10 Solved problems in Alligations and mixture

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1. In what ratio must rice at Rs 9.30 per kg be mixed with rice at Rs 10. 80 per kg,so that the mixture be worth Rs 10 per kg?

(a) 6:7                               (b) 3:7

(c) 5:7                               (d) 8:7

Answer : (d) 8:7

Solution:

The formula for alligation is 

n₁/n₂ = (A₂ – A_w) / (A_w –A₁₎

Where A₁ and A₂ are the price of the two items that are mixed and n1 and n2 are the quantities of each.

We have to find n₁/n₂

Put in the values of A1 = 9.30, A2 = 10.80  and Aw = 10

We have,

n₁/n₂ = (10.80  - 10)/(10  - 9.30) =0 .80/0.70 = 8/7 

2. In what ratio must water be mixed with milk to gain 20 % by selling the mixture at cost price?

(a) 1:5                              (b) 1:9

(c) 1:4                              (d) 1:12

Answer: (a) 1:5

Solution:

For such question where water or some other thing which has negligible cost, a short cut is to divide the percentage profit by 100.

Therefore,

ratio is = 20/100 = 1 :5

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3.  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 a new mixture in vessel C containing half milk and half water?

(a) 7:5                              (b) 3:5

(c) 9:5                              (d) 22:3

Answer: (a) 7:5

 Solution:

Milk  in first vessel = 4/7

And milk in second vessel = 2/5

And after mixing the above mixture the resultant mixture would have ½ milk.

Using allegation formula,

n1/n2 = (2/5 – 1/2)/( ½ - 4/7 )= (1/14 )/( 1/10) = 7:5

4. In what proportion must rice at Rs 3.10/kg be mixed with rice at Rs 3.60 kg so that the mixture be worth Rs 3.25 a kg ?

(a) 11.3                             (b) 25.3

(c) 7. 3                              (d) 22.3

 

Answer: (c) 7:3

Solution:

Let n1 quantity of the rice at 3.10/ Kg be mixed with n2 quantity of rice at 3.60 Kg 

Then

using alligation formula we have,

n1/n2 = (A2 - Aw)/(Aw -A1) = (3.60 - 3.25)/(3.25 - 3.10) =  0.35/0.15 = 7/3

5. How many kg of salt at 42/kg must a man mix with 25 kg of salt at 24 kg so that may he on selling mixture at 40/kg , gain 25 % on the outlay .

(a) 35 kg                           (b) 20 kg

(c) 5 kg                             (d) 15 kg 

Answer: (b) 20 Kg

Solution: 

First we find the cost price / kg of the mixture.

 It is given that selling the mixture at Rs. 40 gains him 25 %, using this to get the cost price.

=> 1.25x = 40

=> X = 32

Using 32 as the cost price /Kg of the mixture in the allegation formula we have,

N1/n2 = (24 – 32)/(32 - 42) = 8/10 = 4/5

Given that n2 = 25 Kg 

Then n1 = (4/5)25 = 20 Kg

6. A mixture of a certain quantity of milk with 16 lit of water is sold at 90 p/lit if the pure milk is worth 108 p/lit, then how much milk is there in the mixture?

(a) 160 liter                        (b) 80 liter

(c) 180 liter                        (d) 220 liters

Answer: (b) 80 liter

Solution:

Let cost of water = 0

Then,

let n1 is the quantity of water and n2 be the quantity if milk then, using alligation equation 

n1/n2 = (108 – 90)/(90 -0)= 18/90 = 2/10= 1/5

16/n2 = 1/5

 n2 = 16 x 5 = 80 liter 

 7.In what proportion must water be mixed with spirit to gain 50/3 % by selling it at C.P ?

(a) 1:10                              (b) 1:20

(c) 1:35                              (d) 1:6

Answer: (d) 1:6

Solution:

For such questions, just divide the percentage profit with 100.

50/3/100 = 50/300 = 1/6

8. 300 gm of sugar solution has 40 % sugar in it how much sugar should be added to make it 50 % in the solution?

(a) 60 gm                           (b) 25 gm

(c) 80 gm                           (d) 10 gm

Answer: (a) 60 gm

Solution:

Let x grams of sugar is added, then

Total sugar in the mixture is 0.40 x 300 + x = (120 + x) grams

And the total solution now becomes (300 + x )grams 

It is given that after adding x grams of sugar in the solution becomes 50 %

Therefore, using percentage formula

(120 + x)/ (300 + x) x 100 = 50

=> (120 + x)/ (300 + x) = (1/2)

=>240 + 2 x = 300 + x

=>x = 60 grams

9. There are 65 students in a class, 39 are distributed among them so that each boy gets 80 p and each girl gets 30 p find the number of boys and girls in the class.

(a) 40, 25                            (b) 35, 30

(c) 39, 26                            (d) 41, 24    

Answer: (c) 39,26

Solution:

Let there are  B boys and G girls,

Then, it is given that,

B + G = 65 ----------------(1)

Since, each boy got 80 p and each girl got 30 p, total amount totaling to 39

Or 0.80 B +0.30 G = 39

8B + 6G = 390 ------------(2)

Solving the above 2 equation to get B = 39 and G = 26

Alternative solution:   

Using allegation equation,

Average money each child got = 39/65 = 0.60 P

Now find n1/n2 = (0.80 – 0.60)/(0.60 – 0.30) = 2/3

where n1 are girls and n2 are boys

This means girls and boys are  in the ratio 2:3

Therefore, boys =( 3/5 )65 = 39 and girls = (2/5)65 = 26

10. A person has a chemical of Rs 25/lit in what ratio should water be mixed in that chemical so that after selling the mixture at Rs 20/lit he may get a profit of 25 % .

(a) 15:7                              (b) 12:5

(c) 33:2                              (d) 16:9

Answer:(d)16:9

Solution:

First find the cost price of the final mixture.

Since, by selling at Rs 20 he got a profit of 25% then , let x be the cost price

1.25x = 20

=> x = 20(4/5) = 16

Now we have cost price at Rs. 16 per liter of the mixture. Let the price of water be 0,

Then using the allegation formula, we have

n1(water)/n2(chemical) = ( 25 – 16)/(16 -0 ) = 9/16

or chemical /water = 16:9



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