alligationproblems.html

aptitude book by r s aggarwal

                                                       Mean Price Rs 62
 
                                3                                                                            2
Required ratio =3:2.

11.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.

13.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. 


14.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:

15.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.
Cost of 1 Kg tea of 1st kind RS 130.50                        Cost of 1 Kg 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.

16.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 陆.
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 陆

             1/10                                                                                            1/14
Required ratio = 1/10  :  1/14 = 7:5.

17.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.

18.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.

19.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 = 戮 lit.
C.P of 1 liter mixture in 1st can =Re 戮
Milk in 1 liter mixture in 2n d  can = 1/2 lit.
C.P of 1 liter mixture in 2n d  can =Re 陆
Milk in 1 liter final mixture  = 5/8 lit.
Mean Price = Re 5/8.

C.P of 1 liter mixture in 1st can Re 戮                          C.P of 1 liter mixture in 2n d  can Re 陆

                                                        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.

20.One quantity of wheat at Rs 9.30 per 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 1 Kg wheat of 1st kind 980p                           C.P of 1 Kg wheat of 2n d kind 100x p

                                                      Mean Price 1000p

          100x-1000 p                                                                  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 

21.8lit 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.

22.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 鈥