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# An alloy contains tin and copper in the ratio of 4:5 . If tin has 20%

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Math Expert
Joined: 02 Sep 2009
Posts: 47019
An alloy contains tin and copper in the ratio of 4:5 . If tin has 20% [#permalink]

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07 Feb 2014, 04:31
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45% (medium)

Question Stats:

65% (01:23) correct 35% (01:48) wrong based on 222 sessions

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An alloy contains tin and copper in the ratio of 4:5 . If tin has 20% impurity and copper has 58% , what is the avg impurity percentage ?

A. 20%

B. 41.1%

C. 35%

D. 38%

E. 43.1%

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Math Expert
Joined: 02 Sep 2009
Posts: 47019
An alloy contains tin and copper in the ratio of 4:5 . If tin has 20% [#permalink]

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07 Feb 2014, 11:22
An alloy contains tin and copper in the ratio of 4:5 . If tin has 20% impurity and copper has 58% , what is the avg impurity percentage ?

A. 20%

B. 41.1%

C. 35%

D. 38%

E. 43.1%

(4*0.2 + 5*0.58)/(4+5) = ~41%.

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Re: An alloy contains tin and copper in the ratio of 4:5 . If tin has 20% [#permalink]

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04 Mar 2014, 01:51
2
Impurity in tin = 20% of 4/9 = 80/900
Impurity in Cu = 58% of 5/9 = 290/900

Total Impurity = 380/900. Percent = 380/9 = 41.2
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Re: An alloy contains tin and copper in the ratio of 4:5 . If tin has 20% [#permalink]

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29 Mar 2015, 07:52
Bunuel wrote:
aimlockfire1 wrote:
An alloy contains tin and copper in the ratio of 4:5 . If tin has 20% impurity and copper has 58% , what is the avg impurity percentage ?

a) 20%

b) 41.1%

c) 35%

d) 38%

e) 43.1%

(4*0.2 + 5*0.58)/(4+5) = ~41%.

Tin and copper are in ratio of 4:5. So let total mixture be 4+5 = 9.
So, tin = 4 and copper = 5.
Now, tin has 20% impurity = 20% * 4 = 0.8
Also, copper has 58% impurity = 58% * 5 = 2.9
Average = (0.8 + 2.9)/(4+5)
= 3.7/9
= 0.411
= 41.1%

Hence option (B).

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Posts: 23
Re: An alloy contains tin and copper in the ratio of 4:5 . If tin has 20% [#permalink]

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18 May 2015, 10:18
An alloy contains tin and copper in the ratio of 4:5 . If tin has 20% impurity and copper has 58% , what is the avg impurity percentage ?

a) 20%

b) 41.1%

c) 35%

d) 38%

e) 43.1%

The problem is on ratio / Proportion but look closely it can be solved with weighted average concept

(4*0.2 + 5*0.58)/(4+5) = ~41%.

Intern
Joined: 21 May 2012
Posts: 2
Re: An alloy contains tin and copper in the ratio of 4:5 . If tin has 20% [#permalink]

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18 May 2015, 10:46
Its a 10sec Question.
Because tin:Copper is in the ratio 4:5, there is more of Copper and hence more impurities from Copper in the final alloy. Since Copper has 58% impurities, the overall average should be closer to 58%, which narrows down the options to B or E. Since T:C is close to 1 (4:5), we want to pick the option closer to the middle but more towards C, which is 42%. Hence option B
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Posts: 980
Re: An alloy contains tin and copper in the ratio of 4:5 . If tin has 20% [#permalink]

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26 May 2016, 19:19
Bunuel wrote:
aimlockfire1 wrote:
An alloy contains tin and copper in the ratio of 4:5 . If tin has 20% impurity and copper has 58% , what is the avg impurity percentage ?

a) 20%

b) 41.1%

c) 35%

d) 38%

e) 43.1%

(4*0.2 + 5*0.58)/(4+5) = ~41%.

Can be done with weighted avg. formula
w1/w2=(a2-avg)/(avg-a1)
4/5=(58-x)/(x-20)
x=41.1
Ans B
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Joined: 01 Jan 2016
Posts: 21
Re: An alloy contains tin and copper in the ratio of 4:5 . If tin has 20% [#permalink]

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09 May 2018, 01:21
Bunuel wrote:
An alloy contains tin and copper in the ratio of 4:5 . If tin has 20% impurity and copper has 58% , what is the avg impurity percentage ?

A. 20%

B. 41.1%

C. 35%

D. 38%

E. 43.1%

(4*0.2 + 5*0.58)/(4+5) = ~41%.

Perhaps this is extremely obvious, but how are you able to "see" the last operation so precisely. I.e. does it not require no division by a decimal without a calculator to determine the ~41%? Best regards.
Math Expert
Joined: 02 Sep 2009
Posts: 47019
Re: An alloy contains tin and copper in the ratio of 4:5 . If tin has 20% [#permalink]

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09 May 2018, 01:28
GMAT2645 wrote:
Bunuel wrote:
An alloy contains tin and copper in the ratio of 4:5 . If tin has 20% impurity and copper has 58% , what is the avg impurity percentage ?

A. 20%

B. 41.1%

C. 35%

D. 38%

E. 43.1%

(4*0.2 + 5*0.58)/(4+5) = ~41%.

Perhaps this is extremely obvious, but how are you able to "see" the last operation so precisely. I.e. does it not require no division by a decimal without a calculator to determine the ~41%? Best regards.

$$\frac{(4*0.2 + 5*0.58)}{(4+5)} = \frac{0.8+2.9}{9}= \frac{3.7}{9}$$.

3.6/9 = 0.4, so 3.7/9 will be a bit larger.
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Re: An alloy contains tin and copper in the ratio of 4:5 . If tin has 20% [#permalink]

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09 May 2018, 05:38
We can also transition the fraction to percent by recognizing that a fraction with denominator of 9 can be multiplied by 11 to generate an approximation.

$$\frac{3.7}{9}*11 = \frac{40.7}{99}$$. As we divide only by 99 and not by 100 the value in percent must be slightly higher so 41,1% is the only reasonable answer.
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Re: An alloy contains tin and copper in the ratio of 4:5 . If tin has 20% [#permalink]

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10 May 2018, 10:33
Bunuel wrote:
An alloy contains tin and copper in the ratio of 4:5 . If tin has 20% impurity and copper has 58% , what is the avg impurity percentage ?

A. 20%

B. 41.1%

C. 35%

D. 38%

E. 43.1%

Let’s re-express the ratio of tin to copper as 4x : 5x. The average impurity is:

[4x(0.2) + 5x(0.58)]/9x = (0.8x + 2.9x)/9x = 3.7/9 = 0.411 = 41.1%.

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Joined: 01 Jan 2016
Posts: 21
Re: An alloy contains tin and copper in the ratio of 4:5 . If tin has 20% [#permalink]

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16 May 2018, 18:37
Bunuel wrote:
An alloy contains tin and copper in the ratio of 4:5 . If tin has 20% impurity and copper has 58% , what is the avg impurity percentage ?

A. 20%

B. 41.1%

C. 35%

D. 38%

E. 43.1%

(4*0.2 + 5*0.58)/(4+5) = ~41%.

Are you completing the (4*0.2 + 5*0.58)/(4+5) = ~41% calculation in your head? Thank you.
Intern
Joined: 01 Jan 2016
Posts: 21
Re: An alloy contains tin and copper in the ratio of 4:5 . If tin has 20% [#permalink]

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16 May 2018, 18:42
ScottTargetTestPrep wrote:
Bunuel wrote:
An alloy contains tin and copper in the ratio of 4:5 . If tin has 20% impurity and copper has 58% , what is the avg impurity percentage ?

A. 20%

B. 41.1%

C. 35%

D. 38%

E. 43.1%

Let’s re-express the ratio of tin to copper as 4x : 5x. The average impurity is:

[4x(0.2) + 5x(0.58)]/9x = (0.8x + 2.9x)/9x = 3.7/9 = 0.411 = 41.1%.

When it comes to the simple arithmetic such as "5x(0.58) = 2.9x" or "3.7/9 = 0.411" what shortcuts are you using to compute these calculations? Thank you.
Math Expert
Joined: 02 Sep 2009
Posts: 47019
Re: An alloy contains tin and copper in the ratio of 4:5 . If tin has 20% [#permalink]

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16 May 2018, 21:49
GMAT2645 wrote:
Bunuel wrote:
An alloy contains tin and copper in the ratio of 4:5 . If tin has 20% impurity and copper has 58% , what is the avg impurity percentage ?

A. 20%

B. 41.1%

C. 35%

D. 38%

E. 43.1%

(4*0.2 + 5*0.58)/(4+5) = ~41%.

Are you completing the (4*0.2 + 5*0.58)/(4+5) = ~41% calculation in your head? Thank you.

Check this: https://gmatclub.com/forum/an-alloy-con ... l#p2058527
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Re: An alloy contains tin and copper in the ratio of 4:5 . If tin has 20%   [#permalink] 16 May 2018, 21:49
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