# 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

=> ^{x}⁄_{y} = ^{3.50}⁄_{1.50}

=> 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 ×

^{100}⁄

_{110 }= 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 ^{10}⁄_{100}(60x + 65y) = 68.20(x + y) – (60x + 65y)

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

or 2.2x = 3.3y

=> ^{x}⁄_{y} = ^{3.3}⁄_{2.2}

=> 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 × ^{100}⁄_{105} = Rs ^{40}⁄_{3}

**RICE-I RICE-II**

10 | 15 | |||

^{40}⁄_{3} |
||||

15 – ^{40}⁄_{3} = ^{5}⁄_{3} |
^{40}⁄_{3} – 10 = ^{10}⁄_{3} |

Hence, ratio of the two quantities = ^{5}⁄_{3} : ^{10}⁄_{3} = 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 = (^{5}⁄_{100})(10x + 15y)

So, (^{5}⁄_{100})(10x + 15y) = 14(x + y) – (10x + 15y)

=> ^{1}⁄_{2}x + ^{3}⁄_{4}y = 4x – y

=> 2x + 3y = 16x – 4y

=> 14x = 7y

=> ^{x}⁄_{y} = ^{7}⁄_{14}

=> 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) 6^{1}⁄_{4}%

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 × ^{100}⁄_{125} = 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 = ^{1}⁄_{5} × 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 × ^{100}⁄_{125} = 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’)