MOA - 1
MIXTURE OF ALLIGATIONS
1.
2.
3.
4.
5.
6.
7.
Gold in 19 times as heavy as water and copper
9 times as heavy as water. In what ratio should these
metal be mixed, so that the mixture may be 15 times
as heavy as water ?
(a)
3:2
(b)
9:5
(c)
2:3
(d)
7:5
Two liquids are mixed in the ratio 3 : 5 and the
mixture is sold at Rs 120 with a profit of 20%. If
the first liquid is costlier than the second by Rs 2
per litre, find the cost of the costlier liquid per litre.
(a)
Rs 92.30
(b)
Rs 74.10
(c)
Rs 101.25
(d)
Rs 99.25
The average weekly salary per head of the entire
staff of a factory consisting of supervisors and the
labourers is Rs 60. The average salary per head of
the supervisors is Rs 400 and that of the labourers
is Rs 56. Find the number of labourers in the
factory.
(a)
986
(b)
1020
(c)
860
(d)
1260
A grocer has 50 kg of rice, a part of which he sells
at 8% profit and the rest at 18%. He gains 14% on
the whole. Find the quantity sold at 18% profit.
(a)
20 kg
(b)
30 kg
(c)
15 kg
(d)
35 kg
8.
9.
10.
A vessel contains milk and water in the ratio 3 : 2.
The volume of the contents is increased by 50% by
adding water to it. From this resultant solution
30 L is withdrawn and then replaced with water.
The resultant ratio of milk to water in the final
solution is 3 : 7. Find the original volume of the
solution.
(a)
80 L
(b)
65 L
(c)
75 L
(d)
82 L
A vessel contain 50 L milk. The milkman delivers
10 L to the first house and adds an equal quantity
of water. He does exactly the same at the second
and third house. What is the ratio of milk and water
when he has finished delivering at the third house.
(a)
61 : 64
(b)
27 : 37
(c)
16 : 9
(d)
none of these
Prabhu purchased 30 kg of rice at the rate of
Rs 17.50 per kg and another 30 kg rice at a certain
rate. He mixed the two rice and sold the entire
quantity at the rate of Rs 18.60 per kg and made
20% over all profit. At what price per kg did he
purchase the lot of another 30 kg rice ?
(a)
Rs 14.50
(b)
Rs 12.50
(c)
Rs 15.50
(d)
Rs 13.50
Sohan lal mixes 80 kg of sugar worth Rs 6.75 per
kg with 120 kg worth Rs 8 per kg. At what rate
shall he sell the mixture to gain 20% ?
(a)
Rs 7.50
(b)
Rs 9
(c)
Rs 8.20
(d)
Rs 8.85
A bucket contains a mixture of two liquids A and B
in the proportion 7 : 5. If 9 L of the mixture is
replaced by 9 L of liquid B, then the ratio of two
liquids become 7 : 9. How much of the liquids A
was there in the bucket.
(a)
36 L
(b)
18 L
(c)
21 L
(d)
30 L
A container contains 240 L of wine. 80 L is taken
out of the container everyday and an equal
quantity of water is put into it. Find the quantity of
the wine that remains in the container at the end of
the fourth day.
(a)
39.2 L
(b)
32 L
(c)
42.5 L
(d)
47.40 L
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
MOA - 2
ANSWERS
1.
a
2.
c
3.
b
8.
a
9.
c
10.
d
4.
b
5.
b
6.
c
7.
d
SOLUTIONS
1.
Using the rule of alligation we find the ratio of gold
and copper = 3 : 2 as calculated below
5.
Total cost price of 200 kg of mixtures
= Rs (80 × 6.75 + 120 × 8) = Rs 1500
Average rate = Rs 7.50 per kg

Required selling price
= 7.50 × 1.2
= Rs 9 per kg
6.
Let the liquid in A is 7x and B is 5x.
When 9L of liquid are taken out from mixture
2.
CP of the mixture = Rs 100

 7 
Then, A remains = 7 x  9

 75
100  x
3
  x  99.25
( x  2)  100 5
= 7x 
Hence, the cost of the costlier liquid
21
4
= Rs 99.25 + 2
15
 5 
B remains = 5x  9
  5x 
7

5
4


= Rs 101.25
3.
Now, when 9 L of liquid B are added

1 : 85

21  
15


 9  7 : 9
 7 x   :  5x 
4
4

 

21
7
4

15

 9
 9
 5x 
4


7x 
Number of supervisor 1. Hence, number
of labour must be multiple of 85 from
option, we check that number is 1020.
4.
84
3
28

x

7x = 7 × 3 = 21 L
4
=2:3

7.
Quantity sold at 18% profit =
50
 3  30 kg
2 3
V
80 


Required quantity = X 1    240 1 

X
240 


1

 240   1  
3


Einstein Classes,
4
240  16
 47.40 L
8
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
4
MOA - 3
8.
Let the original volume be x. Then, quantity of milk
3x
2x
and
respectively. After adding
5
5
water to it to make the volume 150%, the quantity
and water =
of milk and water =
Now,
9.
3x
9x
and
respecively..
5
10
3x
 12
3
5

9x
7
 12
10

14(3x – 60) = 3(9x + 120)

x = 80 L
10 

Final quantity of milk = 50 1  
50 

3
4 4 4
= 50     25.6 L
5 5 5
Hence, quantity of water left in the solution
= 40 – 25.6 = 14.4 L
10.
Now, given
15.5  x 30

2
30

15.5 – x = 2

x = 15.5 – 2 = Rs 13.50
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111