gmat MATH tough problems.doc

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Difficult Problems from the Math Section

1. The sum of the even numbers between 1 and n is 79*80 where n is an odd

number then n!"

Sol: First term a=2, common difference d=2 since even number

therefore sum to first n numbers of Arithmetic progression would be

n/2(2a+(n!"d"

= n/2(2#2+(n!"#2"=n(n+!" and this is e$ual to %&#'

therefore n=%& which is odd)))

#. The $rice of a bushel of corn is currentl% &'.#0 and the $rice of a $ec( of

wheat is &).80. The $rice of corn is increasin at a constant rate of )+ cents

$er da% while the $rice of wheat is decreasin at a constant rate of

cents $er da%. ,hat is the a$$ro+imate $rice when a bushel of corn costs the

same amount as a $ec( of wheat"

(A" *)(-" *)!(." *)(0" *)(1" *)

Soln: 2 + 3 = ' )!34 3 = 5 of da6s after price is same4

solving for 34 3 is appro3imatel6 ', thus the re$uired price is 2 + # ' = cents = *)

'. -ow man% randoml% assembled $eo$le do u need to have a better than )0

$rob. that at least 1 of them was born in a lea$ %ear"

Soln: 7rob) of a randoml6 selected person to have 89 been born in a leap 6r = /a;e 2 people, probabilit6 that none of them was born in a leap = /#/ = &/!) he probabilit6 at least one born in leap = ! &/! = %/! < )a;e people, probabilit6 that none born in leap 6ear = /#/#/ = 2%/) he probabilit6 that at least one born = ! 2%/ = %/ )hus min people are needed)

/. n a bas(etball contest $la%ers must ma(e 10 free throws. ssumin a $la%er

has 90 chance of ma(in each of his shots how li(el% is it that he will ma(e

all of his first 10 shots"

Ans: he probabilit6 of ma;ing all of his first ! shots is given b6

(&/!"# (&/!"# (&/!"# (&/!"# (&/!"# (&/!"# (&/!"# (&/!"# (&/!"# (&/!" =(&/!">! = )' = ?



). 2 3 4D ! where 2 and 4D are two5diit numbers and is a

three diit number6 2 4 and D are distinct $ositive inteers. n the

addition $roblem above what is the value of 4"

(A" !

(-"

(." %

(0" &

(1" .annot be determinedAns: A- + .0 = AAASince A- and .0 are two digit numbers, then AAA must be !!!

herefore !- + .0 = !!!- can assume an6 value between and &@f - = , then .0 = !!!! = &' and . = &@f - = &, then .0 = !!!!& = &2 and . = &So for all - between &, . = &

herefore the correct answer is 0 (. = &"

. and 2 ran a race of /80 m. n the first heat ives 2 a head start of /8 m

and beats him b% 110th of a minute. n the second heat ives 2 a head start

of 1// m and is beaten b% 1'0 th of a minute. ,hat is 2s s$eed in ms"

:; 1#

:2; 1/

:4; 1

:D; 18

:<; #0

Ans: race ! : ta = tb ( because A beats - b6 sec"race 2 : a = tb+2 ( because A looses to - b6 2 sec"

-6 the formula 0= S # we get two e$uations'/Sa = 2/Sb !"'/Sa = /Sb +22"1$uating these two e$uations we get Sb = !2ta,Sa stand for time ta;en b6 A and speed of A resp)

7. certain =uantit% of /0 solution is re$laced with #) solution such that

the new concentration is '). ,hat is the fraction of the solution that was

re$laced"



:; 1/

:2; 1'

:4; 1#

:D; #'

:<; >

Ans: Bet C be the fraction of solution that is replaced)

hen C#2? + (!C"#? = ?

Solving, 6ou get C = !/

8. $erson bu%s a share for & )0 and sells it for & )# after a %ear. ,hat is the

total $rofit made b% him from the share"

:; com$an% $a%s annual dividend

:; The rate of dividend is #)

:; Statement :; ?@A< is sufficient but statement :; alone is not

sufficient:2; Statement :; ?@A< is sufficient but statement :; is not sufficient

:4; 2@T- statements T@B<T-<C are sufficient but A<T-<C statement

alone is sufficient

:D; <ach statement ?@A< is sufficient

:<; Statements :; and :; T@B<T-<C are A@T sufficient

9. ba contains ' red / blac( and # white balls. ,hat is the $robabilit% of

drawin a red and a white ball in two successive draws each ball bein $ut

bac( after it is drawn"

:; ##7

:2; 19

:4; 1'

:D; /#7

:<; #9

ns .ase @: Ded ball first and then white ball

7! = /&#2/&= 2/2%

.ase 2: Ehite ball first and then red ball

72 = 2/&#/& = 2/2%

herefore total probabilit6: p! + p2 = /2%

10. ,hat is the least $ossible distance between a $oint on the circle +E# 3 %E# ! 1

and a $oint on the line % ! '/*+ 5 '"



; 1./

2; s=rt :#;

4; 1.7

D; s=rt :';

<; #.0

Ans: he e$uation of the line will be 3 6 !2 = )his crosses the 3 and 6 a3is at (," and (,"

he circle has the origin at the center and has a radius of ! unit)

So it is closest to the given line when, a perpendicular is drawn to the line, which passesthrough the origin)

his distance of the line from the origin is !2 / s$rt (& + !" which is 2)Bength of perpendicular from origin to line a3 +b6 + c = is

mod (c / s$rt (a>2 + b>2""G

he radius is ! unit)

So the shortest distance is 2) ! unit = !) units

11)

n the s=uare above 1#w = ' x = / y. ,hat fractional $art of the s=uare is shaded"

; #'

2; 1/#)

4; )9

D; 11#)

<; '7



Sol Since !2w=3=6,

w:3=:!2=!: and 3:6=:

so, w = !3 = 6 =

the fractional part of the s$uare is shaded:H(w+3">2 (!/2"w3 + (!/2"w3 +(!/2"36 + (!/2"w(2w"GI/(w+3">2

= H(w+3">2 w3 + (!/2"36 + w>2"GI/(w+3">2G

=(>2" (++!"G/(>2"

= (2 !!"/2

= !/2

1#. The averae of tem$eratures at noontime from Monda% to Frida% is )06 the

lowest one is /) what is the $ossible ma+imum rane of the tem$eratures"

#0 #) /0 /) 7)

ns he answer 2 doesnJt refer to a temperature, but rather to a range of temperatures)

he average of the temps is: (a + b + c + d + e" / = 9ne of these temps is : (a + b + c + d + " / = Solving for the variables: a + b + c + d = 2@n order to find the greatest range of temps, we minimiKe all temps but one) Demember,though, that is the lowest temp possible, so: + + + d = 2Solving for the variable: d = %% = 2

1'. f n is an inteer from 1 to 9 what is the $robabilit% for n*:n31;*:n3#; bein

divisible b% 8" #) )0 #.) 7#.) 7)

Soln: 1 = n#(n+!"#(n+2"

1 is divisible b6 ', if n is even) 8o of even numbers (between ! and &" = '

1 is divisible b6 ', also when n = '; ! (; = !,2,,)))))"Such numbers total = !2(%,!,))))"



Favorable cases = '+!2 = )otal cases = &7 = /& = 2)

Lethod 2:From ! to !, there are sets, which are divisible b6 ')(2##" (#%#" (#%#'"(%#'#&"('#&#!"

So till 9, there will be !2 # such sets = sets

so probabilit6 will be /& = 2)

1/. Gurt a $ainter has 9 Hars of $aint

/ are %ellow

# are redrest are brown

Gurt will combine ' Hars of $aint into a new container to ma(e a new color which he

will name accordinl% to the followin conditions

2run I if the $aint contains # Hars of brown $aint and no %ellow

2run J if the $aint contains ' Hars of brown $aint

Kaune J if the $aint contains at least # Hars of %ellow

Kaune I if the $aint contains e+actl% 1 Har of %ellow

,hat is the $robabilit% that the new color will be Kaune

a; )/#

b; '7/#

c; 1#1

d; /9

e; )9

Sol !) his has at least 2 6ellow meaning))

a there can be all three M = c9D b 2 M and ! out of 2 D and - = c2 3 c!

otal

2)his has e3actl6 ! M and remaining 2 out of = c! 3 c2

otal otal possibilities = (&N/NN" = '



Adding the two probabilities: probabilit6 = %/' = %/2

1). T,@ cou$les and a sinle $erson are to be seated on ) chairs such that no

cou$le is seated ne+t to each other. ,hat is the $robabilit% of the above""

Soln:

Ea6s in which the first couple can sit together = 2#N (! couple is considered one unit"Ea6s for second couple = 2#Nhese cases include an e3tra case of both couples sitting together Ea6s in which both couple are seated together = 2#2#N = N (2 couples considered as 2units so each couple can be arrange between themselves in 2 wa6s and the units in NEa6s"hus total wa6s in which at least one couple is seated together = 2#N + 2#N N = #Notal wa6s to arrange the ppl = Nhus, prob of at least one couple seated together = #N / N = /hus prob of none seated together = ! / = 2/

1. n e+$ress train traveled at an averae s$eed of 100 (ilometers $er hoursto$$in for ' minutes after ever% 7) (ilometers. local train traveled at an averae

s$eed of )0 (ilometers sto$$in for 1 minute after ever% #) (ilometers. f the trains

bean travelin at the same time how man% (ilometers did the local train travel in

the time it too( the e+$ress train to travel 00 (ilometers"

a. '00

b. '0)

c. '07.)

d. 1#00

e. 1#'

Sol the answer is .: %) ;m 13press rain: ;m hours (since ! ;m/h" + stops # min)stops = / % = ', but as it is an integer number, the last stop in ;m is not a realstop, so it would be % stopsso, time= hours + % # min) = hours 2! min

Bocal rain:in hours, it will ma;e ;m (since its speed is ;m/hour"in ;m it will have / 2 stops = !2, !2 stops ! min each = !2 minwe have hours !2 min, but we need to calculate how man6 ;m can it ma;e in hours2! min, so 2!!2 = & minif min it can ma;e ;m, & min it can ma;e %) ;mso, the distance is + %) = %) ;m

17. Matt starts a new Hob with the oal of doublin his old averae commission of

&/00. -e ta(es a 10 commission ma(in commissions of &100.00 &#00.00



&#)0.00 &700.00 and &1000 on his first ) sales. f Matt made two sales on the last

da% of the wee( how much would Matt have had to sell in order to meet his oal"

Sol he two sales on LattJs last da6 of the wee; must total *,)

(! + 2 + 2 + % + ! + 3" / % = '3 = since 3 is LattJs !? commission, the sale is *,)

18. @n how man% wa%s can the letters of the word L4@MPT<CL be arraned"

1; ,ithout an% restrictions.

#; M must alwa%s occur at the third $lace.

'; ll the vowels are toether.

/; ll the vowels are never toether.

); Nowels occu$% the even $ositions.

Sol

1; 8O ! /0'#0

2" %##!####2#!=, '; .onsidering the vowels as ! letter, there are five other letters which areconsonants ., L, 7, , D .L7D (AO1" = letters which can be arranged in p or N Ea6sand A, O, 1 themselves can be arranged in another N Ea6s for a total of N#N Ea6s

" otal combinations all vowels alwa6s together = what u found in !" what u found in "= 'N N #N

); @ thin; it should be # %2there are even positions to be filled b6 three even numbers)

in ###2##!#2#! @t is assumed that Bast even place is 89 filled b6 a vowel)here can be total wa6s to do that)

Pence # %2

19. n the infinite se=uence A

where x is a $ositive inteer constant. For what value of n is the ratio

of to e=ual to "

:; 8

:2; 7

:4;

:D; )

:<; /



Sol he method @ followed was to reduce the Q to (3>/ 3 " # ( M/M"the e$n An= (3 > n !"(!+ 3 + 3>2 + 3> +)))) " (!"

and the e$n 3(!+3(!+3(!+3)))""""which @ call R can be reduced to 3( !+3+3>2+3> ))" (2"

from (!" and (2" we get An / R = 3>(n!" / 3therefore for getting answer 3> (n!" = therefore n=%

Ans: -

#0. n how man% wa%s can one choose cards from a normal dec( of cards so as to

have all suits $resent"

a. :1'E/; + /8 + /7

b. :1'E/; + #7 + /7

c. /84d. 1'E/

e. :1'E/; + /84

Sol: 2 cards in a dec; ! cards per suitFirst card let us sa6 from suit hearts = !.! =!Second card let us sa6 from suit diamonds = !.! =!hird card let us sa6 from suit spade = !.! =!Fourth card let us sa6 from suit clubs = !.! =!Demaining cards in the dec;= 2 = 'Fifth card an6 card in the dec; = '.!Si3th card an6 card in the dec; = %.!

otal number of wa6s = ! # ! # ! # ! # ' # % = !> #'#% choice A

#1. <ach of the inteers from 0 to 9 inclusive is written on a se$arate sli$ of blan(

$a$er and the ten sli$s are dro$$ed into a hat. f the sli$s are then drawn one at a

time without re$lacement how man% must be drawn to ensure that the numbers on

two of the sli$s drawn will have a sum of 10"

'

/

)

7 *

Sol

o; consider this



+ !+ 2 + + + +))Stop 0ont go further) Eh6T Peres wh6)At the worst the order given above is how u could pic; the out the slips) Ontil u add theslip with no) on it, no two slips before that add up to ! (which is what the Q wants"

the best u can approach is a sum of & (slip no) + slip no) "

but as soon as u add slip ) Uoila u get 6our first sum of ! from two slips and that isindeed the answer = % draws ##. Two missiles are launched simultaneousl%. Missile 1 launches at a s$eed of x

miles $er hour increasin its s$eed b% a factor of ever% 10 minutes :so that after

10 minutes its s$eed is after #0 minutes its s$eed is and so forth.

Missile # launches at a s$eed of y miles $er hour doublin its s$eed ever% 10

minutes. fter 1 hour is the s$eed of Missile 1 reater than that of Missile #"

1;

#;

:; Statement :1; ?@A< is sufficient to answer the =uestion but statement :#;

alone is not.

:2; Statement :#; ?@A< is sufficient to answer the =uestion but statement :1;

alone is not.

:4; Statements :1; and :#; TG<A T@B<T-<C are sufficient to answer the

=uestion but A<T-<C statement ?@A< is sufficient.

:D; <4- statement ?@A< is sufficient to answer the =uestion.

:<; Statements :1; and :#; TG<A T@B<T-<C are A@T sufficient to answer the

=uestion.

Sol

Since Lissile !Js rate increases b6 a factor of ever6 ! minutes, Lissile ! will be

traveling at a speed of miles per hour after minutes:

minutes!!22+speed

And since Lissile 2Js rate doubles ever6 ! minutes, Lissile 2 will be traveling at a speed

of after minutes:

minutes!!22+speed

he $uestion then becomes: @s T

Statement (!" tells us that ) S$uaring both sides 6ields ) Ee can substitute



for y: @s T @f we divide both sides b6 , we get: @s T Ee can further

simplif6 b6 ta;ing the s$uare root of both sides: @s T Ee still cannot answer this, sostatement (!" alone is 89 sufficient to answer the $uestion)

Statement (2" tells us that , which tells us nothing about the relationship between x and y) Statement (2" alone is 89 sufficient to answer the $uestion)

a;ing the statements together, we ;now from statement (!" that the $uestion can be

rephrased: @s T From statement (2" we ;now certainl6 that , which is another

wa6 of e3pressing ) So using the information from both statements, we can answerdefinitivel6 that after ! hour, Lissile ! is traveling faster than Lissile 2)

he correct answer is .: Statements (!" and (2" ta;en together are sufficient to answer the$uestion, but neither statement alone is sufficient)

#'. @f , what is the units digit of T

(A" (-" !(." (0" (1" &Sol

he units digit of the left side of the e$uation is e$ual to the units digit of the right sideof the e$uation (which is what the $uestion as;s about") hus, if we can determine theunits digit of the e3pression on the left side of the e$uation, we can answer the $uestion)

Since , we ;now that !N .ontains a factor of !, so its



units digit must be ) Similarl6, the units digit of will also have a units digit of )@f we subtract ! from this, we will be left with a number ending in &)

herefore, the units digit of is &) he correct answer is 1)

#/. The dimensions of a rectanular floor are 1 feet b% #0 feet. ,hen a rectanular

ru is $laced on the floor a stri$ of floor ' feet wide is e+$osed on all sides. ,hat are

the dimensions of the ru in feet"

(A" ! b6 ! (-" ! b6 !% (." ! b6 ! (1" (0" ! b6 !% (1" ! b6 !

Soln he rug is placed in the middle of the room) he rug leaves m on either side of it both lengthwise and breadth wise) 8ow, the dimensions of the rug would be thedimensions of the room the space that it does not occup6) Eith three on either side,

m is not occupied b6 the rug in both dimensions)

So, rug siKe = (!" C (2" = ! C!

#). -ow man% different subsets of the set 101/17#/Q are there that contain an odd

number of elements"

:a; ' :b; :c; 8 :d; 10 : e; 1#

Soln: ' is the answer) he different subsets are

!,!,!%,2,!, !, !%

! !% 2

!% 2 !

2 ! !

#. Seven men and seven women have to sit around a circular table so that no #

women are toether. n how man% different wa%s can this be done"

a.#/ b. c./ d.1# e.'

Soln @ suggest to first arranging men) his can be done in N Ea6s) 8ow to satisf6 abovecondition for women, the6 should sit in spaces between each man) his can be done in %N



Ea6s (because there will be seven spaces between each man on round table"otal wa6s = N # %N

#7. Find the fourth consecutive even number

:; the sum of the last two numbers is '0

:; the sum of the first two numbers is ##

the ans is 0

please e3plain wh6 @ alone is sufficientT

Soln: @ guess, there are consecutive even numbers to start with))Stmt !" Sum of rd even + Sum of th even = = !+! = = th even num = !))Sufficient

Stmt 2" Sum of @st even +Sum of 2nd even = 22 = ! and !2 are the 2 numbers to begin

with, then rd num = ! and th num = !)) Sufficient

Ans V 0

#8. f the sum of five consecutive $ositive inteers is then the sum of the ne+t five

consecutive inteers in terms of is

a; 31

b; 3)

c; 3#)

d; #

e; )

Soln @f 6ou divide the sum obtained b6 adding an6 consecutive numbers b6 JJ, andthen 6ou will get the .enter number of the se$uence itself)

i)e) ! = !/ = ) !, 2, ', ,

so, si3th consecutive number will be JJ more than the JLiddle termJi)e) +=, similarl6 +=%

Pence going b6 this) As;ed sum would be

(A/" + G+(A/" + G+(A/" + G+(A/" + G+(A/" + %G = A + 2

#9. f P re$resents the $roduct of the first 1) $ositive inteers then P is not a

multi$le of

a; 99 b; 8/ c; 7# d; ) e; )7



Solution @f 7 represents the product of the first ! integers, 7 would consist of the primenumbers that are below !)

2,,,%,!!,!

An6 value that has a prime higher than ! would not be a value of 7)

% = , !&

!& is a prime greater than !, so the answer is 1)

'0. ) irls and ' bo%s are arraned randoml% in a row. Find the $robabilit% that

; there is one bo% on each end.

2; There is one irl on each end.

Solution: For the first scenario:A" there is one bo6 on each end)

he first seat can be filled in .! ( bo6s ! seat" wa6s = the last seat can be filled in 2.! (2 bo6s ! seat" wa6s = 2the si3 seats in the middle can be filled in N (! bo6 and girls" wa6sotal possible outcome = 'N7robabilit6= (.! # 2.! # N"/ 'N = /2'

For the second scenario:A" there is one girl on each end)

he first seat can be filled in .! ( girls ! seat" wa6s = the last seat can be filled in .! (2 girls ! seat" wa6s = the si3 seats in the middle can be filled in N ( bo6s and girls" wa6sotal possible outcome = 'N7robabilit6= (.! # .! # N"/ 'N = /!

'1. f 2ob and Ken are two of ) $artici$ants in a race how man% different wa%s can

the race finish where Ken alwa%s finishes in front of 2ob"

Solution approach: first fi3 Wen, and then fi3 -ob) hen fi3 the remaining three)

.ase !: Ehen Wen is in the first place)= -ob can be in an6 of the other four places) = )he remaining can arrange themselves in the remaining places in N Ea6s)Pence total wa6s = #N

.ase 2: Ehen Wen is in the second place)



= -ob can be in an6 of three places = )he remaining can arrange themselves in places in N Ea6s)

.ontinuing the approach)Answer = #N +#N +2#N +!#N = !#N = wa6s)

'#. set of numbers has the $ro$ert% that for an% number t in the set t 3 # is in the

set. f R1 is in the set which of the followin must also be in the set"

. R' . 1 . )

. onl%

2. onl%

4. and onl%

D. and onl%

<. and

Soln Series propert6: t = t+2) (8ote: for an6 given number 8, 98BM 8 + 2 is

compulsor6) 8 2 is not a necessit6 as 8 could be the first term)))this can be used as atrap)"Xiven: ! belongs to the series) = ! = =) 091S 89 impl6 )Pence, @@ and @@@ (0")

''. number is selected at random from first '0 natural numbers. ,hat is the

$robabilit% that the number is a multi$le of either ' or 1'"

:; 17'0

:2; #)

:4; 71)

:D; /1)

:<; 11'0

Solution: otal no from ! to = total no from ! to which r multiple of = ! (eg(,,&,!2,!,!',2!,2,2%,""total no from ! to which r multiple of ! = 2 (eg !,2"7(a or b " = p(a" + p(b" p(a"= !/ p(b"=2/ p(a" + p(b" = !/+2/ = 2/

'/. Two numbers are less than a third number b% '0 and '7 res$ectivel%. -ow

much $ercent is the second number less than the first"

a; 10 b; 7 c; / d; '

Solution: )% )

diff in ?= ()% )"/)% # != )%/)% # != !?



'). f y ' and #+% is a $rime inteer reater than # which of the followin must

be true"

. x = y

II . y ! 1

. x and y are $rime inteers.

:; Aone

:2; onl%

:4; onl%

:D; onl%

:<; and

'7. Someone $assed a certain bride which needs fee. There are # wa%s for him to

choice. &1'month3&0.#time 2 &0.7)time . -e $asses the bride twice a da%.

-ow man% da%s at least he $asses the bride in a month it is economic b% wa%"

; 11 2; 1# 4; 1' D; 1/ <; 1)

Soln: Bet 3 be the no da6s where both are e$ual cost! + )2#23= )%#23!+)3=!)3!=!)!33=!!)'!

b6 plugging in for !2 da6sFor A! +(!2#2"#)2=!+)'=!%)'For -2#)%=!'

therefore answer is -)

'8. Two measure standards C and S. #/ and '0 measured with C are /# and 0 when

the% are measured with S res$ectivel%. f 100 is ac=uired with S what would its

value be measured with C"

'9. <ver% student of a certain school must ta(e one and onl% one elective course. n

last %ear 1# of the students too( biolo% as an elective 1' of the students too(

chemistr% as an elective and all of the other students too( $h%sics. n this %ear 1'

of the students who too( biolo% and 1/ of the students who too( chemistr% left

school other students did not leave and no fresh student come in. ,hat fraction of



all students too( biolo% and too( chemistr%"

. 79 2 .7 4.)7 D./9 <.#)

Soln: @f total =!

Bast 6ear -=!/2.=!/7=!/his 6ear-= !/2 # 2/ = !/.= !/ # /= !/ot !/+!/=%/!2student left this 6ear - = !/2 # !/ = !/.= !/ #!/ = !/!2

tot= !/ +!/!2 = !/So the school has total student this 6ear = Bast 6ear student no total no of student left this 6ear= ! !/=/Answer = %/!2 / /= %/&

/0. f J0.9 which of the followin e=uals to J"

. 0.81E1# 2. 0.9E1# 4. 0.9E# D. 150.01E1#

/1. There are 8 students. / of them are men and / of them are women. f / students

are selected from the 8 students. ,hat is the $robabilit% that the number of men is

e=ual to that of women"

.18') 21') 4.1/') D.1'') <.1#')

Soln: there has to be e$ual no of men women so out of people selected there has to be 2L 2E)otal wa6s of selecting out of ' is '.total wa6s of selecting 2 men out of is .2total wa6s of selecting 2 women out of is .2

so probabilit6 is (.2#.2"/ '.2 = !'/

/#. The area of an e=uilateral trianle is 9.what is the area of it circumcircle.

A)!7@ -)!27@ .)!7@ 0)!7@ 1)!'7@



Soln: area = s$rt(" / # (side">2 = &

so , ( side ">2 = ( &# " / s$rt("

Peight = P = s$rt("/2 # (side" , so P>2 = / # Side>2 = / # (&#" / s$rt("

Dadius of .ircum.ircle = D = 2/ of ( Peight of the 1$uilateral riangle "

so , area of .ircumcircle = 7@ # D>2

= 7@ # /& # P>2

= 7@ # /& # / # (&#" / s$rt("

= 7@ # # s$rt("

/'. rou$ of $eo$le $artici$ate in some curriculums #0 of them $ractice Ioa 10

stud% coo(in 1# stud% weavin ' of them stud% coo(in onl% / of them stud% both

the coo(in and %oa # of them $artici$ate all curriculums. -ow man% $eo$le stud%

both coo(in and weavin"

.1 2.# 4.' D./ <.)

Soln Ee ;now there are ! people who do coo;ing as an activit6)

people who do onl6 coo;ing do coo;ing and Moga2 do all of the activities3 number of people doing coo;ing and weaving

Ehen 6ou sum all this up, we should have !) So ++2+3=! 3=!&=!

//. There ' (inds of boo(s in the librar% fiction non5fiction and biolo%. Catio of

fiction to non5fiction is ' to #6 ratio of non5fiction to biolo% is / to ' and the total of

the boo(s is more than 1000"which one of followin can be the total of the boo("

1001 2. 1009 4.1008 D.1007 <.100

Soln fiction : nonfiction = : 2 = : nonfiction : biolog6 = :

fiction : nonfiction : biolog6 = : :

3 + 3 + 3 = !3 = % !2/!



if we add !, ! will divide evenl6 (!!/! = %%"! + ! = !!

/). n a consumer surve% 8) of those surve%ed li(ed at least one of three

$roducts 1 # and '. )0 of those as(ed li(ed $roduct 1 '0 li(ed $roduct # and

#0 li(ed $roduct '. f ) of the $eo$le in the surve% li(ed all three of the$roducts what $ercentae of the surve% $artici$ants li(ed more than one of the

three $roducts"

; )

2; 10

4; 1)

D; #0

<; #)

Soln: n(!O2O" = n(!" + n(2" + n(" n(!n2" n(2n" n(!n" + n (!n2n"'=++2 n(!n2" n(2n" n(!n"G +n(!n2" n(2n" n(!n"G = 2

/. For a certain com$an% o$eratin costs and commissions totaled &))0 million in

1990 re$resentin an increase of 10 $ercent from the $revious %ear. The sum of

o$eratin costs and commissions for both %ears was

:; &1000 million

:2; &10)0 million

:4; &1100 million

:D; &11)0 million

:<; &11)) million

Solution !&&' = * L!&&& = * L

Sum = * ! LAns:-

/7. Fo+ Heans reularl% sell for &1) a $air and Pon% Heans reularl% sell for &18 a

$air. Durin a sale these reular unit $rices are discounted at different rates so that

a total of &9 is saved b% $urchasin ) $airs of Heans ' $airs of Fo+ Heans and # $airs

of Pon% Heans. f the sum of the two discounts rates is ## $ercent what is the

discount rate on Pon% Heans"

(A" &?(-" !?(." !!?(0" !2?



(1" !?Soln Ans : -

otal discount is 22? = *&

b6 bac; solving, in this case ,the discount percent for pon6 Yeans should be less than !!?(22/2"because the price of this product is more)

ta;e choice -!? of !'= !)'total discount on 2 pon6 Yeans= *)22?!?=!2?!2? of ! = *!)'total discount on fo3 Yeans = *)fo3 Yeans discount +2 pon6 Yeans discount =*&) +)=&

so answer is -)

/8. There are # (inds of staff members in a certain com$an% PCT TM< AD

F?? TM<. #) $ercent of the total members are PCT TM< members others are

F?? TM< members. The wor( time of $art time members is ') of the full time

members. ,ae $er hour is same. ,hat is the ratio of total wae of $art time

members to total wae of all members.

.1/ 2.1) 4 1 D 17 < 18

Soln: Ehat is the ratio of total wage of part time members to total wae of all members)

Mou have calculated ratio of 7art timetoFull time)

7 3/ 6/ 36/#A 3/ M + 3/ 6/ !'36/2

Datio= p/a= 36/!'36= !/

/9. f 7) of a class answered the 1st =uestion on a certain test correctl% ))

answered the #nd =uestion on the test correctl% and #0 answered neither of the

=uestions correctl% what $ercent answered both correctl%"

10

#0

'0

)0

)

Soln: his problem can be easil6 solved b6 Uenn 0iagramsBets thin; the total class consists of ! students

so % students answered $uestion !



and students answered $uestion 2

8ow 2 students not answered an6 $uestion correctl6

herefore out of total ! students onl6 ' students answered either $uestion ! or

$uestion 2 or both the $uestions)))

So %+=! which implies !'=? are the students who answered both correctl6and are counted in both the groups)))thats wh6 the number was more))

Bet me ;now if someone has problems understanding)))

)0. set of data consists of the followin ) numbers 0#/ and 8. ,hich two

numbers if added to create a set of 7 numbers will result in a new standard

deviation that is close to the standard deviation for the oriinal ) numbers"

;. 51 and 92;. / and /

4;. ' and )

D;. # and

<;. 0 and 8

Soln S0 = S$rt(Sum(C3">2/8"Since 8 is changing from to %) Ualue of Sum(C3">2 should change from (current" to') So that S0 remains same)

so due to new numbers it adds ') .hoice 0 onl6 fits here)

)1. -ow man% inteers from 0 to )0 inclusive have a remainder of 1 when divided

b% '"

A)! -)!) .)! 0)!% 1)!'

Soln: if we arrange this in A7, we get+%+!+)))))))+&

so +(n!"=&: n=!. is m6 pic;

)#. f (!m:m3/;:m3); ( and m are $ositive inteers. ,hich of the followin could

divide ( evenl%"

.' ./ .

Soln: he idea is to find what are the common factors that we get in the answer)



m = !, ; = which is divisible b6 !,2,,,,!, ! and m = 2, ; = ' which is divisible b6 !,2,,,, ))))

As can be seen, the common factors are !,2,,

So answer is and

)'. f the $erimeter of s=uare reion S and the $erimeter of circular reion 4 are

e=ual then the ratio of the area of S to the area of 4 is closest to

:; #'

:2; '/

:4; /'

:D; '#

:<; #

Soln and the answer would be -)))here is the e3planation)))

Bet the side of the s$uare be s))then the perimeter of the s$uare is sBet the radius of the circle be r))then the perimeter of the circle is 2#pi#r

it is given that both these $uantities are e$ual))therefore

s=2#pi#r

which is then s/r=pi/2

8ow the ratio of area of s$uare to area of circle would be

s>2/pi#r>2

(!/pi"#(s/r">2

= (!/pi"#(pi/2">2 from the above e$ualit6 relation

pi=22/% or )!

the value of the above e3pression is appro3imate =)%' which is near to answer -

)/. Two $eo$le wal(ed the same distance one $ersons s$eed is between #) and

/)and if he used / hours the s$eed of another $eo$le is between /) and 0and if he

used # hours how lon is the distance"

.11 2.118 4.1#/ D.1' <.1/0

Soln First person speed is between 2mph and mphso for hrs he can travel ! miles if he goes at 2mph speedand for hrs he can travel ! miles if he goes at mph speed



similarl6

Second person speed is between mph and mphso for 2 hrs he can travel & miles if he goes at mph speedand for 2 hrs he can travel !2 miles if he goes at mph speed

So for the first persondistance traveled is greater than ! and Bess than !

and for the second persondistance traveled is greater than & and less than !2

so seeing these conditions we can eliminate ., 0, and 1 answers)))

but @ didnt understand how to select between !! and !!' as both these values aresatisf6ing the conditions)))

)). -ow man% number of ' diit numbers can be formed with the diits 01#'/) if

no diit is re$eated in an% number" -ow man% of these are even and how man%odd"

Soln 9dd: fi3 last as odd, wa6s ZZ ZZ ZZ now, left are , but again leaving , for !st digit again for 2nd digit: ZZ ZZ ZZ=' 9dd)

!'= 2 1ven

). -ow man% '5diit numerals bein with a diit that re$resents a $rime and end

with a diit that re$resents a $rime number"

; 1 2; 80 c; 10 D; 180 <; #/0

Soln he first digit can be 2, , , or % ( choices"he second digit can be , !, 2, , , , , %, ', or & (! choices"he third digit can be 2, , , or % ( choices"

# # ! = !

)7. There are three (inds of business 2 and 4 in a com$an%. #) $ercent of the

total revenue is from business 6 t $ercent of the total revenue is from 2 the others

are from 4. f 2 is &1)0000 and 4 is the difference of total revenue and ##)000

what is t"

.)0 2.70 4.80 D.90 <.100

Soln Bet the total revenue be C)

So C= A + -+ .



which is C= !/ C+ ! + (C22"

C= C+ + C &

Solving for C 6ou get C=

And ! is ? of so the answer is ? (A"

)8. business school club Friends of Foam is throwin a $art% at a local bar. @f

the business school students at the bar /0 are first %ear students and 0 are

second %ear students. @f the first %ear students /0 are drin(in beer /0 are

drin(in mi+ed drin(s and #0 are drin(in both. @f the second %ear students

'0 are drin(in beer '0 are drin(in mi+ed drin(s and #0 are drin(in

both. business school student is chosen at random. f the student is drin(in beer

what is the $robabilit% that he or she is

also drin(in mi+ed drin(s"

. #)

2. /7

4. 1017

D. 7#/

<. 710

Soln The probability of an event A occurring is the number of outcomes that result

in A divided by the total number of possible outcomes.

The total number of possible outcomes is the total percent of students drinking beer.

40% of the students are first year students. 40% of those students are drinking beer.

Thus, the first years drinking beer make up (40% * 40% or !"% of the totalnumber of students.

"0% of the students are second year students. #0% of those students are drinking

beer. Thus, the second years drinking beer make up ("0% * #0% or !$% of the

total number of students.

(!"% !$% or #4% of the group is drinking beer.

The outcomes that result in A is the total percent of students drinking beer and

mi&ed drinks.

40% of the students are first year students. '0% of those students are drinking bothbeer and mi&ed drinks. Thus, the first years drinking both beer and mi&ed drinks

make up (40% * '0% or $% of the total number of students.

"0% of the students are second year students. '0% of those students are drinkingboth beer and mi&ed drinks. Thus, the second years drinking both beer and mi&ed

drinks make up ("0% * '0% or !'% of the total number of students.



($% !'% or '0% of the group is drinking both beer and mi&ed drinks.

f a student is chosen at random is drinking beer, the probability that they are alsodrinking mi&ed drinks is ('0)#4 or !0)!.

)9. merchant sells an item at a #0 discount but still ma(es a ross $rofit of #0$ercent of the cost. ,hat $ercent of the cost would the ross $rofit on the item have

been if it had been sold without the discount"

; #0 2; /0 4; )0 D; 0 <; 7)

Soln: Bets suppose original price is !)

And if it sold at 2? discount then the price would be '

but this ' is !2? of the actual original price)))so )% is the actual price of the item

now if it sold for ! when it actuall6 cost )% then the gross profit would be &)&&?i)e) appro3 ?

0. f the first diit cannot be a 0 or a ) how man% five5diit odd numbers are

there"

. /#)00

2. '7)00

4. /)000

D. /0000

<. )0000

Soln This problem can be solved +ith the ultiplication -rinciple. The ultiplication

-rinciple tells us that the number of +ays independent events can occur together canbe determined by multiplying together the number of possible outcomes for each

event.

There are $ possibilities for the first digit (!, ', #, 4, ", , $, .There are !0 possibilities for the second digit (0, !, ', #, 4, /, ", , $,

There are !0 possibilities for the third digit (0, !, ', #, 4, /, ", , $, There are !0 possibilities for the fourth digit (0, !, ', #, 4, /, ", , $,

There are / possibilities for the fifth digit (!, #, /, ,

sing the ultiplication -rinciple1

2 $ * !0 * !0 * !0 * /2 40,000

61.A bar is creating a new signature drink. There are five possible alcoholicingredients in the drink: rum, vodka, gin, peach schnapps, or whiskey. There



are five possible non-alcoholic ingredients: cranberry uice, orange uice,pineapple uice, limeuice, or lemon uice. !f the bar uses two alcoholic

ingredients and two non-alcoholic ingredients, how many different drinksare possible"

A. 1##

$. %&'. &#

(. )&*. +6##

3oln1 The first step in this problem is to calculate the number of +ays of selectingt+o alcoholic and t+o nonalcoholic ingredients. 3ince order of arrangement does not

matter, this is a combination problem.

The number of combinations of n ob5ects taken r at a time is

6(n,r 2 n7)(r7(nr7

The number of combinations of alcoholic ingredients is

6(/,' 2 /7)('7(#76(/,' 2 !'0)('("

6(/,' 2 !0

The number of combinations of nonalcoholic ingredients is

6(/,' 2 /7)('7(#76(/,' 2 !'0)('("

6(/,' 2 !0

The number of +ays these ingredients can be combined into a drink can bedetermined by the ultiplication -rinciple. The ultiplication -rinciple tells us that the

number of +ays independent events can occur together can be determined bymultiplying together the number of possible outcomes for each event.

The number of possible drinks is

2 !0 * !0

2 !00

6%. The sum of the even numbers between 1 and n is )#, where n is an

odd number. /0"

3oln1 he sum of numbers between ! and n is = (n(n+!""/2

!+2++)))))+n=(n(n+!""/2 HformulaI

we are loo;ing for the sum of the even numbers between ! and n, which means:

2+++)))))+(n!" n is 900



=!#2+2#2+2#+))))))+2#((n!"/2"=2#(!+2++)))))+#((n!"/2""from the formula we obtain :=2#(((n!"/2"#((n!"/2+!""/2=((n!"/2"#((n+!"/2" =%&#'

= (n!"#(n+!"=!'#!= n=!&

'. committee of is chosen from 8 men and ) women so as to contain at least #

men and ' women. -ow man% different committees could be formed if two of the

men refuse to serve toether"

(A" !(-" 22(." !(0" %

(1"

Soln: E 2L == .)('.2!" = )(2%" = !E L == .)('." = !) = total

La3) number of possibilities considering we can choose an6 man 'c2 # . + '.#c= %)consider it this wa6)))) from m6 previous repl6 ma3 possible wa6s considering we canchose an6 man = %

now we ;now that 2 man could not be together))) now thin; opposite))) how man6 wa6sare possible to have these two man alwa6s chosen together)))

since the6 are alwa6s chosen together)))

For chosing 2 men and women for the committee there is onl6 ! wa6 of chosing 2 menfor the committee since we ;now onl6 two specific have to be chosen and there are .wa6s of choosing women

!#. =

For choosing men and women for the committee there are e3actl6 .! wa6s choosing men for the committee since we ;now two specific have to be chosen so from theremaining men we have to chose ! and there are . wa6s of choosing women

.!#. =

So total number of unfavorable cases = + =



8ow since we want to e3clude these cases))) final answer is % =

/. n how man% wa%s can the letters of the word MSSSPP be arraned"

a; 1#0b; 1#000

c; 1#00

d; 1/800

e; #800

Soln otal 5 of alphabets = !so wa6s to arrange them = !N

hen there will be duplicates because !st S is no different than 2nd S)we have @s

Sand 2 7s

Pence 5 of arrangements = !N/N#N#2N

). Boldenrod and Ao -o$e are in a horse race with contestants. -ow man%

different arranements of finishes are there if Ao -o$e alwa%s finishes before

Boldenrod and if all of the horses finish the race"

:; 7#0

:2; '0

:4; 1#0

:D; #/

:<; #1

Soln two horses A and -, in a race of horses))) A has to finish before -

if A finishes !))) - could be in an6 of other positions in wa6s and other horses finishin N Ea6s, so total wa6s #N

if A finishes 2))) - could be in an6 of the last positions in wa6s) -ut the other positions could be filled in N wa6s, so the total wa6s #N

if A finishes rd))) - could be in an6 of last positions in wa6s, but the other positionscould be filled in N wa6s, so total wa6s #N

if A finishes th))) - could be in an6 of last 2 positions in 2 wa6s, but the other positionscould be filled in N wa6s, so total wa6s))) 2 # N

if A finishes th )) - has to be th and the top positions could be filled in N wa6s))



A cannot finish th, since he has to be ahead of -

therefore total number of wa6s

#N + #N + #N + 2#N + N = !2 + & + %2 + ' + 2 =

. On how many ways can the letters of the word "COMPUTER"be arranged?1. M must always occur at the thrd !lace. #owels occu!y the e$en !ostons.

Soln: For !)%##!####2#!=,

For 2)" @ thin; @t should be # %2there are even positions to be filled b6 three even numbers)

in ###2##!#2#! @t is assumed that Bast even place is 89 filled b6 a vowel) herecan be total wa6s to do that)

Pence # %2

7. shi$ment of 10 TN sets includes ' that are defective. n how man% wa%s can a

hotel $urchase / of these sets and receive at least two of the defective sets"

Soln: here are ! U sets4 we have to choose at a time) So we can do that b6 !.wa6s) Ee have % good Us and defective)

8ow we have to choose U sets with at least 2 defective) Ee can do that b6

2 defective 2 good defective ! good

hat stands to .2#%.2 + .#%.! (shows the count"

@f the6 had as;ed probabilit6 for the same $uestion then

.2#%.2 + .#%.! / !.)

8. rou$ of 8 friends want to $la% doubles tennis. -ow man% different wa%s can

the rou$ be divided into / teams of # $eo$le"

. /#0

2. #)#0

4. 18



D. 90

<. 10)

Soln !st team could be an6 of 2 gu6s))) there would be teams (a team of A- is sameas a team of -A"))) possible wa6s '.2 / )

2nd team could be an6 of remaining gu6s) here would be teams (a team of A- issame as a team of -A"))) possible wa6s .2 / rd team could be an6 of remaining 2 gu6s))) there would be 2 teams (a team of A- issame as a team of -A") 7ossible wa6s .2 / 2th team could be an6 of remaining 2 gu6s))) there would be ! such teams))) possiblewa6s 2.2 / !

total number of wa6s)))

'.2#.2#.2#2.2

# # 2 # !

='#%#####2#!##2#!#2#2#2#2

= ! (A8SE1D")))

Another method: sa6 6ou have ' people A-.01FXP

now u can pair A with % others in % wa6s)Demaining now pla6ers)7ic; one and u can pair him with the remaining in wa6s)

8ow 6ou have pla6ers)7ic; one and u can pair him with the remaining in wa6s)

8ow 6ou have 2 pla6ers left) Mou can pair them in ! wa6

so total wa6s is %###! = ! wa6s i)e) 1

9. n how man% wa%s can ) $eo$le sit around a circular table if one should not have

the same neihbors in an% two arranements"

Soln: he wa6s of arranging people in a circle = (!"N = NFor a person seated with 2 neighbors, the number of wa6s of that happening is 2:AC- or -CA, where C is the person in $uestion)So, for each person, we have two such arrangements in N) Since we donJt want to repeatarrangement, we divide N/2 to get !2



70. There are / co$ies of ) different boo(s. n how man% wa%s can the% be arraned

on a shelf"

; #0O/O

2; #0O):/O;

4; #0O:/O;E)

D; #0O

<; )O

Soln copies each of t6pes)

otal = 2 boo;s)otal wa6s to arrange = 2N

a;ing out repeat combos = 2N/(N # N # N # N # N" V each boo; will have copiesthat are duplicate) So we have to divide 2N -6 the repeated copies)

71. n how man% wa%s can ) rins be worn on the four finers of the riht hand"

Soln rings, fingers!st ring can be worn on an6 of the fingers = possibilities2nd ring can be worn on an6 of the fingers = possibilitiesrd ring can be worn on an6 of the fingers = possibilitiesth ring can be worn on an6 of the fingers = possibilitiesth ring can be worn on an6 of the fingers = possibilities

otal possibilities = #### = >)

7#) f both )E# and 'E' are factors of n + :#E); + :E#; + :7E'; what is the smallest

$ossible $ositive value of n"

Soln: Erite down n 3 (2>" 3 (>2" 3 (%>" as= n 3 (2>" 3 (>2" 3 (2>2" 3 (%>",= n 3 (2>%" 3 (>2" 3 (%>"

now at a minimum >2 and a is missing from this to ma;e it completel6 divisible b6>2 3 >

Pence answer = >2 3 = %

7'. @btain the sum of all $ositive inteers u$ to 1000 which are divisible b% ) and

not divisible b% #.



:1; 100)0 :#; )0)0

:'; )000 :/; )0000

Soln: .onsider ! 2 ))) &&

l = a + (n!"#d

l = && = last terma = = first termd = ! = difference

&& = + (n!"#!

thus n = ! = 5 of terms

consider ! ! 2)))) &&

&& = + (n!"#

= n = !&&

Another approach)))

Wust add up && + &' + &% + & + & + & = '2, so it has to be greater than ,and the onl6 possible choices left are !" "

Also, series is ! 2)))) &' &&

5 of terms = !

sum = (!/2"#(2# + (!!"#!" = #! =

7/. f the $robabilit% of rain on an% iven da% in cit% + is )0 what is the

$robabilit% it with rain on e+actl% ' da%s in a five da% $eriod"

81#)

##)

)1

8#)

'/ Soln Ose binomial theorem to solve the problem))))

p = !/2$ = !/25 of favorable cases = = r 5 of unfavorable cases = = 2



total cases = = n

7(probabilit6 of r out of n" = n.r#p>r#$>(nr"

7). The Full -ouse 4asino is runnin a new $romotion. <ach $erson visitin the casino has

the o$$ortunit% to $la% the Tri$ ces ame. n Tri$ ces a $la%er is randoml% dealt three

cards without re$lacement from a dec( of 8 cards. f a $la%er receives ' aces the% will

receive a free tri$ to one of 10 vacation destinations. f the dec( of 8 cards contains ' aces

what is the $robabilit% that a $la%er will win a tri$"

. 1''

2. 11#0

4. 1)

D. 17#0

<. 11//0

he probabilit6 of an event A occurring is the number of outcomes that result in A divided b6 thetotal number of possible outcomes)

here is onl6 one result that results in a win: receiving three aces)

Since the order of arrangement does not matter, the number of possible wa6s to receive cards isa combination problem)

he number of combinations of n obYects ta;en r at a time is

.(n,r" = nN/(rN(nr"N"

.('," = 'N/(N('"N"

.('," = 'N/(N(N""

.('," = 2/((!2""

.('," = 2/%2

.('," =

he number of possible outcomes is )

hus, the probabilit6 of being dealt aces is !/)

7. The Full -ouse 4asino is runnin a new $romotion. <ach $erson visitin the casino has

the o$$ortunit% to $la% the Tri$ ces ame. n Tri$ ces a $la%er is randoml% dealt three

cards without re$lacement from a dec( of 8 cards. f a $la%er receives ' aces the% will

receive a free tri$ to one of 10 vacation destinations. f the dec( of 8 cards contains ' aces

what is the $robabilit% that a $la%er will win a tri$"

. 1''

2. 11#0

4. 1)

D. 17#0

<. 11//0



Soln Since each draw doesnJt replace the cards:7rob) of getting an ace in the first draw = /'getting in the second, after first draw is ace = 2/%getting in the third after the first two draws are aces = !/thus total probabilit6 for these mutuall6 independent events = /'#2/%#!/ = !/

77. Find the $robabilit% that a / $erson committee chosen at random from a rou$

consistin of men 7 women and ) children contains

; e+actl% 1 woman

2; at least 1 woman

4; at most 1 woman

Soln

A)" %.!# !!./ !'.-" ! (!!./!'."

." (!!./!'." + (%.!#!!./!'."

78. rental car service facilit% has 10 forein cars and 1) domestic cars waitin to be

serviced on a $articular Saturda% mornin. 2ecause there are so few mechanics onl%

can be serviced.

:a; f the cars are chosen at random what is the $robabilit% that ' of the cars selected

are domestic and the other ' are forein"

:b; f the cars are chosen at random what is the $robabilit% that at most one domestic

car is selected"

Soln

A" !.#!./2.

-" 7robabilit6 of no domestic car + 7robabilit6 of ! domestic car =!./2. + !.! #!./2.

79. -ow man% $ositive inteers less than )000 are evenl% divisible b% neither 1) nor #1"

. /)1/

2. //7)

4. /)#1

D. //#8

<. /'/9

Soln Ee first determine the number of integers less than , that are evenl6 divisible b6 !)

his can be found b6 dividing ,&&& b6 !:

= ,&&&/!= integers

8ow we will determine the number of integers evenl6 divisible b6 2!:

= ,&&&/2!



= 2' integers

some numbers will be evenl6 divisible b6 -9P ! and 2!) he least common multiple of !and 2! is !) his means that ever6 number that is evenl6 divisible b6 ! will be divisible b6-9P ! and 2!) 8ow we will determine the number of integers evenl6 divisible b6 !:

= ,&&&/!= % integers

herefore the positive integers less than that are not evenl6 divisible b6 ! or 2! are &&&(+2'%"=%

80; Find the least $ositive inteer with four different $rime factors each reater than

#.

Soln ##%#!! = !!

81; From the even numbers between 1 and 9 two different even numbers are to bechosen at random. ,hat is the $robabilit% that their sum will be 8"

Soln @nitiall6 6ou have even numbers (2,,,'"6ou can get the sum of ' in two wa6s = 2 + or + 2so the first time 6ou pic; a number 6ou can pic; either 2 or ' a total of 2 choices out of' = !/2after 6ou have pic;ed 6our first number and since 6ou have alread6 pic;ed ! number 6ouare left with onl6 2 options = either (,,'" or (2,,'" and 6ou have to pic; either fromthe first set or 2 from the second and the probabilit6 of this is !/) Since these two eventshave to happen together we multipl6 them) [ # !/ = !/

8#; ) is $laced to the riht of two R diit number formin a new three R diit

number. The new number is '9# more than the oriinal two5diit number. ,hat was

the oriinal two5diit number"

Soln @f the original number has 3 as the tens digit and 6 as the ones digit (3 and 6 areintegers less than !" then we can set up the e$uation:!3 + !6 + = !3 + 6 + &2&3 + &6 = '%&(!3+6" = '%!3 + 6 = == 3 = , 6 =

the original number is , the new number is

8'; f one number is chosen at random from the first 1000 $ositive inteers what is

the $robabilit% that the number chosen is multi$le of both # and 8"

Soln An6 multiple of ' is also a multiple of 2 so we need to find the multiples of ' from to !the first one is ' and the last one is !



== ((!'"/'" + ! = !2== p (pic;ing a multiple of 2 '" = !2/! = !/'

8/; serial set consists of A bulbs. The serial set lihts u$ onl% if all the A bulbs are

in wor(in condition. <ven if one of the bulbs fails then the entire set fails. The

$robabilit% of a bulb failin is +. ,hat is the $robabilit% of the serial set failin"

Soln 7robabilit6 of 3 to fail) 7robabilit6 of a bulb not failing = !3 probabilit6 that none of the 8 bulbs fail, hence serial set not failing = (!3">8 probabilit6 of serial set failing = !(!3">8

8); 2rad fli$s a two5sided coin 8 times. ,hat is the $robabilit% that he ets tails on

at least 7 of the 8 fli$s"

1'#

11

18

78

none of the above

Soln 8umber of wa6s % tails can turn up = '.%the probabilit6 of those is !/2 eachSince the $uestion as;s for at least %, we need to find the prob of all ' tails the numberof wa6s is '.' = !Add the two probabilities'.%#(!/2">' = '/2>' for getting %7rob of getting ' tails = !/2>'otal prob = '/2>'+!/2>' = &/2>'

Ans is 1)

8. $hotora$her will arrane $eo$le of different heihts for $hotora$h b%

$lacin them in two rows of three so that each $erson in the first row is standin in

front of someone in the second row. The heihts of the $eo$le within each row must

increase from left to riht and each $erson in the second row must be taller than the

$erson standin in front of him or her. -ow man% such arranements of the

$eo$le are $ossible"

. )

2.

4. 9

D. #/



<. '

Soln wa6s

!2 !2 !2 ! !

2 2

87. s a $art of a ame four $eo$le each much secretl% chose an inteer between

1 and / inclusive. ,hat is the a$$ro+imate li(elihood that all four $eo$le will

chose different numbers"

Soln he probabilit6 that the first person will pic; uni$ue number is ! (obviousl6"then the probabilit6 for the second is / since one number is alread6 pic;ed b6 thefirst, then similarl6 the probabilities for the rd and th are !/2 and !/ respectivel6)heir product /#!/2#!/ = /2

88. ,hich of the sets of numbers can be used as the lenths of the sides of atrianle"

. U)71#V

. U#/10V

. U)79V

. onl%

2. onl%

4. and onl%

D. and onl%

<. and onl%

Soln: 8or any side of a triangle. ts length must be greater than the difference

bet+een the other t+o sides, but less than the sum of the other t+o sides.

Answer is -

89. clothin manufacturer has determined that she can sell 100 suits a wee( at a

sellin $rice of #00& each. For each rise of /& in the sellin $rice she will sell # less

suits a wee(. f she sells the suits for +& each how man% dollars a wee( will she

receive from sales of the suits"

Soln Bet 6 be the number of * increases she ma;es, and let S be the number of suits shesells) henC = 2 + 6 == 6 = 3/ S = ! 26 == S = ! 23/ G = ! 3/2 + ! = 2 3/2

so the answer is that the number of suits sheJll sell is 2 3/2

90. certain $ortfolio consisted of ) stoc(s $riced at &#0 &') &/0 &/) and &70

res$ectivel%. @n a iven da% the $rice of one stoc( increased b% 1) while the



$rice of another decreased b% ') and the $rices of the remainin three remained

constant. f the averae $rice of a stoc( in the $ortfolio rose b% a$$ro+imatel% #

which of the followin could be the $rices of the shares that remained constant"

; #0 ') 70

2; #0 /) 70

4; #0 ') /0

D; ') /0 70

<; ') /0 /)

Soln Add the prices together:2 + + + + % = 2!

2? of that is 2! 3 )2 = )2

Bet 3 be the stoc; that rises and 6 be the stoc; that falls)

)!3 )6 = )2 == 3 = (%/"6 + 2%

his tells us that the difference between 3 and 6 has to be at least 2%) Since the answerchoices list the ones that 098 change, we need to loo; for an answer choice in whichthe numbers 89 listed have a difference of at least 2%)

hus the answer is (1"

91. if 5#!W+!W# and 'W!%W!8 which of the followin re$resents the rane of all

$ossible values of %5+"

:; )W!%5+W!

:2; 1W!%5+W!)

:4; 1W!%5+W!

:D; 1W!%5+W!10

:<; 1W!%5+W!10

Soln you can easily solve this by subtracting the t+o ine9ualities. To do this they

need to be in the opposite direction: +hen you subtract them preserve the sign of

the ine9uality from +hich you are subtracting.

# ; y ; $

multiply the second one by (! to reverse the sign' < & < '

3ubtract them to get# ' ; y & ; $ ('

! ; y & ; !0



9#. f a group of %6# people who purchased stocks, 61 purchased A,

purchased $, &6 purchased ', )& purchased (, 6# purchased *. what is thegreatest possible number of the people who purchased both $ and ("

A:2# $:&# ':6# (:)& *:#

3oln: A=3>?@ is (/

since they have asked us to find out the greatest possible number buying both B as

+ell as , the ans+er has to be the smallest no bet+een the t+o +hich is /...as allthe guys purchasing can also buy B and only / out of $$ purchasing B can

simultaneously purchase as +ell....

+. There are +# people and + clubs 4, 3, and 5 in a company. 1# people oined 4, 1% people oined 3 and & people oined 5. !f the members of 4 did

not oin any other club, at most, how many people of the company did not oin any club"

A: 2 $: & ': 6 (: ) *:

Soln total no of people 2 #0

no 5oining 2 !0

no 5oining 3 2 !'no 5oining C 2 /

9uestion asked AT 43T ho+ many people did not 5oin any groupD

solution1 no+ since none of the members of 5oined any other group, the no of

people left 2 #0!0(for 2'0since the 9uestion says at most ho+ many did not 5oin any group, lets assume the all

people +ho 5oin C also 5oin 3. so no of people 5oining group 3 and C are !' (notethat there +ill be / people in group 3 +ho have also 5oined C

therefore no of people not 5oining any group 2 '0!'2$Eence e

9/. Find the numbers of wa%s in which / bo%s and / irls can be seated alternativel%.

1; in a row

#; in a row and there is a bo% named Kohn and a irl named Susan amonst the

rou$ who cannot be $ut in adHacent seats

'; around a table

A:!" N # N # 22" N # N # 2 number of wa6s with Wohn and Susan sitting together = (N # N # 2" (% # N # N #2"

he wa6 that WS arrangements are found is b6



brac;eting W and S and considering it to be a single entit6) So a possible arrangement is (-=bo6, X=girl"

(WS" - X - X - X number of arrangements is % 3 N 3 N = 22(SW" X - X - X - number of arrangements is % 3 N 3 N = 22

" Fi3 one bo6 and arrange the other bo6s in N wa6s) Arrange the girls in N wa6s in the gaps between the bo6s)

otal arrangements = N 3 N

= 3 2

= !

9). From a rou$ of ' bo%s and ' irls / children are to be randoml% selected. ,hatis the $robabilit% that e=ual numbers of bo%s and irls will be selected"

. 110

2. /9

4. 1#

D. ')

<. #'

Soln otal number of wa6s of selecting children = . = !

with e$ual bo6s and girls) = 2 bo6s and 2 girls) = .2 # .2 = &)

Pence p = &/! = /

9. hat is the reminder when 71 8 7% 8 7+ 8 ...... 8 7 is divided by

6"

91 #

9% +9+ 292 %

9& /one of these

3oln: @emainder of *odd )" is #remainder of *even)" is 0

F! F' F# ...... F2*(!F'.....F$



!F'.....F$ is odd.

Thus +e obtain # as a remainder +hen +e divide *(!F'.....F$ by ".

Another way1 >e get the value as *odd)" 2 #*odd)' 3ince #*odd 2 odd: odd)' 2

GGGG./so something divided by ", gives GGGG./, hence remainder is "*0./ 2 #

). 8ind the value of !.!7 8 %.%; 8 +.+; 8 ......8n.n;

91 n; 81

9% 9n81;9+ 9n81;-1

92 9n81;819& /one of these

3oln1 1.1; '.'7 #.#7 ......n.n7

2!.!7 (#!'7 (4!#7 ...... ((n!! n7

2!.!7#7'747#7.......(n!7n7

3o it is (n!7 ! (Ans+er choice 4

. The numbers x and y are three-digit positive integers, and x 8 y is a

four-digit integer. The tens digit of x e<uals ) and the tens digit of y e<uals&. !f x = y , which of the following must be true"

!. The units digit of x 8 y is greater than the units digit of either x or y .!!. The tens digit of x 8 y e<uals %.

!!!. The hundreds digit of y is at least &.

A. !! only

$. !!! only

'. ! and !!

(. ! and !!!

*. !! and !!!

3oln: &2 abcy2 def

& 2 ac

y2 b/f

& < y and &y 2 +&yH.

. The units digit of x y is greater than the units digit of either x or y .

t can carryover one digit. 8alse



. The tens digit of x y e9uals '.

t can be ' or #. 8alse

. The hundreds digit of y is at least /.

ab! <2 !0a <b so a at least /. True.

Ans1 b

. mon ) children there are # siblins. n how man% wa%s can the children be

seated in a row so that the siblins do not sit toether"

:; '8

:2; /

:4; 7#

:D; 8

:<; 10#3oln1 he total number of wa6s of them can sit is !2when the siblings sit together the6 can be counted as one entit6therefore the number of wa6s that the6 sit together is N=2, but sincethe two siblings can sit in two different wa6s e)g) A- and -A we multipl6 2 b6 2 to getthe total number of wa6s in which the children can sit together with the siblings sittingtogether '@n other words 7#272the rest is obvious !2'=%2

1##. There are 70 students in Math or <nlish or Berman. <+actl% /0 are in Math'0 in Berman ') in <nlish and 1) in all three courses. -ow man% students are

enrolled in e+actl% two of the courses" Math <nlish and Berman.

Soln Lu1uX = L + 1 + X Ln1 LnX 1nX 2(Ln1nX"Ln1 + LnX + 1nX = L + 1 + X 2(Ln1nX" Lu1uXLn1 + LnX+ 1nX = + + 2(!" % = ! % =

Ehenever an intersection occurs between 2 sets, (Ln1nX" is counted twice, therefore6ou deduct one of it) @f the intersection occurs between sets, it is counted thrice4therefore 6ou deduct two of it) And so forth)

!f there are four sets, then the formula is A 8 $ 8 ' 8 ( -9two -9three% -9four+ 0 total

101. Kohn can com$lete a iven tas( in #0 da%s. Kane will ta(e onl% 1# da%s to

com$lete the same tas(. Kohn and Kane set out to com$lete the tas( b% beinnin to

wor( toether. -owever Kane was indis$osed / da%s before the wor( ot over. n

how man% da%s did the wor( et over from the time Kohn and Kane started to wor(

on it toether"



5 25 10 45 8 D5 7.) <5 '.)

Soln ogether the6 do !/2+!/!2=/ of the tas;)

Wane and Wohn started the wor; together, but onl6 Wohn finished the wor; because Wanegets sic;) So let 3 be the number of da6s the6 wor;ed together)

3#/+#!/2=!

3/=/ and therefore 3 =

So in total the6 wor;ed da6s on it together and Wohn wor;ed da6s on it) So total da6sspent=!, but if the $uestion is as;ing how man6 time did the6 spend wor;ing on the proYect together, then the answer is )

10#. Kane ave Garen a ) m head start in a 100 race and Kane was beaten b% 0.#)m.

n how man% meters more would Kane have overta(en Garen"

Soln Wane gave \aren a m head start means \aren was m ahead of Wane) So after thelead, \aren ran &m and Wane ran &&)% m when the race ended)

Bet speed of \aren and Wane be \ and W respectivel6 and lets sa6 after C minutes Waneoverta;es \aren)

!st condition: &/\ = &&)%/W2nd condition WC \C = )2

Solving for WC, we get WC=2!/)Pence Wane needs to run )2m more (or total of !m" to overta;e \aren)

10'. ut of %# surveyed students study math and )

study both math and physics. !f 1# students do notstudy either of these subects, how many students

study physics but not math"

9A 19$ %

9' 2

9( &

9* 63oln: total 2 gr! gr' both neither

'0 2 $ - !0- 2 , students study both - and , study both, 2 ' study only -

10/. 4achine A can produce &# components a day while

machine $ only 2#. The monthly maintenance cost formachine A is >1&## while that for machine $ is >&&#. !f



each component generates an income of >1# what is theleast number of days per month that the plant has to work

to ustify the usage of machine A instead of machine $"

9A 69$ )

9' 9( 1#

9* 11

3oln: Iet & be the number of days that need to be +orked/00& !/00 < 400& //0

!00& < /0& < ./ (

10). Four cu$s of mil( are to be $oured into a #5cu$ bottle and a /5cu$ bottle. f

each bottle is to be filled to the same fraction of its ca$acit% how man% cu$s of mil(

should be $oured into the /5cu$ bottle"

. #'

2. 7'

4.)#

D. 8'

<. '

Soln & y 2 4

& 2 4 y

&)' 2 y)44& 2 'y

4(4y 2 'y!" 2 "yy 2 $)# (

10. f 10 $ersons meet at a reunion and each $erson sha(es hands e+actl% once with

each of the others what is the total number of handsha(es"

:; 10O

:2; 10*10

:4; 10*9

:D; /)

:<; '

Soln here are !.2 wa6s to pic; 2 different people out of ! people)



!.2 = !N/2N'N = (0"

107. f o$eration & is defined as

&J ! J 3 # if J is even

&J ! J 5 1 if J is odd

what is &:...&:&:&:1);;;...; 99 times"

:; 1#0

:2; 180

:4; #10

:D; ##)

:<; #)0

Soln: *! gives !

After that it is an A7 with d=2, a=! and n =&&

108. and 2 alternatel% toss a coin. The first one to turn u$ a head wins. if no more

than five tosses each are allowed for a sinle ame.

15 Find the $robabilit% that the $erson who tosses first will win the ame"

#5 ,hat are the odds aainst s losin if she oes first"

Soln look at the conditions: it says that the first person +ho tosses a head +ins.

IetJs say A tosses first.

+hat is the probability that he +ins

E TTE TTTTE TTTTTTE TTTTTTTTE

i.e. either the first toss is head,

or the first time A tosses the coin he gets a tail and B also gets a tail , n in thesecond thro+ A gets a head.....

This continues for a ma& till / thro+s, because the game is for / thro+s only.

So, !. !)' (!)'F# (!)'F/ (!)'F (!)'F

'. (!)'F' (!)'F4 (!)'F" (!)'F$ (!)'F!0

109. -ow man% inteers less than 1000 have no factors :other than 1; in common

with 1000"

:1; /00

:#; /10

:'; /11



:/; /1#

:); Aone of the above

Soln !000 multiples of ' and)or /

multiples of ' 2 /00 (all even K

multiples of / 2 (/ /)!0 ! L sing A- formulaM2 !00

Ans+er 2 !000 (/00 !00

2 400

Nou cannot calculate for all multiples of / because you have already removed all evenintegers (including !0, '0, and #0. The difference in the A- series should be !0

instead of / because youOre looking for the integers that have / as a unitJs digit.Therefore +e divide by !0 and not /.

11#. Two different numbers when divided by the same divisor left

remainders of 11 and %1 respectively. hen the numbers? sum was dividedby the same divisor, the remainder was 2. hat was the divisor"

+6, %, 1%, or none

3oln: Iet the divisor be a.

& 2 a*n !! (!

y 2 a*m '! ('also given, (&y 2 a*p 4 (#

adding the first ' e9uations. (&y 2 a*(nm #' (4

e9uate # and 4.

a*p 4 2 a*(nm #'ora*p 4 2 La*(nm '$M 4

cancel 4 on both sides.u +ill end up +ith.

a*p 2 a*(nm '$.

+hich implies that '$ should be divisible by a. or in short a 2 '$ +orks.

Another method: think the easiest (not necessarily the shortest, +ay to solve this is to use given

ans+er choices. 3ince the remainders are given as !! and '!, therefore the divisor

has to be greater than '! +hich leaves +ith t+o choices '$ and #". Try '$ first: letthe t+o numbers be '$!!2 # and '$'!2 4. 3umming them up and dividing by

'$ gives (4#2$$, $$)'$ remainder is 4, satisfies the given conditions. 6heck for#" +ith same approach, does not +ork, ans+er is '$

111. There are 8 members6 amon them are Gell% and 2en. committee of / is to be

chosen out of the 8. ,hat is the $robabilit% that 2en is chosen to be in the committee

and Gell% is not"



Soln letJs assume Ben has already been chosen. Then have to choose # more

people from the remaining, e&cluding Pelly, that is, three from si& people, thatJs "c#.so the total is (!c!."c#)$c4 +hich is ')

11#. -ow man% )5diit $ositive inteers e+ist where no two consecutive diits are the

same"

.; 9*9*8*7*

2.; 9*9*8*8*8

4.; 9E)

D.; 9*8E/

<.; 10*9E/

Soln: . is correct)he first place has & possibilities, since is not to be counted) All others have & each,since 6ou cannot have the digit, which is same as the preceding one)Pence &>

11+. @ow many five digit numbers can be formed using the digits #, 1, %, +,2 and & which are divisible by +, without repeating the digits"

9A 1&9$ 6

9' %169( 1%#

9* 1#

Soln The sum of digits of a multiple of # should be div by #.

for a / digit number to be div by #, the sum of digits (given the digits here can be

only !' or !/.

8or a sum of !', the digits that can be used 1 0,!,',4,/for a sum of !/1 !,',#,4,/=umber of numbers from the first set 2 4.47 (0 cannot be the first digit in the

numbersfor the second set 1 /7

total 2 /7 4.47 2 47(/4 2 '4* 2 '!"

3ince 0 cannot be the first digit of a number, for the first position, you have 4 choices

(all digits e&cept Hero. =o such constraints e&ist for the rest of the positions: hencethe ne&t choices are 4,#,',! all multiplying up to give a 47. Ead there been no 0

involved, the choices +ouldOve been /7 nstead of 4.47

112. A group of friends want to play doubles tennis. @ow many different

ways can the group be divided into 2 teams of % people"

A. 2%#

$. %&%#'. 16

(. #*. 1#&



3oln: out of $ people one team can be formed in $c' +ays.

$c'*"c'*4c'*'c'2 '/'0.The ans+er is !0/. ivide '/'0 by 47 to remove the multiples ( for e&ample1 (A,B is

same as ( B,A

11&. 4y name is A**T. $ut my son accidentally types the name byinterchanging a pair of letters in my name. hat is the probability that

despite this interchange, the name remains unchanged"

a /% b !0%

c '0% d '/%

3oln1 there are actually '0 +ays to interchange the letters, namely, the first letter

could be one of /, and the other letter could be one of 4 left. 3o total pairs byproduct rule 2 '0.

=o+, there are t+o cases +hen it +ouldnJt change the name. 8irst, keeping them allthe same. 3econd, interchanging the t+o ??s together. Thus ' options +ould leavethe name intact.

-rob 2 ')'0 2 0.!, or !0%.

116. A certain right triangle has sides of length x , y , and z , where x = y = z .

!f the area of this triangular region is 1, which of the following indicates allof the possible values of y "

A. y B CT%

B. ROOT3/2 = y = CT%

C. ROOT2/3 = y = CT+D%

D. ROOT3/4 = y = CT%D+

E. y = CT+D2

Soln: right triangle with sides 3<6<K and area of ! = K = h6potenuse and 36/2 = !i)e 36 = 2

@f 3 were e$ual to 6, we would have had 36 = 6>2 = 2) And 6 = root2

-ut, 3<6 and so 6root2)

117. A certain deck of cards contains 2 blue cards, 2 red cards, 2 yellow cards, and 2 greencards. If two cards are randomly drawn from the deck, what is the probability that they willboth are not blue?

A. 152!

http://businessmajors.about.com/cs/quizzes/a/ps1000000i.htm

http://businessmajors.about.com/cs/quizzes/a/ps1000000a.htm




http://businessmajors.about.com/cs/quizzes/a/ps1000000a.htm





". 1#

$. %1&

'. 1(2

). 11&

3oln1 .hance of drawing a blue on the first draw = 2/', so chance of not drawing a blueon the first draw is /'

similarl6 chance of not drawing blue on second draw = /%

herefore probabilit6 of not drawing blue in 2 draws = /'#/% = !/2'

11!. *ow many integers between 1++ and 15+, inclusie can be eenly diided by neither (

nor 5?

Soln =umber of integers that divide #1

the range is !00!/0@elevant to this case, +e take !0' !/0 (since !0' is the first to div #

!0' 2 #4*#!/02 /0*#, so +e have /0#4! 2 ! multiples of #

8or multiples of /,

!002/*'0!/02/*#0

#0'0! 2!!

=o+ +e have a total of ' integers, but +e double counted the ones that divideBQTE # A= /, ie !/.

!0/ is the first to divide !/.

!0/2!/*!/02!/*!0

!0! 2 4 integers

3o our total is !!!4 2 '4 integers that can be divided by either # or / or both.

/! integers '4 integers 2 ' that cannot be evenly divided.

11. 3eed miEture X is 2# percent ryegrass and 6# percent bluegrass byweightF seed miEture Y is %& percent ryegrass and )& percent fescue. !f a

miEture of X and Y contains +# percent ryegrass, what percent of the weightof this miEture is X "

9A 1#G

9$ ++ 1D+G



9' 2#G

9( &#G

9* 66 %D+G

3oln: (&y#0)!00 2 &*40)!00 y*'/)!00#0& #0y 2 40& '/yy 2 '& or y)& 2 ')! or y1& 2 '1! hence & 2 ## !)#%

1%#. Se=uence and 2. a1!1 b1!(. an!b:n51;5a:n51; bn!b:n51;3a:n51;. ,hat is

a/!"

Soln: a2 = ;! 4 b2 = ;+!

a= (;+!"(;!" = 2 4 b = (;+!"+(;!" = 2;

a = 2; 2 = 2(;!" = 2(b!a!"

1%1. A person put 1### dollars in a bank at a compound interest 6 years

ago. hat percentage of the initial sum is the interest if after the first three

years the accrued interest amounted to 1G of the initial sum"A +G $ 2%G ' 1G ( 2#G

3oln: assume, interest 2 r

so after # years total money 2 !000*(!rF# 2 !000*!.!(!rF# 2 !.!

so after " years total money 2 !000*(!rF" 2 !000*!.!F' 2 !000*!.4'so percentage of interest is 4'%

122. A, " and $ run around a circular track of length 75+m at speeds of ( msec, & msecand 1! msec respectiely. If all three start from the same point, simultaneously and run inthe same direction, when will they meet for the first time after they start the race?

A. 75+ seconds". 5+ seconds$. 25+ seconds'. (75 seconds). 75 seconds

-oln >hen t+o people are running in the same direction the relative speed is a

difference in speeds of the t+o people.

n this case A2# B2" 62!$

3o relative speed of B +rt A is "# 2 #m)s

@elative speed of A +rt to 6 is !$# 2!/m)s

Therefore relative distances +ill be1B +rt A is /0)# 2'/0

6 +rt to A /0)!/ 2 /0



3o they have to bridge this distance of '/0 and /0 bet+een them +hich is the I6of '/0 and /0 +hich is '/0.

Another /ethod 3imply put, @unner AOs time take to run one lap is '/0

@unner BOs time is !'/sand @unner 6Os time is 4!."s

>e can notice that A2'B

and thet B2#6

3o +hen '/0 s elapse, they +ill be at their starting point. A +ill have completed onelap, B ' laps, and 6 " laps

1%+. H percents of the rooms are suits, I percent of the rooms are painted

light blue. hich of the following best represents the least percentage ofthe light blue painted suits"

1 H-I

%I-H 81##

+1##H-I2H8I-1##e1##-HI

3oln: ?9uation from set theory1

n(AB2n(An(Bn(AFB

+here,

A2 % of rooms +hich are suites

B2 % of rooms painted blue

AFB means the intersection of the t+o sets

=o+ in this case, +hat +e need to find is n(AFB, therefore

n(AFB2n(An(Bn(AB2 G N n(AB

=o+ this +ould be least +hen n(AB is ma&imum, +hich +ould happen if these t+o

kinds of rooms are only t+o kinds available, making n(AB2!00

Therefore the ans+er should be GN!00

-'00



PelloL6 name is Ld) Dana, where Ld) is an abbreviated form of Lohammad) L6 previouscertificates (-achelors, Secondar6 and Pigher Secondar6" were issued as Ld) Dana) As itis mandator6 in m6 countr6 to write name in full form on passport, @ had to write theelaborated form of Ld) (Lohammad" on m6 passport, accordingl6 also on m6 official

XLA account) Are the business schools going to accept this slight discrepanc6 @ have between m6 XLA account and certificates nameT L6 e3am is on 2'th Wanuar6 2!)7lease suggest me what @ should do)

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gmat MATH tough problems.doc