
D. 5/32
E. 63/64

Day 1 --> x-x/2 = x/2;
Day 2 --> x/2 - x/4 =x/4;
Day 3 --> x/4-x/8 = x/8;
Day 4 -->x/8 -x/16= x/16;
Day 5 --> x/16-x/32 = x/32;

1/32 of the original remains after 5 days.
Ans:B
Re: X pounds of birdseed are left out in a pile in the forest   [#permalink] 10 Sep 2017, 20:04
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