A 20 kg metal bar made of alloy of tin and silver lost 2 kg

Question

A metal bar weighing 20 kilograms, consisting of tin and silver, has lost 2 kilograms of its weight when placed in water. If 10 kilograms of tin lose 1.375 kilograms in water and 5 kilograms of silver lose 0.375 kilograms in water, what is the proportion of tin to silver in the bar?

A.  \frac{1}{4}

B.  \frac{2}{5}

C.  \frac{1}{2}

D.  \frac{3}{5}

E.  \frac{2}{3}

Bunuel · Accepted Answer

Official Solution:

A metal bar weighing 20 kilograms, consisting of tin and silver, has lost 2 kilograms of its weight when placed in water. If 10 kilograms of tin lose 1.375 kilograms in water and 5 kilograms of silver lose 0.375 kilograms in water, what is the proportion of tin to silver in the bar?

A.  \frac{1}{4}  
B.  \frac{2}{5}  
C.  \frac{1}{2}  
D.  \frac{3}{5}  
E.  \frac{2}{3}

Let  t  be the amount of tin in the bar (in kilograms) and  s  be the amount of silver in the bar (in kilograms). Since 10 kilograms of tin lose 1.375 kilograms in water,  t  kilograms of tin will lose  \frac{t}{10}*1.375=0.1375*t  kilograms in water.

Since 5 kilograms of silver lose 0.375 kilograms in water,  s  kilograms of silver will lose  \frac{s}{5}*0.375= 0.075*s  kilograms in water. 
 
The total weight lost by both tin and silver is  0.1375*t + 0.075*s = 2  kilograms.
 
Since  t  is equal to  20 - s , we can substitute this into the equation to get  0.1375 *(20 - s) + 0.075 * s = 2 . Solving for  s , we find that  s = 12 . Hence,  t = 20 - s = 20 - 12 = 8 .
 
Finally, the ratio of tin to silver in the bar is calculated as  \frac{t}{s} = \frac{8}{12} = \frac{2}{3} .

Answer: E

anirudhoswal · Answer

10 kg tin loses 1.375 kg=> 20 kg loses 2.75 kg
5 kg sliver loses 0.375 kg=> 20 kg loses 1.5 kg
Actual loss is 2 kg.
Apply the alligation rule=>

2.75 1.5
 \ /
 2
/ \
0.5 0.75

ratio of tin/silver => 0.5/0.75 => 1/2/3/4 => 3/8 ???

anirudhoswal · Answer

I have been looking for a good explanation. Unfortunately, I have found none.
There is a brief mention of the method on wikipedia -
https://en.wikipedia.org/wiki/Alligation

Pls let me know if this explanation is satisfactory.
In the menawhile I will look for a better link.

Thanks.

Campanella1989 · Answer

interesting question.

This is my solution for the problem:

10 kg tin loses 1.375 kg => lost 13.75% its weight 
5 kg Silver loses 0.375 kg => lost 7.5% of its weight 
 
Initially, 20 kg of alloy loses 2kg => 10% of its weight

Call X is the weight of Tin the alloy
Y is the weight of Silver in the alloy

We have

=> X/Y = 2.5%/ 3.75% = 2/3

Very nice question!

KarishmaB · Answer

Rather than using the alligation diagram, you can simply use this formula to avoid confusion: w1/w2 = (A2 - Aavg)/(Avg - A1) Weight of Tin/Weight of Silver = (Silver's loss - Avg loss)/(Avg loss - Tin's loss) X/Y = (7.5 - 10)/(10 - 13.75) = 2/3 To check out a detailed discussion on this formula, see these posts: https://anaprep.com/arithmetic-weighted-averages/ https://anaprep.com/arithmetic-mixtures/ and these videos: https://www.youtube.com/watch?v=_GOAU7moZ2Q https://www.youtube.com/watch?v=VdBl9Hw0HBg

GSBae · Answer

Instead of using a formula, just think about it this way:

20kg of tin loses 2.75 kg in water.
20kg of silver loses 1.5kg in water.

Some sort of combination of these metals loses 2 kg in water.

Thus,

2.75(x) + 1.5(1-x) = 2 
 2.75x + 1.5 - 1.5x = 2 
 1.25x = .5 
 x = (1/2)/(5/4) = 2/5.

Thus, the alloy is 2/5 tin and thus 3/5 silver, and therefore the ratio of tin to silver is 2/3.

Answer: E

KarishmaB · Answer

Responding to a pm:

Here is how you will find the values of A1 an A2.

We have an overall loss (average loss). The average loss is 2 kg when 20 kg alloy is immersed. 
This is a loss of (2/20)*100 = 10%. 
This is Aavg

The loss of tin is 1.375 kg for every 10 kg.
This means it loses (1.375/10)*100 = 13.75% of its weight in water.
This is A1.

The loss of silver is .375 kg for every 5 kg.
This means it loses (.375/5)* 100 = 7.5% of its weight in water.
This is A2.

KarishmaB · Answer

Responding to a pm:

A 20 kg metal bar made of tin and silver lost 2 kg of its weight in the water. If 10 kg of tin loses 1.375 kg in the water and 5 kg of silver loses 0.375 kg, what is the ratio of tin to silver in the bar?

A) 1/4
B) 2/5
C) 1/2
D) 3/5
E) 2/3

The bar lost certain percentage of its weight. We don't know how much tin was lost and how much silver was lost but in all 2 kg was lost with is 10% of its overall weight.

Tin loses 1.375 kg in 10 kg so 13.75% of its weight when it is put in water.

Silver loses .375 kg in 5 kg so .375/5 * 100 = 7.5% of its weight in water.

Now, we just need to use weighted averages:

Wt/Ws = (7.5 - 10)/(10 - 13.75) = 2.5/3.75 = 2/3

Answer (E)

For more on weighted averages and mixtures, check:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/03 ... -averages/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/04 ... -mixtures/

EMPOWERgmatRichC · Answer

Hi All,

This question can be solved by TESTing THE ANSWERS.

Since we're told that the total weight of the metal bar is 20 kg, and the additional information is in 10 kg and 5 kg 'increments', I want to do a quick hypothetical calculation of a 20 kg bar that is exactly 10 kg tin and 10 kg silver....

10 kg tin = loses 1.375 kg of water 
10 kg silver = loses 2(.375) = .750 kg of water 
This 'mix' of metals (1:1) will lose... 1.375 + .75 = 2.125 kg of water.... This is pretty close to the 2 kg that it's supposed to be, but it's a little too much. Thus, we need a little more silver and a little less tin. Looking at the answer choices, the closest ratio to 1:1 is Answer E (2:3) - so let's TEST that Answer...

Answer E: 2:3

Tin:Silver in a 2:3 ratio gives us.... 
8 kg of tin and 12 kg of silver.... 
8 kg of tin = (8/10)(1.375) = (8/10)(11/8) = 88/80 = 11/10 kg of water lost 
12 kg of silver = (12/10)(.750) = (12/10)(3/4) = 36/40 = 9/10 kg of water lost 
11/10 + 9/10 = 20/10 = 2 kg... This is an exact match for what we were told, so this MUST be the answer.

Final Answer:  E

GMAT assassins aren't born, they're made, 
Rich

BrentGMATPrepNow · Answer

Given: 10 kg of tin loses 1.375 kg in the water  
This means ONE kg of tin loses 0.1375 kg in the water

Given: 5 kg of silver loses 0.375 kg in the water  
This means ONE kg of silver loses 0.075 kg in the water

Let T = the number of kilograms of TIN in the metal bar
Let V = the number of kilograms of SILVER in the metal bar

Since the metal bar weighs 20 kg, we can write:  T + V = 20

Since the metal bar lost 2 kg of its weight in the water, we can write: (water lost from the TIN) + (water lost from the SILVER) = 2 kg
In other words:  0.1375T + 0.075V = 2

We now have the following system of equations: 
 T + V = 20 
 0.1375T + 0.075V = 2

Take the top equation and multiply both sides by 0.1375 to get: 
 0.1375T + 0.1375V = 2.75 
 0.1375T + 0.075V = 2

Subtract the bottom equation from the top equation to get: 0.0625V = 0.75
Solve: V = 0.75/0.0625 = 12

So, there are  12  kg of silver in the 20 kg metal bar
This means there are  8  kg of tin in the 20 kg metal bar

What is the ratio of tin to silver in the bar?  
Tin/Silver =  8 / 12  = 2/3

Answer: E

Vordhosbn · Answer

We can use the teeter totter approach here:

1.375/10 ----------- 2/20 -------- 0.375/5

1.5/20 --------------2/20 --------- 2.75/20

--> 0.5/0.75 = 2/3.

VikramMarkiv · Answer

Let x be the weight of tin and y be the weight of silver in the bar.
From the problem:
x + y = 20 ..........(1)

Weight lost in water.
 
 Tin loses 1.375 kg for every 10 kg, so tin loses (1.375/10) × x = 0.1375x 
 Silver loses 0.375 kg for every 5 kg, so silver loses (0.375/5) × y = 0.075y  
The total weight lost in water is:
0.1375x + 0.075y = 2 ..........(2)

Solving for the system of equations.
From equation (1):
y = 20−x ......(3)

Substitute y = 20−x into equation (2):
0.1375x + 0.075(20−x)

Expand and simplify:
0.1375x + 1.5 − 0.075x = 20
0.0625x = 0.5 
x = 0.5/0.0625 = 8

Finding y.
From equation (3):
y = 20−x = 20−8 = 12

The proportion of tin to silver is:
x/y = 8/12 = 2/3

Kinshook · Answer

A metal bar weighing 20 kilograms, consisting of tin and silver, has lost 2 kilograms of its weight when placed in water.

If 10 kilograms of tin lose 1.375 kilograms in water and 5 kilograms of silver lose 0.375 kilograms in water, what is the proportion of tin to silver in the bar?

Metal bar , consisting of tin and silver, weight lost percentage in water = 2/20 = 10%

Tin weight lost percentage in water = 1.375/10 = 13.75%

Silver weight lost percentage in water = .375/5 = 7.5%

Attachment:

Screenshot 2024-12-05 at 12.42.54 PM.png [ 19.75 KiB | Viewed 6530 times ]

2.5%: 3.75% = 2:3

IMO E

anirchat · Answer

In stead of using algebra, is there any optimal way of solving this like option elimination or something ?

rohitz98 · Answer

Let x be amount of tin in metal and y be amount of silver in metal

Given 10 kg of tin lose 1.375 kg in water and 5 kg of silver loses 0.375 kg in water 
Hence x kg loses x * (1.375/10) kg and y kg loses y*(0.375/5) kg

Total weight lost = 2 kg;
Hence, 0.1375 x + 0.075 y = 2 kg &
x + y = 20 kg

Solving 2 equation we get x = 8 and y = 12

Hence x/y = 2/3 (Option E)

asharma7899 · Answer

1/2 // 3/4 => 1*4/2*3 => 2/3 which is the correct answer.

sarthak1701 · Answer

Let X = Tin in the 20KG bar, and let 20-X = Silver in the bar. The bar has lost 2 KGS, according to the loss percentages provided:

X*13.75/100 + (20-X)*7.5/100 = 2

(6.25X + 150)/100 = 2, 6.25X = 50, X = 8. Therefore, Tin the Bar = 8 KGS so the Silver in the bar must be 20-8 = 12 KGS. Ratio of Tin to Silver = 8/12 = 2/3.

Ans. E

Diegoram17 · Answer

thanks bro, very helpful explanation

Rishm · Answer

I just used logic and estimation, we see that 1.375 kg was lost by tin from 10kgs and 5kgs was lost by silver in 0.375, total comes to 1.750(lost with 15kgs), which is shy of 0.250kgs(20kgs Target loosing 2kgs), so we need more of tin and silver to ensure they lose total 2kgs. See that there is already less of silver and more of tin, only 2 options are fitting them correctly, D & E. By close look it cannot be 3/5 as tin is already more than 60% in the allow, so we are left with one option that is 2/3 which is around 66-67%, I think this process is good for mind mapping or else weighted average method is best which is mentioned,

A 20 kg metal bar made of alloy of tin and silver lost 2 kg

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A 20 kg metal bar made of alloy of tin and silver lost 2 kg

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