# Mixtures

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1829 Directions: Each of the questions given below is followed by series of options among which you have to choose the best one. You can also review your choices by clicking into answer section.

1. In what ratio must water be mixed with milk costing Rs.12 per liter to obtain a mixture worth of Rs.8 per liter?

A. 3:2

B. 1:2

C. 2:3

D. 2:1

Ans: Option B

Explanation:

12 – 8: 8 – 0 = 1 : 2

2. In what ratio must water be mixed with milk to gain 1623% on selling the mixture at cost price?

A. 6: 1

B. 1: 6

C. 1: 4

D. 4: 1

Ans: Option B

Explanation:       ​

Explanation:

Let CP of 1 liter milk = Rs.1

SP of 1 liter mixture = CP of 1 liter milk = Rs.1Gain

=1623%=503%CP of 1 liter mixture

= 100(100+Gain %) ×SP=100(100+503)×1=100(3503)=300350=67

By the rule of allegation, we have

CP of 1 liter water                                                                                                            CP of 1 liter milk

0                                                                                                                                       1

CP of 1 liter mixture

6/7

1 – 6/7 = 1/7                                                                                                                          6/7 – 0 = 6/7

Quantity of water: Quantity of milk = 1/7 : 6/7 = 1 : 6

3. A bag contains 57 kg of rice with 7.5% contamination. How many kg of non-contaminated rice should be added to the bag so that the resultant mixture contains 6% contamination?

A. 14.25 kg

B. 15 kg

C. 14 kg

D. 14.5 kg

Ans: Option A

Explanation:

​Amount of contamination in the bag = 7.5% of 57 = 7.5*57/100

Let x kg of pure rice be added to the bag.

(7.5*57/100) ÷ (57+x) = 6/100 x = 14.25 kg

4. A dishonest milkman professes to sell his milk at cost price but he mixes it with water and thereby gains 25%. The percentage of water in the mixture is:

A. 4%

B.6 %

C.20%

D.25%

Ans: Option C

Explanation:

Let C.P. of 1 liter milk be Re. 1

Then, S.P. of 1 liter of mixture = Re. 1, Gain = 25%.

C.P. of 1 liter mixture = Re.   (100/125) x 1   = 4/5

By the rule of allegation, we have:

C.P. of 1 liter of milk                                                                                      C.P. of 1 liter of water

Re. 1 Mean Price                                      Mean Price                                                         0

4                                                             4                                                              1

5                                                             5                                                              5

Ratio of milk to water = 4/5: 1/5 = 4: 1.

Hence, percentage of water in the mixture =   (1/5) x 100% = 20%.​

5. The milk and water in two vessels A and B are in the ratio 4:3 and 2:3 respectively. In what ratio the liquids in both the vessels are mixed to obtain a new mixture in vessel c consisting half milk and half water?

A. 4:3

B. 2:3

C. 7:5

D. 8:3

Ans: Option C

Explanation:

Milk in 1 liter mixture of A = 4/7 liter.

Milk in 1 liter mixture of B = 2/5 liter.

Milk in 1 liter mixture of C = 1/2 liter.

By rule of allegation we have required ratio X:Y

X                  :                 Y

4/7                                2/5

\                      /

(Mean ratio)

(1/2)

/                      \

(1/2 – 2/5)     :       (4/7 – 1/2)

1/10                      1/1 4

So Required ratio = X: Y = 1/10 : 1/14 = 7:5

6. In 1 kg mixture of iron and manganese 20% of manganese. How much iron should be added so that the proportion of manganese becomes 10%

A. 1.5 Kg

B. 2 Kg

C. .5 Kg

D. 1 Kg

Ans: Option D

Explanation:

By the rule of allegation, we have

Percentage concentration of                                                                      Percentage concentration of

Manganese in the mixture: 20                                                                   Manganese in the mixture: 0

Percentage concentration of

manganese in the final mixture 10

10 – 0 = 10                                                                                                           20 – 10 = 10

Quantity of the mixture: Quantity of iron = 10: 10 = 1 : 1

Given that Quantity of the mixture = 1 Kg

Hence Quantity of iron to be added = 1 Kg

7. Rahari dal costs Rs.75/kg. Musuri dal which is used as a supplement for Rahari dhal costs Rs.60/kg. A family needs a total of 10 kg of dhal and they spend Rs.660 to buy a few kg of Rahari and musuri. How many kg of Rahari dal and musuri dal have they bought?

A. 4, 6

B. 6, 4

C. 5, 5

D. 3, 7

Ans: Option A

Explanation:

Let the quantity of Rahari dal bought be x kg.

Then, musuri dal = (10-x)kg

75x+60(10-x) = 660

X = 4 ​

8. Two vessels A and B contain milk and water mixed in the ratio 8: 5 and 5 : 2 respectively. The ratio in which these two mixtures be mixed to get a new mixture containing milk and water in the ratio 9: 4?

A. 2:7

B. 3:5

C. 5:2

D. 5:7

Ans: Option A

Explanation:

No explanation​

9. How many kg of Basmati rice costing Rs.42/kg should a shopkeeper mix with 25 kg of ordinary rice costing Rs.24 per kg so that he makes a profit of 25% on selling the mixture at Rs.40/kg?

A. 16 kg

B. 12.5 kg

C. 200 kg

D. 20 kg

Ans: Option D

Explanation:

Let the amount of Basmati rice being mixed be x kg.

As the trader makes 25% profit by selling the mixture at Rs.40/kg, his cost per kg of the mixture = Rs.32/kg.

i.e. (x×42)+(25×24)=32(x+25)

42x+600=32x+800

10x=200

x=20 kg.

10. The cost of Type 1 rice is Rs.15 per kg and Type 2 rice is Rs.20 per kg. If both Type 1 and Type 2 are mixed in the ratio of 2 : 3, then the price per kg of the mixed variety of rice is:

A.Rs. 18

B.Rs. 18.50

C.Rs. 19

D.Rs. 19.50

Ans: Option A

Explanation:

Let the price of the mixed variety be Rs.x per kg.

By rule of allegation, we have:

Cost of 1 kg of Type 1 rice                                                        Cost of 1 kg of Type 2 rice

Rs.15                                                     Mean Price                              Rs.20

Rs.x

(20 – x )                                                                                               (x – 15)

(20 – x )/ (x – 15)= 2/3

60 – 3x = 2x – 30

5x = 90

X = 18.

11. The proportion of milk and water in 3 samples is 4:3, 5:2 and 2:1. A mixture comprising of equal quantities of all 3 samples is made. The proportion of milk and water in the mixture is

A. 11:6

B. 41:21

C. 21:11

D. 41:22

Ans: Option D

Explanation:

​Let us consider 1 liter of each sample.

Proportion of milk = (4/7) + (5/7) + (2/3) = 41/21

Proportion of water = (3/7) + (2/7) + (1/3) = 22/21

Proportion of milk and water = 41:22

12. A merchant has 1000 kg of sugar, part of which he sells at 8% profit and the rest at 18% profit. He gains 14% on the whole. The quantity sold at 18% profit is

A. 640 kg

B. 600 kg

C. 560 kg

D. 400 kg

Ans: Option B

Explanation:

Ratio of first and second part = 4: 6 = 2: 3

Quantity of 2 kind = (1000*3/5) = 600 Kg

13. A can contains a mixture of two liquids A and B is the ratio 7: 5. When 9 liters of mixture are drawn off and the can is filled with B, the ratio of A and B becomes 7: 9. How many liters of liquid A were contained by the can initially?

A.10

B.20

C.21

D.25

Ans: Option C

Explanation:

Suppose the can initially contains 7x and 5x of mixtures A and B respectively.

Quantity of A in mixture left =   7x – (7/12) x 9   liters =   7x – (21/4)   liters.

Quantity of B in mixture left =   5x – (5/12) x 9   liters =   5x – (15/14)   liters.

{[7x – (21/4)] / [5x – (15/4)]} + 9 = 7/9

(28x – 21)/ (20x + 21) = 7/9

252x – 189 = 140x + 147

112x = 336

X = 3.

So, the can contained 21 liters of A.

14. Two glasses A and B contain a mixture of water and honey in the ratio 5:1 and 4:1 respectively. One-fourths of the mixture in glass A is poured out and refilled with the mixture from glass B. Find the ratio of water and honey in glass A.

A. 20:1

B. 33:7

C. 10:1

D. 22:3

Ans: Option B

Explanation:

​The resulting mixture contains A and B in the ratio 3:1(three-fouths of A and one-fourths of B.

Proportion of water = [3*(5/6)+1*(4/5)]/(3+1) = 33/40

Proportion of honey = [3*(1/6)+1*(1/5)]/(3+1) = 7/40

Water and Honey = 33:7

15. We have a 630 ml of mixture of milk and water in the ratio 7:2. How much water must be added to make the ratio 7:3?

A. 70 ml

B. 60 ml

C. 80 ml

D. 50 ml

Ans: Option A

Explanation:

Concentration of water in mixture1 = 2/9 (Since the ratio of milk and water = 7:2) —item (1)

Concentration of water in pure water= 1 —item (2)

Now the above mentioned items are mixed to form a mixture2 where milk and water ratio = 7: 3

Concentration of water in mixture2 = 3/10

By the rule of allegation, we have

Concentration of water                                                                                                                 Concentration of water

In mixture 1: 2/9                                                                                                                                in pure water: 1

Mean concentration

3/10

1 – 3/10 = 7/10                                                                                                                                   3/10 – 2/9 = 7/90

Quantity of mixture1: Quantity of water = 7/10: 7/90 = 1/10: 1/90 = 1 : 1/9

Given that Quantity of mixture1 = 630 ml

630: Quantity of water = 1: 1/9

Quantity of water = 630×19=70 ml

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