ALLEGATION OR MIXTUREARITHMETICMATHS

QUESTIONS ON ALLEGATION OR MIXTURE (PART-III)

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QUESTIONS ON ALLEGATION OR MIXTURE (PART-III)

VERY IMPORTANT: Remember that the weighted average/combined average/mean value is always written in the middle and the averages are written at the top. We write the smaller average/value in the left and greater one in the right. The mean value is subtracted from the value written in the right, and the value written in the left is subtracted from the mean value, figure thus got is considered positive (negative signs are ignored).

But in questions having multiple ratios as are given in queries 1, 2 and 3, we write the value of the thing that is first in the asked ratio in the left and the other value in the right.

QUERY 21

In what ratio a shopkeeper add water to milk so that he gains 25% on selling it at cost price?

A) 2 : 3
B) 3 : 2
C) 3 : 1
D) 4 : 1

MAHA GUPTA
Let cost price of pure milk = Re 1 per litre
As the mixture is sold at the cost price of milk, the selling price of mixture = Rs 1 per litre

As the gain is 25%, cost price of 1 litre of mixture = 1 × 100125 = Rs  45

We know water is free, so cost price of water = Re 0 per litre

MILK                                         WATER

1 0
    45
0 – 4=  45  4– 1 = 15

We see, MILK : WATER = 4: 1= 4 : 1 (option ‘D’)


QUERY 22

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

A) 10
B) 20
C) 21
D) 25

MAHA GUPTA
By allegation method, we can solve this question in the following two ways:
i) By comparing the quantities of liquid B in the solution
ii) By comparing the quantities of liquid A in the solution

i) By comparing the quantities of liquid B in the solution
In this method, we’ll assume that the mixture contains only liquid B and we’ll compare it with the value of liquid B added. Mean value of the allegation method will be quantity of liquid B according to the new ratio given. Addition of the terms of the ratio so got will give us the value of the total quantities of liquid A and liquid B in the mixture.

As whole of the mixture is replaced by liquid B, value of liquid B will be 1. So,

MIXTURE                              LIQUID B

512 1
916
1 –  916 = 716 916512 = 748

Hence, the ratio = 716 : 748 = 3 : 1
Means, added quantity of liquid B = 1 ratio = 1*9 = 9 litres
Total of the terms of the ratio is 3+1 = 4, Thus, total mixture = 4*9 = 36 litres

Therefore, quantity of liquid A initially = 712 × 36 = 21 litres (option ‘C’)

ii) By comparing the quantities of liquid A in the solution
In this method, we’ll assume that the mixture contains only liquid A and we’ll compare it with the value of liquid A added. Mean value of the allegation method will be quantity of liquid A according to the new ratio given. Addition of the terms of the ratio so got will give us the value of the total quantities of liquid A and liquid B in the mixture.

As whole of the mixture is replaced by liquid B, value of liquid A will be 0. So,

MIXTURE                              LIQUID A

712 0
716
0 – 716 = 716 716712 = 748

Hence, the ratio = 716 : 748 = 3 : 1
Means, added quantity of liquid = 1 ratio = 1*9 = 9 litres
Total of the terms of the ratio is 3+1 = 4, Thus, total mixture = 4*9 = 36 litres

Therefore, quantity of liquid A initially = 712 × 36 = 21 litres (option ‘C’)

ANOTHER METHOD
9 litres of mixture replaced contains 9 litres in the ratio of 7 : 5, means 5.25 litres of liquid A and 3.75 litres of liquid B

Let the quantities of liquid A and B in the original mixture be 7x and 5x respectively.
Hence, quantities in the new mixture = (7x – 5.25) : (5x – 3.75 + 9)
=> (7x – 5.25) : (5x + 5.25) = 7 : 9                                         —- as new ratio is given 7 : 9
=> 63x – 47.25 = 35 x + 36.75
=> x = 3

Therefore, quantity of liquid A initially = 7x = 7*3 = 21 litres (option ‘C’)

ONE MORE METHOD
Let the container in the beginning holds the liquids A and B respectively 7x and 5x liters of the mixture
So total mixture = 12x liters
Drawn mixture = 9 litres
Therefore quantity of liquid ‘A’ drawn from it = 9 × 712 = 6312 = 214
So, liquid A in the mixture now left = 7x – 214 = (28x – 21)4

And quantity of liquid ‘B’ drawn from it = 9 × 512 = 4512 = 154
So quantity of liquid B left in the mixture then = 5x – 15= (20x – 15)4

But the whole container was then filled with the liquid B, means 9 liter of liquid B was added to the mixture
So, quantity of liquid B finally in the mixture = (20x – 15)4 + 9 = (20x + 21)4

The new ratio = 7 : 9
Therefore, [(28x – 21)/4]/[(20x + 21)/4] = 7/9
=> (28x – 21)(20x + 21) = 79
=> x = 3

Therefore quantity of liquid A initially in the mixture = 7x = 21 liters (option ‘C’)

TRICK
Better use answer options. Choose such an option first that gives quantity in whole number in liquid B as well. We see by the initial ratio of 7 : 5 that quantity in liquid B is 5/7 of the liquid A.

Now, therefore, take 21 from options, of course liquid B is then 15 litres.

Now liquid A after withdrawn of 9 litres of mixture = 21 – 9*712 = 15.75 litres
Liquid B after withdrawn of 9 litres of mixture = 15 – 9*512 = 11.25 litres

But drawn mixture of 9 litres now is filled in liquid B
Therefore, liquid A after = 15.75 litres
Liquid B after = 11.25 + 9 = 20.25 litres

Hence, 15.7520.25 = 79

1575*9 = 2025*7
= on calculation you’ll find these are equal, so the answer is 21 (option ‘C’)

QUERY 23

A vessel contains mixture of spirit and water. The percentage of spirit in the mixture is 18%. 6 litres of the mixture is replaced with pure water. If the vessel contains 15% spirit now, the capacity of the vessel is?

A) 25 litres
B) 30 litres
C) 36 litres
D) 32 litres

MAHA GUPTA
If you see this question is exactly similar to the above question (QUERY 21). We can re-write this as follows if you want to see.
A vessel contains mixture of spirit and water in the ratio 9 : 41 (The percentage of spirit in the mixture is given 18%). When 6 litres of the mixture is replaced with pure water, the ratio of spirit and water becomes 3 : 17 (the vessel contains 15% spirit now); the capacity of the vessel is?

Well, let’s solve.

By allegation method, we can solve this question in the following two ways:
i) By comparing the quantities of water in the solution
ii) By comparing the quantities of spirit in the solution

i) By comparing the quantities of water in the solution
In this method, we’ll assume that the mixture contains only water and we’ll compare it with the value of water added. Mean value of the allegation method will be quantity of water according to the new ratio given. Addition of the terms of the ratio so got will give us the value of the total quantities of spirit and water in the mixture.

As whole of the mixture is replaced by water, value of water be will be 100%. Mixture will be 100 – 18 = 82%, Mean value of water = 100 – 15 = 85%. So,

MIXTURE                           WATER

82% 100%
85%
100 – 85 = 15 85 – 82 = 3

Hence, the ratio = 15 : 3 = 5 : 1
Means, added quantity of water = 1 ratio = 1*6 = 6 litres
Total of the terms of the ratio is 5+1 = 6, Thus, total mixture = 6*6 = 36 litres

Therefore, capacity of the vessel = 36 litres (option ‘C’)

ii) By comparing the quantities of spirit in the solution
In this method, we’ll assume that the mixture contains only spirit and we’ll compare it with the value of spirit added. Mean value of the allegation method will be quantity of spirit according to the new ratio given. Addition of the terms of the ratio so got will give us the value of the total quantities of spirit and water in the mixture.

As whole of the mixture is replaced by water, value of spirit will be 0. So,

MIXTURE                              SPIRIT

18% 0
15%
0 – 15 = 15 15 – 18 = 3

Hence, the ratio = 15 : 3 = 5 : 1
Means, added quantity of water = 1 ratio = 1*6 = 6 litres
Total of the terms of the ratio is 5+1 = 6, Thus, total mixture = 6*6 = 36 litres

Therefore, capacity of the vessel = 36 litres (option ‘C’)

NOTE: You can also do this question by other methods as described above.

QUERY 24

A mixture contains 80% acid and rest water. Part of the mixture that should be removed and replaced by same amount of water to make the ratio of acid and water 4 : 3 is ?

A) 1/3
B) 3/7
C) 2/3
D) 2/7

MAHA GUPTA
By allegation method, we can solve this question in the following two ways:
i) By comparing the quantities of water in the solution
ii) By comparing the quantities of acid in the solution

i) By comparing the quantities of water in the solution
In this method, we’ll assume that the mixture contains only water and we’ll compare it with the value of water added. Mean value of the allegation method will be quantity of water according to the new ratio given. Addition of the terms of the ratio so got will give us the value of the total quantities of acid and water in the mixture.

As the mixture contains 80% acid, so ratio of acid and water in the mixture = 80 : 20 = 4 : 1

As whole of the mixture is replaced by water, value of water be will be 1. Mixture will be 1/5. Mean value of water = 3/7. So,

MIXTURE                           WATER

15 1
   37
1 – 3= 47 3715= 835

Hence, the ratio = 47 : 835 = 20 : 8 = 5: 2
Therefore, part of the mixture that should be removed and replaced by same amount of water = 27 (option ‘D’)

ii) By comparing the quantities of acid in the solution
Likewise you can also do the sum by comparing quantities of acid.

QUERY 25

A vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?

A) 1:5
B) 1:2
C) 4:1
D) 1:4

RONNIE BANSAL
We do not require the ALLEGATION METHOD in a question when the new mixture has to be in the ratio of 1: 1. Here is the trick:

Suppose there is in total 8 (3+5) liters of mixture in the vessel with 3 liters of water and 5 liters of syrup in it. Now we have to draw the mixture in a way so that water and syrup are equal in the vessel after the replacement with the same quantity that was drawn. Its simple to think if one liter of syrup is drawn from the mixture above and 1 liter of water is added the problem is solved.

In such a question just divide the quantity to be drawn by the total quantity of that item. That number will be the answer. Here we have to draw 1 liter of syrup out of total 5 liters of it.

Therefore 1/5 will be the total quantity to be drawn from whole; hence the answer. (option ‘A’)

Now take another example:
A vessel is filled with liquid, 5 part of which are water and 10 part syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup.

We have to draw 2 + 1/2 i.e. 5/2 liters of syrup. Total syrup is 10 liters. So 5/2 divided by 10 i.e. 1/4 will be the answer.

QUERY 26

A barrel contains a mixture of wine and water in the ratio 3 : 1. How much fraction of the mixture must be drawn off and substituted by water so that the ratio of wine and water in the resultant mixture in the barrel becomes 1 : 1?

A) 1/3
B) 1/4
C) 2/5
D) 2/3

MAHA GUPTA
We do not need to use allegation and mixture method or any other complex one when the new mixture has to be in the ratio of 1:1. Here is the trick:

In such a question just divide the quantity to be drawn by the total quantity of that item. That number will be the answer.

The quantity to be drawn can easily be made out. For example here the mixture of wine and water is in the ratio 3 : 1. To make the ratio equal, obviously one needs to draw 1 part of wine and add 1 part of water in this.

Here we have to draw 1 part of wine out of total 3 parts of it. Therefore 1/3 will be the total quantity to be drawn from whole; hence the answer. (option ‘A’)

QUERY 27

A vessel of 64 litres is filled with milk and water. If 70% of milk and 30% of water is drawn, the vessel is vacated by 55%. Find the initial quantity of the milk.

A) 24 litres
B) 40 litres
C) 45 litres
D) 50 litres

MAHA GUPTA
As the final result of the process is vacation of the vessel we’ll take it as the mean value in the allegation method. So,

Milk                                               Water

 70%  30%
 55%
30 – 55 = 25 55 – 70 = 15

Therefore, initial MILK : WATER = 25 : 15 = 5 : 3

Therefore, initial quantity of milk = 64 × 58

= 40 litres (option ‘B’)

ANOTHER METHOD
Let initial quantity of milk in the mixture = x litres
And initial water in the mixture = y litres

Hence, x + y = 64                                                                   —- (i)

Quantity of milk after it has been withdrawn = x – 70% of x = 3x10 litres
And his monthly savings now = y – 30% of y = 7y10 litres
But the vessel is vacted by 55% after this, so quantity of the mixture after = 64 – 55% of 64 = 28.8 litres

Therefore, 3x10 + 7y10  = 28.8
=> 3x + 7y = 288                                                                    —- (ii)

Solving (i) and (ii)
x = 40

Therefore, Therefore, initial quantity of milk = 40 litres (option ‘B’)

QUERY 28

Two products A and B costing Rs 24 and Rs 18 per kg respectively are mixed. Find the ratio of two quantities if 1% of the quantity is lost in the process of mixing and cost of the mixture after wastage is Rs 20 per kg.

A) 1 : 2
B) 2 : 1
C) 3 : 7
D) 2 : 9

MAHA GUPTA
As the cost of the products is given before wastage, we need to find the cost of one kg of mixture before wastage here. And that only will be our mean value of the allegation method.

Now, the cost of one kg of mixture before wastage = 20 – 1% of 20 = Rs 19.80. So,

PRODUCT A                                            PRODUCT B

24 18
19.80
18 – 19.80 = 1.80 19.80 – 24 = 4.20

Therefore, PRODUCT A : PRODUCT B = 1.80 = 4.20 = 3 : 7 (option ‘C’)

ANOTHER METHOD
As the cost of the products is given before wastage, to know the ratio of the two products, we need to find the cost of one kg of mixture before wastage here.

So, the cost of one kg of mixture before wastage = 20 – 1% of 20 = Rs 19.80.

Now, let the quantity of product A = x kg
And the quantity of product B = y kg

Hence, total cost of the whole mixture = 24x + 18y                                                    —- (i)

Total quantity of the mixture = (x + y) kg
Cost of one kg of mixture = Rs 19.80
Thus, total cost of the whole mixture = (x +y) × 19.80 = 19.80x + 19.80y            —- (ii)

From (i) and (ii)
24x + 18y = 19.80x + 19.80y
=> 4.20x = 1.80y
=> xy 1.804.20
=> x : y = 3 : 7 (option ‘C’)

QUERY 29

Two casks of 48 and 42 litres are filed with mixture of wine and water; the proportions in the two casks being respectively 13 : 7 and 18 : 17. If the contents of the two casks be mixed and 20 litres of water added to the whole, what will be the proportion of wine to water in the result?

A) 13 : 12
B) 12 : 13
C) 21 : 31
D) 31 : 21

RONNIE BANSAL
Proportion of wine to water in the result = Quantity of wine in both of the casks : Quantity of water in both of the casks + 20

= [48*(1320) + 42*(1835)] : [48*(720) + 42*(1735) + 20]

= [31.2 + 21.6] : [16.8 + 20.4 + 20]

= 52.8 : 57.2= 528 : 572

= 12 : 13 (option ‘B’)

TRICK
We, actually, do not need to divide 528 : 572 to arrive at 12 : 13; as in the options only this ratio is with smaller first term.

QUERY 30

Tea worth Rs 126 per kg and Rs 135 per kg are mixed with a third variety in the ratio 1 : 1 : 2. If the mixture is worth Rs 153 per kg, the price of the third variety per kg will be?

A) Rs 169.50
B) Rs 170
C) Rs 175.50
D) Rs 180

MAHA GUPTA
Since first and second varieties are mixed in equal proportions; so, their average price = (126+135)2 = Rs 130.50

So, the mixture is formed by mixing two varieties, one at Rs 130.50 per kg and the other at say, Rs x per kg in the ratio 2 : 2, i.e., 1 : 1. We have to find x.

As the mixture is Rs 153 per kg, therefore price of both must be 153*2 = Rs 306 per two kg

But the price of one of them is Rs 130.50 per kg             —- calculated above
Hence, the price of the other variety i. e. x = 306 – 130.50 = Rs 175.50 (option ‘C’)

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Maha Gupta

Maha Gupta

Founder of www.examscomp.com and guiding aspirants on SSC exam affairs since 2010 when objective pattern of exams was introduced first in SSC. Also the author of the following books:

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