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A bag contains a mixture of beans and pulses. To achieve 20 percent be

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New post 25 Jun 2017, 04:07
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A bag contains a mixture of beans and pulses. To achieve 20 percent beans in the mixture, what percent of the mixture should be taken out and replaced with pulses?

(1) The mixture originally has 40 percent beans and 60 percent pulses.
(2) Total quantity of the mixture is 20 lb.

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Re: A bag contains a mixture of beans and pulses. To achieve 20 percent be  [#permalink]

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New post 27 Jun 2017, 01:23
We want to achieve mixture with 20% beans and 80% pulses. How much mixture to remove and then fill equal amount with pulses?
Statement 1: Bag contain mixture with this proportions 40% beans and 60% pulses
Bag mixture = 40% of B + 60% of P.
Take half out = 20% of B + 30% of P is taken out and replaced by Pulses.
So resultant mixture = 20% of B + 80% of P.
Take 50% of current mixture out i.e., with this you remove 20% of beans and 30% of pulses from the bag. If you replace this completely by pulses then adding replaced pulses to existing 30% of pulses it will become 80%.

Statement 1 Sufficient

Statement 2: Not required or sufficient as weight of the beans, pulse or mixture is never considered. The calculation should be in percentages, so statement 2 is not sufficient.
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Re: A bag contains a mixture of beans and pulses. To achieve 20 percent be  [#permalink]

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New post 27 Jun 2017, 01:53
To achieve 20 percent beans in the mixture, the mixture must be in the ratio 1:4(20% Beans : 80% Pulses)

(1) The mixture originally has 40 percent beans and 60 percent pulses.
According to the original ratio is 2:3(40% Beans : 60% Pulses)
if we replace 50% of the mixture with pure pulses,
we will get 20% of the Beans and 30% of the Pulses,
which when replaced by pure Pulses, makes the mixture 1:4(20% Beans : 80% Pulses) (Sufficient)

(2) Total quantity of the mixture is 20 lb.
Knowing the quantity of the mixture, is not going to be enough for us.
Only if we knew the initial concentration of the Beans:Pulses,
we can give a clear answer about the final concentration of the mixture. Insufficient (Option A)
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A bag contains a mixture of beans and pulses. To achieve 20 percent be  [#permalink]

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New post 22 Aug 2017, 01:48
pushpitkc wrote:
To achieve 20 percent beans in the mixture, the mixture must be in the ratio 1:4(20% Beans : 80% Pulses)

(1) The mixture originally has 40 percent beans and 60 percent pulses.
According to the original ratio is 2:3(40% Beans : 60% Pulses)
if we replace 50% of the mixture with pure pulses,
we will get 20% of the Beans and 30% of the Pulses,
which when replaced by pure Pulses, makes the mixture 1:4(20% Beans : 80% Pulses) (Sufficient)

(2) Total quantity of the mixture is 20 lb.
Knowing the quantity of the mixture, is not going to be enough for us.
Only if we knew the initial concentration of the Beans:Pulses,
we can give a clear answer about the final concentration of the mixture. Insufficient (Option A)


Hi pushpitkc,

Though I got the correct answer, but I am not sure if my approach is correct. Please tell me if there is any flaw in the below mentioned approach.

Initial : 40%(Beans) + 60%(Pulses)
Final : 20%(Beans) + 80%(Pulses)

Using allegation:
(w1/w2)=(100-80)/(80-60) w1-original weight of pulses in the mixture, w2-weight of pulses added to the mixture, concentration of pulses added=100%
(w1/w2)=1:1

Hence w2=1/2=50%

Hence 50% of the mixture should be removed.

Ans. C

Thanks
Dkingdom
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Re: A bag contains a mixture of beans and pulses. To achieve 20 percent be  [#permalink]

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New post 22 Aug 2017, 07:44
Dkingdom wrote:
pushpitkc wrote:
To achieve 20 percent beans in the mixture, the mixture must be in the ratio 1:4(20% Beans : 80% Pulses)

(1) The mixture originally has 40 percent beans and 60 percent pulses.
According to the original ratio is 2:3(40% Beans : 60% Pulses)
if we replace 50% of the mixture with pure pulses,
we will get 20% of the Beans and 30% of the Pulses,
which when replaced by pure Pulses, makes the mixture 1:4(20% Beans : 80% Pulses) (Sufficient)

(2) Total quantity of the mixture is 20 lb.
Knowing the quantity of the mixture, is not going to be enough for us.
Only if we knew the initial concentration of the Beans:Pulses,
we can give a clear answer about the final concentration of the mixture. Insufficient (Option A)


Hi pushpitkc,

Though I got the correct answer, but I am not sure if my approach is correct. Please tell me if there is any flaw in the below mentioned approach.

Initial : 40%(Beans) + 60%(Pulses)
Final : 20%(Beans) + 80%(Pulses)

Using allegation:
(w1/w2)=(100-80)/(80-60) w1-original weight of pulses in the mixture, w2-weight of pulses added to the mixture, concentration of pulses added=100%
(w1/w2)=1:1

Hence w2=1/2=50%

Hence 50% of the mixture should be removed.

Ans. C

Thanks
Dkingdom


Hi Dkingdom,

I don't seem to understand what you have done.
Also, we needn't solve for the quantity of pulses to be added
in order to achieve 20 percent beans in the mixture.

But for all DS questions, if either A or B is sufficient to answer the question
, we do not need to test for C. Option A is the correct answer.
Please ignore this if that was a typo!

Hope that helps
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Re: A bag contains a mixture of beans and pulses. To achieve 20 percent be  [#permalink]

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New post 22 Aug 2017, 08:16
pushpitkc wrote:
Dkingdom wrote:
pushpitkc wrote:
To achieve 20 percent beans in the mixture, the mixture must be in the ratio 1:4(20% Beans : 80% Pulses)

(1) The mixture originally has 40 percent beans and 60 percent pulses.
According to the original ratio is 2:3(40% Beans : 60% Pulses)
if we replace 50% of the mixture with pure pulses,
we will get 20% of the Beans and 30% of the Pulses,
which when replaced by pure Pulses, makes the mixture 1:4(20% Beans : 80% Pulses) (Sufficient)

(2) Total quantity of the mixture is 20 lb.
Knowing the quantity of the mixture, is not going to be enough for us.
Only if we knew the initial concentration of the Beans:Pulses,
we can give a clear answer about the final concentration of the mixture. Insufficient (Option A)


Hi pushpitkc,

Though I got the correct answer, but I am not sure if my approach is correct. Please tell me if there is any flaw in the below mentioned approach.

Initial : 40%(Beans) + 60%(Pulses)
Final : 20%(Beans) + 80%(Pulses)

Using allegation:
(w1/w2)=(100-80)/(80-60) w1-original weight of pulses in the mixture, w2-weight of pulses added to the mixture, concentration of pulses added=100%
(w1/w2)=1:1

Hence w2=1/2=50%

Hence 50% of the mixture should be removed.

Ans. C

Thanks
Dkingdom


Hi Dkingdom,

I don't seem to understand what you have done.
Also, we needn't solve for the quantity of pulses to be added
in order to achieve 20 percent beans in the mixture.

But for all DS questions, if either A or B is sufficient to answer the question
, we do not need to test for C. Option A is the correct answer.
Please ignore this if that was a typo!

Hope that helps


I am sorry! I was supposed to write A.
Using allegation method, I have calculated the additional amount of pulses needed in the mixture to reduce the concentration of Beans from 40% to 20%.
Does it make sense now? I hope I did not make a complete mess of it! :?
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Re: A bag contains a mixture of beans and pulses. To achieve 20 percent be   [#permalink] 15 Sep 2018, 02:47
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