ans for updated patern from 80 to 200

81. a=9,b=8,c=6,d=5,e=4 total sum of fg=32
take f=3 and g=2
fg=10*3+2;
so it will be 2 for g value........
82. as there are 4 fulls weeks i.e 28 days..
so..every day occurs min 4 times.
den remaining 3 days (as jan has 31 days) will be mondaytuesdaywednesday.
so on 31st jan comes wednesday.
so 1st janll be MoNDAY
84. 57
why because ,2(g+m+h)=19/120;
g+m+h=19/240*720 boxes=57.
86. 7.
let born year of r and be x and y.
so, given
x+y=3844;
1994-x=2(1994-y);
from this two: y=1946 and x=1898
so in 1999, R will be 101 years old.
86. If x is year of birth of R and y is year of birth of G, then
x+y= 3844
and
(1994-x)= 2*(1994-y)
solving it, we get
x= 1898 and y = 1946
87. only one
i.e, when f(x)=x^2
88.  since the track is a circular track ,A and B will meet every 50 seconds.i.e.,100/(10-8).
since it is a multiple of 50 they will be meeting at the starting point every 50 seconds.if you multiply 15*50 you will get 750 and after the second 50 it will be 1500.
all of them wil meet at the starting point after 100s
89. 18 (a+b) 1 day work=1/36-1/108=1/54
so time taken by (a+b) is 54 days and c is 108..
108 is the right answer.

q.19 6*5*4*3*2*3=216 numbers
90. wen 2nd lst dig is 2...1st posfilld in 6 ways...2nd pos filled in 5 ways..3 rdposfilld in 4 ways..4 rdposfilld in 3 ways..nd last i.e. 6th posfilld in 2 ways...=6*5*4*3*2=720

91. let G completes the work in x days...
so his son completes it in 5x days...
but according to the give data
5x=x+40
x=10
so G takes 10 days and his son takes 5*10=50 days to finish the work
G's 1 day work=1/10
his son's 1 day work=1/50
together work per day=(1/10)+(1/50)
=6/50=3/25
so they will take 25/3 (or) 8 1/3 days to complete the work together.
92. 7
94. q 23
66.6%
97. x takes 4 days to complete 1/3 rd of the job so
x's work per day is 1/12 and for 3 days =1/4
similarly
y's work per day is 1/18 and for 3 days =1/6
z's work per day is 1/10 and for 3 days =3/10

if all work together their 3 days work=1/4 + 1/6 + 3/10
=(15+10+18)/60 =43/60 work is done
remaining work=1-(43/60)=17/60
for 17/60 th work y takes (17/60)*18= 5.1 days...
[ since y's work per day is 1/18 so y completes the total work in 18 days]

so the answer is 5.1 days...
99. avinit=a
bikash=b
chandan=c
debjit=d; a,teach not spoke with c.
painter with d,a.
b and actor frnd.
so,
avinit-Actor
bikash-painter
100.1 day work for (A+B+C) = 1/36 ............(ii)
        Now, (A+B) = 2C ..........................(i)
       From (i)and(ii): 3C = 1/36
       C = 1/108Thus, C takes 108 days to complete the whole work alone..
101   lcm 0f 1,2,3=6
      10006=166
      so ans in 166
102.  let speed = x kmph
      relative speed=x-2kmph
     distance=0.6km
     time=6min/60min=0.1hr
     d=sXt
     0.6=(x-2)0.1
     x=8kmph
103. remaining work = 1-3/4 = 1/4

  A can do full work in 20 days.
  he wil do 1/4 th of work in 20*1/4= 5 days.
104.Let it be x minutes to 6
  time is (6 hr – x min)
  time before 50 minutes = (6hr – x min – 50 min) = 3 hr + 4x min
  3 hr = (4x + x – 50) min
 180 min = (5x – 50) min
  130 = 5x
  x = 26 min
  time is 6 hr – 26 min = 5 hr 34 min

105. A _ _ _ _ _ =120
so
A E _ _ _ _ =24
A M _ _ _ _ =24
48
A R E M S T
106. A _ _ _ _ _ =120
so
A E _ _ _ _ =24
A M _ _ _ _ =24
48
A R E M S T
107. ans-d)1
square table from 1 to 10
1^2=1
2^2=4
3^2=9
4^2=16 ie last digit 6
5^2=25 ie 5
6^2=36 ie 6
7^2=49 ie 9
8^2=64 ie 4
9^2=81 ie 1
10^2=100 ie 0
so last last digit could be 0,1,4,5,6,9
108. No answer
109. A _ _ _ _ _ =120
so
A E _ _ _ _ =24
A M _ _ _ _ =24
48
A R E M S T
110. No answer
1+3+9.....13
111. No answer
raju's age=x
father's age=y
mother's age =z
y+6=2(x-2)
z=2x
y+z=4x-10
112. S-F-A-Y-T is 19-6-1-25-20
according to alphabetical order A-1,B-2...S-19,F-6...
113. 5C0*11C11+5C1*11C10+5C2*11C9+5C3*11C8=2256
114. Grp Mem
A--> j,k,l,n
B--> j,k,l,m
C--> j,k,m,n
ans is 4
115. when A starts at 12 pm its covers 63*1.5 = 94.50m distance till 1:30 pm ....
relative speed of B = 84-63 = 21m /hr ......
now total distance covered by B is 94.5+34 =128.50 m
time taken by B to reach 34m ahead of A = 128.50/21 = 6.12 hr
so time is =1.30 + 6.12 = 7.42 pm
116.Ans:30
1. N=K+(X+Y-Z) ans = K+11
2. N=K+(X-Y-Z) ans = K+2
3. N=K+X*(Y-Z) ans = K+18
117. if all three work simultaneously, (10+40/3)-5=(70/3)-5=55/3
its given in kilos/minute... in question it will be asked in kilos/hour...
so for 1 hour ther is 60 minutes... so perform (55/3)*60=1100 tonnes filled in one hour
now 1100 filled in 1 hour so 2400 filled in X hour so we have 1100x=2400
so x=2400/1100=24/11.
118. A contains (2/3)*x lt.wine(7/6)*x
B contains (1/4)*2x lt. wine and (3/4)*2x lt. water
if we mix both we will get 3x lts of solution and in that
wine=[(2/3)*x]+[(1/2)*x]=(7/6)*x lt
so proportion of wine=(7/6)*x/3x=7/18....
119. et G completes the work in x days...
so his son completes it in 5x days...
but according to the give data
5x=x+40
x=10
so G takes 10 days and his son takes 5*10=50 days to finish the work
G's 1 day work=1/10
his son's 1 day work=1/50
together work per day=(1/10)+(1/50)
=6/50=3/25
so they will take 25/3 (or) 8 1/3 days to complete the work together..
120. total work done by A+B+C in 1 hr = 37/60
remaining work = 23/60
total work done by B+C in 1 hr = 11/60
total work done by B+C in y hr = 23/60

y=23/11
121. for these problems follow the formulae ((M1)*(D1)*(H1))/(W1)= ((M2)*(D2)*(H2))/(W2)
WHERE M- men or women
H- hours
D-days
W- work
here work is same .therefore W1=W2
M1*D1*H1=M2*D2*H2
30*7*18=21*8*x
implies X=22.5 days

122. In time and work problems the important point to be remebered is (number of MENDAYS is always constant.)i.e product of men into days always remains constant.
if one variable among men or days changes the other varible changes in accordance to make the value of remains constant.
let us assume X be the number of initial persons. somendays = 10* X( as told above), now 5 are absent.i.e (X-5) are present.as i told men days should be always constant
10* X= (X-5)*12
10X= 12X-60
2X=60 implies X=30.

123. 60 men --> 40 days
so they do work in a day is --> 1/40 parts
1 men(1 day) --> 1/(60*40)
so totally they have to do 1200 parts of work
For I five days 5(1/40)+
Next five days 5(55/2400)..... +5(40/2400)+....5(5/2400)=7/8
similarly they do for 60 days for the next 5 days no one will be there to complete the work.
So the job is incomplete and its answer cant be determined.
Incomplete task.

124. Answer: b) 4 : 1
D / v1 = (3/8) x / v2 ... ( 1 )
( D + x ) / v1 = (5/8) x / v2
=> D / v1 = (5/8) x / v2 - x / v1 ... ( 2 )
From equns. ( 1 ) and ( 2 ),
(3/8) x /v2 = (5/8) x / v2 - x / v1
=> 3 / (8v2) = 5 (8v2) - 1/ v1
=> 1 / v1 = ( 5/8 - 3/8) / v2
=> 4v2 = v1
hence, answer will be 4:1

125. series is like 11,23,47.....
then
11*2+1=23
23*2+1=47
47*2+1=95
95*2+1=191
191*2+1=383
hence,last three digit=95,191,383

126. Ans. D) 25
Solution:
(1+2+3)^2 = 36
(2+2+1)^2 = 25

127. heights: 70, 72, 74, 80.
condition: say 4 men, A B C D
if A's height = x
B= A + 2 = x+2
C= B + 2 = x+4
D= C + 6 = x+10
Avg = 74
=> ( x + (x+2) + (x+4) + (x+10) ) / 4 = 74
solving it..
x = 70.
hence, heights:
A, B, C, D = 70, 72, 74, 80.
solve, x=70

129. 57^3661=(57^4)^915*(57)=(49*49)^915*(57)=(2401)^915*9(57)
take tens digit which is 01 and then according to the trick from the link
=10*unit digit of power=01^anuthing is same and 01*57=57
second number 81^3643=8*3=24 so it is 41 and so adding 41+57=98

130. possible case can be
if he bought 8 oranges @ Rs 6 for each orange with an amount of Rs 48 , then
he can buy 12 oranges now @ Rs 4 per orange with same amount.

132. lets say a term 6
Now 6^3=(2^3 )* (3^3)
so, no of factors=(3+1)*(3+1)=16(using prime factorization)
Now similarly
6^4=(2^4)*(3^4)
so factors, (4+1)*(4+1)= 25
So the ans is c)25

133. 137+276=435(given)
=> 1+2=4;
3+7=3;
7+6=5;
so
731+672=534.

134. 4 cases:
{odd,odd,even}
{odd,even,even}
{even,even,even}
{even,odd,even}
3/5*3/5*3/5 + 3/5*2/5*2/5 + 2/5*2/5*2/5 + 2/5*3/5*2/5
=(27+12+8+12)/125
=59/125

136. 17*10=170
170*4=680
21*18=378
378*4=1512
7*1=7
7*4=28
So, 12*25=300
300*4=1200 and is c

137. A will take 3 hrs to complete job.B .. 6 hrs
C-- 2 hrs
If A takes x hrs, B take y7 hrs and C takes x-1 hrs to complete job independently, then
1/x +1/(x-1) +1/y =1 because they together complete 100% job in 1 hour.
1/x +1/y + 3/y = 1
solving these 2 eqns, we get
x=3 and y=6
So C complete total work in 2 hrs
C will complete 50% work in 1 hr.

138. All possibilities:
steps:
_ _ _ _ _ _ _ _ _ _
No 2 steps All 1's:

1 1 1 1 1 1 1 1 1 1 => 1 way//
1 two step:
that is choosing 1 two-step in 9 moves:
9C1 : 9 ways//
2 two-steps:
choosing 2 two-steps in 8 moves:
8C2 = 28 ways//
3 two-steps
7C3 = 35 ways//
4 two-steps//
6C4 = 15 ways//
5 two-steps//
which covers all the 10 stairs.. that means only one way
2 2 2 2 2 = 1 way//
Adding all the ways:
1 + 9 + 28 + 35 + 15 + 1 = 89 ways//

140. here $ represent the - operator so X-X=0 first condition
x$(y$z)=x-(y-z)=x-y+z
it can be written as x$y+z
2012-0+2012-1912=2112 so ans is 2112
141. The answer would be Rs 50 since a single press of the Random button would determine the correct labeling.
Here's the reasoning. Press the Random button. Since all buttons are mislabeled, whatever the Random button produces must be that drinks button (e.g. if tea, label it Tea, otherwise label it Coffee). Now, since the buttons were mislabeled, you need only switch the remaining two buttons labels
142. the ans is 7
the number divisible by 12 should be divisible by 4&3
7 can be written as 5+2 a=5 b=2... so we get is "5231352" it is divisible by 12...
143. the sum of any 2 sides must be greater than 3rd side
so
3 11 13
7 11 13 are only cases
so P= 2/10= 1/5
145. friday..If the Nth year is a non-leap year, then 320th day of year N to 206th day of year N+1 is [(365
– 320) + 206] = 251 days i.e. 35 weeks + 6 days. But this will not make the two days to be
Thursday.
Thus, the Nth year has to be a leap year. In this case 320th day of year N to 206th day of year
N+1 is [(366 – 320) + 206) = 252 days i.e. 36 weeks. Hence both days will be same day of the
week i.e. Thursday as given by the data.

168th day of N-1 year to 320th day of N year is [(365 – 168) + 320) = 517 days i.e. 73 weeks +
6 days. Thus, if the 320th day is Thursday, then the 168th day of year will be Friday, option
146. IT's 19.2
i explain... we have to apply formula 2xy/x+y three timesss
for first 10 &20 avreage speed will be 400/30=40/3 for secound 30& 40
avrage speed=2400/70=240/7... and finally apply avg formula between this two speed
avg speed=2*40/3*240/7 /40/3+240/7 = = =19.2 answer
147. 9+8+7+5+3=32...
so answer is 2
148. Simple case of averaging:
250M + 150W = 400
Thus, 400*12 = 4800
Now, 250*15 = 3750
Now, 4800-3750 = 1050
Thus, for 150W avg prod is : 1050/150 = 7
149. Suppose A takes A hrs,B takes B hrs.C wud take (A-1) hrs
Work done by A in 1/2 hrs=1/2A,B=1/2B,C=1/2(A-1)
Hence (1/2A)+(1/2B)+{1/2(A-1)}=1/2.......(i)
Also,wen A n B work,A works for 1 hr and B for 4 hrs..
Hence...(1/A)+(4/B)=1....(ii)
Solve (2) and get 1/B in terms of A..Put values of 1/B and 1/C in (1) and solve,u vl get a quadratic equation..Solve it and A=3 hrs
C=A-1 hrs= 2 hrs...Hence in 1 hr,C does 1/2 of work or 50%
150. the least possible no of member of members on any one committee = 4
In all 3 committees, say X,Y,Z, 2 persons say A and B are common.
C is common between X and Y.
D is common between Y and Z.
E is common between X and Z.
so X committee has ABCE. Y committee has ABCD. Z committee has ABDE.
151. Work done by Raju in 1 day=>1/10
Work done by Vicky in 1 day=>1/12
Work done by Tinky in 1 day=>1/15
Initialy 3 of them work 2gether for 2 days...
One of them leaves after that working for X days...
at last only tinky remains n works for 3 days...
So
2*(1/10+1/12+1/15) + X*(1/12+1/15) + 3*(1/15)=1
Solving :x=2
Hence 2+2+3=7.
Total work was done in 7 days...:)
152. No. of days is inversely proportional to Speed.
Thus, (son days)/(dad days) = (dad speed)/(son speed) .....(i)
Now given, (Dad speed)/(son speed) = 5/1 ..................(ii)
Now let, son take 'x' days ................................(iii)
Thus, dad takes (x-40) days ...............................(iv)

From (i),(ii),(iii),(iv): x=50
Thus, son takes 50 days and dad takes 10 days.
Hence, son's 1 day work = 1/50
And, dad's 1 day work = 1/10
1/10 + 1/50 = 3/25

Thus, Both takes 25/3 i.e. 8.3 days.....
153. given 6th digit even number , so last digit 2 or 4 or 6-> 3 ways
" 5th digit should be even...so there will be 2 ways(rep. not allowed)
so,therefore we get 5*4*3*2*2*3=720 ways
155. (a+b) 1 day work=1/36-1/108=1/54
so time taken by (a+b) is 54 days and c is 108..
108 is the right answer.

q.19 6*5*4*3*2*3=216 numbers
156. f(0)=2n+3
f(2012)=(2*2012)+3
=4027
161. Initial vol. of Cylinder = pi*(R^2)*H1 = pi*25*20 = 500pi
Now, vol. of Sphere = (4/3)*pi*(R^3) = (500/3)pi ....(Diam have to be assumed as equal)
New vol. of cylinder = Initial cyc. vol + Sphere vol.
pi*25*H2 = 500pi +(1500/3)pi ..(since radius of cycwil remain same)
Thus, H2 = 80/3
Now, dH = H2-H1 = 20/3 = 6.67
163. all sisters r sit together count 3 sisters =1
then total 5!*3!
164. the 30 th term is 780
1,5,6,25,26,30,31,125,126,130,131,150,151,155,156,625,626,630,631,650,651,655,656,750,751,755,756,775,776,780.
the terms are
1*5=5,5+1=6
each is multiplied by 5 and next number is result of that number which is multiplied by 5.
ex: 5*5=25
25+1=26
6*5=30
30+1=31 and so on
168. rule : [any odd no's not ending with 5]^20m thn its last digit will be 01
1941^3480 *1941^3+ 1961^4180*1961= 01*1941^3+01*1961
1941^3=21 as last two digit and when add with 1961 we get 82 as last two digit so 82 is the last two digit..................
169. A+B+C= 155 kg
Average of 7 rounds = (A+B+C+A+B+A+C+B+C+A+B+C)/7 = 4*(A+B+C)/7 = 4*155/7 = 620/7 kg
170. 1 can be arrange in 123, 132
2 can be arrange in 321
3 can be arrange in 213
number of possobolity is 4.
171. Given:
A(1)=1
A(2)=1

A(n)=A(n-1)-A(n-2) & n>=3

THUS,
A(3)= A(2) -A(1)= 1-1= 0
A(4)= A(3) -A(2)= 0-1=-1
A(5)= A(4) -A(3)= -1-0=-1
A(6)= A(5) -A(4)= -1+1= 0
A(7)= A(6) -A(5)= 0+1= 1
A(8)= A(7) -A(6)= 1-0= 1
A(9)= A(8) -A(7)= 1-1= 0
A(10)=A(9) -A(8)= 0-1=-1
A(11)=A(10)-A(9)= -1-0=-1
A(12)=A(11)-A(10)=-1+1= 0
A(13)=A(12)-A(11)= 0+1= 1
.
.
.
this continues till A(1000).
a pattern is formed at an interval of 6 i.e.{1,1,0,-1,-1,0}
So, 1+1+0+(-1)+(-1)+0=0
since there r 1000 terms,1000/6= quotient=166,remainder=4.
thus till the 996th(=166*6) term the sum will be 0 and for the rest of the 4 terms the sum will be {1+1+0-1}=1.
So S(1000)=1
The given options are incorrect.
172. may be answer is 3200 bcz in each 1 hr it doubles the numberhence
3-1000
4-2000
5-4000
6-8000
7-16000
8-32000
173. Ans.B 11110
1234+6543=11110
At unit place 4+3 sum is 0 and carry is 1
at tens place 1+4+3 sum is 1 and carry is 1
at hundreds place 1+5+2 sum is 1 and carry is 1
at thousand place 1+1+6 sum is 1 and carry is 1
hence result will be 11110
176. circular cylinder volume is given...61.6cc=pi*(inner radius)2*h
note all units in above equation are in centimeters
now solving the equation you get inner radius=0.7 cm=7mm
thus thickness= (outer dia/2)-inner radius= 8-7=1 mm
179. 35 percent invest in municipal bonds, 18 percent invest in
oil stocks, and 7 percent invest in both municipal bonds and oil stocks.
so 35-7=28% invest in municipal bonds only.
so the probability t hat the person selected will
be o ne who invests in municipal bonds but not in oil stocks = 0.28
180. 6.66%
Vol of sod chloride = 5*10000/100 = 500 gallons by vol
vol of solution after evaporation of 2500 gallon of water = 7500 gallons.
% of sod chloride in final sol = 100*500/7500= 20/3= 6.66%
181. relative velocity of trains (100+50)kmph
time require to travel 150km=150/150 = 1 hr.
distance travel by the bird in this 1 hr. = 100*1= 100 km
182. distance travel by nav. exp. in 1 hr.= 50*1 = 50 km
distance travel by hw. ahem. exp. in 1/2 hr = 40*1/2= 20 km
distance will remain before the collision = 30 km
relative speeds of trains (50+40)kmph
so time require before collision= 30/90= 1/3 hr. = 20 min.
186. x + 3y = 100
3y = -2x + 100
y = (-2x + 100)/3
since both x and y are positive integers, (-2x + 100)/3 must be an integer too
so -2x + 100 must be a multiply of 3 and x cannot exceed 50

-2(2) + 100 = 96 is a multiply of 3
-2(5) + 100 = 90 is a multiply of 3
-2(8) + 100 = 84 is a multiply of 3
.
.
.
-2(47) + 100 = 6
so x = 2, 5, 8, ..... 47
let a be the first term of the arithmetic sequence and d be the common difference
a + (n - 1)d = 47
2 + (n - 1)3 = 47
3(n - 1) = 45
n - 1 = 15
n = 16
so there are 16 terms
C
187. x+y=17/v;
x-y=1/v;
x=9/v;
y=8/v;
vxy= 72/v;;;;; option isn't correct;;
188. option b... ..because of total 1%gain in 7 day.. so in 1 hour gain 3/5;
and i7 day gain hour is 1+204/300.. and 204/300 hour= (40+(4/5)) minute ... and
4/5 minute is 48 sec...
soans b>..
189. 120 wire cable has laid firmly in the 10 km..soans is 10....
190. (3c1*6c2)+(3c2*6c1)+1=64
191. the volume of sand dig out(volume of well)=volume of embankment formed which is of a hallow cylindrical shape
usually both are cylindrical in shape so volume= pi* radius^2* height(or depth)
since width(radius) of embankment which is out side the well=21
so total radius of cone formed by embankment=21+7(that is inside well)=28
=>pi*7*7*10=[(pi*28*28*h)-(pi*7*7*h)]
=>490=h*(28^2-7^2)
=>h=490/[(7*4)^2-7^2]
=>h=490/[7^2*(4^-1)]
=>h=490/[49*15]
=>h=10/15=2/3 mts...
so option (b)...
192. condition (1)
(PR/Sa)+(3SR/2Sa)=2PR/Sb-------(1)
condition (2)
(PR+RS/2)/Sa+3/2Sa=QT/Sb
(8PR+4RS+6RS+3PR)/8Sa=QT/Sb=2PR/Sb-----(2)
dividing (1) by (2)
(11PR+10RS)/(4PR+6RS)=QT/PR
ON SIMPLIFICATION,(put,PQ=PR+RS+ST+TQ)

(11PR+10RS)/(4PR+6RS)=7/4
SOLVING FURTHER,
RS/PR=8:1
so,option (C)
193. we don't need individual values of x,y,z
so let us consider x be smaller and z is larger
so y is median=5
mean=(x+y+z)/3=(x+z+5)/3
mean=x+10 and z-15
=>(x+z+5)/3=x+10 | (x+z+5)/3=z-45
=>x+z+5=3x+30 | x+z+5=3z-45
=>2x-z+25=0 ---(1) | 2z-x-50=0 ---(2)

from the addition of two equations (1)+(2)
=>(2x-z+25)+(2z-x-50)=0
=>x+z-25=0
=>x+z=25
so x+y+z=(x+z)+y=25+5=30....
195. at each jump it moves 4 slot.
hence total jump * 4 will give at which slot he is
2200 * 4 = 8800
but its given that..there are 60 slots around a circle
so take mod 60
8800 mod 60 = 40
but mod gives and between 0 to 59...but actual slot number are from 1 to 60 hence +1 ans
hence final answer is 40 +1 = 41
196. go by options...at 7.30 a will go 6*7.5=45km
at 7.30 b will go 8*6=48km
so b ewill be 3 km ahead of a
198. only one
i.e, when f(x)=x^2
199. 337
5678%460=158
(460-158)+35=337
200. answer a) 48
if 1 is a root then x-1 is a factor of p(x)
similarly x-2 is a factor,x-3 ,x-4 are factors
but p(x) is 4th degree polynomial therefore it can be in the form
p(x) = k(x-1)(x-2)(x-3)(x-4) , where k is a constant
but given p(0) =48
therefore 24 k =48
k=2
p(x) =2(x-1)(x-2)(x-3)(x-4)
p(5)= 2*4*3*2*1 = 48