GMAT Data Sufficiency Discussions

MBA GMAT
2409
Gail, I differ from your answers on a few of these. Let me know if I got anything wrong.

Thank you,
Hemanth


Originally Posted by Gail.Wynand
1) If the base of triangle PQR is 5, what is the perimeter of the triangle?

(1) The area of triangle PQR is 12.5
(2) The length of a side of triangle PQR is

Lets say base = c = 5. We need the other 2 sides for perimeter.
from (1). area= 12.5 = sqrt(s(s-a)(s-b)(s-c)) where s = (a+b+5)/2..so one eqn 2 variables. not sufficient.
from (2). b=. and c=5 ; again do we Know a? No. Not sufficient.
combining: 12.5 = 12.5 = sqrt(s(s-a)(s-)(s-5)) and s= (a+5+)/2...so one variable, one eqn. Hence sufficient.

So choice C.

I agree.
With 1, we can get the height to be 5 but that doesn't tell us how the other sides are set up. Insufficient.
With 2, We know the other side but don't about the 3rd Side. Insufficinet.
1+2 - It is a right triangle with sides 5,5,5*2^0.5. Sufficient.

-----------------------------------------------------------------

2) If m and q are divisors of c, is mq a divisor of c that is not c or 1?

(1) mq (2) m and q have no common divisors.

Soln:
Lets use plug-in numbers:
mq is a divisor of c that is not c or 1....
C = 28, m = 2, q=7 or m=4, q=7 or m=4, q=2 and so on...
from (1): since mqYes you are correct. It was highly dumb on my part not to simplify the equation. And the 2nd one also seems correct.
------------------------------------------------------------------

4) If the least common multiple of integers x and y is 840, what is the value of x?

(1) The greatest common factor of x and y is 56.
(2) y = 168

from Stem: prime factors of 840 = 2,2,2,5,3,7.
from (1) : Tells us that max 56 is common in both. prime factors of 56 = 2,2,2,7. hence x= 56*A and Y=56*B where A, B are co-prime. Insufficient.
from (2) : prime factors (y) = 2,2,2,3,7. So X can be anything 3 or less 2s, 1 or less 3s, 1 or less 7s and a 5. hence possible options are 5, 10, 15, 35...So Insufficient.
Combining (1) and (2) : y = 168 and x=56*5 or 56*3*5 hence insufficient.

Answer choice E

I disagree. The answer should be C.
Cond1: Given GCF=56. 840 = 7*8*5*3. Also each of x,y
Cond2: y=168 (or 56*3). now x can be 56*5 (280) or 56*15 (840) or ..any other value where LCM=840. Insufficient.
But then combine both of them. Cond2 tells us that Y is 56*3. and if the GCD has to be 56 then X cannot be 56*15 (else the GCD would have been 56*3 not 56). So X has to be 56*5.

------------------------------------------------------------------
5) If c and k are distinct positive integers, is c divisible by k?

(1) 2k > c
(2) k2 + k = c

from (1) : Since both are +ve. c/kfrom (2) : k(k+1)= c hence c/k = k+1 = Integer. Hence c is divisible by K. Sufficient.

Answer choice B

I disagree again. Answer is D.

From 1 you can be certain that it is not divisible since C and K are distinct. Hence it is actually sufficient.

Of course, Condition 2 is also sufficient.
------------------------------------------------------------------

My comments inline in blue...
2410
Originally Posted by Gail.Wynand
1) If the base of triangle PQR is 5, what is the perimeter of the triangle?

(1) The area of triangle PQR is 12.5
(2) The length of a side of triangle PQR is

Lets say base = c = 5. We need the other 2 sides for perimeter.
from (1). area= 12.5 = sqrt(s(s-a)(s-b)(s-c)) where s = (a+b+5)/2..so one eqn 2 variables. not sufficient.
from (2). b=. and c=5 ; again do we Know a? No. Not sufficient.
combining: 12.5 = 12.5 = sqrt(s(s-a)(s-)(s-5)) and s= (a+5+)/2...so one variable, one eqn. Hence sufficient.

So choice C.

I agree.
With 1, we can get the height to be 5 but that doesn't tell us how the other sides are set up. Insufficient.
With 2, We know the other side but don't about the 3rd Side. Insufficinet.
1+2 - It is a right triangle with sides 5,5,5*2^0.5. Sufficient.

-----------------------------------------------------------------

2) If m and q are divisors of c, is mq a divisor of c that is not c or 1?

(1) mq (2) m and q have no common divisors.

Soln:
Lets use plug-in numbers:
mq is a divisor of c that is not c or 1....
C = 28, m = 2, q=7 or m=4, q=7 or m=4, q=2 and so on...
from (1): since mqYes you are correct. It was highly dumb on my part not to simplify the equation. And the 2nd one also seems correct.
------------------------------------------------------------------

4) If the least common multiple of integers x and y is 840, what is the value of x?

(1) The greatest common factor of x and y is 56.
(2) y = 168

from Stem: prime factors of 840 = 2,2,2,5,3,7.
from (1) : Tells us that max 56 is common in both. prime factors of 56 = 2,2,2,7. hence x= 56*A and Y=56*B where A, B are co-prime. Insufficient.
from (2) : prime factors (y) = 2,2,2,3,7. So X can be anything 3 or less 2s, 1 or less 3s, 1 or less 7s and a 5. hence possible options are 5, 10, 15, 35...So Insufficient.
Combining (1) and (2) : y = 168 and x=56*5 or 56*3*5 hence insufficient.

Answer choice E

I disagree. The answer should be C.
Cond1: Given GCF=56. 840 = 7*8*5*3. Also each of x,y Good question. So there can actually be two sets of X and Y whose GCD can be 56. they are (56*3,56*5) or (56*1, 56*15). Agree? Else the GCD will not be 56 given that the LCM is 840.
Cond2: y=168 (or 56*3). now x can be 56*5 (280) or 56*15 (840) or ..any other value where LCM=840. Insufficient.
But then combine both of them. Cond2 tells us that Y is 56*3. and if the GCD has to be 56 then X cannot be 56*15 (else the GCD would have been 56*3 not 56). So X has to be 56*5. So now given that one of the numbers is 56*3 the other number,i.e. X has to be 56*5 otherwise the GCD will not be 56.
There is this other vague formula for positive integers which is GCD(x,y)*LCM(x,y) = X*Y. Using that too we can solve for X which comes to be 56*5. But you can do it without remembering that formula.

------------------------------------------------------------------
5) If c and k are distinct positive integers, is c divisible by k?

(1) 2k > c
(2) k2 + k = c

from (1) : Since both are +ve. c/kfrom (2) : k(k+1)= c hence c/k = k+1 = Integer. Hence c is divisible by K. Sufficient.

Answer choice B

I disagree again. Answer is D.

From 1 you can be certain that it is not divisible since C and K are distinct. Hence it is actually sufficient. That exception is taken care of in the question which reads "DISTINCT POSITIVE INTEGERS" implying that C/K is not 1. Agree?

Of course, Condition 2 is also sufficient.
------------------------------------------------------------------

My comments inline in blue...


Replies in ladies color MAGENTA.
I like the discussion Gail. Feel free to disagree

thank you,
Hemanth
2411
Replies in ladies color MAGENTA.
I like the discussion Gail. Feel free to disagree

thank you,
Hemanth

Originally Posted by Gail.Wynand View Post
Originally Posted by Gail.Wynand
1) If the base of triangle PQR is 5, what is the perimeter of the triangle?

(1) The area of triangle PQR is 12.5
(2) The length of a side of triangle PQR is

Lets say base = c = 5. We need the other 2 sides for perimeter.
from (1). area= 12.5 = sqrt(s(s-a)(s-b)(s-c)) where s = (a+b+5)/2..so one eqn 2 variables. not sufficient.
from (2). b=. and c=5 ; again do we Know a? No. Not sufficient.
combining: 12.5 = 12.5 = sqrt(s(s-a)(s-)(s-5)) and s= (a+5+)/2...so one variable, one eqn. Hence sufficient.

So choice C.

I agree.
With 1, we can get the height to be 5 but that doesn't tell us how the other sides are set up. Insufficient.
With 2, We know the other side but don't about the 3rd Side. Insufficinet.
1+2 - It is a right triangle with sides 5,5,5*2^0.5. Sufficient.

-----------------------------------------------------------------

2) If m and q are divisors of c, is mq a divisor of c that is not c or 1?

(1) mq (2) m and q have no common divisors.

Soln:
Lets use plug-in numbers:
mq is a divisor of c that is not c or 1....
C = 28, m = 2, q=7 or m=4, q=7 or m=4, q=2 and so on...
from (1): since mqYes you are correct. It was highly dumb on my part not to simplify the equation. And the 2nd one also seems correct.
------------------------------------------------------------------

4) If the least common multiple of integers x and y is 840, what is the value of x?

(1) The greatest common factor of x and y is 56.
(2) y = 168

from Stem: prime factors of 840 = 2,2,2,5,3,7.
from (1) : Tells us that max 56 is common in both. prime factors of 56 = 2,2,2,7. hence x= 56*A and Y=56*B where A, B are co-prime. Insufficient.
from (2) : prime factors (y) = 2,2,2,3,7. So X can be anything 3 or less 2s, 1 or less 3s, 1 or less 7s and a 5. hence possible options are 5, 10, 15, 35...So Insufficient.
Combining (1) and (2) : y = 168 and x=56*5 or 56*3*5 hence insufficient.

Answer choice E

I disagree. The answer should be C.
Cond1: Given GCF=56. 840 = 7*8*5*3. Also each of x,y Good question. So there can actually be two sets of X and Y whose GCD can be 56. they are (56*3,56*5) or (56*1, 56*15). Agree? Else the GCD will not be 56 given that the LCM is 840.
Cond2: y=168 (or 56*3). now x can be 56*5 (280) or 56*15 (840) or ..any other value where LCM=840. Insufficient.
But then combine both of them. Cond2 tells us that Y is 56*3. and if the GCD has to be 56 then X cannot be 56*15 (else the GCD would have been 56*3 not 56). So X has to be 56*5. So now given that one of the numbers is 56*3 the other number,i.e. X has to be 56*5 otherwise the GCD will not be 56.
There is this other vague formula for positive integers which is GCD(x,y)*LCM(x,y) = X*Y. Using that too we can solve for X which comes to be 56*5. But you can do it without remembering that formula.

- You got this one too. I must have been drinking

------------------------------------------------------------------
5) If c and k are distinct positive integers, is c divisible by k?

(1) 2k > c
(2) k2 + k = c

from (1) : Since both are +ve. c/kfrom (2) : k(k+1)= c hence c/k = k+1 = Integer. Hence c is divisible by K. Sufficient.

Answer choice B

I disagree again. Answer is D.

From 1 you can be certain that it is not divisible since C and K are distinct. Hence it is actually sufficient. That exception is taken care of in the question which reads "DISTINCT POSITIVE INTEGERS" implying that C/K is not 1. Agree?

- Ohh boy!!! I am so careless. You got it buddy.

Of course, Condition 2 is also sufficient.
------------------------------------------------------------------

I'll try to be more careful in the future.

My comments inline in Eco - Green...
2412

These are real quality questions and would be a fight to do within 2 mins each. VPITC, where are you getting these from?

Gail, lets keep discussing. I learn a lot from you all.

Thank you.
Hemanth

2413
These are real quality questions and would be a fight to do within 2 mins each. VPITC, where are you getting these from?

Gail, lets keep discussing. I learn a lot from you all.

Thank you.
Hemanth


I have been hearing from everyone now-a-days that the QA section is getting tougher by the day. Obviously because of us "IT" fellas. I think quality questions are exactly what we need here. Although I am sure that the GMAT will not go out of its league to come-up with Algebra intensive stuff but they will definitely throw tricky questions in which we will usually miss a case or two as I did in these questions.
More....more....more....
2414

@above 2 post's
GW and PH , I have just started my GMAT preps, I'll be a regular to this thread today onwards are there any more forums for GMAT?

Thanks a lot.
Akshay

2415

Akshay, you are making us sound as though we are the teachers of this thread. :). There are many GMAT related threads, one for each type of section in GMAT. You should be able to locate them easily.

Look forward for some really cruel GMAT level posts from you.

- ph

2416

1)x is a positive integer greater than two; is (x^3 + 19837)(x^2 + 5)(x 3) an odd number?
a)the sum of any prime factor of x and x is even
b)3x is an even number

2)If w, x, y, and z are the digits of the four-digit number N, a positive integer, what is the remainder when N is divided by 9?
a)w + x + y + z = 13
b)N + 5 is divisible by 9

3)x is an integer and x raised to any odd integer is greater than zero; is w - z greater than 5 times the quantity 7^x-1 - 5^x?
a)z b)x = 4

4)If z = x^n - 19, is z divisible by 9?
a)x = 10; n is a positive integer
b)z + 981 is a multiple of 9

5)If b is prime and the symbol # represents one of the following operations: addition, subtraction, multiplication, or division, is the value of b # 2 even or odd?
a) (b # 1) # 2 = 5
b) 4 # b = 3 # (1 # b) and b is even

Thanks.
pH

2418

hey ankit i arrived at similar conclusions for questions 1,2,4 and 5 but i dint get the third one .....would u mind explaining?

2420
1) If the base of triangle PQR is 5, what is the perimeter of the triangle?

(1) The area of triangle PQR is 12.5
(2) The length of a side of triangle PQR is

2) If m and q are divisors of c, is mq a divisor of c that is not c or 1?


(1) mq
(2) m and q have no common divisors.

3) Group A, which is a set of numbers, has an average (arithmetic mean) of 8 while group B, a different set of numbers, has an average of 10. In which group is the sum of the numbers greater?

(1) The average of all numbers in both groups combined is .
(2) The sum of all numbers in both groups is 52

4) If the least common multiple of integers x and y is 840, what is the value of x?

(1) The greatest common factor of x and y is 56.
(2) y = 168

5) If c and k are distinct positive integers, is c divisible by k?

(1) 2k > c
(2) k2 + k = c


The OA s are
1) E 2) C 3) A 4) C 5) D
2421
My answers would be:

1. B
2. D
3. E
4. D
5. D

The explanations are too long to post for wrong answers. Please post the OAs and I will post the explanations to my correct attempts.

For Q3, I have taken the expression as 7^x - 1 - 5^x

If it is 7^(x-1)- 5^x, the answer should be A.


Ankit your assumption for Q3 is correct and you get everything right. Sorry for confusing. The OAs:
1)B
2)D
3)A
4)D
5)D

Thanks and feel free to give feedback on the questions. Like them or not?
2424
Ankit your assumption for Q3 is correct and you get everything right. Sorry for confusing. The OAs:
1)B
2)D
3)A
4)D
5)D

Thanks and feel free to give feedback on the questions. Like them or not?

1)x is a positive integer greater than two; is (x^3 + 19837)(x^2 + 5)(x - 3) an odd number?
a)the sum of any prime factor of x and x is even
b)3x is an even number

2)If w, x, y, and z are the digits of the four-digit number N, a positive integer, what is the remainder when N is divided by 9?
a)w + x + y + z = 13
b)N + 5 is divisible by 9

3)x is an integer and x raised to any odd integer is greater than zero; is w - z greater than 5 times the quantity 7^x-1 - 5^x?
a)z b)x = 4

4)If z = x^n - 19, is z divisible by 9?
a)x = 10; n is a positive integer
b)z + 981 is a multiple of 9

5)If b is prime and the symbol # represents one of the following operations: addition, subtraction, multiplication, or division, is the value of b # 2 even or odd?
a) (b # 1) # 2 = 5
b) 4 # b = 3 # (1 # b) and b is even

Thanks.
pH


1)x is a positive integer greater than two; is (x^3 + 19837)(x^2 + 5)(x - 3) an odd number?
a)the sum of any prime factor of x and x is even
b)3x is an even number


Soln :
from a) X can be {primes > 2, Odd-composites, Even composites with prime factor only 2}.
Hence for primes > 2 expression is even. For odd composites expression is even. For Even composites with prime factor only 2 expression is odd. Hence insufficient.
From b) 3x = even => x is even. Answer Odd. Trying 4 and 6.
(64+19837)(16+5)(1) = odd*odd*odd=odd.
(216+19837)(36+5)(3)= odd*odd*odd = odd. Hence holds. Sufficient.
Answer choice B


2)If w, x, y, and z are the digits of the four-digit number N, a positive integer, what is the remainder when N is divided by 9?
a)w + x + y + z = 13
b)N + 5 is divisible by 9
Soln:
From a) We get no much info as there can be multiple combinations.
From b) If N+5 divided by 9 = remainder 0. Then N+5-5=N will leave remainder 9-5=4 when divided by 9. Hence sufficient.
Try N=67. N+5=72 is divisible by 9. When N/9 remainder = 4. N=76. N+5=81. 76/9 remainder = 4.
Asnwer B.
Got this wrong. @ankitgarg20 can you explain how you arrived at D? Even after checking the answer I am not convinced. Let me know the problem with the reasoning. A million dollar question is why do we need the digits to find the remainder.




3)x is an integer and x raised to any odd integer is greater than zero; is w - z greater than 5 times the quantity 7^x-1 - 5^x?
a)z b)x = 4
Stem : x = Int. x > 0. Is w-z > 5(7x-1 - 5x).
From a) is w-z = 73-24 > 5(72-53). +ve > -ve. True. 7^4-23> 5(7^3- 5^4). The RHS turns out to be -ve and LHS +ve. Hence this always holds. Sufficient.

From b). No info about w and Z hence insufficient.
Answer choice A






4)If z = x^n - 19, is z divisible by 9?
a)x = 10; n is a positive integer
b)z + 981 is a multiple of 9


From a alone. z = x^n - 19 is always divisible by 9. Check for n=1,2,3. Sufficient.
From b alone. Since 981 = 9*109. Hence z+981 = z+9*109 = 9*K. For K to be an integer. Z should be divisible by 9. Answer Yes. Hence sufficient.
Answer choice D



5)If b is prime and the symbol # represents one of the following operations: addition, subtraction, multiplication, or division, is the value of b # 2 even or odd?
a) (b # 1) # 2 = 5
b) 4 # b = 3 # (1 # b) and b is even
From a) The only prime number b that satisfies this is 2 and # = +. 2+1+2=5. Hence B+2 = even. Sufficient.
From b) Since B is even(from stmt) and prime(stem) b=2. And 4+2=3+1+2=6. Hence # = +. B+2=even. Sufficient.

Answer choice D
2425
1)x is a positive integer greater than two; is (x^3 + 19837)(x^2 + 5)(x 3) an odd number?
a)the sum of any prime factor of x and x is even
b)3x is an even number

The expression will not be odd if any of the three expression is even.
from 1-if x is even number, sum will be even or odd/ x is odd, sum will be-even or odd/x is prime, sum will always be even. So cant say
from2-x is even, so the expression will be-
odd*odd*odd=odd
So B

2)If w, x, y, and z are the digits of the four-digit number N, a positive integer, what is the remainder when N is divided by 9?
a)w + x + y + z = 13
b)N + 5 is divisible by 9

From 1- Remider would be 4
From2- Number would be 9*K-5., reminder would be 5.
So D

3)x is an integer and x raised to any odd integer is greater than zero; is w - z greater than 5 times the quantity 7^x-1 - 5^x?
a)z b)x = 4

x is a +ive I.
when x=1, expression would be w-z>5*-4=-20
when x=2,expression would be w-z>5*-18=-90
so the w-z would be -ive always for all x.
May be Not able to understand the question.


4)If z = x^n - 19, is z divisible by 9?
a)x = 10; n is a positive integer
b)z + 981 is a multiple of 9

From 1, 10^n-19=(9+1)^n-9*2-1=9*(xyz......)-always divisible, Suff.
From2, Z will also be a multiple of 9
So D

5)If b is prime and the symbol # represents one of the following operations: addition, subtraction, multiplication, or division, is the value of b # 2 even or odd?
a) (b # 1) # 2 = 5
b) 4 # b = 3 # (1 # b) and b is even

Thanks.
pH

From 1, B has to be 2 and # is a + symbol, so Suff
From2, B is 2 as B is mentioned even as well as prime, so Suff
So D
2426

I have a doubt in this particular kind of question.

What is the price of 3 apples and 4 bananas?
Statement 1. 5 apples and 2 bananas cost Rs.12
Statement 2. 8 bananas and 6 apples cost Rs. 20.

Solution: Statement 2 alone can answer the question and not statement 1.

My argument is, if you look at statement 1, it can be written in the form of,

5x + 2y = 12 where x,y are positive integers.
So, there is only one possible solution set i.e. x = 2, y = 1

Hence, isn't statement 1 enough to answer the question? So wont the solution be "can be answered by either statements".
Can someone please explain as to why my argument is not valid?

2427

It is not mandatory that x and y are only integers unless it is explicitly mentioned.

2428

Yes it is. Lol. Horrible silly mistake dat was. Oops! Neway Thanks a ton.

2429

If u see that question clearly, it was mentioned that 5apples n 2bananas cost Rs.12. Hence, no .of apples and no. of bananas are 5 n 2 respectively. The price of apple can be Rs. 1.80 and banana can be Rs. 1.50.
5*1.80 + 2*1.50 = 9+3 = 12
I hope my explanation is clear to u

2430

Yes it is. Lol. Horrible silly mistake dat was. Oops! Neway Thanks a ton.

2431

Pretty good ones.... :)


1.If n and k are integers and (-2)n^5 > 0, is k^37
(1). (nk)^z > 0, where z is an integer that is not divisible by two
(2). k
A)Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D) EACH statement ALONE is sufficient.
E) Statement (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data are needed.

2. If x and y are integers, what is the ratio of 2x to y?
(1). 8x^3 = 27y^3
(2) 4x^2 = 9y^2

A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D) EACH statement ALONE is sufficient.
E) Statement (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data are needed.

2432
Q. The integers m and p are such that 2

1 ?

1. the greatest common factor of m and p is 2

2. the least common multiple of m and p is 30



IMO...
Facts:
1. p & m are integers greater than 2.
2. m is not factor of p
3. p = K*m + r (r != 0 and integer K > 0)


Now,
1. If GCF of p and m is 2, it implies that both are even numbers. Since m is not a factor of p, r will at the least be 2. (can be more)

So 1 alone is sufficient.

2. LCM of m and p is 30
We can have (1,30) (2,15) (3,10) and (5,6). Out of these combinations (1,30) is eliminated since both m & p > 2
In all other three cases, the remainder (r) is exactly 1 and hence we can answer the question.

So 2 alone is also sufficient.

But both together cannot be used, since according to 1 both are even integers and according to 2, we get odd-even combination.

So either one on it own is sufficient!!