This is a collection of fifty questions from important topics in Aptitude where students should pay more attention and practice. Questions taken from various net sources. Some of the answers were edited. This presentation could be run only in office 2010 or latest.
2. If a person having 1000 rs and he want to distribute this to
hisfive children in the manner that each son having 20 rs more
thanthe younger one , what will be the share of youngest
childx+(x+20)+(x+40)+(x+60)+(x+80)=1000x=160. 3. 35674th term in
12345678910111213....1 to 9=9 no.s 1 digit each no.9 to 99=90no.s 2
digit each,total digit=90*2=180 terms99 to 999=900nos 3 digit
each,total digit=900*3=2700999to 9999=9000nos 4 digit each,total
digit=9000*4=36,000now till 999,we have=9+180+2700=2889 digitstill
9999 we have=2889+36000=38889 digits but it is more than 35675Hence
35674-2889=32785after 999 each no. has 4 digitso 32785/4=8196 with
remainder 1.8196th number after 999 =999+8196=9195next
term=9196...1st digit=9 ..so ans=9 4. 2/3rd of a two digit number
is equal to a number whose ten's placeis three less than the ten's
place of the 2 digit number and unit'splace is one more than the
ten's place of the 2 digit number.let a be tens digit and b be unit
digit.23(10a+b)= 10(a-3)+(b+1)solving we get 10a+b=87;hence the no
is 87 5. no. of prime factors of 307 x 225 x 3411 ?307 * 225 *
3411(2*3*5)7 * (2*11)5 * (2*17)11=> 223 * 37 * 57 * 115 *
1711Total no. of prime factors are=>(23+7+7+5+11) = 53 6. x and
y no. when divided by 6 leaves remainder 4 and 5respectively. so
x+y when divided by 6 leaves remainder
x=6m+4y=6n+5x+y=(6m+4)+(6n+5)6(m+n)+9 = 6(m+n+1)+ 3so,when (x+y)is
divided by 6 gives remainder = 3d)none of the above 7. If i am
twice as old as he was when i was as old as him. sumof our ages is
42. find my present age?let my present age=x;and his present
age=y;x+y=42.........................(1)2{y-(x-y)}=x=>2(2y-x)=x=>4y-3x=0.....(2)3x+3y=126-3x+4y=07y=126=>y=18;then
x=42-18=>24*3 8. A person says that his son is 5 times as old as
his daughter and hiswife is 5 times older than his son and he is
twice the age of his wife .The sum total of all the ages equals the
age of the grand mother whocelebrated her 81st birthday today. How
old was his son?Let the daughter age be 'x'Son age = 5x, Mother age
= 25x, Person age = 50x.Thus we get, x+5x+25x+50x = 8181x = 81x =
1Daughter's age = 1, Son's = 5, Mother's = 25, Person's = 50. 9.
The ratio of the ages of the father and the son is 5:3, After 10
yearsit will be in the ratio 3:2. What will be their ages?let the
present age of father and son be 5x and 3x respectivelythen after
10 years the ratio becomes5+103=3+102so we get x=10thus present age
of father and son are 50 and 30 respectively 10. Three men - Sam,
Cam and Laurie - are married to Carrie, Billy andTina, but not
necessarily in the same order. Sams wife and BillysHusband play
Carrie and Tinas husband at bridge.No wife partners her husband and
Cam does not play bridge.Who is married to Cam?given that Cam does
not play bridgeso Cam is not Tina's Husbandalso he is not Billy's
Husband.therefore only one option is left that is Carrie. 11.
Pointing towards a photo on wall Raju said "His fathers
onlygranddaughter is my wife" how Raju is related to that man in
photoHis father > father of photo person.His fathers only
granddaughter > photo persons daughterIs my wife > raju is
photo persons soninlaw 12. A cycled from P to Q at 10 kmph and
returned at the rate of 9 kmph. Bcycled both ways at 12 kmph. In
the whole journey B took 10 minutesless than A. Find the distance
between P and Q.A -time =10+9=1990hoursB -time =212=6hours10 min
=16hoursThus19906=1619 1590=164d/90 = 1/6thus d= 15/4 km = 3.75km
13. At 10 a.m. two trains started traveling toward each other from
stations287 miles apart.They passed each other at 1:30 p.m. the
same day.If the average speed of the faster train exceeded the
average speed ofthe slower train by 6 miles per hour, which of the
following representsthe speed of the faster train, in miles per
hour?If s is speed of faster train,3.5*s +3.5*(s-6) =287solving it,
we get s=44 mph 14. A man traveled from the village to the
post-office at the rate of 25kmph and walked back at the rate of 4
kmph. If the whole journeytook 5 hours 48 minutes, find the
distance of the post-office from thevillage.Average speed
=2+=225425+4=20029km/hr.Distance covered in 5 hours 48 minutes=
Speed x time =20029X295Distance covered in 5 hours 48 minutes =
40kmsDistance of the post office from the village = (40/2)=20 km.
15. Riya and Priya set on a journey. Riya moves eastward at a speed
of20kmph and Priya moves westward at a speed of 30 kmph. How
farwill be priya from Riya after 30 minutestotal eastward distance=
20kmph*1/2hr=10 kmtotal westward distance= 30kmph*1/2hr=15 kmtotal
distance between them= 10+15=25km 16. I started on my bicycle at 7
a.m. to reach a certain place.After going a certain distance, my
bicycle went out of order.Consequently, I rested for 35 minutes and
came back to my house walking all theway.I reached my house at 1
p.m.If my cycling speed is 10 kmph and my walking speed 1 kmph,then
on my bicycle I covered a distance of:i) 4.92 km ii) 13.44 km
iii)14.375 km iv) 15.476 kmtotal time=6h-35m=5h25m=5+2560=6512hlet
x be the distance from house to placeso10+1=6512Solving x=4.92 17.
In a row of girls, Nivedita is 15th from the left and Vimla is 23rd
from theright.If they interchange their positions, then Nivedita
becomes 18th from theleft.Then at what position will Vimla be from
the right?Total number = 23 + 18 1 = 40.Vimla becomes 15th from the
leftimplies (40 15 +1 ) 26th from the right. 18. Eight friends
Harsha, Fakis, Balaji, Eswar, Dhinesh, Chandra, Geetha,and Ahmed
are sitting in a circle facing the center.Balaji is sitting between
Geetha and Dhinesh.Harsha is third to the left of Balaji and second
to the right of Ahmed.Chandra is sitting between Ahmed and Geetha
andBalaji and Eshwar are not sitting opposite to each other.Who is
third to the left of Dhinesh?consider them as a circle i.e 1-8 in
circle.1 BALAJI 2 GEETA 3 CHANDRA 4 AHMED5 FAKIS 6 HARSHA 7 ESWAR 8
DHINESHBD GE CHFA 19. In a stream running at 2kmph,a motor boat
goes 6km upstream and backagain to the starting point in 33
minutes.find the speed of the motorboatin still water.let speed of
boat in still water is u. distance=6km(given)in upstream
speed=(u-2) where stream speed=2in down stream
speed=(u+2)so,2++2=33606 2+6 + 2=3360solving u=22so speed of boat
in still water=22kmph 20. A boat covers a certain distance
downstream in 1 hour, while itcomes back in 1 hours. If the speed
of the stream be 3 kmph. whatis the speed of the boat in still
Water?A. 12kmph B. 13kmphC. 15kmph D. 16kmphSHORTCUTA man can row a
certain distance downstream in x hoursand returns the same distance
in y hours.If the stream flows at the rate of z km/h,then the speed
of the man in still water is given by + Here x = 1, y = 1.5, z =
3Speed of boat in still water =32.50.5= 15kmph 21. a boat travels
upstream 20km in 6 hrsand 18 km downstream in 4hrsfind the speed of
boat in still watera.1/2kmph b. 7/12kmphc.5kmph d.none of theseB-S
=206=103B+S =184=922B =103+92=476kmB=4712kmph ... none of these 22.
A boat takes 90 minutes less to travel 36 miles downstream than
totravel the same distance upstream.If the speed of the boat in
still water is 10 mph, the speed of thestream isA. 2 mph B. 2.5
mphC. 3 mph D. 4 mphLet the speed of the stream x mph.Speed
downstream = (10 + x) mph.Speed upstream = (10 - x) mph.3610 3610 +
=906072x x 60 = 90(100-x2)x2+48x-100 = 0(x+ 50)(x - 2) = 0x = 2
mph. 23. How many factors of 245374 are odd numbers?A. 20 B.24 C.30
D.36Any factor of this number should be of the form 2a5b7c.For the
factor to be an odd number, a should be 0.b can take values 0,1,
2,3, and c can take values 0, 1, 2,3, 4.Total number of odd factors
=45= 20 24. Company BELIANCE hosted a party for 8 members of
Company AXIAL. In the party nomember of AXIAL had interacted with
more than three members of BELIANCE. Out of all themembers of
BELIANCE, three members each interacted with four members of AXIAL
andthe remaining members each interacted with two members of AXIAL.
The greatestpossible number of members of company BELIANCE in the
party is:A. 9 B. 10C. 11 D. 12no member of AXIAL had interacted
with more than three members ofBELIANCE.Given that there are 8
members of company AXIAL and three members ofcompany BELIANCE
interacted with four members of AXIAL.Therefore, the maximum
possible number of members of companyBELIANCE in the party will be
3+4C2(Since, each of the remaining members of the company BELIANCE
haveinteracted with two members of AXIAL) = 9 25. A seven-digit
number comprises of only 2's and 3's. How many of these are
multiples of12? A. 1 B. 11 C. 21 D.47Number should be a multiple of
3 and 4.So, the sum of the digits should be a multiple of 3.We can
either have all seven digits as 3, or have three 2's and four 3's,
or six 2's and a 3(The number of 2's should be a multiple of 3).For
the number to be a multiple of 4, the last 2 digits should be 32.
Now, let us combinethese two.(i) All seven 3's - No
possibility(ii)Three 2's and four 3's - The first 5 digits should
have two 2's and three 3's in someorder.No of possibilities
=(5!)/(3!2!)=10(iii)Six 2's and one 3 - The first 5 digits should
all be 2's.So, there is - only one number 2222232So we get the
total of seven-digit number comprises of only 2's and 3's which
aremultiples of 12 are = (i) + (ii) + (iii) = 0+10+1=11 26. Some
boys are standing on a circle at distinct points. Each possiblepair
of persons, who are not adjacent, sing a 3 minute song, one
pairafter another. The total time taken by all the pairs to sing is
1 hour.Find the number of boys?A. 6 B. 7C. 8 D. 9Each boy would
pair with n-3 other boysNumber of possible pairs =n(n3)/2 =60/3=20
n(n3)=40 n= 8 27. a, b, c are three distinct integers from 2 to 10
(both inclusive). Exactly oneof ab, bc and ca is odd. abc is a
multiple of 4. The arithmetic mean of aand b is an integer and so
is the arithmetic mean of a, b and c. How manysuch triplets are
possible (unordered triplets) .A. 4 B. 5 C. 6 D. 7Exactly one of
ab, bc & ca is oddTwo are odd and one is evenabc is a multiple
of 4the even number is a multiple of4The arithmetic mean of a and
bis an integer a and b are odd and thearithmetic mean of a, b and
c. a+ b + c is a multiple of 3c can be 4 or 8.c = 4; a, b can be 3,
5 or 5, 9c = 8; a, b can be 3, 7 or 7, 9Four triplets are possible
28. The number of ways of arranging n students in a row such thatno
two boys sit together and no two girls sit together is n(n
>100). If one more student is added, then number of ways
ofarranging as above increases by 200%. The value of n isA. 12 B. 8
C. 9 D. 10If n is even, then the number of boys should be equal to
number ofgirls, let each be 'a'. n=2a.Then the number of
arrangements =2a!a!If one more students is added, then number of
arrangements =a!(a+1)!But this is 200% more than the earlier
3(2a!a!)=a!(a+1)!a+1=6 & a=5 n=10But if n is odd, then number
of arrangements:=a!(a+1)! Where, n=2a+1When one student is
included, number of arrangements: =2(a+1)!(a+1)! By the given
condition, 2(a+1)=3, which is not possible. 29. How many five digit
positive integers that are divisible by 3 canbe formed using the
digits 0, 1, 2, 3, 4 and 5, without any of thedigits getting
repeating A. 15 B. 96 C. 216 D. 120 E. 625Combining the two
criteria that we use only 5 of the 6 digits andpick them in such a
way that the sum is divisible by 3, we shouldnot use either '0' or
'3' while forming the five digit numbers.Case 1 If we do not use
'0', then the remaining 5 digits can be arrangedin 5! ways=120
numbers.Case 2If we do not use '3', then the arrangements should
take into account that'0' cannot be the first digit as a 5-digit
number will not start with '0'.The first digit from the left can be
any of the 4 digits 1,2,4 or 5Then the remaining 4 digits including
'0' can be arranged in the other 4places in 4! ways. So, there will
be 44! numbers =424=96 numbers.Combining Case 1 and Case 2, there
are a total of 120+96= 216 30. If the letters of the word CHASM are
rearranged to form 5 letterwords such that none of the word repeat
and the resultsarranged in ascending order as in a dictionary what
is the rankof the word CHASM? A. 24 B. 31 C. 32 D. 30The 5 letter
word can be rearranged in 5!=120 Ways without anyof the letters
repeating.The first 24 of these words will start with A.Then the
25th word will start will CA _ _ _.The remaining 3 letters can be
rearranged in 3!=6 Ways. i.e. 6words exist that start with CA.The
next word starts with CH and then A, i.e., CHA _ _.The first of the
words will be CHAMS.The next word will be CHASM.Therefore, the rank
of CHASM will be 24+6+2= 32 31. The marks scored in History by
P,Q,R and S form a geometric progression in thatorder. If the marks
scored by R were% less than the sum of the marks scoredby P and Q,
then marks scored by S were what percent more than the marks
scoredby Q? (Assume that everyone scored positive marks).A. 56.25 %
B. 55.55 % C. 64 % D. 67.75 %Let the marks scored by P,Q,R and Sbe
a,ak,ak2 and ak3 .Marks scored by R were% lessthan the marks scored
by P and Q.So, if marks scored by P and Q were36 then those scored
by R is 25.Marks scored by P & Q=a+ak=36Marks scored by R
=ak2=25+2 =3625 k=54P=a,Q=5425165a,R=,S=12564 P=64a,Q=80a,R=100a
& S=125a12580S scored80100% more thanQ4580100=56.25%S scored
56.25 % more than Q. 32. A shopkeeper sells three items P,Q and R
and incurs a loss of 21%, 11% and 10%respectively. The overall loss
percentage on selling P and Q items is 14.33% andthat of Q and R
items is 10.4%. Find the overall loss percentage on selling
thethree items?A. 15% B. 12.16% C. 13.4% D. 12.5%Let the cost of
the item P = Rs pLet the cost of the item Q = Rs qLet the cost of
the item R = Rs rSP of the item P =0.79pSP of the item Q =0.89qSP
of the item R =0.9rOverall loss percentage of the 1sttwo items
=14.33%0.21+0.11+=0.1433=12--------------(i)Overall loss percentage
of the 2ndand 3rd item
=10.4%0.11+0.1+=0.104=23------------(ii)Overall loss
percentage:0.21+0.11+0.1++1001 0.21 +2 0.11 +3(0.1)1+2+3100
0.1216100= 12.16% 33. A shopkeeper gives a discount of 12%, whereas
a customermakes cash payment.Let 'p' denotes the percentage, above
the cost price, that theshopkeeper must mark up the price of the
articles ['p' is aninteger] in order to make a profit of x% (x <
100).Which of the following is the possible value(s) of x?A. 54 B.
76 C. 96 D. 32Let the cost price of one article is Rs r then Marked
price of the article= 1 +100Selling price = 1 +1001 12100Profit
percentage = x% 1 +1001 12100= 1 +100 p=25 12+22As, p is an
integer,12+ x must be multiple of 22.All the possible values x,
less than 100 arek=10,32,54,76,98 Except option 'C' all the options
are possible. 34. Peter got 30% of the maximum marks in an
examination and failed by 10 marks.However, Paul who took the same
examination got 40% of the total marks andgot 15 marks more than
the passing marks. What were the passing marks in theexamination?
A. 35 B. 250 C. 75 D. 85Let x be the maximum marks in the
examination.Peter got 30% of x =0.3x And Paul got 40% of x=0.4x.In
terms of the maximum marks Paul got 0.4x - 0.3x = 0.1x more than
Peter. --------(1)The problem however, states that Paul got 15
marks more than the passing mark and Petergot 10 marks less than
the passing mark.Therefore, Paul has got 15 + 10 = 25 marks more
than Peter.-------- (2)Equating (1) and (2), we get 0.1x=25x=250x
is the maximum mark and is equal to 250 marks.Therefore, Peter got
75 marks.We also know that Peter got 10 marks less than the passing
mark.Therefore, the passing mark will be 10 marks more than what
Peter got = 75 + 10 = 85.A candidate scoring x% in an examination
fails by 'a' marks, while another candidatewho scores y% marks gets
'b' marks more than then the minimum required passmarks. Then the
Maximum marks for the examination are 100+Max marks = 100 x2510=
250 35. A shepherd has 1 million sheep at the beginning of Year
2000.The numbers grow by x% (x > 0) during the year.A famine
hits his village in the next year and many of his sheep die.The
sheep population decreases by y% during 2001 and at the beginning
of 2002the shepherd finds that he is left with 1 million
sheep.Which of the following is correct?A. x > y B. y > x C.
x = y D. Cannot be determinedLet us assume the value of x to be
10%.Therefore, the number of sheep in the herd at the beginning of
year 2001 (end of 2000)will be 1 million + 10% of 1 million = 1.1
millionIn 2001, the numbers decrease by y% and at the end of the
year the number sheep inthe herd = 1 million.i.e., 0.1 million
sheep have died in 2001.In terms of the percentage of the number of
sheep alive at the beginning of 2001, it willbe (0.1/1.1)100 % =
9.09%.From the above illustration it is clear that x > y. 36. A
began business with Rs. 6,250 and is joined afterwards by B with
Rs. 9,375.when did B join if the profit at the end of the year are
divided equally.A. 6 months B. 8 months C. 10 months D. 11
monthsProfit of A = Profit of BCapital of A months = Capital of B
No. of months.6250 12 = 9375 No. of months6250 129375= 8.i.e. B
joined the business after 12 4 = 8 months. 37. Rs. 120 is divided
between x, y and z, So that xs share is 25 morethan y's and Rs. 25
less than that of z.The share of y will be. A. 15 B. 25 C. 35 D.
40let ys share be Rs. yxs share = y + 25zs share = y + 50.xs share
+ ys share + zs share = 120y + 25 + y + y + 50 = 1203y = 45Y =
15i.e. xs share = Rs. 40ys share = Rs. 15zs share = Rs. 75 38. A
sum of money is divided among A, B and C.So, that to each two
rupees A gets, B gets 60 paisa and C gets 80paisa.If Cs share is
Rs. 40 then sum must be: A. 130 B. 155 C. 170 D. 185If A gets 200
paiseB gets 60 paiseC gets 80 paiseHere C is getting Rs. 40Which
implies B is getting Rs. 30 and A is getting Rs. 100.Total amount
40 + 30 + 100 = Rs. 170 39. P & Q entered in to partnership P
investing Rs.8000 and Rs.6000 respectively.after 3 months P
withdrew Rs.2000 while Q invested Rs.2000 more.after 3 more months,
R joins the business with a capital of Rs.11000.The share of Q
exceeds that of R out of the a total profit of Rs.35,100 after one
yearby.A. Rs.2300 B. Rs.3600C. Rs.3800 D. Rs.4200 E. None of theseP
: Q : R = (8000 3 + 6000 9) : (6000 3 + 8000 9) : (11000 6) (24000
+ 54000) : (18000 + 72000) : (6600) 78000 : 90000 : 66000 26 : 30 :
22 13 : 15 : 11 Difference of Q's & R's shares = Rs. 351001539
35100113935100439= 3600 40. If REASON is coded as 5 and BELIEVED as
7, then what is the code forGOVERNMENT?A. 6 B. 8 C. 9 D. 10 E. None
of theseClearly, each word is coded by the numeral which is 1 less
thanthe number of letters in the word.Since, there are 10 letters
in the word GOVERNMENT,so required = 10 - 1 = 9 41. In a certain
code language, col tip mot means singing isappreciable, mot baj min
means dancing is good and tip nopbaj means singing and dancing,
which of the following meansgood in that code language?A. not B.
min C. baj D. Can't be determined E. None of theseStatement 1:
singing is appreciable Code Statement 1: col tip motStatement 2:
dancing is good Code Statement 2: mot baj minStatement 3: singing
and dancing Code Statement 3: tip nop bajBy removing the repeated
code-words; 'mot' and 'baj' from the 2ndstatement, no matter which
word each stands for, we are left with onlythe code-word 'min' that
evidently stands for the word 'good'.The answer, hence, is option
B. 42. If A and B work together, they will complete a job in 7.5
days. However, if A worksalone and completes half the job and then
B takes over and completes theremaining half alone, they will be
able to complete the job in 20 days.How long will B alone take to
do the job if A is more efficient than B?A. 20 days B. 40 days C.
36 days D. 30 days+=As A can complete the job working alone in 'a'
days, he will complete half the job, workingalone, in a/2
days.Similarly, B will complete the remaining half of the job in
b/2 days.Therefore, a/2+b/2=20 a+b=40 or a=40b ...... (2)From (1)
and (2) we get, 140b+1b=215 600=2b(40b) 600=80b2b2 b240b+300=0
(b30)(b10)=0 b=30 or b=10.If b=30, then a=4030=10 or If b=10, then
a=4010=30.As A is more efficient than B, he will take lesser time
to do the job alone.Hence A will take only 10 days and B will take
30 days. 43. A can do a job in 10 days. Due to problematic
relationship with B,his efficiency reduces to 60% and both can
finish the job in 10days. In how many days can B complete the whole
job?A. 15 B. 25 C. 30 D. 35With 100% efficiency A alone can
complete the work in 10 days.10010with 60% efficiency A alone can
complete the work in60=1006daysLet B alone can complete the work in
y days.6100+1=110Solving this,100100+6= 1010y = 100 + 6y4y = 100y =
25. 44. A can complete a project in 20 days and B can complete
thesame project in 30 days. If A and B start working on the
projecttogether and A quits 10 days before the project is
completed, inhow many days will the project be completed?A. 18 B.27
C.26.67 D.16Let the total number of days taken to complete the
project be x days.Then, B would have worked for all x days, while A
would have worked for(x10) days.Therefore, A would have
completed1020of the project and B would havecomplete30of the
project.i.e.,1020+30=1Solving for x, we get x= 18 45. S.I for a sum
of 1550 for 2 years rupees 20 more than the S.I for1450 for the
same duration. Find the rate of interest.A. 5% B. 10% C. 15% D.
24%S.I. on 1550 - S.I on 1450 = 201550210014502100= 201002100=20R =
10. 46. A sum of Rs. 500 amounts to Rs. 650 in 3 years at S.I. if
the interestrate is increased by 3%, it would amount to how much?A.
450 B. 375 C. 695 D. 950S.I. = Rs. 650 - Rs. 500 = Rs. 150Time = 3
yearsR =100 150500 3= 10New Rate = 13%500 3 13S.I =100= 195.Amount
= 695. 47. Geeta borrowed some money at the rate of 6% p.a for the
first two years, at the rateof 9% p.a for the next three years, and
at the rate of 14% p.a for the period beyondfive years. If she pays
a total interest of Rs.11400 at the end of nine years, how muchdid
she borrow?A. Rs. 10,000 B. Rs. 11,000 C. Rs. 12,000 D. Rs.
14,000Let the principal be x. Then, 6 2100+ 9 3100+ 14 4100=
1140095100= 11400 =11400 10095X = 12000. 48. If the diagonal of a
rectangle is 17 cm long and its perimeter is 46cm, find the area of
the rectangle?A. 100 cm2 B. 120 cm2 C. 130 cm2 D. 140 cm2Let length
= x and breadth = y. then,2(x + y) = 46x +y = 23 and x2 + y2 =
(17)2 = 289.Now, (x + y)2 = (23)2 (x2 + y2) + 2xy = 529. 289 + 2xy
= 529 xy = 120.So, Area = xy = 120 cm2. 49. The length of a
rectangle is twice its breadth. If its length is decreased by 5 cm
andbreadth is increased by 5 cm, the area of the rectangle is
increased by 75 sq. cm.find the length of the rectangle.A. 5 cm B.
10 cm C. 15 cm D. 20 cmLet breadth = x. then, length = 2x. then,(2x
- 5) (x +5) 2x x = 75 5x 25 = 75 x = 20.So, length of the rectangle
= 20 cm. 50. In measuring the sides of a rectangle, one side is
taken 5% in excess, andthe other 4% in deficit. Find the error
percent in the area calculated fromthese measurements.A. 0.2 % B.
0.4 % C. 0.8 % D. 0.12 %Let x and y be the sides of the rectangle.
Then, correct area = xyCalculated area =10510096100=504500Error in
measurement = +4500Error % =4500100=45= 0.8%Shortcut:Change in
percentage overall = +5 4 54100= 1 0.2 = 0.8% 51. Statements: All
cups are glasses. Some glasses are bowls. No bowl is a
plate.Conclusions:No cup is a plate.No glass is a plate.Some plates
are bowls.Some cups are not glasses.A.None follows B.Only either I
or III follows C.Only II and III followD.Only III and IV follow
E.None of theseA.No CONCLUSION follows
