General Concept:

(1) Time taken by a train x mt long in passing a signal postor a pole or a standing man = time taken by the train to cover x mt

(2) Time taken by a train x mt long in passing a stationaryobject of length y mt = time taken by the train to cover x+y mt

(3) Suppose two trains or two bodies are moving in the samedirection at u kmph and v kmph such that u > v then theirrelative speed is u-v kmph

(4)If two trains of length x km and y km are moving in oppositediredtions at u kmph and vmph,then time taken by the train tocross each other = (x+y)/(u+v) hr

(5) Suppose two trains or two bdies are moving in opposite directionat u kmph and v kmph then,their relative speed = (u+v) kmph

(6)If two train start at the same time from 2 points A & B towardseach other and after crossing they take a & b hours in reaching B & Arespectively then A’s speed : B’s speed = (b^1/2 : a^1/2 )

Problems

(1)Find the time taken by a train 180m long,running at 72kmph incrossing an electric pole

Solution:Speed of the train =72*5/18m/s =20 m/sDistance move din passing the pole = 180mRequiredtime = 180/20 = 9 seconds

(2)A train 140 m long running at 60kmph.In how much time will itpass a platform 260m long.

Solution:Distance travelled =140 + 260 m =400 m,speed = 60 * 5/18 = 50//3 mtime=400*3 / 50 = 24 Seconds

(3)A man is standing on a railway bridge which is 180 m.He findsthat a train crosses the bridge in 20 seconds but himself in8 sec. Find the length of the train and its sppeed

Solution:

i)D=180+xT = 20 secondsS= 180+x / 20 ———— 1ii)D=xT=8 secondsD=STx=8S ————- 2Substitute 2 in 1S=180 + 8 S / 20S=15 m/sLength of the train,x is 8 *15 = 120 m

(4)A train 150m long is running with a speed of 68 mphIn whttime will it pass a man who is running at a speed of 8kmph inthe same direction in which the train is going

Solution:Relative Speed = 68-8=60kmph*5/18 = 50/3 m/stime= 150 * 3 /50 =9sec

5)A train 220m long is running with a speed of 59 k mph /..Inwhat time will it pass a man who is running at 7 kmph in thedirection opposite to that in which train is going.

Solution:Relative Speed = 59+7=66kmph*5/18 = 55/3 m/stime= 220/55 * 3 =12sec

(6)Two trains 137m and 163m in length are running towards eachother on parallel lines,one at the rate of 42kmph & another at48 mph.In wht time will they be clear of each other from themoment they meet.

Solution:Relative speed =42+48 = 90 *5/18 = 25m/stime taken by the train to pass each other = time taken to cover(137+163)m at 25 m/s= 300 /25 s =12 s

(7)A train running at 54 kmph takes 20 sec to pass a platform.Next it takes 12 sec to pass a man walking at 6kmph in the samedirection in which the train is going.Find length of the trainand length of platform

Solution:Relative speed w.r.t man = 54-6=48kmph

the length of the train is 48 * 5/18 * 12 =160mtime taken to pass platform =20 secSpeed of the train = 54 * 5/18 =15m/s160+x =20 *15x=140mlength of the platform is 140m

(8)A man sitting in a train which is travelling at 50mph observesthat a goods train travelling in opposite irection takes 9 secto pass him .If the goos train is 150m long fin its speed

Solution:Relative speed =150/9 m/s =60 mphspeed of the train = 60-50 =10kmph

(9)Two trains are moving in the sam e direction at 65kmph and45kmph. The faster train crosses a man in slower train in18sec.thelength of the faster train is

Solution:Relative speed =65-45 kmph = 50/9 m/sDistancce covered in18 s =50/9 * 18 = 100mthe length of the train is 100m

(10)Atrain overtakes two persons who are walking in the samedirection in which the train is going at the rate of 2kmph an4kmph and passes them completely in 9 sec an 10 sec respectively.The length of train is

Solution:2kmph = 5/9 m/s4 mph =10/9 m/sLet the length of the trainbe x meters and its speed is y m/sthen x / (y- 5/9) = 9 and x / (y- 10/9) = 109y-5 =x and 10(9y-10)=9x9y-x=5 and 90y-9x=100on solving we get x=50,lenght of trains

(11) Two stations A & B are 110 km apart on a straight line.One train starts from A at 7am and travels towards B at 20kmph.Another train starts from B at 8am an travels toward A at a speedof 25kmph.At what time will they meet

Solution:Suppose the train meet x hr after 7amDistance covered by A in x hr=20x km

20x+25(x-1) = 11045x=135x=3So they meet at 10 am

(12)A traintravelling at 48kmph completely crosses another trainhaving half its length an travelling inopposite direction at 42kmphin12 sec.It also passes a railway platform in 45sec.the length ofplatform is

Solution:Let the length of the first train be x mtthen,the length of second train is x/2 mtrelative speed = 48+42 kmph =90 * 5/18 m/s = 25m/s(x+ x/2)/25 =12x=200Length of the train is 200mLet the length of the platform be y mtspeed f the first train = 48*5/18 m/s = 40/3 m/s200+y * 3/40 = 45y=400m

(13)The length of a running trsain in 30% more than the length ofanother train B runnng in the opposite direction.To find out thespeed of trtain B,which of the following information given in thestatements P & Q is sufficientP : The speed of train A is 80 kmphQ : They too 90 sec to cross each other(a) Either P & Q is sufficient(b)Both P & Q are not sufficient(c)only Q is sufficient(d)Both P & Q are neeedAns: B

Solution:Let the length of th e train A be x mtLength of the train B = 130/100 x mt =13x/10 mtLet the speed of B be y mph,speed of the train A=80mphrelative speed= y+80 * 5/18 m/stime taken by the trains t cross each other is gven by90 = (x + 13x/10)/ (5y+400 / 18)to find y,clearly xis also neededso,both P & Q are not sufficient

(14)The speed of a train A,100m long is 40% more than then the speedof another train B,180m long running in opposite direction.To fin out

the speed of B,which of the information given in statements P & Q issufficientP :The two trains crossed each other in 6 secondsQ : The difference between the spee of the trains is 26kmph(a)Only P is sufficient(b)Only Q is sufficient(c)Both P & Q are needed(d)Both P & Q are not sufficientAns : A

Solution:Let speed of B be x kmphthen,speed of A =140x/100 kmph =7x/5 mphrelative speed = x + 7x/5 =2x/3 m/stime taken to cross each other = (100+180)*3/2x s =420/x snow,420/x = 6x=70 mphthus,only P is sufficient

(15)The train running at certain speed crosses astationary enginein20 seconds.to find out the sped of the train,which of the followinginformation is necessary

(a)Only the length of the train(b)only the length of the engine(c)Either the length of the train or length of engine(d)Both the length of the train or length of engineAns : D

Solution:

Since the sum of lengths of the tran and the engine is needed,so both the length must be known

Formulae:

I)Speed = Distance/Time

II)Time = Distance/speed

III) Distance = speed*time

IV) 1km/hr = 5/18 m/s

V)1 m/s = 18/5 Km/hr

VI)If the ratio of the speed of A and B is a:b,then the ratio ofthe time taken by them to cover the same distance is 1/a : 1/bor b:a

VII) suppose a man covers a distance at x kmph and an equaldistance at y kmph.then the average speed during the wholejourney is (2xy/x+y)kmph

Problems

1)A person covers a certain distance at 7kmph .How many metersdoes he cover in 2 minutes.

Solution::speed=72kmph=72*5/18 = 20m/sdistance covered in 2min =20*2*60 = 2400m

2)If a man runs at 3m/s. How many km does he run in 1hr 40min

Solution::speed of the man = 3*18/5 kmph= 54/5kmphDistance covered in 5/3 hrs=54/5*5/3 = 18km

3)Walking at the rate of 4knph a man covers certain distancein 2hr 45 min. Running at a speed of 16.5 kmph the man willcover the same distance in.

Solution::Distance=Speed* time4*11/4=11kmNew speed =16.5kmphtherefore Time=D/S=11/16.5 = 40min

Complex Problems

1)A train covers a distance in 50 min ,if it runs at a speedof 48kmph on an average.The speed at which the train must runto reduce the time of journey to 40min will be.

Solution::Time=50/60 hr=5/6hrSpeed=48mphdistance=S*T=48*5/6=40kmtime=40/60hr=2/3hrNew speed = 40* 3/2 kmph= 60kmph

2)Vikas can cover a distance in 1hr 24min by covering 2/3 ofthe distance at 4 kmph and the rest at 5kmph.the totaldistance is?

Solution::Let total distance be Stotal time=1hr24minA to T :: speed=4kmphdiistance=2/3ST to S :: speed=5kmdistance=1-2/3S=1/3S21/15 hr=2/3 S/4 + 1/3s /584=14/3S*3S=84*3/14*3= 6km

3)walking at ¾ of his usual speed ,a man is late by 2 ½ hr.the usual time is.

Solution::Usual speed = SUsual time = TDistance = DNew Speed is ¾ SNew time is 4/3 T4/3 T – T = 5/2T=15/2 = 7 ½

4)A man covers a distance on scooter .had he moved 3kmphfaster he would have taken 40 min less. If he had moved2kmph slower he would have taken 40min more.the distance is.

Solution::

Let distance = x mUsual rate = y kmphx/y – x/y+3 = 40/60 hr2y(y+3) = 9x ————–1x/y-2 – x/y = 40/60 hr y(y-2) = 3x —————–2

divide 1 & 2 equationsby solving we get x = 40

5)Excluding stoppages,the speed of the bus is 54kmph andincluding stoppages,it is 45kmph.for how many min does the busstop per hr.

Solution::Due to stoppages,it covers 9km less.time taken to cover 9 km is [9/54 *60] min = 10min

6)Two boys starting from the same place walk at a rate of5kmph and 5.5kmph respectively.wht time will they take to be8.5km apart, if they walk in the same direction

Solution::The relative speed of the boys = 5.5kmph – 5kmph = 0.5 kmphDistance between them is 8.5 kmTime= 8.5km / 0.5 kmph = 17 hrs

7)2 trains starting at the same time from 2 stations 200kmapart and going in opposite direction cross each other atadistance of 110km from one of the stations.what is the ratio oftheir speeds.

Solution::In same time ,they cover 110km & 90 km respectivelyso ratio of their speed =110:90 = 11:9

8)Two trains start from A & B and travel towards each other atspeed of 50kmph and 60kmph resp. At the time of the meeting thesecond train has traveled 120km more than the first.the distancebetween them.

Solution::Let the distance traveled by the first train be x kmthen distance covered by the second train is x + 120kmx/50 = x+120 / 60x= 600so the distance between A & B is x + x + 120 = 1320 km

9)A thief steals a ca r at 2.30pm and drives it at 60kmph.thetheft is discovered at 3pm and the owner sets off in another carat 75kmph when will he overtake the thief

Solution::Let the thief is overtaken x hrs after 2.30pmdistance covered by the thief in x hrs = distance covered bythe owner in x-1/2 hr60x = 75 ( x- ½)x= 5/2 hrthief is overtaken at 2.30 pm + 2 ½ hr = 5 pm

10)In covering distance,the speed of A & B are in the ratioof 3:4.A takes 30min more than B to reach the destion.The timetaken by A to reach the destinstion is.

Solution::Ratio of speed = 3:4Ratio of time = 4:3let A takes 4x hrs,B takes 3x hrsthen 4x-3x = 30/60 hrx = ½ hrTime taken by A to reach the destination is 4x = 4 * ½ = 2 hr

11)A motorist covers a distance of 39km in 45min by moving at aspeed of xkmph for the first 15min.then moving at double thespeed for the next 20 min and then again moving at his originalspeed for the rest of the journey .then x=?

Solution::Total distance = 39 kmTotal time = 45 minD = S*Tx * 15/60 + 2x * 20/60 + x * 10/60 = 39 kmx = 36 kmph

12)A & B are two towns.Mr.Fara covers the distance from A t0 Bon cycle at 17kmph and returns to A by a tonga running at auniform speed of 8kmph.his average speed during the wholejourney is.

Solution::When same distance is covered with different speeds,then theaverage speed = 2xy / x+y=10.88kmph

13)A car covers 4 successive 3km stretches at speed of10kmph,20kmph,30kmph&:60kmph resp. Its average speed is.

Solution::Average speed = total distance / total timetotal distance = 4 * 3 = 12 kmtotal time = 3/10 + 3/20 + 3/30 + 3/60= 36/60 hrspeed =12/36 * 60 = 20 kmph

14)A person walks at 5kmph for 6hr and at 4kmph for 12hr.The average speed is.

Solution::avg speed = total distance/total time= 5*6 + 4*12 / 18=4 1/3 mph

15)A bullock cart has to cover a distance of 80km in 10hrs.If it covers half of the journeyin 3/5th time.wht should be its speed to cover the remaining distance in the time left.

Solution::Time left = 10 – 3/5*10= 4 hrspeed =40 km /4 hr=10 kmph

16)The ratio between the speeds of the A& B is 2:3 antherefore A takes 10 min more than the time taken by B to reachthe destination.If A had walked at double the speed ,he wouldhave covered the distance in ?

Solution::Ratio of speed = 2:3Ratio of time = 3:2A takes 10 min more3x-2x = 10 minA’s time=30 min—>A covers the distance in 30 min ,if its speed is x-> He will cover the same distance in 15 min,if its speeddoubles (i.e 2x)

17)A is twice as fast as B and B is thrice as fast as C is.The journey covered by B in?

Solution::speed’s ratioa : b = 2: 1b : c = 3:1Time’s ratiob : c = 1:3b : c = 18:54(if c covers in 54 min i..e twice to 18 min )

18)A man performed 3/5 of the total journey by ratio 17/20 bybus and the remaining 65km on foot.wht is his total journey.

Solution::Let total distance is xx-(3/5x + 17/20 x) =6.5x- 19x/20 = 6.5x=20 * 6.5=130 km

19)A train M leaves Meerat at 5 am and reaches Delhi at 9am .Another train N leaves Delhi at 7am and reaches Meerut at 1030amAt what time do the 2 trains cross one another

Solution::Let the distance between Meerut & Delhi be xthey meet after y hr after 7amM covers x in 4hrN covers x in 3 ½ i.e 7/2 hrspeed of M =x/4speed of N = 2x/7Distance covered by M in y+2 hr + Distance covered by N iny hr is xx/4 (y+2) +2x/7(y)=xy=14/15hr or 56 min

20)A man takes 5hr 45min in walking to certain place and ridingback. He would have gained 2hrs by riding both ways.The time hewould take to walk both ways is?

Solution::Let x be the speed of walkedLet y be the speed of rideLet D be the distance

Then D/x + D/y = 23/4 hr ——-1D/y + D/y = 23/4 – 2 hr

D/y = 15/8 ——–2substitute 2 in 1D/x + 15/8 = 23/4D/x = 23/4 -15/8 =46-15/8 =31/8Time taken for walk one way is 31/8 hrtime taken to walk to and fro is 2*31/8 = 31/4 hr=7 hr 45 min

Boats and StreamsImportant facts:

1)In water, the direction along the stream is called down stream.

2)Direction against the stream is called upstream.

3)The speed of boat in still water is U km/hr and the speed ofstream is V km/hr then

speed down stream =U + V km/hrspeed up stream = U – V km/hr

Formulae:

If the speed down stream is A km/hr and the speed up stream isB km/hr then speed in still water = ½(A+B) km/hrrate of stream =1/2(A-B) km/hr

Problems:1. In one hour a boat goes 11 km long the stream and 5 kmagainst the stream. The speed of the boat in still water is?Sol:Speed in still water = ½ ( 11+5) km/hr= 8 kmph

2.A man can row 18 kmph in still water. It takes him thriceas long as row up as to row down the river. find the rateof stream.Sol:Let man’s rate up stream be xkmphthen, in still water =1/2[3x+x]=2x kmphso, 2x= 18, x=9

rate upstream =9kmphrate downstream =27 kmphrate of stream = ½ [27-9]= 9kmph

3.A man can row 71/2kmph in still water . if in a river running at 1.5 km an hour, if takes him 50 min to row to place and back. how far off is the place? Sol: speed down stream =7.5+1.5=9kmphspeed upstream =7.5-1.5=6kmphlet the required distence x km. then ,x/9+x/6=50/60 = 2x+3x= 5/6*185x=15, x=3Hence, the required distence is 3 km

4.A man can row 3 quarters of a km aganist the stream is111/4 min. the speed of the man in still water is ?Sol: rate upstream = 750/625 m/sec =10/9 m/secrate downstream =750/450 m/sec = 5/3 m/secrate in still water =1/2[10/9+5/3] = 25/18 m/sec= 25/18*18/5=5 kmph

5.A boat can travel with a speed of 13 kmph in still water.if the speed of stream is 4 kmph,find the time taken by the boat to go 68 km downstream?Sol: Speed down stream = 13+4= 17 kmphtime taken to travel 68km downstream =68/17 hrs= 4 hrs

6.A boat takes 90 min less to travel 36 miles downstream thento travel the same distence upstream. if the speed of the boat in still water is 10 mph . The speed of the stream is :Sol: Let the speed of the stream be x mph .then, speed downstream = [10+x]mphspeed upstream =[10-x] mph36/[10+x] – 36/[10-x] = 90/60 =72x*60= 90[100-x2](x+50)(x-2) =0x=2 kmph

7.At his usual rowing rate, Rahul 12 miles down stream in a certain river in 6 hrs less than it takes him to travel the same distence upstream. but if he could double his usual rowing rate for his 24 miles roundthe down stream 12 miles would then take only one hour less than the up stream 12 miles.what is the speed of the current in miles per hours?

Sol: Let the speed in still water be x mph and the speed ofthe curren be y mph.then, speed upstream = (x-y)speed downstream =(x+y)

12/(x-y) – 12/(x+y) = 66(x2 – y2) m= 2xy => x2 – y2 =4y -(1)and 12/(2x-y) – 12/(2x+y) =1 => 4×2 – y2 = 24yx2= ( 24y + y2)/4 –>(2)from 1 and 2 we have4y+ y2 =( 24y+y2)/4y=8/3 mphy= 22/3 mph

8.There is a road beside a river. two friends started froma place A, moved to a temple situated at another place B and then returned to A again. one of them moves on a cycle at a speed of 12 kmph, while the other sails on a boat at a speed of 10 kmph . if the river flows at the speedof 4 kmph,which of the two friends will return to place A first ?

Sol: Clearly, The cyclist moves both ways at a speed of 12 kmphso, average speed of the cyclist = 12 kmphthe boat sailor moves downstream = (10+4) = 14 kmphupstream =(10-4) = 6 kmphSo, average speed of the boat sailor =[ 2*14*6]/[14+6] kmph=42/5 kmph =8.4 kmphSince, the average speed of the cyclist is greater, he willreturn to A first.

9.A boat takes 19 hrs for travelling downstream from point A topoint B. and coming back to a point C midway between A and B.if the velocity of the sream is 4 kmph. and the speed of theboat in still water is 14 kmph. what is the distence betweenA and B?

Sol:speed downstream =14+4 =18 kmphspeed upstream = 14 -4 = 10 kmphlet the distence between A and B be x km. then,x/18 + (x/2)/10 = 19x/18 + x/20 =1919x/180 =19 =>x = 180kmHence, the distence between A and B bw 180 km

Problems on Ages

Simple problems:

1.The present age of a father is 3 years more than three timesthe age of his son.Three years hence,father’s age will be 10years more than twice the age of the son.Find the present ageof the father.

Solution: Let the present age be ‘x’ years.Then father’s present age is 3x+3 years.Three years hence(3x+3)+3=2(x+3)+10x=10Hence father’s present age = 3x+3 = 33 years.

2. One year ago the ratio of Ramu & Somu age was 6:7respectively. Four years hence their ratio would become 7:8. How old is Somu.

Solution: Let us assume Ramu &Somu ages are x &y respectively.One year ago their ratio was 6:7i.e x-1 / y-1 = 7x-6y=1Four years hence their ratios,would become 7:8i.e x-4 / y-4 = 7 / 88x-7y=-4From the above two equations we get y= 36 years.i.e Somu present age is 36 years.

3. The total age of A &B is 12 years more than the total age ofB&C. C is how many year younger than A.

Solution: From the given dataA+B = 12+(B+C)A+B-(B+C) = 12A-C=12 years.C is 12 years younger than A

4. The ratio of the present age of P & Q is 6:7. If Q is 4 yearsold than P. what will be the ratio of the ages of P & Q after4 years.

Solution: The present age of P & Q is 6:7 i.eP / Q = 6 / 7Q is 4 years old than P i.e Q = P+4.P/ P+4 = 6/77P-6P = 24,

P = 24 , Q = P+4 =24+4 = 28After 4 years the ratio of P &Q isP+4:Q+424+4 : 28+4 = 28:32 = 7:8

5. The ratio of the age of a man & his wife is 4:3.After 4 years this ratio will be 9:7. If the time of marriage the ratio was 5:3,then how many years ago were they married.

Solution: The age of a man is 4x .The age of his wife is 3x.After 4 years their ratio’s will be 9:7 i.e4x+4 / 3x+4 = 9 / 728x-27x=36-28x = 8.Age of a man is 4x = 4*8 = 32 years.Age of his wife is 3x = 3*8 = 24 years.Let us assume ‘y’ years ago they were married ,the ratio was 5:3 ,i.e32-y / 24-y = 5/ 3y=12 yearsi.e 12 years ago they were married

6. Sneh’s age is 1/6th of her father’s age.Sneh’s father’s age will be twice the age of Vimal’s age after 10 years. If Vimal’s eightbirthday was celebrated two years before,then what is Sneh’spresent age.a) 6 2/3 years b) 24 years c) 30 years d) None of the above

Solution: Assume Sneh’s age is ‘x’ years.Assume her fathers age is ‘y’ years.Sneh’s age is 1/6 of her fathers age i.e x = y /6.Father’s age will be twice of Vimal’s age after 10years.i.e y+10 = 2( V+10)( where ‘V’ is the Vimal’s age)Vimal’s eight birthday was celebrated two years before,Then the Vimal’s present age is 10 years.Y+10 = 2(10+10)Y=30 years.Sneh’s present age x = y/6x = 30/6 = 5 years.Sneh’s present age is 5 years.

7.The sum of the ages of the 5 children’s born at the intervals of 3 years each is 50 years what is the age of the youngest child.a) 4 years b) 8 years c) 10 years d)None of the above

Solution: Let the age of the children’s bex ,x+3, x+6, x+9, x+12.x+(x+3)+(x+6)+(x+9)+(x+12) = 505x+30 = 505x = 20x=4.Age of the youngest child is x = 4 years.

8. If 6 years are subtracted from the present age of Gagan andthe remainder is divided by 18,then the present age of his grandson Anup is obtained. If Anup is 2 years younger to Madan whose age is 5 years,then what is Gagan’s present age.a) 48 years b)60 years c)84 years d)65 years

Solution: Let us assume Gagan present age is ‘x’ years.Anup age = 5-2 = 3 years.(x-6) / 18 = 3x-6 = 54x=60 years

9.My brother is 3 years elder to me. My father was 28 years of age when my sister was born while my father was 26 years of age when i was born. If my sister was 4 years of age when my brotherwas born,then what was the age my father and mother respectively when my brother was born.a) 32 yrs, 23yrs b)32 yrs, 29yrs c)35 yrs,29yrs d)35yrs,33 yrs

Solution: My brother was born 3 years before I was born & 4years after my sister was born.Father’s age when brother was born = 28+4 = 32 years.Mother’s age when brother was born = 26-3 = 23 years.

Averages

Formula:

1.Average=Sum of quantities/Number of quantities.

2.Suppose a man covers a certain distance at x kmphand an equal distance at y kmph ,then the average speedduring the whole journey is (2xy/x+y) kmph.

Examples:

1.Find the average of all these numbers.142,147,153,165,157.Solution:142 147 153 165 157Here consider the least number i.e, 142comparing with others,142 147 153 165 157+5 +11 +23 +15Now add 5+11+23+15 = 52/5 = 10.8Now add 10.8 to 142 we get 152.8(Average of all these numbers).Answer is 152.8

2.Find the average of all these numbers.4,10,16,22,28Solution:4,10,16,22,28As the difference of number is 6Then the average of these numbers is central one i.e, 16.Answer is 16.

3.Find the average of all these numbers.4,10,16,22,28,34.Solution:Here also difference is 6.Then middle numbers 16,22 take average of thesetwo numbers 16+22/2=19Therefore the average of these numbers is 19.Answer is 19.

4.The average marks of a marks of a student in 4 Examination is 40.If he got 80 marks in 5th Exam then what is his new average.Solution:4*40+80=240Then average means 240/5=48.Answer is 48.

5.In a group the average income of 6 men is 500 and that of 5 women is 280, then what is average income of the group.

Solution:6*500+5*280=4400then average is 4400/11=400.Another Method: here consider for 6 men6 men – each 500.so 5th women is 280.then 500-280=220.

then 220*6/11=120.therefore 120+280=400.Answer is 400.

6.The average weight of a class of 30 students is 40 kgs if theteacher weight is included then average increases by 2 kgs thenfind the weight of the teacher?

Solution:30 students average weight is 40 kgs.So,when teacher weight is added it increases by 2 kgsso total 31 persons ,therefore 31*2=62.Now add the average weight of all student to itwe get teachers weight i.e, 62+40=102 kgs.Answer is 102 kgs.

7.The average age of Mr and Mrs Sharma 4 years ago is 28 years .If the present average age of Mr and Mrs Sharma and their sonis 22 years. What is the age of their son.

Solution:4 years ago their average age is 28 years.So their present average age is 32 years.32 years for Mr and Mrs Sharma then 32*2=64 years.Then present age including their son is 22 years.So 22*3 =66 years.Therefore son age will be 66-64 = 2 years.Answer is 2 years.

8.The average price of 10 books is increased by 17 Rupees whenone of them whose value is Rs.400 is replaced by a new book.What is the price of new book?Solution:10 books Average increases by 17 Rupeesso 10*17= 170.so the new book cost is more and by adding its cost averageincrease,therefore the cost of new book is 400+170=570Rs.Answer is 570 Rs.

9.The average marks of girls in a class is 62.5. The average marksof 4 girls among them is 60.The average marks of remaining girlsis 63,then what is the number of girls in the class?Solution:Total number of girls be x+4.Average marks of 4 girls is 60.therefore 62.5-60=2.5

then 4*2.5 =10.the average of remaining girls is 63here 0.5 difference therefore 0.5*x=10(since we got from 4 girls)(this is taken becoz both should be equal)x=10/0.5x=20.This clear says that remaining are 20 girlstherefore total is x+4=20+4=24 girlsAnswer is 24 girls.

10.Find the average of first 50 natural numbers.Solution:Sum of the Natural Numbers is n(n+1)/2therefore for 50 Natural numbers 50*51/2=775.the average is 775/50=15.5Answer is 15.5 .

11.The average of the first nine prime number is?Solution:Prime numbers are 2,3,5,7,11,13,17,19,23therefore 2+3+5+7+11+13+17+19+23=100then the average 100/9= 11 1/9.Answer is 11 1/9.

12.The average of 2,7,6 and x is 5 and the average of and theaverage of 18,1,6,x and y is 10 .what is the value of y?Solution:2+7+6+x/4=5=>15+x=20=>x=5.18+1+6+x+y/5=10=>25+5+y=50=>y=20.

13.The average of a non-zero number and its square is 5 times thenumber.The number isSolution:The number be xthen x+x2/2=5x=>x2-9x=0=>x(x-9)=0therefore x=0 or x=9.The number is 9.

14.Nine persons went to a hotel for taking their meals . Eight ofthem spent Rs.12 each on their meals and the ninth spent Rs.8 then

the average expenditure of all the nine. What was the total moneyspent by them?Solution:The average expenditure be x.then 8*12+(x+8)=9x=>96+x+8=9x.=>8x=104=>x=13Total money spent =9x=>9*13=117Answer is Rs.117

15.The average weight of A.B.C is 45 Kgs.If the average weight ofA and B be 40 Kgs and that of Band C be 43 Kgs. Find the weight of B?Solution:The weight of A,B,Care 45*3=135 Kgs.The weight of A,B are 40*2=80 Kgs.The weight of B,C are 43*2=86 Kgs.To get the Weight of B.(A+B)+(B+C)-(A+B+C)=80+86-135B=31 kgs.Answer is 31 Kgs.

16.The sum of three consecutive odd number is 48 more than the average of these number .What is the first of these numbers?Solution:let the three consecutive odd numbers are x, x+2, x+4.By adding them we get x+x+2+x+4=3x+6.Then 3x+6-(3x+6)/3=38(given)=>2(3x+6)=38*3.=>6x+12=114=>6x=102=>x=17.Answer is 17.

17.A family consists of grandparents,parents and three grandchildren.The average age of the grandparents is 67 years,that of parents is 35years and that of the grand children is 6 years . What is the averageage of the family? Solution:grandparents age is 67*2=134parents age is 35*2=70grandchildren age is 6*3=18therefore age of family is 134+70+18=222average is 222/7=31 5/7 years.Answer is 31 5/7 years.

18.A library has an average of 510 visitors on Sundays and 240 onother days .The average number of visitors per day in a month 30days beginning with a Sunday is?Solution:Here specified that month starts with Sundayso, in a month there are 5 Sundays.Therefore remaining days will be 25 days.510*5+240*25=2550+6000=8550 visitors.The average visitors are 8550/30=285.Answer is 285.

19.The average age of a class of 39 students is 15 years .If the age of the teacher be included ,then average increases by 3 months. Find the age of the teacher.Solution: Total age for 39 persons is 39*15=585 years.Now 40 persons is 40* 61/4=610 years(since 15 years 3 months=15 3/12=61/4)Age of the teacher =610-585 years=>25 years.Answer is 25 years.

20.The average weight of a 10 oarsmen in a boat is increases by 1.8 Kgs .When one of the crew ,who weighs 53 Kgs is replaced by new man. Find the weight of the new man.Solution: Weight of 10 oars men is increases by 1.8 Kgsso, 10*1.8=18 Kgstherefore 53+18=71 Kgs will be the weight of the man.Answer is 71 Kgs.

21.A bats man makes a score of 87 runs in the 17th inning and thus increases his average by 3. Find the average after 17th inning.

Solution: Average after 17 th inning =xthen for 16th inning is x-3.Therefore 16(x-3)+87 =17x=>x=87-48=>x=39.Answer is 39.

22.The average age of a class is 15.8 years .The average age of boys in the class is 16.4 years while that of the girls is 15.4 years .What is the ratio of boys to girls in the class.Solution: Ratio be k:1 thenk*16.4 + 1*15.4 = (k+1)*15.8=>(16.4-15.8)k=15.8-15.4=>k=0.4/0.6=>k=2/3therefore 2/3:1=>2:3

Answer is 2:3

23.In a cricket eleven ,the average of eleven players is 28 years .Out of these ,the average ages of three groups of players each are 25 years,28 years, and 30 years respectively. If in these groups ,the captain and the youngest player are not included and the captain iseleven years older than the youngest players , what is the age of the captain?Solution: let the age of youngest player be xthen ,age of the captain =(x+11)therefore 3*25 + 3*28 + 3*30 + x + x+11=11*28=>75+84+90+2x+11=308=>2x=48=>x=24.Therefore age of the captain =(x+11)= 24+11= 35 years.Answer is 35 years.

24.The average age of the boys in the class is twice the number of girls in the class .If the ratio of boys and girls in the class of 36 be 5:1, what is the total of the age (in years) of the boys in the class?Solution: Number of boys=36*5/6=30Number of girls =6Average age of boys =2*6=12 yearsTotal age of the boys=30*12=360 yearsAnswer is 360 years.

25.Five years ago, the average age of P and Q was 15 years ,average age of P,Q, and R today is 20 years,how old will R be after 10 years?Solution: Age of P and Q are 15*2=30 yearsPresent age of P and Q is 30+5*2=40 years.Age of P Q and R is 20*3= 60 years.R ,present age is 60-40=20 yearsAfter 10 years =20+10=30 years.Answer is 30 years.

26.The average weight of 3 men A,B and C is 84 Kgs. Another man D joins the group and the average now becomes 80 Kgs.If another man E whose weight is 3 Kgs more than that of D ,replaces A then theaverage weight B,C,D and E becomes 79 Kgs. The weight of A is.Solution:Total weight of A, B and C is 84 * 3 =252 Kgs.Total weight of A,B,C and Dis 80*4=320 KgsTherefore D=320-252=68 Kgs.E weight (68+3)=71 kgsTotal weight of B,C,D and E = 79*4=316 Kgs(A+B+C+D)-(B+C+D+E)=320-316 =4KgsA-E=4KgsA-71=4 kgsA=75 Kgs

Answer is 75 kgs

27.A team of 8 persons joins in a shooting competition.The best marksman scored 85 points.If he had scored 92 points ,the average score for the team would have been 84.The team scored was.Solution: Here consider the total score be x.therefore x+92-85/8=84=>x+7=672=>x=665.Answer is 665

28.A man whose bowling average is 12.4,takes 5 wickets for 26 runs and there by decrease his average by 0.4. The number of wickets,taken by him before his last match is:Solution: Number of wickets taken before last match be x.therefore 12.4×26/x+5=12(since average decrease by 0.4therefore 12.4-0.4=12)=>12.4x+2612x+60=>0.4x=34=>x=340/4=>x=85.Answer is 85.

29.The mean temperature of Monday to Wednesday was 37 degreesand of Tuesday to Thursday was 34 degrees .If the temperature on Thursday was 4/5th that of Monday. The temperature on Thursday was:Solution:The total temperature recorded on Monday,Wednesday was 37*3=111.The total temperature recorded on Tuesday,Wednesday,Thursday was 34*3=102.and also given that Th=4/5M=>M=5/4Th(M+T+W)-(T+W+Th)=111-102=9M-Th=95/4Th-Th=9Th(1/4)=9=>Th=36 degrees.

30. 16 children are to be divided into two groups A and B of 10 and 6 children. The average percent marks obtained by the children of group A is 75 and the average percent marks of all the 16 children is 76. What is the average percent marks of children of groups B?Solution: Here given average of group A and whole groups .So,(76*16)-(75*10)/6=>1216-750/6=>466/6=233/3=77 2/3Answer is 77 2/3.

31.Of the three numbers the first is twice the second and the second is twice the third .The average of the reciprocal of the numbers is 7/72,the number are.Solution:Let the third number be xLet the second number be 2x.Let the first number be 4x.Therefore average of the reciprocal means1/x+1/2x+1/4x=(7/72*3)7/4x=7/24=>4x=24x=6.ThereforeFirst number is 4*6=24.Second number is 2*6=12Third number is 1*6=6Answer is 24,12,6.

32.The average of 5 numbers is 7.When 3 new numbers are added the average of the eight numbers is 8.5. The average of the three new number is:Solution: Sum of three new numbers=(8*8.5-5*7)=33Their average =33/3=11.Answer is 11.

33.The average temperature of the town in the first four days of a month was 58 degrees. The average for the second ,third,fourth and fifth days was 60 degree .If the temperature of the first andfifth days were in the ratio 7:8 then what is the temperature on the fifth day?Solution :Sum of temperature on 1st 2nd 3rdand 4th days =58*4=232 degrees.Sum of temperature on 2nd 3rd 4thand 5th days =60*4=240 degreesTherefore 5th day temperature is 240-232=8 degrees.The ratio given for 1st and 5th days be 7x and 8x degreesthen 8x-7x=8=>x=8.therefore temperature on the 5th day =8x=8*8=64 degrees.

SURDS AND INDICES

Simple problems:

1. Laws of Indices:

(i) am * an = a(m+n)(ii) am / an = a(m-n)(iii) (am)n = a(m*n)(iv) (ab)n = an * bn(v) (a/b)n = an / bn(vi) a0 = 1

2.Surds :Let ‘a’ be a rational number & ‘n’ be a positive integer such that a1/n = nth root a is irrational.Then nth root a is called ‘a’ surd of ‘n’.

Problems:-(1)(i) (27)2/3 = (33)2/3 = 32 = 9.(ii) (1024)-4/5 = (45)-4/5 = (4)-4= 1/(4)4 = 1/256.(iii)(8/125)-4/3 =((2/5)3)-4/3 = (2/5)-4 = (5/2)4 = 625/16

(2) If 2(x-1)+ 2(x+1) = 1280 then find the value of x .Solution: 2x/2+2x.2 = 12802x(1+22) = 2*12802x = 2560/52x = 512 => 2x = 29x = 9

(3) Find the value of [5[81/3+271/3]3]1/4

Solution: [5[(23)1/3+(33)1/3]3]1/4[5[2+3]3]1/4[54]1/4 => 5.

(4) If (1/5)3y= 0.008 then find the value of (0.25)ySolution: (1/5)3y = 0.008(1/5)3y =[0.2]3(1/5)3y =(1/5)33y= 3 => y=1.(0.25)y = (0.25)1 => 0.25 = 25/100 = 1/4

(5) Find the value of (243)n/5 * 32n+1 / 9n * 3 n-1

Solution: (35)n/5 * 32n +1 / (32)n * 3n-133n+1 / 33n-1 333n+1 * 3-3n+1 => 32 =>9.

(6) Find the value of (21/4-1)( 23/4 +21/2+21/4+1)Solution: Let us say 21/4 = x(x-1)(x3+x2+x+1)(x-1)(x2(x+1)+(x+1))(x-1) (x2+1) (x+1) [(x-1)(x+1) = (x2-1)](x2+1) (x2-1) => (x4-1)((21/4))4 – 1) = > (2-1) = > 1.

(7) If x= ya , y = zb , z = xc then find the value of abc.Solution: z= xcz= (ya)c [ x= ya ]z= (y)acz= (zb)ac [y= zb]z= zabcabc = 1

(8)Simplify (xa/xb)a2+ab+b2*(xb/xc)b2+bc+c2*(xc/xa)c2+ca+a2Solution:[xa-b]a2+ab+b2 * [xb-c]b2+bc+c2 * [xc-a]c2+ca+a2[ (a-b)(a2+ab+b2) = a3-b3]from the above formula=> xa3-b3 xb3-c3 xc3-a3=> xa3-b3+b3-c3+c3-a3=> x0 = 1

(9) (1000)7 /1018 = ?(a) 10 (b) 100 (c ) 1000 (d) 10000Solution: (1000)7 / 1018(103)7 / (10)18 = > (10)21 / (10)18=> (10)21-18 => (10)3 => 1000Ans :( c )

(10) The value of (8-25-8-26) is(a) 7* 8-25 (b) 7*8-26 (c ) 8* 8-26 (d) NoneSolution: ( 8-25 – 8-26 )=> 8-26 (8-1 )=> 7* 8-26Ans: (b)

(11) 1 / (1+ an-m ) +1/ (1+am-n) = ? (a) 0 (b) 1/2 (c ) 1 (d) an+m

Solution: 1/ (1+ an/am) + 1/ ( 1+ am/an)=> am / (am+ an ) + an /(am +an )=> (am +an ) /(am + an)=> 1Ans: ( c)

(12) 1/(1+xb-a+xc-a)+1/(1+xa-b+xc-b)+1/(1+xb-c+xa-c)=?(a) 0 (b) 1 ( c ) xa-b-c (d) None of the aboveSolution: 1/ (1+xb/xa+xc/xa) + 1/(1+xa/xb +xc/xb) +1/(1+xb/xc +xa/xc)=> xa /(xa +xb+xc) + xb/(xa +xb+xc) +xc/(xa +xb+xc)=>(xa +xb+xc) /(xa +xb+xc)=>1Ans: (b)

(13) If x=3+2 √2 then the value of (√x – 1/ √x)is [ √=root](a) 1 (b) 2 (c ) 2√2 ( d) 3√3Solution: (√x-1/√x)2 = x+ 1/x-2=> 3+2√2 + (1/3+2√2 )-2=> 3+2√2 + 3-2√2 -2=> 6-2 = 4(√x-1/√x)2 = 4=>(√x-1/√x)2 = 22(√x-1/√x) = 2.Ans : (b)

(14) (xb/xc)b+c-a (xc/xa)c+a-b (xa/xb)b+a-c = ? (a) xabc (b) 1 ( c) xab+bc+ca (d) xa+b+c

Solution: [xb-c]b+c-a [xc-a]c+a-b [xa-b]a+b-c

=>x(b-c)(b+c-a) x(c-a)(c+a-b) x(a-b)(a+b-c)=>x(b2-c2-ab-ac) x(c2-a2-bc-ab) x(a2-b2-ac-bc)=>x(b2-c2-ab-ac+c2-a2-bc-ab+a2-b2-ac-bc)=> x0=>1Ans: (b)

(15) If 3x-y = 27 and 3x+y = 243 then x is equal to(a) 0 (b) 2 (c ) 4 (d) 6

Solution: 3x-y = 27 => 3x-y = 33x-y= 33x+y = 243 => 3x+y = 35x+y = 5

From above two equations x = 4 , y=1Ans: (c )

(16) If ax = by = cz and b2 = ac then ‘y’equals(a)xz/x+z (b)xz/2(x-z) (c)xz/2(z-x) (d)2xz/x+z

Solution: Let us say ax = by = cz = kax =k => [ax]1/x = k1/x=> a = k1/xSimillarly b = k1/yc = k1/zb2 = ac[k1/y]2=k1/xk1/z=>k2/y = k1/x+1/z=> 2/y = 1/x+1/y=>y= 2xz/x+zAns: (d)

(17) ax = b,by = c ,cz = a then the value of xyz is is(a) 0 (b) 1 (c ) 1/abc (d) abc

Solution: ax = b(cz)x = b [cz = a]by)xz = b [by = c]=>xyz =1Ans: (b)

(18) If 2x = 4y =8z and (1/2x +1/4y +1/6z) =24/7 then the value of ‘z’ is(a) 7/16 (b) 7 / 32 (c ) 7/48 (d) 7/64

Solution: 2x = 4y=8z2x = 22y = 23zx= 2y = 3zMultiply above equation with ‘ 2’2x = 4y= 6z(1/2x+1/4y+1/6z) = 24/7=>(1/6z+1/6z+1/6z) = 24/7=> 3 / 6z = 24/7=> z= 7/48Ans: ( c)

PERCENTAGES

EXAMPLE PROBLEMS:

1 . Express the following as a fraction.a) 56%SOLUTION:56/100=14/25b) 4%SOLUTION:4/100=1/25c) 0.6%SOLUTION:0.6/100=6/1000=3/500d) 0.08%SOLUTION:0.08/100=8/10000=1/1250

2.Express the following as decimalsa) 6%SOLUTION:6% = 6/100=0.06b) 0.04%SOLUTION:0.04% = 0.04/100=0.0004

3 . Express the following as rate percent.i).23/36SOLUTION:= (23/36*100) %= 63 8/9%ii).6 ¾SOLUTION:6 ¾ =27/4(27/4 *100) % =675 %

4.Evaluate the following:28% of 450 + 45% of 280 ?SOLUTION:=(28/100) *450 + (45/100) *280= 28 * 45 / 5= 252

5.2 is what percent of 50?SOLUTION:Formula : (IS / OF ) *100 %

= 2/50 *100= 4%

6.½ is what percent of 1/3?SOLUTION:=( ½) / (1/3) *100 %= 3/2 *100 %= 150 %

7.What percent of 2 Metric tonnes is 40 Quintals?SOLUTION:1 metric tonne =10 QuintalsSo required percentage = (40/(2*10)) *100 %= 200%

8.Find the missing figure .i) ? % of 25 = 2.125SOLUTION :Let x% of 25 = 2.125. then(x/100) *25 =2.125x = 2.125 * 4= 8.5ii) 9% of ? =6.3SOLUTION:Let 9 % of x = 6.3.Then 9/100 of x= 6.3so x = 6.3 *100/7= 70.

9.Which is the greatest in 16 2/3 %, 2/15,0.17?SOLUTION:16 2/3 % = 50/3 %=50/3 * 1/100=1/6= 0.1662 / 15 =0.133So 0.17 is greatest number in the given series.

10.If the sales tax be reduced from 3 ½ % to 3 1/3 % ,then what difference does it make to a person who purchases an article with marked price of RS 8400?SOLUTION:Required difference = 3 ½ % of 8400 – 3 1/3 % of 8400=(7/2-10/3)% of 8400=1/6 % of 8400= 1/6* 1/100* 8400= Rs 14.

11. A rejects 0.08% of the meters as defective .How many will he examine to reject 2?SOLUTION:Let the number of meters to be examined be x.Then 0.08% of x=2.0.08/100*x= 2x= 2 * 100/0.08=2 * 100 * 100/8= 250012.65 % of a number is 21 less than 4/5 of that number. What is the number?SOLUTION: Let the number be x.4/5 x- (65% of x) = 214/5x – 65/100 x=2115x=2100x=140

13. Difference of two numbers is 1660.If 7.5 % of one number is 12.5% of the other number. Find two numbers?SOLUTION:Let the two numbers be x and y.7.5% of x=12.5% of ySo 75x=125 y3x=5yx=5/3y.Now x-y=16605/3y-y=16602/3y=1660y=2490So x= 2490+1660=4150.So the numbers are 4150 , 1660.

14. In expressing a length 81.472 KM as nearly as possible with 3 significant digits ,Find the % error?SOLUTION:Error= 81.5-81.472=0.028So the required percentage = 0.028/81.472*100%= 0.034%

15. In an election between two persons ,75% of the voters cast their votes out of which 2% are invalid. A got 9261 which 75% of the total valid votes. Find total number of votes?SOLUTION:Let x be the total votes.valid votes are 98% of 75% of x.So 75%(98%(75% of x))) = 9261==> 75/100 *98 /100 * 75 100 *x = 9261x= 1029 * 4 *100 *4 / 9

= 16800So total no of votes = 16800

16 . A’s maths test had 75 problems i.e 10 arithmetic, 30 algebra and 35 geometry problems. Although he answered 70% of arithmetic , 40% of algebra and 60 % of geometry problems correctly he didn’t pass the test because he got less than 60% of the problems right. How many more questions he would have needed to answer correctly to get a 60% passing grade.

SOLUTION:70% of 10 =70/100 * 10=740% of 30 = 40 / 100 * 30= 1260 % of 35 = 60 / 100 *35= 21So correctly attempted questions = 7 + 12 + 21=40.Questions to be answered correctly for 60% grade=60% of 75= 60/100 *75=45.So required questions=45-40 = 5

17 . If 50% of (x – y) = 30% of (x + y) then what percent of x is y ?SOLUTION:50/100(x-y) =30/100(x+y)½ (x-y)= 3/10(x+y)5x-5y=3x+3yx=4ySo Required percentage =y/x*100 %=y/4y *100 %= 25%.

18 . If the price of tea is increased by 20% ,find how much percent must a householder reduce her consumption of tea so as not to increase the expenditure?SOLUTION:Reduction in consumption= R/(100+R) *100%=20/120 *100= 16 2/3 %

19.The population of a town is 176400 . If it increases at the rate of 5% per annum ,what will be the population 2 years hence? What was it 2 years ago?SOLUTION:Population After 2 years = 176400[1+5/100]2=176400 * 21/20 *21/20

=194481Population 2 years ago = 176400/(1+5/100)2= 176400 * 20/21 *20/ 21=160000

20.1 liter of water is add to 5 liters of a 20 % solution of alcohol in water . Find the strength of alcohol in new solution?SOLUTION:Alcohol in 5 liters = 20% of 5=1 literAlcohol in 6 liters of new mixture = 1literSo % of alcohol is =1/6 *100=16 2/3%

21. If A earns 33 1/3 more than B .Then B earns less than A by what percent?SOLUTION:33 1/3 =100 / 3Required Percentage = (100/3)/(100 + (100/3)) *100 %= 100/400 *100 = 25 %

22. A school has only three classes which contain 40,50,60 students respectively . The pass percent of these classes are 10, 20 and 10 respectively . Then find the pass percent in the school.SOLUTION:Number of passed candidates =10/100*40+20/100 *50+10/100 * 60=4+10+6=20Total students in school = 40+50+60 =150So required percentage = 20/150 *100= 40 /3=13 1/3 %

23. There are 600 boys in a hostel . Each plays either hockey or football or both .If 75% play hockey and 45 % play football ,Find how many play both?SOLUTION:n(A)=75/100 *600=450n(B) = 45/100 *600= 270n(A^B)=n(A) + n(B) – n(AUB)=450 + 270 -600=120So 120 boys play both the games.

24.A bag contains 600 coins of 25p denomination and 1200 coins of 50p denomination. If 12% of 25p coins and 24 % of 50p coins are removed, Find the percentage of money

removed from the bag ?SOLUTION:Total money = (600 * 25/100 +1200 *50/100)=Rs 75025p coins removed = 12/100 *600=7250p coins removed = 24/100 *1200=288So money removed =72 *1/4 +288 *1/2= Rs 162So required percentage=162/750 *100=21 .6%

25. P is six times as large as Q.Find the percent that Q is less than P?SOLUTION:Given that P= 6QSo Q is less than P by 5Q.Required percentage= 5Q/P*100 %=5/6 * 100 %=83 1/3%

26.For a sphere of radius 10 cm ,the numerical value of surface area is what percent of the numerical value of its volume?SOLUTION:Surface area = 4 *22/7 *r2= 3/r(4/3 * 22/7 * r3)=3/r * VOLUMEWhere r = 10 cmSo we have S= 3/10 V=3/10 *100 % of V= 30 % of VSo surface area is 30 % of Volume.

27. A reduction of 21 % in the price of wheat enables a person to buy 10 .5 kg more for Rs 100.What is the reduced price per kg.SOLUTION:Let the original price = Rs x/kgReduced price =79/100x /kg==> 100/(79x/100)-100/x =10.5==> 10000/79x-100/x=10.5==> 10000-7900=10.5 * 79 x==> x= 2100/10.5 *79So required price = Rs (79/100 *2100/10.5 *79) /kg= Rs 2 per kg.

28.The length of a rectangle is increased by 60 % .By what percent would the width have to be decreased to maintain the same area?SOLUTION:Let the length =l,Breadth= b.Let the required decrease in breadth be x %then 160/100 l *(100-x)/100 b=lb160(100-x)=100 *100or 100-x =10000/160=125/2so x = 100-125/2

APTITUDE-Problems on Numbers

Introduction:

Natural Numbers:

All positive integers are natural numbers.Ex 1,2,3,4,8,……

There are infinite natural numbers and number 1 is the least natural number.Based on divisibility there would be two types of natural numbers. They are Prime and composite.

Prime Numbers:

A natural number larger than unity is a prime number if it does not have other divisors except for itself and unity.

Note:-Unity i e,1 is not a prime number.

Properties Of Prime Numbers:

->The lowest prime number is 2.->2 is also the only even prime number.->The lowest odd prime number is 3.->The remainder when a prime number p>=5 s divided by 6 is 1 or 5.However, if a number on being divided by 6 gives a remainder 1 or 5 need not be prime.->The remainder of division of the square of a prime number p>=5 divide by 24 is 1.->For prime numbers p>3, p²-1 is divided by 24.

->If a and b are any 2 odd primes then a²-b² is composite. Also a²+b²is composite.->The remainder of the division of the square of a prime number p>=5divided by 12 is 1.

Process to Check A Number s Prime or not:

Take the square root of the number.Round of the square root to the next highest integer call this number as Z.Check for divisibility of the number N by all prime numbers below Z. Ifthere is no numbers below the value of Z which divides N then the number will be prime.

Example 239 is prime or not?√239 lies between 15 or 16.Hence take the value of Z=16.Prime numbers less than 16 are 2,3,5,7,11 and 13.239 is not divisible by any of these. Hence we can conclude that 239is a prime number.

Composite Numbers:

The numbers which are not prime are known as composite numbers.

Co-Primes:

Two numbers a an b are said to be co-primes,if their H.C.F is 1.Example (2,3),(4,5),(7,9),(8,11)…..Place value or Local value of a digit in a Number:

place value:

Example 689745132Place value of 2 is (2*1)=2Place value of 3 is (3*10)=30 and so on.Face value:-It is the value of the digit itself at whateverplace it may be.

Example 689745132Face value of 2 is 2.Face value of 3 is 3 and so on.

Tests of Divisibility:

Divisibility by 2:-A number is divisible by 2,if its unit’s digit isany of 0,2,4,6,8.Example 84932 is divisible by 2,while 65935 is not.

Divisibility by 3:-A number is divisible by 3,if the sum of its digits is divisible by 3.

Example 1.592482 is divisible by 3,since sum of its digits5+9+2+4+8+2=30 which is divisible by 3.

Example 2.864329 is not divisible by 3,since sum of its digits8+6+4+3+2+9=32 which is not divisible by 3.

Divisibility by 4:-A number is divisible by 4,if the number formed by last two digits is divisible by 4.

Example 1.892648 is divisible by 4,since the number formed by the lasttwo digits is 48 divisible by 4.

Example 2.But 749282 is not divisible by 4,since the number formed bythe last two digits is 82 is not divisible by 4.

Divisibility by 5:-A number divisible by 5,if its unit’s digit is either 0 or 5.Example 20820,50345

Divisibility by 6:-If the number is divisible by both 2 and 3.example 35256 is clearly divisible by 2sum of digits =3+5+2+5+21,which is divisible by 3Thus the given number is divisible by 6.

Divisibility by 8:-A number is divisible by 8 if the last 3 digits of the number are divisible by 8.

Divisibility by 11:-If the difference of the sum of the digits in the odd places and the sum of the digitsin the even places is zero or divisibleby 11.

Example 4832718(8+7+3+4) – (1+2+8)=11 which is divisible by 11.

Divisibility by 12:-All numbers divisible by 3 and 4 are divisible by 12.

Divisibility by 7,11,13:-The difference of the number of its thousands and the remainder of its division by 1000 is divisible by 7,11,13.

BASIC FORMULAE:

->(a+b)²=a²+b²+2ab->(a-b)²=a²+b²-2ab->(a+b)²-(a-b)²=4ab->(a+b)²+(a-b)²=2(a²+b²)->a²-b²=(a+b)(a-b)

->(a-+b+c)²=a²+b²+c²+2(ab+b c+ca)->a³+b³=(a+b)(a²+b²-ab)->a³-b³=(a-b)(a²+b²+ab)->a³+b³+c³-3a b c=(a+b+c)(a²+b²+c²-ab-b c-ca)->If a+b+c=0 then a³+b³+c³=3a b c

DIVISION ALGORITHM

If we divide a number by another number ,then

Dividend = (Divisor * quotient) + Remainder

MULTIPLICATION BY SHORT CUT METHODS

1.Multiplication by distributive law:

a)a*(b+c)=a*b+a*cb)a*(b-c)=a*b-a*c

Examplea)567958*99999=567958*(100000-1)567958*100000-567958*156795800000-56795856795232042

b)978*184+978*816=978*(184+816)978*1000=978000

2.Multiplication of a number by 5n:-Put n zeros to the right of themultiplicand and divide the number so formed by 2n

Example 975436*625=975436*54=9754360000/16=609647500.

PROGRESSION:

A succession of numbers formed and arranged in a definite order according to certain definite rule is called a progression.

1.Arithmetic Progression:-If each term of a progression differs from itspreceding term by a constant.This constant difference is called the common difference of the A.P.The n th term of this A.P is Tn=a(n-1)+d.The sum of n terms of A.P Sn=n/2[2a+(n-1)d].

Important Results:

a.1+2+3+4+5………………….=n(n+1)/2.b.12+22+32+42+52………………….=n(n+1)(2n+1)/6.c.13+23+33+43+53………………….=n2(n+1)2/4

2.Geometric Progression:-A progression of numbers in which everyterm bears a constant ratio with ts preceding term.i.e a,a r,a r2,a r3……………In G.P Tn=a r n-1Sum of n terms Sn=a(1-r n)/1-r

Problems

1.Simplifya.8888+888+88+8b.11992-7823-456

Solution: a.88888888889872b.11992-7823-456=11992-(7823+456)=11992-8279=3713

2.What could be the maximum value of Q in the following equation?5PQ+3R7+2Q8=1114

Solution: 5 P Q3 R 72 Q 811 1 42+P+Q+R=11Maximum value of Q =11-2=9 (P=0,R=0)

3.Simplify: a.5793405*9999 b.839478*625

Solution:a. 5793405*9999=5793405*(10000-1)57934050000-5793405=57928256595b. 839478*625=839478*54=8394780000/16=524673750.

4.Evaluate 313*313+287*287

Solution:a²+b²=1/2((a+b)²+(a-b)²)

1/2(313+287)² +(313-287)²=1/2(600 ² +26 ² )½(360000+676)=180338

5.Which of the following is a prime number?a.241 b.337 c.391

Solution:a.24116>√241.Hence take the value of Z=16.Prime numbers less than 16 are 2,3,5,7,11 and 13.241 is not divisible by any of these. Hence we canconclude that 241 is a prime number.b. 33719>√337.Hence take the value of Z=19.Prime numbers less than 16 are 2,3,5,7,11,13 and 17.337 is not divisible by any of these. Hence we can concludethat 337 is a prime number.c. 39120>√391.Hence take the value of Z=20.Prime numbers less than 16 are 2,3,5,7,11,13,17 and 19.391 is divisible by 17. Hence we can concludethat 391 is not a prime number.

6.Find the unit’s digit n the product 2467 153 * 34172?

Solution: Unit’s digit in the given product=Unit’s digit in 7 153 * 172Now 7 4 gives unit digit 17 152 gives unit digit 17 153 gives 1*7=7.Also 172 gives 1Hence unit’s digit in the product =7*1=7.

7.Find the total number of prime factors in 411 *7 5 *112 ?

Solution: 411 7 5 112= (2*2) 11 *7 5 *112= 222 *7 5 *112Total number of prime factors=22+5+2=29

8.Which of the following numbers s divisible by 3?a.541326b.5967013

Solution: a. Sum of digits in 541326=5+4+1+3+2+6=21 divisible by 3.b. Sum of digits in 5967013=5+9+6+7+0+1+3=31 not divisible by 3.

9.What least value must be assigned to * so that th number 197*5462 is divisible by 9?Solution: Let the missing digit be x

Sum of digits = (1+9+7+x+5+4+6+2)=34+xFor 34+x to be divisible by 9 , x must be replaced by 2The digit in place of x must be 2.

10.What least number must be added to 3000 to obtain a number exactly divisible by 19?Solution:On dividing 3000 by 19 we get 17 as remainderTherefore number to be added = 19-17=2.

11.Find the smallest number of 6 digits which is exactly divisible by 111?Solution:Smallest number of 6 digits is 100000On dividing 10000 by 111 we get 100 as remainderNumber to be added =111-100=11.Hence,required number =10011.

12.On dividing 15968 by a certain number the quotient is 89 and the remainder is 37.Find the divisor?Solution:Divisor = (Dividend-Remainder)/Quotient=(15968-37) / 89=179.

13.A number when divided by 342 gives a remainder 47.When the same number is divided by 19 what would be the remainder?

Solution:Number=342 K + 47 = 19 * 18 K + 19 * 2 + 9=19 ( 18K + 2) + 9.The given number when divided by 19 gives 18 K + 2 as quotient and 9 as remainder.

14.A number being successively divided by 3,5,8 leaves remainders 1,4,7 respectively. Find the respective remainders if the order of divisors are reversed?

Solution:Let the number be x.

3 x 5 y – 1 8 z – 4 1 – 7 z=8*1+7=15y=5z+4 = 5*15+4 = 79x=3y+1 = 3*79+1=238Now8 2385 29 – 63 5 – 41 – 2Respective remainders are 6,4,2.

15.Find the remainder when 231 is divided by 5?

Solution:210 =1024.unit digit of 210 * 210 * 210 is 4 as4*4*4 gives unit digit 4unit digit of 231 is 8.

Now 8 when divided by 5 gives 3 as remainder.231 when divided by 5 gives 3 as remainder.

16.How many numbers between 11 and 90 are divisible by 7?

Solution:The required numbers are 14,21,28,………..,84This is an A.P with a=14,d=7.Let it contain n termsthen T =84=a+(n-1)d=14+(n-1)7=7+7n7n=77 =>n=11.

17.Find the sum of all odd numbers up to 100?

Solution:The given numbers are 1,3,5………99.This is an A.P with a=1,d=2.Let it contain n terms 1+(n-1)2=99=>n=50Then required sum =n/2(first term +last term)=50/2(1+99)=2500.

18.How many terms are there in 2,4,6,8……….,1024?

Solution:Clearly 2,4,6……..1024 form a G.P with a=2,r=2Let the number of terms be nthen 2*2 n-1=10242n-1 =512=29n-1=9n=10.

19.2+22+23+24+25……….+28=?

Solution:Given series is a G.P with a=2,r=2 and n=8.Sum Sn=a(1-r n)/1-r=Sn=2(1-28)/1-2.=2*255=510.

20.A positive number which when added to 1000 gives a sum ,which is greater than when it is multiplied by 1000.The positive integer is?a.1 b.3 c.5 d.7

Solution:1000+N>1000Nclearly N=1.

21.The sum of all possible two digit numbers formed from threedifferent one digit natural numbers when divided by the sum of the

original three numbers is equal to?a.18 b.22 c.36 d. none

Solution:Let the one digit numbers x,y,zSum of all possible two digit numbers==(10x+y)+(10x+z)+(10y+x)+(10y+z)+(10z+x)+(10z+y)= 22(x+y+z)Therefore sum of all possible two digit numbers when divided by sum ofone digit numbers gives 22.

22.The sum of three prime numbers is 100.If one of them exceeds another by 36 then one of the numbers is?a.7 b.29 c.41 d67.

Solution:x+(x+36)+y=1002x+y=64Therefore y must be even prime which is 22x+2=64=>x=31.Third prime number =x+36=31+36=67.

23.A number when divided by the sum of 555 and 445 gives two times their difference as quotient and 30 as remainder .The number is?a.1220 b.1250 c.22030 d.220030.

Solution:Number=(555+445)*(555-445)*2+30=(555+445)*2*110+30=220000+30=220030.

24.The difference between two numbers s 1365.When the larger number is divided by the smaller one the quotient is 6 and the remainder is 15.The smaller number is?a.240 b.270 c.295 d.360

Solution:Let the smaller number be x, then larger number =1365+xTherefore 1365+x=6x+155x=1350 => x=270Required number is 270.

25.In doing a division of a question with zero remainder,a candidatetook 12 as divisor instead of 21.The quotient obtained by him was 35.The correct quotient is?a.0 b.12 c.13 d.20

Solution:Dividend=12*35=420.Now dividend =420 and divisor =21.

APTITUDE-Problems on RATIO AND PROPORTION

Important Facts:

1.Ratio : The ratio of two qualities a and b in the same units, is the fraction a/b and we write it as a:b. In the ratio, a:b, we call a as the first term of antecedent and b, the second term consequent.Ex: The ratio 5:9 represents 5/9 with antecedent=5 ,consequent=9

3Rule: The multiplication or division of each term of 9 ratio by the same non-zero number does not affect the ratio.

4.Proportion: The equality of two ratios is called proportion. If a:b=c:d, we write a:b::c:d and we say that a,b,c and d are in proportion. Here a and b are called extremes, while b and c are called mean terms.Product of means=product of extremesThus, a:b::c:d => (b*c)=(a*d)

5.Fourth proportional: If a:b::c:d, then d is called the fourth proportional to a,b and c.

6.Third proportional: If a:b::b:c, then c is called third proportional to a and b.

7.Mean proportional: Mean proportional between a and b is SQRT(a*b).

COMPARISION OF RATIOS:

We say that (a:b)>(c:d) => (a/b)>(c/d)

8.Compounded ratio: The compounded ratio of the ratios (a:b), (c:d),(e:f) is (ace:bdf).

9.Duplicate Ratio: If (a:b) is (a2: b2 )

10.Sub-duplicate ratio of (a:b) is (SQRT(a):SQRT(b))

11.Triplicate ratio of (a:b) is (a3: b3 )

12.Sub-triplicate ratio of (a:b) is (a1/3: b1/3 ).

13.If a/b=c/d, then (a+b)/(a-b)=(c+d)/(c-d) (component and dividend o)

VARIATION:

14.we say that x is directly proportional to y, if x=ky for some constant k and we write.

15.We say that x is inversely proportional to y, if xy=k for some constant and we write.

16. X is inversely proportional to y.If a/b=c/d=e/f=g/h=k thenk=(a+c+e+g)/(b+d+f+h)If a1/b1,a2/b2, a3/b3…………..an/bn are unequalfractions then the ratio.

SIMPLE PROBLEMS

1.If a:b =5:9 and b:c=4:7 Find a:b:c?Sol: a:b=5:9 and b:c=4:7=4*9/4:9*4/9=9:63/9a:b:c=5:9:63/9=20:36:63

2.Find the fourth proportion to 4,9,12Sol: d is the fourth proportion to a,b,ca:b=c:d4:9=12:x4x=9*12=>x=27

3.Find third proportion to 16,36Sol: if a:b=b:c then c is the third proportion to a,b16:36=36:x16x=36*36x=81

4.Find mean proportion between 0.08 and 0.18Sol: mean proportion between a and b=square root of abmean proportion =square root of 0.08*0.18=0.12

5.If a:b=2:3 b:c=4:5, c:d=6:7 then a:b:c:d isSol: a:b=2:3 and b:c=4:5=4*3/4:5*3/4=3:15/4c:d=6:7=6*15/24:7*15/24=15/4:35/8a:b:c:d=2:3:15/4:35/8=16:24:30:35

6.2A=3B=4C then A:B:C?Sol: let 2A=3B=4C=k thenA=k/2, B=k/3, C=k/4A:B:C=k/2:k/3:k/4=6:4:3

7.15% of x=20% of y then x:y isSol: (15/100)*x=(20/100)*y3x=4yx:y=4:3

8.a/3=b/4=c/7 then (a+b+c)/c=Sol: let a/3=b/4=c/7=k(a+b+c)/c=(3k+4k+7k)/7k=2

9.Rs 3650 is divided among 4 engineers, 3 MBA’s and 5 CA’s such that 3 CA’s get as much as 2 MBA’s and 3 Eng’s as much as 2 CA’s .Find the share of an MBA.Sol: 4E+3M+5C=36503C=2M, that is M=1.5C3E=2C that is E=.66 CThen, (4*0.66C)+(3*1.5C)+5C=3650C=3650/12.166C=300M=1.5 and C=450.

DIFFICULT PROBLEMS

1.Three containers A,B and C are having mixtures of milk and water in the ratio of 1:5 and 3:5 and 5:7 respectively. If the capacities of the containers are in the ratio of all the three containers are in the ratio 5:4:5, find the ratio of milk to water, if the mixtures of all the three containers are mixed together.Sol: Assume that there are 500,400 and 500 liters respectively in the 3 containers.Then ,we have, 83.33, 150 and 208.33 liters of milk in each of the three containers.Thus, the total milk is 441.66 liters. Hence, the amount of water in the mixture is 1400-441.66=958.33liters.Hence, the ratio of milk to water is 441.66:958.33 => 53:115(using division by .3333)The calculation thought process should be(441*2+2):(958*3+1)=1325:2875Dividing by 25 => 53:115.

2.A certain number of one rupee,fifty parse and twenty five paise coins are in the ratio of 2:5:3:4, add up to Rs 210. How many 50 paise coins were there?Sol: the ratio of 2.5:3:4 can be written as 5:6:8let us assume that there are 5 one rupee coins,6 fifty paise coins and 8 twenty-five paise coins in all.their value=(5*1)+(6*.50)+(8*.25)=5+3+2=Rs 10If the total is Rs 10,number of 50 paise coins are 6.if the total is Rs 210, number of 50 paise coins would be 210*6/10=126.

3.The incomes of A and B are in the ratio of 4:3 and their expenditure are in the ratio of 2:1 . if each one saves Rs 1000,what are their incomes?Sol: Ratio of incomes of A and B=4:3Ratio of expenditures of A and B=2:1Amount of money saved by A=Amount of money saved by B=Rs 1000let the incomes of A and B be 4x and 3x respectivelylet the expense of A and B be 2y and 1yrespectivelyAmount of money saved by A=(income-expenditure)=4x-2y=Rs 1000Amount of money saved by B=3x-y=Rs 1000

this can be even written as 6x-2y=Rs 2000now solve 1 and 3 to getx=Rs 500therefore income of A=4x=4*500=Rs 2000income of B=3x=3*500=Rs 1500

4.A sum of Rs 1162 is divided among A,B and C. Such that 4 times A’s share share is equal to 5 times B’s share and 7 times C’s share . What is the share of C?Sol: 4 times of A’s share =5 times of B’s share=7 times of C’s share=1therefore , the ratio of their share =1/4:1/5:1/7LCM of 4,5,7=140therefore, ¼:1/5:1/7=35:28:20the ratio now can be written as 35:28:20therefore C’s share=(20/83)*1162=20*14=Rs 280.

5.The ratio of the present ages of saritha and her mother is 2:9, mother’s age at the time of saritha’s birth was 28 years , what is saritha’s present age?Sol: ratio of ages of saritha and her mother =2:9let the present age of saritha be 2x years. then the mother’s present age would be 9x yearsDifference in their ages =28 years9x-2x=28 years7x=28=>x=4therefore saritha’s age =2*4=8 years

APTITUDE – problems on SQUARE AND CUBE ROOTS

Formula:

The Product of two same numbers in easiest way as follow.Example:let us calculate the product of 96*96Solution: Here every number must be compare with the 100.See here the given number 96 which is 4 difference with the 100.so subtract 4 from the 96 we get 92 ,then the square of thenumber 4 it is 16 place the 16 beside the 92 we get answeras 9216.

9 6– 4————–9 2————–

4*4=169 2 1 6

therefore square of the two numbers 96*96=9216.

Example: Calculate product for 98*98Solution: Here the number 98 is having 2 difference when compareto 100 subtract 2 from the number then we get 96 square thenumber 2 it is 4 now place beside the 96 as 9604

9 8– 2————-9 6————-2*2=49 6 0 4.so, we get the product of 98*98=9604.

Example: Calculate product for 88*88Solution: Here the number 88 is having 12 difference when compareto 100 subtract 12 from the 88 then we get 76 the square of thenumber 12 is 144 (which is three digit number but should placeonly two digit beside the 76) therefore in such case add one to6 then it becomes 77 now place 44 beside the number 77 we will get7744.88-12————76———–12*12=144

76+ 144——————–7744——————–

Example: Find the product of the numbers 46 *46?Solution:consider the number 50=100/2. Now again go comparision withthe number which gets when division with 100.here consider the number50 which is nearer to the number given. 46 when compared with thenumber 50 we get the difference of 4. Now subtract the number 4 fromthe 46, we get 42. As 50 got when 100 get divided by 2.so, divided the number by 2 after subtraction.

42/2=21 now square the the number 4 i.e, 4*4=16just place the number 16 beside the number 21we get 2116.4 64—————-4 2 as 50 = 100/2

42/2=21now place 4*4=16 beside 212 1 1 6

Example: Find the product of the numbers 37*37Solution:consider the number 50=100/2now again go comparision with the number which gets whendivision with 100.here consider the number 50 which is nearer to the number given.37 when compared with the number 50 we get the difference of 13.now subtract the number 13 from the 37, we get 24.as 50 got when 100 get divided by 2.so, divided the number by 2 after subtraction.24/2=12now square the the number 13 i.e, 13*13=169just place the number 169 beside the number 21now as 169 is three digit number then add 1 to 2 we get1t as 13 then place 69 beside the 13we get 1369.

3 71 3—————–2 4 as 50 = 100/2

24/2=12square 13* 13=169

1 2+ 1 6 9———————–1 3 6 9————————-

Example: Find the product of 106*106Solution: now compare it with 100 ,The given number is more then 100

then add the extra number to the given number.That is 106+6=112then square the number 6 that is 6*6=36just place beside the number 36 beside the 112,thenwe get 11236.1 0 6+ 6———————1 1 2——————–now 6* 6=36 place this beside the number 112, we get1 1 2 3 6

Square root: If x2=y ,we say that the square root of yis x and we write ,√y=x.

Cube root: The cube root of a given number x is the numberwhose cube is x. we denote the cube root of x by x1/3 .

Examples:

1.Evaluate 60841/2 by factorization method.

Solution: Express the given number as the product of primefactors. Now, take the product of these prime factors choosingone out of every pair of the same primes. This product gives thesquare root of the given number.

Thus resolving 6084 in the prime factors ,we get 60842 60242 30423 15213 50713 169136084=21/2 *31/2 *131/260841/2=2*3*13=78.Answer is 78.

2.what will come in place of question mark in each of the followingquestions?

i)(32.4/?)1/2 = 2ii)86.491/2 + (5+?1/2)2 =12.3

Solution: 1) (32.4/x)1/2=2Squaring on both sides we get32.4/x=4=>4x=32.4=>x=8.1

Answer is 8.1

ii)86.491/2 + √(5+x2)=12.3

solutin:86.491/2 + (5+x1/2 )=12.39.3+ √(5+x1/2 )=12.3=> √(5+x1/2 ) =12.3-9.3=> √(5+x1/2 )=3Squaring on both sides we get(5+x1/2 )=9x1/2 =9-5x1/2 =4x=2.Answer is 2.

3.√ 0.00004761 equals:

Solution: √ (4761/108)√4761/√ 108. 69/100000.0069.Answer is 0.0069

4.If √18225=135,then the value of √182.25 + √1.8225 + √ 0.018225 + √0.00018225.

Solution: √(18225/100) +√(18225/10000) +√(18225/1000000) +√(18225/100000000)=√(18225)/10 + (18225)1/2/100 +√(18225)/1000 + √(18225)/10000=135/10 + 135/100 + 135/1000 + 135/10000=13.5+1.35+0.135+0.0135=14.9985.Answer is 14.9985.

5.what should come in place of both the questionmarks in the equation (?/ 1281/2= (162)1/2/?) ?

Solution: x/ 1281/2= (162)1/2/x=>x1/2= (128*162)1/2=> x1/2= (64*2*18*9)1/2

=>x2= (82*62*32)=>x2=8*6*3=>x2=144=>x=12.

6.If 0.13 / p1/2=13 then p equals

Solution: 0.13/p2=13=>p2=0.13/13=1/100p2=√(1/100)=>p=1/10therefore p=0.1Answer is 0.1

7.If 13691/2+(0.0615+x)1/2=37.25 then x is equals to:

Solution37+(0.0615+x)1/2=37.25(since 37*37=1369)=>(0.0615+x)1/2=0.25Squaring on both sides(0.0615+x)=0.0625x=0.001x=10-3.Answer is 10-3.

8.If √(x-1)(y+2)=7 x& y being positive whole numbers thenvalues of x& y are?

Solution: √(x-1)(y+2)=7Squaring on both sides we get(x-1)(y+2)=72x-1=7 and y+2=7therefore x=8 , y=5.Answer x=8 ,y=5.

9.If 3*51/2+1251/2=17.88.then what will be thevalue of 801/2+6*51/2?

Solution: 3*51/2+1251/2=17.883*51/2+(25*5)1/2=17.883*51/2+5*51/2=17.888*51/2=17.8851/2=2.235therefore 801/2+6 51/2=(16*51/2)+6*1/25=4 51/2+6 51/2

=10*2.235=22.35Answer is 22.35

10.If 3a=4b=6c and a+b+c=27*√29 then Find c value is:

Solution: 4b=6c=>b=3/2*c3a=4b=>a=4/3b=>a=4/3(3/2c)=2ctherefore a+b+c=27*291/22c+3/2c+c=27*291/2=>4c+3c+2c/2=27*291/2=>9/2c=27*291/2c=27*291/2*2/9c=6*291/2

11.If 2*3=131/2 and 3*4=5 then value of 5*12 is

Solution:Here a*b=(a2+b2)1/2therefore 5*12=(52+122)1/2=(25+144)1/2=1691/2=13Answer is 13.

12.The smallest number added to 680621 to makethe sum a perfect square is

Solution: Find the square root number whichis nearest to this number8 680621 82464162 4063241644 822165761645therefore 824 is the number ,to get the nearestsquare root number take (825*825)-680621therefore 680625-680621=4hence 4 is the number added to 680621 to make itperfect square.

13.The greatest four digit perfect square number is

Solution: The greatest four digit number is 9999.now find the square root of 9999.9 9999 9981189 18191701198therefore 9999-198=9801 which is required number.Answer is 9801.

14.A man plants 15376 apples trees in his garden and arrangesthem so, that there are as many rows as there are apples treesin each row .The number of rows is.

Solution: Here find the square root of 15376.1 15376 124122 5344244 9769760therefore the number of rows are 124.

15.A group of students decided to collect as many paise fromeach member of the group as is the number of members. If thetotal collection amounts to Rs 59.29.The number of membersin the group is:

Solution: Here convert Money into paise.59.29*100=5929 paise.To know the number of member ,calculate the square root of 5929.7 5929 7749147 102910290Therefore number of members are 77.

16.A general wishes to draw up his 36581 soldiers in the formof a solid square ,after arranging them ,he found that some ofthem are left over .How many are left?

Solution: Here he asked about the left man ,So find thesquare root of given number the remainder will be the left man1 36581 191129 265261381 481381100(since remaining)Therefore the left men are 100.

17.By what least number 4320 be multiplied to obtain numberwhich is a perfect cube?

Solution: find l.c.m for 4320.2 43202 21602 10802 5402 2703 1353 453 1554320=25 * 33 * 5=23 * 33 * 22 *5so make it a perfect cube ,it should be multiplied by 2*5*5=50Answer is 50.

18.3(4*12/125)1/2=?

Solution: 3(512/125)1/23(8*8*8)1/2/(5*5*5)3(83)1/2/(53)((83)/(53))1/3=>8/5 or 1 3/5.

H.C.F AND L.C.M

Facts And Formulae:

1.Highest Common Factor:(H.C.F) or Greatest Common Meaure(G.C.M) :

The H.C.F of two or more than two numbers is the greatest number that divides each of them exactly.There are two methods :1.Factorization method: Express each one of the given numbers as the product of prime factors.The product of least powers of common prime factors gives HCF.Example : Find HCF of 26 * 32*5*74 , 22 *35*52 * 76 , 2*52 *72Sol: The prime numbers given common numbers are 2,5,7Therefore HCF is 22 * 5 *72 .

2.Division Method : Divide the larger number bysmaller one. Now divide the divisor by remainder. Repeat the processof dividing preceding number last obtained till zero is obtained asnumber. The last divisor is HCF.

Example: Find HCF of 513, 1134, 1215Sol:

1134) 1215(11134———-81)1134(1481———–324324———–0———–HCF of this two numbers is 81.

81)513(6486——–27)81(381—–0—

HCF of 81 and 513 is 27.

3.Least common multiple[LCM] : The least number which is divisible by each one of given numbers is LCM.There are two methods for this:1.Factorization method : Resolve each one into product of prime factors.Then LCM is product of highest powers of all factors.

2.Common division method.

Problems:1.The HCF of 2 numbers is 11 and LCM is 693.If one of numbers is 77.find other.Sol: Other number = 11 * 693/77=99.

2.Find largest number of 4 digits divisible by 12,15,18,27Sol: The largest number is 9999.LCM of 12,15,18,27 is 540.on dividing 9999 by 540 we get 279 as remainder.Therefore number =9999 – 279 =9720.

3.Find least number which when divided by 20,25,35,40 leaves remainders 14,19,29,34.

Sol:

20–14=625-19=635-29=640-34=6

Therefore number =LCM of (20,25,35,40) – 6=13944.252 can be expressed as prime as :

2 2522 1263 633 217

prime factor is 2 *2 * 3 * 3 *7

5.1095/1168 when expressed in simple form is

1095)1168(11095——73)1095(1573———

365365———0———-So, HCF is 73Therefore1095/1168 = 1095/73/1168/73= 15/16

6.GCD of 1.08,0.36,0.9 isSol:HCF of 108,36,90

36)90(272—-18)36(236—-0—-HCF is 18.

HCF of 18 and 108 is 1818)108(6108——-0——–Therefore HCF =0.18

7.Three numbers are in ratio 1:2:3 and HCF is 12.Find numbers.Sol:Let the numbers be x.Three numbers are x,2x,3xTherefore HCF is

2x)3x(12x—–x)2x(22x——–0————-

HCF is x so, x is 12Therefore numbers are 12,24,36.

8.The sum of two numbers is 216 and HCF is 27.Sol:Let numbers are

27a + 27 b =216

a + b =216/27=8

Co-primes of 8 are (1,7) and (3,5) numbers=(27 * 1 ), (27 * 7)=27,89

9.LCM of two numbers is 48..The numbers are in ratio 2:3.The sum of numbers isSol:Let the number be x.Numbers are 2x,3xLCM of 2x,3x is 6xTherefore6x=48x=8.Numbers are 16 and 24Sum=16 +24=40.

10.HCF and LCM of two numbers are 84 and 21.If ratio of two numbers is 1:4.Then largest of two numbers is

Sol:Let the numbers be x,4xThen x * 4x = 84 * 21x2 =84 * 21 /4x = 21Largest number is 4 * 21.

11.HCF of two numbers is 23,and other factors of LCM are 13,14.Largest number isSol:23 * 14 is Largest number.

12.The maximum number of students among them 1001 pens and 910 pencils can be distributed in such a way that each student gets same number of pens and pencils is ?Sol:HCF of 1001 and 910

910)1001(1910

————91)910(10910——–0———Therefore HCF=91

13.The least number which should be added to 2497 so that sum is divisible by 5,6,4,3 ?Sol: LCM of 5,6,4,3 is 60.On dividing 2497 by 60 we get 37 as remainder.Therefore number to added is 60 – 37 =23.Answer is 23.

14.The least number which is a perfect square and is divisible by each of numbers 16,20,24 is ?Sol: LCM of 16,20,24 is 240.

2 * 2*2*2*3*5=240To make it a perfect square multiply by 3 * 5Therefore 240 * 3 * 5=3600Answer is 3600.