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30 Sep 2018, 06:34
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Difficulty:

65% (hard)

Question Stats:

57% (02:19) correct 43% (02:23) wrong based on 118 sessions

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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
Math Expert
Joined: 02 Aug 2009
Posts: 7763

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30 Sep 2018, 07:56
1
1
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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Joined: 21 Nov 2016
Posts: 39
Re: A shopkeeper mixes three varieties of rice costing $10,$12, $17 per [#permalink] ### Show Tags 30 Sep 2018, 18:52 1 2 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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Joined: 07 Aug 2018
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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 : $$0$$
$$L.H.S=R.H.S$$