# QUESTIONS ON ALLEGATION OR MIXTURE (PART-IV)

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 31

Zinc and copper are in the ratio 5 : 3 in 400 gm of an alloy. How much of copper (in grams) should be added to make the ratio 5 : 4?

A) 66
B) 72
C) 200
D) 50

MAHA GUPTA
Zinc earlier = 400*(5/8) = 250 gm
Copper earlier = 400 – 250 = 150 gm

Now zinc has to be the same, only quantity of the copper to be changed to make the ratio 5 : 4

So the quantity of copper later = 250*(4/5) = 200 gm
Hence the additional copper needed = 200 – 150 = 50 gm (option ‘D’)

#### QUERY 32

A merchant mixes two varieties of wine containing 25% and 13% alcohol. The resultant mixture contains 17% alcohol. Find the quantity of the second mixture if 8 litres of first mixture is taken.

A) 4 litres
B) 16 litres
C) 24 litres
D) 32 litres

MAHA GUPTA

MIXTURE-I                           MIXTURE-II

 25% 13% 17% 13 – 17 = 4 17 – 25 = 8

So, MIXTURE-I : MIXTURE-II = 4 : 8 = 1 : 2
Quantity of the mixture-I = 1 ratio = 8 litres

Therefore, quantity of mixture-II = 2*8 = 16 litres (option ‘B’)

#### QUERY 33

In what ratio must a grocer mix two varieties of pulses costing Rs 15 and Rs 20 per kg respectively so as to get a mixture worth Rs 16.50 kg?

A) 3 : 7
B) 5 : 7
C) 7 : 3
D) 7 : 5

MAHA GUPTA
PULSES KIND-I                    PULSES KIND-II

 15 20 16.50 20 – 16.50 = 3.50 16.50 – 15 = 1.50

So, PULSES KIND-I : PULSES KIND-II = 3.50 : 1.50 = 7 : 3 (option ‘C’)

ANOTHER METHOD
Let he mixes the two varities of pulses in ratio x : y
Therefore the cost price = 15x + 20y                                                                  —- (i)

But now he mixes these two varieties of pulses worth Rs 16.50 kg
This, the cost price at this rate = 16.50*(x + y)                                                —- (ii)

From (i) and (ii) above
15x + 20y = 16.50*(x + y)
=> 1.5x = 3.50y
=> xy = 3.501.50
=> x : y = 7 : 3 (option ‘C’)

#### QUERY 34

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

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

RICE-I                                 RICE-II

 15 20 x 20 – x x – 15

So, PULSES KIND-I : PULSES KIND-II = (20 – x) : (x – 15) = 2 : 3
By solving, x = 18 (option ‘A’)

ANOTHER METHOD
As the ratio of the two types of quantities is given as 2 : 3, so their total quantity = 2 + 3 = 5 units
Let the price of the mixed variety be Rs. x per kg
Therefore the cost price of the quantity = 5*x = 5x                                                     —- (i)

If taken separately, the cost price = 2*15 + 3*20 = 90                                               —- (ii)

From (i) and (ii) above
5x = 90
=> x = 18 (option ‘A’)

#### QUERY 35

In what ratio must a grocer mix 2 varieties of tea worth Rs. 60 per kg and Rs. 65 per kg so that by selling the mixture at Rs. 68.20 per kg he may gain 10% ?

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

MAHA GUPTA
If the selling price of the tea is Rs 68.20 per kg, for 10% gain the cost price of it = 68.20 × 100110 = Rs 62 per kg.

TEA-I                             TEA-II

 60 65 62 65 – 62 = 3 62 – 60 = 2

Hence, ratio of the two quantities = 3 : 2 (option ‘C’)

[email protected]
Let he mixes the tea in ratio x : y
Therefore the cost price = 60x + 65y
and the selling price = 68.20(x + y)

Profit = SP – CP
So 10100(60x + 65y) = 68.20(x + y) – (60x + 65y)

or 6x + 6.5y = 8.20x + 3.20y

or 2.2x = 3.3y

=> xy = 3.32.2

=> x : y = 3:2 (which is required ratio)               option ‘C’

#### QUERY 36

A shopkeeper mixed 10 kg of rice costing Rs 10 per kg with a certain quantity  of another variety of rice costing Rs 15 per kg. He sold the mixture at the rate of Rs 14 per kg. If he gained by 5%, the quantity of the second variety of rice was?

A) 5 kg
B) 10 kg
C) 20 kg
D) 30 kg

MAHA GUPTA
Cost Price per kg of the mixture = 14 × 100105 = Rs 403
RICE-I                                                                  RICE-II

 10 15 40⁄3 15 – 40⁄3 = 5⁄3 40⁄3 – 10 = 10⁄3

Hence, ratio of the two quantities = 53 : 103 = 5 : 10 = 1 : 2

Quantity of first variety of rice = 1 ratio = 1*10 = 10 kg
Therefore, quantity of second variety of rice = 2*10 = 20 kg (option ‘C’)

ANOTHER METHOD
Let he mixes rice-I and rice-II in the ratio = x : y
Therefore the CP of above = 10x + 15y
And the SP of above= 14(x + y)
Hence, profit = 14(x + y) – (10x + 15y)
But profit also = (5100)(10x + 15y)

So, (5100)(10x + 15y) = 14(x + y) – (10x + 15y)
=> 12x + 34y = 4x – y
=> 2x + 3y = 16x – 4y
=> 14x = 7y
=> xy = 714
=> x : y = 1 : 2

Quantity of first variety of rice = 1 ratio = 1*10 = 10 kg
Therefore, quantity of second variety of rice = 2*10 = 20 kg (option ‘C’)

#### QUERY 37

In what proportion may these kinds of tea of prices @Rs 80, Rs 70 and Rs 50 per kg be mixed to produce a mixture of Rs 60 per kg?

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

MAHA GUPTA
When there are three kinds of things, firstly take any two quantities in a manner that the cost of one of them is above the mean price (here Rs 60 per kg) and cost of the other thing is less than the mean price. Similarly find the ratio between the third quantity and any of the above in the same manner. And then find the combined ratio (total of ratios) of those things.

Now see how it will work when the above question is considered.

TEA-I                             TEA-III

 80 50 60 50 – 60 = 10 60 – 80 = 20

So, TEA-I : TEA-III = 10 : 20 = 1 : 2

TEA-II                           TEA-III

 70 50 60 50 – 60 = 10 60 – 70 = 10

So, TEA-II : TEA-III = 10 : 10 = 1 : 1

We see, TEA-I = 1 ratio, TEA-II = 1 ratio, TEA-III = 2+1 = 3 ratios

Therefore, the combined ratio = 1 : 1 : 3 (option ‘B’)

#### QUERY 38

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) 614%
C) 20%
D) 25%

MAHA GUPTA
Water is free, so cost price of 1 litre water = Rs 0
Let the cost price of 1 litre milk = Rs 100
Thus, as gain is 25%, cost price of 1 litre mixture of milk and water = 100 × 100125 = Rs 80
C.P. of 1 litre of milk       C.P. of 1 litre of water

 100 0 80 0 – 80 = 80 80 – 100 = 20

So, MILK : WATER = 80 : 20 = 4 : 1

Therefore, percentage of water in the mixture = 15 × 100 = 20% (option ‘C’)

ANOTHER METHOD
Let quantity of the mixture = 100 litres
Thus, as gain is 25%, quantity of milk in the mixture (water is free of cost) = 100 × 100125 = 80 litres

Hence, quantity of water in the mixture = 100 – 80 = 20 litres

But this quantity of water is out of 100 litres of the mixture, therefore, percentage of water in the mixture = 20% (option ‘C’)

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