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# Seed mixture X is 40 percent ryegrass and 60 + More Problems

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Seed mixture X is 40 percent ryegrass and 60 + More Problems [#permalink]  29 Nov 2008, 08:32
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Hello GMAT Club'rs

I collected some questions in the Mixture's category of Quant from the GMAT Club posts.
Thought I would put it all together for someone who needs a little more practise in this area.
You can search for all these questions in the Forum.
I have added links for some of them, but not all of them.

Happy Solving...
LS

1.Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weight; seed mixture Y is 25 percent ryegrass and 75 percent fescue. If a mixture of X and Y contains 30 percent ryegrass, what percent of the weight of this mixture is X ?
(A) 10%
(B) 33.33 %
(C) 40%
(D) 50%
(E) 66.66 %

2.A certain shade of gray paint is obtained by mixing 3 parts of white paint with 5 parts of black paint. If 2 gallons of the mixture is needed and the individual colors can be purchased only in one-gallon or half- gallon cans, what is the least amount of paint, in gallons, that must be purchased in order to measure out the portions needed for the mixture?
(A) 2
(B) 2.5
(C) 3
(D) 3.5
(E) 4

3.How much water (in grams) should be added to the 35%-solution of acid to obtain the 10%-solution?
1. There are 50 grams of the 35%-solution
2. In the 35%-solution the ratio of acid to water is 7:13
what is the best way to approach mixture problems ?

4.Are at least 10% of the people in Country X who are 65 years old or older employed?

(1) In Country X, 11% of the population is 65 years old or older.
(2) In Country X, of the population 65 years old or older, 20% of the men and 10% of the women are employed.

5.X grams of water were added to the 80 grams of a strong solution of acid. As a result, the concentration of acid in the solution dropped by Y percent. What was the concentration of acid in the original solution?

1. X = 80
2. Y = 50

6.if 200 lb of a mixture contain 80% husk and 20%sand. Then how much husk needs to be extracted in order to have 75% concentration of Husk?

1/4
20/3
1/2
40
60

7.rabbit on a controlled diet is fed daily 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 the rabbit's diet provides exactly 38 grams of protein daily, how many grams of food X are in the mixture?
A) 100
B) 140
C) 150
D) 160
E) 200
7-t49451?hilit=+mixture

8.Solution Y is 30 percent liquid X and 70 percent water. If 2 kilograms of water evaporate from 8 kilograms of solution Y and 2 kilograms of solution Y are added to the remaining 6 kilograms of liquid, what percent of this new solution is liquid X?
(A) 30%
(B) 33 1/3%
(C) 37 1/2%
(D) 40%
(E) 50%

9.Each piglet in a liiter is fed exactly one-half pound of a mixture of oats and barley. The ratio of the amount of barley to that of oats varies from piglet to piglet, but each piglet is fed some of both grains. how many piglets are there in the litter?

1) Piglet A was fed exactly 1/4 of the oats today
2) Piglet A was fed exactly 1/6 of the barley today

10.Miguel is mixing up a salad dressing. Regardless of the number of servings, the recipe requires that 5/8 of the finished dressing mix be olive oil, 1/4 vinegar, and the remainder an even mixture of salt, pepper and sugar. If Miguel accidentally doubles the vinegar and forgets the sugar altogether, what proportion of the botched dressing will be olive oil?
15/29
5/8
5/16
1/2
13/27

11.A certain type of concrete mixture is to be made of cement, sand and graved in a ratio 1:3:5 by weight. What is the greatest number of kilograms of this mixture that can be made with 5 kilograms of cement ?

1. 13 1/2
2. 15
3. 25
4. 40
5. 45

12.Some of 50%-intensity red paint is replaced with 25% solution of red paint such that the new paint intensity is 30%. What fraction of the original paint was replaced?
1/30
1/5
2/3
3/4
4/5

13.Material A costs $3 per kilogram, and material B costs$5 per kilogram. If
10 kilograms of material K consists of x kilograms of material A and y
kilograms of material B, is x > y?
(1) y > 4
(2) The cost of the 10 kilograms of material K is less than $40. 14.How many liters of pure alcohol must be added to a 100-liter solution that is 20 percent alcohol in order to produce a solution that is 25 percent alcohol? (A) 7/2 (B) 5 (C) 20/3 (D) 8 (E) 39/4 15.Two different solutions of alcohol with respective proportions of alcohol to water of 3:1 and 2:3 were combined. What is the concentration of alcohol in the new solution if the first solution was 4 times the amount of the second solution? 16.10kg of a mixture contains 30% sand and 70% clay. In order to make the mixture contain equal quantities of clay and sand. how much of the mixture is to be removed and replaced with pure sand? 17.Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk? 7-t57862?hilit=+mixture 18.One fourth of a solution that was 10 percent sugar by weight was replaced by a second solution resulting in a solution that was 16 percent sugar by weight. The second solution was what percent sugar by weight ? (A) 34% (B) 24% (C) 22% (D) 18% (E) 8.5% 7-t10946 19.Three grades of milk are 1 percent, 2 percent and 3 percent fat by volume. If x gallon of 1 percent grade, y gallon of 2 percent grade, and z gallon of 3 percent grade are mixed to give x + y + z gallons of 1.5 percent grade, what is x in terms of y and z ? a. y + 3z b. (y+z )/ 4 c. 2y + 3z d. 3y +z e. 3y + 4.5z 7-t40699 20.24 lbs of coffee P and 25 lbs of coffee V are mixed to make coffee X and Y. The ratio of P to V in coffee X is 4 to 1, in Y is 1 to 5. How much of P is contained in the mixture X? 7-t57374?hilit=+mixture 21.A shade of paint is made by evenly mixing m gallons of white paint, costing$12 a gallon, with n gallons of blue paint, costing \$30. What is the cost, in dollars per gallon, of the resulting mixture?
12m +30n
42(12m +30n)
(12m +30n) / 42
(12m +30n) / (M+ N)
42(M+ N) / (12m +30n)
7-t56091?hilit=+mixture

22.Some part of the 50% solution of acid was replaced with the equal amount of 30% solution of acid. As a result, 40% solution f acid was obtained. what part of the original solution was replaced?
1/5
1/4
1/2
3/4
4/5
7-t38736

23.A portion of the 85% solution of chemicals was replaced with an equal amount of 20% solution of chemicals. As a result, 40% solution of chemicals resulted. What part of the original solution was replaced?
7-t55090

24.A contractor combined x tons of gravel mixture that cotained 10 percent gravel G, by weight, with y tons of a mixture that contained 2 percent gravel G, by weight, to produce z tons of a misture that was 5 percent gravel G, by weight. What is the value of X?
1) y=10
2) z=16
7-t54086

25.In a certain senior class, 72 percent of the male students and 80 percent of the female students have applied to college. What fraction of the students in the senior class are male?

1) There are 840 students in the senior class
2) 75 percent of the students in the senior class have applied to college

26.According to the directions on a can of frozen orange juice concentrate, 1 can of concentrate is to be mixed with 3 cans of water to make orange juice. How many 12-ounce cans of the concentrate are required to prepare 200 6-ounce servings of orange juice?

a)25
b)34
c)50
d)67
e)100

27.a 30% alchol mixture is added to a 50% alchol mixture to form a 10 litre mixture of 45% alchol. How much of the 30% mixture was used?
7-t50630

28.There are two bars of gold-silver alloy; one piece has 2 part of gold to 3 parts of silver, and the other has 3 parts of gold to 7 parts of silver. If both bars are melted into an 8 kg bar with final ration of 5:11 of gold to silver, what was the weight of the first bar?
1) 1 kg
2) 3 kg
3) 5 kg
4) 6 kg
5) 7 kg
7-t48285

29.Bob just filled his car's gas tank with 20 gallons of gasohol, a mixture consisting of 5% ethanol and 95% gasoline. If his car runs best on a mixture consisting of 10% ethanol and 90% gasoline, how many gallons of ethanol must he add into the gas tank for his car to achieve optimum performance?
a) 9/10
b) 1
c) 10/9
d) 20/19
e) 2
7-p323094

30.a mixture of nuts is to contain 3 parts cashews to 6 parts almonds to 7 parts walnuts by weight. How many pounds of almonds will be needed to make 5 pounds of the mixture?
a.) 3/8
b.) 8/15
c.) 6/5
d.) 5/3
e.) 15/8
7-t44832?hilit=+mixture
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Re: Mixture Problems - I collected from GMAT Club [#permalink]  23 Mar 2009, 19:22
This is awesome. Thanks for putting it together. +1
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Re: Mixture Problems - I collected from GMAT Club [#permalink]  23 Mar 2009, 20:55
Thank you soooooooooooooo mucch

does any 1 have work and distance problems too
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Re: Mixture Problems - I collected from GMAT Club [#permalink]  29 Mar 2009, 11:52
CAN SOMEONE PLEASE POST ANSWERS/SOLUTIONS WITH EXPLANATIONS (MULTIPLE APPROACH WOULD BE GREAT) TO "ALL" OF THESE PROBLEMS?

TERRIFIC PROBLEMS - ANSWERS WOULD BE GREAT!
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Re: Mixture Problems - I collected from GMAT Club [#permalink]  12 Aug 2009, 23:58
Wanna confirm my answer for 2..(B)?
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Re: Mixture Problems - I collected from GMAT Club [#permalink]  13 Aug 2009, 00:20
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Re: Mixture Problems - I collected from GMAT Club [#permalink]  13 Aug 2009, 00:52
tejal777 wrote:
Wanna confirm my answer for 2..(B)?

Yep, the answer for Q2 is B i.e. 2.5
=> 3*0.25 = 0.75 gallons of white paint and 5*0.25 = 1.25 gallons of black paint are needed. But since only 0.5 and 1 gallon cans are available, 1 gallon of white paint and 1.5 gallon of black paint would be needed => 1 + 1.5 = 2.5
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Re: Mixture Problems - I collected from GMAT Club [#permalink]  13 Aug 2009, 00:57
tejal777 wrote:

3.How much water (in grams) should be added to the 35%-solution of acid to obtain the 10%-solution?
1. There are 50 grams of the 35%-solution
2. In the 35%-solution the ratio of acid to water is 7:13

St1 is sufficient => 125 grams of water needs to be added.
St2 states the question data itself but in a different way.
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Re: Mixture Problems - I collected from GMAT Club [#permalink]  13 Aug 2009, 02:40
Quote:
8.Solution Y is 30 percent liquid X and 70 percent water. If 2 kilograms of water evaporate from 8 kilograms of solution Y and 2 kilograms of solution Y are added to the remaining 6 kilograms of liquid, what percent of this new solution is liquid X?
(A) 30%
(B) 33 1/3%
(C) 37 1/2%
(D) 40%
(E) 50%

I liked this question!!Ans (C)
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Re: Mixture Problems - I collected from GMAT Club [#permalink]  13 Aug 2009, 05:02
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tejal777 wrote:
Quote:
8.Solution Y is 30 percent liquid X and 70 percent water. If 2 kilograms of water evaporate from 8 kilograms of solution Y and 2 kilograms of solution Y are added to the remaining 6 kilograms of liquid, what percent of this new solution is liquid X?
(A) 30%
(B) 33 1/3%
(C) 37 1/2%
(D) 40%
(E) 50%

I liked this question!!Ans (C)

I agree...its C

8 Kg of Y contains:
8*0.3 = 2.4 kg X and
8*0.7 = 5.6 kg water

Y after 2 kg water evaporates:
2.4 kg X and
5.6 - 2 = 3.6 kg water

Now 2 kg of Y contains 2*0.3 = 0.6 kg X and 2*0.7 = 1.4 kg water. Thus on adding this solution to the original solution we get:
2.4 + 0.6 = 3 kg X and
3.6 + 1.4 = 5 kg water

Thus the % of X in this new solution is: (3/(5+3))*100 = 37.5%

ANS: C
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Re: Mixture Problems - I collected from GMAT Club [#permalink]  15 Aug 2009, 19:22
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Last edited by tejal777 on 17 Aug 2009, 23:00, edited 1 time in total.
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Re: Mixture Problems - I collected from GMAT Club [#permalink]  15 Aug 2009, 22:12
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Re: Mixture Problems - I collected from GMAT Club [#permalink]  20 Aug 2009, 01:28
samrus98 wrote:
tejal777 wrote:
Quote:
8.Solution Y is 30 percent liquid X and 70 percent water. If 2 kilograms of water evaporate from 8 kilograms of solution Y and 2 kilograms of solution Y are added to the remaining 6 kilograms of liquid, what percent of this new solution is liquid X?
(A) 30%
(B) 33 1/3%
(C) 37 1/2%
(D) 40%
(E) 50%

I liked this question!!Ans (C)

I agree...its C

8 Kg of Y contains:
8*0.3 = 2.4 kg X and
8*0.7 = 5.6 kg water

Y after 2 kg water evaporates:
2.4 kg X and
5.6 - 2 = 3.6 kg water

Now 2 kg of Y contains 2*0.3 = 0.6 kg X and 2*0.7 = 1.4 kg water. Thus on adding this solution to the original solution we get:
2.4 + 0.6 = 3 kg X and
3.6 + 1.4 = 5 kg water

Thus the % of X in this new solution is: (3/(5+3))*100 = 37.5%

ANS: C

I agree C, i had also done in the same way. but is there any other way?
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Re: Mixture Problems - I collected from GMAT Club [#permalink]  20 Aug 2009, 01:41
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Thanks to Livestronger and samrus98 ....Kudos to both...
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Re: Mixture Problems - I collected from GMAT Club [#permalink]  21 Aug 2009, 05:01
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So what is the OA for Q5? C or E?
I got C.
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Re: Mixture Problems - I collected from GMAT Club [#permalink]  21 Aug 2009, 22:47
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Thanks for such a great post... I am feeling very confident when it comes to mixture problems now!
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Re: Mixture Problems - I collected from GMAT Club [#permalink]  04 Sep 2009, 11:07
Thank you all for this list. Most of all thank you so much KillerSquirrel for coming up with a well... killer approach to solve a majority of these problems.

Tejal missed a few answers though. Can someone confirm that my answers are correct? Not sure if the explanations for these are on here already.

2. B (2.5)
3. A
8. C (37.5)
12. E (4/5)
13. C
14. C (20/3)
15. 68%
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Re: Mixture Problems - I collected from GMAT Club [#permalink]  04 Sep 2009, 17:57
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bharathh: Search the forum,even though a few links are missing (I was being lazy) all the answers are there..most of them started by me cos i sucked at mixture probs
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Re: Mixture Problems - I collected from GMAT Club [#permalink]  11 Nov 2009, 08:22
For solutions please search the forum:
Ans:
1 B
2 B
3A
4B
5C
6D
7B
8C
9C
10A
11E
12E --> not sure
13C --> not sure
14C --> not sure
15 17/25
16 20/7
17 122.5
18 A
19A
20 20
21 D
22C
23 9/13
24 D
25 B
26 A
27 2.5
28 A
29C
30E

Thank you for this thread. I feel confident abt mixtures now.
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Re: Mixture Problems - I collected from GMAT Club [#permalink]  09 Jan 2010, 07:04
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14.How many liters of pure alcohol must be added to a 100-liter solution that is 20 percent alcohol in order to produce a solution that is 25 percent alcohol?
(A) 7/2
(B) 5
(C) 20/3
(D) 8
(E) 39/4

1. (Stronger%-Desired%) * ammount of Stronger = (Desired%-Weaker%) * amount of weaker

2. (1-.25)s = (.25-.2)100

3. (.75)s = (.05)100

4. (.75)s = 5

5. s = 5/.75

6. s = 500/75 => 20/3
Re: Mixture Problems - I collected from GMAT Club   [#permalink] 09 Jan 2010, 07:04

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