GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 20 May 2019, 17:54

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

If Jay has 99 problems, in how many ways can he select k of them to ra

Author Message
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 55188
If Jay has 99 problems, in how many ways can he select k of them to ra  [#permalink]

Show Tags

18 Apr 2017, 04:54
1
27
00:00

Difficulty:

85% (hard)

Question Stats:

41% (01:40) correct 59% (01:26) wrong based on 436 sessions

HideShow timer Statistics

If Jay has 99 problems, in how many ways can he select k of them to rap about?

(1) Jay can select k+1 of his problems in 3764376 different ways.
(2) Jay can select k–1 of his problems in 4851 different ways.

_________________
Math Expert
Joined: 02 Aug 2009
Posts: 7681
If Jay has 99 problems, in how many ways can he select k of them to ra  [#permalink]

Show Tags

18 Apr 2017, 18:37
7
4
Bunuel wrote:
If Jay has 99 problems, in how many ways can he select k of them to rap about?

(1) Jay can select k+1 of his problems in 3764376 different ways.
(2) Jay can select k–1 of his problems in 4851 different ways.

Hi,

We do not and we should not get overwhelmed by the huge numbers here..
the main point is that we require two figures here to get on to value of k, and thus C should be the answer

Why?
Let's try with a smaller number.. 5
Choosing one out of 5 or 4 out of 5 is SAME.
$$5C1=\frac{5!}{1!(5-1)!}=\frac{5!}{1!/4!}=5C4$$....
So it can be 1 or 4, and therefore we require TWO relations of k to get to the value of k...

Let's see the statements..
First let's check the smaller number
II. Choosing k-1 gives 4851 ways..
$$99C(k-1)=4851=99*49=\frac{99*98}{2}=\frac{99*98*97*96*.....*2*1}{2*97*96*..*2*1}=\frac{99!}{2!97!}$$
So k-1 can be 2 or 97, thus k can be 2+1=3 OR 97+1=98..
Insuff alone

I. Choosing k+1is 3764376...
Insufficient

Combined..
If k is 98, k+1 is 99... BUT choosing 99 out of 99 is 1 so eliminate
Therefore k+1 must be 3+1=4 and k MUST be 3
Sufficient

C
_________________
General Discussion
Intern
Joined: 03 Sep 2015
Posts: 18
Re: If Jay has 99 problems, in how many ways can he select k of them to ra  [#permalink]

Show Tags

28 Jul 2018, 07:37
Hi Chetan

Could you please elaborate on the first statement. How did you determine its insufficiency?
Math Expert
Joined: 02 Aug 2009
Posts: 7681
If Jay has 99 problems, in how many ways can he select k of them to ra  [#permalink]

Show Tags

28 Jul 2018, 09:10
Rw27 wrote:
Hi Chetan

Could you please elaborate on the first statement. How did you determine its insufficiency?

Hi when you are looking for combinations..
Choosing k out of say 10 will be same as finding 10-k out of 10 so 10C4 = 10C(10-4)..
Why 10C4=10!/(4!*(10-4)!)=10!/4!6!
And 10C6 = 10!/6!(10-6)!=10!/6!4!
Both same ,

So k can be 4 or 6 as both will give same value
_________________
Intern
Joined: 03 Sep 2015
Posts: 18
Re: If Jay has 99 problems, in how many ways can he select k of them to ra  [#permalink]

Show Tags

28 Jul 2018, 09:25
Ohok, got it. Thanks
Senior PS Moderator
Status: It always seems impossible until it's done.
Joined: 16 Sep 2016
Posts: 750
GMAT 1: 740 Q50 V40
GMAT 2: 770 Q51 V42
Re: If Jay has 99 problems, in how many ways can he select k of them to ra  [#permalink]

Show Tags

10 Jan 2019, 06:42
2
Couldn't help but smile while solving this question. Almost could hear Jay's rap and the tune of the amazing song that this problem subtly refers to

https://youtu.be/6uikJTnmtgw

Posted from my mobile device
_________________
Regards,

“Do. Or do not. There is no try.” - Yoda (The Empire Strikes Back)
GMATH Teacher
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 935
Re: If Jay has 99 problems, in how many ways can he select k of them to ra  [#permalink]

Show Tags

11 Jan 2019, 08:18
Almost could hear Jay's rap and the tune of the amazing song that this problem subtly refers to

https://youtu.be/6uikJTnmtgw

_________________
Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net
GMATH Teacher
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 935
Re: If Jay has 99 problems, in how many ways can he select k of them to ra  [#permalink]

Show Tags

11 Jan 2019, 08:25
Bunuel wrote:
If Jay has 99 problems, in how many ways can he select k of them to rap about?

(1) Jay can select k+1 of his problems in 3764376 different ways.
(2) Jay can select k–1 of his problems in 4851 different ways.

Let me solve a "different" problem. In the end, I am sure you will understand the solution to the original problem throughly!

GMATH wrote:
If Jay has 7 problems, in how many ways can he select k of them to rap about?

(1) Jay can select k+1 of his problems in 35 different ways.
(2) Jay can select k–1 of his problems in 21 different ways.

$$? = C\left( {7,k} \right)\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\boxed{\,? = k\,}$$

$$\left( 1 \right)\,\,\,C\left( {7,k + 1} \right) = 35\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,k + 1 = 4\,\,\,{\text{or}}\,\,\,k + 1 = 3\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,k = 3\,\,\,\,{\text{or}}\,\,\,\,k = 2\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\text{INSUFF}}.$$

$$\left( * \right)\,\,C\left( {7,4} \right) = C\left( {7,7 - 4} \right) = 35\,\,\,\,\,\,$$

$$\left( 2 \right)\,\,\,C\left( {7,k - 1} \right) = 21\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( {**} \right)} \,\,\,\,\,k - 1 = 2\,\,\,{\text{or}}\,\,\,k - 1 = 5\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,k = 3\,\,\,\,{\text{or}}\,\,\,\,k = 6\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\text{INSUFF}}{\text{.}}$$

$$\left( {**} \right)\,\,C\left( {7,2} \right) = C\left( {7,7 - 2} \right) = 21$$

$$\left( {1 + 2} \right)\,\,\,\left\{ \matrix{ \,\left( 1 \right)\,\,\, \Rightarrow \,\,\,k = 3\,\,\,\,{\rm{or}}\,\,\,\,k = 2 \hfill \cr \,\left( 2 \right)\,\,\, \Rightarrow \,\,\,k = 3\,\,\,\,{\rm{or}}\,\,\,\,k = 6 \hfill \cr} \right.\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,k = 3\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\rm{SUFF}}{\rm{.}}\,\,\,\,$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
_________________
Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net
Manager
Joined: 16 Oct 2011
Posts: 94
GMAT 1: 620 Q37 V38
GMAT 2: 640 Q36 V41
GMAT 3: 650 Q42 V38
GMAT 4: 650 Q44 V36
GMAT 5: 570 Q31 V38
Re: If Jay has 99 problems, in how many ways can he select k of them to ra  [#permalink]

Show Tags

11 Jan 2019, 13:48
Bunuel wrote:
If Jay has 99 problems, in how many ways can he select k of them to rap about?

(1) Jay can select k+1 of his problems in 3764376 different ways.
(2) Jay can select k–1 of his problems in 4851 different ways.

One of the things that this problem can test is the concept that factorials can be unpacked into the product of n-1 multiples

(1) This statement tells us that 99C(k+1) =3764376--> 99!/((k+1)!(99-(k+1))! This can be very different numbers depending on the choice of k NS

(2) This statement tells us that 99C(k-1)=4851 -- 99!/(k-1)!(99-(k-1))! = 4851. NS for the same reason as (1)

(1) and (2) if we divide (1) by (2) we get {99!/((k+1)!(99-(k+1)!}/{(99!/((k-1)!(99-(k-1))!} = 3764376/4851. At first glance this seems like it might not help, however we can expand some of our factorial terms to allow us to cancel out numerators and denominators

--> 99!/(k+1)(k)(k-1)!(98-k)! * (k-1)!(100-k)!/99! = 3764376/4851 --> (100-k)!/(k(k+1))(98-k)! = 3764376/4851 -->
(100-k)(100-(k+1))(100-(k+2))!/(k(k+1))(98-k)! = 3764376/4851 --> (100-k)(99-k)(98-k)!/((k(k+1)(98-k)!) = 3764376/4851, allowing our (98-k)! to cancel out.

---> (100-k)(99-k)/(k)(k+1) =3764376/4851 which is messy, but is 1V1E, so suff The answer is C
Intern
Joined: 19 Aug 2017
Posts: 11
GMAT 1: 680 Q51 V30
Re: If Jay has 99 problems, in how many ways can he select k of them to ra  [#permalink]

Show Tags

12 Jan 2019, 10:29
Yes, since the combinatorics formula gives you the same answer for two values of K, then each is insufficient. Then using both you take the value in common. C.
_________________
GMAT Tutor - MBA Prep Tutoring
https://mbapreptutoring.com/

Free resources: Probabilities Seminar - Error_Log - Ultimate guide to understand the GMAT Algorithm
Intern
Joined: 27 Nov 2018
Posts: 30
Re: If Jay has 99 problems, in how many ways can he select k of them to ra  [#permalink]

Show Tags

12 Jan 2019, 14:35
fskilnik wrote:
Bunuel wrote:
If Jay has 99 problems, in how many ways can he select k of them to rap about?

(1) Jay can select k+1 of his problems in 3764376 different ways.
(2) Jay can select k–1 of his problems in 4851 different ways.

Let me solve a "different" problem. In the end, I am sure you will understand the solution to the original problem throughly!

GMATH wrote:
If Jay has 7 problems, in how many ways can he select k of them to rap about?

(1) Jay can select k+1 of his problems in 35 different ways.
(2) Jay can select k–1 of his problems in 21 different ways.

$$? = C\left( {7,k} \right)\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\boxed{\,? = k\,}$$

$$\left( 1 \right)\,\,\,C\left( {7,k + 1} \right) = 35\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,k + 1 = 4\,\,\,{\text{or}}\,\,\,k + 1 = 3\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,k = 3\,\,\,\,{\text{or}}\,\,\,\,k = 2\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\text{INSUFF}}.$$

$$\left( * \right)\,\,C\left( {7,4} \right) = C\left( {7,7 - 4} \right) = 35\,\,\,\,\,\,$$

Hi Fabio, how did you reverse solve for k? I understand the equation and how the answer has two variables, I'm confused at how you actually found the values of the two variables quickly. I tried to solve as a quadratic equation, but that seems to be really inefficient? Thanks for your help!
Senior Manager
Joined: 04 Aug 2010
Posts: 414
Schools: Dartmouth College
Re: If Jay has 99 problems, in how many ways can he select k of them to ra  [#permalink]

Show Tags

14 Jan 2019, 05:35
1
Rule:
nCr = nC(n-r)

Examples:
5C2 = 5C(5-2) = 5C3.
10C3 = 10C(10-3) = 10C7.
99C5 = 99C(99-5) = 99C94.

BTGmoderatorDC wrote:
If Jay has 99 problems, in how many ways can he select k of them to rap about?

(1) Jay can select k+1 of his problems in 3764376 different ways.
(2) Jay can select k–1 of his problems in 4851 different ways.

Statement 2: Jay can select k-1 of his problems in 4851 different ways.
Since 4851 is relatively small, test small values for k-1.
Case 1: k-1 = 2
From 99 problems, the number of ways to choose 2 = 99C2 = (99*98)/(2*1) = 4851.
This works.
Thus, it's possible that k-1=2.
Since 99C2 = 99C97, it's also possible that k-1=97.
k-1=2 implies that k=3.
k-1=97 implies that k=98.
Thus, Statement 1 offers two options:
k=3 or k=98.
INSUFFICIENT.

Rule:
For every DS problem, there must be at least ONE CASE that satisfies both statements.

Statement 1: Jay can select k+1 of his problems in 3764376 different ways.
Only two cases satisfy Statement 2: k=3 and k=98.
At least one of these two cases must also satisfy Statement 1.
Test k=98.

Case 2: k=98, implying that k+1=99
From 99 problems, the number of ways to choose 99 = 99C99 = 1.
k+1=99 does not satisfy the condition that the number of ways to choose k+1 problems is 3764376.
Thus, Case 2 is not viable.

Implication:
Since there must be at least one case that satisfies both statements, it MUST be possible in Statement 1 that k=3, with the result that k+1=4.
Since it's possible in Statement 1 that k+1=4, we know that 99C4 = 3764376.
Since 99C4 = 99C95, it must also be possible in Statement 1 that k+1=95, with the result that k=94.
Thus, Statement 1 offers two options:
k=3 or k=94.
INSUFFICIENT.

Statements combined:
Only the blue option above satisfies both statements, implying that k=3.
SUFFICIENT.

_________________
GMAT and GRE Tutor
Over 1800 followers
GMATGuruNY@gmail.com
New York, NY
If you find one of my posts helpful, please take a moment to click on the "Kudos" icon.
Available for tutoring in NYC and long-distance.
Re: If Jay has 99 problems, in how many ways can he select k of them to ra   [#permalink] 14 Jan 2019, 05:35
Display posts from previous: Sort by