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Bunuel
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M15-20 [#permalink] New post  15 Sep 2014, 23:56
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Tom read a book containing 480 pages by reading the same number of pages each day. If he would have finished the book 5 days earlier by reading 16 pages a day more, how many days did Tom spend reading the book?

A. 10
B. 12
C. 15
D. 16
E. 18
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Bunuel
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Re M15-20 [#permalink] New post  15 Sep 2014, 23:56
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Tom read a book containing 480 pages by reading the same number of pages each day. If he would have finished the book 5 days earlier by reading 16 pages a day more, how many days did Tom spend reading the book?

A. 10
B. 12
C. 15
D. 16
E. 18


Say the number of days Tom spent reading the book was \(d\). Then:

The number of pages he read per day would be \(\frac{480}{d}\);

The number of pages he would read per day with an increased speed would be \(\frac{480}{d-5}\);

We are told that the number of pages for the second case was 16 pages per day more, so \(\frac{480}{d}=\frac{480}{d-5}-16\). At this point it's much better to plug answer choices rather than solve for \(d\). Answer choice \(C\) fits: \(\frac{480}{15}=32=\frac{480}{15-5}-16\).


Answer: C
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Re: M15-20 [#permalink] New post  17 Nov 2015, 14:22
Why cant this be solved by the following equation
\(\frac{480}{N} -\frac{480}{N+16} =5\)
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Re: M15-20 [#permalink] New post  08 May 2016, 11:09
raarun wrote:
Why cant this be solved by the following equation
\(\frac{480}{N} -\frac{480}{N+16} =5\)


A little late to the party but I'll try to answer your question for anyone else facing the same problem Raarun. In the equation above you have \(\frac{480}{N}\) which represents the number of pages divided by the number of days. Now, you also have \(-\frac{480}{N+16}\), where you're going wrong is the 16 in the denominator. By increasing the denominator by 16(which is not a unit of days but rather pages per the prompt) here you're going to incorrectly reduce the number of total pages per day in that part of the equation. Hope this helps.
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Re: M15-20 [#permalink] New post  11 Sep 2016, 11:37
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stov10 wrote:
raarun wrote:
Why cant this be solved by the following equation
\(\frac{480}{N} -\frac{480}{N+16} =5\)


A little late to the party but I'll try to answer your question for anyone else facing the same problem Raarun. In the equation above you have \(\frac{480}{N}\) which represents the number of pages divided by the number of days. Now, you also have \(-\frac{480}{N+16}\), where you're going wrong is the 16 in the denominator. By increasing the denominator by 16(which is not a unit of days but rather pages per the prompt) here you're going to incorrectly reduce the number of total pages per day in that part of the equation. Hope this helps.



Actually the problem CAN be solved this way.

\(\frac{480}{N} -\frac{480}{N+16} =5\)
In the equation above you have \(\frac{480}{N}\). This represents the total number of pages divided by the number of pages read per day.
Gives the same result.
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Re: M15-20 [#permalink] New post  24 Jan 2017, 07:22
Option C it is. The equation is (480/x)-(480/(x+16))=5. We need to find 480/x and not x. Solve for x ; it comes out to be 32. 480/x becomes 15 then. Hence option C.
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Re: M15-20 [#permalink] New post  24 Jan 2017, 07:44
Bunuel wrote:
Tom read a book containing 480 pages by reading the same number of pages each day. If he would have finished the book 5 days earlier by reading 16 pages a day more, how many days did Tom spend reading the book?

A. 10
B. 12
C. 15
D. 16
E. 18


Here is my approach:

Assume that Tom read that book in \(k\) days by \(x\) pages each day.

We have \(kx=480\) and \((k-5)(x+16)=480\)

Hence \(kx=(k-5)(x+16)=kx-5x+16k-80\)
\(\implies 16k-5x-80=0\)

From \(kx=480\) we have \(x=\frac{480}{k}\)

Hence \(16k - 5 \times \frac{480}{k} - 80 = 0\)
\(\iff 16k - \frac{5 \times 16 \times 30}{k} - 16 \times 5 =0\)
\(\iff k - \frac{150}{k} - 5 = 0\)
\(\iff k^2 - 5k - 150 = 0\)
\(\iff (k^2 - 15k) + (10k - 150) = 0\)
\(\iff (k-15)(k+10)=0\)
\(\iff k = 15\)


The answer is C.
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Re: M15-20 [#permalink] New post  10 Oct 2018, 15:49
Correct me if i am wrong here but all i did was take each of the answer choices and added 5 to them. I noticed answer choice C when you add 5 gives you 20. 480 is divisable by 20. The rest of the answer choices when you add 5 do not divide evenly into 480.
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Re: M15-20 [#permalink] New post  09 Nov 2018, 03:12
Bunuel wrote:
Tom read a book containing 480 pages by reading the same number of pages each day. If he would have finished the book 5 days earlier by reading 16 pages a day more, how many days did Tom spend reading the book?

A. 10
B. 12
C. 15
D. 16
E. 18


where am wrong? Bunuel

My method:
Assume that Tom read that book in y days by reading x pages each day.

then yx=480 and (y−5)(x+16)=480
therefore
⟹ yx = (y−5)(x+16)
⟹ 16y - 5x = 80 ---- equation

now instead of solving further, i am putting in numbers to get the value of y
16y - 5x = 80
16(5) - 5(0) = 80 ---- this can't be the solution as number of days i.e. x>0
16(10)-5(16) ----- y can be 10
16(15) - 5(32) --- y can be 15 as well

so my method has more than 2 solutions and both are mentioned in the answer choices. Where am wrong?
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Re: M15-20 [#permalink] New post  09 Nov 2018, 03:16
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harsh8686 wrote:
Bunuel wrote:
Tom read a book containing 480 pages by reading the same number of pages each day. If he would have finished the book 5 days earlier by reading 16 pages a day more, how many days did Tom spend reading the book?

A. 10
B. 12
C. 15
D. 16
E. 18


where am wrong? Bunuel

My method:
Assume that Tom read that book in y days by reading x pages each day.

then yx=480 and (y−5)(x+16)=480
therefore
⟹ yx = (y−5)(x+16)
⟹ 16y - 5x = 80 ---- equation

now instead of solving further, i am putting in numbers to get the value of y
16y - 5x = 80
16(5) - 5(0) = 80 ---- this can't be the solution as number of days i.e. x>0
16(10)-5(16) ----- y can be 10
16(15) - 5(32) --- y can be 15 as well

so my method has more than 2 solutions and both are mentioned in the answer choices. Where am wrong?


You missed that x and y should also satisfy xy = 480. 10 and 15 does not satisfy this.
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Re: M15-20 [#permalink] New post  09 Nov 2018, 03:22
Bunuel wrote:
harsh8686 wrote:
Bunuel wrote:
Tom read a book containing 480 pages by reading the same number of pages each day. If he would have finished the book 5 days earlier by reading 16 pages a day more, how many days did Tom spend reading the book?

A. 10
B. 12
C. 15
D. 16
E. 18


where am wrong? Bunuel

My method:
Assume that Tom read that book in y days by reading x pages each day.

then yx=480 and (y−5)(x+16)=480
therefore
⟹ yx = (y−5)(x+16)
⟹ 16y - 5x = 80 ---- equation

now instead of solving further, i am putting in numbers to get the value of y
16y - 5x = 80
16(5) - 5(0) = 80 ---- this can't be the solution as number of days i.e. x>0
16(10)-5(16) ----- y can be 10
16(15) - 5(32) --- y can be 15 as well

so my method has more than 2 solutions and both are mentioned in the answer choices. Where am wrong?


You missed that x and y should also satisfy xy = 480. 10 and 15 does not satisfy this.


oh! yes, thanks
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Re: M15-20 [#permalink] New post  09 Nov 2018, 11:30
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Testing Answers:

A) days 10:
pages per day= 48
Increase to 48+16= 64
Now Days taken= 10-5= 5
Product= 64*5 is not 480

B) days 12:
Pages per day= 40
Increase to 40+16= 56
Days taken= 7
56*7 is not 480

C) days 15:
Pages per day 32
Increase to 32+16= 48
Days taken= 10
48*10= 480
Ans C

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sanjay1810
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M15-20 [#permalink] New post  09 Nov 2018, 21:50
Hi Bunuel, chetan2u, IanStewart:

Got a silly question but I can't figure it out:
if d is number of days and p is pages read per day, we get d.p = 480 which is clear but finally isn't the question asking for value of d -5 ?
I don't understand why we're finding for d when the question is actually asking for the case when number of days is d-5.
"If he would have finished the book 5 days earlier by reading 16 pages a day more, how many days did Tom spend reading the book?"
Typically,I get these kind of questions wrong because I don't answer for the correct variable.
Please clarify.

Thanks.
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Bunuel
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Re: M15-20 [#permalink] New post  10 Nov 2018, 05:37
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sanjay1810 wrote:
Hi Bunuel, chetan2u, IanStewart:

Got a silly question but I can't figure it out:
if d is number of days and p is pages read per day, we get d.p = 480 which is clear but finally isn't the question asking for value of d -5 ?
I don't understand why we're finding for d when the question is actually asking for the case when number of days is d-5.
"If he would have finished the book 5 days earlier by reading 16 pages a day more, how many days did Tom spend reading the book?"
Typically,I get these kind of questions wrong because I don't answer for the correct variable.
Please clarify.

Thanks.


If he would have finished the book 5 days earlier by reading 16 pages a day more, how many days did Tom spend reading the book?

That IF there describes hypothetical salutation. In reality, Tom read a book containing 480 pages by reading the same number of pages each day (the number of days Tom spent reading the book = d).
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Re: M15-20 [#permalink] New post  29 Nov 2020, 16:00
Bunuel wrote:
sanjay1810 wrote:
Hi Bunuel, chetan2u, IanStewart:

Got a silly question but I can't figure it out:
if d is number of days and p is pages read per day, we get d.p = 480 which is clear but finally isn't the question asking for value of d -5 ?
I don't understand why we're finding for d when the question is actually asking for the case when number of days is d-5.
"If he would have finished the book 5 days earlier by reading 16 pages a day more, how many days did Tom spend reading the book?"
Typically,I get these kind of questions wrong because I don't answer for the correct variable.
Please clarify.

Thanks.


If he would have finished the book 5 days earlier by reading 16 pages a day more, how many days did Tom spend reading the book?

That IF there describes hypothetical salutation. In reality, Tom read a book containing 480 pages by reading the same number of pages each day (the number of days Tom spent reading the book = d).



I got a similar question: I tried to solve this problem, using several equations, all of which didn't work.

I said

\(\frac{480}{x }\) pages = d (days)

Then I said

\(\frac{480}{(x+16)}\) pages = d-5 (days).

I tried to equate these equations but it simply didn't work.

Where did I go wrong here?

Thanks!
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Bunuel
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Re: M15-20 [#permalink] New post  29 Nov 2020, 23:29
Expert Reply
HWPO wrote:
Bunuel wrote:
sanjay1810 wrote:
Hi Bunuel, chetan2u, IanStewart:

Got a silly question but I can't figure it out:
if d is number of days and p is pages read per day, we get d.p = 480 which is clear but finally isn't the question asking for value of d -5 ?
I don't understand why we're finding for d when the question is actually asking for the case when number of days is d-5.
"If he would have finished the book 5 days earlier by reading 16 pages a day more, how many days did Tom spend reading the book?"
Typically,I get these kind of questions wrong because I don't answer for the correct variable.
Please clarify.

Thanks.


If he would have finished the book 5 days earlier by reading 16 pages a day more, how many days did Tom spend reading the book?

That IF there describes hypothetical salutation. In reality, Tom read a book containing 480 pages by reading the same number of pages each day (the number of days Tom spent reading the book = d).



I got a similar question: I tried to solve this problem, using several equations, all of which didn't work.

I said

\(\frac{480}{x }\) pages = d (days)

Then I said

\(\frac{480}{(x+16)}\) pages = d-5 (days).

I tried to equate these equations but it simply didn't work.

Where did I go wrong here?

Thanks!


The equations are correct. When you solve you should get x=32 and d=15.
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Re: M15-20 [#permalink] New post  09 Dec 2020, 12:02
Bunuel wrote:

The equations are correct. When you solve you should get x=32 and d=15.



\(\frac{480}{x}\) = d => 480 = dx

\(\frac{480}{(x+16)}\) pages = d-5 => 480 = xd-5x+16d-80

\(xd = xd-5x+16d-80\)
\(80 = 5x + 16d\)

How?
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Bunuel
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Re: M15-20 [#permalink] New post  09 Dec 2020, 12:20
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HWPO wrote:
Bunuel wrote:

The equations are correct. When you solve you should get x=32 and d=15.



\(\frac{480}{x}\) = d => 480 = dx

\(\frac{480}{(x+16)}\) pages = d-5 => 480 = xd-5x+16d-80

\(xd = xd-5x+16d-80\)
\(80 = 5x + 16d\)

How?


\(\frac{480}{x } = d\)

\(\frac{480}{(x+16)} = d-5 \) --> substitute d: \(\frac{480}{(x+16)}= \frac{480}{x } -5 \) --> \(5 x^2 + 80 x - 7680 = 0\) --> x = 32 --> d = 15.
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Re: M15-20 [#permalink]
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