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Bunuel
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Is there a faster way to do this then to use the weighted average formula?
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A faster way to do this would be to realize that the seller is not going to take a loss in selling, also as profit margin is not mentioned, there is no profit either.
Selling price = Cost price, so the CP has to be a multiple of the three wheats' combined CP.

A)1.27+2.58+3.96=7.81 not a multiple of 1.3
B)1.27+1.29+2.64=5.20 multiple of 1.3 => Our ans is B
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Bunuel
Three types of wheat of $1.27, $1.29 and $1.32 per kg are mixed together to be sold at $1.30 per kg. In what ratio should this wheat be mixed?

A. 1:2:3
B. 1:1:2
C. 2:1:3
D. 2:2:3
E. 2:2:5
­Check using the options answer must be (B) -

\(\frac{1.27*1 + 1.29*1 + 1.32*2}{1 + 1 + 2} = 1.30\)
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Care to explain your last 2 lines? Thank you
sudeshpatodiya

Quote:
Three types of wheat of $1.27, $1.29 and $1.32 per kg are mixed together to be sold at $1.30 per kg. In what ratio should this wheat be mixed?

A. 1:2:3
B. 1:1:2
C. 2:1:3
D. 2:2:3
E. 2:2:5
Average = Avg = 1.30 $/kg
Let three prices be, A=1.27, B=1.29, C=1.32

See how the different parts affect the average. Is it surplus or deficit.

Average Deficit is introduced by A and B
A-Avg = 127-130 = -3
B-Avg = 129-130 = -1

Surplus is introduced by C
C-Avg = 132-130 = 2

Now, for each unit of A and B, total Deficit = -4
For each unit of C, Surplus = 2

So, we need 2 units of C to compensate each unit of A and C combined to get the final average deviation as zero.

So, Ratio of A:B:C = 1:1:2

(B)
­
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Take the ratio of A:C using the allegation method u get:
2:3

Similarly do it for B:C:
2:1

The reason to pick these two ratios is because u need one value always higher that the mean value for weighted averages to work

Now arrange it properly
B:C=2:1
A:C=2:3

We need to add the ratios of C since it is added twice

A:B:C = 2 : 2 : 3+1

Simplifying you get 1:1:2


Bunuel
Three types of wheat of $1.27, $1.29 and $1.32 per kg are mixed together to be sold at $1.30 per kg. In what ratio should this wheat be mixed?

A. 1:2:3
B. 1:1:2
C. 2:1:3
D. 2:2:3
E. 2:2:5
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Usually substituting values and checking will take less time but in this particular case working it out is faster.
Solving with allegations for 2 cases will take max 20 seconds and then just need to arrange and simplify should definitely take less than a minute once the idea is there.
Yeetyeti38
this is how I did it, please let me know if this doesn't hold up

127x+129y+132z = 130(x+y+z)
2z=3x+1y <---- this is essentially a ratio, so the answer would have 2*(last number) be equal to 3*(first number) plus 1*(second number)

1:2:3 -> 3*1+1*2 =/= 2*3 doesn't work
1:1:2 -> 3*1+1*1=2*2 works, so it's B
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A: $1.27
B: $1.29
C: $1.32

You want to maintain average price of $1.30 per kg

For every kg of A, you drive the average price down by $0.03 (i.e. 1.30-1.27 = -0.03)
For every kg of B, you drive the average price down by $0.01 (i.e. 1.30-1.29 = -0.01)
For every kg of C, you drive the average price up by $0.02 (i.e. 1.32-1.30 = +0.02)

What ratio of the 3 quantities will result in zero movement of the average price?

Answer: (1kg*(-0.03) + 1kg*(-0.01) + 2kg*(+0.02)) = 0
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Correct answer is B. This is how I thought about it -

Average of 1.27 and 1.29 is 1.28. Average of this 1.28 and 1.32 is what we require i.e 1.30.

To get 1 kg of 1.28 I need to mix half of 1.27 and 1.29 and then I can mix that with 1kg of 1.32 . the resultant can be sold for 1.3 per kg.

That makes the ratio .5:.5:1

multiplying by 2 gives us 1:1:2
sudeshpatodiya


Average = Avg = 1.30 $/kg
Let three prices be, A=1.27, B=1.29, C=1.32

See how the different parts affect the average. Is it surplus or deficit.

Average Deficit is introduced by A and B
A-Avg = 127-130 = -3
B-Avg = 129-130 = -1

Surplus is introduced by C
C-Avg = 132-130 = 2

Now, for each unit of A and B, total Deficit = -4
For each unit of C, Surplus = 2

So, we need 2 units of C to compensate each unit of A and C combined to get the final average deviation as zero.

So, Ratio of A:B:C = 1:1:2

(B)
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that is a great way of doing it
shreyans0311
A: $1.27
B: $1.29
C: $1.32

You want to maintain average price of $1.30 per kg

For every kg of A, you drive the average price down by $0.03 (i.e. 1.30-1.27 = -0.03)
For every kg of B, you drive the average price down by $0.01 (i.e. 1.30-1.29 = -0.01)
For every kg of C, you drive the average price up by $0.02 (i.e. 1.32-1.30 = +0.02)

What ratio of the 3 quantities will result in zero movement of the average price?

Answer: (1kg*(-0.03) + 1kg*(-0.01) + 2kg*(+0.02)) = 0
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This is the fastest way to do it. Especially when ratios are given
Yeetyeti38
this is how I did it, please let me know if this doesn't hold up

127x+129y+132z = 130(x+y+z)
2z=3x+1y <---- this is essentially a ratio, so the answer would have 2*(last number) be equal to 3*(first number) plus 1*(second number)

1:2:3 -> 3*1+1*2 =/= 2*3 doesn't work
1:1:2 -> 3*1+1*1=2*2 works, so it's B
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"The reason to pick these two ratios is because u need one value always higher that the mean value for weighted averages to work"
Does this line mean that you took the value 1.32 twice because it is higher than 1.30?
Adit_
Take the ratio of A:C using the allegation method u get:
2:3

Similarly do it for B:C:
2:1

The reason to pick these two ratios is because u need one value always higher that the mean value for weighted averages to work

Now arrange it properly
B:C=2:1
A:C=2:3

We need to add the ratios of C since it is added twice

A:B:C = 2 : 2 : 3+1

Simplifying you get 1:1:2



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Let the quantities of the three types of wheat be x, y, and z, respectively.
3x + y = 2z
Used the option to match the equation

[*]A: 3(1)+2=5 ; 2(3)=6
[*]B: 3(1)+1=4; 2(2)=4
[*]=> Ans (LHS=RHS)
[*]C: 3(2)+1=7 ; 2(3)=6
[*]D: 3(2)+2=8; 2(3)=6
[*]E: 3(2)+2=8; 2(5)=10

So the answer is B
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