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

 It is currently 22 Sep 2018, 02:34

### 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

# How many grams of pure iron must be alloyed with 387 grams of pure cop

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 49301
How many grams of pure iron must be alloyed with 387 grams of pure cop  [#permalink]

### Show Tags

25 Jan 2017, 06:56
00:00

Difficulty:

15% (low)

Question Stats:

74% (01:08) correct 26% (00:54) wrong based on 141 sessions

### HideShow timer Statistics

How many grams of pure iron must be alloyed with 387 grams of pure copper to create an alloy that is 90% copper?

A. 38.7
B. 43
C. 129
D. 301
E. 430

_________________
SVP
Joined: 08 Jul 2010
Posts: 2334
Location: India
GMAT: INSIGHT
WE: Education (Education)
Re: How many grams of pure iron must be alloyed with 387 grams of pure cop  [#permalink]

### Show Tags

25 Jan 2017, 07:13
1
4
Bunuel wrote:
How many grams of pure iron must be alloyed with 387 grams of pure copper to create an alloy that is 90% copper?

A. 38.7
B. 43
C. 129
D. 301
E. 430

387 grams of pure copper should serve the purpose of 90% of entire alloy

i.e. 387 = (90/100)*Total Alloy weight

i.e. Total Alloy Weight = 430 Grams

Amount of Iron to be added = 430 - 387 = 43 grams

_________________

Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION

##### General Discussion
Senior Manager
Joined: 06 Jan 2015
Posts: 476
Location: India
Concentration: Operations, Finance
GPA: 3.35
WE: Information Technology (Computer Software)
Re: How many grams of pure iron must be alloyed with 387 grams of pure cop  [#permalink]

### Show Tags

25 Jan 2017, 07:23
Bunuel wrote:
How many grams of pure iron must be alloyed with 387 grams of pure copper to create an alloy that is 90% copper?

A. 38.7
B. 43
C. 129
D. 301
E. 430

Hi,

We need option that gives 10% iron + 90% Copper
So by observation, you can eliminate C, D, & E
Now Coming to option A and B

Testing Option A
387+38.7=425.7(Total alloy)
Here if you subtract 425.7-42.5=383.15(less than 90%)

Hence option B
_________________

आत्मनॊ मोक्षार्थम् जगद्धिताय च

Resource: GMATPrep RCs With Solution

Director
Joined: 21 Mar 2016
Posts: 531
Re: How many grams of pure iron must be alloyed with 387 grams of pure cop  [#permalink]

### Show Tags

25 Jan 2017, 11:25
@vertasprepkarishma

any way to get this done by weighted avg method
Director
Affiliations: CrackVerbal
Joined: 03 Oct 2013
Posts: 522
Location: India
GMAT 1: 780 Q51 V46
Re: How many grams of pure iron must be alloyed with 387 grams of pure cop  [#permalink]

### Show Tags

26 Jan 2017, 12:32
Top Contributor
1
mohshu wrote:
@vertasprepkarishma

any way to get this done by weighted avg method

Hi Mohshu,

The algebraic approach is the best way to solve this problem as it requires a basic understand of percentages. If we keep the the weight of iron as 'x' grams, the total alloy will weigh '387+x' grams. Since copper makes up 90% of the alloy,

387/(387+x) = 90/100

Simplifying we get x = 43grams

The alternate approach is to make use of the alligation approach. To use the alligation approach there are two things that you need to do.

1. Draw the alligation diagram the basic framework of which is given below

Attachment:

Mixtures 1.png [ 8.12 KiB | Viewed 1114 times ]

2. Make sure that the values you substitute as the higher value, lower value and mean value are values associated with the word 'per' (percentages, ratios, averages...)

Now in this question we can form an alligation diagram using the percentages of Copper, we get the higher value as 100% and the lower value as 0%, since we have pure copper and pure iron respectively (the percentage of copper in pure iron is 0%)

Attachment:

Untitled.png [ 7.48 KiB | Viewed 1116 times ]

Since the ratio of copper to iron in the alloy is 9 : 1 and the weight of copper is 387 grams, then this 387 grams represents 9 parts of the ratio. So the weight of iron which is represented by 1 part in the ratio will be 387/9 = 43grams

Hope this helps!

_________________

SVP
Joined: 26 Mar 2013
Posts: 1808
Re: How many grams of pure iron must be alloyed with 387 grams of pure cop  [#permalink]

### Show Tags

27 Jan 2017, 02:15
When looking at this problem, significant amount of the alloy is copper. So we can eliminate choices C, D & E.

When applying allegation method

Iron--------------Alloy---------Copper
0%----90-------90%---10---100%

C/I= 90/10=9............I=387/9..Do not need to divide.. 387 is divisible by 9 So only one answer fits.

Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 3515
Location: United States (CA)
Re: How many grams of pure iron must be alloyed with 387 grams of pure cop  [#permalink]

### Show Tags

27 Jan 2017, 10:46
Bunuel wrote:
How many grams of pure iron must be alloyed with 387 grams of pure copper to create an alloy that is 90% copper?

A. 38.7
B. 43
C. 129
D. 301
E. 430

We need to create an alloy that is 90% copper. In other words:

copper/(iron + copper) = 90%

We can let i = the amount of pure iron in grams and create the following equation:

387/(i + 387) = 9/10

387(10) = 9(i + 387)

3870 = 9i + 3483

387 = 9i

i = 43

_________________

Scott Woodbury-Stewart
Founder and CEO

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

Manager
Joined: 02 Aug 2013
Posts: 65
Location: India
WE: Programming (Consulting)
How many grams of pure iron must be alloyed with 387 grams of pure cop  [#permalink]

### Show Tags

03 May 2017, 23:33
Let's say x grams of pure iron is added to form alloy.

New weight is (387+x).

The % of Iron in alloy = 10% = $$\frac{10}{100}$$

=> $$\frac{x}{(387+x)}$$ =$$\frac{10}{100}$$

=> 10x = 387 + x

=> x=$$\frac{387}{9}$$

=> x=43

Ans: B
VP
Joined: 07 Dec 2014
Posts: 1088
Re: How many grams of pure iron must be alloyed with 387 grams of pure cop  [#permalink]

### Show Tags

04 May 2017, 12:36
b
Bunuel wrote:
How many grams of pure iron must be alloyed with 387 grams of pure copper to create an alloy that is 90% copper?

A. 38.7
B. 43
C. 129
D. 301
E. 430

(387/.9)-387=43
B
Re: How many grams of pure iron must be alloyed with 387 grams of pure cop &nbs [#permalink] 04 May 2017, 12:36
Display posts from previous: Sort by

# Events & Promotions

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.