GMAT Data Sufficiency Discussions

Guys, here is a very simple DS, just want to confirm the answer..

If n is an integer and 100 n n?
(1) N/36 is an odd integer.
(2) N/45 is an even integer.
Another PS (sorry , i know that there is a different thread), i just want to make sure that my answers, which seem to be not matching to OA, are right..
At a certain company, each employee has a salary grade s that is at least 1 and at most 5. Each employee receives an hourly wage p, in dollars, determined by the formula p = 9.50 + 0.25(s 1). An employee with a salary grade of 5 receives how many more dollars per hour than an employee with a salary grade of 1?
A. $0.50
B. $1.00
C. $1.25
D. $1.50
E. $1.75

Cheers


1) Ans B

St 1 :There are 2 odd multiples of 36 between 100 and 200 ..not suff
St 2 : Only one even multiple of 45 between 100 and 200 ..suff

2 ) difference is 0.25 [ (5-1) - (1-1)] = $ 1 ...Ans B
1) Ans B

St 1 :There are 2 odd multiples of 36 between 100 and 200 ..not suff
St 2 : Only one even multiple of 45 between 100 and 200 ..suff

2 ) difference is 0.25 [ (5-1) - (1-1)] = $ 1 ...Ans B



HuH...Thats what i got in 2 mins for both of them , was just giving a test and answers, I found were-> for first showed: C, and the second one 1:25$....feeling relieved...Answers almost shocked me as i felt such lolly-pops are giving me jitters..

Cheers,

Answers posted below.

few DS problems. Please post explanations as well as i need to find the best approach.

Are x and y both positive?
1. 2x - 2y =1
2. x/y > 1

OA: C

Is x-y > |x| -y?
1. y 2. xy
OA: B

The positive integers x, y, and z are such that x is a factor of y and y is a factor of z. is z even?

1. xz is even
2. y is even


OA: D

My answers and explanation in Red. skipping question 1 as i saw your post indicating typo :p. will check it later.

try the following puys:

1) Is x^y
1) aEquation boils down to (a-b)^2 >1
This is possible when absolute value of (a-b) is greater than 1.
1 doesn't give any info. B clearly mentions it hence the ans: B


3) In the fraction x/y, where x and y are +v integers, what is the value of y?
1) the least common denominator of x/y and 1/3 is 6.
2) x=1

x/y is a fraction. y can be 2 or 6. hence a is insuffucient.
2 doesnt give any info.
Ans : E

4)If x is a +ve integer and w is a -ve integer, what is the value of xw?

1) x^w= 1/2
2) w=-1

1. x^w = 1/2 is only possible when x is 2 and w is -1. note that both x & w are integers and its given that x>0 and w 2. w=-1 this value insufficnt to get xw

Ans : A


5) Is x/m (m^2+n^2+k^2) = xm+yn+zk?
1) z/k = x/m
2) x/m = y/n

Simple substitution using both 1 & 2 will give the value.
Ans: C


6)if k and n are integers is n divisible by 7?
1) n-3 =2k
2)2k+4 is divisible by 7.

1. Many combinations of n and k hence insufficnt.
2. doesnt say anything about n hence insufficient.

Using both 1 & 2:
n = 2k + 3 = (2k + 4) -1
since 2k + 4 is divisible by 7, n/7 will have remainder -1 hence not divisible.
Ans : C


OAs later...
enjoy solving.

solve this .My take on the below DS is option C - both statements are suff to solve the q.

If x, y, and z are integers, with x
(1) (x + y)z = 5

(2) x + z

solve this .My take on the below DS is option C - both statements are suff to solve the q.

If x, y, and z are integers, with x
(1) (x + y)z = 5

(2) x + z

IMO E.

we need the value of x+y+z/3

so from 1)
we have 3 unknowns and 1 equation so insuff.

take 2) this is also in suff.

consider 1) and 2) we have 1 equation and 1 inequality and 3 unknowns so insuff to get the avg.

hence E. Whats is the OA?
try the following puys:

1) Is x^y
2) Is 1/(a-b)
1) a


These are all ques from OG quant, the green book. I found them to be tricky so I posted them. I will try to post similar ques for all sections in coming days.

SO the OAs are ( many of the puys got them correct)

1.E
2.A
3.E
4.A
5.C
6.C
solve this .My take on the below DS is option C - both statements are suff to solve the q.

If x, y, and z are integers, with x
(1) (x + y)z = 5

(2) x + z

I believe it must be option A
nuttyvarun Says
I believe it must be option A

do u mind explaining it :biggrin:

If x, y, and z are integers, with x
(1) (x + y)z = 5

Possible values
(x+y)____5___-5___1___-1
(z)_____1___-1___5___-5

So u cant tell if the mean is 2 or -2


(2) x + z
take z =1
then x
take z =-1
then x
take x=5
then x
take x=-5
then x

IMO E is the answer :biggrin:

E should be the ans.

solve this .My take on the below DS is option C - both statements are suff to solve the q.

If x, y, and z are integers, with x
(1) (x + y)z = 5

(2) x + z

if it said x, y, and z are all positive integers i would agree with A.
but thats not the case here. Need your explanation.

nuttyvarun Says
I believe it must be option A

Guys...plz explain the concept behind dis prob..
If k and n are integers, is n divisible by 7?
(1) n - 3 = 2k
(2) 2k - 4 is divisible by 7.

Guys...plz explain the concept behind dis prob..
If k and n are integers, is n divisible by 7?
(1) n 3 = 2k
(2) 2k 4 is divisible by 7.


Here the question is n/7?

so take 1)

n=2k+3 so we know that n is an odd integer, but nothing more than that. so insuff.

take 2)
here there is no info on n so insuff.

take 1) and 2)

here we know that 2k-4 /7 and we know that n= 2k+3
if 2k-4/ 7 then 2k-4+7= 2k+3/7 => n/7

hence C.

Hope this is clear.

Dude i didnt get d part 2k-4/ 7 then 2k-4+7...

Guys...plz explain the concept behind dis prob..
If k and n are integers, is n divisible by 7?
(1) n 3 = 2k
(2) 2k 4 is divisible by 7.


Hi,
Both looks insufficient individually,
Combined them than,

From first
n= 2k+3

From second
Let2k-4/ 7

2k-4+7= 2k+3/7

=> n/7.

Hence Option-C
sdt83 Says
Dude i didnt get d part 2k-4/ 7 then 2k-4+7...



hey sdt83,

if a number x is divisible by 7 then x+7 is also divisible by seven.

Using the same principle here if 2k-4 is div by 7 then 2k-4+7=2k+3 is also div by 7.
hey sdt83,

if a number x is divisible by 7 then x+7 is also divisible by seven.

Using the same principle here if 2k-4 is div by 7 then 2k-4+7=2k+3 is also div by 7.


Seems ok !

thnx dude... got it..

Guys...plz explain the concept behind dis prob..
If k and n are integers, is n divisible by 7?
(1) n 3 = 2k
(2) 2k 4 is divisible by 7.


Either of the statements are not suff but combining both the statements u'll get an answer