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John is reading a 50-page book, starting from 1 page. He places a whit

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John is reading a 50-page book, starting from 1 page. He places a whit [#permalink]

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New post 03 Oct 2016, 00:51
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61% (01:13) correct 39% (01:07) wrong based on 127 sessions

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John is reading a 50-page book, starting from 1 page. He places a white stick per 3 pages and a black stick per 4 pages. How many pages have neither white sticks nor black sticks?
A. 20pages B. 24pages C. 26pages D. 30pages E. 32pages

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John is reading a 50-page book, starting from 1 page. He places a whit [#permalink]

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New post 03 Oct 2016, 01:47
MathRevolution wrote:
John is reading a 50-page book, starting from 1 page. He places a white stick per 3 pages and a black stick per 4 pages. How many pages have neither white sticks nor black sticks?
A. 20pages B. 24pages C. 26pages D. 30pages E. 32pages



Condition 1 : John is placing a white stick per 3 pages...i.e. he'll put at all 3 multiples...i.e. at 3,6,12,.......48 and he can't put the white stick at 51...because the limit is 50 pages...

In total how many multiples of 3 till 50 => 50/3 = max we get 16 .... (count 3,6,12,15,18,21,24,27,30,33,36,39,42,45 and 48 ) => we'll get 16.

Condition 2 : Now John is placing a black stick per 4 pages...i.e..he'll put at 4,8,12,16,20.....48 and he can't put the white stick at 51...because the limit is 50 pages..

In total how many multiples of 4 till 50 => 50/4 = max we get 12 .... (count 4,8,12,16,20,24,28,32,36,40,44 and 48 ) => we'll get 12.

Now if you observe that we have common multiples or numbers for 3 and 4 ( take LCM of 3 and 4..we get 12...if we check multiples of 12 till 50..we get 12,24,36 and 48)...

=> 50/12 - max we get 4 i.e. 12,24,36 and 48 are common in 3 and 4.

we have total 4 numbers common.

Now use formula..

n(WUB) = N(W) + N(B) - N(W∩B) + Neither = > Total pages = only white + only black - both white and black + Neither...

50 = 16+12-4 +Neither.

Neither = 26.
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Re: John is reading a 50-page book, starting from 1 page. He places a whit [#permalink]

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New post 06 Oct 2016, 05:44
==> n(3)=the number of pages with white sticks and n(4)=the number of pages with black sticks. If so, n(3U4)=n(3)+n(4)-n(3∩4)=16+12-4=24. However, the question is about the number of pages without white sticks and black sticks, so 50-24=26. C is the answer
Answer: C
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Re: John is reading a 50-page book, starting from 1 page. He places a whit [#permalink]

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New post 09 Oct 2016, 13:44
The question does not tell us that we can not place white and black sticks on a page.
It is not stated specifically??
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John is reading a 50-page book, starting from 1 page. He places a whit [#permalink]

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New post 10 Oct 2016, 08:31
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MathRevolution wrote:
John is reading a 50-page book, starting from 1 page. He places a white stick per 3 pages and a black stick per 4 pages. How many pages have neither white sticks nor black sticks?

A. 20pages B. 24pages C. 26pages D. 30pages E. 32pages


White sticks = { 3, 6 , 9........48 } - 16 White stickers

Black sticks = { 4,8,12.......48 } - 12 White stickers

White& Black Stick = { 12, 24 , 36, 48 } - 4 Common

Quote:
Pages have neither white sticks nor black sticks


=> 50 - { 16 + 12 - 4 } =26

Hence correct answer will be (C) 26


Tmoni26 wrote:
The question does not tell us that we can not place white and black sticks on a page.
It is not stated specifically??


Dear Tmoni26

You got to calculate all the multiples of 3 , 4 and 12 till 50...

We must account for the multiples of 3 & 4 and subtract it..

Attachment:
Capture.PNG
Capture.PNG [ 11.82 KiB | Viewed 772 times ]


Hope this helps...

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Re: John is reading a 50-page book, starting from 1 page. He places a whit [#permalink]

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New post 25 Dec 2017, 20:31
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Re: John is reading a 50-page book, starting from 1 page. He places a whit   [#permalink] 25 Dec 2017, 20:31
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