GMAT Data Sufficiency Discussions

MBA GMAT
1501

Guys,

Thanks for attempting this question but unfortunately none has got it right yet. I couldnt even do as much as you guys have . I gave it up shortly after I saw this problem

I reason I posted this problem is that i wanted to share this for the benefit of everyone around.

Key to solve this problem is that there is no calculation needed

I will post the ans shortly. Meanwhile , please attempt again. I am sure there will be few amazing solutions.


Car X leaves town A at 2.00pm and drives toward town B at a constant rate of m miles/hr. 15 mins later, car Y begins driving from town B to town A at a constant rate of n miles/hr. If both X and Y drive along the same route, will car X be closer to town A or town B when it passes the car Y?

1. Car X arrives in town B 90 mins after leaving city A
2. Car Y arrived in town A at the same time car X arrived in town B.
1502
From 1) in 90 mins car travels: M + M/2 miles
From 2) in 75 mins car B travels: N + N/4 miles

As the distance is same:

3M/ 2 = 5N / 4

6M = 5N ==> M

Well, I guess yes...none of them has got this right...as said...
lets nail it down till the end...

I guess, if the total distance between city varries...the closness to a city will change...soi in other words if distance varries...the closness of city will varry..

6m = 5n => any multiple of 30 wiil yield the result to satisfy this equation...
m = 5n/6 => n = 1, m = 5/6 miles per hour...

so m will travell...5/6 + 5/12 miles...in 90 mins..so on so forth...

lets say the total distance between the cities is X...and they meet at l distance from City A..

hence we have to prove either..

M/ 4 + (X-L) / M+N > L / (M+N) => L
OR

L> (X + 7M)/ 4...

all the time...but not both...however varying the value of X...you will notice that both holds true...in that case it will BE E)
1503

Guys, I really need your help.
I started doing DS problems few days go and now I realize I really suck at it
Could you please guide me to materials/tips(other than OG, Kaplan) which could help me improve quicker.

You inputs will be received with great deal of appreciation.

Thanks!

1504

well.. the ans is B.

Option 1: it doesnt say anything about Y. Insuff
Option 2:
It says both cars arrive their destinations at the same time. This is the key.
After they meet, time taken by both to reach their respective destination is same and We know car X runs slower than car Y. So what can vary? Its the distance travelled by each car after they cross each other.
Therefore, car X must be closer to its detination that car Y is. Thus car X is closer to B(the point where they meet is certainly after the mid point, far off A).

Above is an official explanation. Any indigenous approach is welcome.


Well, I guess yes...none of them has got this right...as said...
lets nail it down till the end...

I guess, if the total distance between city varries...the closness to a city will change...soi in other words if distance varries...the closness of city will varry..

6m = 5n => any multiple of 30 wiil yield the result to satisfy this equation...
m = 5n/6 => n = 1, m = 5/6 miles per hour...

so m will travell...5/6 + 5/12 miles...in 90 mins..so on so forth...

lets say the total distance between the cities is X...and they meet at l distance from City A..

hence we have to prove either..

M/ 4 + (X-L) / M+N > L / (M+N) => L
OR

L> (X + 7M)/ 4...

all the time...but not both...however varying the value of X...you will notice that both holds true...in that case it will BE E)
1505
well.. the ans is B.

Option 1: it doesnt say anything about Y. Insuff
Option 2:
It says both cars arrive their destinations at the same time. This is the key.
After they meet, time taken by both to reach their respective destination is same and We know car X runs slower than car Y. So what can vary? Its the distance travelled by each car after they cross each other.
Therefore, car X must be closer to its detination that car Y is. Thus car X is closer to B(the point where they meet is certainly after the mid point, far off A).

Above is an official explanation. Any indigenous approach is welcome.

, totally missed that one..both cars arrive their destinations at the same time..
1506
My take C.It would be closer to B.

let the distance travelled by X in the initial 15 mins be l.
total distance between A & B =6l

=> since both the cars meet reach the respective destinations at 3:30
we can find the ratio between m & n
5n=6m

from this we can find out the meeting point in terms of l...which comes out to be 36l/11...which is greater than 3l ...hence it is nearer/closer to city B.

( hopefully,i have not made yet another silly mistake )

PG Rocks


man! I could feel that i was going wrong on this but in the hurry to post the answer i threw it away...is this from one of the kaplan tests?

Thanks.
1507

Pls explain :-

if x^2 + 5y = 49, is y an integer?
(1) 1 (2) x^2 is an integer.

1508

Is y an integer?
A. y^3 is an integer
B. 3y is an integer

1509
Pls explain :-

if x^2 + 5y = 49, is y an integer?
(1) 1 (2) x^2 is an integer.


Lets try it with basics...

A) X = 2; Y=> integer
X = 2.5 Y=> Not an integar

A, D => Gone...

B) X^2 is an integar...so what? X^2...= 2, 3, 4, 5, 6...????no idea..

B) is gone...

Lets try for C:
-------------
X^2 is integar...if X = 1.44.. X^2 = 2, supply it in the equation, we get
Y = 47 / 5 => not an integar..

If X = 2, X^2 = 4..

Y = 45/ 5 => an integar
IMO E)
1510
Is y an integer?
A. y^3 is an integer (Now got the question right)
B. 3y is an integer


A...if y = 3^1/3...then A hold true...so not sufficient..A/D gone

B...Y = 1/3...

I guess (Not able to establish), it shall be C
1511

if x^2 + 5y = 49, is y an integer?

(1) 1 if x=3 then y =8;
if x=1.4 then y =9.408; insuff...

(2) x^2 is an integer.

if x=0 then y=9.8
if x=2 then y=9
if x =squareroot 2 then x^2 is 2 :biggrin: and y =9.4 insuff...

join both but still
1 if x =squareroot 2 then x^2 is 2 :biggrin: and y =9.4 insuff...
if x=2 then y=9

so E:biggrin:

P.S : isn't it so colorfull :biggrin:

1512

Is y an integer?

A. y3 is an integer
y = 3 then y3 =3 :biggrin:
y = 3 then y3 =27
in suff

B. 3y is an integer

y = 1/3 then 3y = 1
y = 1 then 3y = 3 :biggrin: insuff

joining both v can eliminate fractions and cube roots thus leaving behind only integers :

answer C : i still doubt, i fi am right 😉 correct me if i am wrong :biggrin:

1513

wrong post.

1514
Is y an integer?

A. y3 is an integer
y = 3 then y3 =3 :biggrin:
y = 3 then y3 =27
in suff

B. 3y is an integer

y = 1/3 then 3y = 1
y = 1 then 3y = 3 :biggrin: insuff
joining both v can eliminate fractions and cube roots thus leaving behind only integers :
answer C : i still doubt, i fi am right 😉 correct me if i am wrong :biggrin:


In agreement with u buddy .. third of any integer when raised to third power can never be an integer ..

i.e say y=k/3 ...where k is an integer and not a multiple of 3 (i.e y is a fraction)

Then (k/3)^3 cannot be an integer unless k is a multiple of 3 ..or indirectly y is an integer !

Hence, C
1515
Pls explain :-

if x^2 + 5y = 49, is y an integer?
(1) 1 (2) x^2 is an integer.



IMO E.

here x is a real number.

So take 1)

consider x= 1.1

so y cannot be an integer

consider x=2 then y is an integer. Clearly, 1) alone is insuff

Take 2) if x^2= 2 then y is not an integer.

If x^2=4 then y is an integer.

Clearly, 2) alone is insuff.

Now take 1) and 2) here also y can be an integer only if x=2. So, E
1516
IMO E.

here x is a real number.

So take 1)

consider x= 1.1

so y cannot be an integer

consider x=2 then y is an integer. Clearly, 1) alone is insuff

Take 2) if x^2= 2 then y is not an integer.

If x^2=4 then y is an integer.

Clearly, 2) alone is insuff.

Now take 1) and 2) here also y can be an integer only if x=2. So, E


My take is C on this.
Sidd can u give a number,x, between 1 and 4 for which X^2=2.
I think no fraction value squares will result into an integer.
1517

puys help me in solving this .. actually looking for a short cut to solve it..the guess wrk..etc

Store S sold a total of 90 copies of a certain book during the seven days of last week,and it sold different numbers of copies on any two of the days.If for the seven days Store S sold the greatest number of copies on Saturday and the second greatest number of copies on Friday, did the Store S sell more than 11 copies on Friday?

1-Last week store S sold 8 copies of the book on Thursday.
2-Last week store S sold 38 copies of the book on Saturday.

1518

there is a problem given in the gmat prep prac test..

Are positive integers p and q both greater than n?

1-p-q is greater than n
2-q>p

please answer with explanation.

1519

if 500 is the multiple of 100 that is closest to x and 400 is the multiple of 100 that is closest to y,which multiple of 100 is closest to x+y?
1-x2-y

1520
if 500 is the multiple of 100 that is closest to x and 400 is the multiple of 100 that is closest to y,which multiple of 100 is closest to x+y?
1-x2-y

500 is multiple of 100 (closest to X)
&
400 is multiple of 100 (closest to Y)

Ques: Multiple of 100 closest to (X + Y)

Option 1:
X Multiple of 100 less than 500 are 100, 200, 300 & 400

500 can be multiple only of 100. So X must be close to 100

NOT SUFF

Option 2:
y Multiple of 100 less than 400 are 100, 200 & 300

400 can be multiple only of 100 & 200. So Y can be close to 100 or 200.

NOT SUFF

Taking both together.. doesnt give clarity on the value of Y, so cannot be resolved.

I would have marked option E

please share the OA 😉