1.C. Radius is given and revolutions per minute is given.
2.B. if both are even then they do have 2 as common divisor. 1 is not sufficient cos w =1 and k= 4 is an exception case.
3. C.
4.E
5. A
1) If John traveled 9 miles on his unicycle, what was his speed in miles per hour?
(1) The unicycle has spokes that stretch the 23 inches from the center of the wheel to inner edge of the wheel.
(2) John pedals his unicycle at 19 revolutions per minute.
2) If w and k are distinct positive integers, do they have any common divisors other than 1 ?
(1) k w = 3w
(2) k and w are even.
3) If x and y are integers, is xy divisible by x2?
(1) x divides into y2 with no remainder.
(2) x is a prime.
4) Is ax > y?
(1) a = y = x
(2) a
5) Let (k + m) be a prime number where m and k are positive. How many divisors does k2 + mk have?
(1) k has 4 divisors.
(2) k = 6
1.C. Radius is given and revolutions per minute is given.
2.B. if both are even then they do have 2 as common divisor. 1 is not sufficient cos w =1 and k= 4 is an exception case.
3. C.
4.E
5. A
Disagreement on 2 answers. Please let me know what i missed.
4) C - From condition 1: we gather that 'a' is positive and 'y' is negative (but both have the same value). Lets take the case where a=2, y=-2, x=+-2 (we dont know about x). now if X is positive, ax>y. but if x is negative ax
5)D - either condition works here. K+M is prime. So K(K+M) will have all the divisors of K and K+M and K(K+M). So if we know K has 4 factors then we know the number of factors of K(K+M). Take K=15 (1,3,5,15) are the divisors. take M=2. So 15*17 has known number of divisors. Likewise if K=6 we know the divisors of K(K+M).
Thank you,
Hemanth
4. What if a , x and y are all equal to zeroes.
5. Agreed.
Disagreement on 2 answers. Please let me know what i missed.
4) C - From condition 1: we gather that 'a' is positive and 'y' is negative (but both have the same value). Lets take the case where a=2, y=-2, x=+-2 (we dont know about x). now if X is positive, ax>y. but if x is negative axa then, take a=0.5, y=-0.5, x=+-0.5 when this is the case. we know that ax>y is always true. Since we know form condition 1 that a is always positive we need not consider the negative cases.
5)D - either condition works here. K+M is prime. So K(K+M) will have all the divisors of K and K+M and K(K+M). So if we know K has 4 factors then we know the number of factors of K(K+M). Take K=15 (1,3,5,15) are the divisors. take M=2. So 15*17 has known number of divisors. Likewise if K=6 we know the divisors of K(K+M).
Thank you,
Hemanth
4. What if a , x and y are all equal to zeroes.
5. Agreed.
Good point Austin! I did miss that case.
It didn't occur to me to check for 0 where -y was shown in the condition.
Nice!
1) If John traveled 9 miles on his unicycle, what was his speed in miles per hour?
(1) The unicycle has spokes that stretch the 23 inches from the center of the wheel to inner edge of the wheel.
(2) John pedals his unicycle at 19 revolutions per minute.
2) If w and k are distinct positive integers, do they have any common divisors other than 1 ?
(1) k w = 3w
(2) k and w are even.
3) If x and y are integers, is xy divisible by x2?
(1) x divides into y2 with no remainder.
(2) x is a prime.
4) Is ax > y?
(1) a = y = x
(2) a
5) Let (k + m) be a prime number where m and k are positive. How many divisors does k2 + mk have?
(1) k has 4 divisors.
(2) k = 6
My take on the above set:
1. C
2. D
3. C
4. C
5. D
Pls post the OA.I will post the explanations.
-Deepak.
1) Is x 1/16 greater than 5/8?
(1) Four times the value of x is less than three.
(2) One third of two is less than the value of x.
2) Is positive integer n 1 a multiple of 3?
(1) n3 n is a multiple of 3
(2) n3 + 2n2+ n is a multiple of 3
3) 
Quadrilateral ABCD is a rhombus and points C, D, and E are on the same line. Is quadrilateral ABDE a rhombus?
(1) The measure of angle BCD is 60 degrees.
(2) AE is parallel to BD
1) If John traveled 9 miles on his unicycle, what was his speed in miles per hour?
(1) The unicycle has spokes that stretch the 23 inches from the center of the wheel to inner edge of the wheel.
(2) John pedals his unicycle at 19 revolutions per minute.
2) If w and k are distinct positive integers, do they have any common divisors other than 1 ?
(1) k w = 3w
(2) k and w are even.
3) If x and y are integers, is xy divisible by x2?
(1) x divides into y2 with no remainder.
(2) x is a prime.
4) Is ax > y?
(1) a = y = x
(2) a
5) Let (k + m) be a prime number where m and k are positive. How many divisors does k2 + mk have?
(1) k has 4 divisors.
(2) k = 6
Sorry for the late post,
The OAs are 1) E 2) B 3) C 4) A 5) D
1) Is x 1/16 greater than 5/8?
(1) Four times the value of x is less than three.
(2) One third of two is less than the value of x.
2) Is positive integer n 1 a multiple of 3?
(1) n3 n is a multiple of 3
(2) n3 + 2n2+ n is a multiple of 3
3)
(1) The measure of angle BCD is 60 degrees.
(2) AE is parallel to BD
My take :
=======
1. E
2. B
3. B
-Deepak.
Sorry for the late post,
The OAs are 1) E 2) B 3) C 4) A 5) D
I don't agree with the OA for Q.4) A
4) Is ax > y?
(1) a = y = x
(2) a
Statement 1:
==========
Lets take y =-1 ===> a = 1 ===> X can be +/- 1.
If x = +1 then we have 1 > -1 which is true
If x = -1 then we have -1 > -1 which is false.
So,statement 1 is insuff to ans the q.
Pls correct me if there is a flaw in my approach.
-Deepak.
Sorry for the late post,
The OAs are 1) E 2) B 3) C 4) A 5) D
Can you explain the first question. Anyways i don't think OA for 4th one can be A. Cos all variables can take 0 and 0 > 0 will not satisfy in that case.
Also for the 1st question are we taking into consideration that its not the actual width , because what I would believe is that when we need to take the TSD questions we will not consider the width of the tyre including tube etc 😁 so for the first question
1. Using the second choice we calculate circumfrnce, then using the first choice we calculate the speed using 9 miles ....
1) Is x 1/16 greater than 5/8?
(1) Four times the value of x is less than three.
(2) One third of two is less than the value of x.
2) Is positive integer n 1 a multiple of 3?
(1) n3 n is a multiple of 3
(2) n3 + 2n2+ n is a multiple of 3
3)
(1) The measure of angle BCD is 60 degrees.
(2) AE is parallel to BD
1) E
equivalently - Is x-1/16>5/8 i.e or x>0.625+0.0625=0.6875.
Cond1: xCond2: x>2/3=0.667. InSufficient
Both these put together is also inconclusive. As X can be 0.67 or 0.69.
2) B
Cond1: n(n-1)(n+1) is a multiple of 3. Means one of n,n-1 or n+1 is a multiple of 3. This may or may not indicate if (n-1) is a mulitple.
Cond2: n(n+1)(n+1) is a multiple of 3. So either n or (n+1) is a multiple. In this case we can be sure that if n is a multiple, n-1 cannot. or if n+1 is a multiple even then n-1 is not a multiple. So conclusively n-1 is not a multiple.
3) C.
We need both the conditions as with Cond1, the point E can be further away on line CDE in which case it is not a rhombus.
Cond2 tells us that AEBD but without BCD being 60, the sides of ABDE will not be equal.
Guys,
In some explanation to a question, Kaplan people considered 0 a multiple of 5.
Is it correct??
Guys,
In some explanation to a question, Kaplan people considered 0 a multiple of 5.
Is it correct??
I made the same mistake, I guess its safe to assume that 0 is a multiple of 5.
I am doubtful if anyone can tell us what GMAC thinks.
I made the same mistake, I guess its safe to assume that 0 is a multiple of 5.
I am doubtful if anyone can tell us what GMAC thinks.
Yes, 0 is a multiple of 5
Source:
Number properties - is 0 a multiple of every integer? € Manhattan GMAT Forums
1) E
equivalently - Is x-1/16>5/8 i.e or x>0.625+0.0625=0.6875.
Cond1: xCond2: x>2/3=0.667. InSufficient
Both these put together is also inconclusive. As X can be 0.67 or 0.69.
I guess we can conclude if it is greater or less than by either statement, we need not find the exact valuee
2) B
Cond1: n(n-1)(n+1) is a multiple of 3. Means one of n,n-1 or n+1 is a multiple of 3. This may or may not indicate if (n-1) is a mulitple.
Cond2: n(n+1)(n+1) is a multiple of 3. So either n or (n+1) is a multiple. In this case we can be sure that if n is a multiple, n-1 cannot. or if n+1 is a multiple even then n-1 is not a multiple. So conclusively n-1 is not a multiple.
3) C.
We need both the conditions as with Cond1, the point E can be further away on line CDE in which case it is not a rhombus.
Cond2 tells us that AEBD but without BCD being 60, the sides of ABDE will not be equal.
agreed with both you can find
1) E
equivalently - Is x-1/16>5/8 i.e or x>0.625+0.0625=0.6875.
Cond1: xCond2: x>2/3=0.667. InSufficient
Both these put together is also inconclusive. As X can be 0.67 or 0.69.
2) B
Cond1: n(n-1)(n+1) is a multiple of 3. Means one of n,n-1 or n+1 is a multiple of 3. This may or may not indicate if (n-1) is a mulitple.
Cond2: n(n+1)(n+1) is a multiple of 3. So either n or (n+1) is a multiple. In this case we can be sure that if n is a multiple, n-1 cannot. or if n+1 is a multiple even then n-1 is not a multiple. So conclusively n-1 is not a multiple.
3) C.
We need both the conditions as with Cond1, the point E can be further away on line CDE in which case it is not a rhombus.
Cond2 tells us that AE|BD but without BCD being 60, the sides of ABDE will not be equal.
|asbhat Saysagreed with both you can find
Asbhat,
I don't see what you are trying to say with question 1. I am not trying to find the value of X, I am using the inequalities. Care to explain your point?
Thank you,
Hemanth
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
vpitc Says...
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.
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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 mq
hence Answer is choice B.
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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
Soln:
Lets say grp A has x numbers and grp B has y numbers.
Sum of numbers in grp A = 8x
Sum of numbers in grp B = 10y.
We have to find which is greater 8x or 10y?
from (1): (8x+10y)/(x+y) = 8.66..can not find 8x and 10y or any reln b/w them. Not sufficient.
from (2): 8x+10y=52 ...again....can not find 8x and 10y or any reln b/w them. Not sufficient.
combining (1) and (2):52/(x+y) = 8.66...again can not find 8x and 10y or any reln b/w them. Not sufficient.
Answer choice E
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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
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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
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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=
combining: 12.5 = 12.5 = sqrt(s(s-a)(s-
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.
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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 mqm and q have no common divisors; we can choose m = 2, q=7 or m=4, q=7. does mq divide c in both cases? Yes. hence sufficient.
hence Answer is choice B.
Agree. Well done. I didn't get this one right actually.
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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
Soln:
Lets say grp A has x numbers and grp B has y numbers.
Sum of numbers in grp A = 8x
Sum of numbers in grp B = 10y.
We have to find which is greater 8x or 10y?
from (1): (8x+10y)/(x+y) = 8.66..can not find 8x and 10y or any reln b/w them. Not sufficient.
from (2): 8x+10y=52 ...again....can not find 8x and 10y or any reln b/w them. Not sufficient.
combining (1) and (2):52/(x+y) = 8.66...again can not find 8x and 10y or any reln b/w them. Not sufficient.
Answer choice E
I disagree. D is the right answer. You can solve by either of these.
Cond1: Sufficient. As solving the equation we get X/Y=2/1. We know that there are twice the number of elements in A as in B. Lets say if Y=1, then sum of B = 10 and sum A = 16. So for any value of Y, sum of elements in A is always going to be greater than in B.
Cond2: The sum is 8X+10Y = 52 where X and Y are positive integers. The only condition where this will hold true is for X=4 and Y=2. In which case You know that the of each set and hence sufficiency.
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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,yCond2: 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.
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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.
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Gail, I differ from your answers on a few of these. Let me know if I got anything wrong.
Thank you,
Hemanth