The finesse of a gold bar is the weight of the gold present in the bar

Question

The  finesse  of a gold bar is the weight of the gold present in the bar divided by the total weight of the bar. A jeweler has three gold bars. The three bars have identical weights, and the first two bars have identical finesse. If all three bars were melted and combined into one bar, the finesse of the resulting bar would be 0.96.

Select a finesse of the  First two bars  and a finesse of the  Third bar  that are jointly consistent with the given information. Make only two selections, one in each column.

ID: 100426

Sajjad1994 · Accepted Answer

Official Explanation

According to the text, the finesse of a gold bar is equal to the weight of gold in the bar divided by the total weight of the bar. The text also indicates that the three gold bars have the same weight, that the first two bars have the same finesse, and that melting and combining the three bars would result in a gold bar with a finesse of 0.96. Since the first two bars have identical finesse and identical weight, the weight of the gold present in those bars must also be identical.

Let x equal the weight of the gold in each of the first two bars, let y equal the weight of the gold in the third bar, and let w be the total weight of each bar. Then, since the finesse of the combined bar is 0.96, it must be the case that

\frac{2x+y}{3w}=0.96  or  2(\frac{x}{w})+(\frac{y}{w})=(0.96)(3)=2.88 . But then if  F_{1}=\frac{x}{w}  is the finesse of each of the first two bars,

and  F_{3}=\frac{y}{w}  is the finesse of the third bar, then  2F_{1}+F_{3}=2.88 . Among the available options, substituting 0.98

for  F_{1}  and 0.92 for  F_{3}  gives us  2(0.98) + 0.92 = 1.96 + 0.92 = 2.88 .

No other pair of options yields the correct result.

Answer

Third bar: 0.92
First two bars: 0.98

KarishmaB · Answer

Simply imagine the scale of weighted averages. All 3 bars have equal weight. First two bars have same finesse. It simply means that the weight given to the finesse of the first 2 bars is twice the weight given to the finesse of the third bar. So the average finesse of .96 will be at a distance of 1:2 from the finesse of first two bars to the finesse of the third bar. Hence 0.98 could be the finesse of first two bars and .92 the finesses of the third bar. 0.98 - - 0.96 - - - - 0.92 Their distance from the average will be in the ratio 1:2. ANSWER: 0.98, 0.92 Video of weighted averages: https://www.youtube.com/watch?v=_GOAU7moZ2Q

chetan2u · Answer

Since all bars have identical weight and it speaks of average finesse, the question can be solved through weighted average method.

Let the weight be x each, so the ratio of those with higher finesse and lowers finesse will be (x+x):x or 2:1

Thus, the finesse too should be in the same ratio. The distance between the lower finesse and 0.96 should be two times that between higher and 0.96

So, one has to be higher: If it is next higher from 0.96, then 0.98, so (0.98-0.96)=1/2 * (0.96-y)….2*0.02 = 0.96-y
Or y=0.92

Apt0810 · Answer

Weight of gold in 1st and 2nd = x
Since finesse is equal for both
Weight of gold in 3rd bar = y

Total weight of each is =w

Now if they are melted and turned into one bar, resulting finesse is

2x+y/3w = 0.96
2x/w + y/w = 0.96*3
2x/w + y/w = 2.88

Now putting the above options in the equation,so

2*.98+0.92=2.88
So finesse of first and second is 0.98

Finesse of third is 0.92

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mSKR · Answer

Total 0.96*3 = 2.88 (total finness)
2X+Y = 2.88

Prethinking:
Option1: 2 numbers closure above 0.96 and 3rd one far below 0.96 OR 
option2: 2 numbers very near below 0.96 and 3rd one should be much higher than 0.96)

As per all options given higher than 0.96 is very close so it maybe possible than 2 numbers are higher than 0.96 so i chose first nearby option value above 0.96 for X i.e. 0.98 for X

Try Y= 0.98 then Y comes out to be 0.92

A quick check on our answer:
Try X = 0.99 then Y should be 0.90 - not available( Prethinking: option1)
Try Y = 0.98 then X should be 0.95*2 - not available( Prethinking: option2)

Kartik2121 · Answer

This one can be solved pretty quickly using the alligation method. Here is how it can be done:

The average finesse is 0.96. So the value of individual finesse of the two blocks has to be either higher than the average or lower than the average.

The same can be said about the value of finesse of the 3rd bar as well.

Now as there are two gold bars with equal finesse, we can deduce that their impact on the average shall be twice the impact of the 3rd bar. So, the 3rd bar shall be twice as far from the average finesse value as the finesse value of the two bars.

Thus, we are left with only two choices:

Either the finesse of two bars is 0.98 or 0.94 and that of 3rd bar is then 0.92 or 1.00 respectively.

The answer choices are pretty clear:

We have no 1.00. Thus, 0.94 for two bars and 1.00 for 3rd bar is out.

We are left with only one situation:

The finesse value for two bars is 0.98 and that of third bar is 0.92.

Posted from my mobile device

sayan640 · Answer

KarishmaB   chetan2u   GMATinsight  , Would you like to discuss this question ? I derived the equation but could not figure out the value.

Gemmie · Answer

Third bar                           First two bars
1                                      2
 ------------------96--------- 
          2a                  a

x > 96
From the answer choices, x is either 98 or 99

If x = 99 => a = 3 => y = 96 - 2a = 90 => not an answer choice
If x = 98 => a = 2 => y = 96 - 2a = 92 => Ok

First two bars: x = 0.98
Third bar: y = 0.92

Dbrunik · Answer

Very high-quality challenging question

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nikitathegreat · Answer

I have understood the ratio bit of the two bars that have identical weight: third bar will have ratio 1:2. But then how are we finding the values? Can someone provide step by step process for this?

Two bars ------------------------third bar 
Weight = 2 weight = 1
Ratio = 1:2

How are we finding the finesses now with these values?
Thanks

nikiki · Answer

We can solve it with the table we use for mixture problems

Finesse =  \frac{ Weight of gold}{Total Weight}

Finesse * Total Weight = Weight of gold

Finesse 
 Total Weight 
 Weight of Gold 
 
  Bar 1 
 F1 
 W 
 F1 * W 
 
  Bar 2 
 F1 
 W 
 F1 * W 
 
  Bar 3 
 F2 
 W 
 F2 * W

0.96 
 3W 
 2F1*W + F2*W

0.96 * 3W = 2F1*W + F2*W
2F1 + F2 = 2.88

After some trial and error we get
F1 = 0.98
F2 = 0.92

Nsp10 · Answer

do not go by the ratio,

here is how you should proceed,

finesse = Gold_wt/ total_wt

F1= x1/a.........x1=F1*a
F1=F2=x2/a {as finesse of the first 2 gold bar is same hence F1=F2, but we do not know about the wt gold present hence took them as x1 and y1}
F3=x3/a {where a is the total wt of all 3 bars as given same}

now according to Q, all three bars were melted and combined into one bar

hence overall finesse= wt of gold in all 3 bars/total wt of 3 bars
= x1+x2+x3/a+a+a
=(F1*a+F2*a+F3*a)/3a
= F1*a+F1*a+F3*a/3a {as F1=F2}
=we can cancel out a
=(2F1+F3)/3 = 0.96 as given 
hence 
2F1+F2=0.96*3
now we have to check the options and find out the answer.

DhanyaAbhirami · Answer

Finesse is same as concentration of gold
Though 3 ingredients are present in mixture, first two bars have identical weight and identical finesse i.e. both are essentially the same
So we can consider this as a 2 ingredient mixture with 2 units of first bar and 1 unit of third bar
One of the finesse must be more than 0.96 and other less than 0.96
F---1---AVG=0.96----2-----T
If T=0.98, then F=0.95 (Not Present in options)
If T=0.99, then F=0.945 (Not Present in options)
T----2----AVG=0.96---1----F
If F=0.98, then T=0.92 (Present in options)
If F=0.99, then T=0.90 (Not Present in options)

The finesse of a gold bar is the weight of the gold present in the bar

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	Finesse	Total Weight	Weight of Gold
Bar 1	F1	W	F1 * W
Bar 2	F1	W	F1 * W
Bar 3	F2	W	F2 * W
	0.96	3W	2F1W + F2W

GMAT Two-part Analysis (TPA) Questions

The finesse of a gold bar is the weight of the gold present in the bar

Prep Toolkit

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