Numerical ReasoningBy Blue Lotus

Problems on agesProblem 1 : Present ages of X and Y are in the ratio 3:4 respectively. Two years hence this ratio will become 4:5 respectively. What is xs present age in years?

Problems on agesSolution 1 : Let the present ages of X and Y be 3x years and 4x years respectively. Then, 3x + 2 4 -------- = --4x + 2 5 5 (3x + 2) = 4(4x + 2) x=2 X present age = 3* 2 = 6

Problems on agesProblem 2 : Anithas father was 28 years of age when she was born while her mother was 26 years old when her brother two years younger to her was born. What is the difference between the ages of her parents?

Problems on agesSolution 2 : Mothers age when Anithas brother was born = 26 years. Fathers age when Anithas brother was born = (28 + 2) years = 30years Required difference = (30 26) years = 4years

Problems on agesProblem 3 : Three years ago, the average age of a family of 5 members was 17. A baby having born the average of the family is the same today. What is age of child ?

Problems on agesSolution 3 : Present age of 5 members = 5*17+3*5 = 100 years. Also, present age of 5 members + age of baby = 6*17 = 102. Age of baby = 102-100 = 2 years. 102-

Problems on agesProblem 4 : Rajas age after 15 years will be 5 times his age 5 years back. What is the present age of Raja?

Problems on agesSolution 4 : Let Rajas present age be x years. Then, Rajas age after 15 years = (x + 15) years Rajas age 5 years back = (x 5) years x + 15 = 5(x 5) x + 15 = 5x -25 4x = 40 x =10 Hence, Rajas present age = 10 years

Problems on agesProblem 5 : Geethas age after six years will be three-seventh of his fathers age. Ten threeyears ago the ratio of their ages was 1:5. What is Geethas fathers age at present?

Solution 5 : Let the ages of Geetha and his father 10 years ago be x and 5x years respectively. Then, Geethas age after 6 years = (x+10)+6 =(x+16) years. Fathers age after 6 years =(5x+10)+6 =(5x+16) years. (x+16) = 3/7 (5x+16) 7(x+16) =3(5x+16) 7x+112 = 15x+48 x = 8

Problems on ages

Problems on agesProblem 6 : One year ago a father was four times as old as his son. In 6 years time his age exceeds twice his sons age by 9 years. Ratio of their ages.

Problems on agesSolution 6 : Let the present ages of father and son be x and y respectively. ( x-1) = 4( y-1) or 4y x = 3. xy(x+6) -2( y+6) = 9. On solving we get 33:9 or 11:3

Problems on agesProblem 7 : Jayesh is as much younger to anil as he is older to prashant. If the sum of ages of anil and prashant is 48 years, What is age of jayesh ?

Problems on agesSolution 7 : J is x years younger than A and x years older than p. Sum of ages of anil and prashant = 48 years . J + x + J x = 48. J = 24 years.

Problems on agesProblem 8 : A mans age was 125% of what it 125% was 10 years ago , but 83 1/3% of what it will be after 10 years. What is his present years. age ?

Problems on agesSolution 8 : 125%(x-10) 831/ (x+10) 125%(x-10) = 831/3%(x+10) 125/ 125/100 *(x-10) = 250/300 * (x+10) *(x-10) 250/ (x+10) X= 50 years. years.

Problems on agesProblem9 : Present ages of X and Y are in the ratio 5:6 respectively. Seven years hence this ratio will become 6:7 respectively. What is Xs present age in years?

Problems on agesSolution 9 : Let the present ages of X and Y be 5x years and 6x years respectively. Then, 5x + 7 / 4x + 7 = 6/7 7(5x + 7) = 6(6x + 7) x = 7. Xs present age = 5x = 35 years.

Problems on agesProblem10 : Tanyas grandfather was 8 times older to her 16 year ago. He would be 3 times of her age 8 years from now. Eight years ago, what was the ratio of Tanyas age to that of her grandfather?

Problems on agesSolution:10 16 years ago, let T = x years and G = 8x years. After 8 years from now, T = (x+16+8) years and G = (8x+16+8) years. 8x+24 = 3(x+24) 5x =48 8 years ago, T/G = x+8 / 8x + 8 = (48/5 + 8) / (8*48/5+8) = 88/424 = 11/53.

Alligation or MixtureProblem1 : A man who buys two cows for Rs.1350 and sells one as to lose 6% and the other so as to gain 7.5% and on the whole he neither gain nor loses . What does each cow cost ?

Alligation or MixtureSolution 1 : 6% 0 7.5 Thus ratio of two cows is 5 : 4. 6 7.5%

Alligation or mixtureSoln 1 contd : Cost of cow 1 = 1350 * 5/9 = Rs.750. Cost of cow 2 = 1350 * 4/9 = Rs.600.

Alligation or Mixture Problem:2How much water must be added to 14 litres of milk worth Rs.5.40 p a litre so that the value of the mixture may be Rs.4.20 p a litre?

Alligation or MixtureSolution:2 The mean value is Rs.4.20 p a litre and the price of the water is zero paisa (i.e. free of cost) Per litre. By allegation method: Milk / water = Rs.4.20 0 / Rs.5.40 Rs.4.20 = 420 / 120 =7/2 Milk and water must be mixed in the ratio 7:2 Since the milk is 14 litres, so water to be mixed is 4 litres.

Alligation or MixtureProblem3 : A person travels 50 kg of pulses, part of which he sells at 10% profit and the rest at 5% loss. He gains 7% on whole. What is quantity sold at 10% gain and 5% loss ?

Alligation or MixtureSolution: Solution:3 8% profit 14% 14% 4% 6% 18% profit 18%

Thus ratio is 2 : 3.Therefore quantity is

Alligation or MixtureProblem4 : In a zoo there are rabbits and pigeons. If their heads are counted, these are counted, these are 90 while their legs are 224. Find the number of pigeons in the zoo.

Alligation or MixtureSolution 4 : Each pigeon has 2 legs and a rabbit has 4 legs. legs. If there be all pigeons, number of legs are 2 * 90 = 180 Actual number of legs = 224 Difference = 224 180 = 44 When a rabbit in place of pigeon is considered there is a difference of two legs, but actual difference is 44 No. No. of rabbits = 44 / 2 = 22 and No. of pigeons = 136/2= 68 No. 136/

Alligation or MixtureProblem 5 : A mixture of 40 litres of milk and water contains 10% water. How much water should be added to this so that water may be 20% in the new mixture?

Alligation or MixtureSolution 5 : Quantity of milk and water out of 40 litres mixture = 90% * 40 and 10% * 40 i.e. 36 litres and 4 litres respectively. Now the new mixture contains the quantity of milk 36 litres and in it the water required is 20% and milk 80%. 80% of new mixture = 36 litres New mixture = 36 * 100 / 80 litres = 45 litres Thus water added = 45 40 = 5 litres.

Alligation or MixtureProblem 6 : Three utensils contains equal mixtures of milk and water in the ratio of 4:1, 5:2 and 6:1 respectively. If all the solutions are mixed together, what is the ratio of milk to water in the final solution?

Alligation or MixtureSolution:6 In each unit of mixture, milk = 4/5, water = 1/5 in first, milk = 5/7 and water = 2/7 in the second and milk = 6/7 and water = 1/7 in the third utensil. In the final mixture, total contents milk = 4/5 + 5/7 + 6/7 = 83/35 total contents water = 1/5 + 2/7 + 1/7 = 22/35 Ratio of milk : water = 83 : 22

Alligation or MixtureProblem 7 : A man buys milk at Rs.5.00 a litre and after adding water sells it at Rs.6.00 a litre there by making a profit of 25%. Find the proportion of water he has added.

Alligation or MixtureSolution 7 : First find the cost price of the mixture. Since mixture is sold at Rs.6.00 a litre 125% of C.P of mixture = Rs.6.00 C.P of mixture = Rs.6.00 * 100 / 125 = Rs.4.80 cost price of milk = Rs.5.00 a litre cost price of water = 0 cost price of mixture = Rs.4.80 By allegation method; Ratio of milk and water = 4.80 0 / 5.00 4.80 = 480 / 20 = 24 / 1 = 24:1

Alligation or MixtureProblem 8 : A sum of Rs.6.25 p is made up of 80 coins which are either 10 p or 5 p, how many are there of each kind ?

Alligation or MixtureSolution 8 : Let there be x coins of 10 p so that number of 5 p coins be (80 x) Total value of x, 10 p coins = 10x paise Total value of (80 x), 5 p coins = 5(80 x) p 10x + 5(80 x) = 625 x = 45 No. of 10 p coins is 45 and number of 5 p coins be 35.

Alligation or MixtureProblem9 : A jar full of whisky contains 40% alcohol. A part of this whisky is replaced by another containing 19% alcohol and now the percentage of alcohol was found to be 26%. Find the quantity of whisky replaced?

Alligation or MixtureSolution9 : Strength of first jar Strength of 2nd jar 40% 19% Mean strength 26% 7 14

Alligation or MixtureSolution 9 (Cont..) So, ratio of 1st and 2nd quantities = 7:14 = 1:2 Required quantity replaced = 2/3

Alligation or MixtureProblem:10 How many kilograms of sugar costing Rs.9 per kg must be mixed with 27 kg of sugar costing Rs.7 per kg so that there may be a gain of 10% by selling the mixture of Rs.9.24 per kg?

Alligation or MixtureSolution10 : S.P. of 1 kg of mixture = Rs.9.24, Gain = 10% C.P. of 1 kg of mixture = Rs.(100/110 * 9.24) = Rs.8.40

Average

AverageProblem1 : The mean marks of 10 boys in a class is 70% whereas the mean marks of 15 girls is 60%. The mean marks of all the 25 students

AverageSolution1 :

The mean marks of 10 boys = 70% Total marks of 10 boys = 70% * 10 = 700 The mean marks of 15 girls = 60%*15= 900% Sum of the total marks of 25 students = 1600% Mean = 1600/25 = 64%.

AverageProblem2 : The average expenditure of a man for the first five months is Rs.120 and for the next seven months is Rs.130. Find his monthly average income if he saves Rs.290 during the year?.

AverageSolution:2 Average expenditure for 1st five months= Rs.120. So7 So7 1st five months = Rs.600aa Average expenditure for other seven months= Rs.130. So7 So7 1st five months = Rs.910. Total savings = Rs.290. Total income = Rs.600+910+290 = Rs.1800. Avg income per month = Rs.1800/12=Rs.150

Average Problem:3 Problem:A man had 7 children. When their average children. age was 12 years, the child who was 6 years of age, died. What was the average age of died. surviving children 5 years after the death of the above child?

AverageSolution:3 Average age of 7 children = 12 years Total age of 7 Children years = 12 * 7 = 84

Total age of 6 children after the death of a child aged 6 years = 84 6 = 78 Hence the average age of the surviving children = 78/6 = 13 years. After 5 years = 13 + 5 = 18 years.

Average Problem:4The average salary per head of all the workers in a workshop is Rs.95. The average salary of 15 officers is Rs.525 and the average salary per head of the rest Rs.85. Find the total number of workers in the workshop?

AverageSolution:4 S Sum of salaries, x sum of salaries of 15 officers, W no. of workers. Given (S/W)= Rs.95, x/15 = 525 Therefore x= 7875

AverageS-x/(W-15)=85 x/(WS= 95 * w (95*w(95*w-7875) / (W-15)=85 (WW = 660.

Average Problem:5There was one mess for 30 boarders in a certain hostel. On the number of boarders being increased by 10, the expenses of the mess was increased by Rs.40 per month while the average expenditure per head diminished by Rs.2. Find the original monthly expenses.

AverageSolution:5 Let x rupees be the average of 30, boarders 30x + 40 = (x 2) * 20 x = 12 Original expenditure = Rs.12 * 30 = Rs.360

Average Problem:6A man walks from P to Q at the rate of 5 km an hour and returns from Q to P at the rate of 3 km an hour. What is the average rate in km an hour. What is the average rate in km per hour for the whole journey?

AverageSolution:6 Let the length of journey be L.C.M. of 3, 5 =15 km Time taken in going P to Q = 15/5 = 3 Time taken in coming Q to P = 15/3 = 5 Total time in going and coming back =5+3=8 Hence the average speed = 30/8 = 15/4 km/hr.

Average Problem:7A certain factory employed 600 men and 400 women and the average wage was Rs.2.55 per day. If a woman got 50 p. less than a man, what were their daily wages?

AverageSolution:7 Total workers = 600 + 400 = 1000 Total wages per day = 2.55 * 1000 = Rs.2550 Since a man gets 50 p. more, 600 men will get Rs.600 * .50 = Rs.300 more. Hence 2550 300 = 2250 is the daily wages of 1000 women.

Average Problem:8If a train maintains an average speed of 40 km an hour it arrives at its destination punctually. If however the average speed is 35 km an hour it arrives 15 minutes late. Find the length of the journey in km.

AverageSolution: Solution:8 L.C.M. of 40 and 35 = 280. 280. Let the journey be 280 km If the speed is 40 km/hour, time taken = 280/40 = 7 hrs. 280/ hrs. If the speed is 35 km/hour, time taken = 280/35 = 8 hrs. 280/ hrs. Difference of times = 8 7 = 1 hour = 60 minutes If 60 minutes is the difference, length of journey= 280km 280km If 15 minutes is difference, journey = 280/60 * 15 280/ =70km 70km

Average Problem:9The average of first three numbers is double of the fourth number. If the average ofo all four numbers is 12, find the 4th number.

AverageSolution:9 Let the fourth number be x. Then the average of first three = 2x So the sum of three numbers = 3 * 2x = 6x Since the average of 4 numbers = 12 Sum of four numbers = 4 * 12 = 48 So 48 6x = x x = 48/7

Average Problem:10A cyclist rides 24 km at 16 km per hour and a further 36 km at 15 km per hour. Find his average speed for the journey.

AverageSolution:10 Time taken to travel 24 km with the speed of 16 km/hours = 24/16 = 3/2 hours. Similarly time taken to travel 36 km with the speed of 15 km/hour = 36/15 = 12/5 hours. Total distance traveled = 36 + 24 = 60 km and total time taken to travel distance = 3/2 + 12/5 = 39/10 hours. Hence average speed = 60 * 10/39 = 200/13 = 15.38

Boats & StreamProblem:1 Problem:If a man can swim downstream at 6 kmph and upstream at 2 kmph, his speed in still water is?

Boats & StreamSolution : 1 a = 6 km/hr b = 2 km/hr Speed in still water = (a+b) = (6+2) = 4 km/hr

Boats & StreamProblem2 Problem2 : A boat travels upstream from

Q to P and downstream P to Q in 3 hr. If the hr. distance between P to Q is 4 km and the speed of the current is 1 Km/hr, then what is the speed of the boat in still water?

Boats & StreamSolution2 : t1 = 4/(x+1) t2 = 4/(x-1) 4/(xt1+t2 = 3 Sub and solve We will get x = 3 hrs.

Boats & StreamProblem3 Problem3 : A boatman can row 2 km against the stream in 20 minutes and return in 18

minutes. minutes. Find the rate of current ?

Boats & Stream

Solution 3 : Upstream speed = 2/20*60 = 6km/hr Downstream speed = 2/18*60 =20/3 Km/h rate of stream = (20/3)-6 = 1/3Km/hr. (20/3)-

Boats & StreamProblem:4 Problem:A man can row a boat at 10 kmph in still water. water. If the speed of the stream is 6 kmph, the time taken to row a distance of 80 km down the stream is?

Solution : 4

Boats & Stream

(a+b) = 10 km/hr a+b = 20 km/hr (a-b) = 6 km/hr (aa-b = 12 km/hr 2a = 32 km/hr Downstream a = 16 km/hr Upstream b = 4 km/hr Time taken for 80 km downstream = 80/16 km/hr = 5 hrs

Boats & StreamProblem:5 Problem:A boat takes 4 hrs for traveling downstream from point A to point B and coming back to point A upstream. If the velocity of the upstream. stream is 2 kmph and the speed of the boat in still water is 4 kmph, what is the distance between A and B?

Boats & StreamSolution : 5

(a-b) = 2 km/hr (a (a+b) = 4 km/hr Thus a = 6 and b = 2 Form a to b total time = 4 hrs

Boats & StreamProblem:6 Problem:If a mans rate with the current is 11 kmph and the rate of the current is 1.5 kmph, then the mans rate against the current is?

Boats & StreamSolution : 6 Down stream = 11 km/hr Rate of current = 1.5 (11 x) = 1.5 Thus the mans rate against the current x = 8km /hr

Boats & StreamProblem:7 Problem:River is running at 2 kmph. If takes a man kmph. twice as long to row up as to row down the river. river. The rate of the man in still water is?

Solution : 7

Boats & Stream

Speed of stream = 2 km/hr Time for upstream = t1 Time for downstream t2 = t1/2 Distance = x For upstream = x/t1 For downstream = x/t1/2 = 2x/t1 Speed of stream = (2x/t1 x/t1) = 2 km /hr x / t1 = 4 km/hr (2x/t1 + x/t1) = y 3x/t1 = 2y 12 = 2y Thus y = 6 km/hr

Boats & StreamProblem:8 Problem:The speed of a boat in still water is 15 km/ hr and the rate of current is 3 km/hr. the km/hr. distance traveled downstream in 12 minutes is?

Boats & StreamSolution : 8 (d+u) = 15 km/hr = d+u = 30 (d-u) = 3 km/hr = d-u = 6 (ddThus distance = 18/5 kms

Boats & StreamProblem:9 Problem:If a man rows at the rate of 5 kmph in still water and his rate against the current is 3.5 kmph then the mans rate along the current is?

Boats & StreamSolution : 9 (a+b) = 5 km/hr Upstream = 3.5 km/hr d + 3.5 = 10 km/hr Thus distance = 6.5 km/hr

Boats & StreamProblem:10 Problem:A man can row 18 kmph in still water. It water. takes him thrice as long to row up as to row down the river. Find the rate of the stream? river.

Boats & StreamSolution : 10 (a+b) = 18 km/hr (x/t1/3 + x/8t1) = 18 km/hr 4x/t1 = 18 km/hr x / t1 = 18/4 km/hr Now (2x/t1) = y = 2x/t1 = 18/4/2y 4y = 18/4 y = 18/16 = 9/8 Rate of the stream = 9/2 = 4.5 km/hr

CalendarProblem:1What was the day of the week on 28th Feb, 1995?

CalendarSolution:1 A = 1995/7 = 0 B = 1995 / 4 = 498/7 = 8 C = 28/ 7 = 0 D = Feb code = 3 = A + B + C + D 2/ 7 = 0 + 8 + 0 + 3 2/7 = 11-2 / 7 = 2 = Tuesday 11-

Calendar Problem:2What was the day of the week on 1st Jan, 2001?

CalendarSolution:2 A = 2001/7 = 6 B = 2001/4 = 500/ 7 = 3 C = 1/7 = 1 D=0 = (A + B + C + D 2) / 7 = (6 + 3 + 1 + 0 2) / 7 = 8/ 7 = 1 = Monday

Calendar Problem:3What was the day of the week on 30th Sep, 2007?

CalendarSolution:3 A = 2007 / 7 = 5 B = 2007 / 4 = 501/7 = 4 C = 30 / 7 = 2 D=5 = A+B+C+D 2/7 = 5 + 4 + 2 + 5 2/7 = 14/ 7 = 0 = Sunday

Calendar Problem:4What was the day of the week on 29th Dec, 1985?

CalendarSolution:4 A= 1985/ 7 = 4 B = 1985/ 4 = 496 / 7 = 6 C = 29 / 7= 1 D=5 = A + B + C + D 2/7 = 4 + 6 + 1 + 5 2/7 =14/7 = 0 = Sunday

Calendar Problem:5What was the day of the week on 30th May, 1961?

CalendarSolution:5 A = 1961/7 = 1 B = 1961/4 =490/4 = 0 C = 30/7 = 2 D=1 = 1 + 0 + 2 + 1-2/7 1= 2/7 = 2 = Tuesday

Calendar Problem:6What was the day of the week on 15th Aug, 1985?

CalendarSolution:6 A = 1985/ 7 = 4 B = 1985/ 4 = 496/ 7 = 6 C = 15/ 7 = 1 D=2 = A + B + C+ D 2/ 7 = 4 + 6 + 1 + 2 2/7 = 11/7 = 4 = Thursday

Calendar Problem:7What was the day of the week on 13th May, 1984?

Solution:7

Calendar

A = 1984/ 7 = 3 B = 1984/ 4 = 496/ 7 = 6 C = 13/ 7 = 6 D=2 = (A + B + C+ D 2) / 7 = (3 + 6 + 6 + 2 3) /7 = 14/7 = 0 = Sunday

Calendar Problem:8What was the day of the week on 22nd June, 1984?

CalendarSolution:8 A = 1984/7 = 3 B = 1984/4 = 496/ 7 = 6 C = 22/7 = 1 D=5 = (A + B + C + D 3) / 7 = (3 + 6 + 1 + 5 - 3) / 7 = 12/7 = 5 = Friday

Calendar Problem:9What was the day of the week on 25th June, 2007?

CalendarSolution:9 A = 2007/ 7 = 5 B = 2007 / 4 = 501/ 7 = 4 C = 25/ 7 = 4 D=4 = A + B + C + D -2/7 = 5 + 4 + 4 + 4 2/ 7 = 15/7 = 1 = Monday

Calendar Problem:10On what dates of April 1994 did Sunday fall?

Solution:10

Calendar

A = 1994/ 7 = 6 B = 1994 / 4 = 498/ 7= 8 C = 1/7 = 1 D=6 = A + B + C + D 2/7 = 6 + 8 + 1 + 6 2/7 =21=21-2/7 = 19/7 = 5 = Friday April 1st falls on Friday then, Sunday on 3rd So, 3, 10, 17 , 24

Chain RuleBlue Lotus

Chain RuleProblem: Problem:1 If 20 men working together finish a job in 20 days, then how many days will be taken by 25 men of the same capacity to finish the job?

Chain RuleSolution:1 Men Days 20 20 25 ? = x/20 = 20/25 =20 * 20/ 25 = 16 Days

Chain RuleProblem:2 Problem:14 pumps of equal capacity can fill a tank in 6 days. If the tank has to be filled in 4 days, days. how many extra pumps are needed?

Chain RuleSolution:2 Pumps Days 14 6 ? 4 = x/14 = 6/4 x = 6 * 14 /4 = 21 = 21 14 = 7 ( extra)

Chain RuleProblem:3 Problem:If 16 men working 7 hours a day plough a field in 48 days, in how many days will 14 men working 12 hours a day plough the field?

Chain RuleSolution:3 Men Hours Days 16 7 48 14 12 ? = x/ 48 = 7/12 * 16/14 = 7*16*48/12*14 = 5376/168 = 32

Chain RuleProblem:4 Problem:If 12 carpenters, working 6 hours a day can make 460 chairs in 24 days, how many chairs will 18 carpenters make in 36 days, each working 8 hours a day?

Chain RuleSolution:4 Carpenters Hours Chairs Days 12 6 460 24 18 8 ? 36 x/460 = 18/12 * 8/6 * 36/24 x = 18*8*36*460/12*6*24 = 1380

Chain RuleProblem:5 Problem:400 persons working 9 hours per day complete 1/4th of the work in 10 days. Find days. the time required by the additional persons, working 8 hours per day to complete the remaining work in 20 days?

Chain RuleSolution:5 Person Hours work Days 400 9 10 ? 8 20 x/400 = 9/8 * 3/4/1/4 * 10/20 x = 9*3*10*400/ 8*20 = 675 Extra persons = 675 400 =275

Chain RuleProblem:6 Problem:Some persons can do a piece of work in 12 days. days. Two times the number of such persons will do half of that work in how many days?

Chain RuleSolution:6 Time takes to finish the work = 12 days by x men works takes = 6 days by 2x men No of days = 2x = 6 x = 6/2 3 days will take to finish the half of work if workers increase twice.

Chain RuleProblem:7 Problem:If 8 persons can build a wall 140 m long in 42 days, find the number of days that 30 persons will take to complete a similar wall 100 m long?

Chain RuleSolution:7 Person Meter Days 8 140 42 30 100 x X = 42 * 100 * 8/ 140 * 30 =8 It will take 8 days to finish the job

Chain RuleProblem:8 Problem:A certain number of men can finish a piece of work in 100 days. If however, there were days. 10 men less, it would take 10 days more for the work to be finish. How many men were finish. there originally?

Chain RuleSolution:8 Men Days x 100 x-10 110 Less men = More Days ( Indirect Proportion) X/xX/x-10 = 110/100 100x = 110 ( x-10) x110 x = 110x 1100 10x = 1100 X = 1100/10 = 110 Original men at initial stage

Chain RuleProblem:9 Problem:If 18 binder bind 900 books in 10 days. How days. many binders will be required to bind 660 books in 12 days?

Chain RuleSolution:9 Binders 18 x Books 900 660 Days 10 12

x = 18 * 660 * 10 12* 900 = 11 binders

Chain RuleProblem:10 Problem:If 6 men working 8 hours a day earn Rs. 840 Rs. per week, then 9 men working 6 hours a day will earn how much per week?

Chain RuleSolution:10 Men Hours Earn 6 8 840 9 6 ? More men = More earning Less hour = less earning x = 6 * 9*840 6*8 = Rs. 945

ClocksProblem:1 Problem:Find the angle between the hour hand and the minute hand of a clock when the time is 3.25? 25?

ClocksSolution:1 = 11 m/2 30 (h) = 11 25/2 30 (3) = 11*12.5 90 =137.5 90 =47.50

ClocksProblem:2 Problem:Find the reflex angle between the hands of clock at 10.25? 10.25?

ClocksSolution:2 = 11 m/2 30 (h) = 11 25/2 30 (10) = 11*12.5 300 =137.5 300 =(-162.5) =(= 360 162.5 = 197.50

ClocksProblem:3 Problem:Find the angle between the hour hand and the minute hand of a clock when the time is 8.30? 30?

ClocksSolution:3 = 11 m/2 30 (h) = 11 30/2 30 (8) = 11*15 240 =165 240 =-75 = 360 75 = 197.50

ClocksProblem:4 Problem:Find the angle between the hour hand and the minute hand of a clock when the time is 4.20? 20?

ClocksSolution:4 = 11 m/2 30 (h) = 11 20/2 30 (4) = 11*10 120 =110 120 =-10 = 360 10 = 3500

ClocksProblem:5 Problem:How many times in a day, the hands of a clock are straight?

ClocksSolution:5 In 12 hours the hands co inside or are opposite 22 times. Thus 44 times the hands of the clocks are straight.

ClocksProblem:6 Problem:A clock is started at noon. By 10 minutes noon. past 5, the hour hand has turned through?

ClocksSolution:6 Angle traced by hour hand in 12 hours = 3600 5 hours 10 min. = 31 / 6 hrs. = (360 / 12) X (31 / 6) = 1550

ClocksProblem:7 Problem:An accurate clock shows 8 oclock in the morning. morning. Through how many degrees will the hour hand rotate when the clock shows 2 oclock in the afternoon?

ClocksSolution:7 Angle traced by hour hand in 6 hours = 3600 5 hours 10 min. = 31 / 6 hrs. = (360 / 12) X (6) = 1800

ClocksProblem:8 Problem:Find the reflex angle between the hands of clock at 12.20? 12.20?

ClocksSolution:8 = 11 m/2 30 (h) = 11 20/2 30 (12) = 11*10 360 =110 360 =(-250) =(= 360 250 = 1100

ClocksProblem:9 Problem:Find the angle between the hour hand and the minute hand of a clock when the time is 7 oclock 900 seconds?

ClocksSolution:9 = 11 m/2 30 (h) = 11 15/2 30 (7) = 11*7.5 210 =82.5 210 =(-127.5) =(= 360 (127.5) = 232.50

ClocksProblem:10 Problem:Find the angle between the hour hand and the minute hand of a clock when the time is 2 oclock 1800 seconds?

ClocksSolution:10 = 11 m/2 30 (h) = 11 30/2 30 (2) = 11*15 60 11*15 = 165 60 = 1050

Compound InterestProblem:1 Problem:If the difference between CI and SI on a sum of money for 3 years at 10% per annum is 10% Rs. Rs.930 /- Find the sum?

Compound InterestSolution : 1 The difference between CI & SI for 3 yrs at 10% is = 3.1 For Rs. 100 difference is 3.1 At what amount the difference become 930 100 = 3.1 ? = 930 = 930 * 100 / 3.1 = 30000

Compound InterestProblem 2:If the difference between CI and SI on a sum of money for 2 years at 10% per annum is 10% Rs. Rs. 300 /- . Find the sum?

Compound InterestSolution : 2 The difference between CI & SI for 2 yrs at 10% is =1 For Rs. 100 difference is 1 At what amount the difference become 300 100 = 1 ? = 300 = 300 * 100 / 1 = 30000

Compound InterestProblem 3:Mr. Mr. Kumar borrowed Rs.12000 /- at 11% per Rs. 11% annum on SI and immediately he lent the whole sum to Mr. Raj at 11% p.a. on CI. What will be the Mr. 11% CI. gain of Mr. Kumar after 2 years? Mr.

Compound InterestSolution : 3 Simple Interest for 100 100 = 11(R) * 2 (N) 100= 22 For 12000 ( 100 * 120) = 22 * 120 12000 = 2640 Compound Interest for 100 CI = SI + R*R/100 = 11*2 + 11/100 = 23.21

cont

Compound InterestSolution :3(cont.) For 100 = 23.21 Fro 12000 ( 100*120) = 23.21 * 120 12000 = 2785.2 Profit = CI SI = 2785.2 2640 = 145.2

Compound InterestProblem 4:The CI on a sum at 5% per annum for 3 years is Rs. 1261. What will be the SI on the Rs. 1261. same sum for the same period and the same rate?

Compound InterestSolution : 4 CI = SI + R*R/100 CI = 5*3 + 5*5/100 CI = 15.25 for Rs.100 SI = 5*3 = 15 for Rs.100 CI = SI 15.25 = 15 1261 = ? 1261 * 15 / 15.25 = 1240 Simple interest = 1240

Compound InterestProblem:5 Problem:CI on a sum of money at 12% p.a. for 2 12% years is Rs.1272. What will be the SI on the Rs.1272. same sum for the same period and at the same rate?

Compound InterestSolution : 5 CI = SI + R*R/100 CI = 12*2 + 12*12/100 CI = 25.44 for Rs.100 SI = 12*2 = 24 for Rs.100 CI = SI 25.44 = 24 1272 = ? 1272 * 24 / 25.44 = 1200 Simple interest = 1200

Compound InterestProblem:6 Problem:The CI on a sum of at 13% p.a. for the 13% second year is Rs.1469 find the sum? Rs.

Compound InterestSolution : 6 CI = SI + R*R/100 CI = 13*2 + 13*13/100 CI = 27.69 for Rs.100 For Rs. 100 CI is 27.69, at what amount they get 1469 as a interest 100 = 27.69 ? = 1469

Compound InterestProblem:7 Problem:A sum of money at CI amounts to thrice itself in 4 years in how many years will it become 9 times of itself?

Compound InterestSolution : 7 P = 100 Thrice = 300 4 years 900 4 years The amount get 9 times in 8 years

Compound InterestProblem:8 Problem:Rs. Rs.3757 is to be divided between A&B such that A share at the end of 7 years is equal to Bs share at the end of 9 years. If rate years. percent be 10% p.a. CI, Bs share is? 10%

Compound InterestSolution : 8 As share = Rs.x Bs share = 3757 x X (1+10/100)7 = (3757 x)(1+10/100)9 X = Rs. 2057 So, Bs share = 3757 -2057 =1200

Compound InterestProblem:9 Problem:There is 60% increase in an amount in 6 60% years at simple interest. What will be the interest. compound interest of Rs. 12,000 after 3 Rs. 12, years at the same rate?

Compound InterestSolution :9 Let P = Rs.100,S.I= Rs.60,T= 6 years. R= 100 * 60/(100*6)=10% Now take P = Rs.12000,T= 3 years, R= 10% C.I = [12000*(1+10/100)^3]-12000 [12000*(1+10/100)^3]C.I = Rs.3972.

Compound InterestProblem:10 Problem:The difference between CI & SI on a certain sum for 2 years at 4% per annum is Re. 1. Re. Find the sum?

Compound InterestSolution: Solution:10 Let p = x ,R = 4%, T= 2 years. years. C.I = [x*(1+(4/100)^2-x]=51x/625. [x*(1+(4 100)^2 x]=51x/625. S.I = x*4*2/100=2x/25 x*4 100= x/25 C.I-S.I = (51x/625)-(2x/25)=1; x= Rs.625. 51x/625) x/25)=1 Rs.625.

Compound InterestSolution : 10 CI = SI + R*R/100 CI = 4*2 + 4*4/100 CI = 8.16 for Rs.100 SI = 4*2 = 8 for Rs.100 CI SI = R * R/100 8.16 8 = R * R/100 100/.16 = 100 = 100 * 100 /.16 = 625

Compound Interest Problem:1A, B and C enter into partnership. C contributes one third of the capital while B contributes as much as A and C together contribute. If the profit at the end of the year amounts to Rs.840/Rs.840/what would each receive?

PartnershipSolution:1 As A contributes one third of the capital As profit = 840/3 = Rs.280 Now as B contributes as much as A and C So profit of B = Profit of A + Profit of C = Rs.280 + Profit of C Profit of B Profit of C = Rs.280 and Profit of B + Profit of C = Rs.840

PartnershipSolution:1(Cont. . .): Adding 2 Profit of B = Rs.840 Profit of B = Rs.420 Hence Profit of C = 840 420 280 = Rs.140.

Partnership Problem:2If A puts in Rs.400 for 12 months, how much Bs capital must be used for 8 months so that the profits may be divided in the ratio of 2:1?

PartnershipSolution:2 As investment for 1 month in order to have the same profit as before = Rs.400 * 12 = Rs.4800 Now profits are distributed in 2:1 Bs investment * 4800 = Rs.2400 for 1 month Rs.2400 * 1/8 i.e. Rs.300 for 8

Partnership Problem:3A and B enter into partnership and their shares are in the ratio of : 1/3. After 4 months A withdraws half of his capital and after 8 months more, a profit of Rs.500 is divided. What is As share of profit ?

PartnershipSolution:3 Ratio of the profits = : 1/3 = 3:2 They must put their capital in the same ratio. If A puts Rs.300, then B puts Rs.200. As investment for 1 month = Rs.(300*4+150 * 8) = Rs.2400 Bs investment for 1 month = Rs.200 * 12 = Rs.2400 As the investment for each is the same As profit = Rs.1/2 * 500 = Rs.250.

Partnership Problem:4A and B enter into partnership with capitals as 5:6; at the end 8 of months A withdraws. If they receive profits in the ratio of 5:8, find how long Bs capital was used?

PartnershipSolution:4 Let the capital of B be used for x months. Ratio of capitals of A and B is 5:6. Ratio of their periods is 8:x Raito of the their capital for one month is 40:6x As the ratio of the profits = 5:8 40 / 6x = 5 / 8 x = 10 2/3

Partnership Problem:5A began business with Rs.4500 and was joined afterwards by B with Rs.3000. When did B join if the profits at the end of the year were divided in the ratio of 2:1?

PartnershipSolution:5 As investment for 1 month = Rs.12 * 4500 = Rs.54000 As B gets profit, his investment for 1 month = * Rs.54000 = Rs.27000 As B puts in Rs.3000, his investment is used for 27000 / 3000 months = 9 months. Hence B joined A after 3 months of the start of the business.

Partnership Problem:6A and B enter into partnership. A, whose money has been in the business for 4 months claims of the profit. If B has Rs.1560 in the business for 8 months. How much money did A contribute.

PartnershipSolution:6 B claims 1 - 3/8 i.e. 5/8 of the profits. To claim 5/8 of the profit, B has to invest Rs.1560 for 8 months or Rs.1560 *8 i.e. Rs.12480 for 1 month. To claim 3/8 of the profit, A invests Rs.12480 * 8/5 * 3/8 for 1 month or Rs.12480 * 3/5 * for four months. i.e. Rs.1872 for four months.

Partnership Problem:7 Problem:Two partners invest Rs.12,500 and 8,500 Rs.12, respectively in their business and arrange that 60% of the profit should be divided 60% equally between them and the remaining profit treated as interest on the capital. If one capital. partners share is Rs.300 more than that of Rs. the other, find the whole amount of the profit. profit.

PartnershipSolution:7 Ratio of the capitals = 12,500:8,500 = 25:17 The interest is to be divided in the ratio 25:17. Difference = 25 17 = 8. But the actual difference = Rs.300 If Rs.8 is the difference, total interest = 25+17 = Rs.42 Rs.300 is the difference, total interest = 42 * 300 / 8 = Rs.1575 But this interest is 40% of the total income Hence whole profit = 1575 * 100 / 40 = Rs.3937.50

PartnershipProblem : 8 A and B jointly invest Rs.2100 and Rs.3100 respectively in a firm. A is an active partner hence he gets 25% of profit separately if their business yield them total Rs.1040 as profit. What will be the gain for each of them?

PartnershipSolution: 25% of profit = 1040*25/100 = 260 Remaining profit = 1040 -260 = 780 Ratio = 21:31 A share = 260+ 780 * 21/52 = 575 B share = 780 * 31/52 = 465

PartnershipProblem : 9 A, B, C enter into a partnership with shares in the ratio 7/2:4/3:6/5. After 4 months, A increases his share by 50% If total profit at the end of 1 year be Rs. 21600, then Bs share in the profit is?

PartnershipSolution: Given Ratio = 7/2 : 4/3 : 6/5 = 105 : 40: 36 Let their initial investment be Rs. 105 , Rs.40, Rs.36 Ratio of investment = 105*4 + (150 of 105* 8) : 40*12 : 36*12 = 1680 : 480 :432 = 35:10 :9 Bs share = 21600 * 10/54 = 4000

PartnershipProblem 10: A, B and C rented a pasture A puts in 12 oxen for 6 month, B puts 8 oxen for 7 months and C 6 oxen for 8 month. If the rent of the field is Rs. 396 what rent is paid by A?

PartnershipSolution: 10 A, B and C ratio = 12 * 6 : 8*7 : 6*8 = 72 : 56: 48 = 9:7:6 Rent paid by A = 396 * 9/22 = 162

Percentage

PercentageProblem:1A men spent Rs.229.50 p which is 85% of what he earned. How much does he earn?

PercentageSolution:1 85% of earning = Rs.229.50 p His earning = Rs.229.50 * 100/85 = Rs.270

Percentage Problem:2If 97 percent of students are present in a class and 27 students are absent, find the total number of students in the class.

PercentageSolution:2 Let 97% of the students are present so 3% of the students are absent. 3% of the students = 27 Number of students = 27 * 100/3 = 900

Percentage Problem:3In a examination 75 % of the candidates passed in English, 65% in Mathematics and 27% failed in both subjects. Find the pass percentage.

PercentageSolution:3 Those failing in Math = 35% Those failing in English = 30% Failing in both = 27% Failing in one or both the subjects = (35 + 30 27)% =38% Pass% = 62%

Percentage Problem:4Out of a class of 38 girls, 3 were absent. 20% of the remainder failed to do home work. Find the number of girls who did the home work.

PercentageSolution:4 No. of the girls present = 38 -3 =35 No. of girls who did not do home work = 20% * 35 = 7 No. of girls who did their home work = 35 7 = 28

Percentage Problem:5When 10 percent is lost in grinding, a country imports 15 million bags of wheat. When 4% only is lost in grinding, it imports only 6 million bags. Find the produce of the country.

PercentageSolution:5 When 4% is lost instead of 10%, saving = 6% But actual saving of wheat =15 6 = 9 million bags 6% of the total produce = 9 million bags. => Total produce = 9 * 100/6 = 150 million bags.

Percentage Problem:6A rise of 20% in the price of rice compels a person to buy 4 kgs less for Rs.80. Find the increased price per kg.

PercentageSolution:6 Rise in price = 20% of Rs.80 = Rs.16 Increased price of 4 kg = Rs.16 Increased price of rice per kg = Rs.16/4 = Rs.4

Percentage Problem:7An election candidate who gets 35% of the votes is defeated by a majority of 150 votes. Find the total number of votes recorded.

PercentageSolution:7 If the total votes recorded be x, then votes obtained by two candidates are 35% x and 65% x. Now 65% x 35% x = 150 30% x= 150 x = 150 * 100/30 x = 500

Percentage Problem:8 Problem:A reduction of 20 percent in the price of mangoes enables a man to buy 25 mangoes more for Rs.40. Find the Rs.40. reduced price of the basket containing 200 mangoes. mangoes.

PercentageSolution:8 The reduction of 20% on Rs.40 enables the customer to buy 25 mangoes more at the reduced rate. Reduced price of 25 mangoes = 20% of Rs.40= Rs.8 Reduced price of the basket containing 200 mangoes = Rs.8/25 * 200 = Rs.64.

Percentage Problem:9Income of C is 20% more than Bs and income of B is 25% more than As. Find by how much percent Cs income is more than As?

PercentageSolution:9 Let the income of A be Rs.100 Bs income = 125% of Rs.100 = Rs.125. Cs income = 120% of Rs.125 = 120/100 * 125 = Rs.150 Thus income of C is 50% more than As.

Percentage Problem:10A candidate must get 33% marks to pass. He gets 220 marks and fails by 11 marks. What is the maximum number of marks?

PercentageSolution:10

33% x = 220 + 11 + 231 where x be the maximum marks x = 231 * 100/33 x = 700

Permutation & CombinationBlue Lotus

Permutation & CombinationProblem1 : Problem1A code word is to consist of two english alphabets followed by two distinct numbers between 1 and 9. for example, CA23 CA23 is a code word. How many such code word. words are there ?

Permutation & CombinationSolution:1 There are in all 26 alphabets. We have to choose 2 distinct alphabets First alphabet selected in 26 ways Second alphabet selected in 25 ways. Thus out of 9 digits, first can be selected in 9 and second can be selected in 8 ways. = 26 * 25*9* 8=46800.

Permutation & CombinationProblem: Problem:2 The numbers of six digit that are divisible by 10, which can be formed by 10, using the digits 1,2,7,0,9,5 ?

Permutation & CombinationSolution2 : The numbers are divisible by 10 is 0 is in unit place.So only rest of 5 digits can be changed. P( 5,5) = 5! = 120 ways.

Permutation & CombinationProblem: Problem:3 In how many different ways can be the letters of the word RUMOUR be arranged?

Permutation & CombinationSolution3 : Arrangement possible, RUMOUR = 6!/2!*2! = 720/4 = 180

Permutation & CombinationProblem: Problem:4 How many arrangements can be made out of the letters of the word

ENGINEERING?

Permutation & CombinationSolution:4 ENGINEERING (3E, 3N, 2G, 2I) = 11!/3! 3! 2! 2! = 39916800/144 = 277200

Permutation & CombinationProblem:5 Problem:Four alphabets E,K,S and V, one in each , were purchased from a plastic warehouse. How many warehouse. ordered pairs of alphabets , to be used as initials can be formed from them ?

Permutation & CombinationSolution:5 The required number of ordered pair of alphabets , to be used as initials can be formed = P(4,2) = 4!/2! =4*3 = 12.

Permutation & CombinationProblem:6 Problem:In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women?

Permutation & CombinationSolution:6 Required Number of ways = 8C5 * 10 C 6 = 8C5 * 10 C 4 = 8*7 * 6 * 10 * 9 * 8*7 3*2 *1 4*3*2*1 = 11760

Permutation & CombinationProblem7 : Problem7In an examination, yamini has to select 4 questions from each part .There are 6,7,8

questions in part I, part II ,part III respectively. respectively. What is the number of possible combinations in which they can choose questions ?

Permutation & CombinationSolution 7: p = C(6,4) * C(7,4) *C(8,4) = 15*35*70 = 36750 ways.

Permutation & CombinationProblem:8How many words can be formed by using all the letters of the word, ALLAHABAD?

Permutation & CombinationSolution:8 ALLAHABAD = 9! / 4! * 2! = 362880 / 48 = 7560

Permutation & CombinationProblem:9 Problem:How many words can be formed by using all the letters of the word ACTION so that the vowels always come together?

Permutation & CombinationSolution:9 CTN ( AUIO) 3! +1! 4! = 4! * 4! = 24*24 = 576

Permutation & CombinationProblem:10 Problem:How many words can be formed by using all the letters of the word CORPORATION so that the vowels always come together?

Permutation & CombinationSolution:10 CRPRTN (OOAIO) 6!+1! ( R occurred twice) = 7! / 2! = 2520 5 vowels in which 0 occur 3 times = 5! / 3! = 20 ways Required number of ways = 2520 * 20 = 50400

Pipes and Cistern Problem:1A cistern is filled in 9 hours and it takes 10 hours when there is a leak in its bottom. If the cistern is full, in what time shall the leak empty it?

Pipes and CisternSolution:1 Work done in 1 hour by the leak and the filling pipe = 1/9 Work done by the leak in 1 hour=1/9 1/10= 1/90 Hence the leak can empty it in 90 hours.

Pipes and Cistern Problem:2 Problem:Three pipes A, B and C can fill a cistern in 12, 12, 15 an 20 minutes respectively. Pipes A respectively. and C work for one minute and pipes B and C work in next minute. In how many minutes minute. the cistern will be full if this procedure is continued?

Pipes and CisternSolution:2 Work done by pipes A and C in 1 minute = 1/12 + 1/20 = 5+3/60 =8/60 Work done by pipes B and C in the next minute = 1/15 + 1/20 = 4+3/60 = 7/60 Part of the cistern filled in 2 minutes = 8/60 + 7/60 = 15/60 = The cistern shall be full in 8 minutes.

Pipes and CisternProblem:3Two pipes A and B would fill a cistern in 20 and 24 minutes respectively. Both pipes being opened, find when the first pipe must be turned off so that the cistern may be just filled in 12 minutes.

Pipes and CisternSolution: Solution:3 A little thinking will solve this problem quickly. quickly. Since the cistern is to be filled in 12 minutes. minutes. Second pipe shall fill half of it. Rest it. of half of the cistern is to filled by 1st pipe A which it takes 10 minutes. So the 1st pipe minutes. must be opened for 10 minutes i.e. it must be turned off after 10 minutes. minutes.

Pipes and CisternProblem:4 Problem:A person alone can do a piece of work in 10 days. days. B alone can do it in 15 days. Total days. wages for the work are Rs.50. How much A Rs.50. is paid if they both finish the work together?

Pipes and CisternSolution: Solution:4 A can do in 10 days, B can do in 15 days. days. A and B together can do it in 6 days. days. Total wage = Rs.50. Rs.50. As share = 50*6/10 = Rs.30. 50* Rs.30.

Pipes and Cistern Problem:5A vessel is fully filled with a special liquid. Four litres of liquid is drawn out of vessel and is replaced with water. If the ratio of the special liquid to the water becomes 1:2, then what is capacity of vessel.

Pipes and CisternSolution:5 After 1st operation After 2nd operation L xx -4 x+(24/x)x+(24/x)-10 W 4 10 24/x

x+(24/x)-10/{10x+(24/x)-10/{10-(24/x)} = on solving we get x = 12.

Pipes and Cistern Problem6 :A tap can fill a tank in 48 minutes, whereas another tap can empty it in 2 hours. It both the tap are opened at 11.40 A.M., then the tank will be filled at ?

Pipes and CisternSolution:6 In one minute, 1/48 1/120 = 1/80 of tank can be filled. The whole tank can be filled in 80 minutes. 11.40 + 1.20 = 1.00 P.M.

Pipes and Cistern Problem:7Two pipes A and B can fill a tank in 12 minutes and 15 minutes respectively. If both the taps are opened simultaneously, and the tap A is closed after 3 minutes, then how much more time will it take a fill the tank by tap B?

Pipes and CisternSolution:7 Part filled in 3 min. = 3(1/12 + 1/15) = (3*9/60) = 9/20 Remaining part = (1-9/20) = 11/20 (1Part filled by B in 1 min. = 1/15 1/15:11/20::1:x or x = (11/20 * 1 * 15) = 8 = 8min. 15 sec. Remaining part is filled by B in 8 min. 15 sec.

Pipes and Cistern Problem:8Pipe A can fill a tank in 5 hours, pipe B in 10 hours and pipe C in 30 hours. If all the pipes are open, in how many hours will the tank be filled?

Pipes and CisternSolution:8 Part filled by (A+B+C) in 1 hour = (1/5 + 1/10 + 1/30) = 1/3 All the three pipes together will fill the tank in 3 hours.

Pipes and Cistern Problem:9 Problem:A cistern can be filled by a tap in 4 hours while it can be emptied by another tap in 9 hours. If both the taps hours. are opened simultaneously, then after how much time will the cistern get filled?

Pipes and CisternSolution:9 Net part filled in 1 hour = (1/4 1/9) = 5/36 The cistern will be filled in 36/5 hrs i.e.,7.2 hrs.

Pipes and Cistern Problem:10A pump can fill a tank with water in 2 hours. Because of a leak, it took 2 1/3 hrs to fill the tank. The leak can drain all the water of the tank in:

Pipes and CisternSolution:10 Work done by the leak in 1 hour = (1/2 3/7) = 1/14 Leak will empty the tank in 14 hrs.

Problems on Numbers

Problems on Numbers Problem:1Find the divisor when the quotient, dividend and the remainder are 457, 98793 and 81 respectively is. .

Problems on NumbersSolution:1 Divisor = (Dividend Remainder) / Quotient = (98793 83) / 457 = 216

Problems on Numbers Problem:2Find the number nearest to 77993 that is exactly divisible by 718.

Problems on NumbersSolution:2 Dividing 77993 by 718, then the remainder as 449 which is more than half the divisor. Thus the number nearest 77993 is = 77993 + (718 449) = 77993 + 269 = 78262

Problems on Numbers Problem:3A certain number when divided by 899 gives a remainder 63. What is the remainder when the same number is

Problems on NumbersSolution:3 Number = 899 * Quotient + 63 = 29 * 31* Q + 2 * 29 + 5 The remainder obtained by dividing the number by 29 is clearly 5.

Problems on Numbers Problem:4In a division sum the divisor is 10 times the quotient and five times the remainder. Find the dividend when the remainder is 46.

Problems on NumbersSolution:4 If the remainder is 46, the divisor is 5 * 46 i.e. 230 and divisor is 10 times the quotient, therefore, the quotient is (230 / 10) i.e. 23. Hence the number is Divisor * Quotient + remainder 230 *23 + 46 = 5336

Problems on Numbers Problem:5How many numbers between 11 and 90 are divisible by 7?

Problems on NumbersSolution:5 The required numbers are 14, 21, 28, 35, . . ., 77, 84. This is an A.P. with a=14 and d=(21 14) =7 Let it contain n terms. Then, Tn = 84 => a + (n 1) * 7 = 84 n = 11 Required number of terms = 11.

Problems on Numbers Problem:6Find all odd numbers upto 100.

Problems on NumbersSolution:6 The given numbers are 1, 3, 5, 7, . . ., 99. This is an A.P. with a=1 and d=2 Let it contain n terms. Then, 1 + (n 1) * 2 = 99 n = 50.

Problems on Numbers Problem:7Find all 2 digit numbers divisible by 3.

Problems on NumbersSolution:7 All 2 digit numbers divisible by 3 are: This is an A.P. with a=12 and d=3. Let it contain n terms. Then, 12 + (n 1) * 3 = 99 n = 30

Problems on Numbers Problem:8How many numbers between 11 and 90 are divisible by 5?

Problems on NumbersSolution:8 The required numbers are 15, 20, . . ., 90. This is an A.P. with a=15 and d=(20 15)=5 Let it contain n terms. Then, Tn = 90 => 15 + (n 1) * 5 = 90 n = 14

Problems on Numbers Problem:9A number when divided by a certain divisor left remainder 63. When twice the given number was divided by same divisor, remainder was 55. What is the divisor?

Problems on NumbersSolution:9 The require divisor is 2 * 63 55 = 126 55 = 71.

Problems on Numbers Problem:10Two numbers when divided by a certain divisor leave remainder 731 and 483 respectively. When the sum of the numbers be divided by the same divisor, the remainder is 203. What is the divisor?

Problems on NumbersSolution:10 It is clear that divisor is greater than the remainders 731 and 483 both sum of the remainders = 731 + 483 = 1214 The remainder when the sum is divided by the divisor = 203 Divisor = 1214 203 = 1011

Profit and Loss Problem:1An article is bought for Rs.20 and sold for Rs.25. What is the gain %?

Profit and LossSolution:1 Gain = Rs.5, so gain % = 5/20 * 100 = 25%

Profit and Loss Problem:2A book is bought for Rs.80 and sold at a gain of 20%. What is selling price of the book?

Profit and LossSolution:2 S.P. = 120% of the C.P. = 120/100 * Rs.80 = Rs.96.

Profit and Loss Problem:3An article is sold for Rs.60 at a gain of 25%, what is the cost price of the article?

Profit and LossSolution:3 S.P. = 125% of the C.P = Rs.60 C.P. = 60*100/125 = Rs.48

Profit and Loss Problem:4A man buys 5 apples for Rs.3 and sold each for Rs.2. What did he gain or lose?

Profit and LossSolution:4 C.P. of 5 apples = Rs.3 S.P. of 5 apples = Rs.2 * 5 = Rs.10 Gain = Rs.7, so gain % = 7/3 * 100 = 233 1/3%

Profit and Loss Problem:5A man sold two pens at Rs.20 each. He sold one at a loss of 10% and the other at a gain of 10%. His loss or gain % is. . .

Profit and LossSolution:5 Loss % = x% of x where x is a common gain and loss percent. In this question, loss is 10% and gain is also 10%, so common gain or loss% is 10. Loss % = 10% of 10 i.e. 1.

Profit and Loss Problem:6A dishonest dealer professes to sell his goods at cost price, but uses a weight of 900 grams for a kilogram weight. Find his real gain per cent.

Profit and LossSolution:6 For every 1000 grams he charges, he gives only 900 grams i.e. he gains 100 greams on every 900 grams, His gain per cent is 100 / 900 * 100 = 11 1/9%

Profit and Loss Problem:7By selling 11 oranges for a rupee, a man loses 10%. How many for a rupee should he sell to gain 10%?

Profit and LossSolution:7 S.P of 1 orange = Re. 1/11 90% of C.P. = 1/11, so, new S.P. of 1 orange is 110% of C.P. = 1/11 * 100/90 * 110/100 = Re.1/9 He must sell 9 oranges for a rupee.

Profit and Loss Problem:8If cost price of 8 articles is equal to the selling price of 10 articles. Find the gain or loss.

Profit and LossSolution:8 C.P. of 8 articles = S.P. of 10 articles In this case, on every 10 articles sold the cost of 8 article is recovered, so there is a loss of 2 articles Loss % = 2/10 *100 = 20%

Profit and Loss Problem:9If the cost price of 10 articles is equal to the selling price of 8 articles, find the gain of loss per cent.

Profit and LossSolution:9 As C.P of 8 articles = S.P of 10n articles By selling 8 articles, the cost of 10 articles is recovered. Hence on every 8 articles sold, gain is of 2 articles Gain % = 2/8 * 100 = 25%.

Profit and Loss Problem:10A shopkeeper marks his goods 20% above cost price, but allows 10% discount for cash. What per cent profit does he really make?

Profit and LossSolution:10 If the cost is Rs.100, he marks 20% of it more i.e. The marked price = Rs.120 Discount = 10% of Rs.120 = Rs.12. Net S.P. = Rs.120 Rs.12 = Rs.108 so gain = 8%

Ratio and Proportion Problem:1Some men promised to do a job in 18 days, but 6 of them became absent and remaining men did the job in 20 days. What is the original number of men.

Ratio and ProportionSolution:1 Let there be x men in beginning and they were to do the job in 18 days, so one man could do it in 18x days. But now (x -6) could do it in 20 days, hence one man could do it in 20 * (x-6) days. (x18x = 20(x 6) 18x = 20x 120 2x = 120 x = 60 Thus 60 men were there at first.

Ratio and Proportion Problem:2If 8 men can reap 80 hectares, in 24 days; how many hectares can 36 men reap the same field in 36 days.

Ratio and ProportionSolution:2 8 men reap 80 hectares, 36 men shall reap more field. In 24 days, 80 hectares is reaped, in 36 days more field is reaped. No. of hectares reaped = 80 * 36/8 * 36/24

Ratio and Proportion Problem:3What is the ratio between 2 meters 20cm: 15 meters 49 cm.

Ratio and ProportionSolution:3 2 meters 20 cm = 220 cm and 15 meters 40 cm = 1540 cm The ratio = 220/1540 = 1/7 = 1:7

Ratio and Proportion Problem:4A garrions of 750 men has provisions for 20 weeks. If at the end of 4 weeks, they are reinforced by 450 men, how long will the provisions last?

Ratio and ProportionSolution:4 At the end of 4 weeks, the food for 750 men is left for (20 4) i.e.16 weeks. At this stage, the number of men are 750 + 450 = 1200. So the question now takes the shape as follows: If 750 men can have the food in 16 weeks, then how long 1200 men take to consume the same food. If x is the required number. The inverse ratio of men = direct ratio of weeks 1200 : 750 = 16:x x = 750 * 16 / 1200 x = 10 weeks

Ratio and Proportion Problem:5Rs.49 were divided among 150 children. Each girl got 50 paise and a boy 25 paise. How many boys were there.

Ratio and ProportionSolution:5 Let there be x boys, so that there are (150 x)girls (150 x * + (150 x) * = 49 x + 2(150 x) = 49 * 4 x + 300 2x x = 196 = 104

Ratio and Proportion Problem:6If 30 men working 7 hours a day can do a piece of work in 18 days, in how many days will 21 men working 8 hours a day do the same work.

Ratio and ProportionSolution:6 (i) 30 men can do a work in 18 days, 21 men shall do the same work in more days. (ii) If 7 hours a day, the work is done, it takes 18 days, while if 8 hours a day, the work is done, it takes less days. No. of days = 18 * 30/21 * 7/8 = 22 days.

Ratio and Proportion Problem:7Two numbers are in the ratio of 3:5. If 9 be subtracted from each, then they are in the ratio of 12:23. Find the second number.

Ratio and ProportionSolution:7 Let 3x 9 / 5x 9 = 12 / 23, where 3x and 5x be the numbers. => 23(3x 9) = 12(5x 9) 69x 207 = 60x 108 9x = 99 Hence second number = 5 * 11 = 55.

Ratio and Proportion Problem:8Find the fraction which shall bear the same ratio to 4/9 that 3/11 does to 5/33.

Ratio and ProportionSolution:8 Let the required fraction be x; then x : 4/9 = 3/11 : 5/33 => x * 5/33 = 4/9 * 3/11 x = 4/9 * 3/11 * 33/5 x = 4/5

Ratio and Proportion Problem:9A mixture contains alcohol and water in the ratio 5:6. If 5 litres of water is added to the mixture, the ratio becomes 4:5. Find the quantity of alcohol in the given mixture.

Ratio and ProportionSolution:9 Let the quantity of alcohol and water be 4x litres and 3x litres respectively 4x / (3x + 5) = 4 / 5 20x = 4(3x + 5) x = 2.5 Quantity of alcohol = (4 * 2.5) litres = 10 litres.

Ratio and Proportion Problem:10A bag contains 50p, 25 p and 10 p coins in the ratio 6:10:4, amounting to Rs.206. Find the number of coins of type.

Ratio and ProportionSolution:10 Let the number of 50 p, 25 p and 10 p coins be 6x, 10x and 4x respectively. (6x/2) + (10x/4) +(4x/10) = 206 59x = 2060 x = 35. Number of 50 p coins = (6 * 35) = 210 Number of 25 p coins = (10 * 35) = 350 Number of 10 p coins = (4 * 35) = 140

Simple InterestProblem:1 Problem:Mr. Mr. A borrowed Rs. 12000/- at the rate of Rs. 12000/ 10% 10% and lent the same amount to Mr. B Mr. at the rate of 13%. What will be the gain 13% of A, after 5 years?

Simple InterestSolution 1: Amount borrowed = 12000 Borrowed at 10% interest Lent at 13 % interest Gain = Lent Borrowed = 13 10 = 3% 100 = 3*5 100 = 15 100 (120) = 15(120) 12000 = 1800 Gain = 1800 for 12000

Simple InterestProblem:2 Problem:In what time, Rs.6000 gives Rs. 2880 as Rs. Rs. SI at the rate of 12% ? 12%

Simple InterestSolution 2: P = 6000 SI = 2880 R = 12% N=? For 100 = 12 * 1 ( 1 year) For 6000 (100*60) = 12 * 60 1 year for 100 interest is 720 1 year = 720 ? Year = 2880 = 2880 / 720 = 4 years

Simple InterestProblem:3 Problem:Rs. 7000/ Rs. 7000/-amount to Rs. 9,100/- in 3 Rs. 100/ years SI. If the rate of interest is SI. increased by 5%, it would amount to how much?

Simple InterestSolution 3: For 100 5 % increase 3 years 5% means = 15 % increase For 100 = 15% For 7000( 100 *70) = 15*70 = 1050 + 9100 = 10150

Simple InterestProblem:4 Problem:A principal amount to Rs.8880/- in 4 years Rs.8880/ and Rs.9600/- in 5 years. What is the rate Rs.9600/ years. of interest?

Simple InterestSolution 4: S.I for one year = 9600-8880 = Rs 720 9600S.I for four years = 4*720 = 2880 Therefore P = 8880-2880 = Rs 6000 8880SI for 4 years = (6000* 4 * R)/100 = 2880 R= 12%

Simple InterestProblem:5 Problem:A principal amount give Rs. 12000/- as SI Rs. 12000/ for 7 years at the rate of 11%. To get the 11% same amount of interest for 11 years, what will be the rate of interest?

Simple InterestSolution 5: 12000 for 7 years at the rate of 11% If they want to get the same interest in 11 years means reduce the rate of interest (i.e) 12000 = 7 yrs = 11 % 12000 = 11 yrs = 7 %

Simple InterestProblem:6 Problem:A sum of money becomes double in 10 years. years. What is the rate of interest?

Simple InterestSolution 6: Sum of money double in 10 yrs Rate = ? Rate = 100/ N = 100 / 10 = 10%

Simple InterestProblem:7 Problem:A sum of money becomes 5/4 of itself in 5 years at a certain rate of interest. What is interest. the rate of interest?

Simple InterestSolution 7: P = 400 A = 5/4 * 400 ( sum of money 5/4 of itself) = 500 I = 500 400 = 100 100 = 5 years 100 for 5 yrs means 1 yr 20 400 = 20 For 400 20% , then for 100 ? 100 = 5%

Simple InterestProblem:8 Problem:At what rate percent per annum will a sum of money double in 16 years?

Simple InterestSolution 8: Sum of money double in 16 yrs Rate = ? Rate = 100/ N = 100 / 16 = 6.25%

Simple InterestProblem:9 Problem:A sum of Rs. 12,500 amounts to Rs Rs. 12, 15, 15,500 in 4 years at the rate of simple interest. interest. What is the rate of interest?

Simple InterestSolution 9: The sum of 12 500 amount to 15500 in 4 yrs 1 yr = 15500/ 4 = 3875 Difference 15500 12500

3000For 4 yrs1 yr = 3000/4 = 750 = 12500 /750 = 6%

Simple InterestProblem:10 Problem:Simple interest on a certain amount is 9/16 of principal. principal. If the numbers

representing the rate of interest in percent and time in years be equal, then time for which the principal is lent out will be how

Simple InterestSolution 10: If the no. of years = rate% R = 10 I R = 10 9/6 = 10 * = 7 2/4 7 years 6 months

Time and Work Problem:1A can do a piece of work in 6 days, and B can do it in 12 days. What time will they require to do it working together?

Time and WorkSolution:1 Part of the work done by A in one day = 1/6 Part of the work done by B in one day = Part of the work done by A and B in one day = 1/6 + 1/12 = .

Time and Work Problem:2A and B together can do a piece of work in 6 days. A alone can do it in 10 days, what time will B require to do it working alone?

Time and WorkSolution:2 Part of the work done by A and B in 1 day = 1/6 Part of the work done by A in 1 day = 1/10 Part of the work done by B in 1 day = 1/6 1/10 = 1/15 B can work alone at it for 15 days.

Time and Work Problem:3A can do a piece of work in 15 days and B in 20 days. With the help of C, they finish the work in 5 days. How long will it take C to finish the work?

Time and WorkSolution:3 (A+B)s 1 days work = 1/18 (B+C)s 1 days work = 1/24 (C+A)s 1 days work = 1/36; adding, we have 2(A+B+C)s one days work =1/18+1/24+1/36= 1/8 (A+B+C)s 1 days work = * 1/8 = 1/16

Time and Work Problem:4Twenty women can do a work in sixteen days. Sixteen men can complete the same work in fifteen days. What is the ratio between the capacity of a man and a woman?

Time and WorkSolution:4 (20 * 16) women can complete the work in 1 day. 1 womens 1 days work = 1/320 (16 * 115) men can complete the work in 1 day. 1 mans 1 days work = 1/240 So, required ratio = 1`/240:1/320 = 4:3

Time and Work Problem:510 men can finish a work in 10 days, whereas 12 women can finish same work in 10 days. How much time it will take for 15 men and 6 women to complete the work?

Time and WorkSolution:5 1 mans 1 days work = 1/10*10 1 womans 1 days work = 1/12*10 15 mens 1 days work = 15/10*10 = 15/100 6 womens 1 days work = 6/12*10 =6/120 Men & women together = 15/100 + 6/120 = 1/5 5 days

Time and Work Problem:6A does 4/5 of a work in 20 days. He then calls in B and they together finish the remaining work in 3 days. How long B alone would take to do the whole work?

Time and WorkSolution: Solution:6 Whole work is done by A in (20 * 5/4)=25 days. )=25 days. Now, (1-4/5) i.e., 1/5 work is done by A and B in 3 days. days. Whole work will be done by A and B in (3 * 5) = 15 days. days. As 1 days work = 1/25, (A+B)s 1 days work 25, = 1/15 Bs 1 days work = (1/15 1/25) = 4/150 = 2/75 25) So, B alone would do the work in 75/2 = 37 75/ days. days.

Time and Work Problem:72 men undertake to do a piece of work for Rs.1400 first man take 7 days to complete and second man take 8 days to complete. If they work together they complete in 3 days with the help of boy how the amount is

Time and WorkSolution:7 Wages for the first man in 3 days = work done in 3 days * 1400 = 3/7 * 1400 = 600 Wages for second man in 3 days = 3/8 *1400 =525 Boys wage = 1400 (600 +525) = 275

Time and Work Problem:8A can finish a work in 18 days and B can do the same work in 15 days. B worked for 10 days and left the job. In how many days, A alone can finish the remaining work?

Time and WorkSolution:8 Bs 10 days work = (1/15 *10) = 2/3 Remaining work = (1- 2/3) = 1/3. (1Now, 1/18 work is done by A in 1 day. 1/3 work is done by A in (18 * 1/3) = 6 days.S

Time and Work Problem:9X can do a piece of work in 40 days. He works at it for 8 days and they Y finished it in 16 days. How long will they together take to complete the work?

Time and WorkSolution:9 Work done by X in 8 days = (1/40 * 8) = 1/5 Remaining work = (1-1/5) = 4/5 (1Now, 4/5 work is done by Y in 16 days. Whole work will be done by Y in (16 * 5/4) = 20 days. Xs 1 days work =1/40, Ys 1 days work=1/20 (X+Y)s 1 days work = (1/40+1/20) = 3/40. Hence, X and Y will together complete the work in 40/3 = 13 1/3 days.

Time and Work Problem:104 men and 6 women can complete a work in 8 days, while 3 men and 7 women can complete it in 10 days. In how many days will 10 women complete it?

Time and WorkSolution:10 Let 1 mans 1 days work = x and 1 womans 1 days work = y Then, 4x + 6y = 1/8 and 3x + 7y = 1/10 Solving these two equations, we get: x = 11/400, y = 1/40 1 womans 1 days work = 1/400 => 10 womens 1 days work = (1/400 * 10) = 1/40 Hence, 10 women will complete the work in 40 days.

Problems on TrainsBlue Lotus

Problems on TrainsProblem:1 Problem:A train 150m long is running at a speed of 150m 90 kmph. Find the time taken by the train to kmph. cross a tree?

Problems on TrainsSolution:1 90 * 5/18 = 25 m/s 25 m = 1 sec 150 m = ? 150/25 = 6

Problems on TrainsProblem:2 Problem:An express train left Delhi for Howrah at 3 P.M at an average speed of 60 kmph at 1 P.M a goods train also had left Delhi for Howrah on a parallel track at an average of 40 kmph how far from Delhi is the express train overtake goods train. train.

Problems on TrainsSolution:2 At 3 P.M goods train would have traveled 80 km. Relative speed = 60 -40 =20 Time at which they overtake = 3 + 4 = 7 P.M Distance at which they overtake = 60 * 4 = 240km

Problems on TrainsProblem:3 Problem:With a speed of 60 kmph a train crosses a pole in 30 seconds. Find the length of the seconds. train?

Problems on TrainsSolution:3 60 * 5 / 18 = 50/3 m/sec 50/3 m = 1 sec ? = 30 sec 50/3 * 30 = 500

Problems on TrainsProblem:4 Problem:A train 700 m long is running at 72 kmph. If kmph. it crosses a tunnel in 1 minute, find the length of the tunnel?

Problems on TrainsSolution: Solution:4 length of train = 700m,speed = 72kmph = 20m/sec. 700m,speed 72kmph 20m/sec. time to taken by train to enter fully into tunnel = 700/20=35 700/20= sec. sec. Total time to cross tunnel = 1min35sec. min35sec. Length of tunnel = 20 m/sec * 95 sec = 1.9Km. Km.

Problems on TrainsProblem:5 Problem:A train 60 m long passes a platform 90 m long in 10 seconds. Find the speed of the seconds. train?

Problems on TrainsSolution:5 150= 10 sec 15 =1 15 * 18/5 = 54 kmph

Problems on TrainsProblem:6 Problem:A passenger train running at a speed of 80 kmph leaves the railway station after 6 hrs a goods train leaves and it over take in 4 hrs. Find speed of hrs. goods train?

Problems on TrainsSolution:6 Distance Traveled by passenger train to overtake the goods train = 80 * 4 = 320 km Distance traveled by goods train = (6 + 4) x =10x = 320 x = 320 /10 =32 km/ph

Problems on TrainsProblem:7 Problem:If a train 110 m long passes a telegraph pole in 3 seconds, find the time taken by it to cross a railway platform 165 m long?

Problems on TrainsSolution:7 110 m = 35 sec 275 = ? 275 * 3 / 110 = 7.5

Problems on TrainsProblem:8 Problem:The ratio between speeds of two train is 7:8, if the second train runs 400 km in 5 hrs. find hrs. the speed of 1st train?

Problems on TrainsSolution:8 Ratio of speeds = 7 : 8 1st train speed = x 2nd train speed = 400 / 5 = 80 kmph 7/8 = x /80 x = 70

Problems on TrainsProblem:9 Problem:A train 110 m length travel at 60 kmph how much time does that train take in passing a man walking against the train at 6 kmph?

Problems on TrainsSolution : 9 Speed of the train in msec = 60 * 5/18 = 50 /3 Speed of theman in msec = 6 * 5/18 = 5/3 How long it will take to cross the man = 110 / 55/3 = 6 sec

Problems on TrainsProblem:10 Problem:Two train travels in opposite direction at 36 kmph & 45 kmph and a man sitting in slower train passes the faster train in 8 sec. find the sec. length of the faster train?

Problems on TrainsSolution:10 Opposite side = x + y = 36 *5/18 + 45 * 5/18 = 10 + 12.5 =22.5 m/ sec 22.5 m = 1 sec ? = 8 sec 22.5 * 8 = 180 m

Problems on TrainsProblem:11 Problem:Two stations A and B are 110 km apart on a straight line. One train starts from A at 7 a.m line. and towards B at 20 kmph. Another train kmph. starts from B at 8 a.m and travels towards A at a speed of 25 Kmph. At what time will Kmph. they meet?

Problems on TrainsSolution:11 90 km 110 km

A (20) B 7 a.m 8 a.m At 8 a.m the distance is 90 km. Time to meet = 90/ 20 +25 = 2 hrs = 8 + 2 = 10 a.m.

Problems on TrainsProblem:12 Problem:Two trains are running in opposite directions towards each other with a speed of 54 kmph and 48 kmph respectively. If the length of respectively. one train is 250 m and they cross each other in 18 seconds, find the length of the other

Problems on TrainsSolution:12 Opposite Direction = S1 + S2 = 54 + 48 = 102 km/hr 102 * 5/18 = 85 / 3 85/3 = 1 sec 18 = ? 18 * 85 / 3 = 510 510 250 = 260 m

Probability Problem:1 Problem:Four different objects 1,2,3,4 are distributed at random in four places marked 1,2,3,4.What is the probability that none of the objects occupy the place corresponding to the number. number.

ProbabilitySolution1 Solution1 : Let the four places be a , b , c, d b a c d, b a d c, b c a d, b c d a, b d a c, b d c a Here out of six ways only three are permissible because of the non fulfilment of condition . So the required probability is 3/6 = .

Probability Problem:2 Problem:If the probability of rain on any given day in pune city is 50%, then what is the probability that it 50% rains exactly 3 days in a 5-days period ?

ProbabilitySolution2 Solution2 : Prob that it rains on 1st day = . similarly upto 5 days = 1/32. 32. Prob. Prob. That rains on exactly 3 days in a 5 day period = 5C3*(1/32) = 5/16. *(1 32) 16.

Probability Problem3: Problem3What is the possibility of getting at least 6 heads if eight coins are tossed simultaneously ?

ProbabilitySolution: Solution:3 Probability of getting 2 tails + Probability of getting 1 tail + probability of getting no tail = 8C2*(1/256)+8C1 *(1/256)+8C0 (1/256) *(1 256)+8 *(1 256)+8 256) = 37/256. 37/256.

Probability Problem:4 Problem:If the probability that A will 15 years is 7/8 and that B will live 15 years is 9/10, then what is the probability that both 10, will live after 15 years ?

ProbabilitySolution 4:

(7/8) * (9/10) = 63/80. 10) 63/80.

Probability Problem:5 Problem:A man and his wife appear in an interview for two vacancies in the same post. The post. probability of husbands selection is (1/7) and the probability of wifes selection is (1/5). What is the probability that only one of them is selected?

ProbabilitySolution: Solution:5 P(A) = 1/7 P(B) = 1/5 P(A*B) = P(A)*P(B) + P(A)*P(B) = 1/7(1-1/5) + 1/5(1- 1/7) = 4/35 + 6/35 = 10/35 10/ = 2/7

Probability Problem:6 Problem:Four persons are chosen at random from a group of 3 men, 2 women and 4 children. children. What is the chance that exactly 2 of them are children?

ProbabilitySolution: Solution:6 3M, 2N, 4C 9C4 = (18*7) = 126 18* 5C2 * 4C2 / 126 = 60/126 60/

Probability Problem:7 Problem:A box contains 20 electric bulbs, out of which 4 are defective. Two bulbs are chosen defective. at random from this box. What is the box. probability that at least one of these is defective?

ProbabilitySolution: Solution:7 16 bulbs = 16C2 = (16*15) / (2*1) = 120 16C 16*15) = 120/190 = 12/19 120/ 12/ Not defective = 1 - 12/19 12/ = 7/19

Probability Problem:8 Problem:Two cards are drawn from a pack of 52 cards. cards. What is the probability that either both are red or both are kings?

ProbabilitySolution: Solution:8 Red = 26C2 / 52C2 = 26/52 26C 52C 26/ Kings = 4C2 / 52C2 = 4/52 52C P(both red or both king) = (26C2 / 52C2) + (4C2 / 52C2) 1/52 26C 52C 52C = 55/221 55/

Probability Problem:9 Problem:Two cards are drawn from a pack of 52 cards. cards. What is the probability that one is a spade and one is a heart?

ProbabilitySolution: Solution:9 52 cards = 13 spades, 13 hearts P(spade) = 13C2 / 52C2 13C 52C P(heart) = 13C2 / 52C2 13C 52C P(spade & heart) = (13C1 * 13C1) / (52*51/2) 13C 13C 52*51/ =169/1326 169/ = 13/102 13/

Probability Problem:10 Problem:Two cards are drawn from a pack of 52 cards what is the probability that it will be a diamond or a king?

ProbabilitySolution: Solution:10 Total = 52C2 = 1326 52C P(diamonds) = 13C2 / 15C2 13C 15C P(king) = 4C2/ 52C2 52C P(A B) = 1/1326 P(diamond or king) = 83/1326 83/

Area Problem:1 Problem:The length of the room is 5 metres and the width 3 3/4 metres, find the cost of paving the floor by slabs at the rate of Rs.800 per m2. Rs.

AreaSolution: Solution:1 length = 51/2 metres 51/ width 3 3/4 metres L*B = 5.5 * 3.75 Cost = 5.5 * 3.75 * 800 = Rs.16500 Rs.

Area Problem:2 Problem:Find the width of a roller which covers 4.4 kilometers while cutting 2420 m2.

AreaSolution: Solution:2 4.4 km = 4400 metres Thus 2420 / 4400 Width = 0.55 meters

Area Problem:3 Problem:The area of a square field is 6050 m2, find the length of the diagonal. diagonal.

AreaSolution: Solution:3 A = 6050m2 6050m = 77.782 + 77.782 77. 77. Diagonal = 110

Area Problem:4 Problem:How many bricks 20 cm. by 10 cm. Will be needed cm. cm. to have the floor of a room 25 m. long and 16 m. wide?

AreaSolution: Solution:4 0.2 * 0.1 = 0.02m2 02m Area = 25 * 16 = 400 Number of bricks = 20,000 20,

Area Problem:5 Problem:A room is 30 meters long and 18 meters wide, If a carpet whose width is 2 meters and which costs Rs. Rs. 50 per metre is spread on the floor of the room, what will be the total cost?

AreaSolution: Solution:5 Area = 540 Cost for each metre = 540/2 540/ = 270 * 50 = Rs. 13,500 Rs. 13,

Area Problem:6 Problem:A rectangular field 242 meters long has got an area of 4840 m2. What will be the cost of fencing that field on all the four sides, if one metre of fencing costs 50p? 50p?

AreaSolution: Solution:6 Area = 4840m2 4840m B = 4840/242 4840/ = 20m 20m = Rs. 262 * 2 Rs. = 524 * 5p = Rs. 262 Rs.

Area Problem:7 Problem:How long will it take for a boy to run around a square field of area 25 hectares at 10 km. per km. hour?

AreaSolution: Solution:7 Area = 25 hectares Perimeter = 25 * 100 * 100 = 500* 500*4 = 2 km Speed = 10km/hr 10km/hr = 12 minutes

Area Problem:8 Problem:What will be the breadth of a room 25 meters long if it costs Rs. 480 to cover it with tiles each 50 dm2 Rs. dm2 at 60p each? 60p

AreaSolution: Solution:8 Breadth = 25 meters Cost = Rs. 480 Rs. = 480*100 / 60 = 800 tiles 480* = 800 * 0.5 = 400 Area = 400m2 400m Thus breadth = 16m 16m

Area Problem:9 Problem:The cost of carpeting a room 20 meters long is Rs. Rs. 40 had the breadth been 3 meters less, the cost could have been Rs. 25. Find the area of the room. Rs. 25. room.

AreaSolution: Solution:9 Y = 40 Y = 25 = 20 * x = 40 m2 = 20 * (x-3) = 25 m2 (xThus x = 2

Area Problem:10 Problem:The area of circle inscribed in a equilateral triangle of a side 24 cm. cm.

AreaSolution: Solution:10 * 24 * h = *24*24 24* H = 12 3 3r = 12 3 r=43

Area of circle = 48