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One guy has Rs. 100/- in hand. He has to buy 100 balls. One football costs Rs. 15/, One

Cricket ball costs Re. 1/- and one table tennis ball costs Rs. 0.25 He spend the whole Rs. 100/- to buy the

balls. How many of each balls he bought?

F + C + T = 100————-eq1

15F + C + 0.25T = 100————-eq2

eq1=eq2 .solve to get F=3T/56 ;F=3,T=56,C=41

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1.Bird and the train:

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The distance between Station Atena and Station Barcena is 90 miles. A train starts from Atena towards

Barcena. A bird starts at the same time from Barcena straight towards the moving train. On reaching the

train, it instantaneously turns back and returns to Barcena. The bird makes these journeys from Barcena to

the train and back to Barcena continuously till the train reaches Barcena. The bird finally returns to

Barcena and rests. Calculate the total distance in miles the bird travels in the following two cases:

(a) the bird flies at 90 miles per hour and the speed of the train is 60 miles per hour there is no need to consider their meeting pt at all.the train has been running for 90miles/(60miles/hr)=1.5hrs.bird flies till train reaches destination frm strting pt.so bird flies for1.5hrs at the vel given(90).so dist=1.5*90=135miles

(b) the bird flies at 60 miles per hour and the speed of the train is 90 miles per hour time of train=1hr.so dist of bird=60*1=60miles

2.Tennis Championship

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A tennis championship is played on a knock-out basis, i.e., a player is out of the tournament when he loses

a match.

(a) How many players participate in the tournament if 15 matches are totally played?

C u don’t need to sum it up.since it’s a knock out only 1 person emerges winner finally.so15+1=16is

answer.becos after15 matches finally we shud’ve 15losers and 1winner.

(b) How many matches are played in the tournament if 50 players totally participate? -> 49.its always

one less thn no of players as per the idea ive given above.so no need to check okay cos its always true.ans is 49.

old enough:

“When I add 4 times my age 4 years from now to 5 imes my age 5 years from now, I get 10 times my current age. How old will I be 3 years from now?”

Let x= current age

4(x+4)+5(x+5)=10x ;so x=Ò 41 years

3.Gold coins:

A rich merchant had collected many gold coins. He did not want anybody to know about them. One day, his wife

asked, “How many gold coins do we have?” After pausing a moment, he replied, “Well! If I divide the coins into two unequal numbers, then 37 times the difference between the two numbers equals the difference between the squares of the two numbers.” The wife looked puzzled. Can you help the merchant’s wife by finding out how many gold Ò

37(x-y)=x^2-y^2. u no tht x^2-y^2=(x-y)(x+y).so (x-y) cancels on both sides to give x+y=37.so sum of unequal

halves=37 which is the req answer.

4.Football Matches

A set of football matches is to be organized in a “round-robin” fashion, i.e., every participating team plays a match against every other team once and only once. If 21 matches are totally played, how many teams participated?

Ò ans:7 teams okay.for a match u need 2 teams.suppose there r totally ‘n ‘teams. Now uve to choose 2 teams out of ‘n’ teams.so answer

=no of such choices=no. of possible combinations. So we’ve ans = nC2(ncombination2)=21;solve to get n=7.

Sol: n(n-1)/2=21. so n=7.if u don’t understand c the graph below each team plays no. of matches=no of teams ahead of

it. One bar ‘|’ represents one team.

6 5 4 3 2 1 0 ———–21 last team is written as 0 matches becos this team has already played with all other teams-hence sum of matches =6+5+4+3+2+1=21 which is correct only if no of

teams =7

5.Glenn and Jason each have a collection of cricket balls. Glenn said that if Jason would give him 2 of his balls they would have an equal number; but, if Glenn would give Jason 2 of his balls, Jason would have 2 times as many balls as Glenn. How many balls

does Jason have? Ò 14

1. G+2=j-2

2. 2(G-2)=J+2.

solve these 2 to get

J=14

6. Monkeys and bananas:

Suppose 8 monkeys take 8 minutes to eat 8 bananas.

(a) How many minutes would it take 3 monkeys to eat 3 bananas? Sol:each mky takes 8 min to eat a banana

(b) How many monkeys would it take to eat 48 bananas in 48 minutes

ans:8m=48 m=6

7.Vacation time

It was vacation time, and so I decided to visit my cousin’s home. What a grand time we had! In the mornings, we both would go for a jog. The evenings were spent on the tennis court. Tiring as these activities were, we could manage only one per day,

i.e., either we went for a jog or played tennis each day. There were days when we felt lazy and stayed home

all day long. Now, there were 12 mornings when we did nothing, 18 evenings when we stayed at home, and a total of 14

days when we jogged or played tennis. For how many days did I stay at my cousin’s place?

Use sets and venn diagram to solve such questions.a,b

,aub,anb etc.

12=tennis+leave

18=jog +leave

so jog-tennis=6

again jog+tennis=14.so solve and get

jog=10,leave=8,tennis=4.so tot=22

8.The Carpenter and the Nails

A 31″ x 31″ square metal plate needs to be fixed by a carpenter on to a wooden board. The carpenter uses nails all along the edges of the square such that there are 32 nails on each side of the square. Each nail is at the same distance from the neighboring nails. How many nails does the carpenter use?

Ans= 32*2 + 30*2=124

9.Old Man Wrinkle

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Man Wrinkle spent one-fourth of his life as a boy, one-eighth as a youth, and one-half as an active man.

If Man Wrinkle spent 8 years as an old man, then how many years did he spend as an active man?

(1/4)y+(1/8)y+(1/2)y =p; where y =tot life time.

y-p=8.since p=(14/16)y

we’ve y=64

active age=32

10 A farmer has C chickens.A sack of feed comes for 9 days.As the feed cost is increasing the farmer sells some chickens and retains 12 chicken.If he reduces the feed quantity by 10% .Then he observesthat the feed comes for 30 days.What is C?

sol:

C chickens can b fed for 9 days.so C chick can eat 1/9 in 1 day.

so 1 chick can eat

(1/9C)in one day. after his reduction one chick gets to eat only 90% of wht it used to eat in one day. so new feed value=0.9*(1/9C)for 1day.but now he has only 12 chicks. that means as per new rule feed is suitable for 12 chicks for 30 days.

i.e.

0.9*(1/9C)*12*30=1. solve to get C=36

Ans:36

23.

to be 10 min earlier means car has been driven for 10 min less thn usual. i.e 5 min in fwd jrney and 5min in bwd jrny.normally

car reaches station at 6pm. today since it has to drive only 5 min less it reaches at 5:55.since he strtd to walk at 5pm.hes been walking till 5:55 since thts wen he sees the car.so 55 min

27.

m:n t1:t2

ans:(mt2-mt1)/(t1-t2).don’t remember how this eq is obtained but its given in lower class mathematics texts.

38.

its not ur mistake tht u don’t get this answer.this question of shak devi is confusing becos of its grammar. actually its a simple one.in the last sentence of the qs wht she means is tht there r still 6miles to the crossing wen he met train and not tht train has gone 6miles beyond crossing.with this idea in mind lemme explainb the solution. hes late by 25 min.so if his normal time at crossing is 8am; today he reaches at 8:25only .also he meets train 6miles before crossing.tht means at tht time wen they meet he is 6 miles away frm crossing.his speed is 12m/hr. so time at which they must have met shud b 7:55.as per our idea train shud b there at the crossing by 8 am.nw its got only 5min to 8am to b at the crossing.since crossing and road r parallel train also got to cover 6 miles to b at crossing. so its got 5 min to cover 6 miles.so speed of

train=6*12=72m/hr

44.Say he took m kilos of rs32 powder and n kilos of 40 powder.since his profit is 25% we have 125% *cost priceof 1kilo =43.so cost price=rs34.4 for 1kilo of mixture but cp of 1 kilo mixture is (32m+40n)/(m+n)=34.4 solve to get m/n=7/3.so ratio is 7:3.so to get 100kg he shud mix 70kg of rs32 and 30 kg of rs40

77.The main line train starts at 5.00AM and the harbour line train starts at 5.02AM.Each train has the frequency of 10

minutes.If a guy goes in the morning at a random time what is the probability of he getting main line train?

sol:a very easy question but may confuse many suppose the person arrives at station just after 5am.he has 2 min left for the harbr line train. that means he gets hrbr line train if he arrives at any time bet 5 and 5:02am.but if he arrives after 5:02 am next train is at 5:10am(given freq is 10 min).tht means he can arrive at any time bet 5:02 and 5:10am to catch it.so he has 8

minutes(5:10-5:02)at his disposal.s prob of getting main line train=8/freq and tht of harbr train=2/freq.

freq=10.

so prob=0.8–main line

” ” ” =0.2–harb line

Ans: 0.8

98 Wen he covers one circumference of the post he covers a height of 4 feet.but since he covers the circumference spirally,he must have travelled more thn 4feet to reach tht height.apply pythagoras theorem to get root of(4square+3square)=5.i hope u understand this.the dist travelled by squirrel is the hypotenuse of right angled triangle with 3 as base and 4 as height.so he actually climbs 5 feet to reach a height of 4feet.so inorder to reach 16 feet he climbs 20

feet.

ans :20 feet

Qs.101 Start thinking in the right direction. The catch is in the no. 11. The sum has to be a whole no divisible by 11(clr frm

question).moreoverjust assume tht a persons age can’t be more than 150(worst case).so sum can’t b greater than 300 i.e. sum of ages is not more than a 3 digit no.now if a person is to marry his age won’t b a single digit no.that means we can eliminate ages upto

9.i.e we can eliminate the maximum sum obtained from one digit nos i.e 18. Hence we can eliminate the possibility of sum =11

which is less than 18.so focus on 2digit sums first.consider sums divisible by 11 frm 55 only.becos even if they were equal aged we can c they r definitely mature enuf to marry(55/2).our assumption is valid only if a sensible person is putting the

question becos he may be thinking of child marriage also.Since given husband is SENIOR they can’t b equal aged

also.wen we eliminate each possibility starting frm 55 we can fix upon a sum of 99.once we get this its quite

easy na. Note tht the difference bet their ages is a factor of

99.

So we have sum =99 and diff=9

So ages are 54 and 45.

108

let initital length of candles be z rate of burn of each is z/5 and z/4 consider tht each has been burning for ‘t’ hrs wen

light came. at tht time if length remaining of the smaller candle is y then tht of bigger one is 4y.

eq1: (z/5)*t=z-4y

eq2: (z/4)*t=z-y

divide eq1 and eq 2 to liminate ‘t’ and L.H.S ‘z’.now we get a rel bet z,y frm R.H.S now we know z=ky wer k is evaluated frm above.we alkso know z burns in 4 hrs for smaller candle. so z-y i.e z-(z/k) burns in how many hrs.we get 3.75 hrs.

114. one day food of cow and horse is 1/40: c+h=1/40—-1

” ” horse ” sheep is 1/60: h+s=1/60—-2

” ” cow ” sheep is 1/90: c+s=1/90—-3

add each eq to get 2(c+h+s)=(1/40+1/60+1/90). combined one day’s work is c+h+s.time taken to finish=1/(c+h+s).

10

extra distance covered by thief=27* dist covered by him in each stride. given tht dist cov by policeman in 2steps and thief in

5 steps r same.let it b “y”. thief covers y/5 and police y/2 dist in each step respectively. also given: time of 5 steps of police=time of 8 steps of thief.let time taken be 1 unit for this—–(assumption1)

so dist cov in 5steps of police =5y/2 in unit time—-eq(1)

dist ” 8 ” ” thief =8y/5 ” ” ” —-eq(2)

let the common time of run of both(i.e.excluding the time for 27 steps of thief) be “t” units, now we know ;to catchup with the thief police has to cover the same distance totally coverd by thief which is equal to 27steps* dist per step of thief + dist run by him for the time t(which is no of steps it time “t” * y/5) i.e. 27 *(y/5) + (8y/5)*t

——eq(3)

{frm eq(2)}

total dist covered by police=t*(5y/2)—-eq(4)

we know eq(3)=eq(4)

so 27y/5 + 8yt/5 = 5yt/2; y gets cancelled.u get

t=6units. in unit time we know police takes 5 steps and thief tkes 8steps frm (assumption1). so in 6 units of time police takes 5*6=30 steps. “” ” ” thief ” 8*6=48 steps.thief has taken totally 48+27=75steps. so 30 steps is taken by police to catch thief.

Qs.30

let radius of hind wheel(bigger) =R ” ” fore wheel(smaller) =r

first case: let No of rot of hind wheel = N so ” ” fore ” = N+4 therefore 2*pi*R*N = 2*pi*r*(N+4)=96———–eq(1) second case: since radius changes no of rot to cover same dist changes.

let No of rot of hind wheel = n

so ” ” fore ” = n+2

therefore 2*pi*(4/3)*R*n =

2*pi*(3/2)*r*(n+2)=96—–eq(2)

L.H.S of eq(1)/ L.H.S of eq(2) =96/96=1

i.e. R*N=(4/3)*R*n.

so 3N=4n——eq(3)

do the same for R.H.S of eq(1),eq(2) to get

2(N+4)=3(n+2).put eq (3) to get n=6

put in eq(2) to get circumference(i.e 2pi*r)=8,12 for fore and hind wheel.

Qs.37

4:05

B—————–M——-T

<——-y——–><–p—>

B=Bangalore, T= Tumkur,M=meeting point

y= distance BM

p= distance MT

Vbt=velocity of guy frm B to T

Vtb= ” ” T to B

let time at which they reach their destinations be “t”.total dist cov by both is same since route is

same. so

1. Vbt*t=Vtb(t-2hrs) becos TB guy strtd 2hrs after BT guy time bet 2 pm and 4:05pm =2 and 5/60 hrs

” ” 12 o’clk ” ” =4 and 5/60 hrs. consider dist covered by each in reaching M.

2. Vtb * (2 and 5/60)=p and Vbt * (4 and 5/60) =y

also consider the rem dist to b covered by each after meeting.

3. Vtb * (t-2and 5/60))=y and Vbt * (t-4and 5/60)=p so we have

1. Vbt*t=Vtb(t-2hrs)

2. Vtb * (2 and 5/60)=p and Vbt * (4 and 5/60) =y

3. Vtb * (t-2and 5/60))=y and Vbt * (t-4and 5/60)=p

L.H.S of eq(2)=R.H.s of eq(3)

R.H.S of es(2)=L.H.S of eq(3).

3 equations 3 unknowns.solve to get t

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Qs.65 easy one but remmeber this may b difficult for sum people.so u have an advantage

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each shud get the cost of 25 sheeps to make equal money.since i have 30 sheeps i must give radha the price of 5 to make it 25 each. so cost of 5 sheep =150 i.e cost per sheep=30.

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Qs.104

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1. o+y+c=100

2. 3o+2y+0.5c=100