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# A diet contains 300 grams of a mixture of two foods, food X

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Joined: 07 May 2003
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A diet contains 300 grams of a mixture of two foods, food X  [#permalink]

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Updated on: 30 Jul 2012, 08:11
4
21
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Difficulty:

35% (medium)

Question Stats:

73% (02:17) correct 27% (02:29) wrong based on 741 sessions

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A diet contains 300 grams of a mixture of two foods, food X and food Y. Food X contains 10 percent protein and food Y contains 15 percent protein. If a diet provides exactly 38 grams of protein daily, how many grams of food X are in the mixture?

(A) 150
(B) 145
(C) 100
(D) 160
(E) 140

Originally posted by ju_1 on 20 May 2003, 11:49.
Last edited by Bunuel on 30 Jul 2012, 08:11, edited 1 time in total.
Edited the question and added the OA.
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Re: A diet contains 300 grams of a mixture of two foods, food X  [#permalink]

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30 Jul 2012, 08:13
8
3
harshavmrg wrote:
ju_1 wrote:
A diet contains 300 grams of a mixture of two foods, food X and food Y. Food X contains 10 percent protein and food Y contains 15 percent protein. If a diet provides exactly 38 grams of protein daily, how many grams of food X are in the mixture?
(A) 150
(B)145
(C) 100
(D) 160
(E) 140

x(10%) y(15%)
38%
23% 28%

the protein parts in the final mixture for X:Y::23:28

hence

23 X
----------- = --------------
(23+28) 300

23/51 = x/300

x = 23 * 300
------------
51

hence x is approximately equal to 139.8

Keep it simple: $$0.1x+0.15(300-x)=38$$ --> $$x=140$$ (exactly).

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20 May 2003, 21:08
2
1.... 10/100*X + 15/100*Y = 38 (protein)

2.... X + Y = 300 (Total Mix)

Solve for X = 140
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Re: A diet contains 300 grams of a mixture of two foods, food X  [#permalink]

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30 Jul 2012, 07:52
1
ju_1 wrote:
A diet contains 300 grams of a mixture of two foods, food X and food Y. Food X contains 10 percent protein and food Y contains 15 percent protein. If a diet provides exactly 38 grams of protein daily, how many grams of food X are in the mixture?
(A) 150
(B)145
(C) 100
(D) 160
(E) 140

x(10%) y(15%)
38%
23% 28%

the protein parts in the final mixture for X:Y::23:28

hence

23 X
----------- = --------------
(23+28) 300

23/51 = x/300

x = 23 * 300
------------
51

hence x is approximately equal to 139.8

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Re: A diet contains 300 grams of a mixture of two foods, food X  [#permalink]

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13 Aug 2012, 00:41
2
X contains 30 gms of protein (10% of 300 gms)
Y contains 45 gms of protein (15% of 300 gms)
Resultant mixture contains 38 gms of protien.

using scale method :

____8___________7___
30 38 45
w1 w2

therefore,
w1/w2 = 7/8

7x+8x=300 gms
=> x=20

so, w1 = 7x=7(20) = 140 gms
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Re: A diet contains 300 grams of a mixture of two foods, food X  [#permalink]

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18 Apr 2014, 22:12
2
X+Y = 300

0.1X + 0.15Y = 38

X=140
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Re: A diet contains 300 grams of a mixture of two foods, food X  [#permalink]

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07 Aug 2016, 06:02
if x = 100, then y=200
protien from x = 10 and from y = 30 . total protein = 40. greater.. hmm. it means y is somewhat more than it should be. this implies x > 100.
x= 150 , then y=150
protein from x = 15 and from y = 22.5 total = 37.5..lesser
so answer is between 100 and 150 . if we see there are two choices 140 and 145. which one it should be?
well after increasing x by 50 we notice that we reduced total protein from 40 to 37.5 means 2.5. We just need to reduce the protein by 2.

Let us try 140 with this, protein from x = 14 , protein from y = 16+8 = 24. total = 38. this works!
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Re: A diet contains 300 grams of a mixture of two foods, food X  [#permalink]

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07 Aug 2016, 10:53
.10x+.15(300-x)=38
x=140 grams
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A diet contains 300 grams of a mixture of two foods, food X  [#permalink]

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07 Aug 2016, 13:25
1
Take equal proportion to have an idea of ratio,150 each. X will give 15g, Y will give 22.5g which is closer to 38 (37.5)
A little thought on the options D& E will give us E as the answer

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Re: A diet contains 300 grams of a mixture of two foods, food X  [#permalink]

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07 Aug 2016, 21:52
I am a little confused about the wording here.

The question says "a diet provides exactly 38 grams of protein daily..." is it referring to the same diet or some other diet which could be more or less than 300 grams?

I feel if it is referring to the same diet then it should mention "THE diet provides...", don't you think?

ju_1 wrote:
A diet contains 300 grams of a mixture of two foods, food X and food Y. Food X contains 10 percent protein and food Y contains 15 percent protein. If a diet provides exactly 38 grams of protein daily, how many grams of food X are in the mixture?

(A) 150
(B) 145
(C) 100
(D) 160
(E) 140
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Re: A diet contains 300 grams of a mixture of two foods, food X  [#permalink]

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25 Sep 2016, 06:46
10/100 (x) + 15/100 (300-x) = 38

10/100 (x) + 45 - 15/100 (x) = 38

-5/100 (x) = -7 --> x= 140
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Re: A diet contains 300 grams of a mixture of two foods, food X  [#permalink]

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22 Nov 2017, 05:43
Bunuel wrote:
harshavmrg wrote:
ju_1 wrote:
A diet contains 300 grams of a mixture of two foods, food X and food Y. Food X contains 10 percent protein and food Y contains 15 percent protein. If a diet provides exactly 38 grams of protein daily, how many grams of food X are in the mixture?
(A) 150
(B)145
(C) 100
(D) 160
(E) 140

x(10%) y(15%)
38%
23% 28%

the protein parts in the final mixture for X:Y::23:28

hence

23 X
----------- = --------------
(23+28) 300

23/51 = x/300

x = 23 * 300
------------
51

hence x is approximately equal to 139.8

Keep it simple: $$0.1x+0.15(300-x)=38$$ --> $$x=140$$ (exactly).

hi Bunuel
really very simple, man
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Re: A diet contains 300 grams of a mixture of two foods, food X  [#permalink]

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22 Nov 2017, 08:24
Let x, y be the weight of food X, food Y in the mixture, respectively.
We have: x+y = 300 and 0,1x + 015y = 38 => x = 140.
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Re: A diet contains 300 grams of a mixture of two foods, food X  [#permalink]

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22 Nov 2017, 13:19
1
ju_1 wrote:
A diet contains 300 grams of a mixture of two foods, food X and food Y. Food X contains 10 percent protein and food Y contains 15 percent protein. If a diet provides exactly 38 grams of protein daily, how many grams of food X are in the mixture?

(A) 150
(B) 145
(C) 100
(D) 160
(E) 140

We can let x = the amount of X and y = the amount of Y. Let’s create two equations:

x + y = 300

y = 300 - x

And

0.1x + 0.15y = 38

0.1x + 0.15(300 - x) = 38

0.1x + 45 - 0.15x = 38

7 = 0.05x

140 = x

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Re: A diet contains 300 grams of a mixture of two foods, food X  [#permalink]

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28 Mar 2018, 12:16
Alligation:

X Y
10% of 300 =30(value in gms) 15% of 300=45 (value in gms)

Aavg= 38 gms

X or A1 or W1= 7 : Y or A2 or W2= 8

X/Y= 7/8
Hence X = 7/15 of total

To get the amount in gms 7/15 of 300= 140. Hence E is the answer
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Re: A diet contains 300 grams of a mixture of two foods, food X  [#permalink]

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24 Apr 2019, 12:58
bhavinshah5685 wrote:
X contains 30 gms of protein (10% of 300 gms)
Y contains 45 gms of protein (15% of 300 gms)
Resultant mixture contains 38 gms of protien.

using scale method :

____8___________7___
30 38 45
w1 w2

therefore,
w1/w2 = 7/8

7x+8x=300 gms
=> x=20

so, w1 = 7x=7(20) = 140 gms

is the method correct? because X contains 10% protein , 10 % of what ? isn't it 10% of X's contribution? but it is taken 10% of total volume 300 grams
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Re: A diet contains 300 grams of a mixture of two foods, food X  [#permalink]

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25 Apr 2019, 06:04
1
ju_1 wrote:
A diet contains 300 grams of a mixture of two foods, food X and food Y. Food X contains 10 percent protein and food Y contains 15 percent protein. If a diet provides exactly 38 grams of protein daily, how many grams of food X are in the mixture?

(A) 150
(B) 145
(C) 100
(D) 160
(E) 140

Food X has 10% protein
Food Y has 15% protein

The mix has (38/300)*100 = (38/3)% protein (so this is the average protein)

wX/wY = (15 - 38/3) / (38/3 - 10) = 7/8

So food X and food Y are mixed in the ratio 7:8 in the 300 gms of mix.
Food X must be 140 gms and food Y must be 160 gms.

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A diet contains 300 grams of a mixture of two foods, food X  [#permalink]

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25 Apr 2019, 07:15
1
ju_1 wrote:
A diet contains 300 grams of a mixture of two foods, food X and food Y. Food X contains 10 percent protein and food Y contains 15 percent protein. If a diet provides exactly 38 grams of protein daily, how many grams of food X are in the mixture?

(A) 150
(B) 145
(C) 100
(D) 160
(E) 140

For mixture problems we need to check the average in the resultant mixture, in this case we need to check average protein in the resultant mixture and that is 38/3 (38/300*100%).
Now we need to compare it with x: |38/3-10|=8/3 and with Y: |15-38/3|=7/3
x:Y= 7/3:8/3 or 7:8
Proportion of x in the final mixture is therefore =7/15*300 or 140 (Option E)

A diet contains 300 grams of a mixture of two foods, food X   [#permalink] 25 Apr 2019, 07:15
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