GRE Question of the Day: Algebra

avinash
#1. June 26th, 2006, at 1:10 AM.

t cn hve only one value n tht is 0…

dhaval
#2. June 26th, 2006, at 9:07 AM.

answer shud be zero.

Rahul
#4. July 6th, 2006, at 9:02 PM.

‘t’ can have only one value i.e. +1, Hence i’ll go for option A)

neeths
#5. July 6th, 2006, at 9:48 PM.

A)

Gurleen
#6. July 6th, 2006, at 10:56 PM.

A

nutboltoo
#7. July 7th, 2006, at 1:16 AM.

AAA

GRE Tester
#8. July 7th, 2006, at 1:22 AM.

But if you put the 1 in th equation it does not satisfy.
So a can not be the answer.

nami
#9. July 7th, 2006, at 12:23 PM.

if we plug in -1 or 1 either the nr or dr turns out to be 0…choices a,b,c kicked out….
choice E…any infinite values doesnt satisfy the equation to be equal to 2
so according to my work out it has to be choice D..no values….
but i still dont understand even if the value of t is 0 we dont satisfy the equation to be equal to 2…..

can we have the answer……

nutboltoo
#10. July 7th, 2006, at 12:44 PM.

moderator………please help

Devesh
#11. July 7th, 2006, at 4:54 PM.

D is the ans
team confirm it plz

eswar
#12. July 7th, 2006, at 5:17 PM.

A is the answer..

rakeshbusa
#14. July 7th, 2006, at 8:32 PM.

A

nami
#15. July 8th, 2006, at 10:50 AM.

team…can we have te answer

John
#17. July 9th, 2006, at 5:07 AM.

a

John
#18. July 9th, 2006, at 5:08 AM.

oops

PRAVEENKUMAR VANAM
#19. July 9th, 2006, at 11:23 AM.

GOOD SIMPLE TRICKY QUESTION

veenit
#20. August 6th, 2006, at 6:59 PM.

a is answer explain ….t^2-2t+1….solve it ..u get t-1 =0 so 1 is ans

GateKeepeR
#22. May 6th, 2007, at 6:19 PM.

To GRE Team

I think that the question is fundementally wrong since the soultion set of t does not satisfy the equation at all.

junaidali
#23. June 2nd, 2007, at 8:28 PM.

d

JKR
#24. June 7th, 2007, at 11:33 AM.

A

vikas_210
#25. June 15th, 2007, at 10:30 PM.

((t^2)-1)/(t-1)=2
=> (t^2)-1 = 2(t-1)
=> t^2 = 2t-2+1
=> t^2 = 2t -1
=> t^2 - 2t + 1
=> (t-1)^2 = 0
=> t=1

ankur
#26. June 16th, 2007, at 1:16 AM.

only 1 value 1 is possible since (t^2)-1 = (t+1)*(t-1) and since denominator has (t-1) that leaves us with left hand side term being t+1 which should be equal to 2 the (RHS) Therefore t=1
QED

Rishi
#27. June 23rd, 2007, at 9:50 PM.

((t^2)-1)/(t-1)=2
denominator is t-1 => t belongs to R - 1
=> (t+1)(t-1)/(t-1)=2
=> t + 1 = 2
=> t = 1 but t belongs to R - 1
therefore no values of t
ans:- D.

crafter202
#28. July 3rd, 2007, at 9:28 PM.

Good Question ..

yup , The Answer is “D” ..

Geetika
#29. July 16th, 2007, at 9:57 PM.

dear take gre team pls check the question coz according to factorization solving it should take one value n it is not satisfying the equation …….. check your question.

abc
#31. July 17th, 2007, at 10:11 PM.

its A

Geetika
#32. July 19th, 2007, at 10:03 PM.

if question is right then d has to b the answer

jsdhjd
#33. July 20th, 2007, at 9:37 PM.

ans is A coz
(t-1)(t+1)/t-1=2
t+1=2
t=2-1
t=1
so ans is A

l8474545
#34. July 28th, 2007, at 4:29 PM.

d

deepthi
#35. July 31st, 2007, at 10:08 AM.

the main tip we should follow is BODMAS rule
i.e first solve bracket no :1 i.e (t^2-1)i.e t-1* t+1 then follow division so that t-1 cancels so left is t+1=2 so t=1
i think A is option…..

anyl
#36. July 31st, 2007, at 11:33 PM.

answer is D coz if t=1,t-1 will become zero

anyl
#37. July 31st, 2007, at 11:36 PM.

if we follow BODMAS we will get it as 1 but if 1 is substituted in the equation it should satisfy the value 2

amar
#38. August 13th, 2007, at 6:24 PM.

((t^2)-1)/(t-1)=2
t^2-1=2t-2
t^2-2t+1=0
t^2-t-t+1=0
(t(t-1)-(t-1))=0
(t-1)(t-1)=0
so we get….
t=1,1
therefore we have two roots

amar
#39. August 13th, 2007, at 6:27 PM.

((t^2)-1)/(t-1)=2
t^2-1=2t-2
t^2-2t+1=0
t^2-t-t+1=0
(t(t-1)-(t-1))=0
(t-1)(t-1)=0
so we get….
t=1,1
therefore we have two roots
anwser is c

MANAN SANGHVI
#40. August 20th, 2007, at 2:46 AM.

I SUPPOSE THE CORRECT ANS IS
“A”

Kranthi
#41. August 25th, 2007, at 2:48 PM.

A is xact answer

Dhairyashil
#42. August 26th, 2007, at 7:28 PM.

Hi……..
The ans must be [A]

venu mandepudi
#43. September 19th, 2007, at 5:04 PM.

answer is A
t can take only+1

guptarulz
#44. September 25th, 2007, at 3:20 PM.

A

priyamg
#45. October 11th, 2007, at 1:04 PM.

it would be ‘A’ so can u plz conform the answer?

max
#46. October 23rd, 2007, at 11:04 AM.

my answer: A

((t^2)-1) is (t+1)(t-1)

so we have (t+1)=2
hence t= 2-1
thus t=1

masoud
#47. October 23rd, 2007, at 3:37 PM.

if you solve the problem the answer is 1 BUT!!!!
you CAN NOT replace t by 1 because the denominator can not be 0
so the answer is D

Usaf
#48. November 3rd, 2007, at 10:53 AM.

its D …nothing else is right when place back in the equation

Bibek
#49. December 7th, 2007, at 12:14 AM.

D. Qd equation having same root (1 &1)is not possible

siri
#50. December 13th, 2007, at 8:42 PM.

i think the ans is a

sand
#51. December 22nd, 2007, at 3:47 PM.

d

HareRam
#52. December 27th, 2007, at 2:33 AM.

comon guys Its D

mitali dalvi
#53. January 13th, 2008, at 12:08 AM.

i think it’s a.

Deepali
#54. January 28th, 2008, at 9:55 PM.

Answer : A) 1 only.
(t^2 -1) / (t - 1 ) = 2
{ t^2 - 1^2 } / (t-1) = 2
{(t - 1) (t + 1)} / (t - 1 ) = 2
(t - 1 ) gets cancelled
(t + 1) = 2
therefore, t = 1.

Gursimran
#55. January 29th, 2008, at 2:41 PM.

D
as t=1 doesn’t satisfy the LHS of the equation

minesh
#56. January 29th, 2008, at 10:34 PM.

((t^2)-1)/(t-1)=2
t^2-1=2t-2
t^2-2t+1=0
t^2-t-t+1=0
(t(t-1)-(t-1))=0
(t-1)(t-1)=0
so we get….
t=1,1
so ans is A

feng
#57. January 30th, 2008, at 1:30 AM.

okay D,
if t=1 as most of you said, then you got divide by zero error!

samarth
#58. January 31st, 2008, at 1:52 AM.

A.1 only

yami
#59. February 5th, 2008, at 5:57 AM.

of course its d, the only correct would be t=1, but as denominator cannot be zero, d is the correct answer

ajay hulji
#60. February 10th, 2008, at 8:35 PM.

((t^2)-1)/(t-1)=2
(t^2)-1=2t-2
(t^2)-2t+1=0
then look at the eqn. it is of the form
(t-1)^2=0
hence,
t=1
so, the option A is the correct ans. guyz!

malini
#61. March 10th, 2008, at 3:52 PM.

t can take no values ….

(t^2-1)/(t-1)=2
t+1=2
t=1
resubstitute t=1 in the ques
(1-1)/(1-1)=0/0 (is this equal to 2?) this is undefined.
what do u mean by a root..it should satisfy the given eqaution..
so frds…’t’ cannot take any value.

Trainder
#62. March 13th, 2008, at 5:11 PM.

first time i try to calculate, i think T=1
and i bring it back…

Answer is D!

Bhargav@001
#63. March 13th, 2008, at 8:53 PM.

Hey I feel…
The absolute Question is t+1=2. so t=1 (A).
the Query has been masked by giving some other things..as..

(t^2-1)/t-1=2. t^2-1=(t+1)(t-1) isnt it guys?
So, no Question of substituting of t=1 in the denominator..
Am i ryt guys?

Rajaneesh
#64. May 7th, 2008, at 10:08 AM.

its dam sure ans is a

ibrahimfarouk
#65. May 10th, 2008, at 1:17 PM.

of course t has no value
the correct answer is D

neonave
#66. July 19th, 2008, at 12:12 PM.

ans definitely is D
the equation doesnt hold true for t=1 coz substituting t=1 u get LHS as illdefined,whereas RHS is 2……tht shud be the simplest explaination to prove tht 1 is not true…..
the moment t-1 comes to the denominator,equal to 2 dosnt exist any more.

vlakshmilavanya
#67. August 1st, 2008, at 7:47 PM.

Dear take GRE team
Can i know the final correct answer
even i hope that the answer id D
bcoz no other option suits the given equation

justforfun
#68. August 12th, 2008, at 4:17 PM.

1 is the answer
(t^2-1)/(t-1)=
(t-1)(t+1)/(t-1)=
(t+1)=2

hence t=1
Please dont substitue 1 in the equation, that is not logical

it ’s something like doing x=y because it is x*(x-x)=y(x-x) = x*0=y*0
0=0 hence x=y.. the argument does not hold.

justforfun
#69. August 12th, 2008, at 4:19 PM.

GRE TEAM the answer is A, and not D for the argument said above. Please dont mislead people

Karim Yehia
#70. August 13th, 2008, at 9:06 PM.

Hi all,
First I want to say that the “limit” of (t^2-1)/(t-1) as t approaches 1 is equal to 2. that doesnt mean that t^2-1/(t-1)=2 when t=1; becuse when t=1, the answer is 0/0 which is an indefinite number. The answer is D, there is no answer for such question.

Maddy
#71. August 13th, 2008, at 10:48 PM.

the answer is D..
if u argue that the answer is A. thats correct too (when u cross multiply)..
but put the same answer in the original equation, its 0/0 form which is not feasible.
so the one answer which everyone thinks it is, is actually wrong

Aswin
#72. August 13th, 2008, at 11:04 PM.

If the question is right accorind to GRE team then the answer has to be D because if t=1 it wont satisfy the equation as the overall value tends to infinty and if t=-1 the equation value becomes 0….for all other values of t=2,3,4,…etc we will get the value of equation greater than 2…..so the answer has to be D

Ans: D)

naive
#73. August 15th, 2008, at 7:14 AM.

Wow nice question thumbs up to the takegreteam for the effort, keep it up guys

Junaid
#74. August 15th, 2008, at 8:19 AM.

the correct answer is t=1

since the limit of the function is 0/0 , thats indeteminate meaning other than direct substitution, other method be used so that doesnt mean limit does not exist. using different techniques of llimit does exist.

henok
#75. August 16th, 2008, at 9:49 PM.

a simple algebraic manipulation give a fallacious answer A which when you substitute gives us an undefined number 0/0.

so,i think the correct choice is D

Balu
#76. August 17th, 2008, at 8:59 AM.

Goog subterfuge…

The ans should be D.(no ans)

charlee
#77. August 19th, 2008, at 8:27 AM.

Obviously “D” — no answer

Plug in -1 and 1, neither works since both result in 0. Must be no answer.

Chloe Keveryn
#78. August 25th, 2008, at 9:09 PM.

Why are you all trying to factor the the top of the fraction? Only if the “^2″ were on the outside of the parentheses would you factor… Thus, you simply plug in the values given, find that none of them work, and all answers except for “D” are eliminated. Simple Algebra I problem.

jyoti.chhibber
#79. September 13th, 2008, at 2:17 PM.

only one value thats 1

(t^2-1)/(t-1)
= (t-1)(t+1)/(t-1)
= t+1

if t+1=2
then t=1

jyoti.chhibber
#80. September 13th, 2008, at 2:18 PM.

but t can’t be 1 cz then expression tends to infinity..so the ans is D