GMAT Problem Solving Discussions

MBA
4226
How many roots does this equation have?

sq rt (x^2 + 1) + sq rt (x^2 + 2) =2

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

hey buddy u r saying that there is no real solution..but
try in this way...
take "X^2 +1"=Y so X^2 = Y+1
eqn become....sq rt (Y) + sq rt (Y + 1) =2
square both sides...and solve..u gonna find Y=9/16
Y=9/16=X^2 + 1....solve for X, u gonna get 2 rational no's, one with positive sign one wid negative...
4227
hey buddy u r saying that there is no real solution..but
try in this way...
take "X^2 +1"=Y so X^2 = Y+1
eqn become....sq rt (Y) + sq rt (Y + 1) =2
square both sides...and solve..u gonna find Y=9/16
Y=9/16=X^2 + 1....solve for X, u gonna get 2 rational no's, one with positive sign one wid negative...


"Y=9/16=X^2 + 1....solve for X"

X^2= -7/16, n hence no real solution for X
4228

This is the mistake that i made,
without solving, i assumed that x will get 4 values, although x^2 being -ve is impossible !!

x^2+1=9/16, means , x^2= -7/16,
thus no possible value of x satisfies.

hey buddy u r saying that there is no real solution..but
try in this way...
take "X^2 +1"=Y so X^2 = Y+1
eqn become....sq rt (Y) + sq rt (Y + 1) =2
square both sides...and solve..u gonna find Y=9/16
Y=9/16=X^2 + 1....solve for X, u gonna get 2 rational no's, one with positive sign one wid negative...
4229
"Y=9/16=X^2 + 1....solve for X"

X^2= -7/16, n hence no real solution for X

oops...even after solving, i made a silly mistake....

Lesson: silly mistakes kills for sure
4230

During the break of a football match the coach will make 3 substitutions. If the team consists of 11 players among which there are 2 forwards, what is the probability that none of the forwards will be substituted?

A) 21/55
B) 18/44
C) 28/55
D) 28/44
E) 36/55

4231

In the diagram above, AB=CD, From this we can deduce that:
A) AB is parallel to CD.
B) AB is perpendicular to BD.
C) AC=BD
D) Angle ABD equals angle BDC.
E) Triangle ABD is congruent to triangle ACD
4232

In the diagram above, AB=CD, From this we can deduce that:
A) AB is parallel to CD.
B) AB is perpendicular to BD.
C) AC=BD
D) Angle ABD equals angle BDC.
E) Triangle ABD is congruent to triangle ACD

The ans should be OP D
4233

The sum of the even numbers between 1 and k is 79*80, where k is an odd number, then k=?

(A) 79
(B) 80
(C) 81
(D) 157
(E) 159

4234
These type of questions are the easiest. Remember the basic 'formula' of probability.

Probability = Number of favorable outcomes/Number of total possible outcomes

So, in this case,


Probability = C(9,3)/C(11,3) = 28/55

C(9,3), since the favorable (desired outcome) is that 3 people out of the 9 'non-forwards' are chosen for substitution.

C(11,3) is the number of total 3-people substitutions possible in a 11 member team.

-------------------------------------------
Thanks,
Ashish
GMAT Faculty @ EducationAisle
GMAT - 99th Percentile, MBA - ISB



P(none of the forwards sub'bed)= (9/11)*(8/10)*(7/9) = 28/55 where,

9/11 = Probability of the first player subbed who is not a fwd.
8/10 = Prob of second player subbed who is not a fwd, excluding the substitution and likewise 7/9 is the probability of the 3rd and final substitution.
4235

In the diagram above, AB=CD, From this we can deduce that:
A) AB is parallel to CD.
B) AB is perpendicular to BD.
C) AC=BD
D) Angle ABD equals angle BDC.
E) Triangle ABD is congruent to triangle ACD


The ans to this Q is A.
4236

In the diagram above, AB=CD, From this we can deduce that:
A) AB is parallel to CD.
B) AB is perpendicular to BD.
C) AC=BD
D) Angle ABD equals angle BDC.
E) Triangle ABD is congruent to triangle ACD



AB = CD
==> angles ADB = DAC (angles opposite to equal sides are equal).
angle ACD = ABD. (opposite to same side AD).
==> Triangle(ACD) == ABD

Thus angle DAB = ADC.
==> AB CD

Puys pls post the explanation along with your answers. Kinda defeats the purpose of the forum if random answers are posted w/o explanation.
4237
The sum of the even numbers between 1 and k is 79*80, where k is an odd number, then k=?

(A) 79
(B) 80
(C) 81
(D) 157
(E) 159


sum of A.P = n * (a + L) /2

L = a + (n-1) * 2 = 1 * (n-1) * 2 [ a = 1 (given) and d = 2 (even numbers) ]
L = 1 + (n-1) * 2 = 2n + 1

Sum = n * (1 + 2n + 1) / 2 = n * (n+1) = 79 * (79+1)

==> n = 79 option A
4238

In the diagram above, AB=CD, From this we can deduce that:
A) AB is parallel to CD.
B) AB is perpendicular to BD.
C) AC=BD
D) Angle ABD equals angle BDC.
E) Triangle ABD is congruent to triangle ACD

IMO OP D,
HI Switr, m sorry but i think your explanation is completely wrong...
and thats y i think the ans u suggest is also wrong....
pls c the attached word file...i simply reconstructed the diagram as the only
given condition is ab=cd..see the attachment
4239
IMO OP D,
HI Switr, m sorry but i think your explanation is completely wrong...
and thats y i think the ans u suggest is also wrong....
pls c the attached word file...i simply reconstructed the diagram as the only
given condition is ab=cd..see the attachment


Yes, answer is D.. one more way of looking at it is ..

1. ABCD is a parallelogram (square or a rectangle) - in this case AD should pass thru the center but thats not a criteria as per the diagram..
2. If AD is not passing thru center then ABCD can also be a trapezoid..

D is the only valid option which satisfies both conditions.. :cheerio:
4240
Yes, answer is D.. one more way of looking at it is ..

1. ABCD is a parallelogram (square or a rectangle) - in this case AD should pass thru the center but thats not a criteria as per the diagram..
2. If AD is not passing thru center then ABCD can also be a trapezoid..

D is the only valid option which satisfies both conditions.. :cheerio:

Hi vijay...
it is possible to construct a square or a rectangle ABCD and certainly they gonna be one of the possible figures satisfying the given condition AB=CD with AD passing through the centre but in that case....op A,B,C also gonna be correct and so u get multiple answers but since we need to find the ans to which all the possible figures will agree...
4241

D is the only valid option which satisfies both conditions.. :cheerio:

Hi vijay...
it is possible to construct a square or a rectangle ABCD and certainly they gonna be one of the possible figures satisfying the given condition AB=CD with AD passing through the centre but in that case....op A,B,C also gonna be correct and so u get multiple answers but since we need to find the ans to which all the possible figures will agree...



I thought I said that..
4242
bikram4gmat Says
The ans to this Q is A.

The only answer correct is D. As
4243
sum of A.P = n * (a + L) /2

L = a + (n-1) * 2 = 1 * (n-1) * 2 [ a = 1 (given) and d = 2 (even numbers) ]
L = 1 + (n-1) * 2 = 2n + 1

Sum = n * (1 + 2n + 1) / 2 = n * (n+1) = 79 * (79+1)

==> n = 79 option A

@gladvijay: A couple of fundamental mistakes. 1 is not an even number so the sequence has to start from 2. and the number of even numbers between 1 and K is (k-1)/2 ... rest all looks fine going by the above two facts the answer is 159
n(n+1) =79*80
n=79=(k-1)/2 ==> K=159
4244
AB = CD
==> angles ADB = DAC (angles opposite to equal sides are equal).
angle ACD = ABD. (opposite to same side AD).
==> Triangle(ACD) == ABD

Thus angle DAB = ADC.
==> AB CD

Puys pls post the explanation along with your answers. Kinda defeats the purpose of the forum if random answers are posted w/o explanation.


@Switr: the angles extended by a secanyt on a circle is equal only if they are on the sqame side, so your third assumption(
angle ACD = ABD. (opposite to same side AD) is incorrect and i am sorry if i offended you ... just wanted to point ur mistake out
4245
sshekhar18 Says
@switr : where did you get this ******** angle ACD = ABD. (opposite to same side AD)? Think before u post


Irrelevant!