In the first year of a pyramid scheme, John convinced y of

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AAPL: Moderator; Posts: 2685; Joined: Sun Oct 29, 2017 2:08 pmFollowed by: 2 members

by AAPL » Wed Nov 21, 2018 5:12 am

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Veritas Prep

In the first year of a pyramid scheme, John convinced y of his friends to pay 30 dollars each to join a particular website that he created. Each of those y friends then convinced another y people to pay 15 dollars each to join the same website. If no one else joined the website that year and each person joined only once, what was the value of y?

A. The revenue for the website that year was $36,000.
B. The first y friends accounted for 1/25 of the total revenue for the website that year.

OA D.

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Source: Beat The GMAT — Data Sufficiency | Wed Nov 21, 2018 5:12 am

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Wed Nov 21, 2018 6:50 am

AAPL wrote:Veritas Prep

In the first year of a pyramid scheme, John convinced y of his friends to pay 30 dollars each to join a particular website that he created. Each of those y friends then convinced another y people to pay 15 dollars each to join the same website. If no one else joined the website that year and each person joined only once, what was the value of y?

A. The revenue for the website that year was $36,000.
B. The first y friends accounted for 1/25 of the total revenue for the website that year.

OA D.

Target question: What was the value of y?

Statement 1: The revenue for the website that year was $36,000.
First round of "investors": y people paying $30 each = 30y dollars
Second round of "investors": yÂ² people paying $15 each = 15yÂ² dollars
Total revenue = 15yÂ² + 30y
So, 15yÂ² + 30y = 36,000
Set equal to zero: 15yÂ² + 30y - 36,000 = 0

IMPORTANT: This is a quadratic equation, and quadratic equations typically have 2 solutions. IF there are two valid solutions to this equation, then statement 1 is not sufficient.
Divide both side by 15 to get: yÂ² + 2y - 2400 = 0
At this point, we should recognize that we IF we were to factor the left-side, we'd get (y + something)(y - something) = 0, which means one possible value for y is positive, and one possible value for y is negative.
Since the number of friends MUST be positive, we can be certain that the equation has ONLY ONE valid solution.
This means that IF WE WERE to solve the equation, we'd be able to target question with certainty.
So, statement 1 is SUFFICIENT

Statement 2: The first y friends accounted for 1/25 of the total revenue for the website that year
In other words, (1st round of investments) = (1/25)(total investments)
30y = (1/25)(15yÂ² + 30y)
Multiply both sides by 25 to get: 750y = 15yÂ² + 30y
Rearrange: 15yÂ² - 720y = 0
Factor: 15y(y - something) = 0
At this point, we can be certain that there is only 1 valid solution for y (since y = 0 is an invalid solution).
So, IF WE WERE to solve the equation, we'd be able to target question with certainty.
Statement 2 is SUFFICIENT

Answer: D

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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fskilnik@GMATH: GMAT Instructor; Posts: 1449; Joined: Sat Oct 09, 2010 2:16 pmFollowed by: 33 members; Thanked: 59 times

by fskilnik@GMATH » Wed Nov 21, 2018 12:29 pm

AAPL wrote:Veritas Prep

In the first year of a pyramid scheme, John convinced y of his friends to pay 30 dollars each to join a particular website that he created. Each of those y friends then convinced another y people to pay 15 dollars each to join the same website. If no one else joined the website that year and each person joined only once, what was the value of y?

A. The revenue for the website that year was $36,000.
B. The first y friends accounted for 1/25 of the total revenue for the website that year.

$${\rm{Total}}\,\,\,{\rm{ = }}\,\,\,30y + 15{y^2}\,\,\,\,\,\left( \$ \right)$$
$$? = y\,\,\,\,\,\left( {y \ge 1\,\,{\mathop{\rm int}} } \right)$$
$$\left( 1 \right)\,\,\,\,15{y^2} + 30y - 36000 = 0\,\,\,\,\,\mathop \Rightarrow \limits^{{\rm{roots}}} \,\,\,\,\,{y_1} \cdot {y_2} = - {{36000} \over {15}} < 0\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,y\,\,\left( {{\rm{root}}} \right)\,\,\, > 0\,\,\,\,{\rm{unique}}\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,{\rm{SUFF}}.\,$$
$$\left( 2 \right)\,\,\,25 \cdot 30y\,\, = \,\,30y + 15{y^2}\,\,\,\,\,\,\mathop \Rightarrow \limits^{:\,\,15} \,\,\;\,\,\,25 \cdot 2y = {y^2} + 2y\,\,\,\,\,\, \Rightarrow \,\,\,y\left( {y - 48} \right) = 0\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,y\,\,\left( {{\rm{root}}} \right)\,\,\, > 0\,\,\,\,{\rm{unique}}\,\,\,\,\left( { = 48} \right)\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,{\rm{SUFF}}.\,\,\,\,\,$$

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

Regards,
Fabio.

Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

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