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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

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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 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

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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


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. 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

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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 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.

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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 3 60 Cost of 1 Kg tea of 2nd kind 65 2

Mean Price Rs 62

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


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 BACK


Alligation or MixturesImportant Facts and Formulae: 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. 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 Required ratio =3.50 : 1.50 = 35:15 = 7:3. 1.50

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

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6-Aw (6-Aw)/(Aw-5) = 4/8 =1/2 12-2Aw = Aw-5 3Aw = 17 Aw = 5.66 per Kg.

Aw-5

Top6.5Kg 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 Mean Profit 14% 4 Ratio of 1st and 2nd parts =4:6 Quantity of 2nd ind =3/5*1000Kg on 2nd part 18% 6 =2:3. =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

TopComplex 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 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.

Top5.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.

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ALLIGATION OR MIXTURES

SOLVED 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: 7.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. 8.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. 9.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 (20x) /( x-15) = 2/3 => 60-3x = 2x-30 5x = 90=>x=18. 10.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. 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 . By rule of allegation we have: C.P of 1 liter mixture in A C.P of 1 liter mixture

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in B 4/7 2/5 Mean Price 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 = lit. C.P of 1 liter mixture in 1st can =Re Milk in 1 liter mixture in 2n d can = 1/2 lit. C.P of 1 liter mixture in 2n d can =Re Milk in 1 liter final mixture = 5/8 lit. Mean Price = Re 5/8. C.P of 1 liter mixture in 1st can Re C.P of 1 liter mixture in 2n d can Re 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 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. 23.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.

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ALLIGATION OR MIXTURES SOLVED PROBLEMSSimple 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


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 = 40/3C.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) =5/6C.P of 1 liter of water 0 C.P of 1 liter of milk1


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

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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 BACK

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BACK Complex Probems: 1.If the area of a square with side a s equal to the area of a triangle with base a, then the altitude of the triangle is sol: area of a square with side a = a sq unts area of a triangle with base a = * a*h sq unts a =1/2 *a *h => h = 2a altitude of the triangle is 2a 2.An equilateral triangle is described on the diagonal of a square. What is the ratio of the area of the triangle to that of the square? Sol: area of a square = a sq cm length of the diagonal = 2a cm area of equilateral triangle with side 2a = 3/4 * (2a) required ratio = 3a : a = 3 : 2 3.The ratio of bases of two triangles is x:y and that of their areas is a:b. Then the ratio of their corresponding altitudes wll be sol: a/b =( * x*H) /(1/2 * y * h) bxH = ayh =>H/h =ay/bx Hence H:h = ay:bx 4 .A parallelogram has sides 30m and 14m and one of its diagonals is 40m long. Then its area is sol: let ABCD be the given parallelogram area of parallelogram ABCD = 2* (area of triangle ABC) now a = 30m, b = 14m and c = 40m s = (30+14+40) = 42m Area of triangle ABC = [ s(s-a)(s-b)(s-c) = (42*12*28*2 = 168sq m area of parallelogram ABCD = 2 *168 =336 sq m 5.If a parallelogram with area p, a triangle with area R and a triangle with area T are all constructed on the same base and all have the same altitude, then which of the following statements is false? Sol: let each have base = b and height = h then p = b*h, R = b*h and T = * b*h so P = R, P = 2T and T = R are all correct statements 6.If the diagonals of a rhombus are 24cm and 10cm the area and the perimeter of the rhombus are respectively. Sol: area = *diagonal 1 *diagonal 2= * 24 * 10= 120 sq cm * diagonal 1 = * 24 = 12cm * diagonal 2 = *10 =5 cm side of a rhombus = (12) + (5) = 169 => AB = 13cm

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7.If a square and a rhombus stand on the same base, then the ratio of the areas of the square and the rhombus is: sol: A square and a rhombus on the same base are equal in area 8.The area of a field in the shape of a trapezium measures 1440sq m. The perpendicular distance between its parallel sides is 24cm. If the ratio of the sides is 5:3, the length of the longer parallel side is: sol: area of field =1/2 *(5x+3x) *24 = 96x sq m 96x = 1440 => x = 1440 /96 = 15 hence, the length of longer parallel side = 5x = 75m 9.The area of a circle of radius 5 is numerically what percent its circumference? Sol: required percentage = (5)/(2 *5) *100 = 250% 10.A man runs round a circular field of radius 50m at the speed of 12m/hr. What is the time taken by the man to take twenty rounds of the field? Sol: speed = 12 k/h = 12 * 5/18 = 10/3 m/s distance covered = 20 * 2*22/7*50 = 44000/7m time taken = distance /speed = 44000/7 * 3/10 = 220/7min 11.A cow s tethered in the middle of a field with a 14feet long rope.If the cow grazes 100 sq feet per day, then approximately what time will be taken by the cow to graze the whole field? Sol: area of the field grazed = 22/7 * 14 * 14 = 616 sq feet 12.A wire can be bent in the form of a circle of radius 56cm. If it is bent in the form of a square, then its area will be sol: length of wire = 2 r = 2 *22/7 *56 = 352 cm side of the square = 352/4 = 88cm area of the square = 88*88 = 7744sq cm 13.The no of revolutions a wheel of diameter 40cm makes in traveling a distance of 176m is sol: distance covered in 1 revolution = 2 r = 2 *22/7 *20 = 880/7 cm required no of revolutions = 17600 *7/880 = 140 14.The wheel of a motorcycle 70cm in diameter makes 40 revolutions in every 10sec. What is the speed of motorcycle n km/hr? Sol: distance covered in 10sec = 2 *22/7 *35/100 *40 =88m distance covered in 1 sec =88/10m = 8.8m speed =8.8m/s = 8.8 * 18/5 *k/h = 31.68 k/h 15.Wheels of diameters 7cm and 14cm start rolling simultaneously from x & y which are 1980 cm apart towards each other in opposite directions. Both of them make the same number of revolutions per second. If both of them meet after 10seconds. The speed of the smaller wheel is sol: let each wheel make x revolutions per sec. Then (2 *7/2 *x)+(2 * 7*x)*10 = 1980 (22/7 *7 * x) + (2 * 22/7 *7 *x) = 198


66x = 198 => x = 3 distance moved by smaller wheel in 3 revolutions = 2 *22/7 *7/2 *3 = 66cm speed of smaller wheel = 66/3 m/s = 22m/s 16.A circular swimming pool is surrounded by a concrete wall 4ft wide. If the area of the concrete wall surrounding the pool is 11/25 that of the pool, then the radius of the pool is? Sol: let the radius of the pool be R ft radius of the pool including the wall = (R+4)ft area of the concrete wall = [(R+4)2 - R2 ] => = [R+4+R][R+4-R] = 8 (R+2) sq feet 8 (R+2) = 11/25 R2 => 11 R2 = 200 (R+2) Radius of the pool R = 20ft 17.A semicircular shaped window has diameter of 63cm. Its perimeter equals sol: perimeter of window = r +2r = [22/7 * 63/2 +63] = 99+63 = 162 cm 63.The area of the largest triangle that can be inscribed in a semicircle of radius is sol: required area = * base * height = * 2r * r = r 2 18.Three circles of radius 3.5cm are placed in such a way that each circle touches the other two. The area of the portion enclosed by the circles is sol: required area = (area of an equilateral triangle of side 7 cm) - (3 * area of sector with = 6o degrees and r = 3.5cm) = ( * 7 * 7) (3* 22/7 *3.5 *3.5*60/360 ) sq cm = 493/4 11*0.5*3.5 sq cm = 1.967 sq cm 19. Four circular cardboard pieces, each of radius 7cm are placed in such a way that each piece touches two other pieces. The area of the space encosed by the four pieces is sol: required area = 14*14 (4 * * 22/7 * 7 *7) sq cm = 196 154 = 42 sq cm



BACK Medium Problems: 11.Find the ratio of the areas of the incircle and circumcircle of a square. Sol: let the side of the square be x, then its diagonal = 2 x radius of incircle = x/2 and radius of circmcircle =2 x /2 = x/2 required ratio = x/4 : x/2 = : = 1:2 12.If the radius of a circle is decreased by 50% , find the percentage decrease in its area. Sol: let original radius = r and new radius = 50/100 r = r/2 original area = r and new area = (r/2) decrease in area = 3 r/4 * 1/ r * 100 = 75% 13.Two concentric circles form a ring. The inner and outer circumference of the ring are 352/7 m and 528/7m respectively. Find the width of the ring. sol: let the inner and outer radii be r and R meters then, 2 r = 352/7 => r = 352/7 * 7/22 * = 8m 2 R = 528/7 => R= 528/7 * 7/22 * = 12m width of the ring = R-r = 12-8 = 4m 14.If the diagonal of a rectangle is 17cm long and its perimeter is 46 cm. Find the area of the rectangle. sol: let length = x and breadth = y then 2(x+y) = 46 => x+y = 23 x+y = 17 = 289 now (x+y) = 23 =>x+y+2xy= 529 289+ 2xy = 529 => xy = 120 area =xy=120 sq. cm 15.A rectangular grassy plot 110m by 65cm has a gravel path .5cm wide all round it on the inside. Find the cost of gravelling the path at 80 paise per sq.mt sol: area of theplot = 110 * 65 = 7150 sq m area of the plot excluding the path = (110-5)* (65-5) = 6300 sq m area of the path = 7150- 6300 =850 sq m cost of gravelling the path = 850 * 80/100 = 680 Rs 16. The perimeters of ttwo squares are 40cm and 32 cm. Find the perimeter of a third square whose area is equal to the difference of the areas of the two squares. sol: side of first square = 40/4 =10cm side of second square = 32/4 = 8cm area of third squre = 10 8 = 36 sq cm side of third square = 36 = 6 cm required perimeter = 6*4 = 24cm

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17. A room 5m 44cm long and 3m 74cm broad is to be paved with squre tiles. Find the least number of squre tiles required to cover the floor. sol: area of the room = 544 * 374 sq cm size of largest square tile = H.C.F of 544cm and 374cm= 34cm area of 1 tile = 34*34 sq cm no. of tiles required = (544*374) / (34 * 34) = 176 18. The diagonals of two squares are in the ratio of 2:5. Find the ratio of their areas. sol: let the diagonals of the squares be 2x and 5x respectively ratio of their areas = * (2x) : *(5x) = 4:25 19.If each side of a square is increased by 25%. Find the percentage change in its area. sol: let each side of the square be a then area = a new side = 125a/100 = 5a/4 new area = (5a/4) = 25/16 a increase in area = (25/16)a - a = (9/16)a increase % = (9/16)a * (1/a) * 100 = 56.25% 20.The base of triangular field os three times its altitude. If the cost of cultivating the field at Rs. 24.68 per hectare be Rs. 333.18. Find its base and height. sol: area of the field = total cost/ rate = 333.18 /24.68 = 13.5 hectares => = 13.5 * 10000 = 135000 sq m let the altitude = x mt and base = 3x mt then *3x * x = 135000 => x = 90000 => x = 300 base= 900m and altitude = 300m 21.In two triangles the ratio of the areas is 4:3 and the ratio of their heights is 3:4. Find the ratio of their bases? Sol: let the bases of the two triangles be x &y and their heights be 3h and 4h respectively. (1/2*x*3h)/(1/2*y*4h) =4/3 => x/y = 4/3 *4/3 = 16/9 22.Find the length of a rope by which a cow must be tethered in order that it may be able to graze an area of 9856 sq meters. Sol: clearly the cow will graze a circular field of area 9856 sq m and radius equal to the length of the rope. Let the length of the rope be r mts then r=9856 => r=9856*7/22 = 3136 => r=56m 23.The diameter of the driving wheel of a bus is 140cm. How many revolutions per minute must the wheel make inorder to keep a speed of 66 kmph? Sol: Distance to be covered in 1min = (66*1000)/60 m =1100m diameter = 140cm => radius = r =0.7m circumference of the wheel = 2*22/7*0.7 = 4.4m no of revolutions per minute = 1100/4.4 = 250 24.The inner circumference of a circular race track, 14m wide is 440m. Find the radius of the outer circle. Sol: let inner radius be r meters.


Then 2 r =440 => r=440*7/22*1/2 = 70m radius of outer circle = 70+4 =84m 25.A sector of 120 degrees, cut out from a circle, has an area of 66/7 sq cm. Find the radius of the circle. Sol: let the radius of the circle be r cm. Then r/360 =66/7=> 22/7*r*120/360 = 66/7 =>r = 66/7 *7/22*3 =9 radius = 3cm 26.The length of the room is 5.5m and width is 3.75m. Find the cost of paving the floor by slabs at the rate of Rs.800 per sq meter. Sol: l=5.5m w=3.75m area of the floor = 5.5 * 3.75 = 20.625 sq m cost of paving = 800 *20.625 =Rs. 16500 27.A rectangular plot measuring 90 meters by 50 meters is to be enclosed by wire fencing. If the poles of the fence are kept 5 meters apart. How many poles will be needed? Sol: perimeter of the plot = 2(90+50) = 280m no of poles =280/5 =56m 28.The length of a rectangular plot is 20 meters more than its breadth. If the cost of fencing the plot @ 26.50 per meter is Rs. 5300. What is the length of the plot in meter? Sol: let breadth =x then length = x+20 perimeter = 5300/26.50 =200m 2(x+20+x) =200 => 4x+40 =200 x = 40 and length = 40+20 = 60m 29.A rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq feet, how many feet of fencing will be required? Sol: l=20feet and l*b=680 => b= 680/20 = 34feet length of fencing = l+2b = 20+68 =88 feet 30.A rectangular paper when folded into two congruent parts had a perimeter of 34cm foer each part folded along one set of sides and the same is 38cm. When folded along the other set of sides. What is the area of the paper? Sol: when folded along the breadth we have 2(l/2 +b) = 34 or l+2b = 34...........(1) when folded along the length, we have 2(l+b/2)=38 or 2l+b =38............(2) from 1 &2 we get l=14 and b=10 Area of the paper = 14*10 = 140 sq cm 31.A took 15 seconds to cross a rectangular field diagonally walking at the rate of 52 m/min and B took the same time to cross the same field along its sides walking at the rate of 68m/min. The area of the field is? Sol: length of the diagonal = 52*15/60 =13m sum of length and breadth = 68*15/60 = 17m (l+b)=13 or l+b = 17 area =lb = (2lb) = [(l+b) (l+b)] = [17 -169]


=1/2*120 = 60 sq meter 32 . A rectangular lawn 55m by 35m has two roads each 4m wide running in the middle of it. One parallel to the length and the other parallel to breadth. The cost of graveling the roads at 75 paise per sq meter is sol: area of cross roads = 55*4 +35*4-4*4 = 344sq m cost of graveling = 344 *75/100 =Rs. 258 33.The cost of fencing a square field @ Rs. 20 per metre is Rs.10.080.How much will it cost to lay a three meter wide pavement along the fencing inside the field @ Rs. 50 per sq m sol: perimeter = total cost / cost per m = 10080 /20 = 504m side of the square = 504/4 = 126m breadth of the pavement = 3m side of inner square = 126-6 = 120m area of the pavement = (126*126)-(120*120)=246*6 sq m cost of pavement = 246*6*50 = Rs. 73800 34.Amanwalked diagonally across a square plot. Approximately what was the percent saved by not walking along the edges? Sol: let the side of the square be x meters length of two sides = 2x meters diagonal = 2 x = 1.414x m saving on 2x meters = .59x m saving % = 0.59x /2x *100% = 30% (approx) 36.A man walking at the speed of 4 kmph crosses a square field diagonally in 3 meters. The area of the field is sol: speed of the man = 4*5/18 m/sec = 10/9 m/sec time taken = 3*60 sec = 180 sec length of diagonal = speed * time = 10/9 * 180 = 200m Area of the field = *(dioagonal) = * 200*200 sq m = 20000sq m 37.A square and rectangle have equal areas. If their perimeters are p and q respectively. Then sol: A square and a rectangle with equal areas will satisfy the relation p < q 38.If the perimeters of a square and a rectangle are the same, then the area a & b enclosed by them would satisfy the condition: sol: Take a square of side 4cm and a rectangle having l=6cm and b=2cm then perimeter of square = perimeter of rectangle area of square = 16 sq cm area of rectangle = 12 sq cm Hence a >b 39.An error of 2% in excess is made while measuring the side of a square. The percentage of error in the calculated area of the square is sol: 100cm is read as 102 cm a = 100*100 sq cm and b = 102 *102 sq cm then a-b = 404 sq cm


percentage error = 404/(100*100) = 4.04% 40.A tank is 25m long 12m wide and 6m deep. The cost of plastering its walls and bottom at 75 paise per sq m is sol: area to be plastered = [2(l+b)*h]+(l*b) = 2(25+12)*6 + (25*12)= 744 sq m cost of plastering = Rs . 744*75/100 = Rs. 5581 41.The dimensions of a room are 10m*7m*5m. There are 2 doors and 3 windows in the room. The dimensions of the doors are 1m*3m. One window is of size 2m*1.5m and the other 2 windows are of size 1m*1.5m. The cost of painting the walls at Rs. 3 per sq m is sol: Area of 4 walls = 2(l+b)*h =2(10+7)*5 = 170 sq m Area of 2 doors and 3 windows = 2(1*3)+(2*1.5)+2(1*1.5) = 12 sq m area to be planted = 170 -12 = 158 sq m cost of painting = Rs. 158 *3 = Rs. 474 42.The base of a triangle of 15cm and height is 12cm. The height of another triangle of double the area having the base 20cm is sol: a = *15*12 = 90 sq cm b = 2a = 2 * 90 = * 20 *h => h= 18cm 43.The sides of a triangle are in the ratio of :1/3:1/4. If the perimeter is 52cm, then the length of the smallest side is sol: ratio of sides = :1/3 :1/4 = 6:4:3 perimeter = 52 cm, so sides are 52*6/13 =24cm 52*4/13 = 16cm 52 *3/13 = 12cm length of smallest side = 12cm 44.The height of an equilateral triangle is 10cm. Its area is sol: a = (a/2) +(10) a a/4 = 100 =>3a = 100*4 area = 3/4 *a = 3/4*400/3 = 100/3 sq cm 45.From a point in the interior of an equilateral triangle, the perpendicular distance of the sides are 3 cm, 23cm and 53cm. The perimeter of the triangle is sol: let each side of the triangle be a cm then area(AOB) +area(BOC)+area(AOC) = area(ABC) * a *3 +1/2 *a *23 +1/2 * a*53 = 3/4 a a/23(1+2+5) = 3/4 a => a=16 perimeter = 3*16 = 48cm


AreasImportant Facts and Formulae: Results On Triangle 1.Sum of the angles of a triangle is 180 degrees. 2.The sum of any two sides of a triangle is greater than third side. 3.Pythagoras Theorem: In a right angled triangle (Hypotenuse)2 = (Base)2 +(Height)2 4.The line joining the mid point of a side of a triangle to the opposite vertex is called the MEDIAN. 5.The point where the three medians of a triangle meet, is called CENTROID. The centroid divides each of the medians in the ratio 2:1 6.In an isosceles triangle, the altitude from the vertex bisects the base 7.The median of a triangle divides it into two triangles of the same area. 8.The area of the triangle formed by joining the mid points of the sides of a given triangle is one - fourth of the area of the given triangle. Results On Quadrilaterals 1.The diagonals of a Parallelogram bisect each other. 2.Each diagonal of a Parallelogram divides it into two triangles of the same area. 3.The diagonals of a Rectangle are equal and bisect each other 4.The diagonals of a Square are equal and bisect each other at right angles. 5.The diagonals of a Rhombus are unequal and bisect each other at right angles. 6.A Parallelogram and a Rectangle on the same base and between the same parallels are equal in area. 7.Of all he parallelogram of given sides the parallelogram which is a rectangle has the greatest area.

TopFormulae 1.Area of a RECTANGLE = length * breadth Length = (Area/Breadth) and Breadth = (Area/Length) 2.Perimeter of a RECTANGLE = 2(Length + Breadth) 3.Area of a SQUARE = (side)2 = ( Diagonal)2 4.Area of four walls of a room = 2(length + breadth) * height 5.Area of a TRIANGLE = * base * height 6.Area of a TRIANGLE = [s * (s - a) * (s - b) * (s - c)], where a,b,c are the sides of the triangle and s = 1/2(a+b+c) 7.Area of EQUILATERAL TRIANGLE = (3/4)* (side)2 8.Radius of in circle of an EQUILATERAL TRIANGLE of side a = r / 23 9.Radius of circumcircle of an EQUILATERAL TRIANGLE of side a = r / 3 10.Radius of incircle of a triangle of area and semi perimeter S = / s 11.Area of a PARALLELOGRAM = (base * height) 12.Area of RHOMBUS = 1/2 (product of diagonals) 13.Area of TRAPEZIUM = =1/2 * (sum of parallel sides)* (distance between them)

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14.Area of a CIRCLE = r2 where r is the radius 15.Circumference of a CIRCLE = 2r 16.Length of an arc = 2 r / 360, where is central angle 17.Area of a SECTOR = (arc * r) = r2 / 360 18.Area of a SEMICIRCLE = r2 / 2 19.Circumference of a SEMICIRCLE = r

TopSimple Problems 1.One side of a rectangular field is 15m and one of its diagonal is 17m. Find the area of field ? Sol: Other side = [(17*17) (15*15)] = (289 - 225) = 8m Area = 15 * 8 =120 sq. m 2.A lawn is in the form of a rectangle having its sides in the ratio 2:3 The area of the lawn is 1/6 hectares. Find the length and breadth of the lawn. Sol: let length = 2x meters and breadth = 3x mt Now area = (1/6 * 1000)sq m = 5000/3 sq m 2x * 3x = 5000/3 =>x * x =2500 / 9 x = 50/3 length = 2x = 100/3 m and breadth = 3x = 3*(50/3) = 50m

3.Find the cost of carpeting a room 13m long and 9m broad with a carpet 75cm wide at the rate of Rs 12.40 per sq meter Sol: Area of the carpet = Area of the room = 13* 9 =117 sq m length of the carpet = (Area/width) = 117 * (4/3) = 156 m Cost of carpeting = Rs (156 * 12.40) = Rs 1934.40 4.The length of a rectangle is twice its breadth if its length is decreased by 5cm and breadth is increased by 5cm, the area of the rectangle is increased by 75 sq cm. Find the length of the rectangle. Sol: let length = 2x and breadth = x then (2x - 5) (x+5) (2x*x)=75 5x - 25 = 75 => x=20 length of the rectangle = 40 cm

5.In measuring the sides of a rectangle, one side is taken 5% in excess and the other 4% in deficit. Find the error percent in the area, calculate from the those measurements. Sol: let x and y be the sides of the rectangle then correct area = (105/100 * x) * (96 / 100 *y) =(504/500 xy) xy = 4/500 xy Error% = 4/500 xy*(1/xy)*100 % = 4/5% = 0.8%

6.A room is half as long again as it is broad. The cost of carpeting the room at Rs 5 per sq m is Rs 2.70 and the cost of papering the four walls at Rs 10 per sq m is Rs 1720. If a door and 2 windows occupy 8 sq cm. Find the dimensions of the room ? Sol: let breadth=x mt ,length= 3x/2 mt and height=h mt Area of the floor = (total cost of carpeting /rate) = 270/5 sq m = 54 sq m x * 3x/2=54 => x*x= 54*(2/3)=36 => x = 6m so breadth = 6m and length=3/2*6 = 9m now papered area = 1720 /10 = 172 sq m Area of one door and 2 windows =8 sq m total area of 4 walls = 172+8 = 180 sq m 2(9+6)*h = 180 => h=180/30 = 6m 7.The altitude drawn to the base of an isosceles triangle is 8cm and the perimeter is 32cm. Find the area of the triangle ? Sol: let ABC be the isosceles triangle, the AD be the altitude let AB = AC=x then BC= 32 - 2x since in an isoceles triange the altitude bisects the base so BD=DC=16 - x in ADC,(AC) 2 = (AD) 2 + (DC) 2 x*x=(8*8) + (16 - x)*(16 - x) 32x =320 => x = 10 BC = 32-2x = 32-20 = 12 cm Hence, required area = * BC * AD = * 12 * 10 = 60 sq cm 8.If each side of a square is increased by 25%, find the percentage change in its area ? Sol: let each side of the square be a , then area = a * a New side = 125a / 100 = 5a / 4 New area =(5a * 5a)/(4*4) = (25a/16) a = 9a/16 Increase %= 9a/16 * 1/a * 100% = 56.25% 9.Find the area of a Rhombus one side of which measures 20cm and one diagonal 24cm. Sol: Let other diagonal = 2x cm since diagonals of a rhombus bisect each other at right angles, we have 20 = 12 + x => x = [20 - 12]= 256 = 16cm so the diagonal = 32 cm Area of rhombus = * product of diagonals = * 24 * 32 = 384 sq cm


10. The area of a circular field is 13.86 hectares. Find the cost of fencing it at the rate of Rs. 4.40 per meter. Sol: Area = 13.86 * 10000 sq m = 138600 sq m r= 138600 => r = 138600 * 7/22 => 210 m circumference = 2r = 2 * 22/7 * 210m = 1320 m cost of fencing = Rs 1320 * 4.40 = Rs. 5808

TopMedium Problems: 11.Find the ratio of the areas of the incircle and circumcircle of a square. Sol: let the side of the square be x, then its diagonal = 2 x radius of incircle = x/2 and radius of circmcircle =2 x /2 = x/2 required ratio = x/4 : x/2 = : = 1:2 12.If the radius of a circle is decreased by 50% , find the percentage decrease in its area. Sol: let original radius = r and new radius = 50/100 r = r/2 original area = r and new area = (r/2) decrease in area = 3 r/4 * 1/ r * 100 = 75% 13.Two concentric circles form a ring. The inner and outer circumference of the ring are 352/7 m and 528/7m respectively. Find the width of the ring. sol: let the inner and outer radii be r and R meters then, 2r = 352/7 => r = 352/7 * 7/22 * = 8m 2R = 528/7 => R= 528/7 * 7/22 * = 12m width of the ring = R-r = 12-8 = 4m 14.If the diagonal of a rectangle is 17cm long and its perimeter is 46 cm. Find the area of the rectangle. sol: let length = x and breadth = y then 2(x+y) = 46 => x+y = 23 x+y = 17 = 289 now (x+y) = 23 =>x+y+2xy= 529 289+ 2xy = 529 => xy = 120 area =xy=120 sq. cm 15.A rectangular grassy plot 110m by 65cm has a gravel path .5cm wide all round it on the inside. Find the cost of gravelling the path at 80 paise per sq.mt sol: area area cost area of theplot = 110 * 65 = 7150 sq m of the plot excluding the path = (110 - 5)* (65 - 5) = 6300 sq m of the path = 7150 - 6300 =850 sq m of gravelling the path = 850 * 80/100 = 680 Rs

16. The perimeters of ttwo squares are 40cm and 32 cm. Find the perimeter of a third square whose area is equal to the difference of the areas of the two squares. sol: side of first square = 40/4 =10cm side of second square = 32/4 = 8cm area of third squre = 10 8 = 36 sq cm side of third square = 36 = 6 cm required perimeter = 6*4 = 24cm

Top17. A room 5m 44cm long and 3m 74cm broad is to be paved with squre tiles. Find the least number of squre tiles required to cover the floor. sol: area of the room = 544 * 374 sq cm size of largest square tile = H.C.F of 544cm and 374cm= 34cm area of 1 tile = 34*34 sq cm no. of tiles required = (544*374) / (34 * 34) = 176 18. The diagonals of two squares are in the ratio of 2:5. Find the ratio of their areas. sol: let the diagonals of the squares be 2x and 5x respectively ratio of their areas = * (2x) : *(5x) = 4:25 19.If each side of a square is increased by 25%. Find the percentage change in its area. sol: let new side new area increase increase each side of the square be a then area = a = 125a/100 = 5a/4 = (5a/4) = 25/16 a in area = (25/16)a - a = (9/16)a % = (9/16)a * (1/a) * 100 = 56.25%

20.The base of triangular field os three times its altitude. If the cost of cultivating the field at Rs. 24.68 per hectare be Rs. 333.18. Find its base and height. sol: area of the field = total cost/ rate = 333.18 /24.68 = 13.5 hectares => = 13.5 * 10000 = 135000 sq m let the altitude = x mt and base = 3x mt then *3x * x = 135000 => x = 90000 => x = 300 base= 900m and altitude = 300m


21.In two triangles the ratio of the areas is 4:3 and the ratio of their heights is 3:4. Find the ratio of their bases ? Sol: let the bases of the two triangles be x &y and their heights be 3h and 4h respectively. (1/2*x*3h)/(1/2*y*4h) =4/3 => x/y = 4/3 *4/3 = 16/9 22.Find the length of a rope by which a cow must be tethered in order that it may be able to graze an area of 9856 sq meters. Sol: clearly the cow will graze a circular field of area 9856 sq m and radius equal to the length of the rope. Let the length of the rope be r mts then r=9856 => r=9856*7/22 = 3136 => r=56m 23.The diameter of the driving wheel of a bus is 140cm. How many revolutions per minute must the wheel make inorder to keep a speed of 66 kmph? Sol: Distance to be covered in 1min = (66*1000)/60 m =1100m diameter = 140cm => radius = r =0.7m circumference of the wheel = 2*22/7*0.7 = 4.4m no of revolutions per minute = 1100/4.4 = 250 24.The inner circumference of a circular race track, 14m wide is 440m. Find the radius of the outer circle. Sol: let inner radius be r meters. Then 2r =440 => r=440*7/22*1/2 = 70m radius of outer circle = 70+4 =84m 25.A sector of 120 degrees, cut out from a circle, has an area of 66/7 sq cm. Find the radius of the circle. Sol: let the radius of the circle be r cm. Then r/360 =66/7=> 22/7*r*120/360 = 66/7 =>r = 66/7 *7/22*3 =9 radius = 3cm 26.The length of the room is 5.5m and width is 3.75m. Find the cost of paving the floor by slabs at the rate of Rs.800 per sq meter. Sol: l=5.5m w=3.75m area of the floor = 5.5 * 3.75 = 20.625 sq m cost of paving = 800 *20.625 =Rs. 16500

Top27.A rectangular plot measuring 90 meters by 50 meters is to be enclosed by wire fencing. If the poles of the fence are kept 5 meters apart. How many poles will be needed ? Sol: perimeter of the plot = 2(90+50) = 280m no of poles =280/5 =56m 28.The length of a rectangular plot is 20 meters more than its breadth. If the cost of fencing the plot @ 26.50 per meter is Rs. 5300. What is the length of the plot in meter ? Sol: let breadth =x then length = x+20 perimeter = 5300/26.50 =200m 2(x+20+x) =200 => 4x+40 =200 x = 40 and length = 40+20 = 60m 29.A rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq feet, how many feet of fencing will be required ? Sol: l=20feet and l*b=680 => b= 680/20 = 34feet length of fencing = l+2b = 20+68 =88 feet 30.A rectangular paper when folded into two congruent parts had a perimeter of 34cm foer each part folded along one set of sides and the same is 38cm. When folded along the other set of sides. What is the area of the paper ? Sol: when folded along the breadth we have 2(l/2 +b) = 34 or l+2b = 34...........(1) when folded along the length, we have 2(l+b/2)=38 or 2l+b =38.....(2) from 1 &2 we get l=14 and b=10 Area of the paper = 14*10 = 140 sq cm 31.A took 15 seconds to cross a rectangular field diagonally walking at the rate of 52 m/min and B took the same time to cross the same field along its sides walking at the rate of 68m/min. The area of the field is ? Sol: length of the diagonal = 52*15/60 =13m sum of length and breadth = 68*15/60 = 17m (l+b)=13 or l+b = 17 area =lb = (2lb) = [(l+b) (l+b)] = [17 - 169] =1/2*120 = 60 sq meter 32 . A rectangular lawn 55m by 35m has two roads each 4m wide running in the middle of it. One parallel to the length and the other parallel to breadth. The cost of graveling the roads at 75 paise per sq meter is sol: area of cross roads = 55*4 +35*4 - 4*4 = 344sq m cost of graveling = 344 *75/100 =Rs. 258


33.The cost of fencing a square field @ Rs. 20 per metre is Rs.10.080. How much will it cost to lay a three meter wide pavement along the fencing inside the field @ Rs. 50 per sq m sol: perimeter = total cost / cost per m = 10080 /20 = 504m side of the square = 504/4 = 126m breadth of the pavement = 3m side of inner square = 126 - 6 = 120m area of the pavement = (126*126) - (120*120)=246*6 sq m cost of pavement = 246*6*50 = Rs. 73800 34.Amanwalked diagonally across a square plot. Approximately what was the percent saved by not walking along the edges ? Sol: let the side of the square be x meters length of two sides = 2x meters diagonal = 2 x = 1.414x m saving on 2x meters = .59x m saving % = 0.59x /2x *100% = 30% (approx) 36.A man walking at the speed of 4 kmph crosses a square field diagonally in 3 meters.The area of the field is sol: speed of the man = 4*5/18 m/sec = 10/9 m/sec time taken = 3*60 sec = 180 sec length of diagonal = speed * time = 10/9 * 180 = 200m Area of the field = *(dioagonal) = * 200*200 sq m = 20000sq m 37.A square and rectangle have equal areas. If their perimeters are p and q respectively. Then sol: A square and a rectangle with equal areas will satisfy the relation p < q

Top38.If the perimeters of a square and a rectangle are the same, then the area a & b enclosed by them would satisfy the condition: sol: Take a square of side 4cm and a rectangle having l=6cm and b=2cm then perimeter of square = perimeter of rectangle area of square = 16 sq cm area of rectangle = 12 sq cm Hence a >b 39.An error of 2% in excess is made while measuring the side of a square. The percentage of error in the calculated area of the square is sol: 100cm is read as 102 cm a = 100*100 sq cm and b = 102 *102 sq cm then a-b = 404 sq cm percentage error = 404/(100*100) = 4.04% 40.A tank is 25m long 12m wide and 6m deep. The cost of plastering its walls and bottom at 75 paise per sq m is sol: area to be plastered = [2(l+b)*h]+(l*b) = 2(25+12)*6 + (25*12)= 744 sq m cost of plastering = Rs . 744*75/100 = Rs. 5581 41.The dimensions of a room are 10m*7m*5m. There are 2 doors and 3 windows in the room. The dimensions of the doors are 1m*3m. One window is of size 2m*1.5m and the other 2 windows are of size 1m*1.5m. The cost of painting the walls at Rs. 3 per sq m is sol: Area of 4 walls = 2(l+b)*h =2(10+7)*5 = 170 sq m Area of 2 doors and 3 windows = 2(1*3)+(2*1.5)+2(1*1.5) = 12 sq m area to be planted = 170 - 12 = 158 sq m cost of painting = Rs. 158 *3 = Rs. 474 42.The base of a triangle of 15cm and height is 12cm. The height of another triangle of double the area having the base 20cm is sol: a = *15*12 = 90 sq cm b = 2a = 2 * 90 = * 20 *h => h= 18cm 43.The sides of a triangle are in the ratio of :1/3:1/4. If the perimeter is 52cm, then the length of the smallest side is sol: ratio of sides = :1/3 :1/4 = 6:4:3 perimeter = 52 cm, so sides are 52*6/13 =24cm 52*4/13 = 16cm 52 *3/13 = 12cm length of smallest side = 12cm 44.The height of an equilateral triangle is 10cm. Its area is sol: a = (a/2) +(10) a a/4 = 100 =>3a = 100*4 area = 3/4 *a = 3/4*400/3 = 100/3 sq cm 45.From a point in the interior of an equilateral triangle, the perpendicular distance of the sides are 3 cm, 23cm and 53cm. The perimeter of the triangle is sol: let each side of the triangle be a cm then area(AOB) +area(BOC)+area(AOC) = area(ABC) * a *3 +1/2 *a *23 +1/2 * a*53 = 3/4 a


a/23(1+2+5) = 3/4 a => a=16 perimeter = 3*16 = 48cm

TopComplex Probems: 1.If the area of a square with side a s equal to the area of a triangle with base a, then the altitude of the triangle is sol: area of a square with side a = a sq unts area of a triangle with base a = * a*h sq unts a =1/2 *a *h => h = 2a altitude of the triangle is 2a 2.An equilateral triangle is described on the diagonal of a square. What is the ratio of the area of the triangle to that of the square ? Sol: area of a square = a sq cm length of the diagonal = 2a cm area of equilateral triangle with side 2a = 3/4 * (2a) required ratio = 3a : a = 3 : 2 3.The ratio of bases of two triangles is x:y and that of their areas is a:b. Then the ratio of their corresponding altitudes wll be sol: a/b =( * x*H) /(1/2 * y * h) bxH = ayh =>H/h =ay/bx Hence H:h = ay:bx 4 .A parallelogram has sides 30m and 14m and one of its diagonals is 40m long. Then its area is sol: let ABCD be the given parallelogram area of parallelogram ABCD = 2* (area of triangle ABC) now a = 30m, b = 14m and c = 40m s = (30+14+40) = 42m Area of triangle ABC = [ s(s - a)(s - b)(s - c) = (42*12*28*2 = 168sq m area of parallelogram ABCD = 2 *168 =336 sq m 5.If a parallelogram with area p, a triangle with area R and a triangle with area T are all constructed on the same base and all have the same altitude, then which of the following statements is false ? Sol: let each have base = b and height = h then p = b*h, R = b*h and T = * b*h so P = R, P = 2T and T = R are all correct statements 6.If the diagonals of a rhombus are 24cm and 10cm the area and the perimeter of the rhombus are respectively. Sol: area = *diagonal 1 *diagonal 2= * 24 * 10= 120 sq cm * diagonal 1 = * 24 = 12cm * diagonal 2 = *10 =5 cm side of a rhombus = (12) + (5) = 169 => AB = 13cm

Top7.If a square and a rhombus stand on the same base, then the ratio of the areas of the square and the rhombus is: sol: A square and a rhombus on the same base are equal in area 8.The area of a field in the shape of a trapezium measures 1440sq m. The perpendicular distance between its parallel sides is 24cm. If the ratio of the sides is 5:3, the length of the longer parallel side is: sol: area of field =1/2 *(5x+3x) *24 = 96x sq m 96x = 1440 => x = 1440 /96 = 15 hence, the length of longer parallel side = 5x = 75m 9.The area of a circle of radius 5 is numerically what percent its circumference ? Sol: required percentage = (5)/(2*5) *100 = 250% 10.A man runs round a circular field of radius 50m at the speed of 12m/hr. What is the time taken by the man to take twenty rounds of the field ? Sol: speed = 12 k/h = 12 * 5/18 = 10/3 m/s distance covered = 20 * 2*22/7*50 = 44000/7m time taken = distance /speed = 44000/7 * 3/10 = 220/7min 11.A cow s tethered in the middle of a field with a 14feet long rope.If the cow grazes 100 sq feet per day, then approximately what time will be taken by the cow to graze the whole field ? Sol: area of the field grazed = 22/7 * 14 * 14 = 616 sq feet 12.A wire can be bent in the If it is bent in the form of sol: length of wire = 2 r side of the square = 352/4 = area of the square = 88*88 = form of a circle of radius 56cm. a square, then its area will be = 2 *22/7 *56 = 352 cm 88cm 7744sq cm

13.The no of revolutions a wheel of diameter 40cm makes in traveling a distance of 176m is sol:


distance covered in 1 revolution = 2 r = 2 *22/7 *20 = 880/7 cm required no of revolutions = 17600 *7/880 = 140

Top14.The wheel of a motorcycle 70cm in diameter makes 40 revolutions in every 10sec.What is the speed of motorcycle n km/hr? Sol: distance covered in 10sec = 2 *22/7 *35/100 *40 =88m distance covered in 1 sec =88/10m = 8.8m speed =8.8m/s = 8.8 * 18/5 *k/h = 31.68 k/h 15.Wheels of diameters 7cm and 14cm start rolling simultaneously from x & y which are 1980 cm apart towards each other in opposite directions. Both of them make the same number of revolutions per second. If both of them meet after 10seconds.The speed of the smaller wheel is sol: let each wheel make x revolutions per sec. Then (2 *7/2 *x)+(2 * 7*x)*10 = 1980 (22/7 *7 * x) + (2 * 22/7 *7 *x) = 198 66x = 198 => x = 3 distance moved by smaller wheel in 3 revolutions = 2 *22/7 *7/2 *3 = 66cm speed of smaller wheel = 66/3 m/s = 22m/s 16.A circular swimming pool is surrounded by a concrete wall 4ft wide. If the area of the concrete wall surrounding the pool is 11/25 that of the pool, then the radius of the pool is ? Sol: let the radius of the pool be R ft radius of the pool including the wall = (R+4)ft area of the concrete wall = [(R+4)2 - R2 ] => = [R+4+R][R+4 - R] = 8(R+2) sq feet 8(R+2) = 11/25 R2 => 11 R2 = 200 (R+2) Radius of the pool R = 20ft 17.A semicircular shaped window has diameter of 63cm. Its perimeter equals sol: perimeter of window = r +2r = [22/7 * 63/2 +63] = 99+63 = 162 cm 18.Three circles of radius 3.5cm are placed in such a way that each circle touches the other two. The area of the portion enclosed by the circles is sol: required area - (3 * area of = ( * 7 = 493/4 = (area of an equilateral triangle of side 7 cm) sector with = 6o degrees and r = 3.5cm) * 7) (3* 22/7 *3.5 *3.5*60/360 ) sq cm 11*0.5*3.5 sq cm = 1.967 sq cm

19. Four circular cardboard pieces, each of radius 7cm are placed in such a way that each piece touches two other pieces. The area of the space encosed by the four pieces is sol: required area = 14*14 (4 * * 22/7 * 7 *7) sq cm = 196 154 = 42 sq cm

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Important Facts and Formula: RESULTS ON TRIANGLE 1.Sum of the angles of a triangle is 180 degrees. 2.The sum of any two sides of a triangle is greater than third side. 3.PYTHAGORAS Theorem: In a right angled triangle (Hypotenuse)2 = (Base)2 +(Height)2 4.The line joining the mid point of a side of a triangle to the opposite vertex is called the MEDIAN. 5.The point where the three medians of a triangle meet, is called CENTROID. The centroid divides each of the medians in the ratio 2:1 6.In an isosceles triangle, the altitude from the vertex bisects the base 7.The median of a triangle divides it into two triangles of the same area. 8.The area of the triangle formed by joining the mid points of the sides of a given triangle is one-fourth of the area of the given triangle. RESULTS ON QUADRILATERALS 1.The diagonals of a Parallelogram bisect each other. 2.Each diagonal of a Parallelogram divides it into two triangles of the same area. 3.The diagonals of a Rectangle are equal and bisect each other 4.The diagonals of a Square are equal and bisect each other at right angles. 5.The diagonals of a Rhombus are unequal and bisect each other at right angles. 6.A Parallelogram and a Rectangle on the same base and between the same parallels are equal in area.

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7.Of all he parallelogram of given sides the parallelogram which is a rectangle has the greatest area. FORMULAE 1.Area of a RECTANGLE = length * breadth Length = (Area/Breadth) and Breadth = (Area/Length) 2.Perimeter of a RECTANGLE = 2(Length + Breadth) 3.Area of a SQUARE = (side)2 = ( Diagonal)2 4.Area of four walls of a room = 2(length + breadth) * height 5.Area of a TRIANGLE = * base * height 6.Area of a TRIANGLE = [s * (s-a) * (s-b) * (s-c)], where a,b,c are the sides of the triangle and s = 1/2(a+b+c) 7.Area of EQUILATERAL TRIANGLE = (3/4)* (side)2 8.Radius of in circle of an EQUILATERAL TRIANGLE of side a = r / 23 9.Radius of circumcircle of an EQUILATERAL TRIANGLE of side a = r / 3 10.Radius of incircle of a triangle of area and semi perimeter S = / s 11.Area of a PARALLELOGRAM = (base * height) 12.Area of RHOMBUS = 1/2 (product of diagonals) 13.Area of TRAPEZIUM = =1/2 * (sum of parallel sides)* (distance between them) 14.Area of a CIRCLE = r2 where r is the radius

15.Circumference of a CIRCLE = 2 r 16.Length of an arc = 2 r / 360, where is central angle r2 / 360

17.Area of a SECTOR = (arc * r) = 18.Area of a SEMICIRCLE = r2 / 2 r

19.Circumference of a SEMICIRCLE = BACK

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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 BACK Simple Problems:

1.One side of a rectangular field is 15m and one of its diagonal is 17m. Find the area of field? Sol:Other side = [(17*17) (15*15)] = (289-225) = 8m Area = 15 * 8 =120 sq. m

2.A lawn is in the form of a rectangle having its sides in the ratio 2:3 The area of the lawn is 1/6 hectares. Find the length and breadth of the lawn. Sol: let length = 2x meters and breadth = 3x mt Now area = (1/6 * 1000)sq m = 5000/3 sq m 2x * 3x = 5000/3 =>x * x =2500 / 9 x = 50/3 length = 2x = 100/3 m and breadth = 3x = 3*(50/3) = 50m

3.Find the cost of carpeting a room 13m long and 9m broad with a carpet 75cm wide at the rate of Rs 12.40 per sq meter Sol: Area of the carpet = Area of the room = 13* 9 =117 sq m length of the carpet = (Area/width) = 117 * (4/3) = 156 m Cost of carpeting = Rs (156 * 12.40) = Rs 1934.40

4.The length of a rectangle is twice its breadth if its length is decreased by 5cm and breadth is increased by 5cm, the area of the rectangle is increased by 75 sq cm. Find the length of the rectangle. Sol: let length = 2x and breadth = x then

(2x-5) (x+5) (2x*x)=75 5x-25 = 75 => x=20 length of the rectangle = 40 cm

5.In measuring the sides of a rectangle, one side is taken 5% in excess and the other 4% in deficit. Find the error percent in the area, calculate from the those measurements. Sol: let x and y be the sides of the rectangle then

correct area = (105/100 * x) * (96 / 100 *y) =(504/500 xy) xy = 4/500 xy

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Error% = 4/500 xy*(1/xy)*100 % = 4/5% = 0.8%

6.A room is half as long again as it is broad. The cost of carpeting the room at Rs 5 per sq m is Rs 2.70 and the cost of papering the four walls at Rs 10 per sq m is Rs 1720. If a door and 2 windows occupy 8 sq cm. Find the dimensions of the room? Sol: let breadth=x mt ,length= 3x/2 mt and height=h mt

Area of the floor = (total cost of carpeting /rate) = 270/5 sq m = 54 sq m x * 3x/2=54 => x*x= 54*(2/3)=36 => x = 6m

so breadth = 6m and length=3/2*6 = 9m now papered area = 1720 /10 = 172 sq m Area of one door and 2 windows =8 sq m total area of 4 walls = 172+8 = 180 sq m 2(9+6)*h = 180 => h=180/30 = 6m

7.The altitude drawn to the base of an isosceles triangle is 8cm and the perimeter is 32cm. Find the area of the triangle? Sol: let ABC be the isosceles triangle, the AD be the altitude let AB = AC=x then BC= 32-2x since in an isoceles triange the altitude bisects the base so BD=DC=16-x in ADC,(AC) 2 = (AD) 2 + (DC) 2 x*x=(8*8) + (16-x)*(16-x) 32x =320 => x = 10

BC = 32-2x = 32-20 = 12 cm Hence, required area = * BC * AD = * 12 * 10 = 60 sq cm

8.If each side of a square is increased by 25%, find the percentage change in its area? Sol: let each side of the square be a , then area = a * a New side = 125a / 100 = 5a / 4 New area =(5a * 5a)/(4*4) = (25a/16) a = 9a/16 Increase %= 9a/16 * 1/a * 100%

= 56.25%

9.Find the area of a Rhombus one side of which measures 20cm and one diagonal 24cm. Sol: Let other diagonal = 2x cm since diagonals of a rhombus bisect each other at right angles, we have 20 = 12 + x => x = [20 -12]= 256 = 16cm


so the diagonal = 32 cm Area of rhombus = * product of diagonals = * 24 * 32 = 384 sq cm

10. The area of a circular field is 13.86 hectares. Find the cost of fencing it at the rate of Rs. 4.40 per meter. Sol: Area = 13.86 * 10000 sq m = 138600 sq m * 7/22 => 210 m

r= 138600 => r = 138600

circumference = 2 r = 2 * 22/7 * 210m = 1320 m cost of fencing = Rs 1320 * 4.40 = Rs. 5808



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PROBLEMS1.One side of a rectangular field is 15m and one of its diagonal is 17m. Find the area of field? Sol:Other side = [(17*17) (15*15)] = (289-225) = 8m Area = 15 * 8 =120 sq. m 2.A lawn is in the form of a rectangle having its sides in the ratio 2:3 The area of the lawn is 1/6 hectares. Find the length and breadth of the lawn. Sol: let length = 2x meters and breadth = 3x mt Now area = (1/6 * 1000)sq m = 5000/3 sq m 2x * 3x = 5000/3 =>x * x =2500 / 9 x = 50/3 length = 2x = 100/3 m and breadth = 3x = 3*(50/3) = 50m 3.Find the cost of carpeting a room 13m long and 9m broad with a carpet 75cm wide at the rate of Rs 12.40 per sq meter Sol: Area of the carpet = Area of the room = 13* 9 =117 sq m length of the carpet = (Area/width) = 117 * (4/3) = 156 m Cost of carpeting = Rs (156 * 12.40) = Rs 1934.40 4.The length of a rectangle is twice its breadth if its length is decreased by 5cm and breadth is increased by 5cm, the area of the rectangle is increased by 75 sq cm. Find the length of the rectangle. Sol: let length = 2x and breadth = x then (2x-5) (x+5) (2x*x)=75 5x-25 = 75 => x=20 length of the rectangle = 40 cm 5.In measuring the sides of a rectangle, one side is taken 5% in excess and the other 4% in deficit. Find the error percent in the area, calculate from the those measurements. Sol: let x and y be the sides of the rectangle then correct area = (105/100 * x) * (96 / 100 *y) =(504/500 xy) xy = 4/500 xy Error% = 4/500 xy*(1/xy)*100 % = 4/5% = 0.8% 6.A room is half as long again as it is broad. The cost of carpeting the room at Rs 5 per sq m is Rs 2.70 and the cost of papering the four walls at Rs 10 per sq m is Rs 1720.If a door and 2 windows occupy 8 sq cm. Find the dimensions of the room? Sol: let breadth=x mt ,length= 3x/2 mt and height=h mt Area of the floor = (total cost of carpeting /rate) = 270/5 sq m = 54 sq m x * 3x/2=54 => x*x= 54*(2/3)=36 => x = 6m so breadth = 6m and length=3/2*6 = 9m now papered area = 1720 /10 = 172 sq m Area of one door and 2 windows =8 sq m total area of 4 walls = 172+8 = 180 sq m 2(9+6)*h = 180 => h=180/30 = 6m 7.The altitude drawn to the base of an isosceles triangle is 8cm and the perimeter is 32cm. Find the area of the triangle? Sol: let ABC be the isosceles triangle, the AD be the altitude let AB = AC=x then BC= 32-2x since in an isoceles triange the altitude bisects the base so BD=DC=16-x in ADC,(AC) 2 = (AD) 2 + (DC) 2 x*x=(8*8) + (16-x)*(16-x) 32x =320 => x = 10 BC = 32-2x = 32-20 = 12 cm Hence, required area = * BC * AD = * 12 * 10 =