BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
GMATLiveTeach 7 seats left
Chris Peckover
NEXT LIVE COHORT

Oct 13 to Jan 7, 2027

with Chris Peckover

Schedule
Tue, Thu · 8:00 to 10:00 PM ET
Included
40 live hours + 6 months of GMAT OnDemand
  • Live instruction and real-time questions
  • Class recordings and assigned practice
View class & enroll
Limited cohort · enrollment openTarget Test Prep
EALiveTeach 5 seats left
Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
Live EA class + 6 months of EA OnDemand
  • Expert-led weekly online sessions
  • EA Masterclass access between classes
View EA class & enroll
Limited cohort · enrollment openTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

130-point score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions
Start free 5-day trial
Read 559 reviewsTarget Test Prep

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

Rare Coins problem

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Newbie | Next Rank: 10 Posts
Posts: 3
Joined: Mon Oct 13, 2008 5:26 am
Location: Philadelphia, PA
by fk352 » Wed Oct 15, 2008 1:56 pm
Hi All, I took one look at the following problem and was completely stumped. I looked like a deer in headlights. Anyone want to give it a shot? I'll post the OA later:

Rare Coins

In a rare coin collection, one in six coins is gold. If 10 non-gold coins are subsequently traded for 10 gold coins, the ratio of gold coins to non-gold coins will be 1 to 4. Based on this information, how many gold coins are there now in this collection (after the trade)?

A) 50
B) 60
C) 180
D) 200
E) 300

[/u]
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
User avatar
Community Manager
Posts: 1049
Joined: Sun Apr 06, 2008 5:15 pmFollowed by: 27 members
Location: Pittsburgh, PA
Thanked: 113 times
GMAT Score:710

by dmateer25 » Wed Oct 15, 2008 2:27 pm
Ratio is

1 Gold : 5 Non Gold


Let X = Number of Gold Coins Originally
5X = Number of Non-Gold Originally

That means that:

(x+10)/(5x-10) = 1 Gold/ 4 Non Gold

4X + 40 = 5x -10
50 = X

Originally 50 gold coins.

So now there are 60 gold coins and 240 non gold coins (because of the ratio of 4 non gold for every 1 gold)

I go with B
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 574
Joined: Sun Jun 01, 2008 8:48 am
Location: Bangalore
Thanked: 28 times

by vishubn » Wed Oct 15, 2008 5:48 pm
Indeed.
i approached the problem in the same way as dmateer25 did.
60 is my call as well

Vishu

oa ???
Last edited by vishubn on Thu Oct 16, 2008 6:26 pm, edited 1 time in total.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 871
Joined: Wed Aug 13, 2008 7:48 am
Thanked: 48 times

by stop@800 » Wed Oct 15, 2008 8:23 pm
B for me too..

OA?
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 541
Joined: Thu May 31, 2007 6:44 pmFollowed by: 3 members
Location: UK
Thanked: 21 times
GMAT Score:680

by rohangupta83 » Thu Oct 16, 2008 12:29 am
Total # of coins = 5X
gold coins = X

After trade

Total # of coins = 5X - 10
Gold coins = X + 10

New ratio
Coins: Gold = 1:4

Thus, 5X - 10 = 4X + 40
X = 50 (originally)

New # of gold coins = X + 10 or 50 + 10 = 60
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 13
Joined: Thu Mar 27, 2008 8:06 am
Location: Singapore
Thanked: 1 times

by msvmuthu » Thu Oct 16, 2008 12:54 am
let no of gold coins originally be x
and no of non gold coins originally be n

=> one in six coins is gold
no of gold coin originally = 1
no of non gold coins originally = total coins - gold coins = 6-1 =5
there fore forming an equation:
x/n = 1/5
n= 5x
5x-n =0 -> eq1

=>If 10 non-gold coins are subsequently traded for 10 gold coins
current no of gold coins = x+10
current no of non gold coins = n-10

=>the ratio of gold coins to non-gold coins will be 1 to 4
(x+10) / (n-10) = 1/4

4(x+10) = n-10
4x+40 = n-10
4x-n=-50
-4x+n=50 ->eq2

solving eq1 and eq2 will give
x=50

But please remember that x is no of gold coins before the trade.
After the trade 10 gold coins are added to the collection.

There fore no of gold coins in the current collection is 50 + 10 = 60
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Newbie | Next Rank: 10 Posts
Posts: 3
Joined: Mon Oct 13, 2008 5:26 am
Location: Philadelphia, PA

by fk352 » Thu Oct 16, 2008 1:51 pm
haha, sheesh. maybe the problem wasn't as hard as I thought.

Yes, OA is B, 60 gold coins.

Thanks everyone.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 574
Joined: Sun Jun 01, 2008 8:48 am
Location: Bangalore
Thanked: 28 times

by vishubn » Thu Oct 16, 2008 6:30 pm
haha, sheesh. maybe the problem wasn't as hard as I thought.
.... Ya may be wording was bit on tricky side ...

thanks for the question though

Vishu
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1