business mathematics

Allegation or Mixtures problems

Allegation or Mixtures problems


Simple Problems

1.In what ratio must rice at Rs 9.30 per Kg be mixed with rice
at Rs 10.80 per Kg so that the mixture be worth Rs 10 per Kg?
Solution:

C.P of 1 Kg rice of 1st kind 930 p
C.P of 1 Kg rice of 2n d kind 1080p

Mean Price 1000p

80 70
Required ratio=80:70 = 8:7

2.How much water must be added to 60 liters of milk at
11/2 liters for Rs 20 so as to have a mixture worth
Rs 10 2/3 a liter?

Solution: C.P of 1 lit of milk = 20*2/3 = 40/3
C.P of 1 lit of water 0 C.P of 1 lit of milk 40/3
Mean Price 32/3

8/3 32/3
Ratio of water and milk =8/3 : 32/3 = 1:4
Quantity of water to be added to 60 lit of milk
=1/4*60=15 liters.

3.In what ratio must water to be mixed with milk to gain
20% by selling the mixture at cost price?

Solution:Let the C.P of milk be Re 1 per liter
Then S.P of 1 liter of mixture = Re.1
Gain obtained =20%.
Therefore C.P of 1 liter mixture = Rs(100/120*1) =5/6
C.P of 1 liter of water 0 C.P of 1 liter of milk1

Mean Price 5/6

1/6 5/6

Ratio of water and milk =1/6 : 5/6 = 1:5.

4.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 per Kg?

Solution:

Cost of 1 Kg pulses of 1 kind 15 Cost of 1 Kg pulses of
2nd kind 20
Mean Price Rs 16.50

3.50 1.50
Required ratio =3.50 : 1.50 = 35:15 = 7:3.

5. 4Kg s of rice at Rs 5 per Kg is mixed with 8 Kg of rice
at Rs 6 per Kg .Find the average price of the mixture?

Solution:
rice of 5 Rs per Kg rice of 6 Rs per Kg
Average price Aw
6-Aw Aw-5
(6-Aw)/(Aw-5) = 4/8 =1/2
12-2Aw = Aw-5
3Aw = 17
Aw = 5.66 per Kg.

6.5Kg of rice at Rs 6 per Kg is mixed with 4 Kg of rice to
get a mixture costing Rs 7 per Kg. Find the price
of the costlier rice?

Solution: Using the cross method:

rice at Rs 6 per Kg rice at Rs x per Kg
Mean price Rs 7 per Kg
5 4
x-7:1=5:4
4x-28 = 5
4x=33=>x=Rs 8.25.
Therefore price of costlier rice is Rs 8.25 per Kg

Medium Problems

1.A butler stole wine from a butt of sherry which contained
40% of spirit and he replaced,what he had stolen by wine
containing only 16% spirit. The butt was then of 24%
strength only. How much of the butt did he steal?

Solution:
Wine containing 40%spirit Wine containing 16% spirit
Wine containing 24% spirit
8 16
They must be mixed in the ratio of =1:2.
Thus 1/3 of the butt of sherry was left and hence the
butler drew out 2/3 of the butt.

2.The average weekly salary per head of the entire staff
of a factory consisting of supervisors and the laborers
is Rs 60.The average salary per head of the supervisors
is Rs 400 and that of the laborers is Rs 56.Given that
the number of supervisors is 12.Find the number of
laborers in the factory.

Solution:
Average salary of laborer Rs 56 Average salary of
supervisors Rs 400
Average salary of entire staff Rs 60
340 4
Number of laborer / Number of Supervisors = 340 / 4=85/1
Thus,if the number of supervisors is 1,number of
laborers =85.
Therefore if the number of supervisors is 12 number of
laborers 85*12=1020.

3.The cost of type 1 rice is Rs 15 per Kg and type 2 rice
is Rs 20 per Kg. If both type1 and type 2 are mixed in the
ratio of 2:3,then the price per Kg of the mixed variety
of rice is?

Solution:Let the price of the mixed variety be Rs x per Kg.
Cost of 1 Kg of type 1 rice Rs 15 Cost of 1 Kg of type 2
rice Rs 20
Mean Price Rs x
20-x x-15
(20-x) /( x-15) = 2/3
=> 60-3x = 2x-30
5x = 90=>x=18.

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

Solution:S.P of 1 Kg of the mixture = Rs 68.20,gain =10%
S.P of 1 Kg of the mixture = Rs (100/110*68.20)=Rs 62.
Cost of 1 Kg tea of 1st kind 60 Cost of 1 Kg tea of 2nd
kind 65
Mean Price Rs 62
3 2
Required ratio =3:2.

5.A dishonest milkman professes to sell his milk at cost price
but he mixes t with water and there by gains 25% .The
percentage of water in the mixture is?

Solution:Let C. P of 1 liter milk be Re 1.
Then S.P of 1 liter mixture=Re 1. Gain=25%
C.P of 1 liter mixture =Re(100/125*1) = Re 4/5.
C.P of 1 liter milk Re 1 C.P of 1 liter of water 0
Mean Price 4/5
4/5 1/5
Ratio of milk to water =4/5 : 1/5 = 4:1
Hence percentage of water n the mixture=1/5*100=20%.

12.A merchant has 1000Kg of sugar,part of which he sells
at 8% profit and the rest at 18% profit. He gains 14% on the
whole. The quantity sold at 18% profit is?

Solution:
Profit on 1st part 8% Profit on 2nd part 18%
Mean Profit 14%
4 6
Ratio of 1st and 2nd parts =4:6 =2:3.
Quantity of 2nd ind =3/5*1000Kg =600 Kg.

6.A jar full of whiskey contains 40% alcohol. A part of
this whiskey is replaced by another containing 19% alcohol
and now the percentage of alcohol was found to be 26%.
The quantity of whiskey replaced is?

Solution:
Strength of first jar 40% Strength of 2nd jar 19%
Mean Strength 26%
7 14
So,ratio of 1st and 2nd quantities =7:14 =1:2
Therefore required quantity replaced =2/3.

7.A container contains 40lit of milk. From this container
4 lit of milk was taken out and replaced by water.
This process was repeated further two times.
How much milk is now contained by the container?

Solution:Amount of milk left after 3 operations = 40(1-4/40)3lit
=(40*9/10*9/10*9/10)
= 29.16 lit

Complex Problems

1.Tea worth Rs 126 per Kg are mixed with a third variety in
the ratio 1:1:2. If the mixture is worth Rs 153 per Kg ,
the price of the third variety per Kg will be?

Solution:
Since First and second varieties are mixed in equal proportions
so their average price =Rs (126+135)/2 = 130.50.

So the mixture is formed by mixing two varieties ,one at
Rs 130.50 per Kg and the other at say Rs x per Kg in the
ratio 2:2 i e,1:1 we have to find x.
Costof 1Kg tea of 1st kind RS 130.50 Costof 1Kg tea of 2n d
kind Rs x.
Mean Price Rs 153
x-153 22.50
(x=153)/22.5 = 1 =>x-153 = 22.5
x = 175.50.
Price of the third variety =Rs 175.50 per Kg.

2.The milk and water in two vessels A and B are in the ratio 4:3
and 2:3 respectively. In what ratio the liquids in both the
vessels be mixed to obtain a new mixture in vessel c
consisting half milk and half water?

Solution:Let the C.P of milk be Re 1 per liter.
Milk in 1 liter mixture of A = 4/7 liter.
Milk in 1 liter mixture of B = 2/5 liter.
Milk in 1 liter mixture of C = 1/2 liter.
C.P of 1 liter mixture in A=Re 4/7
C.P of 1 liter mixture in B=Re 2/5.
Mean Price = Re 1/2.
By rule of allegation we have:
C.P of 1 liter mixture in A C.P of 1 liter mixture in B
4/7 2/5
Mean Price ½
1/10 1/14
Required ratio = 1/10 : 1/14 = 7:5.

3.How many Kg s of wheat costing him Rs 1.20,Rs 1.44
and Rs 1.74 per Kg so that the mixture may be worth
Rs 1.41 per Kg?

Solution:
Step1:Mix wheat of first and third kind to get a mixture
worth Rs 1.41 per Kg.
C.P of 1 Kg wheat of 1st kind 120p C.P of 1 Kg wheat of
3rd kind 174p
Mean Price 141p
33 21
They must be mixed in the ratio =33:21 = 11:7

Step2:Mix wheats of 1st and 2n d kind to obtain a mixture
worth of 1.41.per Kg.
C.P of 1 Kg wheat of 1st kind 120p C.P of 1 Kg wheat of 2n d
kind 144p
Mean Price 141p
3 21
They must be mixed in the ratio = 3:21=1:7.
Thus,Quantity of 2n d kind of wheat / Quantity of
3rd kind of wheat = 7/1*11/7= 11/1
Quantities of wheat of 1st :2n d:3rd = 11:77:7.

4.Two vessels A and B contain spirit and water mixed in
the ratio 5:2 and 7:6 respectively. Find the ratio n which
these mixture be mixed to obtain a new mixture in vessel
c containing spirit and water in the ratio 8:5?

Solution:Let the C.P of spirit be Re 1 per liter.
Spirit in 1 liter mix of A = 5/7 liter.
C.P of 1 liter mix in A =5/7.
Spirit in 1 liter mix of B = 7/13 liter.
C.P of 1 liter mix in B =7/13.
Spirit in 1 liter mix of C = 8/13 liter.
C.P of 1 liter mix in C =8/13.
C.P of 1 liter mixture in A 5/7 C.P of 1 liter mixture
in B 7/13
Mean Price 8/13
1/13 9/91
Therefore required ratio = 1/13 : 9/91 = 7:9.


5.A milk vendor has 2 cans of milk .The first contains 5% water
and the rest milk. The second contains 50% water. How much milk
should he mix from each of the container so as to get 12 liters
of milk such that the ratio of water to milk is 3:5?

Solution:Let cost of 1 liter milk be Re 1.
Milk in 1 liter mixture in 1st can = 3/4 lit.
C.P of 1 liter mixture in 1st can =Re 3/4
Milk in 1 liter mixture in 2n d can = 1/2 lit.
C.P of 1 liter mixture in 2n d can =Re 1/2
Milk in 1 liter final mixture = 5/8 lit.
Mean Price = Re 5/8.
C.P of 1 lt mix in 1st Re3/4 C.P of 1 lt mix in 2nd Re1/2
Mean Price 5/8
1/8 1/8
There ratio of two mixtures =1/8 :1/8 = 1:1.
So,quantity of mixture taken from each can=1/2*12
= 6 liters.

6.One quantity of wheat at Rs 9.30 p
er Kg are mixed
with another quality at a certain rate in the ratio 8:7.
If the mixture so formed be worth Rs 10 per Kg ,what is
the rate per Kg of the second quality of wheat?

Solution:Let the rate of second quality be Rs x per Kg.
C.P of 1Kg wheat of 1st 980p C.P of 1 Kg wheat of 2nd 100x p
Mean Price 1000p
100x-1000p 70 p
(100x-1000) / 70 = 8/7
700x -7000 = 560
700x = 7560 =>x = Rs 10.80.
Therefore the rate of second quality is Rs10.80

7.8lit are drawn from a wine and is then filled with water.
This operation is performed three more times.The ratio of
the quantity of wine now left in cask to that of the water
is 16:81. How much wine did the cask hold originally?

Solution:
Let the quantity of the wine in the cask originally be
x liters.
Then quantity of wine left in cask after
4 operations = x(1- 8/x)4lit.
Therefore x((1-(8/x))4)/x = 16/81.
(1- 8/x)4=(2/3) 4
(x- 8)/x=2/3
3x-24 =2x
x=24.

8.A can contains a mixture of two liquids A and B in the
ratio 7:5 when 9 liters of mixture are drawn off and the
can is filled with B,the ratio of A and B becomes 7:9.
How many liters of liquid A was contained by the can initially?

Solution:
Suppose the can initially contains 7x and 5x liters
of mixtures A and B respectively .
Quantity of A in mixture left = (7x- (7/12)*9 )lit
= 7x - (21/4) liters.
Quantity of B in mixture left = 5x - 5/12*9
= 5x - (15/4) liters

Therefore (7x – 21/4)/ (5x – 15/4+9)=7/9
(28x-21)/(20x +21)= 7/9
(252x -189)= 140x +147
112x = 336
=> x=3.
So the can contains 21 liters of A.

9.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?

Solution:
Suppose the vessal initially contains 8 liters of liquid.
Let x liters of this liquid be replaced with water
then quantity of water in new mixture
= 3-(3x/8)+x liters.
Quantity of syrup in new mixture = 5 - 5x/8 liters.
Therefore 3 - 3x/8 +x = 5 - 5x/8
5x+24 = 40-5x
10x = 16.
x= 8/5.
So part of the mixture replaced = 8/5*1/8 =1/5.