__Allegations and Mixtures Practice Questions – Set 3__**1. Gold is 19 times as heavy as water and copper 9 times. In what ratio should these metals be mixed that the mixture may be 15 times as heavy as water?**

A) 1:2

B) 2:1

C) 2:3

D) 3 : 2

**Answer: D)**

**Explanation:**

Quantity
of Metal Gold=19, Copper = 9;

Resultant
= 15

Proportion
6 : 4

Gold
: Copper = 6 : 4 = 3 : 2

**2. A mixture has 100 kg of sugar, part of which he sells at 7 p.c. profit and the rest at 17 p.c. profit. He gains 10 p.c. on the whole. Find how much he sold at 7% profit?**

A) 70 kg
and 30 kg

B) 30 kg
and 70 kg

C) 15 kg
and 43 kg

D) 25 kg
and 30 kg

E) None of
these

**Answer: B)**

**Explanation:**

If
the cost price of each kind of sugar be Rs. 100 then the selling price of one
kind is Rs .107 (7% profit),Selling price of the other kind is Rs . 117 (17%
profit).

Selling
price of the mixture = Rs.110

SP
of Each R1=117,R2=107

Mixture
= 110

Proportion
3:7

Let
the quantity of each in 100 kg be 3x & 7x

3x
+ 7x = 100=>x=10.The quantity of each in 100 kg is 30kg and 70 kg

**3. What quantity of sugar costing Rs 6.10 per kg must be mixed with 126 kg of sugar priced at Rs. 2.85 per kg, so that 20% may be gained by selling the mixture at Rs 4.80 per kg?**

A) 49 kg

B) 59 kg

C) 69 kg

D) 68 kg

E) 72 kg

**Answer: C)**

**Explanation:**

R1=Sugar
costing Rs 2.85 per kg

R2=
Sugar costing Rs 6.10 per kg

RM=Average
cost of mixture

We
know that the mixture is being sold at a profit of 20%

ie,
cost price of mixture=(100/120)*4.80=Rs 4

RM
= Rs 4 = 400 paise

R1=285
paise

R2=610
paise

N1=210

N2=115

N1/N2
= 210/115 = 42/23

If
quantity of sugar at 285 paise per kg is 42 kg,

Then
sugar at 610 paise per kg = 23 kg

∴ if sugar
at 285 paise per kg is 126 kg,

Then
sugar at 610 paise per kg=69 kg

**4. 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) 1/3

B) 1/4

C) 1/5

D) 1/6

E) 1/7

**Answer: C)**

**Explanation:**

Suppose
the vessel initially contains 8 litres of liquid.

Let
x litres of this liquid be replaced with water.

Quantity
of water in new mixture = [3 – (3x/8) + x] litres

Quantity
of syrup in new mixture = [5 – (5x/8)] litres

Therefore,
[3 – (3x/8) + x] litres = [5 – (5x/8)] litres

=>
5x + 24 = 40 – 5x => 10x = 16 => x = 8/5

So,
part of the mixture replaced = 8/5 * 1/8 = 1/5

**5. One type of liquid contains 25 % of benzene, the other contains 30% of benzene. A can is filled with 6 parts of the first liquid and 4 parts of the second liquid. Find the percentage of benzene in the new mixture.**

A) 28%

B) 26%

C) 25%

D) 27%

E) 30%

**Answer: D)**

**Explanation:**

Let
the percentage of benzene =X

(30
- X)/(X- 25) = 6/4 = 3/2

=>
5X = 135 or

X
= 27 so,

required
percentage of benzene = 27 %

**6. A container has 40 litres of wine. From this container, 4 litres of wine is taken out and replaced with water. This process is repeated two more times. What will be the final quantity of water (in litres) in the container?**

A) 12

B) 10.84

C) 29.16

D) 28

E) None of
these

**Answer: B)**

**Explanation:**

Final
Qty of wine = starting Qty of container [1- Qty of wine withdrawn/ starting Qty
of container] ^ no.of times

=
40[1- 4/40]3

=
29.16 L

∴ final Qty
of water = 40 – 29.16 = 10.84 L.

**7. An alloy of zinc and tin contains 35% of zinc by weight. What quantity of zinc must be added to 400 lb of this alloy such that there is 60% of zinc by weight in the final mixture?**

A) 400 lb

B) 350 lb

C) 250 lb

D) 450 lb

E) None of
these

**Answer: C)**

**Explanation:**

For
Zinc

35 100

60(Mean)

40 25

Q1
: Q2 = 8: 5

=>
Q1/Q2 = 8/5 => 400 / Q2 = 8/5 => Q2 = 50 × 5 = 250 lb

**8. A painter mixes blue paint with white paint so that the mixture contains 10% blue paint. In a mixture of 40 litres paint how many litres blue paint should be added so that the mixture contains 20% of blue paint.**

A) 2.5
litres

B) 4litres

C) 5litres

D) 2 litres

E) 5.5
litres

**Answer: C)**

**Explanation:**

Percentage
of blue point, in pure Blue point =100%

So,

10 100

20

80 10

Ratio
is 8 : 1

ATQ,
8 = 40 then 1 = 5 litres

**9. Three vessels whose capacities are in the ratio of 6:3:2 are completely filled with milk and water. The ratio of milk and water in the mixture 2:3 , 4:2 and 5:2. Taking ¼ of first, 1/2 of second and ½ of third , new mixture kept in a new vessel. What is the percentage of water in the new mixture ?**

A) 42%

B) 42(2/14)%

C) 43%

D) 40%

E) None of
these

**Answer: B)**

**Explanation:**

Amount
of mixture taken from 3 vessels

1

^{st}vessel = 6 * ¼ = 3/2
2

^{nd}vessel = 3 * ½ = 3/2
3

^{rd}vessel = 2 * ½ = 1
Amount
of water in mix

1

^{st}vessel = 3/2 * 3/5 = 9/10
2

^{nd}vessel = 3/2 * 2/6 = ½
3

^{rd}vessel = 1 * 2/7 = 2/7
Therefore
required % = [(9/10) + (1/2) + (2/7)]/[(3/2) + (3/2) + 1] * 100

=
(236 * 2 * 100)/(140 * 8) = 42 2/14%

**10. 8 litres are drawn from a cask filled with wine and is then filled with water .This operation is performed three more times. The ratio of the quantity of wine now left in the cask to that of the total solution is 16:81.How much wine does the cask originally ?**

A) 20
litres

B) 22
litres

C) 24
litres

D) 26
litres

E) 28
litres

**Answer: C)**

**Explanation:**

When
the final amount of solute that is not replaced calculated as:

**Initial Amount * (Volume After Removal/Volume After Replacing)^n**

Final
ratio of solute not replaced to total is

**Initial Ratio * (Volume After Removal/Volume After Replacing)^n**

Here
let us assume the initial value is 1 then Wine .Let the Quantity of wine in the
cask originally be x

Therefore

1*
[(x-8)^4/x]

=[16/81]

x=24litres

Wine
originally in the cask is 24 Litres