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# The ratio, by weight, of the four ingredients A, B, C, and D of a cert

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Math Expert
Joined: 02 Sep 2009
Posts: 52390
The ratio, by weight, of the four ingredients A, B, C, and D of a cert  [#permalink]

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23 Apr 2015, 02:59
27
00:00

Difficulty:

95% (hard)

Question Stats:

48% (03:28) correct 52% (03:36) wrong based on 276 sessions

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The ratio, by weight, of the four ingredients A, B, C, and D of a certain mixture is 4:7:8:12. The mixture will be changed so that the ratio of A to C is quadrupled and the ratio of A to D is decreased. The ratio of A to B will be held constant. If B will constitute 20% of the weight of the new mixture, by approximately what percent will the ratio of A to D be decreased?

A. 15%
B. 25%
C. 35%
D. 45%
E. 55%

Kudos for a correct solution.

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Re: The ratio, by weight, of the four ingredients A, B, C, and D of a cert  [#permalink]

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23 Apr 2015, 21:45
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2
Bunuel wrote:
The ratio, by weight, of the four ingredients A, B, C, and D of a certain mixture is 4:7:8:12. The mixture will be changed so that the ratio of A to C is quadrupled and the ratio of A to D is decreased. The ratio of A to B will be held constant. If B will constitute 20% of the weight of the new mixture, by approximately what percent will the ratio of A to D be decreased?

A. 15%
B. 25%
C. 35%
D. 45%
E. 55%

Kudos for a correct solution.

Start with what you know so that use of variables is minimized.

A:B:C:D = 4:7:8:12
B is 20% of the weight of new mixture so 7 parts is 20% of the weight. This means total weight of the new mixture is 35 parts.
Ratio of A:B remains constant so 4:7.
Ratio of A:C (4:8) is quadrupled (4*4:8 = 2:1). Since A is 4 parts, C must be 2 parts.

4:7:2:D
Total must add up to 35 parts. So D must be 35 - (4 + 7 + 2) = 22 parts.

A:D ratio has changed from 4:12 (1/3) to 4:22 (2/11). This is a decrease of (1/3 - 2/11)/(1/3) * 100 = 45% approx
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The ratio, by weight, of the four ingredients A, B, C, and D of a cert  [#permalink]

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23 Apr 2015, 13:30
1
Lets say our thing has the weight X(4+7+8+12) = 31*X. When we change mixture our weight changes to X(4 + 7 + 2 + 12*Y) =13*X+12*X*Y
We also know that 7*X = 0,2(13*X+12*X*Y). Need to find Y.
Reduce that last equation by X and lets just solve it for Y: Y = (7 - 0,2*13)/2,4 = 44/24 = 11/6
So the original ratio was 4/12 = 1/3
The new ratio is 1*6/(3*11) = 2/11
The resulting percentages are: (1/3-2/11)/(1/3) = 1 - 6/11 = 5/11 = 0,454 ~ 45%

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Re: The ratio, by weight, of the four ingredients A, B, C, and D of a cert  [#permalink]

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23 Apr 2015, 20:20
1
I guess answer is B .
original ratio A:B:C:D = 4:7:8:12
in ratio , A:C is quardapled , and A:B is constant
Hence ,
New ratio = 16:B:32:D
since ,A:B is constant , Find B from above two equations , which comes as 28
so new ratio =16:28:32:D

Now it is told that B constitutes , 20% of new mixture .
28=20/100 (new mixture)
so New mixtre quantity = 140

so
16:28:32:D =140
Hence D=64
hence ratio of A:D in new mixture= 16:64=25%
In old mixture = 4:12=33.33%

Change = (33.33-25)/33.33 = 8.33/33.33 which is approximately 25%
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The ratio, by weight, of the four ingredients A, B, C, and D of a cert  [#permalink]

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24 Apr 2015, 01:30
2
Bunuel wrote:
The ratio, by weight, of the four ingredients A, B, C, and D of a certain mixture is 4:7:8:12. The mixture will be changed so that the ratio of A to C is quadrupled and the ratio of A to D is decreased. The ratio of A to B will be held constant. If B will constitute 20% of the weight of the new mixture, by approximately what percent will the ratio of A to D be decreased?

A. 15%
B. 25%
C. 35%
D. 45%
E. 55%

Kudos for a correct solution.

Given :
The ratio of A to B will be held constant and B is not changed , so 'A' is also not changed.
If 'A' is not changed then the only way A:C will be quadrupled is by reducing C , from C to C/4 .
A:D is decreased , again 'A' is not changed , so , D is increased.
In new mixture B is 20% , so A+C+Dis 80% .

$$\frac{Old_{A:D } - New_{A:D }}{Old_{A:D }}*100$$

Solution:

A:B:C:D=4:7:8:12
Old : 4x , 7x, 8x, 12x

$$Old_{A:D } = \frac{1}{3}$$

New : 4x, 7x, 2x, UNKNOWN,Say 'Y' (>12x)

we are told that 7x=20%( 4x+7x+2x+y)====>35x=13x+y======>22x=y

$$New_{A:D } = \frac{4x}{22x}$$

$$\frac{Old_{A:D } - New_{A:D }}{Old_{A:D }}*100$$ = 45%

Math Expert
Joined: 02 Sep 2009
Posts: 52390
Re: The ratio, by weight, of the four ingredients A, B, C, and D of a cert  [#permalink]

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27 Apr 2015, 01:25
Bunuel wrote:
The ratio, by weight, of the four ingredients A, B, C, and D of a certain mixture is 4:7:8:12. The mixture will be changed so that the ratio of A to C is quadrupled and the ratio of A to D is decreased. The ratio of A to B will be held constant. If B will constitute 20% of the weight of the new mixture, by approximately what percent will the ratio of A to D be decreased?

A. 15%
B. 25%
C. 35%
D. 45%
E. 55%

Kudos for a correct solution.

First, make a quick table to represent the ratios in the original mixture:

A B C D
4 7 8 12

One way forward is to make the A:C change, holding A:B and A:D constant. (We know that A:D decreases, but we don’t know by how much, so for now, pretend that the ratio stays constant.)

To quadruple the ratio A:C, we can either multiply A by 4 or divide C by 4. Since we want to leave A:B and A:D constant, it’s more efficient to divide C by 4. So if A:D weren’t changing, the new mixture would have these ratios:

A B C D
4 7 2 12

Since A:D decreases, let’s add an unspecified amount to D. (We don’t want to mess with the A side of the ratio, since A is involved in other ratios.) Call that amount x.

A B C D
4 7 2 12 + x

Now we know that B will constitute 20% of the whole, so the ratio of B to the whole in the final mixture will be 20 to 100, or 1:5. If we look at the table, the ratio of B to the whole is 7 to 25 + x. We can equate these proportions:

1/5 = 7/(25 + x)
25 + x = 35
x = 10

So now we know that the final mixture has these proportions:

A B C D
4 7 2 22

The new ratio of A to D is 4:22, or 2:11. As a fraction, this ratio is 2/11.

The original ratio of A to D is 4:12, or 1:3. As a fraction, this ratio is 1/3.

Finally, we are asked this: if 1/3 is decreased to 2/11, what is the percent decrease?

A fast way to compute this number is first to figure out the factor that you multiply 1/3 by to get 2/11. Call that factor y.

(1/3)y = 2/11
y = 6/11

So 2/11 is 6/11 of 1/3. As a percent, 6/11 is approximately 55% (as a decimal, 6/11 = 0.5454…). If the new number (2/11) is 55% of the old number (1/3), then the percent decrease is 100% – 55%, or 45%.

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Re: The ratio, by weight, of the four ingredients A, B, C, and D of a cert  [#permalink]

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29 Jun 2015, 22:02
VeritasPrepKarishma wrote:
Bunuel wrote:
The ratio, by weight, of the four ingredients A, B, C, and D of a certain mixture is 4:7:8:12. The mixture will be changed so that the ratio of A to C is quadrupled and the ratio of A to D is decreased. The ratio of A to B will be held constant. If B will constitute 20% of the weight of the new mixture, by approximately what percent will the ratio of A to D be decreased?

A. 15%
B. 25%
C. 35%
D. 45%
E. 55%

Kudos for a correct solution.

Start with what you know so that use of variables is minimized.

A:B:C:D = 4:7:8:12
B is 20% of the weight of new mixture so 7 parts is 20% of the weight. This means total weight of the new mixture is 35 parts.
Ratio of A:B remains constant so 4:7.
Ratio of A:C (4:8) is quadrupled (4*4:8 = 2:1). Since A is 4 parts, C must be 2 parts.

4:7:2:D
Total must add up to 35 parts. So D must be 35 - (4 + 7 + 2) = 22 parts.

A:D ratio has changed from 4:12 (1/3) to 4:22 (2/11). This is a decrease of (1/3 - 2/11)/(1/3) * 100 = 45% approx

You solved the complete Q there itself, awesome. Thanks for updating your solution.

Regards,
Gaurav
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Joined: 09 Sep 2013
Posts: 9462
Re: The ratio, by weight, of the four ingredients A, B, C, and D of a cert  [#permalink]

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