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A shopkeeper mixes three varieties of rice costing $10, $12, $17 per

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A shopkeeper mixes three varieties of rice costing $10, $12, $17 per  [#permalink]

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New post 30 Sep 2018, 06:34
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A shopkeeper mixes three varieties of rice costing $10, $12, $17 per kg. Which of the following can represent a possible ratio in which the three varieties are mixed, if the trader makes a profit of 20% by selling the mixture at $15.60 per kg?

(A) 9:14:36
(B) 11:14:25
(C) 14:36:43
(D) 2:6:3
(E) None of these
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A shopkeeper mixes three varieties of rice costing $10, $12, $17 per  [#permalink]

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New post 30 Sep 2018, 07:55
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Sriharsh wrote:
A shopkeeper mixes three varieties of rice costing $10, $12, $17 per kg. Which of the following can represent a possible ratio in which the three varieties are mixed, if the trader makes a profit of 20% by selling the mixture at $15.60 per kg?

(A) 9:14:36
(B) 11:14:25
(C) 14:36:43
(D) 2:6:3
(E) None of these



Now the profit is 20%, so let us find the average price the price at which there is no profit and no loss...
15.60=120x/100......x=15.6×100/120=13

So prices are 10:12:17 and average 13..
Let us how far the numbers are from average price..
(13-10):(13-12):(17-13).....3:1:6...
So the ratio *the difference of the price on either side of average price should be equal..
So let us check the choices..
(A) 9:14:36..........so \(9*(13-10)+14*(13-12)=36*(17-13)...9*3+14*1=36*4\).....no
(B) 11:14:25......so \(11*(13-10)+14*(13-12)=25*(17-13)...11*3+14*1=25*4\).....no
(C) 14:36:43......so \(14*(13-10)+36*(13-12)=43*(17-13)...14*3+36*1=43*4\).....no
(D) 2:6:3......so \(2*(13-10)+6*(13-12)=3*(17-13)...2*3+6*1=3*4.....12=12\)...yes
(E) None of these

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Re: A shopkeeper mixes three varieties of rice costing $10, $12, $17 per  [#permalink]

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New post 30 Sep 2018, 07:56
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Sriharsh wrote:
A shopkeeper mixes three varieties of rice costing $10, $12, $17 per kg. Which of the following can represent a possible ratio in which the three varieties are mixed, if the trader makes a profit of 20% by selling the mixture at $15.60 per kg?

(A) 9:14:36
(B) 11:14:25
(C) 14:36:43
(D) 2:6:3
(E) None of these


(10x + 12y + 17z)/ x+y+z = CP ........ (A)

profit % = (SP-CP)/CP

From above formula CP = 13
Putting the value of 13 in equation A , we get
4z =3x+y

Only option D satisfy this equation, Hence correct answer is Option D
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Re: A shopkeeper mixes three varieties of rice costing $10, $12, $17 per  [#permalink]

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New post 30 Sep 2018, 18:52
1
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We have 3 varieties of rice each costing $10, $12, and $17.
The shopkeeper mixes these three varieties and sells the mixture at $15.6 which gets him a profit of 20%.

Now, CP +CP(20/100) = 15.6
Implies, CP = $13.0

So we have x grams of $10 rice, + y grams of $12 rice + z grams of $17 rice, all mixed together to form 1 kg of rice with a cost price of $13.0
Implies , 10x+12y+17z = 13(x+y+z)
Implies, 4z = 3x + y

The above equation shows us that z has to me minimum of all three.
By seeing the options only D has z being minimum.
So lets check for D.
4z = 3x + y
Implies, 4(3) = 3(2)+6
Implies , 12 = 12

So option D is the correct answer.
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Re: A shopkeeper mixes three varieties of rice costing $10, $12, $17 per  [#permalink]

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New post 05 Oct 2018, 03:42
chetan2u wrote:
Sriharsh wrote:
A shopkeeper mixes three varieties of rice costing $10, $12, $17 per kg. Which of the following can represent a possible ratio in which the three varieties are mixed, if the trader makes a profit of 20% by selling the mixture at $15.60 per kg?

(A) 9:14:36
(B) 11:14:25
(C) 14:36:43
(D) 2:6:3
(E) None of these



Now the profit is 20%, so let us find the average price the price at which there is no profit and no loss...
15.60=120x/100......x=15.6×100/120=13

So prices are 10:12:17 and average 13..
Let us how far the numbers are from average price..
(13-10):(13-12):(17-13).....3:1:6...
So the ratio *the difference of the price on either side of average price should be equal..
So let us check the choices..
(A) 9:14:36..........so \(9*(13-10)+14*(13-12)=36*(17-13)...9*3+14*1=36*4\).....no
(B) 11:14:25......so \(11*(13-10)+14*(13-12)=25*(17-13)...11*3+14*1=25*4\).....no
(C) 14:36:43......so \(14*(13-10)+36*(13-12)=43*(17-13)...14*3+36*1=43*4\).....no
(D) 2:6:3......so \(2*(13-10)+6*(13-12)=3*(17-13)...2*3+6*1=3*4.....12=12\)...yes
(E) None of these

D


Hi chetan2u!

I think you made a small mistake with the marked ratio! Shouldn't it be 3:1:4?

Cheers
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A shopkeeper mixes three varieties of rice costing $10, $12, $17 per  [#permalink]

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New post 28 Oct 2018, 23:41
Sriharsh wrote:
A shopkeeper mixes three varieties of rice costing $10, $12, $17 per kg. Which of the following can represent a possible ratio in which the three varieties are mixed, if the trader makes a profit of 20% by selling the mixture at $15.60 per kg?

(A) 9:14:36
(B) 11:14:25
(C) 14:36:43
(D) 2:6:3
(E) None of these


OA:D

Let the weight of each variety in the mixture be \(x,y,\) and \(z\) kg
Weight of the total mixture \(=x+y+z\)
Cost price + Profit = Selling price
Cost price \(= 10x+12y+17z\)
Profit \(=\frac{20}{100}\) Cost price \(= 0.2\) Cost price
Selling price \(= 15.6(x+y+z)\)

\(10x+12y+17z +0.2(10x+12y+17z)= 15.6(x+y+z)\)
\(1.2(10x+12y+17z)=15.6(x+y+z)\)
\(10x+12y+17z = 13(x+y+z)\)
\(3x+y-4z=0\)

Only Option D satisfies \(3x+y-4z=0\)
As per option D, \(x=2m,y=6m\) and \(z=3m\)
L.H.S : \(3x+y-4z = 3*2m+6m-4*3m = 6m+6m-12m=0\)
R.H.S : \(0\)
\(L.H.S=R.H.S\)
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A shopkeeper mixes three varieties of rice costing $10, $12, $17 per   [#permalink] 28 Oct 2018, 23:41
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