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# A lumberjack cuts one-seventh of a log pile in the morning and one-six

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A lumberjack cuts one-seventh of a log pile in the morning and one-six [#permalink]

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09 Sep 2017, 01:04
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5% (low)

Question Stats:

89% (01:10) correct 11% (01:01) wrong based on 76 sessions

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A lumberjack cuts one-seventh of a log pile in the morning and one-sixth of the remaining logs that afternoon. If he still has 60 more logs to cut, how many logs were there originally?

A) 100

B) 95

C) 84

D) 80

E) 66

Source: 800Score

P.S. - Please do not post just the answers saying - "A" for me or "I'll go with B etc.

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A lumberjack cuts one-seventh of a log pile in the morning and one-six [#permalink]

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09 Sep 2017, 02:03
HKD1710 wrote:
A lumberjack cuts one-seventh of a log pile in the morning and one-sixth of the remaining logs that afternoon. If he still has 60 more logs to cut, how many logs were there originally?

A) 100

B) 95

C) 84

D) 80

E) 66

Source: 800Score

P.S. - Please do not post just the answers saying - "A" for me or "I'll go with B etc.

Logs left after morning work $$= 1-\frac{1}{7} = \frac{6}{7}$$
Logs cut in the afternoon $$= \frac{6}{7}*\frac{1}{6} = \frac{1}{7}$$
so logs left after afternoon $$= \frac{6}{7}-\frac{1}{7} = \frac{5}{7}$$
$$\frac{5}{7}$$ of total logs $$= 60$$
Hence total logs $$= 84$$

Option C

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Other methods could be
1.) Assume total logs as $$x$$ and then as per the question formulate an equation and solve for $$x$$
2.) Assume total logs to be a convenient multiple of both $$7$$ & $$6$$ (say $$420$$) and then use equivalence between the arrived number of logs left and original number of logs left
3.) Work backwards through options. Chose a smart number that is multiple of both $$7$$ & $$6$$ from the options. In this case $$84$$ fits the bill perfectly
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A lumberjack cuts one-seventh of a log pile in the morning and one-six [#permalink]

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09 Sep 2017, 08:30
HKD1710 wrote:
A lumberjack cuts one-seventh of a log pile in the morning and one-sixth of the remaining logs that afternoon. If he still has 60 more logs to cut, how many logs were there originally?

A) 100

B) 95

C) 84

D) 80

E) 66

Source: 800Score

P.S. - Please do not post just the answers saying - "A" for me or "I'll go with B etc.

The number of logs must be divisible by 7.

He cuts $$\frac{1}{7}$$ of the logs in the morning.

The original number must be a multiple of 7 in order to subtract 1/7 from it and still have a whole number of logs.

(The remainder will be a multiple of 6, even if the original is smaller than 42. The remainder will also be a multiple of 5. For this question with these answer choices, the latter two facts do not matter.)

Any choice other than C cannot have 1/7 subtracted from it to yield a whole number.

The only choice that is a multiple of 7 is

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Posts: 85
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Re: A lumberjack cuts one-seventh of a log pile in the morning and one-six [#permalink]

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10 Sep 2017, 19:56
1
HKD1710 wrote:
A lumberjack cuts one-seventh of a log pile in the morning and one-sixth of the remaining logs that afternoon. If he still has 60 more logs to cut, how many logs were there originally?

A) 100

B) 95

C) 84

D) 80

E) 66

Source: 800Score

P.S. - Please do not post just the answers saying - "A" for me or "I'll go with B etc.

Let the total number of logs be x.

In the morning he cuts x/7.

In the evening he cuts 1/6*(x-x/7)= x/7.

So he has already cut 2x/7 logs.
Given x-2x/7 logs = 60

5x/7 =60; x = 12*7= 84.

Ans:C
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Re: A lumberjack cuts one-seventh of a log pile in the morning and one-six [#permalink]

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14 Sep 2017, 10:41
HKD1710 wrote:
A lumberjack cuts one-seventh of a log pile in the morning and one-sixth of the remaining logs that afternoon. If he still has 60 more logs to cut, how many logs were there originally?

A) 100

B) 95

C) 84

D) 80

E) 66

We can let n = the total logs, and thus:

(1/7)n + (1/6)(6/7)n + 60 = n

(1/7)n + (1/7)n + 60 = n

2n/7 + 60 = n

60 = 5n/7

420 = 5n

84 = n

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Joined: 21 Sep 2016
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Re: A lumberjack cuts one-seventh of a log pile in the morning and one-six [#permalink]

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16 Sep 2017, 21:27
Let x be the original number of logs.
When he began his afternoon shift, there was still 6/7 of the original stock. In the end, there still were 5/6 of the remaining pile, so:

x*(6/7*(5/6) = 60

x*(5/7) = 60

So, x = 84.
Manager
Joined: 11 Jun 2017
Posts: 80
Re: A lumberjack cuts one-seventh of a log pile in the morning and one-six [#permalink]

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17 Sep 2017, 05:10
Let total number of logs be x
1/7th are cut in the morning , number of logs remaining = x-x/7 = 6x/7
1/6th are cut in the afternoon , remaining logs = 6x/7 - (6x/7*1/6) = 5x/7
5x/7 = 60
x= 84 (C)
Re: A lumberjack cuts one-seventh of a log pile in the morning and one-six   [#permalink] 17 Sep 2017, 05:10
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