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
1st
vessel = 6 * ¼ = 3/2
2nd
vessel = 3 * ½ = 3/2
3rd
vessel = 2 * ½ = 1
Amount
of water in mix
1st
vessel = 3/2 * 3/5 = 9/10
2nd
vessel = 3/2 * 2/6 = ½
3rd
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
