7/23/2019 gmat MATH tough problems.doc http://slidepdf.com/reader/full/gmat-math-tough-problemsdoc 1/52 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$$ro+imate $rice when a bushel of corn costs the same amount as a $ec( of wheat" (A" *) (-" *)! (." *) (0" *) (1" *) Soln: 2 + 3 = ' )!34 3 = 5 of da6s after price is same4 solving for 34 3 is appro3imatel6 ', thus the re$uired price is 2 + # ' = cents = *) '. -ow man% randoml% assembled $eo$le do u need to have a better than )0 $rob. that at least 1 of them was born in a lea$ %ear" Soln: 7rob) of a randoml6 selected person to have 89 been born in a leap 6r = / a;e 2 people, probabilit6 that none of them was born in a leap = /#/ = &/!) he probabilit6 at least one born in leap = ! &/! = %/! < ) a;e people, probabilit6 that none born in leap 6ear = /#/#/ = 2%/) he probabilit6 that at least one born = ! 2%/ = %/ ) hus min people are needed) /. n a bas(etball contest $la%ers must ma(e 10 free throws. ssumin a $la%er has 90 chance of ma(in each of his shots how li(el% is it that he will ma(e all of his first 10 shots" Ans: he probabilit6 of ma;ing all of his first ! shots is given b6 (&/!"# (&/!"# (&/!"# (&/!"# (&/!"# (&/!"# (&/!"# (&/!"# (&/!"# (&/!" = (&/!">! = )' = ?
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1. The sum of the even numbers between 1 and n is 79*80 where n is an odd
number then n!"
Sol: First term a=2, common difference d=2 since even number
therefore sum to first n numbers of Arithmetic progression would be
n/2(2a+(n!"d"
= n/2(2#2+(n!"#2"=n(n+!" and this is e$ual to %&#'
therefore n=%& which is odd)))
#. The $rice of a bushel of corn is currentl% &'.#0 and the $rice of a $ec( of
wheat is &).80. The $rice of corn is increasin at a constant rate of )+ cents
$er da% while the $rice of wheat is decreasin at a constant rate of
cents $er da%. ,hat is the a$$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
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 $robabilit% for n*:n31;*:n3#; bein
divisible b% 8" #) )0 #.) 7#.) 7)
Soln: 1 = n#(n+!"#(n+2"
1 is divisible b6 ', if n is even) 8o of even numbers (between ! and &" = '
1 is divisible b6 ', also when n = '; ! (; = !,2,,)))))"Such numbers total = !2(%,!,))))"
Adding the two probabilities: probabilit6 = %/' = %/2
1). T,@ cou$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
&#)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 $ositive inteer constant. For what value of n is the ratio
Sol he method @ followed was to reduce the Q to (3>/ 3 " # ( M/M"the e$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
+ !+ 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
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
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
'#. 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"
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.
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
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
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
)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+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
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
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!:
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
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 $riced at � &') &/0 &/) and &70
res$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$$ro+imatel% #
which of the followin could be the $rices of the shares that remained constant"
; #0 ') 70
2; #0 /) 70
4; #0 ') /0
D; ') /0 70
<; ') /0 /)
Soln Add the prices together:2 + + + + % = 2!
2? of that is 2! 3 )2 = )2
Bet 3 be the stoc; that rises and 6 be the stoc; that falls)
)!3 )6 = )2 == 3 = (%/"6 + 2%
his tells us that the difference between 3 and 6 has to be at least 2%) Since the answerchoices list the ones that 098 change, we need to loo; for an answer choice in whichthe numbers 89 listed have a difference of at least 2%)
hus the answer is (1"
91. if 5#!W+!W# and 'W!%W!8 which of the followin re$resents the rane of all
$ossible values of %5+"
:; )W!%5+W!
:2; 1W!%5+W!)
:4; 1W!%5+W!
:D; 1W!%5+W!10
:<; 1W!%5+W!10
Soln you can easily solve this by subtracting the t+o ine9ualities. To do this they
need to be in the opposite direction: +hen you subtract them preserve the sign of
the ine9uality from +hich you are subtracting.
# ; y ; $
multiply the second one by (! to reverse the sign' < & < '
9#. f a group of %6# people who purchased stocks, 61 purchased A,
purchased $, &6 purchased ', )& purchased (, 6# purchased *. what is thegreatest possible number of the people who purchased both $ and ("
A:2# $:&# ':6# (:)& *:#
3oln: A=3>?@ is (/
since they have asked us to find out the greatest possible number buying both B as
+ell as , the ans+er has to be the smallest no bet+een the t+o +hich is /...as allthe guys purchasing can also buy B and only / out of $$ 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 $ut in adHacent seats
'; around a table
A:!" N # N # 22" N # N # 2 number of wa6s with Wohn and Susan sitting together = (N # N # 2" (% # N # N #2"
brac;eting W and S and considering it to be a single entit6) So a possible arrangement is (-=bo6, X=girl"
(WS" - X - X - X number of arrangements is % 3 N 3 N = 22(SW" X - X - X - number of arrangements is % 3 N 3 N = 22
" Fi3 one bo6 and arrange the other bo6s in N wa6s) Arrange the girls in N wa6s in the gaps between the bo6s)
otal arrangements = N 3 N
= 3 2
= !
9). From a rou$ 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
. 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(
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 $uestion is as;ing how man6 time did the6 spend wor;ing on the proYect together, then the answer is )
10#. Kane ave Garen a ) m head start in a 100 race and Kane was beaten b% 0.#)m.
n how man% meters more would Kane have overta(en Garen"
Soln Wane gave \aren a m head start means \aren was m ahead of Wane) So after thelead, \aren ran &m and Wane ran &&)% m when the race ended)
Bet speed of \aren and Wane be \ and W respectivel6 and lets sa6 after C minutes Waneoverta;es \aren)
multiples of / 2 (/ /)!0 ! L sing A- formulaM2 !00
Ans+er 2 !000 (/00 !00
2 400
Nou cannot calculate for all multiples of / because you have already removed all evenintegers (including !0, '0, and #0. The difference in the A- series should be !0
instead of / because youOre looking for the integers that have / as a unitJs digit.Therefore +e divide by !0 and not /.
11#. Two different numbers when divided by the same divisor left
remainders of 11 and %1 respectively. hen the numbers? sum was dividedby the same divisor, the remainder was 2. hat was the divisor"
+6, %, 1%, or none
3oln: Iet the divisor be a.
& 2 a*n !! (!
y 2 a*m '! ('also given, (&y 2 a*p 4 (#
adding the first ' e9uations. (&y 2 a*(nm #' (4
e9uate # and 4.
a*p 4 2 a*(nm #'ora*p 4 2 La*(nm '$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 $robabilit% that 2en is chosen to be in the committee
Soln letJs assume Ben has already been chosen. Then have to choose # more
people from the remaining, e&cluding Pelly, that is, three from si& people, thatJs "c#.so the total is (!c!."c#)$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"
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&le1 (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?
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?
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)