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A 20 kilograms mixture of dry fruits

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Joined: 03 Mar 2018
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A 20 kilograms mixture of dry fruits  [#permalink]

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New post 05 Mar 2018, 05:34
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Difficulty:

  75% (hard)

Question Stats:

56% (03:01) correct 44% (04:04) wrong based on 18 sessions

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A 20 kilograms mixture of dry fruits consisting of x percent of almonds is formed by mixing 'a' kilograms of mixture A and 'b' kilograms of mixture B. If mixture A consists of 'y' percent of almonds and mixture B consists of 20% of almonds, what is the weight of almonds in kilograms in mixture A? Assume that each mixture has a uniform composition, that is, samples taken from any part in a mixture will have the same composition as the overall mixture.

(1) The amount of almonds in mixture B is three-fifths of the total weight of mixture A

(2) If the weight of the final mixture is reduced by 25 percent, the almonds in the final mixture would reduce by 1.5 kilograms

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Re: A 20 kilograms mixture of dry fruits  [#permalink]

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New post 05 Mar 2018, 06:31
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itisSheldon wrote:
A 20 kilograms mixture of dry fruits consisting of x percent of almonds is formed by mixing 'a' kilograms of mixture A and 'b' kilograms of mixture B. If mixture A consists of 'y' percent of almonds and mixture B consists of 20% of almonds, what is the weight of almonds in kilograms in mixture A? Assume that each mixture has a uniform composition, that is, samples taken from any part in a mixture will have the same composition as the overall mixture.

(1) The amount of almonds in mixture B is three-fifths of the total weight of mixture A

(2) If the weight of the final mixture is reduced by 25 percent, the almonds in the final mixture would reduce by 1.5 kilograms




so WHAT do we have here..

Final mixture of 20 kgs having x% of almonds
1) a kg of mix A consisting of y% of almonds
2) b kg of B mix containing of 20% of almonds...


weight of almonds in mix A - REQUIRES the value of 'a' and 'y'..

lets see the statements:-

(1) The amount of almonds in mixture B is three-fifths of the total weight of mixture A
this gives me the equation \(0.2b=\frac{3}{5}a........b=3a\)
so I know now that b is 15 and a is 5..
but I still do not know the value of y as there are two variables in % - x and y
insuff

(2) If the weight of the final mixture is reduced by 25 percent, the almonds in the final mixture would reduce by 1.5 kilograms
so 25% reduction means reduction of 5 kgs and in 5 kgs I have 1.5 kg almonds, so % of almonds in final mix or \(x = \frac{1.5}{5}*100\)
here I have x But y will depend on the ratio of weights of a and b..
insuff

combined..
......'a'.........'T'.......'b'
.....y.........30......20
....a=5.....20.......3a=15

so y = (30-20)*3+30=60%
so wight of almond in A is 60% of 5 = 3kgs
suff

c

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Re: A 20 kilograms mixture of dry fruits &nbs [#permalink] 05 Mar 2018, 06:31
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