II wrote:When the positive integer x is divided by 11, the quotient is y and the remainder 3. When x is divided by 19, the remainder is also 3. What is the remainder when y is divided by 19?

A) 0
B) 1
C) 2
D) 3
E) 4

Whats the best way to answer this question ? Please explain your logic.

Thanks in advance.

Since 11 and 19 have no factors in common, the lowest common multiple will be 11 * 19 and the smallest number that gives a remainder of 3 when divided by both 11 and 19 is 11*19 + 3.

We know that y is the biggest number of 11s that goes into 11*19 + 3... so y = 19. Finally, 19/19 gives us a remainder of 0: choose (a).

Note, y could have also equalled other mutliples of 19, but the remainder is always going to be 0.

We know that y is the biggest number of 11s that goes into 11*19 + 3... so y = 19. Finally, 19/19 gives us a remainder of 0: choose (a).

Can't wrap my brain around that statement. Can anyone help please .

mmukher wrote:

We know that y is the biggest number of 11s that goes into 11*19 + 3... so y = 19. Finally, 19/19 gives us a remainder of 0: choose (a).

Can't wrap my brain around that statement. Can anyone help please .

Which part?

The question itself tells us that when x is divided by 11, we get a quotient of y and a remainder of 3.

In other words, x/11 = y + 3/11.

We've determined that one possible value for x is 11*19 + 3.

We know that (11*19 + 3)/11 = (11*19)/11 + 3/11 = 19 + 3/11... so, y=19.

The question asks "What is the remainder when y is divided by 19?"

y=19, so 19/19 = 1. Therefore, the remainder = 0.

[/quote]

Always remember
When X is divided by P & Q, but leaves the same remainder R, then
X = (LCM of P & Q) + R
so in this case
x = (LCM of 11 & 19 ) + 3
x = 199 + 3 = 209 + 3 = 212

you can easily solve many problems.

Guys, thanks !!

'ritz', I see we have the test on the same date.

-- Mayukh

yeah,
i am in cincinnati, 19th april 8 am..
what abt u?

From the GMATPrep Practice Exam-What is this asking? And the answer anyone?


For which of the following functions is f(a+b) = f(a) + f(b) for all positive numbers a and b?

a) f(x) =x^2
b) f(x) =x+1
c) f(x) =square root(x)
d) f(x) = 2/x
e) f(x) = -3x

From the GMATPrep Practice Exam-What is this asking? And the answer anyone?


For which of the following functions is f(a+b) = f(a) + f(b) for all positive numbers a and b?

a) f(x) =x^2
b) f(x) =x+1
c) f(x) =square root(x)
d) f(x) = 2/x
e) f(x) = -3x

m_blooms wrote:From the GMATPrep Practice Exam-What is this asking? And the answer anyone?


For which of the following functions is f(a+b) = f(a) + f(b) for all positive numbers a and b?

a) f(x) =x^2
b) f(x) =x+1
c) f(x) =square root(x)
d) f(x) = 2/x
e) f(x) = -3x

This is a tough tough question. Not because of the math involved, but because of the translation to normal language. Thats the part that test-takers find to be the toughest to master.

I will try explaining this as best as I can:

f(a+b) = f(a) + f(b) is a given condition that tells us whether we should retain an answer choice or not.

Let's try with Option A: f(x) =x^2 We have to test whether this function holds true for f(a+b) = f(a) + f(b)

Substitute a+b for X. Therefore, f(a+b) = (a+b) ^2.
Similarly, f(a) = a^2 ...........and................f(b) = b^2

We have to check now if f(a+b) = f(a) + f(b)
But, (a+b)^2 is clearly not equal to a^2 + b^ 2 ........this is a standard algebric rule.

Now that I have shown what the mathematical approach is, lets look at the test-smart way of doing this.

KAPLAN has a beautiful method for any question with "Which of the Following" as part of the question. Since this question has that phrase, we have to start looking from Option E upwards.

Lets do the same thing as I showed for Option A, except this time we'll do Option E: f(x) = -3x. We get the following from substituting a+b, a, AND b into this function.

f(a+b) = -3 (a+b)
f(a) = -3(a) ...................and f(b) = -3(b)

.
We see clearly that f(a+b) = f(a) + f(b) .............since -3(a+b) is indeed equal to -3(a) + -3(b)

Stop right here. Qa is E.............since its a MUST be true situation. We cannot have 2 answers for this. Saves you a bunch of trouble. Excellent Kaplan technique. Works most of the time.

Sorry for the long explanation, but I tried to really break it down for you, instead of just putting up an answer.

Thanks for detailed explanation.