Replies in ladies color MAGENTA.
I like the discussion Gail. Feel free to disagree
thank you,
Hemanth
Originally Posted by
Gail.Wynand
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.
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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 < c
(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<=840. for the LCM to be 840 x,y should not have any common multiple before 840. Also the given condition is that no common factor after 56. So the possible values of (x,y) are 56*3 and 56*5. We cannot exactly tell what x is is it 56*5 or 56*3?. Insufficient.
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 
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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/k<2. so the only choice is c/k lies b/w 0 and 2...can be anything that is a non integer or 1. Insufficient.
from (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.
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I'll try to be more careful in the future.

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