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Aptitude conc

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APTITUDENumbers H.C.F and L.C.M Decimal Fractions Simplification Square and Cube roots Average Problems on Numbers Problems on Ages Surds and Indices Percentage Profit and Loss Ratio And Proportions Partnership Chain Rule Time and Work Pipes and Cisterns Time and Distance Trains Boats and Streams Alligation or Mixture Simple Interest Compound Interest Logorithms Areas Volume and Surface area Races and Games of Skill Calendar Clocks Stocks ans Shares True Discount Bankers Discount Oddmanout and Series Data Interpretation probability Permutations and Combinations Puzzles

ALLIGATION AND MIXTURESCONCEPT SIMPLE PROBLEMS MEDIUM PROBLEMS COMPLEX PROBLEMS

file:///E|/work/books/placement/09_Aptitude/alligation.html[1/28/2012 12:44:45 AM]

APTITUDENumbers H.C.F and L.C.M Decimal Fractions Simplification Square and Cube roots Average Problems on Numbers Problems on Ages Surds and Indices Percentage Profit and Loss Ratio And Proportions Partnership Chain Rule Time and Work Pipes and Cisterns Time and Distance Trains Boats and Streams Alligation or Mixture Simple Interest Compound Interest Logorithms Areas Volume and Surface area Races and Games of Skill Calendar Clocks Stocks ans Shares True Discount Bankers Discount Oddmanout and Series Data Interpretation probability Permutations and Combinations Puzzles BACK

ALLIGATION OR MIXTURES SOLVED PROBLEMSComplex 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 4/7 C.P of 1 liter mixture in B 2/5

Mean Price 1/10 1/14

file:///E|/work/books/placement/09_Aptitude/alligationcomplex.html[1/28/2012 12:44:47 AM]

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.

file:///E|/work/books/placement/09_Aptitude/alligationcomplex.html[1/28/2012 12:44:47 AM]

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

file:///E|/work/books/placement/09_Aptitude/alligationcomplex.html[1/28/2012 12:44:47 AM]

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. BACK

file:///E|/work/books/placement/09_Aptitude/alligationcomplex.html[1/28/2012 12:44:47 AM]

APTITUDENumbers H.C.F and L.C.M Decimal Fractions Simplification Square and Cube roots Average Problems on Numbers Problems on Ages Surds and Indices Percentage Profit and Loss Ratio And Proportions Partnership Chain Rule Time and Work Pipes and Cisterns Time and Distance Trains Boats and Streams Alligation or Mixture Simple Interest Compound Interest Logorithms Areas Volume and Surface area Races and Games of Skill Calendar Clocks Stocks ans Shares True Discount Bankers Discount Oddmanout and Series Data Interpretation probability Permutations and Combinations Puzzles BACK

ALLIGATION OR MIXTURES

Important Facts and Formula: 1.Allegation:It is the rule that enables us to find the ratio in which two of more ingredients at the given price must be mixed to produce a mixture of a desired price. 2.Mean Price:The cost price of a unit quantity of the mixture is called the mean price. 3.Rule of Allegation:If two ingredients are mixed then Quantity of Cheaper / Quantity of Dearer = (C.P of Dearer Mean Price) /(Mean PriceC.P of Cheaper).C.P of a unit quantity of cheaper(c) C.P of unit quantity of dearer(d)

Mean Price(m) (d-m) (m-c)

Cheaper quantity:Dearer quantity = (d-m):(m-c) 4.Suppose a container contains x units of liquid from which y units are taken out and replaced by water. After n operations the quantity of pure liquid = x (1 y/x)n units. BACK

file:///E|/work/books/placement/09_Aptitude/alligationconcept.html[1/28/2012 12:44:47 AM]

APTITUDENumbers H.C.F and L.C.M Decimal Fractions Simplification Square and Cube roots Average Problems on Numbers Problems on Ages Surds and Indices Percentage Profit and Loss Ratio And Proportions Partnership Chain Rule Time and Work Pipes and Cisterns Time and Distance Trains Boats and Streams Alligation or Mixture Simple Interest Compound Interest Logorithms Areas Volume and Surface area Races and Games of Skill Calendar Clocks Stocks ans Shares True Discount Bankers Discount Oddmanout and Series Data Interpretation probability Permutations and Combinations Puzzles BACK

ALLIGATION OR MIXTURES SOLVED PROBLEMSMedium 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 o

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