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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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New post 03 Jul 2017, 12:05
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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.

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

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New post 03 Jul 2017, 12:32
2
1
(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$ - The 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$), will yield a mix priced at 135$ - The ratio is 10:4:6(5:2:3) - 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 does not give the ratio in which the three varieties were mixed Insufficient

On combining the information from both the statements, we still cannot find out a unique ratio
Mixing 10 portions of the first variety(costing 120$), 3 portions of second variety(costing 135$),
and 6 portions of third variety(costing 160$), will yield a mix priced at 135$ - The 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$), will yield a mix priced at 135$ - The 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
3
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
1
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
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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Re: A person mixed three varieties of tea priced at $120 per pound, $135  [#permalink]

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New post 05 Jun 2018, 17:54
Got it wrong.. but only cz it was from bunuel and i thought 1+2 would be sme kind of uniq although the equation (straight forward as it was) said otherwise :mad: :cry:
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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 28 Jun 2018, 04:05
1
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.



Responding to a pm:

3 varieties - $120 (tea1), $135(tea2), $160(tea3)

Stmnt 1: The price of the mix was $135 per pound.
Note that one of the tea was priced at $135 per pound. So the other two teas would be mixed in a ratio to give an average of 135 too.

w1/w2 = (160 - 135) / (135 - 120) = 5/3

So you need to mix $120 and $160 teas in the ratio 5:3. The tea costing $135 can be added in any quantity, the average will stay the same. We don't know the ratio of the quantity of the three teas so not sufficient.

Stmnt 2: Only 3 pounds of the variety priced at $135 per pound was used.
No ratio of the three. Not sufficient.

Using both together, note that we know that tea1 and tea3 were mixed in the ratio 5:3 but we do not know how much exactly of each was used. Did we use 5 pounds and 3 pounds? Or 10 pounds and 6 pounds? Or 15 pounds and 9 pounds? We don't know. We know that 3 pounds of tea2 was used. Depending on how much of tea1 and tea3 were used, the ratio would be different.
If we used 5 pounds and 3 pounds of tea1 and tea3, ratio tea1:tea2:tea3 would be 5:3:3
If we used 10 pounds and 6 pounds of tea1 and tea3, ratio tea1:tea2:tea3 would be 10:3:6
If we used 15 pounds and 9 pounds of tea1 and tea3, ratio tea1:tea2:tea3 would be 15:3:9 = 5:1:3
and so on...
Not sufficient
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Re: A person mixed three varieties of tea priced at $120 per pound, $135 &nbs [#permalink] 28 Jun 2018, 04:05
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