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A person mixed three varieties of tea priced at $120 per pound, $135

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A person mixed three varieties of tea priced at $120 per pound, $135 [#permalink]

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A person mixed three varieties of tea priced at $120 per pound, $135 per pound and $160 per pound. In what ratio did he mix the three varieties of tea?

(1) The price of the mix was $135 per pound.
(2) Only 3 pounds of the variety priced at $135 per pound was used.
[Reveal] Spoiler: OA

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A person mixed three varieties of tea priced at $120 per pound, $135 [#permalink]

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(1) The price of the mix was $135 per pound.
Mixing 10 portions of the first variety(costing 120$), 2 portions of second variety(costing 135$) and 6 portions of third variety(costing 160$), will yield a mix priced at 135$
Ratio is 10 : 2 : 6(5:1:3)
Mixing 10 portions of the first variety(costing 120$), 4 portions of second variety(costing 135$) and 6 portions of third variety(costing 160$), will yield a mix priced at 135$
Ratio is 10 : 4 : 6(5:2:3)
Hence, insufficient.

(2) Only 3 pounds of the variety priced at $135 per pound was used.
How many pounds of variety priced at $135 was used, cannot give us the ratio in which the three varieties were mixed.(Insufficient)

On combining the information from both the statements,
we still cannot find out the ratio(unique in nature)
Mixing 10 portions of the first variety(costing 120$), 3 portions of second variety(costing 135$) and 6 portions of third variety(costing 160$), will yield a mix priced at 135$
Ratio is 10 : 3 : 6
Mixing 20 portions of the first variety(costing 120$), 3 portions of second variety(costing 135$) and 12 portions of third variety(costing 160$), will yield a mix priced at 135$
Ratio is 20 : 3 : 12 (Insufficient) (Option E)
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Re: A person mixed three varieties of tea priced at $120 per pound, $135 [#permalink]

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New post 03 Jul 2017, 12:35
1st Tea = $ 120 = a pounds mixed
2nd Tea = $ 135 = b pounds mixed
3rd Tea = $ 160 = c pounds mixed

(1) The price of the mix was $135 per pound.

= \(\frac{120a + 135b + 160c}{(a + b + c)}\) = $135 : Not sufficient

(2) Only 3 pounds of the variety priced at $135 per pound was used.

b = 3 ; = \(\frac{120a + 135 * 3 + 160c}{(a + b + c)}\) : Not sufficient

[1] + [2]

=> \(\frac{120a + 135 * 3 + 160c}{(a + b + c)}\) = $135 :
=>120 a + 135 * 3 + 160c = 135a + 135 * 3 + 135c
=> 25 c = 15 a
=> 5c = 3a

still we do not know the relatioship between a & b or b & c

hence E
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Re: A person mixed three varieties of tea priced at $120 per pound, $135 [#permalink]

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New post 03 Jul 2017, 12:51
Bunuel wrote:
A person mixed three varieties of tea priced at $120 per pound, $135 per pound and $160 per pound. In what ratio did he mix the three varieties of tea?

(1) The price of the mix was $135 per pound.
(2) Only 3 pounds of the variety priced at $135 per pound was used.


(1) New, but insufficient information
(2) New, but insufficient information

(1) and (2)
2 new information, left with 2 unrelated variables and a constant. Hence insufficient and E.
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Re: A person mixed three varieties of tea priced at $120 per pound, $135 [#permalink]

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New post 03 Jul 2017, 13:02
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Bunuel wrote:
A person mixed three varieties of tea priced at $120 per pound, $135 per pound and $160 per pound. In what ratio did he mix the three varieties of tea?

(1) The price of the mix was $135 per pound.
(2) Only 3 pounds of the variety priced at $135 per pound was used.


Tea A = $120 per round
Tea B = $135 per round
Tea C = $160 per round
Tea N = A + B + C
A:B:C = ?

1) Price of Tea N = $135
(120a+135b+160c)/(a+b+c)= 135
Insufficient. Multiple values possible.

2) B = 3 pounds
No relation to A or C. Insufficient.

(1+2)
(120a+405+160c) / (a+3+c) = 135
a:c = 5:3 but we do not know the relationship with B.
Insufficient.
Hence should be E.
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Re: A person mixed three varieties of tea priced at $120 per pound, $135 [#permalink]

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New post 21 Aug 2017, 09:22
Bunuel wrote:
A person mixed three varieties of tea priced at $120 per pound, $135 per pound and $160 per pound. In what ratio did he mix the three varieties of tea?

(1) The price of the mix was $135 per pound.
(2) Only 3 pounds of the variety priced at $135 per pound was used.


Tea A = x pounds
Tea B = y pounds
Tea C = z pounds

We need to find x:y:z

(1) ((120*x + 135*y + 160*z) / (x+y+z) ) = 135
Insufficient

(2) z=3
Insufficient

(1) + (2)
Put z=3 in the weighted average formula above.
Insufficient

Ans. E
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Re: A person mixed three varieties of tea priced at $120 per pound, $135 [#permalink]

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New post 04 Jan 2018, 13:11
Hi Bunuel,

I have one question, as per statement 1, The price of the mix was $135 per pound. should we assume the person will sell at weighted price of three mixtures. or can we also assume the person might markup the weighted price and sell?

Please clarify

Thanks
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Re: A person mixed three varieties of tea priced at $120 per pound, $135 [#permalink]

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New post 04 Jan 2018, 20:34
Expert's post
hellosanthosh2k2 wrote:
Hi Bunuel,

I have one question, as per statement 1, The price of the mix was $135 per pound. should we assume the person will sell at weighted price of three mixtures. or can we also assume the person might markup the weighted price and sell?

Please clarify

Thanks


From (1) is follows that the mix was selling at $135 per pound.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: A person mixed three varieties of tea priced at $120 per pound, $135   [#permalink] 04 Jan 2018, 20:34
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